TEST BANK for Essential Cell Biology 5th Edition by Bruce Alberts, Alexander D Johnson, David Morgan

CHAPTER 1

Cells: The Fundamental Units of Life UNITY AND DIVERSITY OF CELLS

1.1.a Compare, with examples, some ways in which cells may vary in appearance and function.

1.1.b Outline, with examples, ways in which cells share a basic fundamental chemistry.

1.1.c Explain how the relationship between DNA, RNA, and protein—as laid out in the central dogma—makes the selfreplication of living cells possible.

1.1.d Summarize how the processes of mutation and selection promote the gradual evolution of individuals best suited for survival in a wide range of habitats.

1.1.e Explain how differentiated cell types can vary widely in form and function despite having the same genome sequence.

CELLS UNDER THE MICROSCOPE 1.2.a List the three tenets of cell theory and explain their ramifications for the study of cell biology.

1.2.b Contrast light microscopy, super-resolution fluorescence light microscopy, and electron microscopy in terms of the cell components that can generally be distinguished using each.

1.2.c Compare how samples are prepared for light versus electron microscopy and explain how these preparations affect whether the technique can be used for viewing living cells or tissues.

THE PROKARYOTIC CELL 1.3.a Describe the structural differences between prokaryotes and eukaryotes.

1.3.b Analyze how eukaryotic cells and organisms rely on the function of prokaryotic cells and their descendants.

1.3.c Compare prokaryotes and eukaryotes in terms of their relative preponderance on Earth, their range of habitat, and their tendency toward multicellularity.

1.3.d Justify the division of prokaryotes into bacteria and archaea.

THE EUKARYOTIC CELL 1.4.a State the function of the nucleus and describe its structural features.

1.4.b Explain how the structure of the mitochondrion supports its function.

1.4.c Outline the evolution of mitochondria and chloroplasts and cite the evidence for these origins.

1.4.d Explain how chloroplasts and mitochondria cooperate as plant cells convert light energy into chemical energy.

1.4.e Compare the function of lysosomes and peroxisomes.

1.4.f Compare the structure, location, and function of the endoplasmic reticulum and Golgi apparatus.

1.4.g Outline the role that transport vesicles play in endocytosis, exocytosis, and the movement of materials between one membrane-enclosed organelle and another.

1.4.h Relate the location of the cytosol with respect to the cell’s membrane-enclosed organelles.

1.4.i List the three major filaments of the cytoskeleton and contrast the roles they have in animal cells.

1.4.j Outline the role the cytoskeleton has in plant cells.

1.4.k Describe the ancestral cell that likely engulfed the aerobic bacteria that gave rise to mitochondria and explain why this event is thought to have preceded the acquisition of chloroplasts.

MODEL ORGANISMS 1.5.a Review why scientists study model organisms.

1.5.b Compare E. coli, S. cerevisiae, and A. thaliana and list the types of discoveries made by studying each.

1.5.c Compare flies, worms, fish, and mice as model organisms and name a benefit of studying each.

1.5.d Review the benefits of studying cultured human cells.

1.5.e Assess the relationship between genome size and gene number.

1.5.f Explain the significance of homologous genes and proteins.

1.5.g Summarize the roles played by the nucleotide sequences contained in an organism’s genome.

1.5.h Outline an experiment that would allow investigators to determine whether proteins from different eukaryotes are functionally interchangeable.

MULTIPLE CHOICE 1. Living systems are incredibly diverse in size, shape, environment, and behavior. It is estimated that there are between 10 million and 100 million different species. Despite this wide variety of organisms, it remains difficult to define what it means to say something is alive. Which of the following can be described as the smallest living unit? a. DNA b. cell c. organelle d. protein ANS: b DIF: Easy REF: 1.1 OBJ: 1.1.b Outline, with examples, ways in which cells share a basic fundamental chemistry. MSC: Understanding 2. The central dogma provides a framework for thinking about how genetic information is copied and used to produce structural and catalytic components of the cell. From the choices below, select the order of biochemical processes that best correlates with the tenets of the central dogma. a. replication, transcription, translation b. replication, translation, transcription c. translation, transcription, replication d. translation, replication, transcription

ANS: A DIF: Easy REF: 1.1 OBJ: 1.1.c Explain how the relationship between DNA, RNA, and protein—as laid out in the central dogma—makes the self-replication of living cells possible. MSC: Understanding 3. Proteins are important architectural and catalytic components within the cell, helping to determine its chemistry, its shape, and its ability to respond to changes in the environment. Remarkably, all of the different proteins in a cell are made from the same 20 __________. By linking them in different sequences, the cell can make protein molecules with different conformations and surface chemistries, and therefore different functions. a. nucleotides b. sugars c. amino acids d. fatty acids ANS: C DIF: Easy REF: 1.1 OBJ: 1.1.b Outline, with examples, ways in which cells share a basic fundamental chemistry. MSC: Remembering 4. Which statement is NOT true about mutations? a. A mutation is a change in the DNA that can generate offspring less fit for survival than their parents. b. A mutation can be a result of imperfect DNA duplication. c. A mutation is a result of sexual reproduction. d. A mutation is a change in the DNA that can generate offspring that are as fit for survival as their parents are.

ANS: C DIF: Easy REF: 1.1 OBJ: 1.1.d Summarize how the processes of mutation and selection promote the gradual evolution of individuals best suited for survival in a wide range of habitats. MSC: Analyzing 5. Changes in DNA sequence from one generation to the next may result in offspring that are altered in fitness compared with their parents. The process of change and selection over the course of many generations is the basis of a. mutation. b. evolution. c. heredity. d. reproduction.

ANS: B DIF: Easy REF: 1.1 OBJ: 1.1.d Summarize how the processes of mutation and selection promote the gradual evolution of individuals best suited for survival in a wide range of habitats. MSC: Understanding 6. Select the option that BEST finishes the following statement: Evolution is a process a. that can be understood based on the principles of mutation and selection. b. that results from repeated cycles of adaptation over billions of years. c. by which all present-day cells arose from 4–5 different ancestral cells. d. that requires hundreds of thousands of years.

ANS: A DIF: Moderate REF: 1.1 OBJ: 1.1.d Summarize how the processes of mutation and selection promote the gradual evolution of individuals best suited for survival in a wide range of habitats. MSC: Analyzing 7. Select the option that correctly finishes the following statement: A cell’s genome a. is defined as all the genes being used to make protein. b. contains all of a cell’s DNA.

c. constantly changes, depending upon the cell’s environment. d. is altered during embryonic development.

ANS: B DIF: Easy REF: 1.1 OBJ: 1.1.b Outline, with examples, ways in which cells share a basic fundamental chemistry. MSC: Remembering 8. Which statement is NOT true about the events/conclusions from studies during the mid-1800s surrounding the discovery of cells? a. Cells came to be known as the smallest universal building block of living organisms. b. Scientists came to the conclusion that new cells can form spontaneously from the remnants of ruptured cells. c. Light microscopy was essential in demonstrating the commonalities between plant and animal tissues. d. New cells arise from the growth and division of previously existing cells.

ANS: B DIF: Easy REF: 1.2 OBJ: 1.2.a List the three tenets of cell theory and explain their ramifications for the study of cell biology. MSC: Remembering 9. What unit of length would you generally use to measure a typical plant or animal cell? a. centimeters b. nanometers c. millimeters d. micrometers ANS: D DIF: Easy REF: 1.1 OBJ: 1.1.a Compare, with examples, some ways in which cells may vary in appearance and function. | 1.2.a List the three tenets of cell theory and explain their ramifications for the study of cell biology.

MSC: Remembering 10. Cell biologists employ targeted fluorescent dyes or modified fluorescent proteins in both standard fluorescence microscopy and confocal microscopy to observe specific details in the cell. Even though fluorescence permits better visualization, the resolving power is essentially the same as that of a standard light microscope because the resolving power of a fluorescent microscope is still limited by the __________ of visible light. a. absorption b. intensity c. filtering d. wavelength ANS: D DIF: Moderate REF: 1.2 OBJ: 1.2.b Contrast light microscopy, super-resolution fluorescence light microscopy, and electron microscopy in terms of the cell components that can generally be distinguished using each.

MSC: Understanding 11. What is the smallest distance two points can be separated and still resolved using light microscopy? a. 20 nm b. 0.2 μm c. 2 μm d. 200 μm ANS: B DIF: Moderate REF: 1.2 OBJ: 1.2.b Contrast light microscopy, super-resolution fluorescence light microscopy, and electron microscopy in terms of the cell components that can generally be distinguished using each.

MSC: Understanding 12. Prokaryotic cells do not possess a. a nucleus. b. replication machinery. c. ribosomes. d. membrane bilayers.

ANS: A DIF: Easy REF: 1.3 OBJ: 1.3.a Describe the structural differences between prokaryotes and eukaryotes.

MSC: Remembering 13. Which three characteristics best support the rapid evolution of prokaryotic populations? a. microscopic, motile, anaerobic b. aerobic, motile, rapid growth c. no organelles, cell wall, can exchange DNA d. large population, rapid growth, can exchange DNA ANS: D DIF: Easy REF: 1.3 OBJ: 1.3.c Compare prokaryotes and eukaryotes in terms of their relative preponderance on Earth, their range of habitat, and their tendency toward multicellularity. MSC: Analyzing 14. The world of prokaryotes is divided into two domains (bacteria and archaea), each as different from each other as from eukaryotes. Select the observable characteristic that BEST separates archaea from bacteria. a. can metabolize inorganic substances b. are found in extremely harsh environments c. thrive in anaerobic conditions d. are photosynthetic organisms ANS: B DIF: Easy REF: 1.3 OBJ: 1.3.d Justify the division of prokaryotes into bacteria and archaea.

MSC: Remembering 15. The __________ __________ is made up of two concentric membranes and is continuous with the membrane of the endoplasmic reticulum. a. plasma membrane b. Golgi network c. mitochondrial membrane d. nuclear envelope ANS: D DIF: Easy REF: 1.4 OBJ: 1.4.a State the function of the nucleus and describe its structural features.

MSC: Remembering 16. The nucleus, an organelle found in eukaryotic cells, confines the __________, keeping them separated from other components of the cell. a. lysosomes b. chromosomes c. peroxisomes d. ribosomes ANS: B DIF: Easy REF: 1.4 OBJ: 1.4.a State the function of the nucleus and describe its structural features.

MSC: Remembering 17. Which of the following organelles has both an outer and an inner membrane? a. endoplasmic reticulum b. mitochondrion c. lysosome d. peroxisome ANS: B DIF: Easy REF: 1.4 OBJ: 1.4.b Express how the structure of the mitochondrion supports its function.

MSC: Remembering 18. Mitochondria perform cellular respiration, a process that uses oxygen, generates carbon dioxide, and produces chemical energy for the cell. Which answer below indicates a correct pairing of the material “burned” and the form of energy produced during cellular respiration? a. fat, ADP b. sugar, fat c. sugar, ATP d. fat, protein ANS: C DIF: Easy REF: 1.4 OBJ: 1.4.d Explain how chloroplasts and mitochondria cooperate as plant cells convert light energy into chemical energy. MSC: Understanding 19. Mitochondria contain their own genome, are able to duplicate, and actually divide on a different time line from the rest of the cell. Nevertheless, mitochondria cannot function for long when isolated from the cell because they are a. viruses. b. parasites. c. endosymbionts. d. anaerobes.

ANS: C DIF: Easy REF: 1.4 OBJ: 1.4.b Explain how the structure of the mitochondrion supports its function.

MSC: Remembering 20. The mitochondrial proteins found in the inner membrane are involved in the conversion of ADP to ATP, a source of energy for the cell. This process consumes which of the following substances? a. oxygen b. nitrogen c. sulfur d. carbon dioxide ANS: A DIF: Easy REF: 1.4 OBJ: 1.4.b Explain how the structure of the mitochondrion supports its function.| 1.4.d Explain how chloroplasts and mitochondria cooperate as plant cells convert light energy into chemical energy.

MSC: Remembering 21. Chloroplasts are complex organelles present in all photosynthesizing eukaryotes. Three different membranes create chemically different environments inside the chloroplast. Precisely where are chlorophyll molecules localized in the chloroplast? a. in the first, outer membrane

b. in the space between the first and second membranes c. in the second, inner membrane d. in the third, innermost membrane ANS: D DIF: Easy REF: 1.4 OBJ: 1.4.d Explain how chloroplasts and mitochondria cooperate as plant cells convert light energy into chemical energy. MSC: Remembering 22. Photosynthesis enables plants to capture the energy from sunlight. In this essential process, plants incorporate the carbon from CO2 into high-energy __________ molecules, which the plant cell mitochondria use to produce ATP. a. fat b. sugar c. protein d. fiber ANS: B DIF: Easy REF: 1.4 OBJ: 1.4.d Explain how chloroplasts and mitochondria cooperate as plant cells convert light energy into chemical energy. MSC: Remembering 23. Which of the following choices BEST describes the role of the lysosome? a. transport of material to the Golgi apparatus b. clean-up, recycling, and disposal of macromolecules c. sorting of transport vesicles d. the storage of excess macromolecules ANS: B DIF: Easy REF: 1.4 OBJ: 1.4.e Compare the function of lysosomes and peroxisomes. MSC: Remembering 24. The cell constantly exchanges materials by bringing nutrients in from the external environment and shuttling unwanted by-products back out. Which term describes the process by which external materials are captured inside transport vesicles and brought into the cell? a. degradation b. exocytosis c. phagocytosis d. endocytosis ANS: D DIF: Easy REF: 1.4 OBJ: 1.4.g Outline the role that transport vesicles play in endocytosis, exocytosis, and the movement of materials between one membrane-enclosed organelle and another. MSC: Remembering 25. Eukaryotic cells are able to trigger the release of material from secretory vesicles to the extracellular space using a process called exocytosis. An example of materials commonly released this way is a. hormones. b. nucleic acids. c. sugars. d. cytosolic proteins.

ANS: A DIF: Easy REF: 1.4 OBJ: 1.4.g Outline the role that transport vesicles play in endocytosis, exocytosis, and the movement of materials between one membrane-enclosed organelle and another. MSC: Remembering 26. __________ are fairly small organelles that provide a safe place within the cell to carry out certain biochemical reactions that generate harmful, highly reactive oxygen species. These chemicals are both generated and broken

down in the same location. a. Nucleosomes b. Lysosomes c. Peroxisomes d. Endosomes ANS: C DIF: Easy REF: 1.4 OBJ: 1.4.e Compare the function of lysosomes and peroxisomes. MSC: Remembering 27. The cytoskeleton provides support, structure, motility, and organization, and it forms tracks to direct organelle and vesicle transport. Which of the cytoskeletal elements listed below is the thickest? a. actin filaments b. microtubules c. intermediate filaments d. none of the above (all have the same thickness) ANS: B DIF: Easy REF: 1.4 OBJ: 1.4.i List the three major filaments of the cytoskeleton and contrast the roles they have in animal cells. MSC: Remembering 28. Despite the differences between eukaryotic and prokaryotic cells, prokaryotes have proteins that are distantly related to eukaryotic actin filaments and microtubules. What is likely to be the most ancient function of the cytoskeleton? a. cell motility b. vesicle transport c. membrane support d. cell division ANS: D DIF: Moderate REF: 1.4 OBJ: 1.4.i List the three major filaments of the cytoskeleton and contrast the roles they have in animal cells. MSC: Understanding 29. Choose the phrase that best completes this sentence: Microtubules __________ and are required to pull duplicated chromosomes to opposite poles of dividing cells. a. generate contractile forces b. are intermediate in thickness c. can rapidly reorganize d. are found in especially large numbers in muscle cells ANS: C DIF: Easy REF: 1.4 OBJ: 1.4.i List the three major filaments of the cytoskeleton and contrast the roles they have in animal cells. MSC: Understanding 30. Which pair of values best fills in the blanks in this statement: On average, eukaryotic cells are __________ times longer and have __________ times more volume than prokaryotic cells. a. 5; 100 b. 10; 200 c. 10; 100 d. 10; 1000 ANS: D DIF: Easy REF: 1.4 OBJ: 1.3.a Describe the structural differences between prokaryotes and eukaryotes.

MSC: Remembering

31. Cells that are specialized for the secretion of proteins are likely to have which of the following features? a. long bundles of actin/myosin proteins b. small volume of cytoplasm c. large population of mitochondria d. enlarged endoplasmic reticulum ANS: D DIF: Easy REF: 1.4 OBJ: 1.4.f Compare the structure, location, and function of the endoplasmic reticulum and Golgi apparatus. MSC: Understanding 32. Scientists learned that cell death is a normal and even important part of life by studying the development of the nematode worm C. elegans. What was the most important feature of C. elegans for the study of programmed cell death? a. The nematode is smaller and simpler than the fruit fly. b. Seventy percent of C. elegans genes have homologs in humans. c. The developmental pathway of each cell in the adult worm was known. d. Its genome was partially sequenced.

ANS: C This is the best answer because it was the prior developmental studies tracing cell lineages from the egg to the adult that allowed scientists to identify the precise time and location of cells that were being targeted for cell death.

It was observed that this cell death was a normal and necessary part of the developmental pathway in the worm. Programmed cell death has since become known to be an important process in all multicellular eukaryotic organisms.

DIF: Moderate REF: 1.5 OBJ: 1.5.a Review why scientists study model organisms. MSC: Understanding 33. Biologists cannot possibly study all living species. Instead, they try to understand cell behavior by studying a select subset of species. Which of the following characteristics are useful in an organism chosen for use as a model in laboratory studies? a. amenability to genetic manipulation b. ability to grow under controlled conditions c. rapid rate of reproduction d. all of the above ANS: D DIF: Easy REF: 1.5 OBJ: 1.5.a Review why scientists study model organisms. MSC: Understanding 34. Which species was the key model organism for the advancement of molecular biology (understanding DNA replication, decoding the DNA to make proteins, etc.)? a. E. coli b. D. melanogaster c. S. pombe d. C. elegans ANS: A DIF: Easy REF: 1.5 OBJ: 1.5.b Compare E. coli, S. cerevisiae, and A. thaliana and list the types of discoveries made by studying each. MSC: Remembering 35. A. thaliana, or Arabidopsis, is a common weed. Biologists have selected it over hundreds of thousands of other flowering plant species to serve as an experimental model organism because a. it can withstand extremely cold climates. b. it can reproduce in 8–10 weeks.

c. it produces thousands of offspring per plant. d. Both (B) and (C) are true.

ANS: B DIF: Easy REF: 1.5 OBJ: 1.5.a Review why scientists study model organisms. | 1.5.c Compare flies, worm, fish, and mice as model organisms and name a benefit of studying each. MSC: Remembering 36. Drosophila melanogaster is a/an __________. This type of animal is the most abundant of all animal species, making it an appropriate choice as an experimental model. a. insect b. bird c. amphibian d. mammal ANS: A DIF: Easy REF: 1.5 OBJ: 1.5.a Review why scientists study model organisms. | 1.5.c Compare flies, worm, fish, and mice as model organisms and name a benefit of studying each. MSC: Remembering 37. Caenorhabditis elegans is a nematode. During its development, it produces more than 1000 cells. However, the adult worm has only 959 somatic cells. The process by which 131 cells are specifically targeted for destruction is called a. directed cell pruning. b. programmed cell death. c. autophagy. d. necrosis.

ANS: B DIF: Easy REF: 1.5 OBJ: 1.5.c Compare flies, worm, fish, and mice as model organisms and name a benefit of studying each. MSC: Remembering 38. Zebrafish (Danio rerio) are especially useful in the study of early development because their embryos a. are exceptionally large. b. develop slowly. c. are transparent. d. are pigmented.

ANS: C DIF: Easy REF: 1.5 OBJ: 1.5.a Review why scientists study model organisms. | 1.5.c Compare flies, worm, fish, and mice as model organisms and name a benefit of studying each. MSC: Remembering 39. Brewer’s yeast, apart from being an irreplaceable asset in the brewery and in the bakery, is an experimental organism used to study eukaryotic cells. However, it does have some limitations. Select all the processes below that CANNOT be studied in yeast. a. differentiation b. motility c. exocytosis d. cell division ANS: A, B DIF: Easy REF: 1.5 OBJ: 1.5.a Review why scientists study model organisms. | 1.5.c Compare flies, worm, fish, and mice as model organisms and name a benefit of studying each. MSC: Understanding MATCHING

1. Match the following types of microscopy with the corresponding description provided below them. There is one best match for each.

A. confocal B. transmission electron C. fluorescence D. phase-contrast E. scanning electron F. bright-field 1. ____ uses a light microscope with an optical component to take advantage of the different refractive indices of light passing through different regions of the cell.

2. ____ employs a light microscope and requires that samples be fixed and stained in order to reveal cellular details.

3. ____ requires the use of two sets of filters. The first filter narrows the wavelength range that reaches the specimen and the second blocks out all wavelengths that pass back up to the eyepiece except for those emitted by the dye in the sample.

4. ____ scans the specimen with a focused laser beam to obtain a series of two-dimensional optical sections, which can be used to reconstruct an image of the specimen in three dimensions. The laser excites a fluorescent dye molecule, and the emitted light from each illuminated point is captured through a pinhole and recorded by a detector.

5. ____ has the ability to resolve cellular components as small as 2 nm.

6. ____ requires coating a sample with a thin layer of a heavy metal to produce three-dimensional images of the sample surface.

1. ANS: D DIF: Moderate REF: 1.2 OBJ: 1.2.b Contrast light microscopy, super-resolution fluorescence light microscopy, and electron microscopy in terms of the cell components that can generally be distinguished using each. | 1.2.c Compare how samples are prepared for light versus electron microscopy and explain how these prepa rations affect whether the technique can be used for viewing living cells or tissues. MSC: Analyzing 2. ANS: F DIF: Moderate REF: 1.2 OBJ: 1.2.b Contrast light microscopy, high-resolution fluorescence light microscopy, and electron microscopy in terms of the cell components that can generally be distinguished using each. | 1.2.c Compare how samples are prepared for light versus electron microscopy and explain how these preparations affect whether the technique can be used for viewing living cells or tissues. MSC: Analyzing 3. ANS: C DIF: Moderate REF: 1.2 OBJ: 1.2.b Contrast light microscopy, high-resolution fluorescence light microscopy, and electron microscopy in terms of the cell components that can generally be distinguished using each. | 1.2.c Compare how samples are prepared for light versus electron microscopy and explain how these preparations affect whether the technique can be used for viewing living cells or tissues. MSC: Analyzing 4. ANS: A DIF: Moderate REF: 1.2 OBJ: 1.2.b Contrast light microscopy, high-resolution fluorescence light microscopy, and electron microscopy in terms of the cell components that can generally be distinguished using each. | 1.2.c Compare how samples are prepared for light versus electron microscopy and explain how these preparations affect whether the technique can be used for viewing living cells or tissues. MSC: Analyzing 5. ANS: B DIF: Moderate REF: 1.2 OBJ: 1.2.b Contrast light microscopy, high-resolution fluorescence light microscopy, and electron microscopy in terms of the cell components that can generally be distinguished using each. | 1.2.c Compare how samples are prepared for light versus electron microscopy and explain how these prepa rations affect whether the technique can be used for viewing living cells or tissues. MSC: Analyzing 6. ANS: E DIF: Moderate REF: 1.2 OBJ: 1.2.b Contrast light microscopy, high-resolution fluorescence light microsco-

py, and electron microscopy in terms of the cell components that can generally be distinguished using each. | 1.2.c Compare how samples are prepared for light versus electron microscopy and e xplain how these preparations affect whether the technique can be used for viewing living cells or tissues. MSC: Analyzing 2. Match a structure from the list below (A–G) with the labels 1–7 in the schematic drawing of an animal cell in Figure 1-41.

Figure 1-41

A. plasma membrane B. nuclear envelope C. cytosol D. Golgi apparatus E. endoplasmic reticulum F. mitochondrion G. transport vesicles 1. Label 1 ANS: C DIF: Easy REF: 1.4 OBJ: 1.4.f Compare the structure, location, and function of the endoplasmic reticulum and Golgi apparatus. | 1.4.h Relate the location of the cytosol with respect to the cell’s membrane-enclosed organelles.

MSC: Remembering 2. Label 2 ANS: D DIF: Easy REF: 1.4 OBJ: 1.4.f Compare the structure, location, and function of the endoplasmic reticulum and Golgi apparatus. | 1.4.h Relate the location of the cytosol with respect to the cell’s membrane-enclosed organelles.

MSC: Remembering 3. Label 3 ANS: A DIF: Easy REF: 1.4 OBJ: 1.4.f Compare the structure, location, and function of the endoplasmic reticulum and Golgi apparatus. | 1.4.h Relate the location of the cytosol with respect to the cell’s membrane-enclosed organelles.

MSC: Remembering

4. Label 4 ANS: E DIF: Easy REF: 1.4 OBJ: 1.4.f Compare the structure, location, and function of the endoplasmic reticulum and Golgi apparatus. | 1.4.h Relate the location of the cytosol with respect to the cell’s membrane-enclosed organelles.

MSC: Remembering 5. Label 5 ANS: B DIF: Easy REF: 1.4 OBJ: 1.4.f Compare the structure, location, and function of the endoplasmic reticulum and Golgi apparatus. | 1.4.h Relate the location of the cytosol with respect to the cell’s membrane-enclosed organelles.

MSC: Remembering 6. Label 6 ANS: G DIF: Easy REF: 1.4 OBJ: 1.4.f Compare the structure, location, and function of the endoplasmic reticulum and Golgi apparatus. | 1.4.h Relate the location of the cytosol with respect to the cell’s membrane-enclosed organelles.

MSC: Remembering 7. Label 7 ANS: F DIF: Easy REF: 1.4 OBJ: 1.4.f Compare the structure, location, and function of the endoplasmic reticulum and Golgi apparatus. | 1.4.h Relate the location of the cytosol with respect to the cell’s membrane-enclosed organelles.

MSC: Remembering 3. Select all the cell types in which the listed structure or molecule can be found. Note that the structure or molecule can be found in more than one type of cell.

Figure 1-42

A. animal B. plant C. bacterial D. animal and plant E. plant and bacterial F. animal and bacterial G. animal, plant, and bacterial 1. DNA ANS: G DIF: Easy REF: 1.3 OBJ: 1.3.a Describe the structural differences between prokaryotes and eukaryotes.

MSC: Analyzing

2. nucleus ANS: D DIF: Easy REF: 1.3 OBJ: 1.3.a Describe the structural differences between prokaryotes and eukaryotes.

MSC: Analyzing 3. plasma membrane ANS: G DIF: Easy REF: 1.3 OBJ: 1.3.a Describe the structural differences between prokaryotes and eukaryotes.

MSC: Analyzing 4. chloroplast ANS: E DIF: Easy REF: 1.3 OBJ: 1.3.a Describe the structural differences between prokaryotes and eukaryotes.

MSC: Analyzing 5. cell wall ANS: E DIF: Easy REF: 1.3 OBJ: 1.3.a Describe the structural differences between prokaryotes and eukaryotes.

MSC: Analyzing 6. lysosome ANS: D DIF: Easy REF: 1.3 OBJ: 1.3.a Describe the structural differences between prokaryotes and eukaryotes.

MSC: Analyzing 7. mitochondrion ANS: D DIF: Easy REF: 1.3 OBJ: 1.3.a Describe the structural differences between prokaryotes and eukaryotes.

MSC: Analyzing 8. Golgi apparatus ANS: D DIF: Easy REF: 1.3 OBJ: 1.3.a Describe the structural differences between prokaryotes and eukaryotes.

MSC: Analyzing

SHORT ANSWER 1. Indicate whether the following statements are TRUE or FALSE. If the statement is false, explain why it is false.

A. The Paramecium is a multicellular microorganism covered with hair-like cilia.

B. Cells of different types can have different chemical requirements.

C. The branchlike extensions that sprout from a single nerve cell in a mammalian brain can extend over several hundred micrometers.

ANS: A. False. The Paramecium is a single-celled organism.

B. True C. True DIF: Easy REF: 1.1 OBJ: 1.1.a Compare, with examples, some ways in which cells may vary in appearance and function. MSC: Evaluating 2. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. amino acids

micrometer(s)

viruses

DNA

millimeter(s)

yeast

fatty acids

plants

meter

plasma membranes

Cells can be very diverse: superficially, they come in various sizes, ranging from bacterial cells such as Lactobacillus, which is a few __________ in length, to larger cells such as a frog’s egg, which has a diameter of about 1 __________.

Despite the diversity, cells resemble each other to an astonishing degree in their chemistry. For example, the same 20 __________ are used to make proteins. Similarly, the genetic information of all cells is stored in their __________.

Although __________ contain the same types of molecules as cells, their inability to reproduce themselves by their own efforts means that they are not considered living matter.

ANS: Cells can be very diverse: superficially, they come in various sizes, ranging from bacterial cells such as Lactobacillus, which is a few micrometers in length, to larger cells such as a frog’s egg, which has a diameter of about 1 millimeter. Despite the diversity, cells resemble each other to an astonishing degree in their chemistry. For example, the same 20 amino acids are used to make proteins. Similarly, the genetic information of all cells is stored in their DNA. Although viruses contain the same types of molecules as cells, their inability to reproduce themselves by their own efforts means that they are not considered living matter.

DIF: Easy REF: 1.1 OBJ:1.1.a Compare, with examples, some ways in which cells may vary in appearance and function. | 1.1.b Outline, with examples, ways in which cells share a basic fundamental chemistry.

MSC: Understanding 3. Cellular populations in bacterial colonies and the human body both start with single founder cells, yet these two collections of cells are vastly different. What is the primary reason for these differences?

ANS: In humans and other higher-order species, multiple cell types arise from the single founder cell to perform specialized functions for the organism. In bacterial colonies, cells do communicate, but there is no observed development into different cell types.

DIF: Difficult REF: 1.1 OBJ: 1.1.a Compare, with examples, some ways in which cells may vary in appearance and function. MSC: Evaluating 4. Indicate whether the following statements are TRUE or FALSE. If the statement is false, explain why it is false.

A. The nucleus of an animal cell is round, small, and difficult to distinguish using light microscopy.

B. The presence of the plasma membrane can be inferred by the well-defined boundary of the cell.

C. The cytosol is fairly empty, containing a limited number of organelles, which allows room for rapid movement via diffusion.

D. The extracellular matrix is the space between individual cells and is exclusively composed of proteins.

ANS: A. False. The nucleus is one of the largest organelles and is the easiest organelle to discern within a typical cell.

B. True C. False. The cytosol is actually brimming with individual proteins, protein fibers, extended membrane systems, transport vesicles, and small molecules. And although cellular components do move by diffusion, the rate of movement is limited by the space available and the size of the component in question.

D. False. The extracellular matrix contains small molecules, protein fibers, and a dense matrix of sugar chains.

DIF: Easy REF: 1.4 OBJ: 1.4.a State the function of the nucleus and describe its structural features. | 1.4.h Relate the location of the cytosol with respect to the cell’s membrane-enclosed organelles. MSC: Evaluating

5. Describe the benefits and limitations of utilizing electron microscopy to understand biological systems.

ANS: Electron microscopy can provide biologists with the highest sample magnification and resolution in the n anometer range. Samples must be fixed, preventing researchers from observing living cells.

DIF: Easy REF: 1.2 OBJ: 1.2.b Contrast light microscopy, high-resolution fluorescence light microscopy, and electron microscopy in terms of the cell components that can generally be distinguished using each. | 1.2.c Compare how samples are prepared for light versus electron microscopy and explain how these preparations affect whether the technique can be used for viewing living cells or tissues. MSC: Understanding 6. Indicate whether the following statements are TRUE or FALSE. If the statement is false, explain why it is false.

A. The terms “prokaryote” and “bacterium” are synonyms.

B. Prokaryotes can adopt several different basic shapes, including spherical, rod-shaped, and spiral.

C. Some prokaryotes have cell walls surrounding the plasma membrane.

D. Oxygen is toxic to certain prokaryotic organisms.

E. Mitochondria are thought to have evolved from anaerobic bacteria.

F. Photosynthetic bacteria contain chloroplasts.

ANS: A. False. Archaea make up a class of prokaryotic organisms that are significantly different from bacteria.

B. True C. True D. True E. False. Mitochondria use oxygen to generate energy and are thought to have evolved from aerobic bacteria.

F. False. Photosynthetic bacteria have enzyme systems similar to those found in chloroplasts, which allow them to harvest light energy to fix carbon dioxide.

DIF: Easy REF: 1.3 OBJ: 1.3.a Express the structural differences between prokaryotes and eukaryotes. | 1.3.d Summarize the division of prokaryotes into bacteria and archaea. | MSC: Evaluating 7. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. chloroplast

cytosol

nucleus

chromosome

endoplasmic reticulum

ribosomes

cytoskeleton

mitochondrion

Eukaryotic cells are bigger and more elaborate than prokaryotic cells. By definition, all eukaryotic cells have a __________, usually the most prominent organelle. Another organelle found in essentially all eukaryotic cells is the __________, which generates the chemical energy for the cell. In contrast, the __________ is a type of organelle found only in the cells of plants and algae, and performs photosynthesis. If we were to strip away the plasma membrane from a eukaryotic cell and remove all of its membrane-enclosed organelles, we would be left with the __________, which contains many long, fine filaments of protein that are responsible for cell shape and structure, and thereby form the cell’s __________.

ANS: Eukaryotic cells are bigger and more elaborate than prokaryotic cells. By definition, all eukaryotic cells have a nucleus, usually the most prominent organelle. Another organelle found in essentially all eukaryotic cells is the mitochondrion, which generates the chemical energy for the cell. In contrast, the chloroplast is a type of organelle found

only in the cells of plants and algae, and performs photosynthesis. If we were to strip away the plasma membrane from a eukaryotic cell and remove all of its membrane-enclosed organelles, we would be left with the cytosol, which contains many long, fine filaments of protein that are responsible for cell shape and structure and thereby form the cell’s cytoskeleton.

DIF: Easy REF: 1.4 OBJ: 1.4.a State the function of the nucleus and describe its structural features. | 1.4.d Explain how chloroplasts and mitochondria cooperate as plant cells convert light energy into chemical energy. | 1.4.h Relate the location of the cytosol with respect to the cell’s membrane-enclosed organelles. MSC: Understanding 8. You fertilize egg cells from a healthy plant with pollen (which contains the male germ cells) that has been treated with DNAdamaging agents. You find that some of the offspring have defective chloroplasts, and that this characteristic can be passed on to future generations. This surprises you at first because you happen to know that the male germ cell in the pollen grain contributes no chloroplasts to the fertilized egg cell and thus to the offspring. What can you deduce from these results?

ANS: Your results show that not all of the information required for making a chloroplast is encoded in the chloroplast’s own DNA; some, at least, must be encoded in the DNA carried in the nucleus. The reasoning is as follows. Genetic information is carried only in DNA, so the defect in the chloroplasts must be due to a mutation in DNA. But all of the chloroplasts in the offspring (and thus all of the chloroplast DNA) must derive from those in the female egg cell, since chloroplasts only arise from other chloroplasts. Hence, all of the chloroplasts contain undamaged DNA from the female parent’s chloroplasts. In all the cells of the offspring, however, half of the nuclear DNA will have come from the male germ-cell nucleus, which combined with the female egg nucleus at fertilization. Since this DNA has been treated with DNA-damaging agents, it must be the source of the heritable chloroplast defect. Thus, some of the information required for making a chloroplast is encoded by the nuclear DNA. DIF: Difficult REF: 1.1 OBJ: 1.1.d Summarize how the processes of mutation and selection promote the gradual evolution of individuals best suited for survival in a wide range of habitats. MSC: Applying 9. Indicate whether the following statements are TRUE or FALSE. If the statement is false, explain why it is false.

A. With respect to cellular respiration, the only organelles used by animal cells are mitochondria, while plant cells use both mitochondria and chloroplasts.

B. The number of mitochondria inside a cell remains constant over the life of the cell.

C. Membrane components in the cell are made in the endoplasmic reticulum.

D. The Golgi apparatus is made up of a series of membrane-enclosed compartments through which materials destined for secretion must pass.

E. Lysosomes are small organelles where fatty acid synthesis occurs.

ANS: A. False. In plants, only mitochondria perform cellular respiration (using oxygen to break down organic molecules to produce carbon dioxide) just as in animal cells. Chloroplasts perform photosynthesis in which water molecules are split to generate oxygen and fix carbon dioxide molecules.

B. False. Mitochondria have their own division cycle and their numbers change based on the rate of division.

C. True D. True E. False. Lysosomes house enzymes that break down nutrients for use by the cell and help recycle materials that cannot be used, which will later be excreted from the cell.

DIF: Moderate REF: 1.4 OBJ: 1.4.d Explain how chloroplasts and mitochondria cooperate as plant cells convert light energy into chemical energy. | 1.4.e Compare the function of lysosomes and peroxisomes. | 1.4.f Compare the structure, location, and function of the endoplasmic reticulum and Golgi apparatus. MSC: Evaluating 10. Number the following cells based on when they are believed to have evolved. ___ anaerobic eukaryote ___ aerobic prokaryote ___ anaerobic prokaryote ___ photosynthetic eukaryote ___ aerobic eukaryote ___ photosynthetic prokaryote ANS: _2_ anaerobic eukaryote _4_ aerobic prokaryote _1_ anaerobic prokaryote _6_ photosynthetic eukaryote _5_ aerobic eukaryote _3_ photosynthetic prokaryote DIF: Difficult REF: 1.3 OBJ: 1.3.b Analyze how eukaryotic cells and organisms rely on the function of prokaryotic cells and their descendants. MSC: Analyzing 11. The protozoan Didinium feeds on other organisms by engulfing them. Describe the cellular characteristics necessary to feed this way and explain why this is not a mechanism by which bacteria can acquire energy. Are bacteria, in general, unable to feed on other cells in this way?

ANS: For Didinium to engulf its prey, it needs to be larger than the cellular target, it needs to be mobile, and it needs to be able to change its shape. Movement and changing shape requires a flexible membrane and a dynamic cytoskeleton.

By comparison, Bacteria are small, have no cytoskeleton and cannot easily change their shape because they are generally surrounded by a rigid cell wall.

DIF: Difficult REF: 1.4 OBJ: 1.3.a Express the structural differences between prokaryotes and eukaryotes.

MSC: Analyzing 12. Indicate whether the following statements are TRUE or FALSE. If the statement is false, explain why it is false.

A. Plants do not require a cytoskeleton because they have a cell wall that lends structure and support to the cell.

B. The cytoskeleton is used as a transportation grid for the efficient, directional movement of cytosolic components.

C. Thermal energy promotes random movement of proteins, vesicles, and small molecules in the cytosol.

ANS: A. False. Although plant cells do have a cell wall that lends structure and support, they still need a cytoskeleton, which also helps with connections between cells and the transport of vesicles inside the cell.

B. True C. True DIF: Easy REF: 1.4 OBJ: 1.4.j Outline the role the cytoskeleton plays in plant cells. MSC: Evaluating 13. Indicate whether the following statements are TRUE or FALSE. If the statement is false, explain why it is false.

A. Primitive plant, animal, and fungal cells probably acquired mitochondria after they diverged from a common ancestor.

B. Protozoans are single-celled eukaryotes with cell morphologies and behaviors that can be as complex as those of some multicellular organisms.

C. The first eukaryotic cells on Earth must have been aerobic; otherwise, they would not have been able to survive when the planet’s atmosphere became oxygen-rich.

ANS: A. False. The mitochondria in modern plant, animal, and fungal cells are very similar, implying that these lines diverged after the mitochondrion was acquired by the ancestral eukaryote.

B. True C. False. The first eukaryotic cells likely contained a nucleus but no mitochondria. These ancestral eukaryotes subsequently adapted to survive in a world filled with oxygen by engulfing primitive aerobic prokaryotic cells.

DIF: Easy REF: 1.4 OBJ: 1.4.k Describe the ancestral cell that likely engulfed the aerobic bacteria that gave rise to mitochondria and explain why this event is thought to have preceded the acquisition of chloroplasts.

MSC: Evaluating 14. Given what you know about the differences between prokaryotic cells and eukaryotic cells, rate the following biological processes as “suitable” or “unsuitable” for study using E. coli as the model organism. Provide a short explanation for each answer.

1. formation of the endoplasmic reticulum 2. DNA replication 3. how the actin cytoskeleton contributes to cell shape 4. how cells decode their genetic instructions to make proteins 5. how mitochondria get distributed to cells during cell division ANS: 1. Unsuitable because prokaryotes do not have organelles.

2. Suitable because DNA replication is a process that that occurs in all cells.

3. Unsuitable because prokaryotes lack cytoskeletons.

4. Suitable because the genetic code and the processes of transcription and translation is highly conserved across all species.

5. Unsuitable because prokaryotes do not have organelles—mitochondria are thought to be derived from early aerobic bacteria, and may be instead suitable for studying the mechanism for consuming oxygen to generate ATP. DIF: Moderate REF: 1.3 OBJ: 1.3.a Express the structural differences between prokaryotes and eukaryotes.

MSC: Evaluating 15. For each process (A–D), identify the simplest model organism from the list of three that would be best used for investigation:

Figure 1-57

ANS: (A) C. elegans (B) Arabidopsis (C) mouse (D) Drosophila DIF: Easy REF: 1.5 OBJ: 1.5.a Review why scientists study model organisms. | 1.5.c Compare flies, worm, fish, and mice as model organisms and name a benefit of studying each. MSC: Analyzing 16. You wish to explore how mutations in specific genes affecting sugar metabolism might alter tooth development. Which organism is likely to provide the best model system for your studies, and why? a. horses b. mice c. E. coli d. Arabidopsis ANS: (B) Mice are likely to provide the best model system. Mice have teeth and have long been used as a model organism. Mice reproduce relatively rapidly, and the extensive scientific community that works with mice has developed techniques to facilitate genetic manipulations. E. coli (a bacterium) and Arabidopsis (a plant) do not have teeth. Horses like sugar and have big teeth, but they would not be a good model organism. There is not an extensive scientific community working on the molecular and biochemical mechanisms of cell behaviors in horses; they are expensive and have a long reproduction time, which makes genetic studies costly and slow; and tools for genetic manipulation (other than traditional breeding) have not been developed.

DIF: Easy REF: 1.5 OBJ: 1.5.c Compare flies, worm, fish, and mice as model organisms and name a benefit of studying each. MSC: Evaluating 17. Indicate whether the following statements are TRUE or FALSE. If the statement is false, explain why it is false.

A. The human genome is roughly 30 times larger than the Arabidopsis genome, but contains approximately the same number of protein-coding genes.

B. The variation in genome size among protozoans is larger than that observed across all species of mammals, birds, and reptiles.

C. The vast majority of our genome encodes functional RNA molecules or proteins and most of the intervening DNA is nonfunctional.

ANS: A. True B. True

C. False. It is a relatively small proportion of our DNA that encodes RNA and protein molecules. The majority of nonencoding sequences is probably involved in critical regulatory processes.

DIF: Difficult REF: 1.5 OBJ: 1.5.g Summarize the roles played by the nucleotide sequences contained in an organism’s genome. | 1.5.e Assess the relationship between genome size and gene number. MSC: Evaluating 18. Genes that have homologs in a variety of species have been discovered through the analysis of genome sequences.

In fact, it is not uncommon to find a family of homologous genes encoding proteins that are unmistakably similar in amino acid sequence in organisms as diverse as budding yeast, archaea, plants, and humans. Even more remarkably, many of these proteins can substitute functionally for their homologs in other organisms (i.e., if we delete a native gene and replace it with a homologous gene from another species, the protein will be produced and its function preserved). Use your understanding of evolution to explain how this is possible.

ANS: All living beings on Earth (and thus, all cells) are thought to be derived from a common ancestor. Solutions to many of the essential challenges that face a cell (such as the synthesis of proteins, lipids, and DNA) seem to have been achieved in this ancient common ancestor. The ancestral cell therefore possessed sets of proteins to carry out these essential functions. Many of the essential challenges facing modern-day cells are the same as those facing the ancestral cell, and the ancient solutions are often still effective. Thus, it is not uncommon for organisms to use proteins and biochemical pathways inherited from their ancestors. Although these proteins usually show some species-specific diversification, they still retain the basic biochemical characteristics of the ancestral protein. For example, homologous proteins often retain their ability to interact with a specific protein target, even in cells of diverse species. Because the basic biochemical characteristics are retained, homologous proteins are often capable of functionally substituting for one another.

DIF: Moderate REF: 1.5 OBJ: 1.5.f Classify the nature of homologous genes and proteins. MSC: Evaluating 19. Match each biological process with the model organism that is best suited or most specifically useful for its study, based on information provided in your textbook. You may use individual processes more than once and more than one process can be matched with each organism.

A. cell division B. development (multicellular) C. programmed cell death D. photosynthesis E. immunology 1. _____ A. thaliana (Arabidopsis) 2. _____ M. musculus (mouse) 3. _____ S. pombe 4. _____ C. elegans 5. _____ S. cerevisiae 6. _____ D. rerio (zebrafish) 7. _____ D. melanogaster 1. ANS: B, D A. thaliana (Arabidopsis) 2. ANS: B, E M. musculus (mouse) 3. ANS: A S. pombe

4. ANS: C C. elegans 5. ANS: A S. cerevisiae 6. ANS: B D. rerio (zebrafish) 7. ANS: B D. melanogaster DIF: Easy REF: 1.5 OBJ: 1.5.a Review why scientists study model organisms | 1.5.c Compare flies, worm, fish, and mice as model organisms and name a benefit of studying each. MSC: Analyzing 20. Generate an analogy to describe the cell as the simplest unit of life to somebody who has not studied biology. Be sure to incorporate the observable features of living organisms, which distinguishes them from inanimate objects.

ANS: Cities, factories, other types of human communities provide good models for analogical thinking about cells and organisms. Key elements that should be included are: Organization: Cities have organizational requirements for delivery of water, heat, and light, and for determining how things are moving/transported.

Homeostasis: An example would include snow removal from city streets and thermostat-controlled temperatures inside buildings. Homeostasis also can include waste removal (sewers and garbage trucks).

Reproduction: Blueprints and structural information for an entire city could be used to create a replica of any city.

Growth and development: Cities grow in response to increases in population. Resources are expended to build new dwellings and increase the capacity to provide energy to them.

DIF: Moderate REF: 1.1 OBJ: 1.1.b Outline, with examples, ways in which cells share a basic fundamental chemistry.

MSC: Creating 21. Employ the principles of evolution discussed in this chapter to explain how the specific features and predatory behaviors of some primitive eukaryotes may have given them a selective advantage over others 1.5 billion years ago.

ANS: The Earth’s atmosphere became oxygen-rich roughly 1.5 billion years ago. If some primitive predatory eukaryotic cells were similar to modern-day protozoans, they may have been mobile and able to engulf other cells. These characteristics would have been advantageous in the face of a changing atmosphere, and the establishment of a symbiotic relationship with an engulfed aerobe would have been selected for in the eukaryotic cell populations.

DIF: Difficult REF: 1.4 OBJ: 1.4.k Describe the ancestral cell that likely engulfed the aerobic bacteria that gave rise to mitochondria and explain why this event is thought to have preceded the acquisition of chloroplasts. MSC: Applying 22. Evolutionary biologists have always used a broad range of modern organisms to infer the characteristics that ancestral organisms may have possessed. Genomic sequences are now available for an increasing number of species, and scientists studying evolutionary processes can take advantage of this enormous amount of data to bring evolution into the arena of molecular studies. By aligning the sequences of homologous genes and looking for regions of similarity and where changes have occurred, it is possible to infer the sequence of the ancestral gene.

A. What term is used to describe the changes in gene sequences that have occurred? How can we use what we know about this process to construct a time line showing when various sequence changes occurred and when they led to the modern sequences that we know today?

B. It is possible to express an ancestral gene sequence in modern organisms and subsequently compare the function of its product with that of the modern protein. Why might this approach give misleading conclusions?

ANS: A. Changes in gene sequence occur through mutation. Mutations accumulate over time, occurring independently and at different sites in each gene lineage. Homologous genes that diverged recently will differ only slightly;

genes that diverged long ago will differ more. Knowing the average mutation rate, you can estimate the time that has elapsed since the different versions of the gene diverged. By seeing how closely the various members of th e family of homologous genes resemble one another, you can draw up a family tree, showing the sequence of lineage splits that lead from the ancestral gene to its many modern descendants. Suppose this family tree shows that family members A and B diverged from one another long ago, but that C diverged from B more recently; and suppose that at a certain site in the gene, A and B have the same sequence but C is different. Then, it is likely that the sequence of A and B is ancestral, while that of C reflects a recent mutation that has occurred in the lineage of C alone.

B. Although an inferred ancestral sequence can be reconstructed, and the protein expressed, you would be placing an inferred, ancient protein in the context of a modern cell. If there are important interacting partners for the modern protein, there is a chance they may not recognize the ancestral protein, and therefore any information about its function may be inaccurate.

DIF: Difficult REF: 1.5 OBJ: 1.5.f Classify the nature of homologous genes and proteins. | 1.5.g Summarize the roles played by the nucleotide sequences contained in an organism’s genome. MSC: Evaluating 23. The antibiotic streptomycin inhibits protein synthesis in bacteria. If this antibiotic is added to a culture of animal cells, protein synthesis in the cytosol continues normally. However, over time, the population of mitochondria in the cell becomes depleted. Specifically, it is observed that the protein-synthesis machinery inside the mitochondria is inhibited.

A. Explain this observation based on what you know about the origins of the modern eukaryote.

B. What do you expect to observe if, in a new experiment, animal cells are treated with diphtheria toxin, a compound that is known to block cytosolic protein synthesis but does not have any impact on bacterial growth?

ANS: A. If the mitochondria originated from an ancient aerobic bacterium that was engulfed by an ancient eukaryote, as postulated, it is possible that an antibiotic that inhibits protein synthesis in bacteria could also block that process in mitochondria.

B. We would expect that although cytosolic protein synthesis would stop, mitochondrial protein synthesis should still occur normally (at least for a little while). This result would lend further support to the idea that mitochondria are derived from a noneukaryotic organism. If this were not the case, these compounds would be expected to affect protein synthesis at both locations.

DIF: Difficult REF: 1.4 OBJ: 1.4.k Describe the ancestral cell that likely engulfed the aerobic bacteria that gave rise to mitochondria and explain why this event is thought to have preceded the acquisition of chloroplasts.

MSC: Applying 24. You have been following the recent presidential elections and have heard some candidates disparaging excessive and “unnecessary” federal government expenditures. One particular candidate asks: “Why are we spending millions of dollars studying fruit flies? How can that possibly help us find a cure for cancer?” Use your knowledge of model organisms to explain why studies in D. melanogaster (the fruit fly) are actually an excellent use of research funding.

ANS: Funding research on D. melanogaster is a worthwhile investment for several reasons: (1) working with insect animal models is relatively inexpensive; (2) fruit flies have historically proven useful in helping understand eukaryotic chromosome behavior; and (3) many of the genes in Drosophila are highly similar in sequence to the homologous human genes, and thus can be used to study human diseases.

DIF: Moderate REF: 1.5 OBJ: 1.5.a Review why scientists study model organisms. | 1.5.c Compare flies, worm, fish, and mice as model organisms and name a benefit of studying each. MSC: Evaluating 25. Cellular processes are often regulated by unknown mechanisms. In many cases, biologists work backward in an attempt to understand a process in which they are interested. This was the case when Nurse and Hartwell were trying to understand how cell division is controlled in yeast. Describe the process by which they “broke” the system and then supplied the “missing parts” to get the cell cycle running again. What further evidence did they collect to show that human cells and yeast cells regulate the cell cycle using a similar mechanism?

ANS: Nurse and Hartwell first treated yeast cells with a chemical mutagen. The mutated population of cells was then grown and observed. Cells that demonstrated defects in cell-cycle regulation (characterized by cell-cycle arrest, largerthan-normal cells, and smaller-than-normal cells) were then isolated. The use of a library of plasmids that each express a normal gene from yeast cells allowed the scientists to identify exactly which gene could be used to “rescue” the mutant, because when the normal gene is expressed again, the cells return to a normal cell cycle. After this big result, the scientists went on to show that the homologous gene from other organisms could also rescue the mutant phenotype.

The most exciting result was obtained with the human version of the cdc2 gene, which demonstrated that there are common principles underlying cell-cycle regulation across a large range of eukaryotic organisms.

DIF: Moderate REF: 1.5 OBJ: 1.5.a Review why scientists study model organisms. | 1.5.f Classify the nature of homologous genes and proteins. | 1.5.h Outline an experiment that would allow investigators to determine whether proteins from different eukaryotes are functionally interchangeable. MSC: Understanding 26. Your friend has just returned from a deep-sea mission and claims to have found a new single-celled life form. He believes this new life form may not have descended from the common ancestor that all types of life on Earth share. You are convinced that he must be wrong, and you manage to extract DNA from the cells he has discovered. He says that the mere presence of DNA is not enough to prove the point: his cells might have adopted DNA as a useful molecule quite independently of all other known life-forms. What could you do to provide additional evidence to support your argument?

ANS: You could use modern technology to discover the sequence of the DNA. If you are right, you would expect to find parts of this sequence that are unmistakably similar to corresponding sequences in other, familiar, living organisms; it would be highly improbable that such similar sequences would have evolved independently. You could, of course, also analyze other features of the chemistry of his cells; for example, do they contain proteins made of the same set of 20 amino acids? This could all be supporting evidence that this newly discovered species arose from the same common ancestral cells as all other life on Earth.

DIF: Easy REF: 1.5 OBJ: 1.5.f Classify the nature of homologous genes and proteins. | 1.5.g Summarize the roles played by the nucleotide sequences contained in an organism’s genome. MSC: Applying 27. Draw and annotate a diagram that explains how living systems are autocatalytic.

ANS: Use Figure 4.1 for reference. Important points for student annotations should include: • Nucleotides are building blocks for DNA and RNA. • RNA polynucleotides contain ordered information templating the assembly of amino acids into functional proteins.

• Some proteins are catalysts necessary to produce DNA and RNA from individual nucleotides.

DIF: Easy REF: 1.1 OBJ: 1.1.c Explain how the relationship between DNA, RNA, and protein—as laid out in the central dogma—makes the self-replication of living cells possible. MSC: Creating

28. Use the list of structures below to label the schematic drawing of an animal cell in Figure 1-41.

Figure 1-41

A. plasma membrane B. nuclear envelope C. cytosol D. Golgi apparatus E. endoplasmic reticulum F. mitochondrion G. transport vesicles ANS: A. plasma membrane—3 B. nuclear envelope—5 C. cytosol—1 D. Golgi apparatus—2 E. endoplasmic reticulum—4 F. mitochondrion—7 G. transport vesicles—6 DIF: Easy REF: 1.4 OBJ: 1.4.f Compare the structure, location, and function of the endoplasmic reticulum and Golgi apparatus. | 1.4.h Relate the location of the cytosol with respect to the cell’s membrane-enclosed organelles.

MSC: Remembering 29. Identify the appropriate cell type in which the listed structure or molecule can be found. Note that the structure or molecule can be found in more than one type of cell.

Figure 1-71

ANS:

Figure 1-71A

DIF: Easy REF: 1.3 OBJ: 1.3.a Describe the structural differences between prokaryotes and eukaryotes.

MSC: Analyzing

CHAPTER 2

Chemical Components of Cells CHEMICAL BONDS

2.1.a Review the properties that distinguish the chemistry of living things from the abiotic chemistry of solids, liquids, and gases.

2.1.b Distinguish between elements, atoms, ions, isotopes, molecules, and salts.

2.1.c State the location, charge, and relative size of protons, neutrons, and electrons and their numbers in an atom of carbon, hydrogen, oxygen, and nitrogen.

2.1.d Differentiate between atomic number and atomic weight and state how each is estimated.

2.1.e Express the concept of a mole and explain how to prepare a 100 mM solution.

2.1.f Relate how the behavior of atoms corresponds with their position in the periodic table and infer what this means for the chemical reactivity of the elements commonly found in living organisms.

2.1.g State the number of covalent bonds that can be formed by atoms of hydrogen, oxygen, nitrogen, and carbon and express how the electronic configuration of carbon, in particular, dictates the three-dimensional shape of organic molecules.

2.1.h Identify what determines the polarity of a chemical bond and summarize the consequences of this property for the solubility and hydrophobicity of a molecule or salt, as well as its ability to act as an acid or base.

2.1.i Recall the characteristics of a hydrophilic molecule.

2.1.j Distinguish between acids and bases and demonstrate how they cooperate to maintain the pH of cells.

2.1.k Differentiate between covalent and ionic bonds in terms of their electronic configuration, strength and stability, and their prevalence and role in biological systems.

2.1.l Contrast hydrogen bonds, electrostatic attractions, van der Waal’s attractions, and the hydrophobic force in terms of how and when they form and the role they play in cell biology.

SMALL MOLECULES IN CELLS 2.2.a Distinguish between organic and inorganic compounds.

2.2.b Express how the chemical and physical properties of methyl groups (−CH3), hydroxyl groups (−OH), carboxyl groups (−COOH), phosphate groups (−PO32–), and amino groups (−NH2) influence the behavior of molecules in which these groups typically occur.

2.2.c Illustrate how the processes of condensation and hydrolysis are central to the synthesis and breakdown of the large organic molecules of the cell from sets of smaller organic building blocks.

2.2.d Relate the different roles that sugars can play in the cell.

2.2.e Propose how a monosaccharide can form a ring structures in an aqueous solution.

2.2.f Evaluate the role that isomers play in biological systems.

2.2.g Define the terms fatty acid, steroid, phospholipid, and triacylglycerol and describe the structure, properties, and function of each.

2.2.h Summarize why the amphipathic nature of phospholipids is crucial for cell biology.

2.2.i Predict how the saturation of fatty acid tails affects the fluidity of cell membranes.

2.2.j Identify the features that all amino acids have in common.

2.2.k Summarize what makes amino acids chemically unique and categorize the 20 amino acids commonly found in proteins on the basis of their chemical properties.

2.2.l Express the difference between nucleotides and nucleosides.

2.2.m Contrast RNA and DNA and evaluate why these nucleic acids play different roles in cells.

2.2.n Outline the roles that phosphoanhydride, phosphodiester, and hydrogen bonds play in the actions of nucleotides and nucleic acids.

MACROMOLECULES IN CELLS 2.3.a Relate how the repetitive polymerization of monomers into polymers can yield macromolecules with diverse properties and functions.

2.3.b Assess the role that covalent and noncovalent bonds play in the three-dimensional conformation of macromolecules.

2.3.c Explain how weak, noncovalent bonds can lead to strong and specific associations between macromolecules or between an enzyme and its substrate.

2.3.d Review how investigators can experimentally differentiate between a macromolecule, such as a purified protein, and a loosely associated aggregate of heterogeneous, small organic molecules.

MULTIPLE CHOICE 1. Select the answer that BEST completes the following statement: Chemical reactions in living systems occur in an __________ environment, within a narrow range of temperatures. a. optimal b. organic c. extracellular d. aqueous ANS: D DIF: Easy REF: 2.1 OBJ: 2.1.a Review the properties that distinguish the chemistry of living things from the abiotic chemistry of solids, liquids, and gases. MSC: Remembering 2. Which subatomic particles contribute to the atomic number for any given element? a. protons b. protons and neutrons c. neutrons d. protons and electrons

ANS: A DIF: Easy REF: 2.1 OBJ: 2.1.d Differentiate between atomic number and atomic weight and state how each is estimated. MSC: Remembering 3. Which subatomic particles contribute to the atomic mass for any given element? a. protons b. protons and neutrons c. neutrons d. protons and electrons ANS: B DIF: Easy REF: 2.1 OBJ: 2.1.d Differentiate between atomic number and atomic weight and state how each is estimated. MSC: Remembering 4. Which subatomic particles can vary between isotopes of the same element, without changing the observed chemical properties? a. electrons b. protons and neutrons c. neutrons d. neutrons and electrons ANS: C DIF: Easy REF: 2.1 OBJ: 2.1.b Distinguish between elements, atoms, ions, isotopes, molecules, and salts.

MSC: Remembering 5. Figure 2-5 depicts the structure of carbon. Use the information in the diagram to choose the correct atomic number and atomic weight, respectively, for an atom of carbon.

Figure 2-5

a. 6; 12 b. 12; 12 c. 6; 18 d. 12; 6 ANS: A DIF: Easy REF: 2.1 OBJ: 2.1.d Differentiate between atomic number and atomic weight and state how each is estimated. | 2.1.c State the position, size and charge, of protons, neutrons, and electrons—and compute their relative numbers in an atom of carbon, hydrogen, oxygen, and nitrogen. MSC: Understanding 6. Carbon 14 is an unstable isotope of carbon that decays very slowly. Compared to the common, stable carbon 12

isotope, carbon 14 has two additional a. electrons. b. neutrons. c. protons. d. ions.

ANS: B DIF: Easy REF: 2.1 OBJ: 2.1.b Distinguish between elements, atoms, ions, isotopes, molecules, and salts.

MSC: Remembering 7. If the isotope 32S has 16 protons and 16 neutrons, how many protons, neutrons, and electrons will the isotope 35S have, respectively? a. 16; 20; 15 b. 16; 19; 15 c. 16; 19; 16 d. 16; 19; 17 ANS: C DIF: Easy REF: 2.1 OBJ: 2.1.b Distinguish between elements, atoms, ions, isotopes, molecules, and salts.

MSC: Understanding 8. Avogadro’s number was established as the total number of units (atoms or molecules) in a mole, and a mole of any substance is X grams of it, where X is equal to the substance’s molecular weight. A standard unit, the mole, allows scientists to calculate concentrations of materials dissolved in solutions.

Example: Sulfur has a molecular weight of 32. Therefore, 32 g of sulfur = 1 mole of sulfur = 6 × 1023 sulfur atoms.

How many moles and atoms, respectively, are there in 120 grams of sulfur? a. 3.75; 6 × 1023 b. 32; 6 × 1023 c. 1.75; 1.05 × 1024 d. 3.75; 2.25 × 1024 ANS: D DIF: Moderate REF: 2.1 OBJ: 2.1.e Express the concept of a mole and explain how to prepare a 100 mM solution. MSC: Applying 9. The first task you are assigned in your summer laboratory job is to prepare a concentrated NaOH stock solution. The molecular weight of NaOH is 40. How many grams of solid NaOH will you need to weigh out to obtain a 500 mL solution that has a concentration of 10 M? a. 800 g b. 200 g c. 400 g d. 160 g ANS: B DIF: Moderate REF: 2.1 OBJ: 2.1.e Express the concept of a mole and explain how to prepare a 100 mM solution. MSC: Applying 10. You have a concentrated stock solution of 10 M NaOH and want to use it to produce a 150 mL solution of 3 M NaOH. What volume of water and stock solutions will you measure out to make this new solution? a. 135 mL of water; 15 mL of NaOH stock

b. 115 mL of water; 35 mL of NaOH stock c. 100 mL of water; 50 mL of NaOH stock d. 105 mL of water; 45 mL of NaOH stock ANS: D DIF: Moderate REF: 2.1 OBJ: 2.1.e Express the concept of a mole and explain how to prepare a 100 mM solution. MSC: Applying 11. There are 90 naturally occurring elements on the earth, __________ of which compose 96% of the mass of living organisms. a. 4 b. 6 c. 7 d. 8 ANS: A DIF: Easy REF: 2.1 OBJ: 2.1.a Review the properties that distinguish the chemistry of living things from the abiotic chemistry of solids, liquids, and gases. MSC: Remembering 12. A covalent bond between two atoms is formed as a result of the a. sharing of electrons. b. loss of electrons from both atoms. c. loss of a proton from one atom. d. transfer of electrons from one atom to the other.

ANS: A DIF: Easy REF: 2.1 OBJ: 2.1.k Differentiate between covalent and ionic bonds in terms of their electronic configuration, strength and stability, and their prevalence and role in biological systems.

MSC: Remembering 13. An ionic bond between two atoms is formed as a result of the a. sharing of electrons. b. loss of electrons from both atoms. c. loss of a proton from one atom. d. transfer of electrons from one atom to the other.

ANS: D DIF: Easy REF: 2.1 OBJ: 2.1.k Differentiate between covalent and ionic bonds in terms of their electronic configuration, strength and stability, and their prevalence and role in biological systems.

MSC: Remembering 14. Table 2-14 indicates the number and arrangement of electrons in the first four atomic electron shells for selected elements. On the basis of the information in the chart and what you know about atomic structure, which elements are chemically inert?

Table 2-14

a. carbon; sulfur b. helium; neon c. sodium; potassium d. magnesium; calcium ANS: B DIF: Easy REF: 2.1 OBJ: 2.1.f Relate how the behavior of atoms corresponds with their position in the periodic table and infer what this means for the chemical reactivity of the elements commonly found in living organisms.

MSC: Understanding 15. Table 2-14 indicates the number and arrangement of electrons in the first four atomic electron shells for selected elements. On the basis of the information in the chart and what you know about atomic structure, which elements will form ions with a net charge of +1 in solution?

Table 2-14

a. carbon; sulfur b. helium; neon c. sodium; potassium

d. magnesium; calcium ANS: C DIF: Easy REF: 2.1 OBJ: 2.1.f Relate how the behavior of atoms corresponds with their position in the periodic table and infer what this means for the chemical reactivity of the elements commonly found in living organisms.

MSC: Understanding 16. Table 2-14 indicates the number and arrangement of electrons in the first four atomic electron shells for selected elements. On the basis of the information in the chart and what you know about atomic structure, which elements will form ions with a net charge of +2 in solution?

Table 2-14

a. carbon; sulfur b. helium; neon c. sodium; potassium d. magnesium; calcium ANS: D DIF: Easy REF: 2.1 OBJ: 2.1.f Relate how the behavior of atoms corresponds with their position in the periodic table and infer what this means for the chemical reactivity of the elements commonly found in living organisms.

MSC: Understanding 17. Table 2-14 indicates the number and arrangement of electrons in the first four atomic electron shells for selected elements. On the basis of the information in the chart and what you know about atomic structure, which elements form stable but reactive diatomic gases?

Table 2-14

a. nitrogen; oxygen b. helium; neon c. sodium; potassium d. magnesium; calcium ANS: A An oxygen atom with six outer electrons needs two more to attain a stable outer shell. This can be achieved by forming two covalent bonds with a second oxygen, as shown on the right. Similarly, a nitrogen atom needs three more electrons and makes three covalent bonds with another nitrogen atom.

DIF: Moderate REF: 2.1 OBJ: 2.1.f Relate how the behavior of atoms corresponds with their position in the periodic table and infer what this means for the chemical reactivity of the elements commonly found in living organisms.

MSC: Understanding 18. Which of the following factors DO NOT influence the length of a covalent bond? a. the tendency of atoms to fill the outer electron shells b. the attractive forces between negatively charged electrons and positively charged nuclei c. the repulsive forces between the positively charged nuclei d. the minimization of repulsive forces between the two nuclei by the cloud of shared electrons ANS: A The tendency to complete the outer electron shell is the reason covalent bonds form, but it does not address the length of the bond that is formed.

DIF: Difficult REF: 2.1 OBJ: 2.1.k Differentiate between covalent and ionic bonds in terms of their electronic configuration, strength and stability, and their prevalence and role in biological systems. MSC: Analyzing 19. Double covalent bonds are both shorter and stronger than single covalent bonds, but they also limit the geometry of the molecule because they a. create a new arrangement of electron shells. b. change the reactivity of the bonded atoms. c. limit the rotation of the bonded atoms.

d. prevent additional bonds from being formed with the bonded atoms.

ANS: C DIF: Moderate REF: 2.1 OBJ: 2.1.k Differentiate between covalent and ionic bonds in terms of their electronic configuration, strength and stability, and their prevalence and role in biological systems. MSC: Understanding 20. Polar covalent bonds are formed when the electrons in the bond are not shared equally between the two nuclei.

Which one of these molecules contains polar bonds? a. molecular oxygen b. methane c. propane d. water ANS: D DIF: Easy REF: 2.1 OBJ: 2.1.h Identify what determines the polarity of a chemical bond and summarize the consequences of this property for the solubility and hydrophobicity of a molecule or salt, as well as its ability to act as an acid or base. MSC: Understanding 21. Carbon atoms can form double bonds with other carbon atoms, nitrogen atoms, and oxygen atoms. Double bonds can have important consequences for biological molecules because they are __________ compared to single covalent bonds. a. long b. rigid with respect to rotation c. weak d. unstable ANS: B DIF: Easy REF: 2.1 OBJ: 2.1.g Calculate the number of bonds that can be formed by atoms of hydrogen, oxygen, nitrogen, and carbon—and express how the electronic configuration of carbon, in particular, dictates the threedimensional shape of organic molecules. MSC: Remembering 22. Which combination of answers best completes the following statement: When atoms are held together by __________ __________, they are typically referred to as __________. a. hydrogen bonds; molecules b. ionic interactions; salts c. ionic interactions; molecules d. double bonds; nonpolar ANS: B DIF: Easy REF: 2.1 OBJ: 2.1.k Differentiate between covalent and ionic bonds in terms of their electronic configuration, strength and stability, and their prevalence and role in biological systems. MSC: Understanding 23. Although covalent bonds are 10–100 times stronger than noncovalent interactions, many biological processes depend upon the number and type of noncovalent interactions between molecules. Which of the noncovalent interactions below will contribute most to the strong and specific binding of two molecules, such as a pair of proteins? a. electrostatic attractions b. hydrogen bonds c. hydrophobic interactions d. van der Waals attractions ANS: A DIF: Easy REF: 2.1 OBJ: 2.1.l Contrast hydrogen bonds, electrostatic attractions, van der Waal’s attractions,

and the hydrophobic force in terms of how and when they form and the role they play in cell biology.

MSC: Analyzing 24. Which of the following expressions accurately describes the calculation of pH? a. pH = −log10[H+] b. pH = log10[H+] c. pH = −log2[H+] d. pH = −log10[OH−] ANS: A DIF: Easy REF: 2.1 OBJ: 2.1.j Distinguish between acids and bases and demonstrate how they cooperate to maintain the pH of cells. MSC: Remembering 25. The pH of an aqueous solution is an indication of the concentration of available protons. However, you should not expect to find lone protons in solution; rather, the proton is added to a water molecule to form a/an __________ ion. a. hydroxide b. ammonium c. chloride d. hydronium ANS: D DIF: Moderate REF: 2.1 OBJ: 2.1.j Distinguish between acids and bases and demonstrate how they cooperate to maintain the pH of cells. MSC: Understanding 26. Larger molecules have hydrogen-bonding networks that contribute to specific, high-affinity binding. Smaller molecules such as urea can also form these networks. How many hydrogen bonds can urea (Figure 2 -26) form if dissolved in water?

Figure 2-26

a. 6 b. 5 c. 3 d. 4 ANS: A Urea can form at least six hydrogen bonds in water: two from the oxygen atom and one from each hydrogen atom.

DIF: Easy REF: 2.1 OBJ: 2.1.g Calculate the number of bonds that can be formed by atoms of hydrogen, oxygen, nitrogen, and carbon—and express how the electronic configuration of carbon, in particular, dictates the threedimensional shape of organic molecules. MSC: Applying 27. Aromatic carbon compounds such as benzene are planar and very stable. Double-bond character extends around the

entire ring, which is why it is often drawn as a hexagon with a circle inside. This characteristic is caused by electron a. resonance. b. pairing. c. partial charge. d. stacking.

ANS: A DIF: Easy REF: 2.1 OBJ: 2.1.c State the position, size and charge, of protons, neutrons, and electrons—and compute their relative numbers in an atom of carbon, hydrogen, oxygen, and nitrogen. MSC: Understanding 28. __________ play an important role in organizing lipid molecules with long hydrocarbon tails into biological membranes. a. Hydrogen bonds b. Ionic bonds c. Hydrophobic forces d. Van der Waals attractions ANS: C DIF: Easy REF: 2.1 OBJ: 2.1.l Contrast hydrogen bonds, electrostatic attractions, van der Waal’s attractions, and the hydrophobic force in terms of how and when they form and the role they play in cell biology.

MSC: Understanding 29. Substances that release protons when they dissolve in water are acids. Which of the following household substance is acidic? a. coffee b. bleach c. hand soap d. water ANS: A DIF: Easy REF: 2.1 OBJ: 2.1.j Distinguish between acids and bases and demonstrate how they cooperate to maintain the pH of cells. MSC: Understanding 30. Cells contain buffers that help maintain a neutral pH. Which of the following statements is not relevant to how buffers work? a. Buffers are mixtures of weak acids and bases. b. Buffers can accept protons from acids. c. Buffers can donate protons to bases. d. Buffers catalyze oxidation-reduction reactions..

ANS: D DIF: Easy REF: 2.1 OBJ: 2.1.j Distinguish between acids and bases and demonstrate how they cooperate to maintain the pH of cells. MSC: Analyzing 31. Which of the following monomer building blocks is necessary to assemble selectively permeable boundaries around and inside cells? a. sugars b. fatty acids c. amino acids d. nucleotides

ANS: B DIF: Easy REF: 2.2 OBJ: 2.2.h Summarize why the amphipathic nature of fatty acids and phospholipids is crucial for cell biology. MSC: Remembering 32. There is incredible chemical diversity even in the simplest of cells. A typical bacterial cell contains more than 6000 different types of molecules. From the list below, select the class of molecules with the largest number of different types. a. nucleotides and precursors b. sugars and precursors c. amino acids and precursors d. fatty acids and precursors ANS: B DIF: Easy REF: 2.2 OBJ: 2.2.d Relate the different roles that sugars can play in the cell.

MSC: Remembering 33. Which of the following are examples of isomers? a. glucose and galactose b. alanine and glycine c. adenine and guanine d. glycogen and cellulose ANS: A Glucose and galactose are both six-carbon sugars and thus both have the formula C6H12O6. They are therefore isomers of each other. Adenine and guanine are bases containing different numbers of nitrogen and oxygen atoms. Glycogen and cellulose are different polymers of glucose. Alanine and glycine are amino acids with quite different side chains: a methyl group and a hydrogen atom, respectively.

DIF: Difficult REF: 2.2 OBJ: 2.2.f Evaluate the role that isomers play in biological systems. MSC: Understanding 34. Oligosaccharides are short sugar polymers that can become covalently linked to proteins and lipids through condensation reactions. These modified proteins and lipids are called glycoproteins and glycolipids, respectively.

Within a protein, which of the amino acids (shown in Figure 2-34) is the most probable target for this type of modification?

Figure 2-34

a. serine b. glycine c. phenylalanine d. methionine

ANS: A DIF: Moderate REF: 2.2 OBJ: 2.2.k Summarize what makes amino acids chemically unique and categorize the 20 amino acids commonly found in proteins on the basis of their chemical properties. MSC: Applying 35. Many types of cells have stores of lipids in their cytoplasm, usually seen as fat droplets. What is the lipid most commonly found in these droplets? a. cholesterol b. palmitic acid c. isoprene d. triacylglycerol ANS: D DIF: Easy REF: 2.2 OBJ: 2.2.g Define the terms fatty acid, steroid, phospholipid, and triacylglycerol and describe the structure, properties, and function of each. MSC: Remembering 36. Choose the answer that best fits the following statement: Cholesterol is an essential component of biological membranes. Although it is much smaller than the typical phospholipids and glycolipids in the membrane, it is a/an __________ molecule, having both hydrophilic and hydrophobic regions. a. polar b. oxygen-containing c. hydrophobic d. amphipathic ANS: D DIF: Easy REF: 2.2 OBJ: 2.2.g Define the terms fatty acid, steroid, phospholipid, and triacylglycerol and describe the structure, properties, and function of each. | 2.2.h Summarize why the amphipathic nature of phospholipids is crucial for cell biology. MSC: Understanding 37. Most types of molecules in the cell have asymmetric (chiral) carbons. Consequently there is the potential to have two different molecules that look much the same but are mirror images of each other and therefore not equivalent. These special types of isomers are called stereoisomers. Which of the four carbons circled in Figure 2-37 is the asymmetric carbon that determines whether the amino acid (threonine in this case) is a ᴅ or an ʟ stereoisomer?

Figure 2-37

a. 1 b. 2 c. 3 d. 4 ANS: C Two of the carbon atoms of threonine are asymmetric (numbered 2 and 3 in Figure 2-37) but by convention it is the αcarbon (number 3) that determines whether the amino acid is the ᴅ or ʟ isomer.

DIF: Difficult REF: 2.2 OBJ: 2.2.f Evaluate the role that isomers play in biological systems. MSC: Applying

38. The amino acids glutamine and glutamic acid are shown in Figure 2-38. They differ only in the structure of part of their side chains (circled). At pH7, what type of interactions are possible for glutamic acid but not for glutamine?

Figure 2-38

a. ionic bonds b. hydrogen bonds c. van der Waals interactions d. covalent bonds ANS: A DIF: Easy REF: 2.2 OBJ: 2.2.k Summarize what makes amino acids chemically unique and categorize the 20 amino acids commonly found in proteins on the basis of their chemical properties. MSC: Applying 39. Cells require one particular monosaccharide as a starting material to synthesize nucleotide building blocks. Which of the monosaccharides below fills this important role? a. glucose b. fructose c. ribulose d. ribose ANS: D DIF: Easy REF: 2.2 OBJ: 2.2.d Relate the different roles that sugars can play in the cell.

MSC: Remembering 40. DNA and RNA are different types of nucleic acid polymer. Which of the following is true of DNA but NOT true of RNA? a. It contains uracil. b. It contains thymine. c. It is single-stranded. d. It has 5′-to-3′ directionality.

ANS: C DIF: Easy REF: 2.2 OBJ: 2.2.m Contrast RNA and DNA and evaluate why these nucleic acids play different roles in cells. MSC: Remembering 41. The variety and arrangement of chemical groups on monomer subunits contribute to the conformation, reactivity, and surface of the macromolecule into which they become incorporated. What type of chemical group is circled on the nucleotide shown in Figure 2-41?

Figure 2-41

a. pyrophosphate b. phosphoryl c. carbonyl d. carboxyl ANS: B DIF: Easy REF: 2.2 OBJ: 2.2.n Outline the roles that phosphoanhydride, phosphodiester, and hydrogen bonds play in the actions of nucleotides and nucleic acids. MSC: Remembering 42. Both DNA and RNA are synthesized by covalently linking a nucleoside triphosphate to the previous nucleotide, constantly adding to a growing chain. In the case of DNA, the new strand becomes part of a stable helix. The two strands are complementary in sequence and antiparallel in directionality. What is the principal force that holds these two strands together? a. ionic interactions b. hydrogen bonds c. covalent bonds d. van der Waals interactions ANS: B DIF: Easy REF: 2.3 OBJ: 2.3.b Assess the role that covalent and noncovalent bonds play in the threedimensional conformation of macromolecules. MSC: Remembering 43. Each nucleotide in DNA and RNA has an aromatic base. What is the principal force that keeps the bases in a polymer from interacting with water? a. hydrophobic interactions b. hydrogen bonds c. covalent bonds d. van der Waals interactions ANS: A DIF: Easy REF: 2.3 OBJ: 2.3.b Assess the role that covalent and noncovalent bonds play in the threedimensional conformation of macromolecules. MSC: Remembering 44. Because there are four different monomer building blocks that can be used to assemble RNA polymers, the number of possible sequence combinations that can be created for an RNA molecule made of 100 nucleotides is a. 1004. b. 4100. c. 4 × 100.

d. 100/4.

ANS: B DIF: Easy REF: 2.3 OBJ: 2.3.a Relate how the repetitive polymerization of monomers into polymers can yield macromolecules with diverse properties and functions. MSC: Applying 45. There are 20 100 different possible sequence combinations for a protein chain with 100 amino acids. In addition to the amino acid sequence of the protein, what other factors INCREASE the potential for diversity in these macromolecules? a. free rotation around single bonds during synthesis b. noncovalent interactions sampled as protein folds c. the directionality of amino acids being added d. the planar nature of the peptide bond ANS: B DIF: Easy REF: 2.3 OBJ: 2.3.a Relate how the repetitive polymerization of monomers into polymers can yield macromolecules with diverse properties and functions. MSC: Analyzing 46. Macromolecules in the cell can often interact transiently as a result of noncovalent interactions. These weak interactions also produce stable, highly specific interactions between molecules. Which of the factors below is the most significant in determining whether the interaction will be transient or stable? a. the size of each molecule b. the concentration of each molecule c. the rate of synthesis d. surface complementarity between molecules ANS: D DIF: Easy REF: 2.3 OBJ: 2.3.c Explain how weak, noncovalent bonds can lead to strong and specific associations between macromolecules or between an enzyme and its substrate. MSC: Understanding 47. Table 2-14 indicates the number and arrangement of electrons in the first four atomic electron shells for selected elements. On the basis of the information in the chart and what you know about atomic structure, which elements form stable but reactive diatomic gases?

Table 2-14

a. nitrogen; oxygen

b. helium; neon c. sodium; potassium d. magnesium; calcium ANS: A An oxygen atom with six outer electrons needs two more to attain a stable outer shell. This can be achieved by forming two covalent bonds with a second oxygen, as shown on the right. Similarly, a nitrogen atom needs three more electrons and makes three covalent bonds with another nitrogen atom.

Figure 2-47A

DIF: Moderate REF: 2.1 OBJ: 2.1.f Relate how the behavior of atoms corresponds with their position in the periodic table and infer what this means for the chemical reactivity of the elements commonly found in living organisms.

MSC: Understanding

MATCHING 1. Oxygen, hydrogen, carbon, and nitrogen atoms are enriched in the cells and tissues of living organisms. The covalent bond geometries for these atoms influence the structure of larger biomolecules. Match the elements on the left with the bond geometries shown in Figure 2-48.

Figure 2-48

A. oxygen B. carbon C. nitrogen 1. ANS: B DIF: Easy REF: 2.1 OBJ: 2.1.g Calculate the number of bonds that can be formed by atoms of hydr ogen, oxygen, nitrogen, and carbon—and express how the electronic configuration of carbon, in particular, dictates the three-dimensional shape of organic molecules. MSC: Understanding 2. ANS: C DIF: Easy REF: 2.1 OBJ: 2.1.g Calculate the number of bonds that can be formed by atoms of hydr ogen, oxygen, nitrogen, and carbon—and express how the electronic configuration of carbon, in particular, dictates the three-dimensional shape of organic molecules. MSC: Understanding 3. ANS: B DIF: Easy REF: 2.1 OBJ: 2.1.g Calculate the number of bonds that can be formed by atoms of hydr ogen, oxygen, nitrogen, and carbon—and express how the electronic configuration of carbon, in particular, dictates the three-dimensional shape of organic molecules. MSC: Understanding 4. ANS: A DIF: Easy REF: 2.1 OBJ: 2.1.g Calculate the number of bonds that can be formed by atoms of hydrogen, oxygen, nitrogen, and carbon—and express how the electronic configuration of carbon, in particular, dictates the three-dimensional shape of organic molecules. MSC: Understanding 5. ANS: C DIF: Easy REF: 2.1 OBJ: 2.1.g Calculate the number of bonds that can be formed by atoms of hydrogen, oxygen, nitrogen, and carbon—and express how the electronic configuration of carbon, in particular, dictates the three-dimensional shape of organic molecules. MSC: Understanding 6. ANS: A DIF: Easy REF: 2.1 OBJ: 2.1.g Calculate the number of bonds that can be formed by atoms of hydrogen, oxygen, nitrogen, and carbon—and express how the electronic configuration of carbon, in particular, dictates the three-dimensional shape of organic molecules. MSC: Understanding 2. Indicate whether the molecules below are inorganic (A) or organic (B).

1. glucose 2. ethanol

3. sodium chloride 4. water 5. cholesterol 6. adenosine 7. calcium 8. glycine 9. oxygen 10. iron 11. phospholipid 1. ANS: B DIF: Easy REF: 2.2 OBJ: 2.2.a Distinguish between organic and inorganic compounds. MSC: Understanding 2. ANS: B DIF: Easy REF: 2.2 OBJ: 2.2.a Distinguish between organic and inorganic compounds. MSC: Understanding 3. ANS: A DIF: Easy REF: 2.2 OBJ: 2.2.a Distinguish between organic and inorganic compounds. MSC: Understanding 4. ANS: A DIF: Easy REF: 2.2 OBJ: 2.2.a Distinguish between organic and inorganic compounds. MSC: Understanding 5. ANS: B DIF: Easy REF: 2.2 OBJ: 2.2.a Distinguish between organic and inorganic compounds. MSC: Understanding 6. ANS: B DIF: Easy REF: 2.2 OBJ: 2.2.a Distinguish between organic and inorganic compounds. MSC: Understanding 7. ANS: A DIF: Easy REF: 2.2 OBJ: 2.2.a Distinguish between organic and inorganic compounds. MSC: Understanding 8. ANS: B DIF: Easy REF: 2.2 OBJ: 2.2.a Distinguish between organic and inorganic compounds. MSC: Understanding 9. ANS: A DIF: Easy REF: 2.2 OBJ: 2.2.a Distinguish between organic and inorganic compounds. MSC: Understanding 10. ANS: A DIF: Easy REF: 2.2 OBJ: 2.2.a Distinguish between organic and inorganic compounds. MSC: Understanding 11. ANS: B DIF: Easy REF: 2.2 OBJ: 2.2.a Distinguish between organic and inorganic compounds. MSC: Understanding 3. On the phospholipid molecule in Figure 2-50, label each numbered line with the correct term selected from the list below.

Figure 2-50

1. ANS: D DIF: Easy REF: 2.2 OBJ: 2.2.g Define the terms fatty acid, steroid, phospholipid, and triacylglycerol and describe the structure, properties, and function of each. MSC: Remembering 2. ANS: A DIF: Easy REF: 2.2 OBJ: 2.2.g Define the terms fatty acid, steroid, phospholipid, and triacylglycerol and describe the structure, properties, and function of each. MSC: Remembering 3. ANS: C DIF: Easy REF: 2.2 OBJ: 2.2.g Define the terms fatty acid, steroid, phospholipid, and triacylglycerol and describe the structure, properties, and function of each. MSC: Remembering 4. ANS: J DIF: Easy REF: 2.2 OBJ: 2.2.g Define the terms fatty acid, steroid, phospholipid, and triacylglycerol and describe the structure, properties, and function of each. MSC: Remembering 5. ANS: I DIF: Easy REF: 2.2 OBJ: 2.2.g Define the terms fatty acid, steroid, phospholipid, and triacylglycerol and describe the structure, properties, and function of each. MSC: Remembering 6. ANS: H DIF: Easy REF: 2.2 OBJ: 2.2.g Define the terms fatty acid, steroid, phospholipid, and triacylglycerol and describe the structure, properties, and function of each. MSC: Remembering 7. ANS: E DIF: Easy REF: 2.2 OBJ: 2.2.g Define the terms fatty acid, steroid, phospholipid, and triacylglycerol and describe the structure, properties, and function of each. MSC: Remembering 4. Match each term related to the structure of nucleic acids (A–I) with one of the descriptions provided.

A. base B. glycosidic bond C. nucleoside D. nucleotide E. phosphoanhydride bond F. phosphoester bond G. ribose

H. phosphodiester bond I. deoxyribose 1. ____ the linkage between two nucleotides 2. ____ the linkage between the 5′ sugar hydroxyl and a phosphate group 3. ____ the nitrogen-containing aromatic ring 4. ____ five-carbon sugar found in DNA 5. ____ sugar unit linked to a base 6. ____ linkage between the sugar and the base 7. ____ linkages between phosphate groups 8. ____ sugar linked to a base and a phosphate 9. ____ five-carbon sugar found in RNA 1. ANS: H DIF: Moderate REF: 2.2 OBJ: 2.2.l Express the difference between nucleotides and nucleosides. | 2.2.m Contrast RNA and DNA and evaluate why these nucleic acids play different roles in cells. | 2.2.n Outline the roles that phosphoanhydride, phosphodiester, and hydrogen bonds play in the actions of nucleotides and nucleic acids.

MSC: Analyzing 2. ANS: F DIF: Moderate REF: 2.2 OBJ: 2.2.l Express the difference between nucleotides and nucleosides. | 2.2.m Contrast RNA and DNA and evaluate why these nucleic acids play different roles in cells. | 2.2.n Outline the roles that phosphoanhydride, phosphodiester, and hydrogen bonds play in the actions of nucleotides and nucleic acids. MSC: Analyzing 3. ANS: A DIF: Moderate REF: 2.2 OBJ: 2.2.l Express the difference between nucleotides and nucleosides. | 2.2.m Contrast RNA and DNA and evaluate why these nucleic acids play different roles in cells. | 2.2.n Outline the roles that phosphoanhydride, phosphodiester, and hydrogen bonds play in the actions of nucleotides and nucleic acids. MSC: Analyzing 4. ANS: I DIF: Moderate REF: 2.2 OBJ: 2.2.l Express the difference between nucleotides and nucleosides. | 2.2.m Contrast RNA and DNA and evaluate why these nucleic acids play different roles in cells. | 2.2.n Outline the roles that phosphoanhydride, phosphodiester, and hydrogen bonds play in the actions of nucleotides and nucleic acids. MSC: Analyzing 5. ANS: C DIF: Moderate REF: 2.2 OBJ: 2.2.l Express the difference between nucleotides and nucleosides. | 2.2.m Contrast RNA and DNA and evaluate why these nucleic acids play different roles in cells. | 2.2.n Outline the roles that phosphoanhydride, phosphodiester, and hydrogen bonds play in the actions of nucleotides and nucleic acids. MSC: Analyzing 6. ANS: B DIF: Moderate REF: 2.2 OBJ: 2.2.l Express the difference between nucleotides and nucleosides. | 2.2.m Contrast RNA and DNA and evaluate why these nucleic acids play different roles in cells. | 2.2.n Outline the roles that phosphoanhydride, phosphodiester, and hydrogen bonds play in the actions of nucleotides and nucleic acids.

MSC: Analyzing 7. ANS: E DIF: Moderate REF: 2.2 OBJ: 2.2.l Express the difference between nucleotides and nucleosides. | 2.2.m Contrast RNA and DNA and evaluate why these nucleic acids play different roles in cells. | 2.2.n Outline the roles that phosphoanhydride, phosphodiester, and hydrogen bonds play in the actions of nucleotides and nucleic acids.

MSC: Analyzing 8. ANS: D DIF: Moderate REF: 2.2 OBJ: 2.2.l Express the difference between nucleotides and nucleosides. | 2.2.m Contrast RNA and DNA and evaluate why these nucleic acids play different roles in cells. | 2.2.n Outline the roles that phosphoanhydride, phosphodiester, and hydrogen bonds play in the actions of nucleotides and nucleic acids.

MSC: Analyzing 9. ANS: G DIF: Moderate REF: 2.2 OBJ: 2.2.l Express the difference between nucleotides and nucleosides. | 2.2.m Contrast RNA and DNA and evaluate why these nucleic acids play different roles in cells. | 2.2.n Outline the roles that phosphoanhydride, phosphodiester, and hydrogen bonds play in the actions of nucleotides and nucleic acids.

MSC: Analyzing

SHORT ANSWER 1. Indicate whether the statements below are TRUE or FALSE. If a statement is false, explain why it is false.

A. The chemistry of life is carried out and coordinated primarily by the action of small molecules.

B. Carbon-based compounds make up the vast majority of molecules found in cells.

C. The chemical reactions in living systems are loosely regulated, allowing for a wide range of products and more rapid evolution.

ANS: A. False. Although small molecules are important in many processes, the chemical reactions in living systems are regulated by the coordinated action of large polymeric molecules.

B. True C. False. The chemical reactions in living systems are very tightly controlled, ensuring that events occur at the proper time and at the proper location inside the cell.

DIF: Easy REF: 2.1 OBJ: 2.1.a Review the properties that distinguish the chemistry of living things from the abiotic chemistry of solids, liquids, and gases. MSC: Evaluating 2. A. If 0.5 mole of glucose weighs 90 g, what is the molecular mass of glucose?

B. How much glucose do you have to add to water to produce 1 liter of a 0.25 M solution of glucose?

C. How many molecules are there in 1 mole of glucose?

ANS: A. 180 daltons. A mole of a substance has a mass equivalent to its molecular weight expressed in grams.

B. 45 g/L C. 6 × 1023 molecules DIF: Moderate REF: 2.1 OBJ: 2.1.e Express the concept of a mole and explain how to prepare a 100 mM solution.

MSC: Applying 3. You explain to a friend what you have learned about Avogadro’s number. Your friend thinks the number is so large that he doubts there is even a mole of living cells on the Earth. You have recently heard that there are about 50 trillion (5 × 1013) human cells in each adult human body and that each human body carries more bacterial cells (the microbiome) than human cells, and the human population is approximately 7.6 billion (7.6 × 109). Armed with this information, you bet your friend $5 that there is more than a mole of cells on Earth. Write out the calculation that proves you are correct.

ANS: Avogadro’s number, or 6 × 1023, is the number of atoms or molecules in a mole. If you multiply the number of people on Earth by the number of cells in the human body, then double it to account for the bacteria, you will calculate: (7 × 109) × (1 × 1014) = 7 × 1023. Thus, even when only considering the human population and the associated microbial populations, you can estimate more than a mole of living cells on Earth. You win $5.

DIF: Difficult REF: 2.1 OBJ: 2.1.e Express the concept of a mole and explain how to prepare a 100 mM solution.

MSC: Applying 4. Indicate whether the statements below are TRUE or FALSE. If a statement is false, explain why it is false.

A. Electron shells fill discrete regions around the nucleus of the atom and limit the number of electrons that can occupy a specific orbit.

B. H, C, O, and N are the most common elements in biological molecules because they are the most stable.

C. Some atoms are more stable when they lose one or two electrons, even though this means they will have a net positive charge.

ANS: A. True B. False. H, C, N, and O are the most common elements in biological molecules because their outer shells are unfilled, making them highly reactive.

C. True DIF: Easy REF: 2.1 OBJ: 2.1.a Review the properties that distinguish the chemistry of living things from the abiotic chemistry of solids, liquids, and gases. | 2.1.b Distinguish between elements, atoms, ions, isotopes, molecules, and salts.

MSC: Evaluating 5. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. charge

length

polar

covalent

molecule

salt

double bond

noncovalent

single bond

ionic

nonpolar

weight

Whereas ionic bonds form a/an __________, covalent bonds between atoms form a/an __________. These covalent bonds have a characteristic bond __________ and become stronger and more rigid when two electrons are shared in a/an __________. Equal sharing of electrons yields a/an __________ covalent bond. If one atom participating in the bond has a stronger affinity for the electron, this produces a partial negative charge on one atom and a partial positive charge on the other. These __________ covalent bonds should not be confused with the weaker __________ bonds that are critical for the three-dimensional structure of biological molecules and for interactions between these molecules.

ANS: Whereas ionic bonds form a salt, covalent bonds between atoms form a molecule. These covalent bonds have a characteristic bond length and become stronger and more rigid when two electrons are shared in a double bond. Equal sharing of electrons yields a nonpolar covalent bond. If one atom participating in the bond has a stronger affinity for the electron, this produces a partial negative charge on one atom and a partial positive charge on the other. These polar covalent bonds should not be confused with the weaker noncovalent bonds that are

critical for the three-dimensional structure of biological molecules and for interactions between these molecules.

DIF: Easy REF: 2.1 OBJ: 2.1.k Differentiate between covalent and ionic bonds in terms of their electronic configuration, strength and stability, and their prevalence and role in biological systems. MSC: Understanding 6. Indicate whether the statements below are TRUE or FALSE. If a statement is false, explain why it is false.

A. Electrons are constantly moving around the nucleus of the atom, but they can move only in discrete regions.

B. There is no limit to the number of electrons that can occupy the fourth electron shell.

C. Atoms with unfilled outer electron shells are especially stable and are therefore less reactive.

ANS: A. True B. False. The fourth electron shell has the capacity to hold 18 electrons.

C. False. Atoms that have their outer electron shells filled are the most stable and least reactive. Atoms with unfilled outer shells are more reactive because they tend to share or transfer electrons to fill and therefore stabilize the outer shell.

DIF: Easy REF: 2.1 OBJ: 2.1.k Differentiate between covalent and ionic bonds in terms of their electronic configuration, strength and stability, and their prevalence and role in biological systems. MSC: Evaluating 7. Table 2-14 indicates the number and arrangement of electrons in the first four atomic electron shells for selected elements. Use the information in the table to fill in the blanks for A–E. There may be more than one answer for each.

Table 2-14

A. __________ are chemically inert.

B. __________ form ions with a net charge of +1 in solution.

C. __________ form stable but reactive diatomic gases.

D. __________ form ions with a net charge of −1 in solution.

E. __________ form ions with a net charge of +2 in solution.

ANS: A. Helium and neon B. Sodium and potassium C. Nitrogen and oxygen

D. Chlorine E. Calcium and magnesium DIF: Easy REF: 2.1 OBJ: 2.1.f Relate how the behavior of atoms corresponds with their position in the periodic table and infer what this means for the chemical reactivity of the elements commonly found in living organisms.

MSC: Understanding 8. Indicate whether the statements below are TRUE or FALSE. If a statement is false, explain why it is false.

A. H, C, N and O constitute 99% of the total number of atoms found in the human body.

B. Copper, zinc, and manganese are among 11 nonessential trace elements that contribute less than 0.1% of all the atoms in the human body.

C. Approximately 0.9% of the atoms in the human body come from seven essential elements—Na, Mg, K, Ca, P, S, and Cl—all of which form stable ions in aqueous solution.

ANS: A. True B. False. Cu, Zn, and Mn are essential trace elements in the human body.

C. False. Na, Mg, K, Ca, and Cl form ions in aqueous solution, but P and S form covalent bonds in order to fill their outer electron shells.

DIF: Moderate REF: 2.1 OBJ: 2.1.a Review the properties that distinguish the chemistry of living things from the abiotic chemistry of solids, liquids, and gases. MSC: Evaluating 9. A. In which scientific unit is the strength of a chemical bond usually expressed?

B. If 0.5 kilocalories of energy are required to break 6 × 1023 bonds of a particular type, what is the strength of this bond?

ANS: A. kilocalories per mole (or kilojoules per mole) B. 0.5 kcal/mole DIF: Moderate REF: 2.1 OBJ: 2.1.k Differentiate between covalent and ionic bonds in terms of their electronic configuration, strength and stability, and their prevalence and role in biological systems. MSC: Understanding 10. The relative strengths of covalent bonds and van der Waals interactions remain the same when tested in a vacuum or in water. However, this is not true of hydrogen bonds or ionic bonds, whose bond strengths are lowered considerably in the presence of water. Explain these observations.

ANS: We estimate bond strengths by measuring the amount of energy needed to break them. As explained in Panel 2–7 in the text (pp. 78–79), in an aqueous solution, water can form hydrogen bonds with any polar molecules that are capably of forming hydrogen bonds with each other. This formation of bonds with water takes away from the net energy that would be gained from the molecules forming hydrogen bonds with each other, as they would in a vacuum.

Similarly, water forms favorable electrostatic interactions with ions, thereby greatly weakening the ionic bonds that form between positive and negative ions in a vacuum (see Panel 2–7). Thus, for example, solid table salt (NaCl) readily dissociates in water, producing separate Na+ and Cl− ions as it dissolves. In contrast, covalent bonds and van der Waals attractions have an intrinsic bond strength that is independent of the aqueous environment, because changes in water molecule associations are not involved in the formation of these two types of bonds.

DIF: Difficult REF: 2.1 OBJ: 2.1.l Contrast hydrogen bonds, electrostatic attractions, van der Waal’s attractions, and the

hydrophobic force in terms of how and when they form and the role they play in cell biology. MSC: Understanding 11. Indicate whether the statements below are TRUE or FALSE. If a statement is false, explain why it is false.

A. Any covalently bonded H atom can participate in a hydrogen bond if it comes in close proximity with an oxygen atom that forms part of a water molecule.

B. Protons are constantly moving between water molecules, which means there is an overall equilibrium between hydroxyl ions and hydronium ions in aqueous solutions.

C. A strong base is defined as a molecule that can readily remove protons from water.

D. Electrons are always shared equally between covalently bonded atoms.

ANS: A. False. Hydrogen atoms that are covalently bonded to carbon atoms do not participate in hydrogen bonds because these hydrogens have almost no net positive charge.

B. True C. True D. False. There are many covalent bonds in which the electrons are shared unequally between bonded atoms. This occurs when one atom is more electronegative than the other. These covalent bonds are referred to as polar covalent bonds.

DIF: Easy REF: 2.1 OBJ: 2.1.j Distinguish between acids and bases and demonstrate how they cooperate to maintain the pH of cells. | 2.1.k Differentiate between covalent and ionic bonds in terms of their electronic configuration, strength and stability, and their prevalence and role in biological systems. | 2.1.l Contrast hydrogen bonds, electrostatic attractions, van der Waal’s attractions, and the hydrophobic force in terms of how and when they form and the role they play in cell biology. MSC: Evaluating 12. A. What is the pH of pure water?

B. What concentration of hydronium ions does a solution of pH8 contain?

C. Complete the following reaction: CH3COOH + H2O ↔ __________.

D. Will the reaction in (C) occur more readily (be driven to the right) if the pH of the solution is high?

ANS: A. pH7 B. 10−8 M C. CH3COO− + H3O+ D. Yes. If the pH is high, then the concentration of hydronium ions will be low. Therefore the rightward reaction, which produces hydronium ions, will be favored.

DIF: Easy REF: 2.1 OBJ: 2.1.j Distinguish between acids and bases and demonstrate how they cooperate to maintain the pH of cells. MSC: Remembering 13. Indicate whether the statements below are TRUE or FALSE. If a statement is false, explain why it is false.

A. A disaccharide consists of a sugar covalently linked to another molecule such as an amino acid or a nucleotide.

B. The hydroxyl groups on monosaccharides are reaction hot spots and can be replaced by other functional groups to produce derivatives of the original sugar.

C. The presence of double bonds in the hydrocarbon tail of a fatty acid does not greatly influence its structure.

ANS: A. False. A disaccharide consists of two sugar molecules that undergo a condensation reaction to form a covalent bond (known as a glycosidic linkage).

B. True C. False. The presence of a double bond in the hydrocarbon chain of a fatty acid causes a kink in the chain, decreasing its flexibility and packing with neighboring hydrocarbon chains.

DIF: Easy REF: 2.2 OBJ: 2.2.b Express how the chemical and physical properties of methyl groups (−CH3), hydroxyl groups (−OH), carboxyl groups (−COOH), phosphate groups (−PO32–), and amino groups (−NH2) influence the behavior of molecules in which these groups typically occur. | 2.2.c Illustrate how the processes of condensation and hydrolysis drive the synthesis and breakdown of the large organic molecules of the cell from sets of smaller organic building blocks. | 2.2.i Predict how the saturation of fatty acid tails affects the fluidity of cell membranes. MSC: Evaluating 14. A. How many carbon atoms does the molecule represented in Figure 2-65 have?

B. How many hydrogen atoms does it have?

C. What type of molecule is it?

Figure 2-65

ANS: A. 20 carbon atoms B. 31 hydrogen atoms C. a fatty acid DIF: Easy REF: 2.2 OBJ: 2.2.g Define the terms fatty acid, steroid, phospholipid, and triacylglycerol and describe the structure, properties, and function of each. MSC: Understanding 15. 1. Write out the sequence of amino acids in the following peptide, using the full names of the amino acids: Pro-ValThr-Gly-Lys-Cys-Glu. 2. Write the same sequence with the single-letter code for amino acids.

3. According to the conventional way of writing the sequence of a peptide or a protein, which is the C-terminal amino acid and which is the N-terminal amino acid in the above peptide?

ANS: 1. proline-valine-threonine-glycine-lysine-cysteine-glutamic acid (or glutamate) 2. PVTGKCE 3. The C-terminal is glutamic acid (or glutamate); the N-terminal is proline.

DIF: Easy REF: 2.2 OBJ: 2.2.k Summarize what makes amino acids chemically unique and categorize the 20 amino acids commonly found in proteins on the basis of their chemical properties. MSC: Remembering 16. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. amino

ionized

polypeptides

α-carbon

length

protein

carbon

noncovalent

R group

carboxyl

peptide bonds

side chains

hydroxide Proteins are __________ built from amino acids, which each have an amino group and a __________ group attached to the central __________. There are 20 possible __________ that differ in structure and are generally referred to as “R.” In solutions of neutral pH, amino acids are __________, carrying both a positive and a negative charge. When a protein is made, amino acids are linked together through __________, which are formed by condensation reactions between the carboxyl end of the last amino acid and the __________ end of the next amino acid to be added to the growing chain.

ANS: Proteins are polypeptides built from amino acids, which each have an amino group and a carboxyl group attached to the central α-carbon. There are twenty possible side chains that differ in structure and are generally referred to as “R.” In solutions of neutral pH, amino acids are ionized, carrying both a positive and negative charge.

When a protein is made, amino acids are linked together through peptide bonds, which are formed by condensation reactions between the carboxyl end of the last amino acid and the amino end of the next amino acid to be added to the growing chain.

DIF: Easy REF: 2.2 OBJ: 2.2.j Identify the features that all amino acids have in common. | 2.2.k Summarize what makes amino acids chemically unique and categorize the 20 amino acids commonly found in proteins on the basis of their chemical properties. MSC: Understanding 17. Indicate whether the statements below are TRUE or FALSE. If a statement is false, explain why it is false.

A. “Nonpolar interactions” is simply another way of saying “van der Waals attractions.”

B. Condensation reactions occur in the synthesis of all the macromolecules found in cells.

C. All proteins and RNAs pass through many unstable conformations as they are folded, finally settling on one single, preferred conformation.

ANS: A. False. Van der Waals attractions describe the general attractive forces between all atoms. The contact distance between any two nonbonded atoms is the sum of the van der Waals radii. Nonpolar interactions are based on the exclusion of hydrophobic molecules from a hydrophilic environment.

B. True C. True DIF: Easy REF: 2.2 OBJ: 2.2.c Illustrate how the processes of condensation and hydrolysis drive the synthesis and breakdown of the large organic molecules of the cell from sets of smaller organic building blocks.

MSC: Evaluating 18. A protein chain folds into its stable and unique three-dimensional structure, or conformation, by making many noncovalent bonds between different parts of the chain. Such noncovalent bonds are also critical for interactions with other proteins and cellular molecules. From the list provided, choose the class or classes of amino acids that are most important for the interactions detailed below.

A. forming ionic bonds with negatively charged DNA B. forming hydrogen bonds to aid solubility in water C. binding to another water-soluble protein D. localizing an “integral membrane” protein that spans a lipid bilayer E. tightly packing the hydrophobic interior core of a globular protein

acidic

nonpolar

basic

uncharged polar

ANS: A. basic B. uncharged polar C. uncharged polar, basic, and acidic D. nonpolar E. nonpolar DIF: Moderate REF: 2.2 OBJ: 2.2.k Summarize what makes amino acids chemically unique and categorize the 20 amino acids commonly found in proteins on the basis of their chemical properties. MSC: Analyzing 19. The amino acid histidine is often found in enzymes. Depending on the pH of its environment, sometimes histidine is neutral and at other times it acquires a proton and becomes positively charged. Consider an enzyme with a histidine side chain that is known to have an important role in the function of the enzyme. It is not clear whether this histidine is required in its protonated or its unprotonated state. To answer this question, you measure enzyme activity over a range of pH, with the results shown in Figure 2-70. Which form of histidine is necessary for the active enzyme?

Figure 2-70

ANS: Assuming that the change in enzyme activity is due to the change in the protonation state of histidine, the enzyme must require histidine in the protonated, charged state. The enzyme is active only at low, acidic pH, where the proton (or hydronium ion) concentration is high; thus, the loss of a proton from histidine will be disfavored so that histidine is likely to be protonated.

DIF: Difficult REF: 2.1 OBJ: 2.1.j Distinguish between acids and bases and demonstrate how they cooperate to maintain the pH of cells. MSC: Applying 20. Silicon is an element that, like carbon, has four vacancies in its outer electron shell and therefore has the same bonding chemistry as carbon. Silicon is not found to any significant degree in the molecules found in living systems,

however. Does this difference arise because elemental carbon is more abundant than silicon? What other explanations are there for the preferential selection of carbon over silicon as the basis for the molecules of life?

ANS: According to Figure 2-4 in your textbook, silicon is actually more abundant in the Earth’s crust than carbon, so this is not likely to be the reason that carbon was used preferentially. Carbon might have been the element of choice in living systems because it is lighter than silicon and forms shorter covalent bonds with other elements. Shorter bonds are typically stronger and more stable.

DIF: Difficult REF: 2.1 OBJ: 2.1.a Review the properties that distinguish the chemistry of living things from the abiotic chemistry of solids, liquids, and gases. MSC: Applying 21. Selenium (Se) is an element required in the human body in trace amounts. Selenium is obtained through the diet and levels of selenium found in food depend greatly on the soil where it is grown. Once ingested and absorbed as selenate, it can become incorporated into a small number of polypeptides. These selenoproteins are formed when selenium replaces an element that is found in 2 of the 20 “standard” amino acids. Using your knowledge of atomic structure, the periodic table in Figure 2-7, and the structure of amino acids found in Panel 2–5, deduce which two amino acids may be converted to “seleno” amino acids and used to make selenoproteins.

ANS: Sulfur is the only element found exclusively in 2 of the 20 amino acids. This element is located directly above selenium in the periodic table, indicating that these elements have the same number of electrons in their outer shell and both prefer to form bonds with other atoms to fill their outer orbital. If selenium instead of sulfur is incorporated into cysteine or methionine, the altered “seleno” amino acids will be produced (selenocysteine and selenomethionine). We can expect that this substitution will alter the nature of the proteins in which these amino acids are incorporated because selenium is a larger atom than sulfur.

DIF: Difficult REF: 2.2 OBJ: 2.2.k Summarize what makes amino acids chemically unique and categorize the 20 amino acids commonly found in proteins on the basis of their chemical properties. MSC: Applying 22. The cell is able to harvest energy from various processes in order to generate ATP molecules. These ATPs represent a form of stored energy that can be used later to drive other important processes. Explain how the cell can convert the chemical energy stored in ATP to generate mechanical energy; for example, changing the shape of a protein.

ANS: The terminal phosphate group is typically hydrolyzed and the energy released from this chemical bond is often “reinvested” to generate a new bond that links the phosphate group to a protein. This addition of a phosphate group can cause a change in the protein’s conformation. This conformational change is usually associated with change in function or transient interactions with other macromolecules, generating a domino effect within the cell.

DIF: Easy REF: 2.2 OBJ: 2.2.b Express how the chemical and physical properties of methyl groups (−CH3), hydroxyl groups (−OH), carboxyl groups (−COOH), phosphate groups (−PO32–), and amino groups (−NH2) influence the behavior of molecules in which these groups typically occur. MSC: Understanding 23. It is now a routine task to determine the exact order in which individual subunits have been linked together in polynucleotides (DNA) and polypeptides (proteins). However, it remains difficult to determine the arrangement of monomers in a polysaccharide. Explain why this is the case.

ANS: Nucleotides and amino acids have an intrinsic directionality, and the mechanism by which monomers are added into a growing polymer is always the same. This yields a linear polymer with the same directionality as the monomers.

Polysaccharides are produced by linking monosaccharides together. The monosaccharides can be either added directly or modified to produce various derivatives before addition. Beyond this, there are multiple sites on each mono-

saccharide where addition can occur, producing highly complex, branched polymers.

DIF: Moderate REF: 2.2 OBJ: 2.2.c Illustrate how the processes of condensation and hydrolysis drive the synthesis and breakdown of the large organic molecules of the cell from sets of smaller organic building blocks.

MSC: Evaluating 24. As a protein is made, the polypeptide is in an extended conformation, with every amino acid exposed to the aqueous environment. Although both polar and charged side chains can mix readily with water, this is not the case for nonpolar side chains. Explain how hydrophobic interactions may play a role in the early stages of protein folding, and have an influence on the final protein conformation.

ANS: One reason that nonpolar groups are excluded from an aqueous environment is that a hydrophobic surface would organize water into a highly structured network of hydrogen bonds, which is energetically unfavorable. Thus, you would expect that nonpolar amino acids would group together early, forming “hydrophobic pockets,” while the polar and charged side chains remain at the interface of the surrounding solution. In the final, folded protein, most of the nonpolar amino acids will remain buried inside the protein. This fold is more stable because nonpolar atoms are prevented from contact with water and remain in contact with each other.

DIF: Moderate REF: 2.3 OBJ: 2.3.b Assess the role that covalent and noncovalent bonds play in the threedimensional conformation of macromolecules. MSC: Applying 25. You are trying to make a synthetic copy of a particular protein but accidentally join the amino acids together in exactly the reverse order. One of your classmates says the two proteins must be identical, and bets you $20 that your synthetic protein will have exactly the same biological activity as the original. After having read this chapter, you have no hesitation in staking your $20 that it will not. What particular feature of a polypeptide chain makes you sure your $20 is safe and that the project must be restarted?

ANS: As a peptide bond has a distinct chemical polarity, a polypeptide chain also has a distinct polarity. The reversed protein chain cannot make the same noncovalent interactions during folding and thus will not adopt the same threedimensional structure as the original protein. The activities of these two proteins will definitely be different, because the activity of a protein depends on its three-dimensional structure. It is unlikely that the reverse chain will fold into any well-defined, and hence functionally useful, structure at all, because it has not passed the stringent selective pressures imposed during evolution.

DIF: Easy REF: 2.3 OBJ: 2.3.a Relate how the repetitive polymerization of monomers into polymers can yield macromolecules with diverse properties and functions. MSC: Evaluating 26. Your lab director requests that you add new growth medium to the mammalian cell cultures before heading home from the lab on a Friday night. Unfortunately, you need to make fresh medium because all the premixed bottles of medium have been used. One of the ingredients you know you need to add is a mix of the essential amino acids (those that cannot be made by the cells, but are needed in proteins). On the shelf of dry chemicals you find the amino acids you need, and you mix them into your medium, along with all the other necessary nutrients, and replace the old medium with your new medium. On Sunday, you come in to the lab just to check on your cells and find that the cells have not grown. You are sure you made the medium correctly, but on checking you see that somebody wrote a note on the dry mixture of amino acids you used: “Note: this mixture contains only ᴅ-amino acids.”

A. What is the meaning of the note and how does it explain the lack of cell growth in your culture?

B. Are there any organisms that could grow using this mixture? Justify your answer.

ANS: A. The note indicates that the mixture contains only one of the two possible stereoisomers (ʟ or ᴅ). Because mammalian cells use only the ʟ stereoisomer, the ᴅ-amino acid mixture could not be used and therefore it is as though no amino acids were added at all.

B. Not unless ʟ-amino acids were also mixed in. Certain types of bacteria use ᴅ-amino acids to produce their cell walls, but they would still require ʟ-amino acids for the rest of the proteins they make.

DIF: Difficult REF: 2.2 OBJ: 2.2.f Evaluate the role that isomers play in biological systems. MSC: Evaluating 27. Eukaryotic cells have their DNA molecules inside their nuclei. However, to package all the DNA into such a small volume requires the cell to use specialized proteins called histones. Histones have amino acid sequences enriched for lysines and arginines.

A. What problem might a cell face in trying to package DNA into a small volume without histones, and how do these special packaging proteins alleviate the problem?

B. Lysine side chains are substrates for enzymes called acetylases. A diagram of an acetylated lysine side chain is shown in Figure 2-78. How do you think the acetylation of lysines in histone proteins will affect the ability of a histone to perform its role (refer to your answer in part A)?

Figure 2-78

ANS: A. DNA is a nucleic acid polymer in which each monomer has a negatively charged phosphate group. The negative charges will naturally repel each other, so in order to wrap the high density of negative charges into a small space, positively charged molecules must be present. Histones accomplish this because they are rich in lysines and arginines, which are positively charged in solution at pH7.

B. A histone with acetylated lysine residues will not be as good at packaging the DNA. The addition of the acetyl group to the terminal amino on the lysine side chain lowers the histone’s net positive charge, which makes it less effective at buffering the negative charges on the DNA backbone.

DIF: Difficult REF: 2.3 OBJ: 2.3.c Explain how weak, noncovalent bonds can lead to strong and specific associations between macromolecules or between an enzyme and its substrate. MSC: Understanding 28. A. Sketch three different ways in which three water molecules could be held together by hydrogen-bonding.

B. On a sketch of a single water molecule, indicate the distribution of positive and negative charge (using the symbols δ+ and δ−).

C. How many hydrogen bonds can a hydrogen atom in a water molecule form? How many hydrogen bonds can the oxygen atom in a water molecule form?

ANS: A. See Figure 2-79 A. B. See Figure 2-79 B.

C. Hydrogen can form one; oxygen can form two.

Figure 2-79

DIF: Easy REF: 2.1 OBJ: 2.1.l Contrast hydrogen bonds, electrostatic attractions, van der Waal’s attractions, and the hydrophobic force in terms of how and when they form and the role they play in cell biology. MSC: Creating 29. You are trying to make a synthetic copy of a particular protein but accidentally join the amino acids together in exactly the reverse order. One of your classmates says the two proteins must be identical, and bets you $20 that your synthetic protein will have exactly the same biological activity as the original. After having read this chapter, you have no hesitation in staking your $20 that it will not. What particular feature of a polypeptide chain makes you sure your $20 is safe and that the project must be restarted?

ANS: As a peptide bond has a distinct chemical polarity, a polypeptide chain also has a distinct polarity (see Figure A2-80). The reversed protein chain cannot make the same noncovalent interactions during folding and thus will not adopt the same three-dimensional structure as the original protein. The activities of these two proteins will definitely be different, because the activity of a protein depends on its three-dimensional structure. It is unlikely that the reverse chain will fold into any well-defined, and hence functionally useful, structure at all, because it has not passed the stringent selective pressures imposed during evolution.

Figure 2-80A

DIF: Easy REF: 2.3 OBJ: 2.3.a Relate how the repetitive polymerization of monomers into polymers can yield macromolecules with diverse properties and functions. MSC: Evaluating

CHAPTER 3

Energy, Catalysis, and Biosynthesis THE USE OF ENERGY BY CELLS

3.1.a Relate anabolic, catabolic, and metabolic reactions.

3.1.b State how living systems can generate and maintain order without violating the second law of thermodynamics.

3.1.c Summarize how living systems take advantage of the first law of thermodynamics.

3.1.d Express why sunlight provides the ultimate source of energy for nearly all living things.

3.1.e Contrast photosynthesis and cellular respiration and evaluate how the processes complement one another.

3.1.f Differentiate between oxidation and reduction in terms of the movement of electrons, protons, and oxygen atoms.

3.1.g State why CO2 is an end product of cellular respiration.

FREE ENERGY AND CATALYSIS 3.2.a Recall the relationship between the free energy change and the entropy change of the universe.

3.2.b Articulate how enzymes increase the speed of a chemical reaction.

3.2.c Relate how the free-energy change of a reaction determines its energetic favorability.

3.2.d Review the relationship between /deltaG, equilibrium, and the concentrations of a reaction’s substrates and products.

3.2.e Relate /deltaG to /deltaG0.

3.2.f Formulate how the equilibrium constant, K, governs the direction of a reaction and the amount of product it generates.

3.2.g Express how the equilibrium constant, K, reflects the strength of noncovalent binding interactions between two molecules.

3.2.h Explain how the pathway that converts sugars into CO2 and H2O proceeds rapidly to completion despite the fact that several of its constituent reactions are energetically unfavorable.

3.2.i Summarize how substrate concentration affects an enzyme’s activity.

3.2.j Describe how noncovalent interactions allow enzymes to interact with specific substrates.

3.2.k Explain why substrates for different enzymes can coexist in the same compartment.

3.2.l Illustrate how enzymes affect the equilibrium point of chemical reactions.

ACTIVATED CARRIERS AND BIOSYNTHESIS 3.3.a Review how activated carriers link catabolic and anabolic reactions.

3.3.b Evaluate why ATP hydrolysis has such a large negative /deltaG0 and an even larger negative /deltaG.

3.3.c Relate how the energy of ATP hydrolysis can be harnessed to drive an energetically unfavorable condensation reaction.

3.3.d Delineate the role played by NADPH in oxidation–reduction reactions.

3.3.e State why the oxidation of NADPH is energetically favorable.

3.3.f Relate the structures of NADH and NADPH and their different roles within cells.

3.3.g Express how the relative concentrations of NADPH/NADP+ and NADH/NAD+ in cells influence each carrier’s net affinity for electrons.

3.3.h Name 3 activated carriers and state the role each plays in a cell.

3.3.i Assess the structural similarities possessed by the activated carriers that transfer chemical groups during biosynthetic reactions.

3.3.j Evaluate the energetic favorability of condensation and hydrolysis reactions and state which type of reaction generally requires linkage to ATP hydrolysis to occur.

3.3.k Outline how ATP hydrolysis provides the energy needed for highly unfavorable reactions, such as the biosynthesis of nucleic acids.

3.3.l Review how the inhibition of fermentation revealed that phosphate bonds provide energy to power cell processes.

MULTIPLE CHOICE 1. Chemical reactions carried out by living systems depend on the ability of some organisms to capture and use atoms from nonliving sources in the environment. The specific subset of these reactions that break down nutrients in food can be described as a. metabolic. b. catabolic. c. anabolic. d. biosynthetic.

ANS: B DIF: Easy REF: 3.1 OBJ: 3.1.a Relate anabolic, catabolic, and metabolic reactions. MSC: Remembering 2. When there is an excess of nutrients available in the human body, insulin is released to stimulate the synthesis of glycogen from glucose. This is a specific example of a/an __________ process, a general process in which larger molecules are made from smaller molecules. a. metabolic b. catabolic c. anabolic d. biosynthetic ANS: C DIF: Easy REF: 3.1 OBJ: 3.1.a Relate anabolic, catabolic, and metabolic reactions. MSC: Remembering 3. The second law of thermodynamics states that the disorder in any system is always increasing. In simple terms, you can think about dropping NaCl crystals into a glass of water. The solvation and diffusion of ions is favored because there is an increase in

a. pH. b. entropy. c. ions. d. stored energy.

ANS: B DIF: Easy REF: 3.1 OBJ: 3.1.b State how living systems can generate and maintain order without violating the second law of thermodynamics. MSC: Applying 4. The energy used by the cell to generate specific biological molecules and highly ordered structures is stored in the form of a. Brownian motion. b. heat. c. light waves. d. chemical bonds.

ANS: D DIF: Easy REF: 3.1 OBJ: 3.1.c Summarize how living systems take advantage of the first law of thermodynamics. | 3.1.b State how living systems can generate and maintain order without violating the second law of thermodynamics. MSC: Remembering 5. At first glance, it may seem that living systems are able to defy the second law of thermodynamics. However, on closer examination, it becomes clear that although cells create organization from raw materials in the environment, they also contribute to disorder in the environment by releasing a. water. b. radiation. c. heat. d. proteins.

ANS: C DIF: Easy REF: 3.1 OBJ: 3.1.b State how living systems can generate and maintain order without violating the second law of thermodynamics. MSC: Remembering 6. Unlike what occurs when fuel is burned to make a fire, all living systems use the energy from heat-generating reactions to create and maintain a. movement. b. order. c. light. d. electricity.

ANS: B DIF: Easy REF: 3.1 OBJ: 3.1.b State how living systems can generate and maintain order without violating the second law of thermodynamics. | 3.1.c Summarize how living systems take advantage of the first law of thermodynamics. MSC: Understanding 7. During respiration, energy is retrieved from the high-energy bonds found in certain organic molecules. Which of the following, in addition to energy, are the ultimate products of respiration? a. CO2; H2O b. CH3; H2O c. CH2OH; O2 d. CO2; O2

ANS: A DIF: Easy REF: 3.1 OBJ: 3.1.g State why CO2 is an end product of cellular respiration. MSC: Remembering 8. Your body extracts energy from the food you ingest by catalyzing reactions that essentially “burn” the food molecules in a stepwise fashion. What is another way to describe this process? a. reduction b. oxidation c. dehydration d. solvation ANS: B DIF: Easy REF: 3.1 OBJ: 3.1.f Differentiate between oxidation and reduction in terms of the movement of electrons, protons, and oxygen atoms. MSC: Understanding 9. Oxidation is a favorable process in an aerobic environment, which is the reason cells are able to derive energy from the oxidation of macromolecules. Once carbon has been oxidized to __________, its most stable form, it can only cycle back into the organic portion of the carbon cycle through __________. a. CO2; photosynthesis. b. CH3; combustion. c. CO2; respiration. d. CO; reduction.

ANS: A DIF: Easy REF: 3.1 OBJ: 3.1.e Contrast photosynthesis and cellular respiration and evaluate how the processes complement one another. | 3.1.f Differentiate between oxidation and reduction in terms of the movement of electrons, protons, and oxygen atoms. | 3.1.g State why CO 2 is an end product of cellular respiration.

MSC: Remembering 10. Oxidation is the process by which oxygen atoms are added to a target molecule. Generally, the atom that is oxidized will experience which of the following with respect to the electrons in its outer shell? a. a net gain b. a net loss c. no change d. an equal sharing ANS: B DIF: Easy REF: 3.1 OBJ: 3.1.f Differentiate between oxidation and reduction in terms of the movement of electrons, protons, and oxygen atoms. MSC: Remembering 11. When elemental sodium is added to water, the sodium atoms ionize spontaneously. Uncharged Na becomes Na+.

This means that the Na atoms have been a. protonated. b. oxidized. c. hydrogenated. d. reduced.

ANS: B DIF: Easy REF: 3.1 OBJ: 3.1.f Differentiate between oxidation and reduction in terms of the movement of electrons, protons, and oxygen atoms. MSC: Applying 12. The reduction of a molecule can sometimes result in the acquisition of a proton, a reaction referred to as a. protonation.

b. neutralization. c. hydrogenation. d. isomerization.

ANS: C DIF: Easy REF: 3.1 OBJ: 3.1.f Differentiate between oxidation and reduction in terms of the movement of electrons, protons, and oxygen atoms. MSC: Remembering 13. The best way to know if an organic molecule has been reduced is to see if there was an increase in the number of __________ bonds. a. C–H b. C–O c. H–H d. C–N ANS: A DIF: Easy REF: 3.1 OBJ: 3.1.f Differentiate between oxidation and reduction in terms of the movement of electrons, protons, and oxygen atoms. MSC: Understanding 14. Seed oils are often dehydrogenated and added back into processed foods. The new fatty acids have an increased number of carbon–carbon double bonds. The dehydrogenation reaction could also be described as a/an __________ reaction. a. isomerization. b. oxidation. c. reduction. d. protonation.

ANS: B DIF: Easy REF: 3.1 OBJ: 3.1.f Differentiate between oxidation and reduction in terms of the movement of electrons, protons, and oxygen atoms. MSC: Applying 15. Chemical reactions that lead to a release of free energy are referred to as “energetically favorable.” Another way to describe these reactions is a. uphill. b. uncatalyzed. c. spontaneous. d. activated.

ANS: C DIF: Easy REF: 3.2 OBJ: 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. MSC: Remembering 16. Even though cellular macromolecules contain a large number of carbon and hydrogen atoms, they are not all spontaneously converted into CO2 and H2O. This absence of spontaneous combustion is due to the fact that biological molecules are relatively __________ and an input of energy is required to reach lower energy states. a. large b. polar c. stable d. unstable ANS: C DIF: Easy REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction.

MSC: Understanding 17. ΔG° indicates the change in the standard free energy as a reactant is converted to product. Given what you know about these values, which reaction below is the most favorable? a. ADP + Pi → ATP

ΔG° = +7.3 kcal/mole

b. glucose 1-phosphate → glucose 6-phosphate

ΔG° = −1.7 kcal/mole

c. glucose + fructose → sucrose

ΔG° = +5.5 kcal/mole

d. glucose → CO2 + H2O

ΔG° = −686 kcal/mole

ANS: D DIF: Easy REF: 3.2 OBJ: 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. | 3.2.e Relate delta G to delta G0. MSC: Analyzing 18. Catalysts are molecules that lower the activation energy for a given reaction. Cells produce their own catalysts, called a. proteins. b. enzymes. c. cofactors. d. complexes.

ANS: B DIF: Easy REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction.

MSC: Remembering 19. A chemical reaction is defined as spontaneous if there is a net loss of free energy during the reaction process.

However, spontaneous reactions do not always occur rapidly. Favorable biological reactions require __________ to selectively speed up reactions and meet the demands of the cell. a. heat b. ATP c. ions d. enzymes ANS: D DIF: Easy REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction. | 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. MSC: Understanding 20. Figure 3-20 is an energy diagram for the reaction X→Y. Which equation below provides the correct calculation for the amount of free-energy change when X is converted to Y? a. a + b − c b. a − b c. a − c d. c − a

Figure 3-20

ANS: D DIF: Easy REF: 3.2 OBJ: 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. MSC: Applying 21. Enzymes facilitate reactions in living systems. Figure 3-21 presents an energy diagram for the reaction X→Y. The solid line in the energy diagram represents changes in energy as the reactant is converted to product under standard conditions. The dashed line shows changes observed when the same reaction takes place in the presence of a dedicated enzyme. Which equation below indicates how the presence of an enzyme affects the activation energy of the reaction (catalyzed versus uncatalyzed)? a. d − c versus b − c b. d − a versus b − a c. a + d versus a + b d. d − c versus b − a

Figure 3-21

ANS: B DIF: Easy REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction.

MSC: Applying 22. ΔG measures the change of free energy in a system as it converts reactant (Y) into product (X). When [Y] = [X], ΔG is equal to a. ΔG° + RT. b. RT. c. ln [X]/[Y]. d. ΔG°.

ANS: D DIF: Moderate REF: 3.2 OBJ: 3.2.d Review the relationship between delta G, equilibrium, and the concentra-

tions of a reaction’s substrates and products. | 3.2.e Relate delta G to delta G0. MSC: Understanding 23. For the reaction Y→X at standard conditions with [Y] = 1 M and [X] = 1 M, ΔG is initially a large negative number. As the reaction proceeds, [Y] decreases and [X] increases until the system reaches equilibrium. How do the values of ΔG and ΔG° change as the reaction equilibrates? a. ΔG becomes less negative and ΔG° stays the same. b. ΔG becomes positive and ΔG° becomes positive. c. ΔG stays the same and ΔG° becomes less negative. d. ΔG reaches zero and ΔG° becomes more negative.

ANS: A DIF: Moderate REF: 3.2 OBJ: 3.2.d Review the relationship between delta G, equilibrium, and the concentrations of a reaction’s substrates and products. | 3.2.e Relate delta G to delta G0. | 3.2.f Formulate how the equilibrium constant, K, governs the rate of a reaction and the amount of product it generates. MSC: Applying 24. The equilibrium constant (K) for the reaction Y→X can be expressed with respect to the concentrations of the reactant and product molecules. Which of the expressions below shows the correct relationship between K, [Y], and [X]? a. K = [Y]/[X] b. K = [Y] * [X] c. K = [X]/[Y] d. K = [X] − [Y] ANS: C DIF: Easy REF: 3.2 OBJ: 3.2.f Formulate how the equilibrium constant, K, governs the rate of a reaction and the amount of product it generates. MSC: Understanding 25. Isomerization of glucose 1-phosphate to glucose 6-phosphate is energetically favorable. At 37°C, ΔG° = −1.42 log10K. What is the equilibrium constant for this reaction if ΔG° = −1.74 kcal/mole at 37°C? a. 16.98 b. 0.09 c. −0.09 d. 0.39 ANS: A DIF: Easy REF: 3.2 OBJ: 3.2.d Review the relationship between delta G, equilibrium, and the concentrations of a reaction’s substrates and products. | 3.2.e Relate delta G to delta G0. | 3.2.f Formulate how the equilibrium constant, K, governs the rate of a reaction and the amount of product it generates. MSC: Applying 26. The potential energy stored in high-energy bonds is commonly harnessed when the bonds are split by the addition of __________ in a process called __________. a. ATP; phosphorylation b. water; hydrolysis c. hydroxide; hydration d. acetate; acetylation ANS: B DIF: Easy REF: 3.2 OBJ: 3.2.a Recall the relationship between free energ y and entropy.

MSC: Remembering 27. When the polymer X-X-X . . . is broken down into monomers, it is “phosphorylyzed” rather than hydrolyzed, in the following repeated reaction:

X-X-X . . . + P → X-P + X-X . . .

(reaction 1)

Given the ΔG° values of the reactions listed in Table 3-27, what is the expected ratio of X-phosphate (X-P) to free phosphate (P) at equilibrium for reaction 1? a. 1:106 b. 1:104 c. 1:1 d. 104:1 X-X-X . . . + H2O → X + X-X . . .

ΔG° = −4.5 kcal/mole

X + ATP → X-P + ADP

ΔG° = −2.8 kcal/mole

ATP + H2O → ADP + P

ΔG° = −7.3 kcal/mole Table 3-27

ANS: C Reaction 1 can be written as the sum of the three reactions given, because the ATP used in step 2 is restored in step 3.

X-X-X . . . + H2O → X + X-X . . . ΔG° = −4.5 kcal/mole X + ATP → X-P + ADP

ΔG° = −2.8 kcal/mole

ADP + P → ATP + H2O

ΔG° = +7.3 kcal/mole

Because ΔG° values are additive, ΔG°total = 0, and if ΔG° = 0, Keq = 1, meaning that [products]/[reactants] = 1, and the ratio of X-P to P is 1:1.

DIF: Moderate REF: 3.2 OBJ: 3.2.d Review the relationship between delta G, equilibrium, and the concentrations of a reaction’s substrates and products. | 3.2.e Relate delta G to delta G0. MSC: Applying 28. In the case of a simple conversion reaction such as X→Y, which value of ΔG° is associated with a larger concentration of X than Y at equilibrium? (Hint: How is ΔG° related to K?) a. ΔG° = −5 b. ΔG° = −1 c. ΔG° = 0 d. ΔG° = 1 ANS: D DIF: Easy REF: 3.2 OBJ: 3.2.d Review the relationship between delta G, equilibrium, and the concentrations of a reaction’s substrates and products. | 3.2.e Relate delta G to delta G0. | 3.2.f Formulate how the equilibrium constant, K, governs the rate of a reaction and the amount of product it generates. MSC: Understanding 29. If proteins A and B have complementary surfaces, they may interact to form the dimeric complex AB. Which of the following is the correct way to calculate the equilibrium constant for the association between A and B? a. kon/koff = K b. K = [A][B]/[AB] c. K = [AB]/[A][B] d. (A) and (C) ANS: D DIF: Moderate REF: 3.2 OBJ: 3.2.f Formulate how the equilibrium constant, K, governs the rate of a reaction and the amount of product it generates. | 3.2.g Express how the equilibrium constant, K, dictates the strength of non-

covalent binding interactions between two molecules. MSC: Applying 30. The equilibrium constant for complex formation between molecules A and B will depend on their relative concentrations, as well as the rates at which the molecules associate and dissociate. The association rate will be larger than the dissociation rate when complex formation is favorable. The energy that drives this process is referred to as __________ energy. a. dissociation b. association c. binding d. releasing ANS: C DIF: Easy REF: 3.2 OBJ: 3.2.g Express how the equilibrium constant, K, dictates the strength of noncovalent binding interactions between two molecules. MSC: Understanding 31. Which of the following statements is FALSE for a favorable binding reaction? a. The free-energy change is negative for the system. b. The concentration of the complex remains lower than the concentration of the unbound components. c. The complex dissociation rate is slower than the rate for component association. d. The binding energy for the association is large and negative.

ANS: B DIF: Easy REF: 3.2 OBJ: 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. MSC: Analyzing 32. The net distance a molecule travels through the cytosol via diffusion is relatively short in comparison with the total distance it may need to travel. This is because movement governed by diffusion alone is a __________ process that is most effective for the dispersion of small molecules over short distances. a. slow b. random c. regulated d. complicated ANS: A DIF: Easy REF: 3.2 OBJ: 3.2.i Summarize how substrate concentration affects an enzyme’s activity.

MSC: Understanding 33. The small molecule cyclic AMP (cAMP) takes about 0.2 second to diffuse 10 μm, on average, in a cell. Suppose that cAMP is produced near the plasma membrane on one end of the cell; how long will it take for this cAMP to diffuse through the cytosol and reach the opposite end of a very large cell, on average? Assume that the cell is 200 μm in diameter. a. 4 seconds b. 16 seconds c. 80 seconds d. 200 seconds ANS: A DIF: Easy REF: 3.2 OBJ: 3.2.i Summarize how substrate concentration affects an enzyme’s activity.

MSC: Applying 34. The graph in Figure 3-34 illustrates the relationship between reaction rates and substrate concentration for an enzyme-

catalyzed reaction. What does the Km value indicate with respect to enzyme–substrate interactions?

Figure 3-34

a. the maximum rate of catalysis b. the number of enzyme active sites c. the enzyme–substrate binding affinity d. the equilibrium rate of catalysis ANS: C DIF: Moderate REF: 3.2 OBJ: 3.2.i Summarize how substrate concentration affects an enzyme’s activity.

MSC: Understanding 35. The graph in Figure 3-34 illustrates the change in the rate of an enzyme-catalyzed reaction as the concentration of substrate is increased. Which of the values listed below is used to calculate the enzyme turnover number?

Figure 3-34

a. ½Vmax b. Km c. Vmax d. Vmax − Km ANS: C DIF: Easy REF: 3.2 OBJ: 3.2.i Summarize how substrate concentration affects an enzyme’s activity.

MSC: Remembering 36. Protein E can bind to two different proteins, S and I. The binding reactions are described by the following equations and values: E + S → ES

Keq for ES = 10

E + I → EI

Keq for EI = 2

Given the equilibrium constant values, which one of the following statements is TRUE? a. E binds I more tightly than S. b. When S is present in excess, no I molecules will bind to E. c. The binding energy of the ES interaction is greater than that of the EI interaction. d. Changing an amino acid on the binding surface of I from a basic amino acid to an acidic one will probably make the free energy of association with E more negative.

ANS: C The binding energy is the standard free energy of the binding reaction, and thus is proportional to ln Keq. As the binding energy increases, the equilibrium constant for the association reaction becomes larger.

DIF: Easy REF: 3.2 OBJ: 3.2.i Summarize how substrate concentration affects an enzyme’s activity.

MSC: Analyzing 37. The study of enzyme kinetics is usually performed with purified components and requires the characterization of several aspects of the reaction, including the rate of association with the substrate, the rate of catalysis, and a. the enzyme’s structure. b. the optimal pH of the reaction. c. the subcellular localization of the enzyme. d. the regulation of the enzyme activity.

ANS: D DIF: Easy REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction.

MSC: Remembering 38. The maximum velocity (Vmax) of an enzymatic reaction is an important piece of information regarding how the enzyme works. What series of measurements can be taken in order to infer the maximum velocity of an enzyme-catalyzed reaction? a. the rate of substrate consumption after the system reaches equilibrium, for several reactant concentrations b. the rate of product consumption shortly after mixing the enzyme and substrate c. the rate of substrate consumption at high levels of enzyme concentration d. the rate of substrate consumption shortly after mixing the enzyme and substrate, for several substrate concentrations ANS: D DIF: Easy REF: 3.2 OBJ: 3.2.i Summarize how substrate concentration affects an enzyme’s activity.

MSC: Analyzing 39. What information regarding an enzyme-catalyzed reaction is obtained in a plot of the inverse of the initial velocities against the inverse of the corresponding substrate concentrations? a. 1/Vmax and 1/Km b. 1/V and 1/[S] c. Vmax and Km d. V and [S] ANS: A DIF: Easy REF: 3.2 OBJ: 3.2.i Summarize how substrate concentration affects an enzyme’s activity.

MSC: Understanding 40. The study of enzymes also includes an examination of how the activity is regulated. Molecules that can act as competitive inhibitors for a specific reaction are often similar in shape and size to the enzyme’s substrate. Which

variable or variables used to describe enzyme activity will remain the same in the presence and absence of a competitive inhibitor? a. Vmax b. V c. Vmax and Km d. Km ANS: A DIF: Easy REF: 3.2 OBJ: 3.2.i Summarize how substrate concentration affects an enzyme’s activity. | 3.2.j Describe how noncovalent interactions allow enzymes to interact with specific substrates.

MSC: Understanding 41. Activated carriers are small molecules that can diffuse rapidly and be used to drive biosynthetic reactions in the cell.

Their energy is stored in a readily transferable form such as high-energy electrons or chemical groups. Which of the molecules below donates a chemical group rather than electrons? a. FADH2 b. NADH c. NADPH d. ATP ANS: D DIF: Easy REF: 3.3 OBJ: 3.3.a Review how activated carriers link catabolic and anabolic reactions.

MSC: Understanding 42. Energy cannot be created or destroyed, but it can be converted into other types of energy. Cells harvest some of the potential energy in the chemical bonds of foodstuffs to generate stored chemical energy in the form of activated carrier molecules, which are often employed to join two molecules together in __________ reactions. a. oxidation b. hydrolysis c. condensation d. dehydrogenation ANS: C DIF: Easy REF: 3.3 OBJ: 3.3.c Relate how the energy of ATP hydrolysis can be harnessed to drive an energetically unfavorable condensation reaction. MSC: Remembering 43. You are studying a biochemical pathway that requires ATP as an energy source. To your dismay, the reactions soon stop, partly because the ATP is rapidly used up and partly because an excess of ADP builds up and inhibits the enzymes involved. You are about to give up when the following table from a biochemistry textbook catches your eye.

Table 3-43

Which of the following reagents is/are most likely to revitalize your reaction?

a. a vast excess of ATP b. glucose 6-phosphate and enzyme D c. creatine phosphate and enzyme A d. pyrophosphate ANS: C An excess of ATP will initially restore the reactions, but as ATP is hydrolyzed, ADP will build up and inhibit the enzymes again. Pyrophosphate does not look like ATP and is therefore unlikely to be used by the enzymes as an alternative energy source. Pyrophosphate + enzyme C will just heat things up. What you need is a high-energy source of phosphate that can convert ADP back to ATP. Because the ΔG° of the reaction ATP + creatine → ADP + creatine phosphate catalyzed by enzyme A is greater than zero, the addition of creatine phosphate and enzyme A can be used to form ATP from ADP, regenerating the ATP while also forming creatine as a waste product.

DIF: Difficult REF: 3.3 OBJ: 3.3.k Outline how ATP hydrolysis provides the energy needed for highly unfavorable reactions, such as the biosynthesis of nucleic acids. MSC: Applying 44. All of the biological molecules listed below contain high-energy phosphate bonds. Which one is the key driver of most phosphorylation reactions and the transfer of metabolic energy? a. glucose-P b. creatine-P c. acetyl-P d. adenosine-P3 ANS: D DIF: Easy REF: 3.3 OBJ: 3.3.j Outline how ATP hydrolysis provides the energy needed for highly unfa vorable reactions, such as the biosynthesis of nucleic acids. MSC: Remembering 45. The anhydride formed between a carboxylic acid and a phosphate (Figure 3 -45A) is a high-energy intermediate for some reactions in which ATP is the energy source. Arsenate can also be incorporated into a similar high-energy intermediate in place of the phosphate (Figure 3-45B). Figure 3-45C shows the reaction profiles for the hydrolysis of these two high-energy intermediates. What is the effect of substituting arsenate for phosphate in this reactio n?

Figure 3-45

a. It forms a high-energy intermediate of lower energy. b. It forms a high-energy intermediate of the same energy. c. It decreases the stability of the high-energy intermediate. d. It increases the stability of the high-energy intermediate.

ANS: C The activation energy of the arsenate compound is extremely low, as can be seen from the reaction profile, meaning that its high-energy intermediate is very unstable and will be spontaneously hydrolyzed more rapidly than the pho sphate compound. In fact, this hydrolysis occurs rapidly without enzyme catalysis, even in cellular conditions. Arsenate is therefore quite deleterious for living organisms.

DIF: Difficult REF: 3.3 OBJ: 3.3.j Outline how ATP hydrolysis provides the energy needed for highly unfavorable reactions, such as the biosynthesis of nucleic acids. MSC: Applying 46. The synthesis of glutamine from glutamic acid requires the production of an activated intermediate followed by a condensation step that completes the process. Both amino acids are shown in Figure 3-46.

Figure 3-46

Which molecule is added to glutamic acid in the activation step? a. phosphate

b. NH3 c. ATP d. ADP ANS: A DIF: Moderate REF: 3.3 OBJ: 3.3.a Review how activated carriers link catabolic and anabolic reactions.

MSC: Remembering 47. The synthesis of glutamine from glutamic acid requires the production of an activated intermediate followed by a condensation step that completes the process. Both amino acids are shown in Figure 3-47.

Figure 3-47

In the condensation step, __________ is displaced by __________. a. OH; NH3. b. ADP; NH2. c. ATP; NH3. d. phosphate; NH3.

ANS: D DIF: Easy REF: 3.3 OBJ: 3.3.a Review how activated carriers link catabolic and anabolic reactions. MSC: Remembering 48. NADH and NADPH are activated carrier molecules that function in completely different metabolic reactions. Both carry two additional __________ and one additional __________. This combination can also be referred to as a hydride ion. a. protons; electron. b. electrons; phosphate. c. hydrogens; electron. d. electrons; proton.

ANS: D DIF: Easy REF: 3.3 OBJ: 3.3.f Relate the structures of NADH and NADPH and correlate with their roles within cells. MSC: Remembering

MULTIPLE SELECT 1. In the first stage of photosynthesis, light energy can be converted into what other form, or forms, of energy? There may be more than one answer. a. electrical b. chemical c. potential d. kinetic

ANS: A, D DIF: Moderate REF: 3.1 OBJ: 3.1.d Express why sunlight provides the ultimate source of energy for most all living things. MSC: Understanding

MATCHING 1. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

A. activation B. chemical bond C. completely D. favorable E. free F. kinetic G. rapidly H. selectively I. slowly J. unfavorable 1. By definition, catalysis allows a reaction to occur more __________.

2. Chemical reactions occur only when there is a loss of __________ energy.

3. Enzymes act more __________ than other catalysts.

4. A catalyst decreases the __________ energy of a reaction.

1. ANS: G DIF: Easy REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction. | 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. MSC: Understanding 2. ANS: E DIF: Easy REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction. | 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. MSC: Understanding 3. ANS: H DIF: Easy REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction. | 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. MSC: Understanding 4. ANS: A DIF: Easy REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction. | 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. MSC: Understanding 2. Activated carriers store energy in different types of high energy bonds. Match the groups used in high-energy linkages (A–F) for each of the of the activated carriers (1–8). Group letters may be used more than once.

A. glucose B. electrons and hydrogens C. carboxyl group D. phosphate E. methyl group F. acetyl group 1. _____ ATP 2. _____ NADH

3. _____ acetyl CoA 4. _____ carboxylated biotin 5. _____ NADPH 6. _____ S-adenosylmethionine 7. _____ FADH2 8. _____ uridine diphosphate glucose 1. ANS: D DIF: Easy REF: 3.3 OBJ: 3.3.h Name 3 activated carriers and state the role each plays in a cell. MSC: Remembering 2. ANS: B DIF: Easy REF: 3.3 OBJ: 3.3.h Name 3 activated carriers and state the role each plays in a cell. MSC: Remembering 3. ANS: F DIF: Easy REF: 3.3 OBJ: 3.3.h Name 3 activated carriers and state the role each plays in a cell. MSC: Remembering 4. ANS: C DIF: Easy REF: 3.3 OBJ: 3.3.h Name 3 activated carriers and state the role each plays in a cell. MSC: Remembering 5. ANS: B DIF: Easy REF: 3.3 OBJ: 3.3.h Name 3 activated carriers and state the role each plays in a cell. MSC: Remembering 6. ANS: E DIF: Easy REF: 3.3 OBJ: 3.3.h Name 3 activated carriers and state the role each plays in a cell. MSC: Remembering 7. ANS: B DIF: Easy REF: 3.3 OBJ: 3.3.h Name 3 activated carriers and state the role each plays in a cell. MSC: Remembering 8. ANS: A DIF: Easy REF: 3.3 OBJ: 3.3.h Name 3 activated carriers and state the role each plays in a cell. MSC: Remembering

SHORT ANSWER 1. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. The second law of thermodynamics states that the total amount of energy in the universe remains constant.

B. The source of all energy for living systems is chlorophyll.

C. CO2 gas is fixed in a series of reactions that are light-dependent.

D. H2 is the most stable and abundant form of hydrogen in the environment.

ANS: A. False. The second law of thermodynamics states that components of any system move toward greater disorder. It is the first law of thermodynamics that states that energy is neither created nor destroyed.

B. False. The source of all energy for living organisms is sunlight.

C. False. The fixation of carbon from CO2 occurs independently of light.

D. False. The most stable form of hydrogen is H2O. DIF: Moderate REF: 3.1 OBJ: 3.1.b State how living systems can generate and maintain order without violating the second law of thermodynamics. | 3.1.d Express why sunlight provides the ultimate source of energy for most all living

things. MSC: Evaluating 2. Fill in the blanks, selecting from the choices below.

CO,

CO2,

O2,

H2 ,

H2O,

N2,

NO

Light + _________ + _________ → _________ + heat + sugars ANS: H2O, CO2, O2 DIF: Easy REF: 3.1 OBJ: 3.1.e Contrast photosynthesis and cellular respiration and evaluate how the processes complement one another. MSC: Remembering 3. For each of the pairs A–D in Figure 3-54, pick the more reduced member of the pair.

Figure 3-54

ANS: A—ii; B—ii; C—i; D—ii. “More reduced” means having more electrons; a gain of electrons can result in an increased negative charge or a decreased positive charge and can be due to an increase in the number of hydrogen atoms in a molecule. DIF: Moderate REF: 3.1 OBJ: 3.1.f Differentiate between oxidation and reduction in terms of the movement of electrons, protons, and oxygen atoms. MSC: Applying 4. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Photosynthetic organisms release only O 2 into the atmosphere, while non-photosynthetic organisms release only CO2.

B. The cycling of carbon through the biosphere first requires the incorporation of inorganic CO2 into organic molecules.

C. The oxidation of one molecule is always coupled to the reduction of a second molecule.

D. During cellular respiration, carbon-containing molecules become successively more oxidized until they reach their most oxidized form, as CO2.

ANS: A. False. Plants, as well as photosynthetic algae and bacteria, perform both photosynthesis and respiration. This means that photosynthetic organisms release both O2 and CO2 into the atmosphere.

B. True C. True. This forms the basis for redox pairs.

D. True DIF: Easy REF: 3.1 OBJ: 3.1.e Contrast photosynthesis and cellular respiration and evaluate how the processes complement one another. | 3.1.f Differentiate between oxidation and reduction in terms of the movement of ele ctrons, protons, and oxygen atoms. | 3.1.g State why CO 2 is an end product of cellular respiration.

MSC: Evaluating 5. Arrange the following molecules in order with respect to their relative levels of oxidation (assign 5 to the most oxidized and 1 to the most reduced). _______ CH2O (formaldehyde) _______ CH4 (methane) _______ CHOOH (formic acid) _______ CH3OH (methanol) _______ CO2 (carbon dioxide) ANS: __3__ CH2O (formaldehyde) __1__ CH4 (methane) __4__ CHOOH (formic acid) __2__ CH3OH (methanol) __5__ CO2 (carbon dioxide) DIF: Difficult REF: 3.1 OBJ: 3.1.f Differentiate between oxidation and reduction in terms of the movement of electrons, protons, and oxygen atoms. MSC: Applying 6. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Reactions that are energetically favorable occur rapidly because there is no energy of activation.

B. Enzymes accelerate reactions by transferring energy to reactant molecules.

C. Spontaneous reactions are characterized by a negative change in free energy.

ANS: A. False. Reactions that are energetically favorable are spontaneous, but do not always occur rapidly. And, even though there is a net negative change in free energy for the reaction, there is still activation energy required for substrate molecules to react and yield products.

B. False. Enzymes catalyze reactions by reducing the activation energy required for reactants to yield products.

C. True DIF: Easy REF: 3.2 OBJ: 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. | 3.2.b Articulate how enzymes increase the speed of a chemical reaction. MSC: Evaluating 7. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Enzymes lower the free energy released by the reaction that they facilitate.

B. Enzymes lower the activation energy for a specific reaction.

C. Enzymes increase the probability that any given reactant molecule will be converted to product.

D. Enzymes increase the average energy of reactant molecules.

ANS: A. False. Enzymes do not affect the initial energy of the reactants nor the final energy of the products after the reaction is complete, which are the values that determine the change in free energy of a reaction.

B. True C. True D. False. By lowering the energy of activation, enzymes increase the number of molecules in a population that can overcome the activation barrier.

DIF: Easy REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction.

MSC: Evaluating 8. On the basis of the two reactions below, indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

1: ATP + Y → Y-P + ADP

ΔG = −100 kcal/mole

2: Y-P + A → B

ΔG = 50 kcal/mole

A. Reaction 1 is favorable because of the large negative ΔG associated with the hydrolysis of ATP. B. Reaction 2 is an example of an unfavorable reaction.

C. Reactions 1 and 2 are coupled reactions, and when they take place together, reaction 2 will proceed in the forward direction.

D. Reaction 2 can be used to drive reaction 1 in the reverse direction.

ANS: A. True B. True C. True D. False. This is false for two reasons: (1) reaction 2 is unfavorable, as indicated by the positive free-energy change associated with the reaction; (2) the reverse reaction, although possibly more favorable, will yield the product for reaction 1, not reactants to help drive it forward.

DIF: Easy REF: 3.2 OBJ: 3.2.c Relate how the free energy change of a reaction determines its energetic favorability.

MSC: Evaluating 9. Consider the reaction X→Y in a cell at 37°C. At equilibrium, the concentrations of X and Y are 50 μM and 5 μM, respectively. Use this information and the equations below to answer questions 1–5. ΔG° = −0.616 ln Keq ΔG = ΔG° + 0.616 ln [Y]/[X] Recall that the natural log of a number z will have a negative value when z < 1, positive when z > 1, and 0 when z = 1.

1. What is the value of Keq for this reaction?

2. Is the standard free-energy change of this reaction positive or negative? Is the reaction X→Y an energetically favorable or unfavorable reaction under standard conditions?

3. What is the value of the standard free energy? Refer to Table 3-1 in the textbook or use a calculator.

4. Imagine circumstances in which the concentration of X is 1000 μM and that of Y is 1 μM. Is conversion of X to Y favorable? Will it happen quickly?

5. Imagine starting conditions in which the reaction X→Y is unfavorable, yet the cell needs to produce more Y. Describe two ways in which this may be accomplished.

ANS: 1. Keq = [Y]/[X] = 5 μM/50 μM = 0.1 2. The standard free-energy change, ΔG°, is positive because Keq is less than 1. Under standard conditions (equal concentrations of X and Y), the reaction X→Y is unfavorable.

3. ΔG° = −0.616 ln Keq = −0.616 ln 0.1 = (−0.616) (−2.3) = 1.4 kcal/mole.

4. Yes, the conversion is favorable because the value of [Y]/[X] is less than the equilibrium value. However, the speed of the reaction cannot be determined from the free-energy difference.

5. The cell may directly couple the unfavorable reaction to a second, energetically favorable reaction whose negative ΔG has a value larger than the positive ΔG of the X→Y reaction; the coupled reaction will have a ΔG equal to the sum of the component reactions. Alternatively, more X will be converted to Y if the concentration of Y drops; this may happen if Y is converted to Z in a second reaction or if Y is exported from the cell or compartment where the X→Y reaction occurs.

DIF: Moderate REF: 3.2 OBJ: 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. | 3.2.d Review the relationship between delta G, equilibrium, and the concentrations of a reaction’s substrates and products. | 3.2.e Relate delta G to delta G0. | 3.2.f Formulate how the equilibrium constant, K, governs the rate of a reaction and the amount of product it generates. MSC: Applying 10. Hydrolysis reactions are commonly used inside the cell to split high-energy covalent bonds. For each of the three reactions below, use the ΔG° for each reaction to determine the equilibrium constants (K). Assume that each reaction occurs independently of the other two. ΔG° (kcal/mole) Reaction 1: acetyl-P → acetate + P

−10.3

Reaction 2: ATP → ADP + P

−7.3

Reaction 3: glucose 6-P → glucose + P −3.3 ANS: K can be easily calculated from the standard free-energy values by solving the standard free-energy equation (ΔG° = −1.42 log K) for K (K = 10ΔG°/−1.42).

Reaction 1: K = 107.25 Reaction 2: K = 105.14 Reaction 3: K = 102.32 DIF: Difficult REF: 3.2 OBJ: 3.2.d Review the relationship between delta G, equilibrium, and the concentrations of a reaction’s substrates and products. | 3.2.e Relate delta G to delta G0. | 3.2.f Formulate how the equilibrium constant, K, governs the rate of a reaction and the amount of product it generates. MSC: Applying 11. Indicate whether the following statements are true or false. If a statement is false, explain why it is false.

A. When two macromolecules form a complex, the free energy of the system increases because there is a net increase of order within the cell.

B. Sequential pathways can help drive unfavorable reactions by siphoning off the products into the next energetically

favorable reaction in the series.

C. The cytosol is densely packed with molecules, creating what is more an aqueous gel than a solution.

D. The diffusion rates for smaller molecules in the cytosol are much lower than what is observed for the same molecules in water.

ANS: A. False. Even nonspecific interactions between macromolecules can be favorable if there is a large number of water molecules and ions displaced at the interaction interface. This would lead to an overall increase in disorder, even though the two larger molecules become associated and more ordered.

B. True C. True D. False. Small molecules diffuse through the cytosol nearly as rapidly as they diffuse in water.

DIF: Easy REF: 3.2 OBJ: 3.2.a Recall the relationship between free energy and entropy. | 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. MSC: Evaluating 12. Indicate whether the following statements about enzymes are TRUE or FALSE. If a statement is false, explain why it is false.

A. Enzymes alter the equilibrium point of a reaction.

B. Vmax can be determined by measuring the amount of product accumulated late in the reaction.

C. Competitive inhibitors bind irreversibly to the enzyme active site, lowering Vmax.

ANS: A. False. An enzyme catalyzes its reaction in both directions, lowering the energy of activation for both the forward and reverse reactions. Enzymes do not affect the free energy of the reactants and products, and thus they do not affect the reaction equilibrium.

B False. Initial reaction velocities are measured to determine Vmax.

C. False. Competitive inhibitors bind reversibly to an enzyme’s active site.

DIF: Moderate REF: 3.2 OBJ: 3.2.i Summarize how substrate concentration affects an enzyme’s activity. | 3.2.j Describe how noncovalent interactions allow enzymes to interact with specific substrates. | 3.2.l Illustrate how enzymes affect the equilibrium point of chemical reactions. MSC: Evaluating 13. A. You are measuring the effect of temperature on the rate of an enzyme-catalyzed reaction. If you plot reaction rate against temperature, which of the graphs in Figure 3-64 would you expect your plot to resemble?

B. Explain why temperature has this effect.

Figure 3-64

ANS: A. Graph 1 is correct.

B. An increase in temperature increases the number of collisions of sufficient energy to overcome the activation energy. A moderate increase in temperature could therefore increase the reaction rate. At higher temperatures enzymes will denature and no longer function as reaction catalysts.

DIF: Moderate REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction.

MSC: Applying 14. Consider a description of an enzymatic reaction pathway that begins with the binding of substrate S to enzyme E, and ends with the release of product P from the enzyme.

E + S → ES → EP → E + P In many circumstances, Km = [E][S]/[ES] A. What proportion of enzyme molecules is bound to substrate when [S] = Km? B. Recall that when [S] = Km, the reaction rate is ½Vmax. Does your answer to part A make sense in the light of this rate information?

ANS: A. When [S] is substituted for Km in the equation, it becomes clear that [E] = [ES]. Thus, half of the enzyme molecules are free and half are bound to the substrate.

B. Yes. If half of the enzyme molecules are bound to the substrate, it makes intuitive sense that the reaction rate is half of the maximum possible rate, or half of the rate observed when all of the enzyme molecules are bound to the substrate.

DIF: Difficult REF: 3.2 OBJ: 3.2.i Summarize how substrate concentration affects an enzyme’s activity.

MSC: Evaluating 15. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Activated carriers can provide energy and donate chemical groups to substrates in biosynthetic reactions.

B. ATP is a very common activated carrier molecule that donates electrons rather than a chemical group.

C. Activated carriers provide energy for unfavorable reactions and are “reloaded” when coupled to favorable reactions.

D. Activated carriers supply energy for catabolic reactions and are regenerated when coupled to anabolic reactions.

ANS: A. True B. False. When ATP supplies energy for a biosynthetic reaction, it is hydrolyzed and donates a phosphate group to the substrate.

C. True D. False. Catabolic reactions are the favorable reactions occurring during the breakdown of food molecules, the energy from these reactions is stored in activated carriers. The activated carriers then supply energy for anabolic reactions, which are unfavorable reduction reactions.

DIF: Easy REF: 3.3 OBJ: 3.3.a Review how activated carriers link catabolic and anabolic reactions. | 3.3.j Outline how ATP hydrolysis provides the energy needed for highly unfavorable reactions, such as the biosynthesis of nucleic acids. MSC: Evaluating 16. If you weigh yourself on a scale one morning, then eat four pounds of food during the day, will you weigh four pounds more the next morning? Why or why not? (Hint: What happens to the atoms from the food you ingested?)

ANS: No, you will not weigh four pounds more the next morning because only a small portion of the mass of the food will form components of the body. Much of the mass of food is either released as CO 2 and H2O that are breathed out into the atmosphere or converted into materials excreted as waste products.

DIF: Easy REF: 3.1 OBJ: 3.1.a Relate anabolic, catabolic, and metabolic reactions. MSC: Evaluating 17. In the cytoplasm, materials are organized, separated, and sorted by membranes. Cells exploit the selective permeability of these membranes to partition populations of molecules and generate chemical energy for the cell. Use the principles of the first and second laws of thermodynamics to explain how membranes can be used to produce chemical energy.

ANS: When one type of molecule accumulates at a higher concentration on one side of the membrane, the molecules become “organized” by having their movement limited to the space they occupy. The second law of thermodynamics indicates that, if allowed to do so, the molecules would move across the membrane until there is an equal distribution of molecules on either side. The accumulation of molecules on one side of the membrane represents a store of potential energy. The first law of thermodynamics tells us that this energy will not be lost, but rather converted into a different type of energy. First, when the molecules are then allowed to move across the membrane, this potential energy is converted into kinetic energy. The kinetic energy of molecules moving through a protein channel to cross the membrane is often linked to conformational changes in the protein, promoting specific chemical reactions.

DIF: Difficult REF: 3.1 OBJ: 3.1.b State how living systems can generate and maintain order without violating the second law of thermodynamics. | 3.1.c Summarize how living systems take advantage of the first law of thermodynamics.

MSC: Evaluating 18. Two college roommates do not agree on the best way to handle the clutter piled up in your dorm room. Roommate 1 explains that chaos is inevitable, so why fight it? Roommate 2 counters that maintaining an organized environment makes life easier in many ways, and that chaos is not inevitable. What law of thermodynamics drives the thinking of Roommate 1? What thermodynamic argument can be used to support Roommate 2?

ANS: The second law of thermodynamics supports Roommate 1’s view. It is favorable for a system to become less ordered. However, if the energy used to create an ordered environment in the room is accompanied by enough release of heat, the Universe will become more disordered as the room becomes more organized. Thus, while increasing chaos is inevitable, the room—like a cell—can be kept highly organized, by interconverting types of energy.

DIF: Easy REF: 3.1 OBJ: 3.2.a Recall the relationship between free energy and entropy. | 3.1.b State how living systems can generate and maintain order without violating the second law of thermodynamics. MSC: Applying 19. Assume that the average human adult requires 2000 kilocalories per day to sustain all normal processes and maintain a constant weight. If manufactured solar panels could somehow provide power directly to the human body, what size solar panel would be required (in cm 2)? Assume there are 10 hours of sunlight per day, and that the usable energy output for a typical solar panel is 850 kJ/ft2 per hour.

Note: 1 kcal = 4.184 kJ 1 ft2 = 929.03 cm2 ANS: Conversion factors: 1 kcal = 4.184 kJ 1 ft2 = 929.03 cm2 If there are 10 hours of sunlight each day hitting the solar panel, there are 8500 kJ/ft2 produced per day. The average adult human requires 8368 kJ per day; thus, with a conventional solar panel, we would require a surface area of about a square foot, or more precisely in cm 2: 8368 kJ/X = 8500 kJ/929.03 cm 2 929.03 × 8368/8500 = X X = 914.57 cm2 DIF: Moderate REF: 3.1 OBJ: 3.1.d Express why sunlight provides the ultimate source of energy for most all living things. MSC: Applying 20. Although the biochemical study of reaction rates and free energies is important for understanding each biological reaction individually, these studies do not provide an accurate picture of what is happening to reactants and products inside the cell. Why not?

ANS: Chemical reactions inside the cell do not reach a state of equilibrium because both reactants and products are typically used in more than one set of reactions, which means their concentrations are constantly fluctuating. As a result, the forward and reverse reaction rates are almost never identical.

DIF: Moderate REF: 3.2 OBJ: 3.2.i Summarize how substrate concentration affects an enzyme’s activity.

MSC: Evaluating 21. Figure 3-72 illustrates the amount of energy per molecule for a population in a contained, controlled environment.

Most molecules will have the average energy of the population, shown in region 1. The number of molecules in the population with enough energy to be converted to product is shown in region 2. The number of molecules with enough energy to react in the presence of enzyme is shown in region 3. Use this information to explain how enzymes catalyze reactions.

Figure 3-72

ANS: The presence of enzyme in the mixture of reactant molecules does not change the energy distribution of the population of molecules. Their average energy will remain the same, and there still will be only a very small proportion of the molecules with high energy. Enzyme catalysis increases the total number of molecules that have sufficient energy to participate in the reaction because the total energy required per molecule that reacts is lowered.

DIF: Difficult REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction.

MSC: Understanding 22. Chemical reactions are reversible; they can proceed in both the forward and reverse directions. If the ΔG° for the reaction Y→X is energetically favorable, how can you explain the fact that not all of the Y molecules will be converted to X molecules?

ANS: Even when the forward reaction is highly favorable, it is important to keep in mind that molecules exist as part of a population, and each member of a given population has a varying level of energy per molecule. Statistically speaking, there will always be some molecules that have sufficient energy to reach the energy of activation for the back reaction X→Y, even though the proportion of molecules with this energy will be much lower than that for the forward reaction Y→X. As more and more X molecules are converted to Y molecules, eventually the Y molecules in the mixture outnumber the X molecules to such a large extent that the fluxes in the backward and forward directions become equal; it is here that the reaction reaches its equilibrium point.

DIF: Moderate REF: 3.2 OBJ: 3.2.d Review the relationship between delta G, equilibrium, and the concentrations of a reaction’s substrates and products. | 3.2.e Relate delta G to delta G0. MSC: Understanding 23. Enzymes A and B catalyze different reactions, but use the same reactant molecule as a substrate. The graph in Figure 3-74 presents the reaction rates observed when enzyme A and enzyme B are mixed together in a single test tube containing molecule X. What are the Vmax and the apparent Km values for each enzyme under these conditions? How might these values change for enzyme B if it were analyzed in the absence of enzyme A? Explain your answer.

Figure 3-74

ANS: Under the mixed conditions, enzyme A has a Vmax of 6 and an apparent Km of 1 μM X; enzyme B has a Vmax of 10 and an apparent Km of 3 μM. Because enzyme A has a higher affinity for substrate, it more quickly binds to reactant X and converts it into the product Y, lowering the effective concentration of X reactant available for enzyme B. If enzyme B were tested separately, the Vmax should stay the same, but the Km might be smaller and be a more accurate reflection of the binding affinity of enzyme B for the reactant molecule X. DIF: Difficult REF: 3.2 OBJ: 3.2.j Describe how noncovalent interactions allow enzymes to interact with specific substrates. MSC: Applying 24. Consider an analogy between reaction-coupling and money. In a simple economy, barter provides a means of direct exchange of material goods. For example, the owner of a cow may have excess milk and need eggs, whereas a chicken owner has excess eggs and needs milk. Provided that these two people are in close proximity and can communicate, they may exchange or barter eggs for milk. But in a more complex economy, money serves as a mediator for the exchanges of goods or services. For instance, the cow owner with excess milk may not need other goods until three months from now, or may want goods from someone who does not need milk. In this case, the “energy” from providing milk to the economy can be temporarily “stored” as money, which is a form of “energy” used for many transactions in the economy. Using barter and money as analogies, describe two mechanisms that can serve to drive an unfavorable chemical reaction in the cell.

ANS: Barter is analogous to the direct coupling of a favorable to an unfavorable reaction by a single enzyme. Money is analogous to the storage of energy from a favorable reaction in the form of high-energy bonds in an activated carrier molecule. Such activated carrier molecules can later be used to drive a huge variety of other unfavorable reactions in the cell, either by being hydrolyzed to provide the needed energy for a reaction or by transferring an activated chemical group to another molecule.

DIF: Moderate REF: 3.3 OBJ: 3.3.a Review how activated carriers link catabolic and anabolic reactions.

MSC: Understanding 25. In general, there is a positive change in free energy associated with reduction reactions, and most of them are coupled with oxidation reactions. The last step in the biosynthesis of cholesterol involves the reduction of a carbon– carbon double bond. What activated carrier molecule is used in this reaction (and generally for the reduction of lipids) and how would this reaction be influenced by the levels of available ATP? ANS: NADPH is the activated carrier used in the final reduction reaction to produce cholesterol. The rate of this reaction will depend upon the concentration of NADPH, the regeneration of which depends upon ATP hydrolysis. If ATP levels are low, then we can expect that the levels of NADPH will be correspondingly low. The rate of cholesterol synthesis will be lower than if the cell is in a high-energy state with abundant levels of ATP, and consequently high levels of NADPH. DIF: Easy REF: 3.3 OBJ: 3.3.f Relate the structures of NADH and NADPH and correlate with their roles within cells. | 3.3.j Outline how ATP hydrolysis provides the energy needed for highly unfavorable reactions, such as the biosynthesis of nucleic acids. MSC: Understanding 26. Coenzyme A can be converted to acetyl CoA, which is an important activated carrier molecule that has a central role in metabolism and can be used to add two carbons in each successive cycle of fatty acid synthesis.

A. Identify the location and type of high-energy bond in the acetyl CoA molecule shown in Figure 3-77.

B. How does the bond energy help promote the synthesis of fatty acids?

C. What function does the rest of the coenzyme A molecule serve in these reaction pathways?

Figure 3-77

ANS: A. The thioester bond is the high-energy bond B. The generation of long-chain fatty acids requires the generation of new carbon–carbon bonds and the reduction of carbon–oxygen bonds. The energy stored in the acetyl CoA thioester bond promotes the addition of new carbons to an elongating fatty acid chain, two carbons at a time. However, the high-energy electrons required to reduce the carbonyl bond are derived from a second type of molecule, NADPH. C. The rest of the acetyl CoA molecule provides a specific surface recognized by the enzymes that catalyze reactions in which acetyl CoA is required.

DIF: Difficult REF: 3.3 OBJ: 3.3.a Review how activated carriers link catabolic and anabolic reactions.

MSC: Understanding 27. The addition of a new deoxynucleotide to a growing DNA chain requires more energy than can be obtained by the hydrolysis of ATP to ADP + Pi. What alternative series of reactions is used, and how does this help overcome the energy barrier for DNA synthesis?

ANS: The hydrolysis of ATP to ADP is favorable, with a ΔG between −11 and −13 kcal/mole. However, this is not sufficient to drive the addition of a nucleotide to the end of a growing DNA strand. Instead, two reactions are used. The first reaction converts ATP to a DNA-linked AMP residue when a phosphodiester bond is formed during DNA synthesis; simultaneously, a pyrophosphate molecule (PPi) is released. In the second reaction, thePPi is hydrolyzed to form two molecules of Pi. This second reaction is also favorable, providing roughly another −13 kcal/mole. Adding up the ΔG for the entire process, there will be about −26 kcal/mole to drive the addition of the nucleotide to the growing DNA chain, which is sufficient to drive the reaction strongly in one direction.

DIF: Difficult REF: 3.3 OBJ: 3.3.j Outline how ATP hydrolysis provides the energy needed for highly unfavorable reactions, such as the biosynthesis of nucleic acids. MSC: Understanding 28. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. activation

free

selectively

chemical bond

kinetic

slowly

completely

rapidly

unfavorable

favorable By definition, catalysts allow a reaction to occur more __________.

Chemical reactions occur only when there is a loss of __________ energy. Enzymes act more __________ than other catalysts. A catalyst decreases the __________ energy of a reaction.

ANS: By definition, catalysts allow a reaction to occur more rapidly. Chemical reactions occur only when there is a loss of free energy. Enzymes act more selectively than other catalysts. A catalyst decreases the activation energy of a reaction.

DIF: Easy REF: 3.2 OBJ: 3.2.b Articulate how enzymes increase the speed of a chemical reaction. | 3.2.c Relate how the free energy change of a reaction determines its energetic favorability. MSC: Understanding 29. Activated carriers store energy in different types of high energy bonds. Match the groups used in high-energy linkages (right column) for each of the of the activated carriers (left column). Each group number may be used more than once. _____ ATP

1. glucose

_____ NADH

2. electrons and hydrogens

_____ acetyl CoA

3. carboxyl group

_____ carboxylated biotin

4. phosphate

_____ NADPH

5. methyl group

_____ S-adenosylmethionine

6. acetyl group

_____ FADH2 _____ uridine diphosphate ANS: ___4__ATP ___2__NADH ___6__acetyl CoA ___3__carboxylated biotin ___2__NADPH

___5__S-adenosylmethionine ___2__FADH2 ___1__uridine diphosphate glucose DIF: Easy REF: 3.3 OBJ: 3.3.h Name 3 activated carriers and state the role each plays in a cell.

MSC: Remembering

CHAPTER 4

Protein Structure and Function THE SHAPE AND STRUCTURE OF PROTEINS

4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions.

4.1.b Describe the relationship between free energy and protein conformation.

4.1.c Explain how chaperone proteins guide the folding of a polypeptide chain—and why some proteins can fold without chaperone assistance.

4.1.d Contrast the hydrogen bonding patterns that give rise to alpha helices with those that produce beta sheets.

4.1.e Explain how amyloid structures form and discuss some of the consequences of their formation.

4.1.f Describe the role that protein domains play within a protein’s three-dimensional structure.

4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures.

4.1.h Explain the role that unstructured sequences play in protein function and how amino acid changes in these regions may affect these functions.

4.1.i Explain how binding sites allow the assembly of multisubunit proteins and multiprotein complexes.

4.1.j Compare how covalent crosslinks and noncovalent bonds help to establish protein structure.

HOW PROTEINS WORK 4.2.a Summarize the roles that noncovalent interactions and exact protein conformation play in allowing proteins to recognize and bind specifically to their ligands.

4.2.b Explain how antibodies, which share the same basic structure, can recognize a limitless diversity of antigens.

4.2.c State the significance of an enzyme’s Michaelis constant, KM, and explain how this value influences which biochemical pathway a substrate might follow.

4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions.

HOW PROTEINS ARE CONTROLLED 4.3.a Explain how and why different forms of feedback control might be used to regulate enzyme activity.

4.3.b Explain how the binding of a ligand at a regulatory site can alter the activity of a protein or enzyme.

4.3.c Explain how chemical modification such as phosphorylation can influence a protein’s location and interactions.

4.3.d Contrast how protein activity is regulated by phosphorylation or by the binding of nucleotides such as GTP or ATP. 4.3.e Explain how the hydrolysis of ATP or GTP can produce the directional movement of motor proteins or coordinate the activity of large protein machines.

4.3.f Describe how scaffold proteins aid in the assembly of protein complexes.

4.3.g Explain how intracellular condensates can form biochemical subcompartments in a cell.

HOW PROTEINS ARE STUDIED 4.4.a Contrast chromatography and electrophoresis as methods for protein separation.

4.4.b Explain how mass spectrometry allows the identification of proteins.

4.4.c Compare X-ray crystallography, NMR spectrometry, and cryoelectron microscopy as methods for determining the three-dimensional structure of proteins.

4.4.d Describe how the existence of protein families affects the determination of protein structure.

MULTIPLE CHOICE 1. Polypeptides are synthesized from amino acid building blocks. The condensation reaction between the growing polypeptide chain and the next amino acid to be added involves the loss of a. a water molecule. b. an amino group. c. a carbon atom. d. a carboxylic acid group.

ANS: A DIF: Easy REF: 4.1 OBJ: 4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures.

MSC: Remembering 2. The variations in the physical characteristics between different proteins are influenced by the overall amino acid compositions, but even more important is the unique amino acid a. number. b. sequence. c. bond. d. orientation.

ANS: B DIF: Easy REF: 4.1 OBJ: 4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures.

MSC: Remembering 3. Complete the sentence with the best option provided below. The primary structure of a protein is the a. amino acid composition. b. amino acid sequence. c. average size of amino acid side chains. d. lowest energy conformation.

ANS: B DIF: Easy REF: 4.1 OBJ: 4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures.

MSC: Remembering 4. Fully folded proteins typically have polar side chains on their surfaces, where electrostatic attractions and hydrogen bonds can form between the polar group on the amino acid and the polar molecules in the solvent. In contrast, some proteins have a polar side chain in their hydrophobic interior. Which of the following would NOT occur to help

accommodate an internal, polar side chain? a. A hydrogen bond forms between two polar side chains. b. A hydrogen bond forms between a polar side chain and the protein backbone. c. A hydrogen bond forms between a polar side chain and an aromatic side chain. d. Hydrogen bonds form between polar side chains and a buried water molecule.

ANS: C DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Analyzing 5. To study how proteins fold, scientists must be able to purify the protein of interest, use solvents to denature the folded protein, and observe the process of refolding at successive time points. What is the effect of the solvents used in the denaturation process? a. The solvents break all covalent interactions. b. The solvents break all noncovalent interactions. c. The solvents break some of the noncovalent interactions, resulting in a misfolded protein. d. The solvents create a new protein conformation.

ANS: B DIF: Easy REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Understanding 6. Which of the following statements is TRUE? a. Peptide bonds are the only covalent bonds that can link together two amino acids in proteins. b. There is free rotation around all covalent bonds in the polypeptide backbone. c. Nonpolar amino acids tend to be found in the interior of proteins. d. The sequence of the atoms in the polypeptide backbone varies between different proteins.

ANS: C DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. 4.1.j Compare how covalent crosslinks and noncovalent bonds help to establish protein structure. MSC: Analyzing 7. Protein folding can be studied using a solution of purified protein and a denaturant (urea), a solvent that interferes with noncovalent interactions. Which of the following is observed after the denaturant is removed from the protein solution? a. The polypeptide returns to its original conformation. b. The polypeptide remains denatured. c. The polypeptide forms solid aggregates and precipitates out of solution. d. The polypeptide adopts a new, stable conformation.

ANS: A DIF: Easy REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Understanding 8. The correct folding of proteins is necessary to maintain healthy cells and tissues. The presence of unfolded proteins are associated with some neurodegenerative disorders as Alzheimer’s disease, Huntington’s disease, and Creutzfeldt– Jakob disease (the specific faulty protein is different for each disease). What happens to these disease-causing,

unfolded proteins? a. They are degraded. b. They bind a different target protein. c. They form structured filaments. d. They form protein aggregates.

ANS: D DIF: Easy REF: 4.1 OBJ: 4.1.e Explain how amyloid structures form and discuss some of the consequences of their formation. MSC: Remembering 9. Which of the following is FALSE about molecular chaperones? a. They assist polypeptide folding by helping the folding process follow the most energetically favorable pathway. b. They can isolate proteins from other components of the cells until folding is complete. c. They can interact with unfolded polypeptides in a way that changes the final fold of the protein. d. They help streamline the protein-folding process by making it a more efficient and reliable process inside the cell.

ANS: C DIF: Easy REF: 4.1 OBJ: 4.1.c Explain how chaperone proteins guide the folding of a polypeptide chain—and why some proteins can fold without chaperone assistance. MSC: Analyzing 10. Molecular chaperones can work by creating an “isolation chamber.” What is the purpose of this chamber? a. The chamber acts as a garbage disposal, degrading improperly folded proteins so that they do not interact with properly folded proteins. b. This chamber is used to increase the local protein concentration, which will help speed up the folding process. c. This chamber serves to transport unfolded proteins out of the cell. d. This chamber serves to protect unfolded proteins from interacting with other proteins in the cytosol, until protein folding is completed.

ANS: D DIF: Easy REF: 4.1 OBJ: 4.1.c Explain how chaperone proteins guide the folding of a polypeptide chain—and why some proteins can fold without chaperone assistance. MSC: Understanding 11. One of the key features of living systems is the use of energy to create and maintain order. A good example is found in the folding of newly synthesized proteins. Which activated carrier molecule is used by chaperone proteins to support protein folding? a. FADH2 b. ATP c. NADPH d. NADH ANS: B DIF: Easy REF: 4.1 OBJ: 4.1.c Explain how chaperone proteins guide the folding of a polypeptide chain—and why some proteins can fold without chaperone assistance. MSC: Remembering 12. The three-dimensional coordinates of atoms within a folded protein are determined experimentally. After researchers obtain a protein’s structural details, they can use different techniques to highlight particular aspects of the structure.

What visual model best displays a protein’s secondary structures (α helices and β sheets)? a. ribbon b. space-filling c. backbone

d. wire ANS: A Space-filling and wire models illustrate all atoms, which makes it difficult to see the secondary structure. Backbone models are better but not as good as the ribbon models, which are stylized and make it easy for even the untrained eye to see secondary structures.

DIF: Easy REF: 4.1 OBJ: 4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures.

MSC: Remembering 13. Complete the sentence with the best option provided below. The secondary structures of a protein are the a. regular, repeated folds present in a lowest energy conformation. b. temporary, unstable protein folding conformations. c. interactions between polar amino acid side chains. d. chemical modifications of amino acid side chains.

ANS: A DIF: Easy REF: 4.1 OBJ: 4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures.

4.1.b Describe the relationship between free energy and protein conformation. MSC: Remembering 14. Some proteins have α helices, some have β sheets, and still others have a combination of both. What makes it possible for proteins to have these common structural elements? a. specific amino acid sequences b. side-chain interactions c. the hydrophobic-core interactions d. hydrogen bonds along the protein backbone ANS: D DIF: Easy REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Analyzing 15. Which of the following is NOT a feature commonly observed in α helices? a. left-handedness b. 1 helical turn every 3.6 amino acids c. cylindrical shape d. amino acid side chains that point outward ANS: A DIF: Moderate REF: 4.1 OBJ: 4.1.d Contrast the hydrogen bonding patterns that give rise to alpha helices with those that produce beta sheets. MSC: Analyzing 16. Which of the following is NOT a feature commonly observed in β sheets? a. antiparallel regions b. coiled-coil patterns c. extended polypeptide backbone d. parallel regions ANS: B DIF: Moderate REF: 4.1 OBJ: 4.1.d Contrast the hydrogen bonding patterns that give rise to alpha helices with those that produce beta sheets. MSC: Analyzing 17. Two or three α helices can sometimes wrap around each other to form coiled-coils. The stable wrapping of one helix

around another is typically driven by __________ interactions. a. hydrophilic b. hydrophobic c. van der Waals d. ionic ANS: B DIF: Moderate REF: 4.1 OBJ: 4.1.d Contrast the hydrogen bonding patterns that give rise to alpha helices with those that produce beta sheets. MSC: Remembering 18. Coiled-coils are typically found in proteins that require an elongated structural framework. Which of the following proteins do you expect to have a coiled-coil domain? a. insulin b. collagen c. myoglobin d. porin ANS: B DIF: Moderate REF: 4.1 OBJ: 4.1.f Describe the role that protein domains play within a protein’s threedimensional structure. 4.1.d Contrast the hydrogen bonding patterns that give rise to alpha helices with those that produce beta sheets. MSC: Applying 19. β sheets can participate in the formation of amyloid fibers, which are insoluble protein aggregates. What drives the formation of amyloid fibers? a. denaturation of proteins containing β sheets b. extension of β sheets into much longer β strands c. formation of biofilms by infectious bacteria d. β-sheet stabilization of abnormally folded proteins ANS: D DIF: Moderate REF: 4.1 OBJ: 4.1.d Contrast the hydrogen bonding patterns that give rise to alpha helices with those that produce beta sheets. MSC: Understanding 20. Protein structures have several different levels of organization. The primary structure of a protein is its amino acid sequence. The secondary and tertiary structures are more complicated. Consider the definitions below and select the one that best fits the term “protein domain.” a. a small cluster of α helices and β sheets b. the tertiary structure of a substrate-binding pocket c. a complex of more than one polypeptide chain d. a protein segment that folds independently ANS: D DIF: Moderate REF: 4.1 OBJ: 4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures.

4.1.f Describe the role that protein domains play within a protein’s three-dimensional structure.

MSC: Analyzing 21. Globular proteins fold up into compact, spherical structures that have uneven surfaces. They tend to form multisubunit complexes, which also have a rounded shape. Fibrous proteins, in contrast, span relatively large distances within the cell and in the extracellular space. Which of the proteins below is NOT classified as a fibrous protein? a. elastase b. collagen

c. keratin d. elastin ANS: A DIF: Easy REF: 4.1 OBJ: 4.1.i Explain how binding sites allow the assembly of multi-subunit proteins and multi-protein complexes. MSC: Understanding 22. Which of the following globular proteins is used to form filaments as an intermediate step to assembly into hollow tubes? a. tubulin b. actin c. keratin d. collagen ANS: A DIF: Easy REF: 4.1 OBJ: 4.1.i Explain how binding sites allow the assembly of multi-subunit proteins and multi-protein complexes. MSC: Remembering 23. Which of the following statements is TRUE? a. Disulfide bonds are formed by the cross-linking of methionine residues. b. Disulfide bonds are formed mainly in proteins that are retained within the cytosol. c. Disulfide bonds stabilize but do not change a protein’s final conformation. d. Disulfide bonds are more common for intracellular proteins, compared to extracellular proteins.

ANS: C DIF: Moderate REF: 4.1 OBJ: 4.1.j Compare how covalent crosslinks and noncovalent bonds help to establish protein structure. MSC: Understanding 24. Proteins bind selectively to small-molecule targets called ligands. The selection of one ligand out of a mixture of possible ligands depends on the number of weak, noncovalent interactions in the protein’s ligand-binding site. Where is the binding site typically located in the protein structure? a. on the surface of the protein b. inside a cavity on the protein surface c. buried in the interior of the protein d. forms on the surface of the protein in the presence of ligand ANS: B DIF: Easy REF: 4.2 OBJ: 4.2.a Summarize the role that noncovalent interactions and conformation play in allowing proteins to recognize and bind to their ligands. MSC: Remembering 25. Cyclic AMP (cAMP) is a small molecule that associates with its binding site with a high degree of specificity. Which types of noncovalent interactions are the most important for providing the “hand in a glove” binding of cAMP? a. hydrogen bonds b. electrostatic interactions c. van der Waals interactions d. hydrophobic interactions ANS: A DIF: Easy REF: 4.2 OBJ: 4.2.a Summarize the role that noncovalent interactions and conformation play in allowing proteins to recognize and bind to their ligands. MSC: Remembering 26. The process of generating monoclonal antibodies is labor-intensive and expensive. An alternative is to use polyclonal antibodies. A subpopulation of purified polyclonal antibodies that recognize a particular antigen can be isolated by

chromatography. Which type of chromatography is used for this purpose? a. affinity b. ion-exchange c. gel-filtration d. all of these answers are correct ANS: A DIF: Difficult REF: 4.2 OBJ: 4.2.b Explain how antibodies, which share the same basic structure, can recognize an infinitely diverse array of antigens. MSC: Applying 27. Antibody production is an indispensable part of our immune response, but it is not the only defense our bodies have.

Which of the following is observed during an infection that is NOT a result of antibody–antigen interactions? a. B cell proliferation b. aggregation of viral particles c. systemic temperature increase d. antibody secretion ANS: C DIF: Moderate REF: 4.2 OBJ: 4.2.b Explain how antibodies, which share the same basic structure, can recognize an infinitely diverse array of antigens. MSC: Analyzing 28. Lysozyme is an enzyme that specifically recognizes bacterial polysaccharides, which renders it an effective antibacterial agent. Into what classification of enzymes does lysozyme fall? a. isomerase b. protease c. nuclease d. hydrolase ANS: D DIF: Moderate REF: 4.2 OBJ: 4.2.a Summarize the role that noncovalent interactions and conformation play in allowing proteins to recognize and bind to their ligands. MSC: Understanding 29. Which of the following mechanisms best describes the manner in which lysozyme lowers the energy required for its substrate to reach its transition-state conformation? a. by binding two molecules and orienting them in a way that favors a reaction between them b. by altering the shape of the substrate to mimic the conformation of the transition state c. by speeding up the rate at which water molecules collide with the substrate d. by binding irreversibly to the substrate so that it cannot dissociate ANS: C DIF: Easy REF: 4.2 OBJ: 4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions. MSC: Understanding 30. Studies conducted with a lysozyme mutant that contains an Asp→Asn change at position 52 and a Glu→Gln change at position 35 exhibited almost a complete loss in enzymatic activity. What is the most likely explanation for the decrease in enzyme activity in the mutant? a. increased affinity for substrate b. absence of negative charges in the active site c. change in the active-site scaffold d. larger amino acids in the active site decreases the affinity for substrate ANS: B

The negatively charged amino acids aspartic acid and glutamic acid are required to attack the sugar bonds being cleaved by lysozyme. Replacing these with side chains that are the same length and are polar, but uncharged, would most probably affect only the catalysis, not the binding of substrate or the stability of the protein.

DIF: Moderate REF: 4.2 OBJ: 4.2.a Summarize the role that noncovalent interactions and conformation play in allowing proteins to recognize and bind to their ligands. MSC: Applying 31. In some cases, small molecules are integral to the function of enzymes, and are dubbed “coenzymes.” Which of the following is a coenzyme for the enzyme carboxypeptidase? a. retinal b. biotin c. zinc d. heme ANS: C DIF: Easy REF: 4.2 OBJ: 4.2.a Summarize the role that noncovalent interactions and conformation play in allowing proteins to recognize and bind to their ligands. MSC: Remembering 32. Which of the following statements about allostery is TRUE? a. Allosteric regulators are often products of other chemical reactions in the same biochemical pathway. b. Allosteric regulation is always used for negative regulation of enzyme activity. c. Enzymes are the only types of proteins that are subject to allosteric regulation. d. Binding of allosteric molecules usually locks an enzyme in its current conformation, such that the enzyme cannot adopt a different conformation.

ANS: A DIF: Easy REF: 4.3 OBJ: 4.2.a Summarize the role that noncovalent interactions and conformation play in allowing proteins to recognize and bind to their ligands. MSC: Analyzing 33. The biosynthetic pathway for the two amino acids E and H is shown schematically in Figure 4-33. You are able to show that E inhibits enzyme V, and H inhibits enzyme X. Which biosynthetic product is most likely the inhibitor of enzyme T?

Figure 4-33

a. H b. B c. C d. E ANS: C DIF: Moderate REF: 4.3 OBJ: 4.3.a Explain how and why different forms of feedback control might be used to regulate enzyme activity. MSC: Applying

34. The Ras protein is a GTPase that functions in many growth factor–signaling pathways. In its active form, with GTP bound, it transmits a downstream signal that leads to cell proliferation; in its inactive form, with GDP bound, the signal is not transmitted. Mutations in the gene for Ras are found in many cancers. Of the choices below, which alteration of Ras activity is most likely to contribute to the uncontrolled growth of cancer cells? a. a change that prevents Ras from being made b. a change that increases the affinity of Ras for GDP c. a change that decreases the affinity of Ras for GTP d. a change that decreases the rate of hydrolysis of GTP by Ras ANS: D Ras is a proto-oncogene. When it is active, it promotes cell growth.

DIF: Moderate REF: 4.3 OBJ: 4.3.d Contrast how protein activity is regulated by phosphorylation or by the binding of nucleotides such as GTP or ATP. 4.3.e Explain how the hydrolysis of ATP or GTP can produce the directional movement of motor proteins or coordinate the activity of large protein machines. MSC: Analyzing 35. Motor proteins use the energy in ATP to transport organelles, rearrange elements of the cytoskeleton during cell migration, and move chromosomes during cell division. Which of the following mechanisms is sufficient to ensure the unidirectional movement of a motor protein along its substrate? a. A conformational change is coupled to the release of a phosphate (Pi). b. The substrate on which the motor moves has a conformational polarity. c. A conformational change is coupled to the binding of ADP. d. A conformational change is coupled to ATP hydrolysis.

ANS: D DIF: Difficult REF: 4.3 OBJ: 4.3.e Explain how the hydrolysis of ATP or GTP can produce the directional movement of motor proteins or coordinate the activity of large protein machines. MSC: Analyzing 36. Proteins can assemble to form large complexes that work coordinately, like moving parts inside a single machine.

Which of the following steps in modulating the activity of a complex protein machine is LEAST likely to be directly affected by ATP or GTP hydrolysis? a. translation of protein components b. conformational change of protein components c. complex assembly d. complex disassembly ANS: A DIF: Easy REF: 4.3 OBJ: 4.3.e Explain how the hydrolysis of ATP or GTP can produce the directional movement of motor proteins or coordinate the activity of large protein machines. MSC: Evaluating 37. The phosphorylation of a protein is typically associated with a change in activity, the assembly of a protein complex, or the triggering of a downstream signaling cascade. The addition of ubiquitin, a small polypeptide, is another type of covalent modification that can affect the protein function. Ubiquitylation often results in a. membrane association. b. protein degradation. c. protein secretion. d. nuclear translocation.

ANS: B DIF: Easy REF: 4.3 OBJ: 4.3.c Explain how chemical modification such as phosphorylation can influence a

protein’s location and interactions. MSC: Remembering 38. Energy required by the cell is generated in the form of ATP. ATP is hydrolyzed to power many of the cellular processes, increasing the pool of ADP. As the relative amount of ADP molecules increases, they can bind to glycolytic enzymes, which will lead to the production of more ATP. The best way to describe this mechanism of regulation is a. feedback inhibition. b. oxidative phosphorylation. c. allosteric activation. d. substrate-level phosphorylation.

ANS: C DIF: Difficult REF: 4.3 OBJ: 4.3.a Explain how and why different forms of feedback control might be used to regulate enzyme activity. MSC: Analyzing 39. Which of the following methods would be the most suitable to assess the relative purity of a protein in a sample you have prepared? a. gel-filtration chromatography b. gel electrophoresis c. western blot analysis d. ion-exchange chromatography ANS: B DIF: Easy REF: 4.4 OBJ: 4.4.a Contrast chromatography and electrophoresis as methods for protein separation. MSC: Applying 40. Which of the following methods would be the most suitable to assess whether your protein exists as a monomer or in a complex? a. gel-filtration chromatography b. gel electrophoresis c. western blot analysis d. ion-exchange chromatography ANS: A DIF: Moderate REF: 4.4 OBJ: 4.4.a Contrast chromatography and electrophoresis as methods for protein separation. MSC: Applying 41. Which of the following methods would be the most suitable to assess levels of expression of your target protein in different cell types? a. gel-filtration chromatography b. gel electrophoresis c. western blot analysis d. ion-exchange chromatography ANS: C DIF: Moderate REF: 4.4 OBJ: 4.4.a Contrast chromatography and electrophoresis as methods for separating and analyzing proteins. MSC: Applying 42. Which of the following methods used to study proteins is limited to proteins with a molecular mass of 50 kD or less? a. X-ray crystallography b. fingerprinting c. nuclear magnetic resonance d. mass spectroscopy

ANS: C DIF: Easy REF: 4.4 OBJ: 4.4.c Compare X-ray crystallography, NMR spectrometry, and cryoelectron microscopy as methods for determining the three-dimensional structure of proteins. MSC: Remembering 43. Determining a protein’s sequence, site of covalent modification, or entire three-dimensional structure requires the careful analysis of complex data sets. Which of the data sets below would you have to interpret to solve the structure of a protein by using X-ray crystallography?

Figure 4-43

ANS: B DIF: Difficult REF: 4.4 OBJ: 4.4.c Compare X-ray crystallography, NMR spectrometry, and cryo-electron microscopy as methods for determining the three-dimensional structure of proteins.

MSC: Evaluating 44. Instead of studying one or two proteins or protein complexes present in the cell at any given time, we can now look at a snapshot of all proteins being expressed in cells being grown in specific conditions. This large-scale, systematic approach to the study of proteins is called a. proteomics. b. structural biology. c. systems biology. d. genomics.

ANS: A DIF: Easy REF: 4.4 OBJ: 4.4.b Explain how mass spectrometry allows the identification of proteins.

MSC: Understanding

MULTIPLE SELECT 1. Fibrous proteins serve many important functions in multicellular organisms. Select all functional roles you expect to be filled by fibrous proteins.

A. cytoskeleton B. immunity C. metabolism D. connective tissue ANS: A, D and enzymes important for metabolism are globular proteins.

DIF: 4.1 REF: Easy OBJ: 4.1.i Explain how binding sites allow the assembly of multi-subunit proteins and multi-protein complexes. MSC: Understanding

MATCHING 1. Match the basic protein functions labeled A–I with a specific example of that type of protein (numbered 1–9).

A. insulin B. carboxylase C. rhodopsin D. hemoglobin

E. ferritin F. myosin G. green fluorescent protein H. tubulin I. homeodomain proteins 1. ___ gene regulatory 2. ___ motor 3. ___ storage 4. ___ enzyme 5. ___ transport 6. ___ structural 7. ___ special purpose 8. ___ receptor 9. ___ signal 1. ANS: I DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Analyzing 2. ANS: F DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Analyzing 3. ANS: E DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Analyzing 4. ANS: B DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Analyzing 5. ANS: D DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Analyzing 6. ANS: H DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Analyzing 7. ANS: G DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Analyzing 8. ANS: C DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions.

MSC: Analyzing 9. ANS: A DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions.

MSC: Analyzing 2. For each polypeptide sequence listed (1–3), choose from the options (A–E) given below to indicate which secondary structure the sequence is most likely to form upon folding. The nonpolar amino acids are italicized.

A. amphipathic α helix B. amphipathic β sheet C. hydrophilic α helix D. hydrophilic β sheet E. E. hydrophobic α helix 1. Leu-Gly-Val-Leu-Ser-Leu-Phe-Ser-Gly-Leu-Met-Trp-Phe-Phe-Trp-Ile 2. Leu-Leu-Gln-Ser-Ile-Ala-Ser-Val-Leu-Gln-Ser-Leu-Leu-Cys-Ala-Ile 3. Thr-Leu-Asn-Ile-Ser-Phe-Gln-Met-Glu-Leu-Asp-Val-Ser-Ile-Arg-Trp 1. ANS: E Nearly all of the amino acid side chains in this sequence are nonpolar or hydrophobic, which favors the only hydrophobic option given in the list.

DIF: Difficult REF:4.1 OBJ: 4.1.d Contrast the hydrogen bonding patterns that give rise to alpha helices with those that produce beta sheets. | 4.1.a Explain how noncovalent interactions —including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. | 4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures. MSC: Evaluating 2. ANS: A In an ideal α helix, there are 3.6 residues per complete turn. Thus, an amphipathic helix with one hydrophobic side and one hydrophilic side will have, minimally, nonpolar side chains (N) repeating every third then next fourth amino acid: NxxNxxxNxxNxxxN. Polar side chains (P) will have the same pattern but shifted relative to the nonpolar side chains; for example, xxPxxxPxxPxxxPxxP. DIF: Difficult REF: 4.1 OBJ: 4.1.d Contrast the hydrogen bonding patterns that give rise to alpha helices with those that produce beta sheets. | 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. | 4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures. MSC: Evaluating 3. ANS: B Because of the zigzag-like structure of a β sheet, a sequence with alternating nonpolar and polar side chains may form an amphipathic β sheet that is hydrophobic on one side and hydrophilic on the other.

DIF: Difficult REF: 4.1 OBJ: 4.1.d Contrast the hydrogen bonding patterns that give rise to alpha helices with those that produce beta sheets. | 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how

the polypeptide backbone and amino acid side chains participate in these interactions. | 4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures. MSC: Evaluating 3. Examine the three protein monomers in Figure 4-48. Use the 2D depiction of complementary binding surfaces to match each monomer to the likely higher-order protein complexes they will form.

1. dimers and/or tetramers 2. filaments 3. sheets

Figure 4-48

1. ANS: A DIF: Difficult REF: 4.1 OBJ: 4.1.i Explain how binding sites allow the assembly of multi-subunit proteins and multi-protein complexes. MSC: Evaluating 2. ANS: C DIF: Difficult REF: 4.1 OBJ: 4.1.i Explain how binding sites allow the assembly of multi-subunit proteins and multi-protein complexes. MSC: Evaluating 3. ANS: B DIF: Difficult REF: 4.1 OBJ: 4.1.i Explain how binding sites allow the assembly of multi-subunit proteins and multi-protein complexes. MSC: Evaluating 4. Match the general type of biochemical reaction catalyzed (1–9) with the class of enzyme (A–I).

A. ATPase B. polymerase C. ligase D. kinase E. isomerase F. nuclease G. oxido-reductase H. protease I. phosphatase 1. removes a phosphate group from a molecule 2. hydrolyzes ATP 3. hydrolyzes bonds between nucleotides 4. adds phosphate groups to molecules 5. catalyzes reactions in which one molecule is oxidized and another is reduced 6. hydrolyzes peptide bonds 7. joins two ends of DNA together 8. catalyzes the synthesis of polymers such as RNA and DNA 9. rearranges bonds within a single molecule

1. ANS: I DIF: Moderate REF: 4.2 OBJ: 4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions. MSC: Analyzing 2. ANS: A DIF: Moderate REF: 4.2 OBJ: 4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions. MSC: Analyzing 3. ANS: F DIF: Moderate REF: 4.2 OBJ: 4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions. MSC: Analyzing 4. ANS: D DIF: Moderate REF: 4.2 OBJ: 4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions. MSC: Analyzing 5. ANS: G DIF: Moderate REF: 4.2 OBJ: 4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions. MSC: Analyzing 6. ANS: H DIF: Moderate REF: 4.2 OBJ: 4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions. MSC: Analyzing 7. ANS: C DIF: Moderate REF: 4.2 OBJ: 4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions. MSC: Analyzing 8. ANS: B DIF: Moderate REF: 4.2 OBJ: 4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions. MSC: Analyzing 9. ANS: E DIF: Moderate REF: 4.2 OBJ: 4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions. MSC: Analyzing

SHORT ANSWER 1. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Generally, the total number of nonpolar amino acids has a greater effect on protein structure than the exact order of amino acids in a polypeptide chain.

B. The “polypeptide backbone” refers to all atoms in a polypeptide chain, except for those that form the peptide bonds.

C. The chemical properties of amino acid side chains include charged, uncharged polar, and nonpolar.

D. The relative distribution of polar and nonpolar amino acids in a folded protein is determined largely by hydrophobic interactions, which favor the clustering of nonpolar side chains in the interior.

ANS: A. False. The order in which amino acids are linked is unique for each protein and is the most important factor in determining overall protein structure.

B. False. Peptide bonds are planar amide bonds that are central to the polypeptide backbone formation. The atoms in the amino acid side chains are not considered to be part of the backbone.

C. True D. True DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Evaluating

2. Use your knowledge of amino acid characteristics to order the peptides below by their degree of polarity. Each peptide contains eight amino acids. Use the single-letter amino acid designations to generate your list, placing the most polar peptide on the left and the most nonpolar peptide on the right.

A. SGAKKRAH B. CATWNGQV C. FWGTSILA D. DDAEIHWA E. SSTAMYRK ANS: E, A, D, B, C DIF: Easy REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Understanding 3. Use your knowledge of amino acid characteristics to order the peptides below according to the net charge contributed by their side chains at physiological pH (~pH7). Each peptide contains eight amino acids. Use the single-letter amino acid designations to generate your list, placing the most negatively charged peptide on the left and the most positively charged peptide on the right. In addition, for each peptide, list the total number of positive and negative charges.

Remember that, at neutral pH, the amino terminus carries a positive charge and the carboxyl terminus carries a negative charge.

A. YGAKKRA B. ARRKSTRK C. DERKQNST D. DDAEIYSA E. NQSTYEEG ANS: (D) This peptide has 4 negative charges and 1 positive charge. The peptide in (E) has 3 negative charges and 1 positive charge. The peptide in (C) has 3 negative charges and 3 positive charges. The peptide in (A) has 4 positive charges and 1 negative charge. The peptide in (B) has 6 positive charges and 1 negative charge.

DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Understanding 4. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Van der Waals interactions and hydrophobic interactions are two ways to describe the same type of weak forces that help proteins fold.

B. A large number of noncovalent interactions is required to hold two regions of a polypeptide chain together in a stable conformation.

C. A single polypeptide tends to adopt three or four different conformations, which all have equivalent free-energy values (G).

ANS: A. False. Van der Waals attractions are weakly attractive forces that occur between all atoms. Hydrophobic interactions are only observed between nonpolar molecules in the context of an aqueous solution.

B. True C. False. There is a single most stable fold for every polypeptide. The fold adopted has the single, lowest free-energy

value.

DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. 4.1.b Describe the relationship between free energy and protein conformation. MSC: Evaluating 5. The sequences for three different tripeptides are written out below. Indicate whether you expect to find them in the inner core or on the surface of a cytosolic protein, and explain your answer.

A. serine-threonine-tyrosine B. alanine-glycine-leucine C. proline-serine-alanine ANS: A. This tripeptide is made up of entirely polar amino acids, which means it will most likely be found on the surface of the protein, interacting with the aqueous environment of the cytosol.

B. This peptide is made up of nonpolar amino acids. The side chains are most likely buried in the interior of the protein, which would promote interactions with other hydrophobic side chains and avoid unfavorable interactions with the aqueous environment of the cytosol.

C. This peptide is made up of both polar and nonpolar amino acids, and one of the nonpolar amino acids is proline.

Proline residues have a restricted degree of conformational freedom because the side chain is covalently linked to the backbone nitrogen as well as the α-carbon. So, a likely scenario is that the proline is at the surface of the protein, providing a structural turn between two secondary structure elements (β strands or α helices), the serine is still close enough to the surface to interact with water, and the alanine is close enough to the interior of the protein to interact with other hydrophobic side chains.

DIF: Difficult REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Evaluating 6. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. composition

irreversible

reversible

covalent

lowest

sequence

denatured

noncovalent

stable

highest

renatured

unstable

A newly synthesized protein generally folds up into a __________ conformation. All the information required to determine a protein’s conformation is contained in its amino acid __________. On being heated, a protein molecule will become __________ as a result of breakage of __________ bonds. On removal of urea, an unfolded protein can become __________. The final folded conformation adopted by a protein is that of __________ energy.

ANS: A newly synthesized protein generally folds up into a stable conformation. All the information required to determine a protein’s conformation is contained in its amino acid sequence. On being heated, a protein molecule will become denatured as a result of breakage of noncovalent bonds. On removal of urea, an unfolded protein can become renatured. The final folded conformation adopted by a protein is that of lowest energy.

DIF: Easy REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen

bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. | 4.1.b Describe the relationship between free energy and protein conformation. MSC: Understanding 7. Typical folded proteins have a stability ranging from 7 to 15 kcal/mole at 37°C. Stability is a measure of the equilibrium between the folded (F) and unfolded (U) forms of the protein. For a protein with a stability of 7.1 kcal/mole, calculate the fraction of protein molecules that would be unfolded at equilibrium at 37°C. The equilibrium constant (Keq) is related to the standard free energy (ΔG°) by the equation Keq = 10−ΔG°/1.42.

Figure 4-56

ANS: 1/100,000 The ΔG° of the unfolding reaction is equal to the stability of the protein, 7.1 kcal/mole. At equilibrium, the ratio of unfolded to folded protein is Keq = [U]/[F] = 10−ΔG°/1.42 = 10−7.1/1.42 = 10−5.

Thus, 1 molecule in 100,000 is unfolded.

DIF: Moderate REF: 4.1 OBJ: 4.1.b Describe the relationship between free energy and protein conformation.

MSC: Applying 8. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. allosteric

ligand

secondary

domain

primary

subunit

helix

quaternary

tertiary

The α helices and β sheets are examples of protein __________ structure. A protein such as hemoglobin, which is composed of more than one protein __________, has __________ structure. A protein’s amino acid sequence is known as its __________ structure. A protein __________ is the modular unit from which many larger single-chain proteins are constructed. The three-dimensional conformation of a protein is its __________ structure.

ANS: The α helices and β sheets are examples of protein secondary structure. A protein such as hemoglobin, which is composed of more than one protein subunit, has quaternary structure. A protein’s amino acid sequence is known as its primary structure. A protein domain is the modular unit from which many larger single-chain proteins are constructed. The three-dimensional conformation of a protein is its tertiary structure.

DIF: Easy REF: 4.1 OBJ: 4.1.f Describe the role that protein domains play within a protein’s three-dimensional structure. | 4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures. MSC: Understanding 9. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Collagen is a protein that participates in both the cytoskeleton and the extracellular matrix.

B. Collagen fibers and elastin fibers serve similar functions, which is expected because the structure of these two types of fibers is quite similar.

C. The assembly of both collagen and elastin fibers requires the formation of disulfide bonds.

ANS: A. False. Collagen is not used inside the cell; it is secreted and incorporated into the existing collagen fibers in the extracellular matrix.

B. False. Collagen fibers and elastin fibers are very different in structure and function. Collagen fibers are highly organized, triple-strand coiled-coils that provide strength to hold tissue together. Elastin molecules are linked t ogether in a loose network with disulfide bonds; this allows the fibers (and tissues) to stretch without tearing.

C. True DIF: Easy REF: 4.1 OBJ: 4.1.i Explain how binding sites allow the assembly of multi-subunit proteins and multi-protein complexes. 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Understanding 10. For each of the following, indicate whether the individual folded polypeptide chain forms a globular (G) or fibrous (F) protein molecule.

A. keratin B. lysozyme C. elastin D. collagen E. hemoglobin F. actin ANS: A—(F); B—(G); C—(F); D—(F); E—(G); F—(G) DIF: Easy REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Analyzing 11. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. affinity

billions

ligands

antibodies

coiled-coils

loops

antigens

hundreds

size-exclusion

β strands

ion-exchange

The human immune system produces __________ of different immunoglobulins, also called __________, which enable the immune system to recognize and fight germs by specifically binding one or a few related __________. The hypervariable structural element that forms the ligand-binding site is comprised of several __________. Purified antibodies are useful for a variety of experimental purposes, including protein purification using __________ chromatography.

ANS: The human immune system produces billions of different immunoglobulins, also called antibodies, which enable the immune system to recognize and fight germs by specifically bindin g one or a few related antigens.

The hypervariable structural element that forms the ligand-binding site is comprised of several loops. Purified antibodies are useful for a variety of experimental purposes, including protein purification using affinity chromatog-

raphy.

DIF: Easy REF: 4.2 OBJ: 4.2.b Explain how antibodies, which share the same basic structure, can recognize an infinitely diverse array of antigens. MSC: Remembering 12. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. The amino acids in the interior of a protein do not interact with the ligand and do not play a role in selective binding.

B. Antibodies are Y-shaped and are composed of six different polypeptide chains.

C. ATPases generate ATP for the cell.

D. Hexokinase recognizes and phosphorylates only one of the glucose stereoisomers.

ANS: A. False. The interior amino acids form a structural scaffold that maintains the specific orientation for those that directly interact with the ligand. Changes to these interior amino acids can change the protein shape and render it nonfunctional.

B. False. Although antibodies are Y-shaped, they are composed of four, not six, polypeptide chains. There are two heavy chains and two light chains.

C. False. ATPases hydrolyze ATP; they do not produce it. These enzymes enable the cell to harness the chemical energy stored in the high-energy phosphate bonds.

D. True DIF: Difficult REF: 4.2 OBJ: 4.2.a Summarize the role that noncovalent interactions and conformation play in allowing proteins to recognize and bind to their ligands. | 4.2.b Explain how antibodies, which share the same basic structure, can recognize an infinitely diverse array of antigens. MSC: Evaluating 13. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. activation energy

inhibitors

products

active site

isomers

substrates

free energy

ligand

transition state

high-energy

low-energy

Any substance that will bind to a protein is known as its __________. Enzymes bind their __________ at the __________. The enzyme hexokinase is so specific that it reacts with only one of the two __________ of glucose.

Enzymes catalyze a chemical reaction by lowering the __________, because they provide conditions favorable for the formation of a __________ intermediate called the __________. Once the reaction is completed, the enzyme releases the __________ of the reaction.

ANS: Any substance that will bind to a protein is known as its ligand. Enzymes bind their substrates (or inhibitors) at the active site. The enzyme hexokinase is so specific that it reacts with only one of the two isomers of glucose.

Enzymes catalyze a chemical reaction by lowering the activation energy, because they provide conditions favorable for the formation of a high-energy intermediate called the transition state. Once the reaction is completed, the enzyme releases the products of the reaction.

DIF: Easy REF: 4.2 OBJ: 4.2.a Summarize the role that noncovalent interactions and conformation play in allowing proteins to recognize and bind to their ligands. | 4.2.d Describe how enzymes can reduce the activation energy needed

to catalyze chemical reactions. MSC: Understanding 14. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Feedback inhibition is defined as a mechanism of down-regulating enzyme activity by the accumulation of a product earlier in the pathway.

B. If an enzyme’s allosteric binding site is occupied, the enzyme may adopt an alternative conformation that is not optimal for catalysis.

C. Protein phosphorylation is another way to alter the conformation of an enzyme and serves exclusively as a mechanism to increase enzyme activity.

D. GTP-binding proteins typically have GTPase activity, and the hydrolysis of GTP transforms them to the “off” conformation.

ANS: A. False. Feedback inhibition occurs when an enzyme acting early in a metabolic pathway is inhibited by the accumulation of a product late in the pathway. The inhibitory product binds to a site on the enzyme that lowers its cat alytic activity.

B. True C. False. Although phosphorylation of a protein can change its conformation, this modification may be either as a positive or a negative regulator of enzyme activity, depending on the protein in question.

D. True DIF: Difficult REF: 4.3 OBJ: 4.3.a Explain how and why different forms of feedback control might be used to regulate enzyme activity. | 4.3.c Explain how chemical modification such as phosphorylation can influence a protein’s location and interactions. | 4.3.d Contrast how protein activity is regulated by phosphorylation or by the binding of nucleotides such as GTP or ATP. MSC: Evaluating 15. Chromatography is frequently used to purify proteins from cellular extracts. There are various strategies that can be used, depending on the tools and reagents available. You are interested in isolating additional proteins that interact with your protein target, but your lab mates have used all the purified protein stocks.

1. Why would you need purified target protein to do this experiment?

2. What other strategies/tools could you use to carry out the affinity chromatography?

3. What are the limitations to the method you described in part B that would not be a concern if you could use the purified protein directly?

ANS: 1. The target protein can be covalently linked to the resin used to make the affinity column. When cell extracts are applied to the column, any proteins that associate with high affinity to your target will be bound until they are eluted in the presence of high-salt buffer.

2. You could instead use antibodies specific for your target protein. This is very similar to the first procedure, but instead of your protein being directly linked to the column resin, the antibodies are bound instead. The antibody will bind to your target, which should be bound to any associated proteins in the cell extracts.

3. One important limitation to recognize is that the antibodies could block the binding surface typically recognized by the other proteins that would normally bind to your target, reducing the number of binding partners isolated by the affinity chromatography. You would also need to have an antibody that recognizes your target protein that could be attached to the column in the first place.

DIF: Difficult REF: 4.4 OBJ: 4.4.a Contrast chromatography and electrophoresis as methods for protein separation.

MSC: Evaluating 16. You wish to produce a human enzyme, protein A, by introducing its gene into bacteria. The genetically engineered bacteria make large amounts of protein A, but it is in the form of an insoluble aggregate with no enzymatic activity.

Which of the following procedures might help you to obtain soluble, enzymatically active protein? Select all options that may be useful. Explain your reasoning.

A. Make the bacteria synthesize protein A in smaller amounts.

B. Dissolve the protein aggregate in urea, then dilute the solution and gradually remove the urea.

C. Treat the insoluble aggregate with a protease.

D. Make the bacteria overproduce chaperone proteins in addition to protein A. E. Heat the protein aggregate to denature all proteins, then cool the mixture.

ANS: Choices A, B, and D are all worth trying. Some proteins require molecular chaperones if they are to fold properly within the environment of the cell. In the absence of chaperones, a partly folded polypeptide chain has exposed amino acids that can form noncovalent bonds with other regions of the protein itself and with other proteins, thus causing nonspecific aggregation of proteins.

A. Because the protein you are expressing in bacteria is being made in large quantities, it is possible that there are not enough chaperone molecules in the bacterium to fold the protein. Expressing the protein at lower levels might increase the amount of properly folded protein.

B. Urea should solubilize the protein and completely unfold it. Removing the urea slowly and gradually often allows the protein to refold. Presumably, under less crowded conditions, the protein should be able to refold into its proper conformation.

C. Treating the aggregate with a protease, which cleaves peptide bonds, will probably solubilize the protein by trimming it into pieces that do not interact as strongly with one another; however, chopping up the protein will also destroy its enzymatic activity.

D. Overexpressing chaperone proteins might increase the amount of properly folded protein.

E. Heating can lead to the partial denaturation and aggregation of proteins to form a solid gelatinous mass, as when cooking an egg white, and rarely helps solubilize proteins.

DIF: Difficult REF: 4.1 OBJ: 4.1.c Explain how chaperone proteins guide the folding of a polypeptide chain—and why some proteins can fold without chaperone assistance. | 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions.

MSC: Evaluating 17. Knowing that there are 20 different possible amino acids that can be used at each position in a polypeptide, calculate the number of different polypeptides that could theoretically be produced for a protein that is 180 amino acids in length. Do you expect to find all of these possible protein sequences produced in living systems? Explain your answer.

ANS: There are 20180 possible sequences for a 180 amino acid polypeptide (20 different possible amino acids for each position). No, we would not expect all of the theoretically possible proteins to be made. A much smaller subset can be expected in living systems because it is not likely that all sequences would lead to a stably folded protein. Natural selection favors the retention of genes that encode proteins with stable conformations.

DIF: Moderate REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydro-

gen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. | 4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures. MSC: Evaluating 18. In an attempt to define the protein domains of protein X, you treat it with a protease and use polyacrylamide gel electrophoresis to analyze the peptides produced. In the past, you have used chymotrypsin to perform this experiment, but the stock of this enzyme has been used up. You find a stock of elastase and decide to use it instead of waiting for a new stock of chymotrypsin to arrive.

A. Give two reasons why elastase is a good substitute for chymotrypsin in this assay.

B. Why might proteolysis of the same substrate by chymotrypsin or elastase yield different results?

ANS: A. You might assume that chymotrypsin and elastase would yield the same results because (1) they are both serine proteases and (2) they have a high degree of structural similarity.

B. The slight structural differences of the substrates cause the enzymatic activities of the proteases to differ. As a result, they have different substrate affinities and cleave the bond between a different set of amino acids.

DIF: Moderate REF: 4.4 OBJ: 4.4.a Contrast chromatography and electrophoresis as methods for protein separation.

MSC: Evaluating 19. Protein families arise when a protein sequence that generates a stable fold diverges over many generations and acquires new functions. One example of this can be seen in the globin family. Myoglobin is a stable monomeric protein that can help carry oxygen using a heme molecule. Hemoglobin is stable as a tetramer. It also carries oxygen through the use of heme groups, but it is useful over a much more dynamic range of oxygen than myoglobin. The “globin fold” is structurally conserved across these proteins, but the ability to tetramerize arose through genetic drift and natural selection. Provide an explanation for how the globin sequence can change and still produce the same overall fold.

Support your explanation by suggesting the location and type of sequence alterations that might have little effect on the overall protein fold, but may favor the formation of a multi-subunit protein.

ANS: Amino acids that are found on the surface of a folded monomeric protein are the best candidates for mutations.

Because they are on the surface of the protein, the side-chain interactions are not important for forming the structural core. If alternative amino acids are polar, they can interact equally well with the aqueous environment. It is likely that the sequence comparisons between myoglobin and the α/β globins will reflect changes of surface residues. Furthermore, we may predict that if the surface amino acids changed from polar amino acids to nonpolar amino acids, this would promote multimerization. Nonpolar residues would interact with each other rather than the polar molecules in the cytosol.

DIF: Difficult REF: 4.4 OBJ: 4.4.d Describe how the existence of protein families affects the determination of protein structure. MSC: Evaluating 20. You have produced a monoclonal antibody that binds to the protein actin. To be sure that the antibody does not crossreact with other proteins, you test your antibody in a western blot assay on whole-cell lysates that have been subjected to electrophoresis under nondenaturing conditions (shown in Figure 4-69A) and denaturing conditions (shown in Figure 4-69B). Does the antibody cross-react with other proteins? If so, does this explain the results in the two western blots? If not, how do you explain the difference observed?

Figure 4-69

ANS: No, the antibody does not seem to cross-react with other proteins. In each western blot, there is only one band, indicating that only one protein is bound by the monoclonal antibody. Actin is a protein that forms long filaments. Under the nondenaturing conditions of the first gel (A), the filaments remain intact and, as a multi-protein complex, actin migrates very slowly through the polyacrylamide matrix. In the case where sodium dodecyl sulfate (SDS) is added (denaturing conditions), actin filaments dissociate into monomers. Thus, the band is lower in panel (B) because the monomers have a lower molecular weight and migrate faster through the gel.

DIF: Difficult REF: 4.4 OBJ: 4.4.a Contrast chromatography and electrophoresis as methods for protein separation.

MSC: Evaluating 21. Protein Y is a globular protein that normally assembles as a tetramer. You are examining the interactions between the subunits by changing the amino acids on the surface of the protein. You compare the wild-type (nonmutated) protein and a mutant version with a single amino acid substitution. When washed through the same gel-filtration column, mutant protein Y runs through the column more slowly than the normal protein. Which of the following changes in the mutant protein is most likely to explain this result? Explain your choice.

A. the loss of a binding site on the mutant-protein surface through which protein Y normally forms dimers B. a change that results in the mutant protein acquiring an overall positive instead of a negative charge C. a change that results in the mutant protein being larger than the wild-type protein D. a change that results in the mutant protein having a slightly different shape from the wild-type protein ANS: (A) Dimers formed by a normal protein will run through the gel-filtration column faster than a mutant protein Y monomer. Choice (B) is unlikely, because gel-filtration columns separate proteins on the basis of size, not charge or affinity for small molecules. Choice (C) is unlikely, because if the mutant protein were larger than normal it would be less able to enter the porous beads and would run through the column faster than the normal protein. Choice (D) is unlikely, because a small change in shape without a change in size would be unlikely to have a major effect on the behavior of a protein in a gel-filtration column.

DIF: Difficult REF: 4.4 OBJ: 4.4.a Contrast chromatography and electrophoresis as methods for protein separation.

MSC: Analyzing 22. Enzymes generally make good drug targets because a specific reaction of interest can be targeted with a high degree of selectivity. Consider the following three drugs and explain why, although reaction-specific, the first two produce side effects, while the third does not.

A. Statins inhibit HMG CoA reductase to block intracellular cholesterol synthesis.

B. Methotrexate inhibits dihydrofolate reductase, which subsequently leads to blocked DNA replication.

C. Gleevec® inhibits BCR, a kinase that is only produced in certain types of leukemia cells.

ANS: The first two inhibitors (statins and methotrexate) are blocking reactions that are very important to all cells, not just the cells affected by the illness in question. Thus, drug side effects can be Difficult to predict, and may be very specific to the patient being treated. In the case of chronic myeloid leukemia (CML), the mutant enzyme is specific to the leukemia cells, and no other cells in the body are affected by this drug. This also means that the usefulness of this drug is limited to those CML patients that have this specific mutation.

DIF: Moderate REF: 4.2 OBJ: 4.2.a Summarize the role that noncovalent interactions and conformation play in allowing proteins to recognize and bind to their ligands. | 4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions. MSC: Evaluating 23. One way in which an enzyme can lower the activation energy required for a reaction is to bind the substrate(s) and distort its structure so that the substrate more closely resembles the transition state of the reaction. This mechanism will be facilitated if the shape and chemical properties of the enzyme’s active site are more complementary to the transition state than to the undistorted substrate; in other words, if the enzyme were to have a higher affinity for the transition state than for the substrate. Knowing this, your friend looked in an organic chemistry textbook to identify a stable chemical that closely resembles the transition state of a reaction that converts X into Y. She generated an antibody against this transition-state analog and mixed the antibody with chemical X. What do you think might happen?

ANS: If your friend was lucky, she made a “catalytic antibody” that catalyzed the conversion of X into Y. Such cat alytic antibodies have been isolated and shown to catalyze a variety of reactions, but with lower efficiency than genuine enzymes.

DIF: Difficult REF: 4.2 OBJ: 4.2.d Describe how enzymes can reduce the activation energy needed to catalyze chemical reactions. MSC: Evaluating 24. Some of the enzymes that oxidize sugars to yield usable cellular energy (for example, ATP) are regulated by phosphorylation. For these enzymes, would you expect the inactive form to be the phosphorylated form or the dephosphorylated form? Explain your answer.

ANS: In general, the inactive form is the phosphorylated form. The main purpose of glycolysis and the citric acid cycle is to generate ATP; thus, the enzymes are inactive when the concentration of ATP is high and active when it is low. It makes sense that cells would not want to have to phosphorylate their enzymes to turn them on when ATP levels are already low, because phosphorylation requires ATP. DIF: Moderate REF: 4.3 OBJ: 4.3.c Explain how chemical modification such as phosphorylation can influence a protein’s location and interactions. MSC: Applying 25. Using the example of the p53 protein, postulate how different combinations of covalent modifications can lead to a wide variety of protein functions.

ANS: In a protein with a complex regulatory protein code, such as p53, the covalent attachment or removal of modifying groups can change the protein’s behavior, its activity or stability, its binding partners, or its location within a cell. In the case of p53, there are at least 20 different locations (amino acids) that can be modified through such processes as phosphorylation, ubiquitylation, and acetylation.

DIF: Easy REF: 4.3 OBJ: 4.3.c Explain how chemical modification such as phosphorylation can influence a protein’s location and interactions. MSC: Applying 26. Fill in the blank spaces in the table below. The first row has been completed for you.

Table 4-75A

ANS:

Table 4-75B

DIF: Easy

REF: 4.1 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds,

van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide

backbone and amino acid side chains participate in these interactions. MSC: Remembering 27. The protein structure in Figure 4-76 contains four α helices arranged in a bundle. Label each helix by number (1 to 4) starting from the N-terminus and going to the C-terminus, both of which are labeled. List the six possible pairings of these helices, and indicate within each pair whether the helices are parallel or antiparallel.

Figure 4-76

ANS:

DIF: Moderate REF: 4.1 OBJ: 4.1.f Describe the role that protein domains play within a protein’s three-dimensional structure. MSC: Applying 28. For each polypeptide sequence listed, choose from the options given below to indicate which secondary structure the sequence is most likely to form upon folding. The nonpolar amino acids are italicized.

A. Leu-Gly-Val-Leu-Ser-Leu-Phe-Ser-Gly-Leu-Met-Trp-Phe-Phe-Trp-Ile B. Leu-Leu-Gln-Ser-Ile-Ala-Ser-Val-Leu-Gln-Ser-Leu-Leu-Cys-Ala-Ile C. Thr-Leu-Asn-Ile-Ser-Phe-Gln-Met-Glu-Leu-Asp-Val-Ser-Ile-Arg-Trp amphipathic α helix

hydrophilic β sheet

amphipathic β sheet

hydrophobic α helix

hydrophilic α helix ANS: A. Hydrophobic α helix. Nearly all of the amino acid side chains in this sequence are nonpolar or hydrophobic, which favors the only hydrophobic option given in the list.

B. Amphipathic α helix. In an ideal α helix, there are 3.6 residues per complete turn. Thus, an amphipathic helix with one hydrophobic side and one hydrophilic side will have, minimally, nonpolar side chains (N) repeating every third

then next fourth amino acid: NxxNxxxNxxNxxxN. Polar side chains (P) will have the same pattern but shifted relative to the nonpolar side chains; for example, xxPxxxPxxPxxxPxxP. C. Amphipathic β sheet. Because of the zigzag-like structure of a β sheet, a sequence with alternating nonpolar and polar side chains may form an amphipathic β sheet that is hydrophobic on one side and hydrophilic on the other.

DIF: Difficult REF: 4.1 OBJ: 4.1.d Contrast the hydrogen bonding patterns that give rise to alpha helices with those that produce beta sheets. | 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. | 4.1.g Define a protein’s primary, secondary, tertiary, and quaternary structures. MSC: Evaluating 29. Figure 4-78 shows a fatty-acid-binding protein from two different angles. Apart from its two short α helices, its structural elements are extended strands that form a curved β sheet, which is called a β barrel.

A. Draw arrows on the six top strands in panel (A) (those running horizontally) to determine whether the β barrel is made up of parallel or antiparallel strands.

B. Look at panel (B) and predict the relative distribution of polar and nonpolar side chains (inside the barrel or outside the barrel) and explain your answer.

Figure 4-78

ANS: A. The strands form an antiparallel β sheet.

Figure 4-78A

B. The prediction is that nonpolar side chains are on the inside of the barrel, and polar side chains are distributed to the outside of the barrel. The outside of the barrel is completely exposed to the aqueous solvent and is stabilized by solvent–protein hydrogen bonds. And, although the barrel seems open and accessible to solvent molecules, this protein binds fatty acids and would most probably do that by enclosing them inside the barrel, away from the

aqueous solvent and close to nonpolar amino acid side chains lining the inside of the barrel.

DIF: Difficult REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. MSC: Applying 30. Drawn below are segments of β sheets, which are rigid pleated structures held together by hydrogen bonds between the peptide backbones of adjacent strands (Figure 4-79). The amino acid side chains attached to the α-carbons are omitted for clarity.

Figure 4-79

A. For panel (A) and for panel (B), indicate whether the structure is parallel or antiparallel.

B. Draw the hydrogen bonds as dashed lines (||||||).

ANS: A. (A) is parallel and (B) is antiparallel.

B. See Figure 4-79A.

Figure 4-79A

DIF: Easy REF: 4.1 OBJ: 4.1.a Explain how noncovalent interactions—including electrostatic attractions, hydrogen bonds, van der Waals attractions, and hydrophobic forces—influence the shape of a folded protein and how the polypeptide backbone and amino acid side chains participate in these interactions. | 4.1.d Contrast the hydrogen bonding patterns that give rise to alpha helices with those that produce beta sheets. MSC: Understanding 31. Fill in the blanks with the labels in the list below to identify various parts of the antibody structure in Figure 4-80.

A. constant domain of the light chain B. constant domain of the heavy chain C. antigen-binding site D. variable domain of the heavy chain E. variable domain of the light chain

Figure 4-80

ANS:

Figure 4-80A

DIF: Easy REF: 4.2 OBJ: 4.2.b Explain how antibodies, which share the same basic structure, can recognize an infinitely diverse array of antigens. MSC: Remembering

CHAPTER 5

DNA and Chromosomes THE STRUCTURE OF DNA

5.1.a Contrast the functions of the DNA and protein components of chromosomes.

5.1.b Explain why biologists initially thought that proteins were the most likely carriers of genetic information.

5.1.c Describe how, experimentally, researchers demonstrated that DNA carries genetic information.

5.1.d Distinguish between the bonds that link together the subunits in a single strand of DNA and those that hold together the two strands in a DNA double helix, and summarize how these bonds affect the behavior of the DNA molecule.

5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. 5.1.f Describe the chemical differences that dictate the polarity of a DNA strand.

5.1.g Explain how the structure of DNA carries information for producing proteins.

5.1.h Explain how the structure of DNA suggests a mechanism by which genetic information can be copied.

THE STRUCTURE OF EUKARYOTIC CHROMOSOMES 5.2.a Contrast prokaryotic and eukaryotic chromosomes in terms of structure and specialized sequence elements.

5.2.b Describe how human chromosomes can be distinguished from one another and how such information can be of value.

5.2.c Recall how many molecules of DNA are in each eukaryotic chromosome.

5.2.d Describe a full complement of human chromosomes in a diploid somatic cell, including sex chromosomes.

5.2.e Define the terms “gene” and “genome.”

5.2.f Describe the relationship among gene number, genome size, and organismal complexity.

5.2.g Explain why much “junk DNA” is thought to serve a biological function.

5.2.h Compare the roles played by centromeres, telomeres, and replication origins.

5.2.i Explain the organization and attachments that keep interphase chromosomes from becoming extensively entangled.

5.2.j Describe the structure and function of the nucleolus.

5.2.k Contrast the extents of compression in interphase and mitotic chromosomes.

5.2.l Compare the roles played by non-histone proteins and histone proteins (including histone H1) in the packaging of chromatin.

5.2.m Distinguish between a nucleosome and a nucleosome core particle.

5.2.n Explain how histone proteins are able to bind tightly to DNA.

THE REGULATION OF CHROMOSOME STRUCTURE 5.3.a Explain how chromatin-remodeling complexes and histone-modifying enzymes regulate the accessibility of DNA. 5.3.b Explain why a cell might decondense a particular segment of DNA. 5.3.c Contrast euchromatin and heterochromatin in terms of structure, gene activity, and location along an interphase chromosome.

5.3.d Explain how heterochromatin is established and spreads.

5.3.e Explain how heterochromatin participates in gene silencing and provide an example.

MULTIPLE CHOICE 1. Mitotic chromosomes were first visualized with the use of very simple tools: a basic light microscope and some dyes.

Which of the following characteristics of mitotic chromosomes reflects how they were named? a. motion b. color c. shape d. location ANS: B DIF: Easy REF: 5.1 OBJ: 5.1.c Describe how, experimentally, researchers demonstrated that DNA carries genetic information. MSC: Remembering 2. In a DNA double helix, a. the two DNA strands are identical. b. purines pair with purines. c. thymine pairs with cytosine. d. the two DNA strands run antiparallel.

ANS: D DIF: Easy REF: 5.1 OBJ: 5.1.f Describe the chemical differences that dictate the polarity of a DNA strand.

MSC: Remembering 3. Which of the following chemical groups is NOT used to construct a DNA molecule? a. five-carbon sugar b. phosphate c. nitrogen-containing base d. six-carbon sugar ANS: D DIF: Easy REF: 5.1 OBJ: 5.1.f Describe the chemical differences that dictate the polarity of a DNA strand.

MSC: Understanding 4. Which of the following structural characteristics is NOT normally observed in a DNA duplex? a. purine–pyrimidine pairs b. external sugar–phosphate backbone c. uniform left-handed twist d. antiparallel strands

ANS: C DIF: Easy REF: 5.1 OBJ: 5.1.d Distinguish between the bonds that link together the subunits in a single strand of DNA and those that hold together the two strands in a DNA double helix, and summarize how these bonds affect the behavior of the DNA molecule. | 5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. | 5.1.f Describe the chemical differences that dictate the polarity of a DNA strand. MSC: Understanding 5. Which of the following DNA strands can form a DNA duplex by pairing with itself at each position? a. 5′-AAGCCGAA-3′ b. 5′-AAGCCGTT-3′ c. 5′-AAGCGCAA-3′ d. 5′-AAGCGCTT-3′ ANS: D DIF: Moderate REF: 5.1 OBJ: 5.1.d Distinguish between the bonds that link together the subunits in a single strand of DNA and those that hold together the two strands in a DNA double helix, and summarize how these bonds affect the behavior of the DNA molecule. MSC: Applying 6. The DNA from two different species can often be distinguished by a difference in the a. ratio of A + T to G + C. b. ratio of A + G to C + T. c. ratio of sugar to phosphate. d. presence of bases other than A, G, C, and T. ANS: A DIF: Easy REF: 5.1 OBJ: 5.1.g Explain how the structure of DNA carries information for producing proteins.

MSC: Applying 7. Which DNA base pair is represented in Figure 5-7? a. A-T b. T-A c. G-C d. C-G

Figure 5-7

ANS: C DIF: Easy REF: 5.1 OBJ: 5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. MSC: Understanding 8. The complete set of information in an organism’s DNA is called its a. genetic code. b. coding sequence.

c. gene. d. genome.

ANS: D DIF: Easy REF: 5.2 OBJ: 5.2.e Define the terms “gene” and “genome.” MSC: Remembering 9. The relationship between the nucleic acid sequence over a stretch of DNA and the order of amino acids in the resulting protein is referred to as the __________ code. a. protein b. genetic c. translational d. expression ANS: B DIF: Easy REF: 5.1 OBJ: 5.1.g Explain how the structure of DNA carries information for producing proteins.

MSC: Understanding 10. DNA is an information storage molecule, whose sequences serve as a template to make a. lipids. b. RNA. c. polypeptides. d. carbohydrates.

ANS: B DIF: Easy REF: 5.1 OBJ: 5.1.g Explain how the structure of DNA carries information for producing proteins.

MSC: Remembering 11. The human genome is divided into linear segments and packaged into structures called chromosomes. What is the total number of chromosomes found in each of the somatic cells in your body? a. 22 b. 23 c. 44 d. 46 ANS: D DIF: Easy REF: 5.2 OBJ: 5.2.d Describe a full complement of human chromosomes in a diploid somatic cell, including sex chromosomes. MSC: Remembering 12. The human genome is a diploid genome. However, when germ-line cells produce gametes, these specialized cells are haploid. What is the total number of chromosomes found in each of the gametes (egg or sperm) in your body? a. 22 b. 23 c. 44 d. 46 ANS: B DIF: Easy REF: 5.2 OBJ: 5.2.d Describe a full complement of human chromosomes in a diploid somatic cell, including sex chromosomes. MSC: Remembering 13. What type of macromolecule helps package DNA in eukaryotic chromosomes? a. lipids b. carbohydrates c. proteins

d. RNA ANS: C DIF: Easy REF: 5.2 OBJ: 5.2.n Explain how histone proteins are able to bind tightly to DNA. MSC: Understanding 14. The process of sorting human chromosome pairs by size and morphology is called karyotyping. A modern method employed for karyotyping is called chromosome painting. How are individual chromosomes “painted”? a. with a laser b. using fluorescent antibodies c. using fluorescent DNA molecules d. using green fluorescent protein ANS: C DIF: Easy REF: 5.2 OBJ: 5.2.b Describe how human chromosomes can be distinguished from one another and how such information can be of value. MSC: Remembering 15. Which of the following questions would NOT be answered by using karyotyping? a. Is the individual genetically female or male? b. Do any of the chromosomes contain pieces that belong to other chromosomes? c. Does the individual have an extra chromosome? d. Do any chromosomes contain point mutations?

ANS: D DIF: Moderate REF: 5.2 OBJ: 5.2.b Describe how human chromosomes can be distinguished from one another and how such information can be of value. MSC: Applying 16. The chromosomes we typically see in images are isolated from mitotic cells. These mitotic chromosomes are in the most highly condensed form. Interphase cells contain chromosomes that are less densely packed and a. occupy discrete territories in the nucleus. b. share the same nuclear territory as their homolog. c. are restricted to the nucleolus. d. are completely tangled with other chromosomes.

ANS: A DIF: Easy REF: 5.2 OBJ: 5.2.k Contrast the extent of compression of interphase and mitotic chromosomes.

MSC: Understanding 17. Specific regions of eukaryotic chromosomes contain sequence elements that are absolutely required for the proper transmission of genetic information from a mother cell to each daughter cell. Which of the following is NOT known to be one of these required elements in eukaryotes? a. protein-coding regions b. origins of replication c. telomeres d. centromeres ANS: A DIF: Easy REF: 5.2 OBJ: 5.2.h Compare the roles played by centromeres, telomeres, and replication origins.

MSC: Understanding 18. Mitotic chromosomes are __________ times more compact than a DNA molecule in its extended form. a. 10,000 b. 100,000

c. 1000 d. 100 ANS: A DIF: Easy REF: Remembering OBJ: 5.2.k Contrast the extent of compression of interphase and mitotic chromosomes. MSC: Remembering 19. Interphase chromosomes are about __________ times less compact than mitotic chromosomes, but still are about __________ times more compact than a DNA molecule in its extended form. a. 10; 1000 b. 20; 500 c. 5; 2000 d. 50; 200 ANS: B DIF: Easy REF: 5.2 OBJ: 5.2.k Contrast the extent of compression of interphase and mitotic chromosomes.

MSC: Remembering 20. The classic “beads-on-a-string” structure is the most decondensed chromatin structure possible and is produced experimentally. Which chromatin components are NOT retained when this structure is generated? a. linker histones b. linker DNA c. nucleosome core particles d. core histones ANS: A DIF: Easy REF: 5.2 OBJ: 5.2.m Distinguish between a nucleosome and a nucleosome core particle. | 5.2.n Explain how histone proteins are able to bind tightly to DNA. MSC: Understanding 21. Nucleosomes are formed when DNA wraps __________ times around the histone octamer in a __________ coil. a. 2.0; right-handed b. 2.5; left-handed c. 1.7; left-handed d. 1.3; right-handed ANS: C DIF: Easy REF: 5.2 OBJ: 5.2.m Distinguish between a nucleosome and a nucleosome core particle. | 5.2.n Explain how histone proteins are able to bind tightly to DNA. MSC: Remembering 22. The octameric histone core is composed of four different histone proteins, assembled in a stepwise manner. Once the core octamer has been formed, DNA wraps around it to form a nucleosome core particle. Which of the following histone proteins does NOT form part of the octameric core? a. H4 b. H2A c. H3 d. H1 ANS: D DIF: Easy REF: 5.2 OBJ: 5.2.m Distinguish between a nucleosome and a nucleosome core particle.

MSC: Remembering 23. The core histones are small, basic proteins that have a globular domain at the C-terminus and a long, extended conformation at the N-terminus. Which of the following is NOT true of the N-terminal “tail” of these histones?

a. It is subject to covalent modifications. b. It extends out of the nucleosome core. c. It binds to DNA in a sequence-specific manner. d. It helps DNA pack tightly.

ANS: C DIF: Moderate REF: 5.2 OBJ: 5.2.n Explain how histone proteins are able to bind tightly to DNA. MSC: Understanding 24. Stepwise condensation of linear DNA happens in five different packing processes. Which of the following four processes has a direct requirement for histone H1? a. formation of “beads-on-a-string” b. formation of the 30-nm fiber c. looping of the 30-nm fiber d. packing of loops to form interphase chromosomes ANS: B DIF: Easy REF: 5.2 OBJ: 5.2.n Explain how histone proteins are able to bind tightly to DNA. MSC: Understanding 25. Although the chromatin structure of interphase and mitotic chromosomes is very compact, DNA -binding proteins and protein complexes must be able to gain access to the DNA molecule. Chromatin-remodeling complexes provide this access by a. recruiting other enzymes. b. modifying the N-terminal tails of core histones. c. using the energy of ATP hydrolysis to move nucleosomes. d. denaturing the DNA by interfering with hydrogen-bonding between base pairs.

ANS: C DIF: Easy REF: 5.3 OBJ: 5.3.a Explain how chromatin-remodeling complexes and histone-modifying enzymes regulate the accessibility of DNA. MSC: Remembering 26. The N-terminal tail of histone H3 can be extensively modified, and depending on the number, location, and combination of these modifications, these changes may promote the formation of heterochromatin. What is the result of heterochromatin formation? a. increase in gene expression b. gene silencing c. recruitment of remodeling complexes d. displacement of histone H1 ANS: B DIF: Easy REF: 5.3 OBJ: 5.3.d Explain how heterochromatin is established and spreads. | 5.3.e Explain how heterochromatin participates in gene silencing and provide an example. MSC: Underst anding 27. Methylation and acetylation are common changes made to histone H3, and the specific combination of these changes is sometimes referred to as the “histone code.” Which of the following patterns will probably lead to gene silencing? a. lysine 9 methylation b. lysine 4 methylation and lysine 9 acetylation c. lysine 14 acetylation d. lysine 9 acetylation and lysine 14 acetylation

ANS: A DIF: Moderate REF: 5.3 OBJ: 5.3.e Explain how heterochromatin participates in gene silencing and provide an example. MSC: Applying 28. Which of the following best describes the mechanism by which chromatin-remodeling complexes “loosen” the DNA wrapped around the core histones? a. They use energy derived from ATP hydrolysis to change the relative position of the DNA and the core histone octamer. b. They chemically modify the DNA, changing the affinity between the histone octamer and the DNA. c. They remove histone H1 from the linker DNA adjacent to the core histone octamer. d. They chemically modify core histones to alter the affinity between the histone octamer and the DNA. ANS: A DIF: Moderate REF: 5.3 OBJ: 5.3.a Explain how chromatin-remodeling complexes and histone-modifying enzymes regulate the accessibility of DNA. MSC: Analyzing 29. Which of the following is NOT a chemical group commonly found on core histone N-terminal tails for chromatin regulation? a. methyl b. sulfhydryl c. phosphoryl d. acetyl ANS: B DIF: Easy REF: 5.3 OBJ: 5.3.a Explain how chromatin-remodeling complexes and histone-modifying enzymes regulate the accessibility of DNA. MSC: Remembering 30. How do changes in histone modifications lead to changes in chromatin structure? a. They directly lead to changes in the positions of the core histones. b. They change the affinity between the histone octamer and the DNA. c. They help recruit other proteins to the chromatin. d. They cause the histone N-terminal tails to become hyperextended.

ANS: C DIF: Easy REF: 5.3 OBJ: 5.3.a Explain how chromatin-remodeling complexes and histone-modifying enzymes regulate the accessibility of DNA. MSC: Understanding 31. Most eukaryotic cells only express 20–30% of the genes they possess. The formation of heterochromatin maintains the other genes in a transcriptionally silent (unexpressed) state. Which histone modification is associated with the formation of the most common type of heterochromatin? a. H3 lysine 4 methylation b. H3 lysine 9 methylation c. H3 lysine 14 methylation d. H3 lysine 27 methylation ANS: B DIF: Easy REF: 5.3 OBJ: 5.3.e Explain how heterochromatin participates in gene silencing and provide an example. MSC: Remembering 32. The inactivation of one X chromosome is established by the directed spreading of heterochromatin. The silent state of this chromosome is __________ in the subsequent cell divisions. a. completed

b. switched c. erased d. maintained ANS: D DIF: Easy REF: 5.3 OBJ: 5.3.d Explain how heterochromatin is established and spreads. | 5.3.e Explain how heterochromatin participates in gene silencing and provide an example. MSC: Remembering 33. Many of the breakthroughs in modern biology came after Watson and Crick published their model of DNA in 1953.

However, chromosomes were identified earlier. In what decade did scientists first identify chromosomes? a. 1880s b. 1920s c. 1940s d. 1780s ANS: A DIF: Easy REF: 5.1 OBJ: 5.1.c Describe how, experimentally, researchers demonstrated that DNA carries genetic information. MSC: Remembering 34. In the 1940s, proteins were thought to be the more likely molecules to house genetic information. What was the primary reason that DNA was not originally believed to be the genetic material? a. DNA has a high density of negative charges. b. Nucleotides were known to be a source of chemical energy for the cell. c. Both protein and nucleic acids were found to be components of chromosomes. d. DNA was found to contain only four different chemical building blocks.

ANS: D DIF: Easy REF: 5.1 OBJ: 5.1.b Explain why biologists initially thought that proteins were the most likely carriers of genetic information. MSC: Remembering 35. You are a virologist interested in studying the evolution of viral genomes. You are studying two newly isolated viral strains and have sequenced their genomes. You find that the genome of strain 1 contains 25% A, 55% G, 20% C, and 10% T. You report that you have isolated a virus with a single-stranded DNA genome. Based on what evidence can you make this conclusion? a. because single-stranded genomes always have a large percentage of purines b. by using the formula G − A = C + T c. because double-stranded genomes have equal amounts of A and T d. because single-stranded genomes have a higher rate of mutation ANS: C DIF: Moderate REF: 5.1 OBJ: 5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. MSC: Applying 36. Several experiments were required to demonstrate how traits are inherited. Which scientist or team of scientists first demonstrated that cells contain some component that can be transferred to a new population of cells and permanently cause changes in the new cells? a. Griffith b. Watson and Crick c. Avery, MacLeod, and McCarty d. Hershey and Chase ANS: A DIF: Easy REF: 5.1 OBJ: 5.1.c Describe how, experimentally, researchers demonstrated that DNA carries ge-

netic information. MSC: Remembering 37. Several experiments were required to demonstrate how traits are inherited. Which scientist or team of scientists obtained definitive results demonstrating that DNA is the genetic molecule? a. Griffith b. Watson c. Crick d. Hershey and Chase ANS: D DIF: Easy REF: 5.1 OBJ: 5.1.c Describe how, experimentally, researchers demonstrated that DNA carries genetic information. MSC: Remembering 38. Fred Griffith studied two strains of Streptococcus pneumonia, one that causes a lethal infection when injected into mice, and a second that is harmless. He observed that pathogenic bacteria that have been killed by heating can no longer cause an infection. But when these heat-killed bacteria are mixed with live, harmless bacteria, this mixture is capable of infecting and killing a mouse. What did Griffith conclude from this experiment? a. The infectious strain cannot killed by heating. b. The heat-killed pathogenic bacteria “transformed” the harmless strain into a lethal one. c. The harmless strain somehow revived the heat-killed pathogenic bacteria. d. The mice had lost their immunity to infection with S. pneumoniae.

ANS: B DIF: Easy REF: 5.1 OBJ: 5.1.c Describe how, experimentally, researchers demonstrated that DNA carries genetic information. MSC: Understanding 39. Hershey and Chase used radiolabeled macromolecules to identify the material that contains heritable information.

What radioactive material was used to track DNA during this experiment? a. 3H b. 14C c. 35S d. 32P ANS: D DIF: Easy REF: 5.1 OBJ: 5.1.c Describe how, experimentally, researchers demonstrated that DNA carries genetic information. MSC: Remembering

SHORT ANSWER 1. Using terms from the list below, fill in the blanks in the following brief description of the experiment with Streptococcus pneumoniae that identified which biological molecule carries heritable genetic information. Some terms may be used more than once. carbohydrate

lipid

R-strain

DNA

nonpathogenic

RNA

identify

pathogenic

S-strain

label

purify

transform

Cell-free extracts from S-strain cells of S. pneumoniae were fractionated to __________ DNA, RNA, protein, and other cell components. Each fraction was then mixed with __________ cells of S. pneumoniae. Its ability to change

these into cells with __________ properties resembling the __________ cells was tested by injecting the mixture into mice. Only the fraction containing __________ was able to __________ the __________ cells to __________ (or __________) cells that could kill mice.

ANS: Cell-free extracts from S-strain cells of S. pneumoniae were fractionated to purify DNA, RNA, protein, and other cell components. Each fraction was then mixed with R-strain cells of S. pneumoniae. Its ability to change these into cells with pathogenic properties resembling the S-strain cells was tested by injecting the mixture into mice. Only the fraction containing DNA was able to transform the R-strain cells to pathogenic (or S-strain) cells that could kill mice.

DIF: Easy REF: 5.1 OBJ: 5.1.c Describe how, experimentally, researchers demonstrated that DNA carries genetic information. MSC: Understanding 2. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. DNA molecules, like proteins, consist of a single, long polymeric chain that is assembled from small monomeric subunits.

B. The polarity of a DNA strand results from the polarity of the nucleotide subunits.

C. There are five different nucleotides that become incorporated into a DNA strand.

D. Hydrogen bonds between each nucleotide hold individual DNA strands together.

ANS: A. False. DNA is double-stranded. It is actually made of two polymers that are complementary in sequence.

B. True C. False. There are four different nucleotides that are used to make a DNA polymer: adenine, thymine, guanine, and cytosine. A fifth nucleotide, uracil, is found exclusively in RNA molecules, replacing thymine nucleotides in the DNA sequence.

D. False. Nucleotides are linked covalently through phosphodiester bonds. Hydrogen-bonding between nucleotides from opposite strands holds the DNA molecule together.

DIF: Easy REF: 5.1 OBJ: 5.1.d Distinguish between the bonds that link together the subunits in a single strand of DNA and those that hold together the two strands in a DNA double helix, and summarize how these bonds affect the behavior of the DNA molecule. | 5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. | 5.1.f Describe the chemical differences that dictate the polarity of a DNA strand. MSC: Evaluating 3. The structures of the four bases in DNA are given in Figure 5-42.

Figure 5-42

A. Which are purines and which are pyrimidines?

B. Which bases pair with each other in double-stranded DNA? ANS: A. Adenine and guanine are purines; cytosine and thymine are pyrimidines.

B. Cytosine pairs with guanine, and adenine with thymine.

DIF: Easy REF: 5.1 OBJ: 5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. MSC: Remembering 4. Because hydrogen bonds hold the two strands of a DNA molecule together, the strands can be separated without breaking any covalent bonds. Every unique DNA molecule “melts” at a different temperature. In this context, Tm (melting temperature) is the point at which two strands separate, or become denatured. Order the DNA sequences listed below according to relative melting temperatures (from lowest Tm to highest Tm). Assume that they all begin as stable double-stranded DNA molecules. Explain your answer.

A. GGCGCACC B. TATTGTCT C. GACTCCTG D. CTAACTGG ANS: The order in which the DNA molecules would denature as the temperature increases is: 1—B; 2—D; 3—C; 4—A All the DNA molecules are the same length, so only the A + T and G + C content determines their relative Tm. Molecules with higher G + C content will be more stable than molecules with a high A + T content. This is because there are three hydrogen bonds between each G-C base pair but only two between each A-T base pair. More energy (heat) is required to disrupt a larger number of hydrogen bonds.

DIF: Moderate REF: 5.1 OBJ: 5.1.d Distinguish between the bonds that link together the subunits in a single strand of DNA and those that hold together the two strands in a DNA double helix, and summarize how these bonds affect the

behavior of the DNA molecule. | 5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. MSC: Evaluating 5. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Each strand of DNA contains all the information needed to create a new double-stranded DNA molecule with the same sequence information.

B. All functional DNA sequences inside a cell code for protein products.

C. Gene expression is the process of duplicating genes during DNA replication.

D. Gene sequences correspond exactly to the respective protein sequences produced from them.

ANS: A. True B. False. Some sequences encode only RNA molecules, some bind to specific regulatory proteins, and others are sites where specific chromosomal protein structures are built (for example, centromeric and telomeric DNA).

C. False. Gene expression is the process of going from gene sequence to RNA sequence, to protein sequence.

D. False. This statement is false for two reasons. First, genes often contain intron sequences. Second, genes always contain nucleotides flanking the protein-coding sequences that are required for the regulation of transcription and translation.

DIF: Easy REF: 5.2 OBJ: 5.2.g Explain why “junk DNA” is thought to serve a biological function. | 5.2.h Compare the roles played by centromeres, telomeres, and replication origins. MSC: Evaluating 6. Given the sequence of one strand of a DNA helix (below), provide the sequence of the complementary strand and label the 5′ and 3′ ends.

5′-GCATTCGTGGGTAG-3′ ANS: 5′-CTACCCACGAATGC-3′ DIF: Moderate REF: 5.1 OBJ: 5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. | 5.1.f Describe the chemical differences that dictate the polarity of a DNA strand. MSC: Creating 7. In principle, what is the minimum number of nucleotides necessary to code for each amino acid? Keep in m ind there are twenty different naturally-occurring amino acids, and that each amino acid is encoded by the same number of nucleotides.

ANS: Because there are 20 amino acids used in proteins, each amino acid would have to be encoded by a min imum of three nucleotides. For example, a code of two consecutive nucleotides could specify a maximum of 16 (4 2) different amino acids, excluding stop and start signals. A code of three consecutive nucleotides has 64 (4 3) different members, and thus can easily accommodate the 20 amino acids plus a signal to stop protein synthesis.

DIF: Moderate REF: 5.1 OBJ: 5.1.g Explain how the structure of DNA carries information for producing proteins.

MSC: Applying 8. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. bands

extended

kinetochore

chromatin

homologous

nonhomologous

chromosomes

hybridization

condensation

karyotype

In eukaryotic __________, DNA is complexed with proteins to form __________. The paternal and maternal copies of human Chromosome 1 are __________, whereas the paternal copy of Chromosome 1 and the maternal copy of Chromosome 3 are __________. Cytogeneticists can determine large-scale chromosomal abnormalities by looking at a patient’s __________. Fluorescent molecules can be used to paint a chromosome, by a technique that employs DNA __________, and thereby to identify each chromosome by microscopy.

ANS: In eukaryotic chromosomes, DNA is complexed with proteins to form chromatin. The paternal and maternal copies of human Chromosome 1 are homologous, whereas the paternal copy of Chromosome 1 and the maternal copy of Chromosome 3 are nonhomologous. Cytogeneticists can determine large-scale chromosomal abnormalities by looking at a patient’s karyotype. Fluorescent molecules can be used to paint a chromosome, by a technique that employs DNA hybridization, and thereby to identify each chromosome by microscopy.

DIF: Easy REF: 5.2 OBJ: 5.2.b Describe how human chromosomes can be distinguished from one another and how such information can be of value. | 5.2.d Describe a full complement of human chromosomes in a diploid somatic cell, including sex chromosomes. MSC: Understanding 9. A. Define a “gene.”

B. Consider two different species of yeast that have similar genome sizes. Is it likely that they contain the same number of genes? A similar number of chromosomes?

ANS: A. A gene is a segment of DNA that stores the information required to specify the particular sequence found in a protein (or, in some cases, the sequence of a structural or catalytic RNA).

B. A similar genome size indicates relatively little about the number of genes and virtually nothing about the number of chromosomes. For example, the commonly studied yeasts Saccharomyces cerevisiae (Sc) and Schizosaccharomyces pombe (Sp) are separated by roughly 400 million years of evolution, and both have a genome of 14 million base pairs. However, Sc has 6500 genes packaged into 16 chromosomes and Sp has 4800 genes in 3 chromosomes.

DIF: Difficult REF: 5.2 OBJ: 5.2.e Define the terms “gene” and “genome.” | 5.2.f Describe the relationship between gene number, genome size, and organismal complexity. MSC: Applying 10. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. For the majority of genes in the eukaryotic genome, the final product is an RNA molecule.

B. Eukaryotic genes are distributed across multiple DNA molecules, which can be packaged into linear or circular chromosomes.

C. Chromosomes of different length emit different wavelengths of light, making them appear colored under the microscope.

D. In many eukaryotic genomes, genes contain additional, interspersed, noncoding sequences.

ANS: A. False. For the majority of genes in the eukaryotic genome, the final product is a protein molecule.

B. False. Eukaryotic genes are distributed among multiple DNA molecules, and these are linear chromosomes. Prokaryotic cells typically have circular chromosomes.

C. False. Chromosome “painting” can be done with dyes that are linked to DNA sequences that are chromosome-

specific. The color is not due to an intrinsic property of the DNA itself.

D. True DIF: Moderate REF: 5.2 OBJ: 5.2.a Contrast prokaryotic and eukaryotic chromosomes in terms of structure and the specialized sequence elements they contain. | 5.2.e Define the terms “gene” and “genome.” | 5.2.g Explain why “junk DNA” is thought to serve a biological function. MSC: Evaluating 11. The human genome comprises 23 pairs of chromosomes found in nearly every cell in the body. Answer the quantitative questions below by choosing one of the numbers in the following list: 23

69

>200

46

92

>109

A. How many centromeres are in each cell? What is the main function of the centromere?

B. How many telomeres are in each cell? What is their main function?

C. How many replication origins are in each cell? What is their main function?

ANS: A. There are 46 centromeres per cell, one on each chromosome. The centromeres have a key role in the distribution of chromosomes to daughter cells during mitosis.

B. There are 92 telomeres per cell, two on each chromosome. Telomeres serve to protect the ends of chromosomes and to enable complete replication of the DNA of each chromosome all the way to its tips.

C. There are far more than 200 replication origins in a human cell, probably about 10,000. These DNA sequences direct the initiation of DNA synthesis needed to replicate chromosomes.

DIF: Difficult REF: 5.2 OBJ: 5.2.h Compare the roles played by centromeres, telomeres, and replication origins.

MSC: Applying 12. Explain the differences between chromosome painting and the older, more traditional method of staining chromosomes being prepared for karyotyping. Highlight the way in which each method identifies chromosomes by the unique sequences they contain.

ANS: Chromosome painting relies on the specificity of DNA complementarity. Because unique sequences for each chromosome are known, short DNA molecules matching a set of these sites can be designed for each chromosome.

Each set is labeled with a specific combination of fluorescent dyes and then allowed to hybridize (form base pairs) with the two homologous chromosomes that contain the unique sequences being targeted.

Giemsa stain is a nonfluorescent dye that has a high affinity for DNA, and specifically accumulates in regions that are rich in A-T nucleotide pairs. This dye produces a pattern of dark and light bands, which differ for each chromosome on the basis of the distribution of AT-rich regions.

DIF: Moderate REF: 5.2 OBJ: 5.2.b Describe how human chromosomes can be distinguished from one another and how such information can be of value. MSC: Evaluating 13. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Comparing the relative number of chromosome pairs is a good way to determine whether two species are closely related.

B. Chromosomes exist at different levels of condensation, depending on the stage of the cell cycle.

C. Eukaryotic chromosomes contain many different sites where DNA replication can be initiated.

D. The telomere is a specialized DNA sequence where microtubules from the mitotic spindle attach to the

chromosome so that duplicate copies move to opposite ends of the dividing cell.

ANS: A. False. There are several examples of closely related species that have a drastically different number of chromosome pairs. Two related species of deer—Chinese and Indian muntjac—have 23 and 3, respectively.

B. True C. True D. False. The telomere is a specialized DNA sequence, but not for the attachment of spindle microtubules. Telomeres form special caps that stabilize the ends of linear chromosomes.

DIF: Easy REF: 5.2 OBJ: 5.2.f Describe the relationship between gene number, genome size, and organismal complexity. | 5.2.h Compare the roles played by centromeres, telomeres, and replication origins. | 5.2.k Contrast the extent of compression of interphase and mitotic chromosomes. MSC: Evaluating 14. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

10,000

chromosome

mitosis

100

different

nuclear envelope

1000

DNA

nucleolus

cell cycle

dynamic

proteins

cell wall

interphase

similar

chromatin

lipids

static

Each chromosome is a single molecule of __________ whose extraordinarily long length can be compacted by as much as __________-fold during __________ and tenfold more during __________. This is accomplished by binding to __________ that help package the DNA in an orderly manner so it can fit in the small space delimited by the __________. The structure of the DNA–protein complex, called __________, is highly __________ over time.

ANS: Each chromosome is a single molecule of DNA whose extraordinarily long length can be compacted by as much as 1000-fold during interphase and tenfold more during mitosis. This is accomplished by binding to proteins that help package the DNA in an orderly manner so it can fit in the small space delimited by the nuclear envelope. The structure of the DNA–protein complex, called chromatin, is highly dynamic over time.

DIF: Easy REF: 5.2 OBJ: 5.2.k Contrast the extent of compression of interphase and mitotic chromosomes.

MSC: Understanding 15. Origins of replication typically have a relatively high number of A-T base pairs. How does this sequence feature relate to the function of these DNA regions?

ANS: A-T-rich regions of the DNA are thermodynamically easier to “open” than G-C-rich base pairs because A-T base pairs are held together by two hydrogen bonds, rather than three, as found in C-G base pairs. DNA replication begins at origins of replication, and pulling apart the DNA double helix at these positions is needed to assemble the replication machinery to begin bidirectional replication.

DIF: Moderate REF: 5.2 OBJ: 5.2.h Compare the roles played by centromeres, telomeres, and replication origins.

MSC: Applying 16. For each of the following sentences, choose one of the options enclosed in square brackets to make a correct statement about nucleosomes.

A. Nucleosomes are present in [prokaryotic/eukaryotic] chromosomes, but not in [prokaryotic/eukaryotic] chromosomes.

B. A nucleosome contains two molecules each of histones [H1 and H2A/H2A and H2B] as well as of histones H3 and H4.

C. A nucleosome core particle contains a core of histone with DNA wrapped around it approximately [twice/three times/four times].

D. Nucleosomes are aided in their formation by the high proportion of [acidic/basic/polar] amino acids in histone proteins.

E. Nucleosome formation compacts the DNA into approximately [one-third/one-hundredth/one-thousandth] of its original length.

ANS: A. Nucleosomes are present in eukaryotic chromosomes, but not in prokaryotic chromosomes.

B. A nucleosome contains two molecules each of histones H2A and H2B as well as of histones H3 and H4.

C. A nucleosome core particle contains a core of histone with DNA wrapped around it approximately twice.

D. Nucleosomes are aided in their formation by the high proportion of basic amino acids in histone proteins.

E. Nucleosome formation compacts DNA into approximately one-third of its original length.

DIF: Moderate REF: 5.2 OBJ: 5.2.a Contrast prokaryotic and eukaryotic chromosomes in terms of structure and the specialized sequence elements they contain. | 5.2.m Distinguish between a nucleosome and a nucleosome core particle.

MSC: Analyzing 17. Early studies of chromosome structure relied on isolating chromatin from cells followed by nuclease treatment in vitro, followed by analysis of the products of this treatment. Demonstrate your understanding of this process by answering the questions below: A. How can nucleosome core particles be isolated from chromatin?

B. What molecular components were identified after this treatment was complete?

C. What portion of the nucleosome was destroyed/removed during this treatment and what function does it normally serve?

ANS: A. In a test tube, the nucleosome core particle can be released from chromatin by treatment with a nuclease that degrades the exposed, linker DNA, but not the DNA wrapped around the nucleosome core.

B. The core nucleosome was revealed to contain two molecules of the histones H2A, H2B, H3, and H4, as well as a 147-base-pair (bp) fragment of DNA. C. Nuclease treatment degrades linker DNA, which can be up to 80 bp in length. This region of DNA is typically bound to linker histones (H1), which are involved in higher-level packing of the chromatin.

DIF: Difficult REF: 5.2 OBJ: 5.2.m Distinguish between a nucleosome and a nucleosome core particle. | 5.2.n Explain how histone proteins are able to bind tightly to DNA. MSC: Understanding 18. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

30-nm fiber

heterochromatin

linker

active chromatin

histone H1

loops

axis

histone H3

more

beads-on-a-string

histone H4

synaptic complex

euchromatin

less

zigzag

Interphase chromosomes contain both darkly staining __________ and more lightly staining __________. Genes that are being transcribed are thought to be packaged in a __________ condensed type of euchromatin. Nucleosome core particles are separated from each other by stretches of __________ DNA. A string of nucleosomes coils up with the help of __________ to form the more compact structure of the __________. A __________ model describes the structure of the 30-nm fiber. The 30-nm chromatin fiber is further compacted by the formation of __________ that emanate from a central __________.

ANS: Interphase chromosomes contain both darkly staining heterochromatin and more lightly staining euchromatin.

Genes that are being transcribed are thought to be packaged in a less condensed type of euchromatin. Nucleosome core particles are separated from each other by stretches of linker DNA. A string of nucleosomes coils up with the help of histone H1 to form the more compact structure of the 30-nm fiber. A zigzag model describes the structure of the 30-nm fiber. The 30-nm chromatin fiber is further compacted by the formation of loops that emanate from a central axis.

DIF: Moderate REF: 5.3 OBJ: 5.3.c Contrast euchromatin and heterochromatin in terms of structure, gene activity, and location along an interphase chromosome. MSC: Understanding 19. Evidence suggests that the replication of DNA packaged into heterochromatin occurs later than the replication of other chromosomal DNA. What is the simplest possible explanation for this phenomenon?

ANS: The DNA double helix in heterochromatin may be so tightly packed and condensed that it is inaccessible to the proteins that bind replication origins, including the DNA replication machinery. It may take extra time to remodel the chromatin to make it more accessible to the proteins required to initiate and perform DNA replication.

DIF: Moderate REF: 5.3 OBJ: 5.3.a Explain how chromatin-remodeling complexes and histone-modifying enzymes regulate the accessibility of DNA. | 5.3.b Explain why a cell might decondense a particular segment of DNA. MSC: Understanding 20. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. The histone proteins that constitute the core nucleosome include tetramers of histones H2A, H2B, H3, and H4.

B. Linker histones help compact genomic DNA by influencing the path of the DNA after it has wrapped about the nucleosome core.

C. Histone proteins have a lower-than-average number of lysines and arginines in their polypeptide chains.

D. Interphase chromosomes represent a physical state of the chromatin with the highest order of packaging.

ANS: A. False. When the core nucleosome is analyzed, it is revealed that there are H2A/H2B tetramers and H3/H4 tetramers in solution. Each of the tetramers has two subunits of the respective histone proteins.

B. True C. False. Histones have a higher number of lysines and arginines than most proteins. These amino acids are positively charged and help to increase the nonspecific affinity between the histones and the negatively charged phosphates in the DNA backbone.

D. False. When cells enter mitosis, the interphase chromosomes undergo at least one more level of packaging, which facilitates the segregation of sister chromatids.

DIF: Moderate REF: 5.2 OBJ: 5.2.k Contrast the extent of compression of interphase and mitotic chromosomes. | 5.2.m

Distinguish between a nucleosome and a nucleosome core particle. | 5.2.n Explain how histone proteins are able to bind tightly to DNA. MSC: Evaluating 21. Gene A, which is normally expressed, has been moved by DNA recombination near an area of heterochromatin. None of the daughter cells produced after this recombination event express gene A, even though its DNA sequence is unchanged. Explain this observation.

ANS: Regions of heterochromatin are usually transcriptionally silent. And well-established segments of heterochromatin are typically contained within special barrier sequences, which prevent the heterochromatin from expanding along the entire chromosome. When Gene A was moved into a region of heterochromatin, it is likely the gene became incorporated into heterochromatin, and no longer expressed. Because regions of heterochromatin are maintained through cell division, this silencing is now inherited by daughter cells.

DIF: Moderate REF: 5.3 OBJ: 5.3.a Explain how chromatin-remodeling complexes and histone-modifying enzymes regulate the accessibility of DNA. | 5.3.e Explain how heterochromatin participates in gene silencing and provide an example. MSC: Evaluating 22. Describe the mechanism by which heterochromatin can spread, once it has been established in one region of the chromosome.

ANS: Once the initial H3 lysine 9 methylation is established on core histone octamers in one region, the modification attracts a specific set of proteins and other histone-methylating enzymes. These enzymes create the same modification on adjacent histone octamers, which continue to recruit more heterochromatin-specific proteins and enzymes, creating a wave of heterochromatin spreading along the chromosome.

DIF: Difficult REF: 5.3 OBJ: 5.3.d Explain how heterochromatin is established and spreads.

MSC: Understanding 23. Avery, MacLeod, and McCarty carried out experiments to identify the class of biological molecule that carries heritable information. Explain how they identified the “transforming principle” that could convert a harmless strain of bacteria to a pathogenic one.

ANS: They prepared extracts from the infectious, pathogenic bacterial strain, and separated the different types of macromolecules (RNA, DNA, protein, lipids, and carbohydrates). Each of these materials was incubated separately with the noninfectious strain. The researchers were able to conclude that DNA was the “transforming principle” because it was the only macromolecule isolated from the pathogenic strain that was able to convert the noninfectious strain into an infectious one.

DIF: Moderate REF: 5.1 OBJ: 5.1.c Describe how, experimentally, researchers demonstrated that DNA carries genetic information. MSC: Understanding 24. When double-stranded DNA is heated, the two strands separate into single strands in a process called melting or denaturation. The temperature at which half of the duplex DNA molecules are intact and half have melted is defined as the Tm. A. Do you think Tm is a constant, or can it depend on other small molecules in the solution? Do you think high salt concentrations increase, decrease, or have no effect on Tm? B. Under standard conditions, the expected melting temperature in degrees Celsius can be calculated from the equation Tm = 59.9 + [0.41 × %(G + C)] − (675/length of duplex). Does the Tm increase or decrease if there are more G + C (and thus fewer A + T) base pairs? Does the Tm increase or decrease as the length of DNA increases?

Why? C. Calculate the predicted Tm for a stretch of double helix that is 100 nucleotides long and contains 50% G + C content.

ANS: A. Tm depends on the identity and concentration of other molecules in the solution. High salt concentrations are more effective at shielding the two negatively charged sugar–phosphate backbones in the double helix from each other, so the two strands repel each other less strongly. Thus, a high salt concentration stabilizes the duplex and increases the melting temperature.

B. The Tm increases as the proportion of G + C bases increases and as the length increases. The thermal energy required for melting depends on how many hydrogen bonds between the strands must be broken. Each G-C base pair contributes three hydrogen bonds, whereas an A-T base pair contributes only two.

C. Inserting values into the equation in part B gives Tm = 59.9 + (0.41 × 50) − (675/100) = 73.65°C, which is about twice the normal temperature of the human body and nearly too hot to touch.

DIF: Difficult REF: 5.1 OBJ: 5.1.d Distinguish between the bonds that link together the subunits in a single strand of DNA and those that hold together the two strands in a DNA double helix, and summarize how these bonds affect the behavior of the DNA molecule. | 5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. MSC: Applying 25. Consider the structure of the DNA double helix.

A. You and a friend want to split a double-stranded DNA molecule so you each have half. Is it better to cut the length of DNA in half so each person has a shorter length, or to separate the strands and each take one strand?

Explain.

B. In the original 1953 publication describing the discovery of the structure of DNA, Watson and Crick wrote, “It has not escaped our notice that the specific pairings we have postulated immediately suggest a possible copying mechanism for the genetic material.” What did they mean?

ANS: A. It is better to separate the strands and each take a single strand, because all of the information found in the original molecule is preserved in a full-length single strand but not in a half-length double-stranded molecule.

B. Watson and Crick meant that the complementary base-pairing of the strands allows a single strand to contain all of the information necessary to direct the synthesis of a new complementary strand.

DIF: Moderate REF: 5.1 OBJ: 5.1.d Distinguish between the bonds that link together the subunits in a single strand of DNA and those that hold together the two strands in a DNA double helix, and summarize how these bonds affect the behavior of the DNA molecule. | 5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. MSC: Applying 26. A. Explain the reason why the cell requires a mechanism for identifying specific sequences of DNA. B. On average, how often would the nucleotide sequence CGATTG be expected to occur in a DNA strand 4000 bases long? Show your work and explain your answer.

C. Molecular processes depend upon sequence-specific interactions of proteins with DNA. Recognition sequences can be 4, 5, 6, 7, or even 8 base pairs in length for a single protein. What might be the advantages of a short recognition sequence? What might be the advantage of a longer recognition sequence?

ANS:

A. Sequence information contains indicators important for the regulation of gene expression and DNA packaging.

Examples include sequence indicators for where a gene starts and ends, where transcription begins, and where to assemble specific protein complexes at specialized sequences such as those found in telomeric or centromeric DNA. B. Because 46 (= 4096) different sequences of six nucleotides can occur in DNA, any given sequence of six nucleotides would be expected to occur on average once in a DNA strand 4000 bases long, assuming a random distribution of sequences.

C. Short recognition sequences do not have as many sequence-specific contacts (which means they don’t bind as tightly to the binding site in question), and they are more likely to be found randomly throughout the genome. Using the same type of calculation from part B, there are 256 possible combinations for a 4-base-pair recognition sequence, which could be found 15–16 times over a 4000-base-pair segment by random chance. This could be useful for proteins that need to bind to a large number of sites with low affinity. If we take the case of the 8-base-pair sequence, there are 65,536 different possible sequences. Therefore, not only do they represent high-affinity binding sites, they are much less likely to be found by random chance.

DIF: Difficult REF: 5.1 OBJ: 5.1.g Explain how the structure of DNA carries information for producing proteins.

MSC: Applying 27. The number of cells in an average-sized adult human is on the order of 1014. Use this information, and the estimate that the length of DNA contained in each cell is 2 m, to do the following calculations (look up the necessary distances and show your working): A. Over how many miles would the total DNA from the average human stretch?

B. How many times would the total DNA from the average human wrap around the planet Earth at the Equator?

C. How many times would the total DNA from the average human stretch from Earth to the Sun and back?

D. How many times would the total DNA from the average human stretch from the Earth to Pluto and back?

ANS: A. 2 × 1014 m = 124,274,238,447 miles.

B. The Earth’s circumference at the Equator is 24,902 miles. The length of DNA from the average human body could wrap around the Earth 4,990,532 times.

C. The average distance from the Earth to the Sun is 93,000,000 miles. So, the round-trip distance is 186,000,000 miles. The length of DNA from the average human body could stretch from the Earth to the Sun and back 668 times.

D. The distance from the Earth to Pluto is, on average, about 39 × 93,000,000 miles. So, the round trip distance is 78 × 93,000,000 miles. The length of DNA from the average human body could stretch from the Earth to Pluto and back 17 times.

DIF: Easy REF: 5.1 OBJ: 5.1.g Explain how the structure of DNA carries information for producing proteins.

MSC: Applying 28. You are studying a newly identified chromatin-remodeling complex, which you call NICRC. You decide to run an in vitro experiment to characterize the activity of the purified complex. Your molecular toolbox includes: (1) a 400-basepair DNA molecule that has a single recognition site for the restriction endonuclease EcoRI, an enzyme that cleaves internal sites on double-stranded DNA (dsDNA); (2) purified EcoRI enzyme; (3) purified DNase I, a DNA endonuclease that will cleave dsDNA at nonspecific sites if they are exposed; and (4) core octamer histones. You are able to assemble core nucleosomes on this DNA template and test for NICRC activity. Figure 5-67A illustrates

the DNA template used and indicates both the location of the EcoRI cleavage site and the size of the DNA fragments that are produced when it cuts. Figure 5-67B illustrates how the DNA molecules in your experiment looked after separation according to size by using gel electrophoresis. Your experiment had a total of six samples, each of which was treated according to the legend below the gel. The sizes of the DNA fragments observed are indicated on the left side of the gel.

Figure 5-67

A. Explain the results in lanes 1–4 and why it is important to have this information before you begin to test your remodeling complex.

B. What can you conclude about your purified remodeling complex from the results in lanes 5 and 6?

ANS: A. Sample 1 confirms the location of the EcoRI restriction site and shows what those fragments should look like when separated on the gel. The cleavage is the readout that will tell us whether the remodeling complex is wor king. Sample 2 demonstrates that DNA that is not assembled into nucleosomes can be cut into many small fragments by DNaseI. That is why we do not see discrete bands. Sample 3 demonstrates that when nucleosomes are assembled on the DNA, DNase I cuts only in one place, presumably in the linker region between two a ssembled nucleosomes. Sample 4 demonstrates that EcoRI cannot access its cleavage site when nucleosomes are assembled over it.

B. Sample 5 demonstrates that our purified complex is working. It must be moving the nucleosomes, providing a ccess for EcoRI, such that it can now cleave at its restriction site, which was not possible in the absence of NICRC. Sample 6 shows that NICRC will function only if ATP is available as an energy source for the remode ling process.

DIF: Difficult REF: 5.3 OBJ: 5.3.a Explain how chromatin-remodeling complexes and histone-modifying enzymes regulate the accessibility of DNA. MSC: Applying

29. For a better understanding of DNA structure, it helps to be able to compare physical characteristics evident from a side view of double-stranded DNA with those of individual base pairs.

A. Use brackets to designate the major and minor grooves on Figure 5-68A and shade in the surface that will be exposed in the major groove in Figure 5-68B. B. If base pairs were aligned and stacked directly on top of each other, the major and minor grooves would be linear depressions all along the DNA. Explain why this is not the actual conformation of a DNA molecule.

Figure 5-68

ANS: A. See Figure 5-68A below.

Figure 5-68A

B. The DNA base pairs are rotated with respect to each other. For a double-stranded DNA molecule with 10 base

pairs, a full 360° rotation has occurred. This is referred to as one turn of the helix, and it can be seen in Figure A568A in the alignment of the A-T base pair at the bottom with the G-C base pair at the top.

DIF: Moderate REF: 5.1 OBJ: 5.1.d Distinguish between the bonds that link together the subunits in a single strand of DNA and those that hold together the two strands in a DNA double helix, and summarize how these bonds affect the behavior of the DNA molecule. | 5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. MSC: Evaluating 30. Use the terms listed to fill in the blanks in Figure 5-69.

A. A-T base pair B. G-C base pair C. deoxyribose D. phosphodiester bonds E. purine base F. pyrimidine base

Figure 5-69

ANS: See Figure 5-69A below.

Figure 5-69A

DIF: Easy REF: 5.1 OBJ: 5.1.d Distinguish between the bonds that link together the subunits in a single strand of DNA and those that hold together the two strands in a DNA double helix, and summarize how these bonds affect the behavior of the DNA molecule. | 5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. MSC: Analyzing 31. Using the structures in Figure 5-70 as a guide, sketch the hydrogen bonds between the base pairs in DNA. Hint: The bases in the figure are all drawn with the −NH− that attaches to the sugar at the bottom of the structure.

Figure 5-70

ANS: See Figure 5-70A below.

Figure 5-70A

DIF: Moderate REF: 5.1 OBJ: 5.1.d Distinguish between the bonds that link together the subunits in a single strand of DNA and those that hold together the two strands in a DNA double helix, and summarize how these bonds affect the behavior of the DNA molecule. | 5.1.e Describe complementary base-pairing and explain how this arrangement gives rise to the twisting, consistently proportioned, double helical structure of DNA. MSC: Creating

CHAPTER 6 DNA Replication and Repair

DNA REPLICATION 6.1.a Explain how a DNA double helix provides a template for its own replication, and describe the resulting daughter helices in terms of their sequence and the distribution of parental and newly synthesized DNA strands.

6.1.b Describe the experiment that revealed the semiconservative nature of DNA replication.

6.1.c Recall where along a chromosome DNA synthesis begins, and explain what characterizes these nucleotide sequences in simple cells such as bacteria and yeast.

6.1.d Compare the direction in which replication forks move with the direction in which the new DNA strands are synthesized.

6.1.e Compare the bonds that link together nucleotides in a DNA strand with the bonds that hold together the two strands of DNA in a double helix.

6.1.f Explain how nucleoside triphosphates provide the energy for DNA synthesis.

6.1.g Explain why an asymmetrical replication fork poses a challenge for DNA polymerization and how DNA polymerase solves this problem to keep the replication fork moving forward.

6.1.h Explain how DNA polymerase contributes to the accuracy of DNA replication.

6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands.

6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand.

6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication.

6.1.l Describe the problem created by a moving replication fork, and explain how DNA topoisomerases relieve this difficulty.

6.1.m Describe the “end replication problem” and explain how telomerase solves this dilemma.

DNA REPAIR 6.2.a List some of the causes of DNA damage.

6.2.b Name some of the types of damage that can alter DNA. 6.2.c List the three main steps involved in repairing damage that affects only one strand of the DNA double helix.

6.2.d Explain how the mismatch repair system recognizes and corrects replication errors.

6.2.e Contrast nonhomologous end joining and homologous recombination as mechanisms for repairing double-stranded DNA breaks.

6.2.f Describe the consequences of a failure to repair damaged DNA.

MULTIPLE CHOICE 1. The process of DNA replication requires that each of the parental DNA strands be used as a __________ to produce a duplicate of the opposing strand. a. catalyst b. competitor c. template d. copy ANS: C DIF: Easy REF: 6.1 OBJ: 6.1.a Explain how a DNA double helix provides a template for its own replication and describe the resulting daughter helices in terms of their sequence and their inclusion of parental and newly synthesized DNA strands. MSC: Understanding 2. DNA replication is considered semiconservative because a. after many rounds of DNA replication, the original DNA double helix is still intact. b. each daughter DNA molecule consists of two new strands copied from the parent DNA molecule. c. each daughter DNA molecule consists of one strand from the parent DNA molecule and one new strand. d. new DNA strands must be copied from a DNA template.

ANS: C DIF: Easy REF: 6.1 OBJ: 6.1.a Explain how a DNA double helix provides a template for its own replication and describe the resulting daughter helices in terms of their sequence and their inclusion of parental and newly synthesized DNA strands. MSC: Understanding 3. The classic experiments conducted by Meselson and Stahl used isotopes of which element to distinguish newly synthesized DNA from parental DNA? a. carbon b. nitrogen c. oxygen d. hydrogen ANS: B DIF: Easy REF: 6.1 OBJ: 6.1.b Describe the experiment that revealed the semiconservative nature of DNA replication. MSC: Remembering 4. Initiator proteins bind to replication origins and disrupt hydrogen bonds between the two DNA strands being copied.

Which of the factors below does not contribute to the relative ease of strand separation by initiator proteins? a. replication origins are rich in A-T base pairs b. the reaction can occur at room temperature c. they only separate a few base pairs at a time d. once opened, other proteins of the DNA replication machinery bind to the origin ANS: B DIF: Easy REF: 6.1 OBJ: 6.1.e Compare the bonds that link together nucleotides in a DNA strand with the bonds that hold together the two DNA strands of DNA in a double helix. MSC: Understanding 5. If the genome of the bacterium E. coli requires about 20 minutes to replicate itself, how can the genome of the fruit fly Drosophila be replicated in only 3 minutes? a. The Drosophila genome is smaller than the E. coli genome. b. Eukaryotic DNA polymerase synthesizes DNA at a much faster rate than prokaryotic DNA polymerase.

c. The nuclear membrane keeps the Drosophila DNA concentrated in one place in the cell, which increases the rate of polymerization. d. Drosophila DNA contains more origins of replication than E. coli DNA. ANS: D DIF: Difficult REF: 6.1 OBJ: 6.1.c Explain where along a chromosome DNA synthesis begins and what makes these nucleotide sequences unique. MSC: Understanding 6. How many replication forks are formed when an origin of replication is opened? a. 1 b. 2 c. 3 d. 4 ANS: B DIF: Easy REF: 6.1 OBJ: 6.1.d Compare the direction in which replication forks move from the origin of replication to the direction in which DNA synthesis proceeds. MSC: Understanding 7. How does the total number of replication origins in bacterial cells compare to the number of origins in human cells? a. 1 versus 100 b. 5 versus 500 c. 10 versus 1000 d. 1 versus 10,000 ANS: D DIF: Easy REF: 6.1 OBJ: 6.1.c Explain where along a chromosome DNA synthesis begins and what makes these nucleotide sequences unique. MSC: Remembering 8. Which of the following statements correctly explains what it means for DNA replication to be bidirectional? a. The replication fork can open or close, depending on the conditions. b. The DNA replication machinery can move in either direction on the template strand. c. Replication-fork movement can switch directions when the fork converges on another replication fork. d. The replication forks formed at the origin move in opposite directions.

ANS: D DIF: Easy REF: 6.1 OBJ: 6.1.d Compare the direction in which replication forks move from the origin of replication with the direction in which DNA synthesis proceeds. MSC: Understanding 9. The chromatin structure in eukaryotic cells is much more complicated than that observed in prokaryotic cells. This is thought to be the reason why DNA replication occurs much faster in prokaryotes. How much faster is it? a. 2× b. 5× c. 10× d. 100× ANS: C DIF: Easy REF: 6.1 OBJ: 6.1.f Explain how nucleoside triphosphates provide the energy for DNA synthesis.

MSC: Remembering 10. DNA polymerase catalyzes the joining of a nucleotide to a growing DNA strand. What prevents this enzyme from catalyzing the reverse reaction? a. hydrolysis of pyrophosphate (PPi) to inorganic phosphate (Pi) + Pi b. release of PPi from the nucleotide

c. hybridization of the new strand to the template d. loss of ATP as an energy source ANS: A DIF: Moderate REF: 6.1 OBJ: 6.1.h Explain how DNA polymerase maintains the accuracy of DNA replication.

MSC: Understanding 11. You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin.

Figure 6-11

Which of the following statements is TRUE with respect to this in vitro replication system? a. There will be only one leading strand and one lagging strand produced using this template. b. The leading and lagging strands compose one half of each newly synthesized DNA strand. c. The DNA replication machinery can assemble at multiple places on this plasmid. d. One daughter DNA molecule will be slightly shorter than the other.

ANS: B Leading and lagging strands are synthesized bidirectionally from the replication origin, and are joined together by DNA ligase where the two replication forks meet at the termination site.

DIF: Moderate REF: 6.1 OBJ: 6.1.c Explain where along a chromosome DNA synthesis begins and what makes these nucleotide sequences unique. | 6.1.d Compare the direction in which replication forks move from the origin of replication with the direction in which DNA synthesis proceeds. MSC: Analyzing 12. You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin.

Figure 6-11

You decide to use different bacterial strains (each having one protein of the replication machinery mutated) in order to examine the role of individual proteins in the normal process of DNA replication.

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking primase were used to make the cell extracts? a. initiation of DNA synthesis b. Okazaki fragment synthesis c. leading-strand elongation d. lagging-strand completion ANS: A DNA synthesis cannot begin without the initial primers. DIF: Difficult REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. MSC: Applying 13. You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin.

Figure 6-11

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking the exonuclease activity of DNA polymerase were used to make the cell extracts? a. initiation of DNA synthesis

b. Okazaki fragment synthesis c. leading-strand elongation d. lagging-strand completion ANS: D DIF: Moderate REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. MSC: Applying 14. You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin.

Figure 6-11

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking helicase were used to make the cell extracts? a. initiation of DNA synthesis b. Okazaki fragment synthesis c. leading-strand elongation d. lagging-strand completion ANS: A Because helicase unwinds the two DNA template strands, replication of both strands depends upon the activity of helicase at the time of initiation.

DIF: Easy REF: 6.1 OBJ: 6.1.e Compare the bonds that link together nucleotides in a DNA strand with the bonds that hold together the two DNA strands of DNA in a double helix. MSC: Applying 15. You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin.

Figure 6-11

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking single-strand binding protein were used to make the cell extracts? a. initiation of DNA synthesis b. Okazaki fragment synthesis c. leading-strand elongation d. lagging-strand completion ANS: B DIF: Moderate REF: 6.1 OBJ: 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Applying 16. You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin.

Figure 6-11

What part of the DNA replication process would be most directly affected if a strain of bacteria lacking DNA ligase were used to make the cell extracts? a. initiation of DNA synthesis b. Okazaki fragment synthesis c. leading-strand elongation

d. lagging-strand completion ANS: D DIF: Easy REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. MSC: Applying 17. Which of the following statements about the newly synthesized strand of a human chromosome is TRUE? a. It was synthesized from a single origin solely by continuous DNA synthesis. b. It was synthesized from a single origin by a mixture of continuous and discontinuous DNA synthesis. c. It was synthesized from multiple origins solely by discontinuous DNA synthesis. d. It was synthesized from multiple origins by a mixture of continuous and discontinuous DNA synthesis.

ANS: D Each newly synthesized strand in a daughter duplex was synthesized by a mixture of continuous and discontinuous DNA synthesis from multiple origins. Consider a single replication origin. The fork moving in one direction synthesizes a daughter strand continuously as part of leading-strand synthesis; the fork moving in the opposite direction synthesizes a portion of the same daughter strand discontinuously as part of lagging-strand synthesis.

DIF: Difficult REF: 6.1 OBJ: 6.1.c Explain where along a chromosome DNA synthesis begins and what makes these nucleotide sequences unique. | 6.1.d Compare the direction in which replication forks move from the origin of replication with the direction in which DNA synthesis proceeds. MSC: Analyzing 18. You have discovered an “Exo–” mutant form of DNA polymerase in which the 3′-to-5′ exonuclease function has been destroyed but the ability to join nucleotides together is unchanged. Which of the following properties do you expect the mutant polymerase to have? a. It will polymerize in both the 5′-to-3′ direction and the 3′-to-5′ direction. b. It will polymerize more slowly than the normal Exo+ polymerase. c. It will fall off the template more frequently than the normal Exo+ polymerase. d. It will be more likely to generate mismatched base pairs.

ANS: D DIF: Easy REF: 6.1 OBJ: 6.1.h Explain how DNA polymerase maintains the accuracy of DNA replication.

MSC: Applying 19. Which diagram accurately represents the directionality of DNA strands at one side of a replication fork?

Figure 6-19

a. Diagram A b. Diagram B c. Diagram C d. Diagram D ANS: D DIF: Easy REF: 6.1 OBJ: 6.1.d Compare the direction in which replication forks move from the origin of replication with the direction in which DNA synthesis proceeds. MSC: Understanding 20. DNA polymerases are processive, which means that they remain tightly associated with the template strand while moving rapidly and adding nucleotides to the growing daughter strand. Which piece of the replication machinery accounts for this characteristic? a. helicase

b. sliding clamp c. single-strand binding protein d. primase ANS: B DIF: Easy REF: 6.1 OBJ: 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Remembering 21. Which of the following statements about sequence proofreading during DNA replication is FALSE? a. The exonuclease activity is in a different domain of the DNA polymerase. b. The exonuclease activity cleaves DNA in the 5′-to-3′ direction. c. The DNA proofreading activity occurs concomitantly with strand elongation. d. If an incorrect base is added, it is “unpaired” before removal.

ANS: B DIF: Easy REF: 6.1 OBJ: 6.1.h Explain how DNA polymerase maintains the accuracy of DNA replication.

MSC: Understanding 22. The DNA duplex consists of two long covalent polymers wrapped around each other many times over their entire length. The separation of the DNA strands for replication causes the strands to be “overwound” in front of the replication fork. How does the cell relieve the torsional stress created along the DNA duplex during replication? a. Nothing needs to be done because the two strands will be separated after replication is complete. b. Topoisomerases break the covalent bonds of the backbone, allowing the local unwinding of DNA ahead of the replication fork. c. Helicase unwinds the DNA and rewinds it after replication is complete. d. DNA repair enzymes remove torsional stress as they replace incorrectly paired bases.

ANS: B DIF: Moderate REF: 6.1 OBJ: 6.1.l Describe the problem created by the activity of the helicase and explain how DNA topoisomerases relieve this difficulty. MSC: Understanding 23. Telomeres serve as caps protecting the ends of linear chromosomes. Which of the following is FALSE regarding the replication of telomeric sequences? a. The lagging-strand telomeres are not completely replicated by DNA polymerase. b. Telomeres are made of repeating sequences. c. Additional repeated sequences are added to the template strand. d. The leading strand doubles back on itself to form a primer for the lagging strand.

ANS: D DIF: Easy REF: 6.1 OBJ: 6.1.m Describe the “end replication problem” and explain how telomerase helps solve this dilemma. MSC: Understanding 24. Even though DNA polymerase has a proofreading function, it still introduces errors in the newly synthesized strand at a rate of 1 per 107 nucleotides. To what degree does the mismatch repair system decrease the error rate arising from DNA replication? a. 2-fold b. 5-fold c. 10-fold d. 100-fold ANS: D DIF: Moderate REF: 6.2 OBJ: 6.2.d Explain how the mismatch repair system recognizes and corrects replica-

tion errors. MSC: Understanding 25. Which of the choices below represents the correct way to repair the mismatch shown in Figure 6-25?

Figure 6-25

a. Choice A b. Choice B c. Choice C d. Choice D ANS: A DIF: Easy REF: 6.2 OBJ: 6.2.d Explain how the mismatch repair system recognizes and corrects replication errors. MSC: Applying 26. Beside the distortion in the DNA backbone caused by a mismatched base pair, what additional mark is there on eukaryotic DNA to indicate which strand needs to be repaired? a. a nick in the template strand b. a chemical modification of the new strand c. a nick in the new strand d. a sequence gap in the new strand ANS: C DIF: Easy REF: 6.2 OBJ: 6.2.d Explain how the mismatch repair system recognizes and corrects replication errors. MSC: Remembering 27. The repair of mismatched base pairs or damaged nucleotides in a DNA strand requires a multistep process. Which choice below describes the known sequence of events in this process? a. DNA damage is recognized, the newly synthesized strand is identified by an existing nick in the backbone, a segment of the new strand is removed by repair proteins, the gap is filled by DNA polymerase, and the strand is

sealed by DNA ligase. b. DNA repair polymerase simultaneously removes bases ahead of it and polymerizes the correct sequence behind it as it moves along the template. DNA ligase seals the nicks in the repaired strand. c. DNA damage is recognized, the newly synthesized strand is identified by an existing nick in the backbone, a segment of the new strand is removed by an exonuclease, and the gap is repaired by DNA ligase. d. A nick in the DNA is recognized, DNA repair proteins switch out the wrong base and insert the correct base, and DNA ligase seals the nick.

ANS: A DIF: Easy REF: 6.2 OBJ: 6.2.d Explain how the mismatch repair system recognizes and corrects replication errors. MSC: Understanding 28. Human beings with the inherited disease xeroderma pigmentosum have serious problems with lesions on their skin and often develop skin cancer with repeated exposure to sunlight. What type of DNA damage is not being recognized in the cells of these individuals? a. chemical damage b. X-ray irradiation damage c. mismatched bases d. ultraviolet irradiation damage ANS: D DIF: Easy REF: 6.2 OBJ: 6.2.b Name the types of damage that can alter DNA. MSC: Remembering 29. You are examining the DNA sequences that code for the enzyme phosphofructokinase in skinks and Komodo dragons. You notice that the coding sequence that actually directs the sequence of amino acids in the enzyme is very similar in the two organisms but that the surrounding sequences vary quite a bit. What is the most likely explanation for this? a. Coding sequences are repaired more efficiently. b. Coding sequences are replicated more accurately. c. Coding sequences are packaged more tightly in the chromosomes to protect them from DNA damage. d. Mutations in coding sequences are more likely to be deleterious to the organism than mutations in noncoding sequences.

ANS: D DIF: Moderate REF: 6.2 OBJ: 6.2.f Describe the consequences of a failure to repair damaged DNA. MSC: Applying 30. In somatic cells, if a base is mismatched in one new daughter strand during DNA replication, and is not repaired, what fraction of the DNA duplexes will have a permanent change in the DNA sequence after the second round of DNA replication? a. 1/2 b. 1/4 c. 1/8 d. 1/16 ANS: B DIF: Easy REF: 6.2 OBJ: 6.2.f Describe the consequences of a failure to repair damaged DNA. MSC: Applying 31. Sometimes, chemical damage to DNA can occur just before DNA replication begins, not giving the repair system enough time to correct the error before the DNA is duplicated. This gives rise to mutation. If the cytosine in the

sequence TCAT is deaminated and not repaired, which of the following is the point mutation you would observe after this segment has undergone two rounds of DNA replication? a. TTAT b. TUAT c. TGAT d. TAAT ANS: A DIF: Moderate REF: 6.2 OBJ: 6.2.f Describe the consequences of a failure to repair damaged DNA. MSC: Applying 32. Sometimes, chemical damage to DNA can occur just before DNA replication begins, not giving the repair system enough time to correct the error before the DNA is duplicated. This gives rise to mutation. If the adenosine in the sequence TCAT is depurinated and not repaired, which of the following is the point mutation that you would observe after this segment has undergone two rounds of DNA replication? a. TCGT b. TAT c. TCT d. TGTT ANS: C DIF: Moderate REF: 6.2 OBJ: 6.2.f Describe the consequences of a failure to repair damaged DNA. MSC: Applying 33. Which of the following statements is NOT an accurate statement about thymine dimers? a. Thymine dimers can cause the DNA replication machinery to stall. b. Thymine dimers are covalent links between thymidines on opposite DNA strands. c. Prolonged exposure to sunlight causes thymine dimers to form. d. Repair proteins recognize thymine dimers as a distortion in the DNA backbone.

ANS: B DIF: Easy REF: 6.2 OBJ: 6.2.b Name the types of damage that can alter DNA. MSC: Understanding 34. Several members of the same family were diagnosed with the same kind of cancer when they were unusually young.

Which one of the following is the most likely explanation for this phenomenon? It is possible that the individuals with the cancer have inherited a. a cancer-causing gene that suffered a mutation in an ancestor’s somatic cells. b. a mutation in a gene required for DNA synthesis. c. a mutation in a gene required for mismatch repair. d. a mutation in a gene required for the synthesis of purine nucleotides.

ANS: C Individuals in some families with a history of early-onset colon cancer have been found to carry mutations in mismatch repair genes. Mutations arising in somatic cells are not inherited.

DIF: Difficult REF: 6.2 OBJ: 6.2.f Describe the consequences of a failure to repair damaged DNA. MSC: Applying 35. Sickle-cell anemia is an example of an inherited disease. Individuals with this disorder have misshapen (sickleshaped) red blood cells caused by a change in the sequence of the β-globin gene. What is the nature of the change? a. chromosome loss

b. base-pair change c. gene duplication d. base-pair insertion ANS: B DIF: Easy REF: 6.2 OBJ: 6.2.f Describe the consequences of a failure to repair damaged DNA. MSC: Understanding 36. Select the option that best completes the following statement: Nonhomologous end joining is a process by which a double-stranded DNA end is joined a. to a similar stretch of sequence on the complementary chromosome. b. after repairing any mismatches. c. to the nearest available double-stranded DNA end. d. after filling in any lost nucleotides, helping to maintain the integrity of the DNA sequence.

ANS: C DIF: Easy REF: 6.2 OBJ: 6.2.e Contrast nonhomologous end-joining and homologous recombination as mechanisms for repairing double-strand DNA breaks. MSC: Remembering 37. Nonhomologous end joining can result in all but which of the following? a. the recovery of lost nucleotides on a damaged DNA strand b. the interruption of gene expression c. loss of nucleotides at the site of repair d. translocations of DNA fragments to an entirely different chromosome ANS: A DIF: Easy REF: 6.2 OBJ: 6.2.e Contrast nonhomologous end-joining and homologous recombination as mechanisms for repairing double-strand DNA breaks. MSC: Understanding 38. Homologous recombination is an important mechanism in which organisms use a “backup” copy of the DNA as a template to fix double-strand breaks without loss of genetic information. Which of the following is NOT necessary for homologous recombination to occur? a. 3′ DNA strand overhangs b. 5′ DNA strand overhangs c. a long stretch of sequence similarity d. nucleases ANS: B DIF: Easy REF: 6.2 OBJ: 6.2.e Contrast nonhomologous end-joining and homologous recombination as mechanisms for repairing double-strand DNA breaks. MSC: Understanding 39. In addition to the repair of DNA double-strand breaks, homologous recombination is a mechanism for generating genetic diversity by swapping segments of parental chromosomes. During which process does swapping occur? a. DNA replication b. DNA repair c. meiosis d. transposition ANS: C DIF: Easy REF: 6.2 OBJ: 6.2.e Contrast nonhomologous end-joining and homologous recombination as mechanisms for repairing double-strand DNA breaks. MSC: Remembering 40. Recombination has occurred between the chromosome segments shown in Figure 6-40. The genes A and B, and the

recessive alleles a and b, are used as markers on the maternal and paternal chromosomes, respectively. After alignment and homologous recombination, the specific arrangements of A, B, a, and b have changed.

Figure 6-40

Which of the choices below correctly indicates the gene combination from the replication products of the maternal chromosome? a. AB and aB b. ab and Ab c. AB and Ab d. aB and Ab ANS: D DIF: Moderate REF: 6.2 OBJ: 6.2.e Contrast nonhomologous end-joining and homologous recombination as mechanisms for repairing double-strand DNA breaks. MSC: Applying 41. The events listed below are all necessary for homologous recombination to occur properly: A. Holliday junction cut and ligated B. strand invasion C. DNA synthesis D. DNA ligation E. double-strand break F. nucleases create uneven strands Which of the following is the correct order of events during homologous recombination?

a. E, B, F, D, C, A b. B, E, F, D, C, A c. C, E, F, B, D, A d. E, F, B, C, D, A ANS: D DIF: Easy REF: 6.2 OBJ: 6.2.e Contrast nonhomologous end-joining and homologous recombination as mechanisms for repairing double-strand DNA breaks. MSC: Understanding

MATCHING 1. Because all DNA polymerases synthesize DNA in the 5′-to-3′ direction, and the parental strands are antiparallel, DNA replication is accomplished with the use of two mechanisms: continuous and discontinuous replication. Indicate whether the following items relate to (A) continuous replication, (B) discontinuous replication, or (C) both modes of replication.

1. ______ primase 2. ______ single-strand binding protein 3. ______ sliding clamp 4. ______ RNA primers 5. ______ leading strand 6. ______ lagging strand 7. ______ Okazaki fragments 8. ______ DNA helicase 9. ______ DNA ligase 1. ANS: C DIF: Difficult REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Analyzing 2. ANS: B DIF: Difficult REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Analyzing 3. ANS: C DIF: Difficult REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Analyzing 4. ANS: C DIF: Difficult REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Analyzing 5. ANS: A DIF: Difficult REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication

machine and state the role each plays in DNA replication. MSC: Analyzing 6. ANS: B DIF: Difficult REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Analyzing 7. ANS: B DIF: Difficult REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Analyzing 8. ANS: C DIF: Difficult REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Analyzing 9. ANS: B DIF: Difficult REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Analyzing 2. Use the components in the list below to label the diagram of a replication fork in Figure 6-43.

Figure 6-43

1. DNA polymerase 2. single-strand binding protein 3. Okazaki fragment 4. primase 5. sliding clamp 6. RNA primer 7. DNA helicase 1. ANS: B DIF: Easy REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to

produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Remembering 2. ANS: F DIF: Easy REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Remembering 3. ANS: D DIF: Easy REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Remembering 4. ANS: G DIF: Easy REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Remembering 5. ANS: A DIF: Easy REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Remembering 6. ANS: E DIF: Easy REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Remembering 7. ANS: C DIF: Easy REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Remembering

SHORT ANSWER 1. Which scientists first proposed a general strategy for DNA replication. How did they imagine it would work and what was their reasoning?

ANS: Watson and Crick proposed the first model for making new copies of DNA when cells divide. The y imagined that the two strands would unwind and each would serve as a template for a new, complementary strand. The resulting DNA duplexes would contain one original strand and one newly synthesized strand. Their reasoning was based on the proposed DNA structure, in which they observed complementarity between the A/T and C/G base pairs.

DIF: Easy REF: 6.1 OBJ: 6.1.a Explain how a DNA double helix provides a template for its own replication and describe the resulting daughter helices in terms of their sequence and their inclusion of parental and newly synthesized DNA strands. MSC: Remembering 2. How were Meselson and Stahl able to separate “light” DNA from “heavy” DNA. Explain how this experimental approach was used to rule out two of the three models for DNA replication.

ANS: Light and heavy DNA samples were separated using a cesium chloride density gradient and high-speed centrifugation. The heavy and light DNA products separate in the centrifuge tube, corresponding to the differences in their densities. The conservative model was eliminated with the observation that the daughter DNA molecules have an intermediate density between fully heavy and fully light DNA. The dispersive model was ruled out by using heat to denature the daughter DNA duplexes and then comparing the densities of the single-stranded DNA. If the dispersive model had been correct, individual strands should have had an intermediate density. However, this was not the case; only heavy strands and light strands were observed, which convincingly supported the semiconservative model for DNA replication.

DIF: Moderate REF: 6.1 OBJ: 6.1.b Describe the experiment that revealed the semiconservative nature of DNA replication. MSC: Understanding 3. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. When DNA is being replicated inside a cell, local heating occurs, allowing the two strands to separate.

B. DNA replication origins are typically rich in G-C base pairs.

C. Meselson and Stahl ruled out the dispersive model for DNA replication.

D. DNA replication is a bidirectional process that is initiated at multiple locations along chromosomes in eukaryotic cells.

ANS: A. False. The two strands do need to separate for replication to occur, but this is accomplished by the binding of initiator proteins at the origin of replication.

B. False. DNA replication origins are typically rich in A-T base pairs, which are held together by only two hydrogen bonds (instead of three for C-G base pairs), making it easier to separate the strands at these sites.

C. True D. True DIF: Easy REF: 6.1 OBJ: 6.1.a Explain how a DNA double helix provides a template for its own replication and describe the resulting daughter helices in terms of their sequence and their inclusion of parental and newly synthesized DNA strands. | 6.1.b Describe the experiment that revealed the semiconservative nature of DNA replication. | 6.1.c Explain where along a chromosome DNA synthesis begins and what makes these nucleotide sequences unique.

MSC: Evaluating 4. Use your knowledge of how a new strand of DNA is synthesized to explain why DNA replication must occur in the 5′to-3′ direction. In other words, what would be the consequences of 3′–to-5′ strand elongation?

ANS: There would be several detrimental consequences to 3′–to-5′ strand elongation. One of those most directly linked to the processes of DNA replication involves synthesis of the lagging strand. After the RNA primers are degraded, the DNA segments remaining will have 5′ ends with a single phosphate group. The incoming nucleotide will have a 3′-OH group. Without the energy provided by the release of PPi from the 5′ end, the process of elongation would no longer be energetically favorable.

DIF: Moderate REF: 6.1 OBJ: 6.1.e Compare the bonds that link together nucleotides in a DNA strand with the bonds that hold together the two DNA strands of DNA in a double helix. MSC: Applying 5. You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on

the opposite side of the plasmid from the origin.

Figure 6-11A

In addition to the extracts and the plasmid DNA, are there any additional materials you should add to this in vitro replication system? Explain your answer.

ANS: You will probably add exogenous nucleoside triphosphates to serve as the building blocks needed to make new strands of DNA. Although these monomers will be present in the extracts, they will be present at lower concentrations than are normally found inside the cell. They may also be subject to hydrolysis, and the nucleoside diphosphates that are the products of this hydrolysis are not usable substrates for DNA replication. For both these reasons, it is important to add excess nucleotides to the reaction mixture for efficient DNA replication to occur.

DIF: Moderate REF: 6.1 OBJ: 6.1.f Explain how nucleoside triphosphates provide the energy for DNA synthesis. MSC: Applying 6. A bacterial chromosome introduced into a yeast cell will not be duplicated along with yeast DNA chromosomes during cell division. Explain this observation.

ANS: DNA from all organisms is chemically identical, what differs is the exact sequence of nucleotides. Replication initiator proteins recognize specific DNA sequences at the origins of replication, and these sequences differ between bacteria and yeast. Without the initiator to help assemble the replication machinery on the bacterial chromosome, none of the downstream reactions can take place.

DIF: Moderate REF: 6.1 OBJ: 6.1.c Explain where along a chromosome DNA synthesis begins and what makes these nucleotide sequences unique. MSC: Applying 7. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Primase is needed to initiate DNA replication on both the leading strand and the lagging strand.

B. The sliding clamp is loaded once on each DNA strand, where it remains associated until replication is complete.

C. Telomerase is a DNA polymerase that carries its own RNA molecule to use as a template at the end of the lagging strand.

D. Primase requires a proofreading function that ensures there are no errors in the RNA primers used for DNA replication.

ANS: A. True B. False. Although the sliding clamp is only loaded once on the leading strand, the lagging strand needs to unload the clamp once the polymerase reaches the RNA primer from the previous segment and then reload it where a new

primer has been synthesized.

C. True D. False. Primase does not have a proofreading function, nor does it need one because the RNA primers are not a permanent part of the DNA. The primers are removed, and a DNA polymerase that does have a proofreading function fills in the remaining gaps.

DIF: Easy REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. | 6.1.m Describe the “end replication problem” and explain how telomerase helps solve this dilemma. MSC: Evaluating 8. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. The repair polymerase is the enzyme that proofreads the newly synthesized strands to ensure the accuracy of DNA replication.

B. There is a single enzyme that degrades the RNA primers and lays down the corresponding DNA sequence behind it.

C. DNA ligase is required to seal the sugar–phosphate backbone between all the DNA fragments on the lagging strand.

D. The repair polymerase does not require the aid of the sliding clamp, because it is only synthesizing DNA over very short stretches.

ANS: A. False. The repair polymerase is used to fill in the spaces left vacant after the RNA primers are degraded.

B. False. This is a two-step process that requires two different enzymes. First, a nuclease removes the RNA primers.

Then, the repair polymerase fills in the complementary DNA sequence.

C. True D. True DIF: Moderate REF: 6.1 OBJ: 6.1.h Explain how DNA polymerase maintains the accuracy of DNA replication. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Evaluating 9. The sliding clamp complex must be reloaded for the synthesis of each Okazaki fragment on the lagging strand, while on the leading strand the clamp complex remains assembled over much longer stretches of DNA. Considering this information, explain how the cell is able to replicate the leading and lagging strands at the same rate.

ANS: The cell makes use of a pool of clamp complexes and a protein called the clamp loader, which harnesses energy from ATP hydrolysis to catalyze the loading of a sliding clamp complex around the DNA. This happens each time a new Okazaki fragment is copies, and makes replication on the lagging strand more efficient.

DIF: Moderate REF: 6.1 OBJ: 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Understanding 10. DNA helicase has an essential role in DNA replication, but its unwinding activity also creates problems ahead of the replication fork, requiring the function of another class of enzymes. Describe the problem caused by DNA helicase, name the enzymes that resolve the problem, and explain the mechanism of resolution.

ANS: As DNA helicase unwinds DNA strands to make them accessible for DNA polymerase and all the other machinery required for DNA replication, the DNA ahead of the fork becomes overwound. This creates torsional stress, only some of which can be relieved by DNA supercoiling. DNA topoisomerases make single-stranded breaks in the backbone for rapid relief of the tension, and then reseals the backbone with its intrinsic DNA ligase activity.

DIF: Easy REF: 6.1 OBJ: 6.1.l Describe the problem created by the activity of the helicase and explain how DNA topoisomerases relieve this difficulty. MSC: Understanding 11. In terms of its function, is telomerase more like DNA polymerase or Primase? Explain your reasoning.

ANS: Primase is a type of RNA polymerase, making RNA primers necessary for strand elongation by DNA polymerase.

Telomerase is more like DNA polymerase and is highly specialized to help complete DNA replication on the lagging strand. Telomerase carries its own RNA primer, which pairs to repeated sequences in the telomeric region of the chromosome. It then synthesizes new DNA to extend the repeats.

DIF: Moderate REF: 6.1 OBJ: 6.1.m Describe the “end replication problem” and explain how telomerase helps solve this dilemma. MSC: Evaluating 12. A mismatched base pair causes a distortion in the DNA backbone. If this were the only indication of an error in replication, the overall rate of mutation would be much higher. Explain why.

ANS: The distortion in the DNA backbone is insufficient information for the mismatch repair system to identify which base is incorrect and which was originally part of the chromosome when replication began. Without additional marks that identify the difference between the newly synthesized strand and the template strand, the repair would be corrected only 50% of the time by random chance. The error rate (and therefore the mutation rate) would still be less than in a system that lacked the mismatch repair enzymes (1 mistake per 10 7 base pairs), but greater than the error rate in a system that accurately identifies the newly synthesized strand (1 mistake per 10 9 base pairs).

DIF: Moderate REF: 6.2 OBJ: 6.2.d Explain how the mismatch repair system recognizes and corrects replication errors. MSC: Evaluating 13. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Ionizing radiation and oxidative damage can cause DNA double-strand breaks.

B. After damaged DNA has been repaired, nicks in the phosphate backbone are maintained as a way to identify the strand that was repaired.

C. Depurination of DNA is a rare event that is caused by ultraviolet irradiation.

D. Nonhomologous end joining is a mechanism that ensures that DNA double-strand breaks are repaired with a high degree of fidelity to the original DNA sequence.

ANS: A. True B. False. It is believed that the nicks are generated during DNA replication as a means of easy identification of the newly synthesized strand but are sealed by DNA ligase shortly after replication is completed.

C. False. Depurination occurs constantly in our cells through spontaneous hydrolysis of the bond linking the DNA base to the deoxyribose sugar.

D. False. Homologous recombination can repair double-strand breaks without any change in DNA sequence, but nonhomologous end joining always involves a loss of genetic information because the ends are degraded by nucleases before they can be ligated back together.

DIF: Easy REF: 6.2 OBJ: 6.2.a List the causes of DNA damage. | 6.2.b Name the types of damage that can alter DNA. | 6.2.e Contrast nonhomologous end-joining and homologous recombination as mechanisms for repairing double-strand DNA breaks. MSC: Evaluating 14. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Homologous recombination cannot occur in prokaryotic cells, because they are haploid, and therefore have no extra copy of the chromosome to use as a template for repair.

B. The first step in repair requires a nuclease to remove a stretch of base pairs from the 5′ end of each strand at the site of the break.

C. The 3′ overhang “invades” the homologous DNA duplex, which can be used as a primer for the repair DNA polymerase.

D. The DNA template used to repair the broken strand is the homologous chromosome inherited from the other parent.

ANS: A. False. Homologous recombination also occurs in prokaryotic cells, and typically occurs very shortly after DNA replication, when the newly replicated duplexes are in close proximity.

B. True C. True D. False. Although it is called homologous recombination, this is not a process that depends on the proximity of parental homologs. When used as a mechanism for DNA repair, homologous recombination uses the sister chromatids in an undamaged, newly replicated (homologous) DNA helix as a template.

DIF: Easy REF: 6.2 OBJ: 6.2.e Contrast nonhomologous end-joining and homologous recombination as mechanisms for repairing double-strand DNA breaks. MSC: Evaluating 15. Most cells in the body of an adult human lack the telomerase enzyme because its gene is turned off and is therefore not expressed. An important step in the conversion of a normal cell into a cancer cell, which circumvents normal growth control, is the resumption of telomerase expression. Explain why telomerase might be necessary for the ability of cancer cells to divide over and over again.

ANS: In the absence of telomerase, the life-span of a cell and its progeny cells is limited. With each round of DNA replication, the length of telomeric DNA will shrink, until finally all the telomeric DNA has disappeared. Without telomeres capping the chromosome ends, the ends might be treated like breaks arising from DNA damage, or crucial genetic information might be lost. Cells whose DNA lacks telomeres will stop dividing or die. However, if telomerase is provided to cells, they may be able to divide indefinitely because their telomeres will remain a constant length despite repeated rounds of DNA replication.

DIF: Moderate REF: 6.1 OBJ: 6.1.m Describe the “end replication problem” and explain how telomerase helps solve this dilemma. MSC: Applying 16. Researchers have isolated a mutant strain of E. coli that carries a temperature-sensitive variant of the enzyme DNA ligase. At the permissive temperature, the mutant cells grow just as well as the wild-type cells. At the nonpermissive temperature, all of the cells in the culture tube die within 2 hours. DNA from mutant cells grown at the nonpermissive temperature for 30 minutes is compared with the DNA isolated from cells grown at the permissive temperature. The results are shown in Figure 6-59, where DNA molecules have been separated by size by means of electrophoresis (P,

permissive; NP, nonpermissive). Explain the appearance of a distinct band with a size of 200 base pairs (bp) in the sample collected at the nonpermissive temperature.

Figure 6-59

ANS: DNA ligase has an important role in DNA replication. After Okazaki fragments are synthesized, they must be ligated (covalently connected) to each other so that they finally form one continuous strand. At the nonpermissive temperature this does not happen, and although there may be a range of fragments, the notable band at 200 base pairs is the typical size of an individual Okazaki fragment.

DIF: Difficult REF: 6.1 OBJ: 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Applying 17. During DNA replication in a bacterium, a C is accidentally incorporated instead of an A into one newly synthesized DNA strand. Imagine that this error was not corrected and that it has no effect on the ability of the progeny to grow and reproduce.

A. After this original bacterium has divided once, what proportion of its progeny would you expect to contain the mutation?

B. What proportion of its progeny would you expect to contain the mutation after three more rounds of DNA replication and cell division?

ANS: A. One-half, or 50%. DNA replication in the original bacterium will create two new DNA molecules, one of which will now carry a mismatched C-T base pair. So one daughter cell of that cell division will carry a completely normal DNA molecule; the other cell will have the molecule with the mutation mispaired to a correct nucleotide.

B. One-quarter, or 25%. At the next round of DNA replication and cell division, the bacterium carrying the mismatched C-T will produce and pass on one normal DNA molecule from the undamaged strand containing the T and one mutant DNA molecule with a fully mutant C-G base pair. So at this stage, one out of the four progeny of the original bacterium is mutant. Subsequent cell divisions of these mutant bacteria will give rise only to mutant bacteria, whereas the other bacteria will give rise to normal bacteria. The proportion of progeny containing the mutation will therefore remain at 25%.

DIF: Moderate REF: 6.2 OBJ: 6.2.f Describe the consequences of a failure to repair damaged DNA. MSC: Applying 18. You have made a collection of mutant fruit flies that are defective in various aspects of DNA repair. You test each mutant for its hypersensitivity to three DNA-damaging agents: sunlight, nitrous acid (which causes deamination of cytosine), and formic acid (which causes depurination). The results are summarized in Table 6-61, where a “yes”

indicates that the mutant is more sensitive than a normal fly, and blanks indicate normal sensitivity.

Table 6-61

A. Which mutant is most likely to be defective in the DNA repair polymerase? Why? B. What aspect of repair is most likely to be affected in the other mutants?

ANS: A. Mr. Self-Destruct is more likely than the other mutants to be defective in the DNA repair polymerase because Mr. Self-Destruct is defective in the repair of all three kinds of DNA damage. The repair pathways for all three kinds of damage are similar in the later steps, including a requirement for the DNA repair polymerase.

B. The other mutants are specific for a particular type of damage. Thus, the mutations are likely to be in genes required for the first stage of repair—the recognition and excision of the damaged bases. Dracula and Mole are likely to be defective in the recognition or excision of thymidine dimers; Faust is likely to be defective in the recognition or excision of U-G mismatched base pairs; and Marguerite is likely to be defective in the recognition or excision of abasic sites.

DIF: Moderate REF: 6.2 OBJ: 6.2.a List the causes of DNA damage. | 6.2.f Describe the consequences of a failure to repair damaged DNA. MSC: Applying 19. The deamination of cytosine generates a uracil base. This is a naturally occurring nucleic acid base, and so does not represent a DNA lesion caused by damage due to chemicals or irradiation. Why is this base recognized as “foreign” and why is it important for cells to have a mechanism to recognize and remove uracil when it is found in the DNA duplex?

ANS: Uracil is an RNA base and it is recognized as a mutational lesion because, as it is formed from the deamination of cytosine, it will be paired with a guanine in the context of the DNA duplex. Uracil pairs by forming two hydrogen bonds, similar to thymine, and is thus a poor partner for guanine, which forms three hydrogen bonds with c ytosine. The mismatch causes a distortion of the DNA backbone, allowing the repair machinery to recognize the uracil as a lesion. Because uracil pairs preferably with adenine (its partner in double-stranded RNA), the deamination of cytosine to uracil is highly mutagenic. If unrepaired, it can result in the transition of a C-G base pair to a T-A base pair.

DIF: Moderate REF: 6.2 OBJ: 6.2.f Describe the consequences of a failure to repair damaged DNA. MSC: Evaluating 20. Figure 6-63 shows a replication bubble.

Figure 6-63

A. On the figure, indicate where the origin of replication was located (use O).

B. Label the leading-strand template and the lagging-strand template of the right-hand fork [R] as X and Y, respectively.

C. Indicate by arrows the direction in which the newly made DNA strands (indicated by dark lines) were synthesized.

D. Number the Okazaki fragments on each strand as 1, 2, and 3 in the order in which they were synthesized.

E. Indicate where the most recent DNA synthesis has occurred (use S).

F. Indicate the direction of movement of the replication forks with arrows.

ANS: See Figure 6-63A.

Figure 6-63A

DIF: Difficult REF: 6.1 OBJ: 6.1.a Explain how a DNA double helix provides a template for its own replication and describe the resulting daughter helices in terms of their sequence and their inclusion of parental and newly synthesized DNA strands. | 6.1.d Compare the direction in which replication forks move from the origin of replication with the direction in which DNA synthesis proceeds. | MSC: Applying 21. Use the components in the list below to label the diagram of a replication fork in Figure 6-64.

A. DNA polymerase B. single-strand binding protein C. Okazaki fragment D. primase E. sliding clamp F. RNA primer G. DNA helicase

Figure 6-64

22. See Figure 6-65A.

Figure 6-65A

DIF: Easy REF: 6.1 OBJ: 6.1.i Describe the primers required for DNA replication and compare how they are used in synthesizing the leading and lagging strands. | 6.1.j Explain how primers are removed and replaced to produce a continuous newly synthesized DNA strand. | 6.1.k Name five proteins that form part of the replication machine and state the role each plays in DNA replication. MSC: Remembering 23. Homologous recombination is initiated by double-strand breaks (DSBs) in a chromosome. DSBs arise from DNA damage caused by harmful chemicals or by radiation (for example, X-rays). During meiosis, the specialized cell division that produces gametes (sperm and eggs) for sexual reproduction, the cells intentionally cause DSBs so as to stimulate crossover homologous recombination. If there is not at least one occurrence of crossing-over within each pair of homologous chromosomes during meiosis, those noncrossover chromosomes will not segregate properly.

Figure 6-66

A. Consider the copy of Chromosome 3 that you received from your mother. Is it identical to the Chromosome 3 that she received from her mother (her maternal chromosome) or identical to the Chromosome 3 she received from her father (her paternal chromosome), or neither? Explain.

B. Starting with the representation in Figure 6-66 of the double-stranded maternal and paternal chromosomes found in your mother, draw two possible chromosomes you may have received from your mother.

C. What does this indicate about your resemblance to your grandfather and grandmother?

ANS: A. Neither. The copy of Chromosome 3 you received from your mother is a hybrid of the ones she received from her mother and her father.

B. See Figure 6-66A. The correct answers include any chromosome in which a portion matches the information from the paternal chromosome and the remainder matches the information from the maternal chromosome.

Figure 6-66A

C. As a result of extensive crossing-over, you resemble both your grandmother and your grandfather. If there were no crossing-over, you might have a much stronger resemblance to one than the other.

DIF: Difficult REF: 6.2 OBJ: 6.2.e Contrast nonhomologous end-joining and homologous recombination as mechanisms for repairing double-strand DNA breaks. MSC: Creating 24. The synthesis of DNA in living systems occurs in the 5′-to-3′ direction. However, scientists synthesize short DNA sequences needed for their experiments on an instrument dedicated to this task.

A. The chemical synthesis of DNA by this instrument proceeds in the 3′-to-5′ direction. Draw a diagram to show how

this is possible and explain the process.

B. Although 3′-to-5′ synthesis of DNA is chemically possible, it does not occur in living systems. Why not?

ANS: A. The actual chemical reaction in DNA synthesis is the same regardless of whether going in the 5 ′-to-3′ or in the 3′-to-5′ direction. The most important distinction between these two options is that if DNA is synthesized in the 3′to-5′ direction, the 5′ end of the elongating strand, rather than the 3′ end, will have a nucleoside triphosphate.

Figure 6-67

B. DNA synthesis from 3′ to 5′ does not allow proofreading. If the last nucleotide added is mispaired and is removed, the last nucleotide on the growing strand is a nucleoside monophosphate and the nucleotide coming in only has a hydroxyl group on the 3′ end. Thus, there is no favorable hydrolysis reaction to drive the addition of new nucleotides.

DIF: Difficult REF: 6.1 OBJ: 6.1.h Explain how DNA polymerase maintains the accuracy of DNA replication.

MSC: Creating

CHAPTER 7 From DNA to Protein

FROM DNA TO RNA 7.1.a Recall the central dogma and explain how its steps relate to gene expression depending on whether the final product of the gene is an RNA or a protein.

7.1.b Explain how cells can produce large quantities of one protein and tiny quantities of another.

7.1.c Compare RNA and DNA in terms of chemical composition, base-pairing properties, and overall structure.

7.1.d Compare the reactions catalyzed by RNA and DNA polymerases in terms of templates, substrates, directionality, and sources of energy to drive the reactions.

7.1.e Explain why RNA and DNA polymerases differ in their fidelity.

7.1.f Explain how the behavior of RNA polymerase allows multiple RNA transcripts to be made from a single gene at once.

7.1.g List the most common types of RNA produced by transcription and identify those that represent the final product of gene expression.

7.1.h Explain how a bacterial RNA polymerase recognizes where transcription will begin, which DNA strand to transcribe, and when to stop.

7.1.i Contrast transcription initiation in bacteria and eukaryotes.

7.1.j Explain how the eukaryotic general transcription factors assemble on a promoter, form a transcription initiation complex, and release RNA polymerase to begin transcription.

7.1.k Contrast the structures of bacterial and eukaryotic mRNAs and compare how these transcripts are handled as they are being synthesized.

7.1.l Describe how RNA splicing is carried out largely by RNA molecules.

7.1.m State the potential benefits of the presence of introns in eukaryotic genes.

7.1.n Describe the assembly of the molecular aggregates that act as “factories” for the synthesis and processing of RNA. 7.1.o Explain how cells control the quality of the RNAs that are exported from the nucleus.

7.1.p Describe how cells control the lifetime of an mRNA molecule.

FROM RNA TO PROTEIN 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence.

7.2.b Explain how a ribosome determines which reading frame in an mRNA specifies a protein, and describe the consequences of translating an incorrect reading frame.

7.2.c Explain how wobble base-pairing underlies the redundancy in the genetic code.

7.2.d Explain how each tRNA is charged with the correct amino acid, and describe the consequences of a tRNA carrying an incorrect amino acid.

7.2.e Explain how investigators used synthetic polynucleotides and trinucleotides to decipher the genetic code.

7.2.f Delineate the roles the small and large ribosomal subunits play in translation.

7.2.g Contrast the roles that ribosomal proteins and ribosomal RNAs play in the structure and activity of the ribosome.

7.2.h Summarize the cycle by which amino acids are covalently linked to a growing polypeptide chain.

7.2.i State the directionality of replication, of transcription, and of translation.

7.2.j Contrast translation initiation in eukaryotes and bacteria.

7.2.k Describe how translation is terminated in eukaryotes and in bacteria.

7.2.l Explain the benefits of polyribosomes in protein translation.

7.2.m

Explain why antibiotics that interfere with the synthesis of RNA or proteins eliminate bacterial infections without

harming the patient.

7.2.n Explain how and why proteins are targeted for degradation in proteasomes.

7.2.o Describe the structure of proteasomes and summarize how these molecular chambers destroy proteins marked for elimination.

7.2.p Outline the steps at which the final concentrations of a functional protein can be regulated by the cell.

RNA AND THE ORIGINS OF LIFE 7.3.a Describe the properties of RNA that allow it to store genetic information and, potentially, to catalyze its own synthesis.

7.3.b Explain why RNA is thought to predate DNA in evolution.

7.3.c Explain the features that make DNA a better molecule for the permanent storage of genetic information than RNA.

MULTIPLE CHOICE 1. Consider two genes that are next to each other on a chromosome, as arranged in Figure 7-1.

Figure 7-1

Which of the following statements is TRUE? a. The two genes must be transcribed into RNA using the same strand of DNA. b. If gene A is transcribed in a cell, gene B cannot be transcribed. c. Gene A and gene B can be transcribed at different rates, producing different amounts of RNA within the same cell. d. If gene A is transcribed in a cell, gene B must be transcribed.

ANS: C DIF: Moderate REF: 7.1 OBJ: 7.1.a Recall the central dogma and explain how its steps relate to gene expres-

sion depending on whether the final product of the gene is an RNA or a protein. 7.1.b Explain how cells can produce large quantities of one protein and tiny quantities of another. MSC: Applying 2. RNA in cells differs from DNA in that a. it contains the base uracil, which pairs with cytosine. b. it is single-stranded and cannot form base pairs. c. it is single-stranded and can fold up into a variety of structures. d. the sugar ribose contains fewer oxygen atoms than does deoxyribose.

ANS: C Although RNA contains uracil, uracil pairs with adenine, not cytosine. RNA can form base pairs with a complementary RNA or DNA sequence. Ribose contains one more oxygen atom than deoxyribose.

DIF: Easy REF: 7.1 OBJ: 7.1.c Compare RNA and DNA in terms of chemical composition, base-pairing properties, and overall structure. MSC: Analyzing 3. Figure 7-3 shows a ribose sugar. RNA bases are added to the part of the ribose sugar pointed to by arrow

Figure 7-3

a. 3. b. 4. c. 5. d. 6.

ANS: A DIF: Easy REF: 7.1 OBJ: 7.1.c Compare RNA and DNA in terms of chemical composition, base-pairing properties, and overall structure. MSC: Remembering 4. Figure 7-3 shows a ribose sugar. The part of the ribose sugar that is different from the deoxyribose sugar used in DNA is pointed to by arrow

Figure 7-3

a. 1. b. 4. c. 5. d. 6.

ANS: C DIF: Easy REF: 7.1 OBJ: 7.1.c Compare RNA and DNA in terms of chemical composition, base-pairing properties, and overall structure. MSC: Remembering 5. Figure 7-3 shows a ribose sugar. The part of the ribose sugar where a new ribonucleotide will attach in an RNA molecule is pointed to by arrow

Figure 7-3

a. 1. b. 3. c. 4. d. 5.

ANS: C DIF: Easy REF: 7.1 OBJ: 7.1.c Compare RNA and DNA in terms of chemical composition, base-pairing properties, and overall structure. MSC: Remembering 6. Transcription is similar to DNA replication in that a. an RNA transcript is synthesized discontinuously and the pieces are then joined together. b. it uses the same enzyme as that used to synthesize RNA primers during DNA replication. c. the newly synthesized RNA remains paired to the template DNA.

d. nucleotide polymerization occurs only in the 5′-to-3′ direction.

ANS: D Both RNA and DNA molecules are made in a 5′-to-3′ direction. An RNA transcript is made by a single polymerase molecule that proceeds from the start site to the termination site without falling off. The enzyme used to make primers during DNA synthesis is indeed an RNA polymerase, but it is a special enzyme, primase, and not the enzyme that is used for transcription.

DIF: Easy REF: 7.1 OBJ: 7.1.d Compare the reactions catalyzed by RNA and DNA polymerases in terms of templating, substrate, directionality, and source of energy to drive the reactions. MSC: Understanding 7. Which of the following statements is FALSE? a. A new RNA molecule can begin to be synthesized from a gene before the previous RNA molecule’s synthesis is completed. b. If two genes are to be expressed in a cell, these two genes can be transcribed with different efficiencies. c. RNA polymerase is responsible for both unwinding the DNA helix and catalyzing the formation of the phosphodiester bonds between nucleotides. d. Unlike DNA, RNA uses a uracil base and a deoxyribose sugar.

ANS: D RNA nucleotides contain the sugar ribose.

DIF: Easy REF: 7.1 OBJ: 7.1.a Recall the central dogma and explain how its steps relate to gene expression depending on whether the final product of the gene is an RNA or a protein. | 7.1.c Compare RNA and DNA in terms of chemical composition, base-pairing properties, and overall structure. | 7.1.f Explain how the behavior of RNA polymerase allows multiple RNA transcripts to be made from a single gene at once. MSC: Understanding 8. Unlike DNA, which typically forms a helical structure, different molecules of RNA can fold into a variety of threedimensional shapes. This is largely because RNA a. contains uracil and uses ribose as the sugar. b. bases cannot form hydrogen bonds with each other. c. nucleotides use a different chemical linkage between nucleotides compared to DNA. d. is single-stranded.

ANS: D Although RNA contains uracil and uses ribose as the sugar, this is not the main reason why different RNA molecules can form different three-dimensional structures (although ribose does increase potential hydrogen-bonding potentials compared to deoxyribose).

DIF: Easy REF: 7.1 OBJ: 7.1.c Compare RNA and DNA in terms of chemical composition, base-pairing properties, and overall structure. MSC: Understanding 9. Which of the following molecules of RNA would you predict to be the most likely to fold into a specific structure as a result of intramolecular base-pairing? a. 5′-CCCUAAAAAAAAAAAAAAAAUUUUUUUUUUUUUUUUAGGG-3′ b. 5′-UGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUG-3′ c. 5′-AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA-3′ d. 5′-GGAAAAGGAGAUGGGCAAGGGGAAAAGGAGAUGGGCAAGG-3′

ANS: A In order to base pair, intrastrand complementarity is required (C-G pairs and U-A pairs). More complementarity will lead to a greater likelihood for the formation of a specific structure. Without complementarity, base pairing cannot occur.

DIF: Moderate REF: 7.1 OBJ: 7.1.c Compare RNA and DNA in terms of chemical composition, base-pairing properties, and overall structure. MSC: Applying 10. Which of the following statements is FALSE? a. RNA polymerase can start making a new RNA molecule without a primer; DNA polymerase cannot. b. RNA polymerase does not proofread its work; DNA polymerase does. c. RNA polymerase catalyzes the linkage of ribonucleotides while DNA polymerase catalyzes the linkage of deoxyribonucleotides. d. RNA polymerase adds bases in a 3′-to-5′ direction; DNA polymerase adds bases in a 5′-to-3′ direction.

ANS: D Both RNA and DNA polymerases add bases in a 5′-to-3′ direction.

DIF: Easy REF: 7.1 OBJ: 7.1.d Compare the reactions catalyzed by RNA and DNA polymerases in terms of templating, substrate, directionality, and source of energy to drive the reactions. MSC: Understanding 11. Which one of the following is the main reason that a typical eukaryotic gene is able to respond to a far greater variety of regulatory signals than a typical prokaryotic gene or operon? a. Eukaryotes have three types of RNA polymerase. b. Eukaryotic RNA polymerases require general transcription factors. c. The transcription of a eukaryotic gene can be influenced by proteins that bind far from the promoter. d. Prokaryotic genes are packaged into nucleosomes.

ANS: C DIF: Easy REF: 7.1 OBJ: 7.1.i Contrast transcription initiation in bacteria and eukaryotes. MSC: Understanding 12. You have a piece of DNA that includes the following sequence: 5′-ATAGGCATTCGATCCGGATAGCAT-3′ 3′-TATCCGTAAGCTAGGCCTATCGTA-5′ Which of the following RNA molecules could be transcribed from this piece of DNA? a. 5′-UAUCCGUAAGCUAGGCCUAUGCUA-3′ b. 5′-AUAGGCAUUCGAUCCGGAUAGCAU-3′ c. 5′-UACGAUAGGCCUAGCUUACGGAUA-3′ d. none of these answers are correct.

ANS: B RNA molecules are synthesized from a 5′-to-3′ fashion and must have the correct polarity.

DIF: Difficult REF: 7.1 OBJ: 7.1.c Compare RNA and DNA in terms of chemical composition, base -pairing properties, and overall structure. | 7.1.d Compare the reactions catalyzed by RNA and DNA polymerases in terms of templating, substrate, directionality, and source of energy to drive the reactions. MSC: Applying 13. You have a segment of DNA that contains the following sequence:

5′-GGACTAGACAATAGGGACCTAGAGATTCCGAAA-3′ 3′-CCTGATCTGTTATCCCTGGATCTCTAAGGCTTT-5′ You know that the RNA transcribed from this segment contains the following sequence: 5′-GGACUAGACAAUAGGGACCUAGAGAUUCCGAAA–3′ Which of the following choices best describes how transcription occurs? a. The top strand is the template strand; RNA polymerase moves along this strand from 5′ to 3′. b. The top strand is the template strand; RNA polymerase moves along this strand from 3′ to 5′. c. The bottom strand is the template strand; RNA polymerase moves along this strand from 5′ to 3′. d. The bottom strand is the template strand; RNA polymerase moves along this strand from 3′ to 5′.

ANS: D The bottom strand can hybridize with the RNA molecule and thus is the template strand. The polymerase moves along the DNA in a 3′-to-5′ direction, because the RNA nucleotides are joined in a 5′-to-3′ polarity.

DIF: Difficult REF: 7.1 OBJ: 7.1.c Compare RNA and DNA in terms of chemical composition, base -pairing properties, and overall structure. | 7.1.d Compare the reactions catalyzed by RNA and DNA polymerases in terms of templating, substrate, directionality, and source of energy to drive the reactions. MSC: Applying 14. The sigma subunit of bacterial RNA polymerase a. contains the catalytic activity of the polymerase. b. remains part of the polymerase throughout transcription. c. recognizes promoter sites in the DNA. d. recognizes transcription termination sites in the DNA. ANS: C DIF: Easy REF: 7.1 OBJ: 7.1.h Explain how a bacterial RNA polymerase recognizes where transcription will begin, which DNA strand to transcribe, and when to stop. MSC: Remembering 15. Which of the following might decrease the transcription of only one specific gene in a bacterial cell? a. a decrease in the amount of sigma factor b. a decrease in the amount of RNA polymerase c. a mutation that introduced a stop codon into the DNA that precedes the gene’s coding sequence d. a mutation that introduced extensive sequence changes into the DNA that precedes the gene’s transcription start site ANS: D Changes in the DNA in the region that precedes the gene’s transcription site would probably destroy the function of the promoter, making RNA polymerase unable to bind to it. Decreasing the amount of sigma factor or RNA polymerase would affect the transcription of most of the genes in the cell, not just one specific gene. Introducing a stop codon before the coding sequence would have no effect on transcription of the gene, because the transcription machinery does not recognize translational stops.

DIF: Easy REF: 7.1 OBJ: 7.1.h Explain how a bacterial RNA polymerase recognizes where transcription will begin, which DNA strand to transcribe, and when to stop. MSC: Applying 16. There are several reasons why the primase used to make the RNA primer for DNA replication is not suitable for gene transcription. Which of the statements below is NOT one of those reasons?

a. Primase initiates RNA synthesis on a single-stranded DNA template. b. Primase can initiate RNA synthesis without the need for a base-paired primer. c. Primase synthesizes only RNAs of about 5–20 nucleotides in length. d. The RNA synthesized by primase remains base-paired to the DNA template.

ANS: B Both primase and RNA polymerase can initiate RNA synthesis without a base-paired primer, so it does not describe why primase cannot be used for gene transcription.

DIF: Moderate REF: 7.1 OBJ: 7.1.d Compare the reactions catalyzed by RNA and DNA polymerases in terms of templating, substrate, directionality, and source of energy to drive the reactions. MSC: Analyzing 17. You have a bacterial strain with a mutation that removes the transcription termination signal from the Abd operon.

Which of the following statements describes the most likely effect of this mutation on Abd transcription? a. The Abd RNA will not be produced in the mutant strain. b. The Abd RNA from the mutant strain will be longer than normal. c. Sigma factor will not dissociate from RNA polymerase when the Abd operon is being transcribed in the mutant strain. d. RNA polymerase will move in a backward fashion at the Abd operon in the mutant strain.

ANS: B With a mutation in the termination signal, the Abd RNA from the mutant strain will be longer. Without the termination signal, the polymerase will not halt and release from the DNA template at the normal location when transcribing the Abd operon. Most probably, the polymerase will continue to transcribe RNA until it reaches a sequence in the DNA that can serve as a termination sequence, either from the next downstream operon or in the intervening sequence between the Abd operon and the next operon. Dissociation of sigma factor occurs once an approximately 10-nucleotide length of RNA has been synthesized by RNA polymerase, and it should not be affected by the lack of a termination signal.

DIF: Easy REF: 7.1 OBJ: 7.1.h Explain how a bacterial RNA polymerase recognizes where transcription will begin, which DNA strand to transcribe, and when to stop. MSC: Applying 18. Transcription in bacteria differs from transcription in a eukaryotic cell because a. RNA polymerase (along with its sigma subunit) can initiate transcription on its own. b. RNA polymerase (along with its sigma subunit) requires the general transcription factors to assemble at the promoter before polymerase can begin transcription. c. the sigma subunit must associate with the appropriate type of RNA polymerase to produce mRNAs. d. RNA polymerase must be phosphorylated at its C-terminal tail for transcription to proceed.

ANS: A Eukaryotic cells, but not bacteria, require general transcription factors. There is only a single type of RNA polymerase in bacterial cells. The general transcription factor TFIIH phosphorylates the C-terminal tail of RNA polymerase in eukaryotic cells but not in bacteria.

DIF: Easy REF: 7.1 OBJ: 7.1.i Contrast transcription initiation in bacteria and eukaryotes. MSC: Analyzing 19. Which of the following does not occur before a eukaryotic mRNA is exported from the nucleus? a. The ribosome binds to the mRNA.

b. The mRNA is polyadenylated at its 3′ end. c. A guanine nucleotide with a methyl group is added to the 5′ end of the mRNA. d. RNA polymerase dissociates.

ANS: A Ribosomes are in the cytosol and will bind to the mRNA once it has been exported from the nucleus.

DIF: Easy REF: 7.1 OBJ: 7.1.o Explain how cells control the quality of the RNAs that are exported from the nucleus.

MSC: Remembering 20. Total nucleic acids are extracted from a culture of yeast cells and are then mixed with resin beads to which the polynucleotide 5′-TTTTTTTTTTTTTTTTTTTTTTTTT-3′ has been covalently attached. After a short incubation, the beads are then extracted from the mixture. When you analyze the cellular nucleic acids that have stuck to the beads, which of the following is most abundant? a. DNA b. tRNA c. rRNA d. mRNA ANS: D mRNA is the only type of RNA that is polyadenylated, and its poly-A tail would be able to base-pair with the strands of poly T on the beads and thus stick to them. DNA would not be found in the sample, because the poly-A tail is not encoded in the DNA and long runs of T are rare in DNA. DIF: Moderate REF: 7.1 OBJ: 7.1.o Explain how cells control the quality of the RNAs that are exported from the nucleus. MSC: Applying 21. Which of the following statements about RNA splicing is FALSE? a. Conventional introns are not found in bacterial genes. b. For a gene to function properly, every exon must be removed from the primary transcript in the same fashion on every mRNA molecule produced from the same gene. c. Small RNA molecules in the nucleus perform the splicing reactions necessary for the removal of introns. d. Splicing occurs after the 5′ cap has been added to the end of the primary transcript.

ANS: B The primary transcript of a gene can sometimes be spliced differently so that different exons can be stitched together to produce distinct proteins in a process called alternative splicing.

DIF: Easy REF: 7.1 OBJ: 7.1.l Describe how RNA splicing is carried out largely by RNA molecules. | 7.1.o Explain how cells control the quality of the RNAs that are exported from the nucleus. MSC: Understanding 22. Genes in eukaryotic cells often have intronic sequences coded for within the DNA. These sequences are ultimately not translated into proteins. Why? a. Intronic sequences are removed from RNA molecules by the spliceosome, which works in the nucleus. b. Introns are not transcribed by RNA polymerase. c. Introns are removed by catalytic RNAs in the cytoplasm. d. The ribosome will skip over intron sequences when translating RNA into protein.

ANS: A DIF: Easy REF: 7.1 OBJ: 7.1.o Explain how cells control the quality of the RNAs that are exported from the

nucleus. MSC: Understanding 23. Which of the following statements is FALSE? a. Information determining the lifetimes of mRNA molecules is encoded in the nucleotide sequences of the mRNA molecule. b. An mRNA molecule in bacteria will typically have a shorter lifetime compared to an mRNA molecule in a eukaryotic cell. c. Eukaryotic mRNA molecules do not have lifetimes longer than 30 minutes. d. mRNA molecules are degraded into nucleotides by ribonucleases in the cytosol.

ANS: C There is a wide range of lifetimes for eukaryotic mRNAs. Abundant proteins tend to be translated from mRNAs with long lifetimes, which can be more than 10 hours!

DIF: Easy REF: 7.1 OBJ: 7.1.p Describe how cells control the lifetime of an mRNA molecule. MSC: Remembering 24. snRNAs a. are translated into snRNPs. b. are important for producing mature mRNA transcripts in bacteria. c. are removed by the spliceosome during RNA splicing. d. can bind to specific sequences at intron–exon boundaries through complementary base-pairing.

ANS: D snRNAs are part of the snRNPs, which include proteins and RNA molecules. The proteins within the snRNPs are encoded by their own genes and not translated from snRNPs. Bacteria do not have introns.

DIF: Easy REF: 7.1 OBJ: 7.1.n Describe the assembly of the molecular aggregates that act as “factories” for the synthesis and processing of RNA. MSC: Remembering 25. Which of the following statements about the genetic code is CORRECT? a. All codons specify more than one amino acid. b. The genetic code is redundant. c. All amino acids are specified by more than one codon. d. All codons specify an amino acid.

ANS: B Most amino acids can be specified by more than one codon. Each codon specifies only one amino acid. Tryptophan and methionine are encoded by only one codon. Some codons specify translational stop signals.

DIF: Easy REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.c Explain how wobble base-pairing accommodates a redundant genetic code.

MSC: Remembering 26. The ribosome is important for catalyzing the formation of peptide bonds. Which of the following statements is TRUE? a. The number of rRNA molecules that make up a ribosome greatly exceeds the number of protein molecules found in the ribosome. b. The large subunit of the ribosome is important for binding to the mRNA. c. The catalytic site for peptide bond formation is formed primarily from an rRNA. d. Once the large and small subunits of the ribosome assemble, they will not separate from each other until degraded

by the proteasome.

ANS: C A ribosome is built from many more proteins than rRNA molecules, although the ribosome is about two-thirds RNA and one-third protein by weight. The small subunit binds to mRNA. The assembly and disassembly of the small subunit with the large subunit occurs every time a protein is produced from an mRNA. When release factor binds to an mRNA, the ribosome will release the mRNA and dissociate into its two subunits, to be recycled during another round of protein synthesis.

DIF: Moderate REF: 7.2 OBJ: 7.2.g Contrast the roles that ribosomal proteins and ribosomal RNAs play in the conformation and activity of the ribosome. MSC: Understanding 27. Which of the following statements is TRUE? a. Ribosomes are large RNA structures composed solely of rRNA. b. Ribosomes are synthesized entirely in the cytoplasm. c. rRNA contains the catalytic activity that joins amino acids together. d. A ribosome binds one tRNA at a time.

ANS: C Ribosomes contain proteins as well as rRNA. rRNA is synthesized in the nucleus, and ribosomes are partly assembled in the nucleus. A ribosome must be able to bind two tRNAs at any one time.

DIF: Easy REF: 7.2 OBJ: 7.2.g Contrast the roles that ribosomal proteins and ribosomal RNAs play in the conformation and activity of the ribosome. MSC: Understanding 28. You have discovered a gene (Figure 7-28A) that is alternatively spliced to produce several forms of mRNA in various cell types, three of which are shown in Figure 7-28B. The lines connecting the exons that are included in the mRNA indicate the splicing. From your experiments, you know that protein translation begins in exon 1. For all forms of the mRNA, the encoded protein sequence is the same in the regions of the mRNA that correspond to exons 1 and 10.

Exons 2 and 3 are alternative exons used in different mRNA, as are exons 7 and 8. Which of the following statements about exons 2 and 3 is the most accurate?

Figure 7-28

a. Exons 2 and 3 must have the same number of nucleotides. b. Exons 2 and 3 must contain an integral number of codons (that is, the number of nucleotides divided by 3 must be an integer). c. Exons 2 and 3 must contain a number of nucleotides that when divided by 3, leaves the same remainder (that is, 0, 1, or 2). d. Exons 2 and 3 must have different numbers of nucleotides.

ANS: C Exons 2 and 3 must have a number of nucleotides that when divided by 3, leaves the same remainder. Although the other choices could be true, they do not have to be. Because the protein sequence is the same in segments of the mRNA corresponding to exons 1 and 10, the choice of either exon 2 or exon 3 would not alter the reading frame. To maintain the normal reading frame, whatever it is, the alternative exons must have a number of nucleotides that when divided by 3 (the number of nucleotides in a codon) gives the same remainder.

DIF: Difficult REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.b Explain how a ribosome determines which reading frame in an mRNA specifies a protein, and describe the consequences of translating an incorrect reading frame MSC: Applying 29. The piece of RNA below includes the region that codes for the binding site for the initiator tRNA needed in translation.

5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′

Table 7-29

Which amino acid will be on the tRNA that is the first to bind to the A site of the ribosome? a. methionine b. arginine

c. cysteine d. valine ANS: B The initiator methionine is underlined on the RNA molecule below.

5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′ The first tRNA to bind at the A site is the second codon of the protein, because the initiator tRNA is already bound to the P site when translation begins. The codon that follows the binding site for the initiator tRNA is CGU, which codes for arginine.

DIF: Moderate REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.b Explain how a ribosome determines which reading frame in an mRNA contains a protein-coding sequence, and describe the consequences of translating an incorrect reading frame. | 7.2.d Explain how each tRNA is charged with the correct amino acid, and describe the consequences of a tRNA carrying an incorrect amino acid. MSC: Applying 30. Which amino acid would you expect a tRNA with the anticodon 5′-CUU-3′ to carry?

Table 7-29

a. lysine b. glutamic acid c. leucine d. phenylalanine ANS: A As is conventional for nucleotide sequences, the anticodon is given reading from 5 ′ to 3′. The complementary basepairing occurs between antiparallel nucleic acid sequences, and the codon recognized by this anticodon will ther efore be 5′-AAG-3′.

DIF: Moderate REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.d Explain how each tRNA is charged with the correct amino acid, and describe the consequences of a tRNA carrying an incorrect amino acid. MSC: Applying

31. Which of the following pairs of codons might you expect to be read by the same tRNA as a result of wobble?

Table 7-29

a. CUU and UUU b. GAU and GAA c. CAC and CAU d. AAU and AGU ANS: C These two codons differ only in the third position and also encode the same amino acid, which is the definition of wobble. Although the codons GAU and GAA also differ only in the third position, they are unlikely in normal circumstances to be read by the same tRNA, because they encode different amino acids.

DIF: Easy REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.c Explain how wobble base-pairing accommodates a redundant genetic code. | 7.2.d Explain how each tRNA is charged with the correct amino acid, and describe the consequences of a tRNA carrying an incorrect amino acid. MSC: Understanding 32. Below is the sequence from the 3′ end of an mRNA. 5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′

Table 7-29

If you were told that this sequence contains the stop codon for the protein encoded by this mRNA, what is the anticodon on the tRNA in the P site of the ribosome when release factor binds to the A site? a. 5′-CCA-3′ b. 5′-CCG-3′ c. 5′-UGG-3′ d. 5′-UUA-3′ ANS: A The stop codon (UAA) is underlined in the mRNA sequence below; this is the only stop codon on this piece of mRNA. The codon (UGG) preceding the stop codon will be binding to a tRNA in the P site of the ribosome when release factor binds to the A site. The anticodon of the tRNA will bind to the codon UGG and will be CCA. 5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′ DIF: Difficult REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.d Explain how each tRNA is charged with the correct amino acid, and describe the consequences of a tRNA carrying an incorrect amino acid. MSC: Applying 33. A strain of yeast translates mRNA into protein inaccurately. Individual molecules of a particular protein isolated from this yeast have variations in the first 11 amino acids compared with the sequence of the same protein isolated from normal yeast cells, as listed in Figure 7-33. What is the most likely cause of this variation in protein sequence?

Figure 7-33

Table 7-29

a. a mutation in the DNA coding for the protein b. a mutation in the anticodon of the isoleucine-tRNA (tRNAIle) c. a mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between different amino acids d. a mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between different tRNA molecules ANS: C A mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between amino acids would allow an assortment of amino acids to be attached to the tRNAIle. These assorted aminoacyl-tRNAs would then base-pair with the isoleucine codon and cause a variety of substitutions at positions normally occupied by isoleucine. A mutation in the gene encoding the protein would cause only a single variant protein to be made. A mutation in the anticodon loop of tRNAIle or a mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between different tRNA molecules would cause the substitution of isoleucine for some other amino acid (which is the opposite of what is observed).

DIF: Moderate REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.d Explain how each tRNA is charged with the correct amino acid, and describe

the consequences of a tRNA carrying an incorrect amino acid. MSC: Applying 34. A mutation in the tRNA for the amino acid lysine results in the anticodon sequence 5′-UAU-3′ (instead of 5′-UUU-3′).

Which of the following aberrations in protein synthesis might this tRNA cause?

Table 7-29

a. read-through of stop codons b. substitution of lysine for isoleucine c. substitution of lysine for tyrosine d. substitution of lysine for phenylalanine ANS: B The mutant tRNALys will be able to pair with the codon 5′-AUA-3′, which codes for isoleucine.

DIF: Moderate REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.d Explain how each tRNA is charged with the correct amino acid, and describe the consequences of a tRNA carrying an incorrect amino acid. MSC: Applying 35. What do you predict would happen if you created a tRNA with an anticodon of 5′-CAA-3′ that is charged with methionine, and added this modified tRNA to a cell-free translation system that has all the normal components required for translating RNAs?

Table 7-29

a. Methionine would be incorporated into proteins at some positions where glutamine should be. b. Methionine would be incorporated into proteins at some positions where leucine should be. c. Methionine would be incorporated into proteins at some positions where valine should be. d. Translation would no longer be able to initiate.

ANS: B The 5′-CAA-3′ anticodon binds to the 5′-UUG-3′ codon, which normally codes for leucine.

DIF: Difficult REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.d Explain how each tRNA is charged with the correct amino acid, and describe the consequences of a tRNA carrying an incorrect amino acid. MSC: Applying 36. A poison added to an in vitro translation mixture containing mRNA molecules with the sequence 5′AUGAAAAAAAAAAAAUAA-3′ has the following effect: the only product made is a Met-Lys dipeptide that remains attached to the ribosome. What is the most likely way in which the poison acts to inhibit protein synthesis? a. It inhibits peptidyl transferase activity. b. It inhibits movement of the small subunit relative to the large subunit. c. It inhibits release factor. d. It mimics release factor.

ANS: B Inhibition of peptidyl transferase activity would prevent all peptide bond formation. Inhibiting release factor would have no effect on translation until the stop codon was reached. Mimicking release factor would be likely to result in a mixture of polypeptides of various lengths; a poison mimicking a release factor could conceivably cause only Met-Lys to be made, but this dipeptide would not remain bound to the ribosome.

DIF: Moderate REF: 7.2 OBJ: 7.2.f Delineate the roles the small and large ribosomal subunits play in translation.

MSC: Applying 37. In eukaryotes, but not in prokaryotes, ribosomes find the start site of translation by a. binding directly to a ribosome-binding site preceding the initiation codon. b. scanning along the mRNA from the 5′ end.

c. recognizing an AUG codon as the start of translation. d. binding an initiator tRNA. ANS: B In prokaryotes, ribosomes will bind directly to a ribosome-binding site preceding the initiation codon while eukaryotes will scan along the mRNA from the 5′ end to find the translational start site. In both prokaryotes and eukaryotes, the AUG codon must be recognized by the ribosome, which leads to the binding of an initiator tRNA. DIF: Easy REF: 7.2 OBJ: 7.2.j Contrast the translation initiation in eukaryotes and in bacteria.

MSC: Remembering 38. Which of the following statements about prokaryotic mRNA molecules is FALSE? a. A single prokaryotic mRNA molecule can be translated into several proteins. b. Ribosomes must bind to the 5′ cap before initiating translation. c. mRNAs are not polyadenylated. d. Ribosomes can start translating an mRNA molecule before transcription is complete.

ANS: B Bacterial mRNAs do not have 5′ caps. Instead, ribosome-binding sites upstream of the start codon tell the ribosome where to begin searching for the start of translation.

DIF: Easy REF: 7.2 OBJ: 7.2.j Contrast the translation initiation in eukaryotes and in bacteria.

MSC: Understanding 39. You have discovered a protein that inhibits translation. When you add this inhibitor to a mixture capable of translating human mRNA and centrifuge the mixture to separate polyribosomes and single ribosomes, you obtain the results shown in Figure 7-39. Which of the following interpretations is consistent with these observations?

Figure 7-39

a. The protein binds to the small ribosomal subunit and increases the rate of initiation of translation. b. The protein binds to sequences in the 5′ region of the mRNA and inhibits the rate of initiation of translation. c. The protein binds to the large ribosomal subunit and slows down elongation of the polypeptide chain. d. The protein binds to sequences in the 3′ region of the mRNA and prevents termination of translation.

ANS: B The results in Figure 7-39 show a marked decrease in the number of polyribosomes formed relative to normal. Poly-

ribosomes form because the initiation of translation is fairly rapid: ribosomes can bind successively to the free 5′ end of an mRNA molecule and start translation before the first ribosome has had a chance to finish translating the message.

Therefore, inhibition of the rate of initiation will tend to decrease the number of ribosomes in the polyribosome, and in the extreme case there will be only one ribosome per mRNA. Conversely, increasing the rate of initiation or slowing the rate of elongation would result in an increased number of ribosomes per polyribosome (up to a maximum point). Preventing termination would prevent release of the ribosomes at the end of the coding sequence and would be expected to “freeze” the assembled polyribosomes, so that the ratio of polyribosomes to ribosomes would be much as normal.

DIF: Difficult REF: 7.2 OBJ: 7.2.l Explain the benefits of involving polyribosomes in protein translation.

MSC: Applying 40. The concentration of a particular protein, X, in a normal human cell rises gradually from a low point, immediately after cell division, to a high point, just before cell division, and then drops sharply. The level of its mRNA in the cell remains fairly constant throughout this time. Protein X is required for cell growth and survival, but the drop in its level just before cell division is essential for division to proceed. You have isolated a line of human cells that grow in size in culture but cannot divide, and on analyzing these mutants, you find that levels of X mRNA in the mutant cells are normal. Which of the following mutations in the gene for X could explain these results? a. the introduction of a stop codon that truncates protein X at the fourth amino acid b. a change of the first ATG codon to CCA c. the deletion of a sequence that encodes sites at which ubiquitin can be attached to the protein d. a change at a splice site that prevents splicing of the RNA ANS: C The decrease in the level of protein X in the normal cell is most probably due to protein degradation, because levels of mRNA remain constant. The inability of the mutant cell to divide could be due to a mutation that inhibits protein degradation. This would be achieved by the removal of sites for attachment of ubiquitin, which targets proteins for destruction. The other choices would probably not produce the results described, because without the production of a functional protein X, the mutant cells could not grow in size.

DIF: Moderate REF: 7.2 OBJ: 7.2.n Explain how and why proteins are targeted for degradation in proteasomes.

MSC: Understanding 41. Which of the following methods is not used by cells to regulate the amount of a protein in the cell? a. Genes can be transcribed into mRNA with different efficiencies. b. Many ribosomes can bind to a single mRNA molecule. c. Proteins can be tagged with ubiquitin, marking them for degradation. d. Nuclear pore complexes can regulate the speed at which newly synthesized proteins are exported from the nucleus into the cytoplasm.

ANS: D Proteins are synthesized in the cytoplasm and therefore newly synthesized proteins would not be exported from the nucleus into the cytoplasm.

DIF: Easy REF: 7.2 OBJ: 7.2.n Explain how and why proteins are targeted for degradation in proteasomes.

MSC: Understanding 42. Which of the following statements about the proteasome is FALSE? a. Ubiquitin is a small protein that is covalently attached to proteins to mark them for delivery to the proteasome.

b. Proteases reside in the central cylinder of a proteasome. c. Misfolded proteins are delivered to the proteasome, where they are sequestered from the cytoplasm and can attempt to refold. d. The protein stoppers that surround the central cylinder of the proteasome use the energy from ATP hydrolysis to move proteins into the proteasome inner chamber.

ANS: C Once proteins are sent to the proteasome, proteases degrade them. Chaperone proteins provide a place for misfolded proteins to attempt to refold.

DIF: Easy REF: 7.2 OBJ: 7.2.o Describe the structure of proteasomes and summarize how these molecular chambers destroy proteins marked for elimination. MSC: Understanding 43. Which of the following molecules is thought to have arisen first during evolution? a. protein b. DNA c. RNA d. All came to be at the same time.

ANS: C Because RNA is known to catalyze reactions within the cell, because the components of RNA are thought to be more readily formed in the conditions on primitive Earth, and because RNA can contain genetic information, it is the most likely of the three molecules to have arisen first in evolution.

DIF: Easy REF: 7.3 OBJ: 7.3.b Explain why RNA is thought to predate DNA in evolution. MSC: Remembering 44. According to current thinking, the minimum requirement for life to have originated on Earth was the formation of a a. molecule that could provide a template for the production of a complementary molecule. b. double-stranded DNA helix. c. molecule that could direct protein synthesis. d. molecule that could catalyze its own replication.

ANS: D Although providing a template is an important step in self-replication, it would not by itself be sufficient. Being a double-stranded DNA helix and being a molecule that could direct protein synthesis are likely stages in the evolution of the cell that must have succeeded the formation of the first self-replicating molecules.

DIF: Easy REF: 7.3 OBJ: 7.3.a Describe the properties possessed by RNA that allows the molecule to store information and, potentially, to catalyze its own synthesis. MSC: Understanding 45. Ribozymes are known to catalyze which of the following reactions in cells? a. DNA synthesis b. transcription c. RNA splicing d. protein hydrolysis ANS: C DIF: Easy REF: 7.1 OBJ: 7.1.n Describe the assembly of the molecular aggregates that act as “factories” for the synthesis and processing of RNA. MSC: Remembering 46. You are studying a disease that is caused by a virus, but when you purify the virus particles and analyze them you

find they contain no trace of DNA. Which of the following molecules are likely to contain the genetic information of the virus? a. high-energy phosphate groups b. RNA c. lipids d. carbohydrates ANS: B DIF: Easy REF: 7.3 OBJ: 7.3.a Describe the properties possessed by RNA that allows the molecule to store information and, potentially, to catalyze its own synthesis. MSC: Understanding 47. When using a repeating trinucleotide sequence (such as 5′-AAC-3′) in a cell-free translation system, you will obtain a. three different types of peptides, each made up of a single amino acid. b. peptides made up of three different amino acids in random order. c. peptides made up of three different amino acids, each alternating with each other in a repetitive fashion. d. polyasparagine, as the codon for asparagine is AAC. ANS: A An mRNA composed of a trinucleotide repeat of AAC can be “read” in three different frames: AAC, ACA, and CAA. Thus, this mRNA will yield polyasparagine (codon = AAC), polythreonine (codon = ACA), and polyglutamine (codon = CAA).

DIF: Moderate REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.b Explain how a ribosome determines which reading frame in an mRNA contains a protein-coding sequence, and describe the consequences of translating an incorrect reading frame. | 7.2.e Explain how investigators used synthetic polynucleotides and trinucleotides to decipher the genetic code.

MSC: Applying 48. You have discovered an alien life-form that surprisingly uses DNA as its genetic material, makes RNA from DNA, and reads the information from RNA to make protein using ribosomes and tRNAs, which read triplet codons. Because it is your job to decipher the genetic code for this alien, you synthesize some artificial RNA molecules and examine the protein products produced from these RNA molecules in a cell-free translation system using purified alien tRNAs and ribosomes. You obtain the results shown in Table 7-48.

Table 7-48

From this information, which of the following peptides can be produced from poly UAUC? a. Ile-Phe-Val-Tyr b. Tyr-Ser-Phe-Ala c. Ile-Lys-His-Tyr d. Cys-Pro-Lys-Ala ANS: D All other answers are not possible, because poly UAUC cannot code for Tyr. Tyr must be encoded by AUA, because both poly AUA and poly UA lead to the synthesis of Tyr. DIF: Difficult REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.b Explain how a ribosome determines which reading frame in an mRNA contains a protein-coding sequence, and describe the consequences of translating an incorrect reading frame. | 7.2.e Explain how investigators used synthetic polynucleotides and trinucleotides to decipher the genetic code.

MSC: Applying 49. An extraterrestrial organism (ET) is discovered whose basic cell biology seems pretty much the same as that of terrestrial organisms except that it uses a different genetic code to translate RNA into protein. You set out to break the code by translation experiments using RNAs of known sequence and cell-free extracts of ET cells to supply the necessary protein-synthesizing machinery. In experiments using the RNAs below, the following results were obtained when the 20 possible amino acids were added either singly or in different combinations of two or three: RNA 1: 5′-GCGCGCGCGCGCGCGCGCGCGCGCGCGC-3′ RNA 2: 5′-GCCGCCGCCGCCGCCGCCGCCGCCGCCGCC-3′ Using RNA 1, a polypeptide was produced only if alanine and valine were added to the reaction mixture. Using RNA 2, a polypeptide was produced only if leucine and serine and cysteine were added to the reaction mixture.

Assuming that protein synthesis can start anywhere on the template, that the ET genetic code is nonoverlapping and linear, and that each codon is the same length (like the terrestrial triplet code), how many nucle otides does an

ET codon contain? a. 2 b. 3 c. 4 d. 5 ANS: D An organism having codons with an even number of nucleotides (such as 2, 4, or 6) could read 5′-GCGCGCGCGC-3′ (RNA 1) in either of two ways, namely “GC GC GC GC . . .” or “CG CG CG CG . . .” Either of the two amino acids alone could have supported protein synthesis, so you would not need them in combination [thus eliminating 2, 4, and 6 as possible answers]. An organism having three bases per codon could read the sequence 5′GCCGCCGCCGCCGCC-3′ (RNA 2) in one of three ways, namely “GCC GCC GCC GCC . . . ,” “CCG CCG CCG CCG . . . ,” or “CGC CGC CGC CGC . . . ,” and so again, any one of the three amino acids could have supported synthesis of a polypeptide, and you would not need to add all three amino acids to produce a polypeptide chain, thus 3 could not be a possible answer. Only a five-nucleotide code gives you two different consecutive codons for RNA 1 and three different consecutive codons for RNA 2.

DIF: Difficult REF: 7.2 OBJ: 7.2.e Explain how investigators used synthetic polynucleotides and trinucleotides to decipher the genetic code. MSC: Applying 50. You have discovered an alien life-form that surprisingly uses DNA as its genetic material, makes RNA from DNA, and reads the information from RNA to make protein using ribosomes and tRNAs, which read triplet codons. Because it is your job to decipher the genetic code for this alien, you synthesize some artificial RNA molecules and examine the protein products produced from these RNA molecules in a cell-free translation system using purified alien tRNAs and ribosomes. You obtain the results shown in Table 7-48.

Table 7-48

From this information, which of the following peptides can be produced from poly UAUC? a. Ile-Phe-Val-Tyr b. Tyr-Ser-Phe-Ala

c. Ile-Lys-His-Tyr d. Cys-Pro-Lys-Ala ANS: D All other answers are not possible, because poly UAUC cannot code for Tyr. Tyr must be encoded by AUA, because both poly AUA and poly UA lead to the synthesis of Tyr (see Table 7-48A).

Table 7-48A

DIF: Difficult REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.b Explain how a ribosome determines which reading frame in an mRNA contains a protein-coding sequence, and describe the consequences of translating an incorrect reading frame. | 7.2.e Explain how investigators used synthetic polynucleotides and trinucleotides to decipher the genetic code.

MSC: Applying

MATCHING 51. Match the following names with their structures.

Figure 7-51

1. ANS: C DIF: Easy REF: 7.1 OBJ: 7.1.c Compare RNA and DNA in terms of chemical composition, base-pairing properties, and overall structure. MSC: Remembering 2. ANS: B DIF: Easy REF: 7.1 OBJ: 7.1.c Compare RNA and DNA in terms of chemical composition, base-pairing properties, and overall structure. MSC: Remembering 3. ANS: D DIF: Easy REF: 7.1 OBJ: 7.1.c Compare RNA and DNA in terms of chemical composition, base-pairing properties, and overall structure. MSC: Remembering 4. ANS: A DIF: Easy REF: 7.1 OBJ: 7.1.c Compare RNA and DNA in terms of chemical composition, base-pairing properties, and overall structure. MSC: Remembering 52. Figure 7-52 shows an mRNA molecule.

Figure 7-52

Match the labels given in the list below with the label lines in Figure 7-52.

A. ribosome-binding site B. initiator codon C. stop codon D. untranslated 3′ region E. untranslated 5′ region F. protein-coding region 1. ANS: E DIF: Easy REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.j Contrast the translation initiation in eukaryotes and in bacteria. | 7.2.k Describe how translation is terminated in eukaryotes and in bacteria. MSC: Understanding 2. ANS: B DIF: Easy REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information con-

tained in an mRNA into a protein sequence. | 7.2.j Contrast the translation initiation in eukaryotes and in bacteria. | 7.2.k Describe how translation is terminated in eukaryotes and in bacteria. MSC: Understanding 3. ANS: A DIF: Easy REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.j Contrast the translation initiation in eukaryotes and in bacteria. | 7.2.k Describe how translation is terminated in eukaryotes and in bacteria. MSC: Understanding 4. ANS: C DIF: Easy REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.j Contrast the translation initiation in eukaryotes and in bacteria. | 7.2.k Describe how translation is terminated in eukaryotes and in bacteria. MSC: Understanding 5. ANS: F DIF: Easy REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.j Contrast the translation initiation in eukaryotes and in bacteria. | 7.2.k Describe how translation is terminated in eukaryotes and in bacteria. MSC: Understanding 6. ANS: D DIF: Easy REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.j Contrast the translation initiation in eukaryotes and in bacteria. | 7.2.k Describe how translation is terminated in eukaryotes and in bacteria. MSC: Understanding

SHORT ANSWER 1. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.

4

gene

proteasome

20

Golgi

replisome

109

kinase

sugar-phosphate

128

nuclear pore

transcribed

cytoplasm

nucleus

transferase

exported

polymerase

translated

The instructions specified by the DNA will ultimately specify the sequence of proteins. This process involves DNA, made up of __________ different nucleotides, which gets __________ into RNA, which is then __________ into proteins, made up of __________ different amino acids. In eukaryotic cells, DNA gets made into RNA in the __________, while proteins are produced from RNA in the __________. The segment of DNA called a __________ is the portion that is copied into RNA; this process is catalyzed by RNA __________.

ANS: The instructions specified by the DNA will ultimately specify the sequence of proteins. This process involves DNA, made up of 4 different nucleotides, which gets transcribed into RNA, which is then translated into proteins, made up of 20 different amino acids. In eukaryotic cells, DNA gets made into RNA in the nucleus, while proteins are produced from RNA in the cytoplasm. The segment of DNA called a gene is the portion that is copied into RNA; this process is catalyzed by RNA polymerase.

DIF: Easy REF: 7.1 OBJ: 7.1.a Explain how the central dogma relates to gene expression. MSC: Understanding 2. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.

For a cell’s genetic material to be used, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called __________. Various kinds of RNA are produced, each with different functions. __________

molecules code for proteins, __________ molecules act as adaptors for protein synthesis, __________ molecules are integral components of the ribosome, and __________ molecules are important in the splicing of RNA transcripts. incorporation

rRNA

translation

mRNA

snRNA

transmembrane

pRNA

transcription

tRNA

proteins ANS: For a cell’s genetic material to be used, the information is first copied from the DNA into the nucleotide s equence of RNA in a process called transcription. Various kinds of RNA are produced, each with different functions. mRNA molecules code for proteins, tRNA molecules act as adaptors for protein synthesis, rRNA molecules are integral components of the ribosome, and snRNA molecules are important in the splicing of RNA transcripts.

DIF: Easy REF: 7.1 OBJ: 7.1.a Recall the central dogma and explain how its steps relate to gene expression d epending on whether the final product of the gene is an RNA or a protein. | 7.1.g List the most common types of RNA produced by transcription and identify those that represent the final product of gene expression.

MSC: Understanding 3. Match the following types of RNA with the main polymerase that transcribes them in eukaryotic cells.

Table 7-55

ANS: A—1; B—3; C—3; D—2; E—2 DIF: Easy REF: 7.1 OBJ: 7.1.g List the most common types of RNA produced by transcription and identify those that represent the final product of gene expression. MSC: Remembering 4. List three ways in which the process of eukaryotic transcription differs from the process of bacterial transcription.

ANS: Any three of the following are acceptable.

1. Bacterial cells contain a single RNA polymerase, whereas eukaryotic cells have three.

2. Bacterial RNA polymerase can initiate transcription without the help of additional proteins, whereas eukaryotic RNA polymerases need general transcription factors.

3. In eukaryotic cells, transcription regulators can influence transcriptional initiation thousands of nucleotides away from the promoter, whereas bacterial regulatory sequences are very close to the promoter.

4. Eukaryotic transcription is affected by chromatin structure and nucleosomes, whereas bacteria lack nucleosomes.

DIF: Easy REF: 7.1 OBJ: 7.1.h Explain how a bacterial RNA polymerase recognizes where transcription will begin, which DNA strand to transcribe, and when to stop. | 7.1.i Contrast transcription initiation in bacteria and eukaryotes.

MSC: Remembering 5. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. activator

lac

TFIIA

CAP

ligase

TFIID

TFIIH

enhancer

TATA

In eukaryotic cells, general transcription factors are required for the activity of all promoters transcribed by RNA polymerase II. The assembly of the general transcription factors begins with the binding of the factor __________ to DNA, causing a marked local distortion in the DNA. This factor binds at the DNA sequence called the __________ box, which is typically located approximately 30 nucleotides upstream from the transcription start site. Once RNA polymerase II has been brought to the promoter DNA, it must be released to begin making transcripts. This release process is facilitated by the addition of phosphate groups to the tail of RNA polymerase by the factor __________. It must be remembered that the general transcription factors and RNA polymerase are not sufficient to initiate transcription in the cell and are affected by proteins bound thousands of nucleotides away from the promoter.

ANS: In eukaryotic cells, general transcription factors are required for the activity of all promoters transcribed by RNA polymerase II. The assembly of the general transcription factors begins with the binding of the factor TFIID to DNA, causing a marked local distortion in the DNA. This factor binds at the DNA sequence called the TATA box, which is typically located approximately 30 nucleotides upstream from the transcription start site. Once RNA polymerase II has been brought to the promoter DNA, it must be released to begin making transcripts. This release process is facilitated by the addition of phosphate groups to the tail of RNA polymerase by the factor TFIIH. It must be remembered that the general transcription factors and RNA polymerase are not sufficient to initiate transcription in the cell and are affected by proteins bound thousands of nucleotides away from the promoter.

DIF: Easy REF: 7.1 OBJ: 7.1.j Explain how eukaryotic general transcription factors assemble on a promoter, form a transcription initiation complex, and release RNA polymerase to begin transcription. MSC: Understanding 6. The following segment of DNA is from a transcribed region of a chromosome. You know that RNA polymerase moves from left to right along this piece of DNA, that the promoter for this gene is to the left of the DNA shown, and that this entire region of DNA is made into RNA. 5′-GGCATGGCAATATTGTAGTA-3′ 3′-CCGTACCGTTATAACATCAT-5′ Given this information, a student claims that the RNA produced from this DNA is: 3′-GGCATGGCAATATTGTAGTA-5′ Give two reasons why this answer is incorrect.

ANS: First, the RNA molecule should have uracil instead of thymine bases. Second, the polarity of the molecule is incorrectly labeled. The correct RNA molecule produced, using the bottom strand of the DNA duplex as a template, would be: 5′-GGCAUGGCAAUAUUGUAGUA-3′ DIF: Difficult REF: 7.1 OBJ: 7.1.d Compare the reactions catalyzed by RNA and DNA polymerases in terms of templating, substrate, directionality, and source of energy to drive the reactions. MSC: Evaluating 7. Imagine that an RNA polymerase is transcribing a segment of DNA that contains the following sequence: 5′-AGTCTAGGCACTGA-3′ 3′-TCAGATCCGTGACT-5′ A. If the polymerase is transcribing from this segment of DNA from left to right, which strand (top or bottom) is the

template?

B. What will be the sequence of that RNA (be sure to label the 5′ and 3′ ends of your RNA molecule)?

ANS: A. The bottom strand.

B. 5′-AGUCUAGGCACUGA-3′ DIF: Moderate REF: 7.1 OBJ: 7.1.d Compare the reactions catalyzed by RNA and DNA polymerases in terms of templating, substrate, directionality, and source of energy to drive the reactions. MSC: Applying 8. Name three covalent modifications that are made to most eukaryotic RNA molecules before the RNA molecule becomes a mature mRNA. ANS: 1. A poly-A tail is added.

2. A 5′ cap is added.

3. Introns can be spliced out.

DIF: Easy REF: 7.1 OBJ: 7.1.o Explain how cells control the quality of the RNAs that are exported from the nucleus.

MSC: Remembering 9. The length of a particular gene in human DNA, measured from the start site for transcription to the end of the proteincoding region, is 10,000 nucleotides, whereas the length of the mRNA produced from this gene is 4000 nucleotides.

What is the most likely reason for this difference?

ANS: The gene contains one or more introns. DIF: Easy REF: 7.1 OBJ: 7.1.m State the potential benefits of the presence of introns in eukaryotic genes. MSC: Evaluating 10. Why is the old dogma “one gene—one protein” not always true for eukaryotic genes?

ANS: The transcripts from some genes can be spliced in more than one way to give mRNAs containing different sequences, thus encoding different proteins. A single eukaryotic gene may therefore encode more than one protein.

DIF: Moderate REF: 7.1 OBJ: 7.1.a Explain how the central dogma relates to gene expression. | 7.1.m State the potential benefits of the presence of introns in eukaryotic genes. MSC: Understanding 11. Is the following statement TRUE or FALSE? Explain your answer.

Since introns do not contain protein-coding information, they do not have to be removed precisely from the primary transcript during RNA splicing. In other words, leaving in an extra nucleotide or two should not make a difference to the protein produced.

ANS: False, although it is true that the sequences within the introns are mostly dispensable, the introns must still be removed precisely because an error of one or two nucleotides would shift the reading frame of the resulting mRNA molecule and change the protein it encodes.

DIF: Moderate REF: 7.1 OBJ: 7.1.m State the potential benefits of the presence of introns in eukaryotic genes. MSC: Evaluating 12. The following RNA sequence includes the beginning of a sequence coding for a protein. What would be the result of a mutation that changed the C that is bolded and marked by an asterisk to an A? 5′-AGGCUAUGAAUCGACACUGCGAGCCC . . .

Table 7-29

ANS: The change creates a stop codon (TGA, or UGA in the mRNA) very near the beginning of the protein-coding sequence and in the correct reading frame (the beginning of the coding sequence is indicated by the ATG). Thus, translation would terminate after only four amino acids had been joined together, and the complete protein would not be made.

DIF: Difficult REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. MSC: Applying 13. Below is a segment of RNA from the middle of an mRNA. 5′- . . . UAGUCUAGGCACUGA . . . -3′

Table 7-29

If you were told that this segment of RNA was part of the coding region of an mRNA for a large protein, give the amino acid sequence for the protein that is encoded by this segment of mRNA. Write your answer using the one-letter amino acid code.

ANS: SLGT is the correct answer. (Reading frame two is the only reading frame that does not contain a stop codon.)

DIF: Difficult REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.b Explain how a ribosome determines which reading frame in an mRNA contains a protein-coding sequence, and describe the consequences of translating an incorrect reading frame. MSC: Applying 14. One strand of a section of DNA isolated from the bacterium E. coli reads: 5′-GTAGCCTACCCATAGG-3′

Table 7-29

A. Suppose that an mRNA is transcribed from this DNA using the complementary strand as a template. What will be the sequence of the mRNA in this region (make sure you label the 5′ and 3′ ends of the mRNA)?

B. How many different peptides could potentially be made from this sequence of RNA, assuming that translation initiates upstream of this sequence?

C. What are these peptides? (Give your answer using the one-letter amino acid code.)

ANS: A. 5′-GUAGCCUACCCAUAGG-3′ B. Two. (There are three potential reading frames for each RNA. In this case, they are GUA GCC UAC CCA UAG . . .

UAG CCU ACC CAU AGG . . .

AGC CUA CCC AUA GG? . . .

The center one cannot be used in this case, because UAG is a stop codon.)

C. VAYP SLPIG Note: PTHR will not be a peptide because it is preceded by a stop codon.

DIF: Difficult REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an

mRNA into a protein sequence. | 7.2.b Explain how a ribosome determines which reading frame in an mRNA contains a protein-coding sequence, and describe the consequences of translating an incorrect reading frame. MSC: Applying 15. After treating cells with a mutagen, you isolate two mutants. One carries alanine and the other carries methionine at a site in the protein that normally contains valine. After treating these two mutants again with mutagen, you isolate mutants from each that now carry threonine at the site of the original valine (see Figure 7-67). Assuming that all mutations caused by the mutagen are due to single nucleotide changes, deduce the codons that are used for valine, alanine, methionine, and threonine at the affected site.

Figure 7-67

Table 7-29

ANS: Given that there are only single nucleotide changes, the only codons consistent with the changes are GUG for valine, GCG for alanine, AUG for methionine, and ACG for threonine.

DIF: Difficult REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. MSC: Applying 16. In a diploid organism, the DNA encoding one of the tRNAs for the amino acid tyrosine is mutated such that the sequence of the anticodon is now 5′-CTA-3′ instead of 5′-GTA-3′. What kind of aberration in protein synthesis will this tRNA cause? Explain your answer.

Table 7-29

ANS: If the DNA sequence specifying the anticodon is changed from 5′-GTA-3′ to 5′-CTA-3′, this tRNA will now pair with the 5′-UAG-3′ codon (instead of 5′ -UAC-3′). The UAG codon normally serves as a stop codon. Thus, this change will result in the amino acid tyrosine being incorrectly incorporated where there is a stop codon, resulting in the addition of amino acids at the end of proteins that normally would come to a stop because of the UAG codon in the mRNA. (Note that the tyrosine codons will NOT cause premature termination of translation, as tyrosine should continue to be incorporated into proteins, as there are additional tyrosine-tRNA genes in the cell that will provide a normal supply of tyrosine-tRNAs.) DIF: Difficult REF: 7.2 OBJ: 7.2.a Explain how the genetic code allows translation of the information contained in an mRNA into a protein sequence. | 7.2.d Explain how each tRNA is charged with the correct amino acid, and describe the consequences of a tRNA carrying an incorrect amino acid. MSC: Applying 17. Figure 7-52 shows an mRNA molecule.

Figure 7-52

Is the mRNA shown prokaryotic or eukaryotic? Explain your answer.

ANS: The mRNA is prokaryotic. It contains coding regions for more than one protein, as shown by the multiple init iation codons, each preceded by a ribosome-binding site. It contains an unmodified 5′ end, as shown by the three phosphate groups, and an unmodified 3′ end, as shown by the absence of a poly-A tail.

DIF: Easy REF: 7.2 OBJ: 7.2.j Contrast the translation initiation in eukaryotes and in bacteria.

MSC: Understanding 18. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.

A

medium

protein

central

P

RNA

DNA

peptidyl transferase

small

E

polymerase

T

large

proteasome

ubiquitin

Once an mRNA is produced, its message can be decoded on ribosomes. The ribosome is composed of two subunits: the __________ subunit, which catalyzes the formation of the peptide bonds that link the amino acids together into a polypeptide chain, and the __________ subunit, which matches the tRNAs to the codons of the mRNA. During the chain elongation process of translating an mRNA into protein, the growing polypeptide chain attached to a tRNA is bound to the __________ site of the ribosome. An incoming aminoacyl-tRNA carrying the next amino acid in the chain will bind to the __________ site by forming base pairs with the exposed codon in the mRNA. The __________ enzyme catalyzes the formation of a new peptide bond between the growing polypeptide chain and the newly arriving amino acid. The end of a protein-coding message is signaled by the presence of a stop codon, which binds the __________ called release factor. Eventually, most proteins will be degraded by a large complex of proteolytic enzymes called the __________.

ANS: Once an mRNA is produced, its message can be decoded on ribosomes. The ribosome is composed of two subunits: the large subunit, which catalyzes the formation of the peptide bonds that link the amino acids together into a polypeptide chain, and the small subunit, which matches the tRNAs to the codons of the mRNA. During the chain elongation process of translating an mRNA into protein, the growing polypeptide chain attached to a tRNA is bound to the P site of the ribosome. An incoming aminoacyl-tRNA carrying the next amino acid in the chain will bind to the A site by forming base pairs with the exposed codon in the mRNA. The peptidyl transferase enzyme catalyzes the formation of a new peptide bond between the growing polypeptide chain and the newly arriving amino acid. The end of a protein-coding message is signaled by the presence of a stop codon, which binds the protein called release factor. Eventually, most proteins will be degraded by a large complex of proteolytic enzymes called the proteasome.

DIF: Easy REF: 7.2 OBJ: 7.2.d Explain how each tRNA is charged with the correct amino acid, and describe the consequences of a tRNA carrying an incorrect amino acid. | 7.2.f Delineate the roles the small and large ribosomal subunits play in translation. | 7.2.h Summarize the cycle by which amino acids are covalently linked to a growing polypeptide chain. | 7.2.n Explain how and why proteins are targeted for degradation in proteasomes.

MSC: Understanding 19. Give a reason why DNA makes a better material than RNA for the storage of genetic information, and explain your answer.

ANS: Three possible answers are: 1. The deoxyribose sugar of DNA makes the molecule much less susceptible than RNA to breakage, because of the lack of the hydroxyl group on carbon 2 of the ribose sugar.

2. DNA is double-stranded and therefore the complementary strand provides a template from which damage can be repaired accurately.

3. The use of “T” in DNA instead of “U” (as in RNA) protects against the effect of deamination, a common form of damage. Deamination of T produces an aberrant base (methyl C), whereas deamination of U generates C, a normal base. The presence of an abnormal base eases the cell’s job of recognizing the damaged strand.

DIF: Moderate REF: 7.1 OBJ: 7.3.c Explain the features that make DNA a better molecule for the permanent storage of genetic information than RNA. MSC: Evaluating 20. NASA has discovered an alien life-form. You are called in to help NASA scientists to deduce the genetic code for this alien. Surprisingly, this alien life-form shares many similarities with life on Earth: this alien uses DNA as its genetic material, makes RNA from DNA, and reads the information from RNA to make protein using ribosomes and tRNAs. Even more amazing, this alien uses the same 20 amino acids, like the organisms found on Earth, and also codes for each amino acid by a triplet codon. However, the scientists at NASA have found that the genetic code used by the alien life-form differs from that used by life on Earth. NASA scientists drew this conclusion after creating a cell-free protein-synthesis system from alien cells and adding an mRNA made entirely of uracil (poly U). They found that poly U directs the synthesis of a peptide containing only glycine. NASA scientists have synthesized a poly AU mRNA and observe that it codes for a polypeptide of alternating serine and proline amino acids. From these experiments, can you determine which codons code for serine and proline? Explain.

Bonus question: Can you propose a mechanism for how the alien’s physiology is altered so that it uses a different genetic code from life on Earth, despite all the similarities?

ANS: No, you cannot definitively determine the codons that code for serine or proline, because it could be either UAU or AUA. Bonus. The alien aminoacyl-tRNA synthetases could adapt a different amino acid to each tRNA, thus matching an amino acid with a different codon compared with the codons used by life on Earth.

DIF: Difficult REF: 7.2 OBJ: 7.2.e Explain how investigators used synthetic polynucleotides and trinucleotides to decipher the genetic code. MSC: Applying 21. Use the numbers in the choices below to indicate where in the schematic diagram of a eukaryotic cell (Figure 7-73) those processes take place.

Figure 7-73

1. transcription 2. translation 3. RNA splicing 4. polyadenylation 5. RNA capping ANS: See Figure 7-73A.

Figure 7-73A

DIF: Easy REF: 7.1 OBJ: 7.1.a Explain how the central dogma relates to gene expression. | 7.1.o Explain how cells control the quality of the RNAs that are exported from the nucleus. MSC: Remembering 22. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. incorporation

rRNA

translation

mRNA

snRNA

transmembrane

pRNA

transcription

tRNA

proteins For a cell’s genetic material to be used, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called __________. Various kinds of RNA are produced, each with different functions. __________ molecules code for proteins, __________ molecules act as adaptors for protein synthesis, __________ molecules are integral components of the ribosome, and __________ molecules are important in the splicing of RNA transcripts.

ANS: For a cell’s genetic material to be used, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called transcription. Various kinds of RNA are produced, each with different functions. mRNA molecules code for proteins, tRNA molecules act as adaptors for protein synthesis, rRNA molecules are integral components of the ribosome, and snRNA molecules are important in the splicing of RNA transcripts.

DIF: Easy REF: 7.1 OBJ: 7.1.a Recall the central dogma and explain how its steps relate to gene expression d epending on whether the final product of the gene is an RNA or a protein. | 7.1.g List the most common types of RNA produced by transcription and identify those that represent the final product of gene expression.

MSC: Understanding 23. Figure 7-75A shows the stage in translation when an incoming aminoacyl-tRNA has bound to the A site on the

ribosome. Using the components shown in Figure 7-75A as a guide, show on Figures 7-75B and 7-75C what happens in the next two stages to complete the addition of the new amino acid to the growing polypeptide chain.

Figure 7-75

ANS: See Figure 7-75A.

Figure 7-75A

DIF: Moderate REF: 7.2 OBJ: 7.2.h Summarize the cycle by which amino acids are covalently linked to a growing polypeptide chain. MSC: Creating

CHAPTER 8 Control of Gene Expression

AN OVERVIEW OF GENE EXPRESSION 8.1.a Review how nuclear transplantation experiments demonstrated that the different specialized cell types in a multicellular organism contain the same genetic instructions.

8.1.b Present and evaluate several methods for assessing the types of proteins produced by different cell types.

8.1.c Define and provide examples of housekeeping proteins.

8.1.d Contrast how liver cells and fat cells alter their gene expression in response to the hormone cortisol.

8.1.e Articulate the steps at which gene expression can be regulated and identify the step that, for most genes, is the main point of control.

HOW TRANSCRIPTION SWITCHES WORK 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them.

8.2.b Illustrate how transcription regulators recognize and bind to regulatory DNA sequences in a DNA double helix.

8.2.c Describe the structure of an operon and note the type of organism in which this arrangement is most common.

8.2.d Review how, in bacteria, the amino acid tryptophan shuts down production of the enzymes responsible for its biosynthesis.

8.2.e Compare how the bacterial promoters that interact with transcriptional activators and repressors differ in terms of how efficiently they bind and position RNA polymerase.

8.2.f Outline how the Lac operon allows bacteria to efficiently utilize the alternative carbon sources lactose and glucose.

8.2.g Explain how eukaryotic activator proteins can enhance transcription even when bound to sequences hundreds or thousands of nucleotide pairs away from a gene’s promoter.

8.2.h Summarize how eukaryotic repressor proteins decrease transcription.

8.2.i Compare how eukaryotic activator and repressor proteins exploit the mechanisms that regulate chromatin packaging to enhance or suppress transcription.

8.2.j Describe how enhancers are prevented from inappropriately activating the transcription of nearby genes.

GENERATING SPECIALIZED CELL TYPES 8.3.a Differentiate between general transcription factors and transcription regulators and compare how they participate in gene expression.

8.3.b Define combinatorial control as it relates to the regulation of gene expression.

8.3.c Compare how bacteria and eukaryotes coordinate the expression of multiple genes.

8.3.d Articulate how a reporter gene can be used to determine where or when a particular gene is expressed during embryonic development.

8.3.e Explain how a combination of regulatory sequences could produce a well-defined “stripe” of gene expression at a given time and in a consistent location in a developing fruit fly embryo.

8.3.f Summarize how a limited number of transcription regulators can direct the differentiation of specific cell types.

8.3.g Recall how a master transcription regulator can direct the formation of an entire organ.

8.3.h Outline how transcription regulators can be used to artificially reprogram cells in culture, including induced plurip otent stem cells and other specialized cell types.

8.3.i Relate how positive feedback, DNA methylation, and histone modification allow differentiated cells to maintain their identity and to pass this information to daughter cells during cell division.

8.3.j Define epigenetic inheritance.

POST-TRANSCRIPTIONAL CONTROLS 8.4.a Define post-transcriptional control.

8.4.b Compare how sequences in bacterial and eukaryotic mRNAs regulate whether their message will be translated into protein.

8.4.c Differentiate between noncoding RNAs and regulatory RNAs. 8.4.d Outline how microRNAs can block the production of a particular protein and specify how the mRNAs targeted by this mechanism can be destroyed.

8.4.e Summarize the role that RNAi plays in cell defense.

8.4.f Compare microRNAs (miRNAs) and small interfering RNAs (siRNAs) in terms of their generation, processing, structure, and mechanism of destroying target RNA molecules.

8.4.g Contrast the roles played by RISC and RITS in RNAi. 8.4.h Articulate how long noncoding (lnc) RNAs can silence genes on the X chromosome.

8.4.i Recall how lnc RNAs can serve as scaffolds and provide an example of an lnc RNA that serves this function.

MULTIPLE CHOICE 1. A neuron and a white blood cell have very different functions. For example, a neuron can receive and respond to electrical signals, while a white blood cell defends the body against infection. This is because a. all of the proteins found in a neuron are completely different from the proteins found in a white blood cell. b. the neuron and the white blood cell within an individual have the same genome. c. the neuron expresses some mRNAs that the white blood cell does not. d. neurons and white blood cells are differentiated cells and thus no longer need to transcribe and translate genes.

ANS: C

Different cell types express different mRNAs, leading to differences in protein expression. There are proteins common to all cells in multicellular organisms. Although it is true that the neuron and white blood cell within an individual have the same genome, this does not explain why these two cells have different functions. Differentiated cells still need to transcribe and translate genes.

DIF: Easy REF: 8.1 OBJ: 8.1.c Define and provide examples of housekeeping proteins. MSC: Understanding 2. Investigators performed nuclear transplant experiments to determine whether DNA is altered irreversibly during development. Which of the following statements about these experiments is TRUE? a. Because the donor nucleus is taken from an adult animal, the chromosomes from the nucleus must undergo recombination with the DNA in the egg for successful development to occur. b. The cells in the embryo that develop from the nuclear transplant experiment have DNA that is identical to the donor of the nucleus. c. The meiotic spindle of the egg must interact with the chromosomes of the injected nuclei for successful nuclear transplantation to occur. d. Although nuclear transplantation has been successful in producing embryos in some mammals with the use of foster mothers, evidence of DNA alterations during differentiation has not been obtained for plants.

ANS: B The meiotic spindle and the nucleus in the egg are usually removed in these experiments. Some differentiated plant cells have the ability to dedifferentiate and regenerate an entire adult plant, providing evidence that DNA is not irreversibly altered during development.

DIF: Easy REF: 8.1 OBJ: 8.1.a Review how nuclear transplantation demonstrated that the various specialized cell types in a multicellular organism contain the same genetic instructions. MSC: Understanding 3. The distinct characteristics of different cell types in a multicellular organism result mainly from the differential regulation of the a. replication of specific genes. b. transcription of genes transcribed by RNA polymerase II. c. transcription of housekeeping genes. d. proteins that directly bind the TATA box of eukaryotic genes.

ANS: B The major cause of differences between different cell types is in the differential expression of protein-coding genes transcribed by RNA polymerase II, because these genes encode, not only the specific proteins characteristic of different cell types, but also the transcription regulators required to maintain and control this pattern of expression. All genes are replicated equally when cells divide. The expression of housekeeping genes does not differ much from cell to cell, because they mainly encode the proteins that are necessary for all cells to live. The general transcription factors bind to the TATA box and do not mediate differential gene expression.

DIF: Easy REF: 8.1 OBJ: 8.1.e Articulate the steps at which gene expression can be regulated and identify the step that, for most genes, is the main point of control. MSC: Understanding 4. The human genome encodes about 21,000 protein-coding genes. Approximately how many such genes does the typical differentiated human cell express at any one time? a. 21,000—all of them

b. between 18,900 and 21,000—at least 90% of the genes c. between 5000 and 15,000 d. less than 2100 ANS: C DIF: Easy REF: 8.1 OBJ: 8.1.e Articulate the steps at which gene expression can be regulated and identify the step that, for most genes, is the main point of control. MSC: Remembering 5. Which of the following is not a good example of a housekeeping protein? a. DNA repair enzymes b. histones c. ATP synthase d. hemoglobin ANS: D DIF: Easy REF: 8.1 OBJ: 8.1.c Define and provide examples of housekeeping proteins.

MSC: Understanding 6. Which of the following statements about differentiated cells is TRUE? a. Cells of distinct types express nonoverlapping sets of transcription regulators. b. Once a cell has differentiated, it can no longer change its gene expression. c. Once a cell has differentiated, it will no longer need to transcribe RNA. d. Some of the proteins found in differentiated cells are found in all cells of a multicellular organism.

ANS: D The housekeeping proteins are proteins common to all cells and are used in processes that are important to the basic function of cells. Some transcription regulators, particularly those used to regulate housekeeping genes, are found in all cell types. Even though a cell is differentiated, it may be able to respond to changes in its environment by changing its gene expression pattern. A differentiated cell usually continues to express genes and make new proteins.

DIF: Easy REF: 8.1 OBJ: 8.1.c Define and provide examples of housekeeping proteins. MSC: Understanding 7. Which of the following statements about transcriptional regulators is FALSE? a. Transcription regulators interact only with the sugar–phosphate backbone on the outside of the double helix to determine where to bind on the DNA helix. b. Transcription regulators will form hydrogen bonds, ionic bonds, and hydrophobic interactions with DNA. c. The DNA-binding motifs of transcription regulators usually bind in the major groove of the DNA helix. d. The binding of transcription regulators generally does not disrupt the hydrogen bonds that hold the double helix together.

ANS: A Transcription regulators generally recognize specific sequences in the DNA by binding to the edges of the bases.

DIF: Easy REF: 8.2 OBJ: 8.2.b Illustrate how transcription regulators recognize and bind to regulatory DNA sequences along a DNA double helix. MSC: Analyzing 8. Operons a. are commonly found in eukaryotic cells. b. are transcribed by RNA polymerase II. c. contain a cluster of genes transcribed as a single mRNA. d. can only be regulated by gene activator proteins.

ANS: C DIF: Easy REF: 8.2 OBJ: 8.2.c Describe the structure of an operon and note the type of organism in which this arrangement is most common. MSC: Remembering 9. The tryptophan operator a. is an allosteric protein. b. binds to the tryptophan repressor when the repressor is bound to tryptophan. c. is required for production of the mRNA encoded by the tryptophan operon. d. is important for the production of the tryptophan repressor.

ANS: B DIF: Easy REF: 8.2 OBJ: 8.2.d Review how, in bacteria, the amino acid tryptophan shuts down production of the enzymes responsible for its biosynthesis. MSC: Remembering 10. Which of the following statements about the Lac operon is FALSE? a. The Lac repressor binds when lactose is present in the cell. b. Even when the CAP activator is bound to DNA, if lactose is not present, the Lac operon will not be transcribed. c. The CAP activator can only bind DNA when it is bound to cAMP. d. The Lac operon only produces RNA when lactose is present and glucose is absent.

ANS: A The Lac repressor binds when lactose is not present in the cell.

DIF: Easy REF: 8.2 OBJ: 8.2.f Outline how the Lac operon allows bacteria to utilize the alternative carbon sources lactose and glucose. MSC: Understanding 11. What do you predict would happen if you replace the Lac operator DNA from the Lac operon with the DNA from the operator region from the tryptophan operon? a. The presence of lactose will not cause allosteric changes to the Lac repressor. b. The Lac operon will not be transcribed when tryptophan levels are high. c. The lack of glucose will no longer allow CAP binding to the DNA. d. RNA polymerase will only bind to the Lac promoter when lactose is present.

ANS: B Replacing the Lac operator with the tryptophan operon operator will now allow binding of the tryptophan repressor instead of the Lac repressor. Since the tryptophan repressor binds to DNA when tryptophan levels are high, the tryptophan repressor is predicted to block mRNA production from the Lac operon when tryptophan is abundant in the cell.

Changing the operator sequences at the Lac operon should not affect how the presence of lactose changes the Lac repressor, nor should it affect the ability of CAP to bind to the DNA at low glucose levels, since the CAP-binding site is on the other side of the RNA polymerase binding site. Removing the Lac operator will make it so that the Lac promoter is no longer responsive to lactose levels in the cell.

DIF: Moderate REF: 8.2 OBJ: 8.2.f Outline how the Lac operon allows bacteria to utilize the alternative carbon sources lactose and glucose. | 8.2.d Review how, in bacteria, the amino acid tryptophan shuts down production of the enzymes responsible for its biosynthesis. MSC: Applying 12. You are interested in examining the Psf gene. It is known that Psf is normally produced when cells are exposed to high levels of both calcium (Ca2+) and magnesium (Mg2+). MetA, MetB, and MetC are important for binding to the promoter of the Psf gene and are involved in regulating its transcription. MetA binds to the “A” site in the promoter region, MetB to the “B” site, and MetC to the “C” site. You create binding-site mutations in the A, B, and C sites and

observe what happens to transcription of the Psf gene. Your results are summarized in Table 8-12.

Table 8-12

Which of the following proteins are likely to act as gene activators? a. MetA only b. MetB only c. MetC only d. both MetA and MetC ANS: D DIF: Moderate REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 13. You are interested in examining the Psf gene. It is known that Psf is normally produced when cells are exposed to high levels of both calcium (Ca2+) and magnesium (Mg2+). MetA, MetB, and MetC are important for binding to the promoter of the Psf gene and are involved in regulating its transcription. MetA binds to the “A” site in the promoter region, MetB to the “B” site, and MetC to the “C” site. You create binding-site mutations in the A, B, and C sites and observe what happens to transcription of the Psf gene. Your results are summarized in Table 8-12.

Table 8-12

Which of the following proteins are likely to act as gene repressors? a. MetA only b. MetB only c. MetC only d. both MetA and MetC ANS: B DIF: Moderate REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 14. You are interested in examining the Psf gene. It is known that Psf is normally produced when cells are exposed to high levels of both calcium (Ca2+) and magnesium (Mg2+). MetA, MetB, and MetC are important for binding to the promoter of the Psf gene and are involved in regulating its transcription. MetA binds to the “A” site in the promoter region, MetB to the “B” site, and MetC to the “C” site. You create binding-site mutations in the A, B, and C sites and observe what happens to transcription of the Psf gene. Your results are summarized in Table 8-12.

Table 8-12

Which transcription regulator(s) are normally bound to the Psf promoter in the presence of Mg2+ only? a. none b. MetA only c. MetA and Met B d. MetA, MetB, and MetC ANS: C Both MetA and MetB must be bound, because Psf is not normally transcribed in the presence of Mg2+ only. When the binding site for MetA is present without any other binding sites, low levels of Psf transcription occur in the presence of Mg2+ only, suggesting that the gene activator MetA binds in the presence of Mg 2+. Because no transcription occurs normally, the gene repressor MetB must also bind to DNA in the presence of Mg 2+.

DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying

15. You are interested in examining the Psf gene. It is known that Psf is normally produced when cells are exposed to high levels of both calcium (Ca 2+) and magnesium (Mg 2+). MetA, MetB, and MetC are important for binding to the promoter of the Psf gene and are involved in regulating its transcription. MetA binds to the “A” site in the promot er region, MetB to the “B” site, and MetC to the “C” site. You create binding -site mutations in the A, B, and C sites and observe what happens to transcription of the Psf gene. Your results are summarized in Table Q8-12.

Table 8-12

Which transcription factors are normally bound to the Psf promoter in the presence of both Mg2+ and Ca2+? a. MetA and MetB b. MetB and MetC c. MetA and MetC d. MetA, MetB, and MetC ANS: C MetA and MetC are both gene activators. MetA binds in the presence of Ca2+, whereas MetC binds in the presence of Mg2+. Because MetB is a gene repressor, it is unlikely to be bound when both Mg2+ and Ca2+ are present.

DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles

they play in gene expression and the proteins that bind to them. MSC: Applying 16. You are interested in understanding the gene regulation of Lkp1, a protein that is normally produced in liver and kidney cells in mice. Interestingly, you find that the LKP1 gene is not expressed in heart cells. You isolate the DNA upstream of the LKP1 gene, place it upstream of the gene for green fluorescent protein (GFP), and insert this entire piece of recombinant DNA into mice. You find GFP expressed in liver and kidney cells but not in heart cells, an expression pattern similar to the normal expression of the LKP1 gene. Further experiments demonstrate that there are three regions in the promoter, labeled A, B, and C in Figure 8-16, that contribute to this expression pattern. Assume that a single and unique transcription factor binds each site such that protein X binds site A, protein Y binds site B, and protein Z binds site C. You want to determine which region is responsible for tissue-specific expression, and create mutations in the promoter to determine the function of each of these regions. In Figure 8-16, if the site is missing, it is mutated such that it cannot bind its corresponding transcription factor.

Figure 8-16

Which of the following proteins is likely to act as a gene repressor? a. factor X b. factor Y c. factor Z d. none of these answers are correct.

ANS: C DIF: Moderate REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 17. You are interested in understanding the gene regulation of Lkp1, a protein that is normally produced in liver and kidney cells in mice. Interestingly, you find that the LKP1 gene is not expressed in heart cells. You isolate the DNA upstream of the LKP1 gene, place it upstream of the gene for green fluorescent protein (GFP), and insert this entire piece of recombinant DNA into mice. You find GFP expressed in liver and kidney cells but not in heart cells, an expression pattern similar to the normal expression of the LKP1 gene. Further experiments demonstrate that there are three regions in the promoter, labeled A, B, and C in Figure 8 -16, that contribute to this expression pattern. Assume that a single and unique transcription factor binds each site such that protein X binds site A, protein Y binds site B, and protein Z binds site C. You want to determine which region is responsible for tissue specific expression, and create mutations in the promoter to determine th e function of each of these regions. In

Figure 8-16, if the site is missing, it is mutated such that it cannot bind its corresponding transcription factor.

Figure 8-16

Which of the following proteins are likely to act as gene activators? a. factors X and Y b. factors X and Z c. factors Y and Z d. factor X only ANS: A DIF: Moderate REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 18. You are interested in understanding the gene regulation of Lkp1, a protein that is normally produced in liver and kidney cells in mice. Interestingly, you find that the LKP1 gene is not expressed in heart cells. You isolate the DNA upstream of the LKP1 gene, place it upstream of the gene for green fluorescent protein (GFP), and insert this entire piece of recombinant DNA into mice. You find GFP expressed in liver and kidney cells but not in heart cells, an expression pattern similar to the normal expression of the LKP1 gene. Further experiments demonstrate that there are three regions in the promoter, labeled A, B, and C in Figure 8-16, that contribute to this expression pattern. Assume that a single and unique transcription factor binds each site such that protein X binds site A, protein Y binds site B, and protein Z binds site C. You want to determine which region is responsible for tissue-specific expression, and create mutations in the promoter to determine the function of each of these regions. In Figure 8-16, if the site is

missing, it is mutated such that it cannot bind its corresponding transcription factor.

Figure 8-16

Experiment 1 in Figure 8-18 is the positive control, demonstrating that the region of DNA upstream of the gene for GFP results in a pattern of expression that we normally find for the LKP1 gene. Experiment 2 shows what happens when the sites for binding factors X, Y, and Z are removed. Which experiment above demonstrates that factor X alone is sufficient for expression of LPK1 in the kidney? a. experiment 3 b. experiment 5 c. experiment 6 d. experiment 7 ANS: A Experiment 5 does not say anything about factor X, because factor X binds to site A. Experiment 6 and experiment 7 both examine the effect of factor X in combination with another factor, and thus would not demonstrate that factor X alone is sufficient for expression of the gene encoding Lkp1 in the kidney.

DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Evaluating 19. You are interested in understanding the gene regulation of Lkp1, a protein that is normally produced in liver and kidney cells in mice. Interestingly, you find that the LKP1 gene is not expressed in heart cells. You isolate the DNA upstream of the LKP1 gene, place it upstream of the gene for green fluorescent protein (GFP), and insert this entire piece of recombinant DNA into mice. You find GFP expressed in liver and kidney cells but not in heart cells, an expression pattern similar to the normal expression of the LKP1 gene. Further experiments demonstrate that there are three regions in the promoter, labeled A, B, and C in Figure 8-16, that contribute to this expression pattern. Assume that a single and unique transcription factor binds each site such that protein X binds site A, protein Y binds site B, and protein Z binds site C. You want to determine which region is responsible for tissue-specific expression, and create mutations in the promoter to determine the function of each of these regions. In Figure 8-16, if the site is missing, it is mutated such that it cannot bind its corresponding transcription factor.

Figure 8-16

In what tissue is factor Z normally present and bound to the DNA? a. kidney b. liver c. heart d. none of these answers are correct.

ANS: C Experiments 1 and 2 demonstrate that without factors X, Y, or Z, the gene encoding Lkp1 will be switched on in the heart, where it should not be on. Experiment 5 shows that when factor Z can bind, there is no longer expression in the heart. We confirm this hypothesis by comparing the results of experiment 6 with those of experiments 7 and 8. Taken together, these results suggest that factor Z expression is normally needed in the heart to repress the expression of LKP1.

DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 20. You are interested in the regulation of gene Q. Proteins G, H, and J are proteins that are important for regulating gene Q, and bind to its promoter region in a sequence-specific fashion. Proteins G and H both bind to site “A” but cannot bind to site “A” at the same time. Protein J binds to site “B” on the promoter. The promoter region is diagrammed in Figure 8-20.

Figure 8-20

You develop a cell-free transcriptional system to study the effects of proteins G, H, and J on the transcription of gene Q. Using this system, you can examine the effects of adding these proteins to the transcriptional system in equal amounts and measuring how much gene Q is produced. When you add these proteins to the system, you get the results shown in Table 8-20.

Table 8-20

Which proteins are likely to act as gene activators? a. G b. H c. J d. both H and J ANS: D DIF: Moderate REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 21. You are interested in the regulation of gene Q. Proteins G, H, and J are proteins that are important for regulating gene Q, and bind to its promoter region in a sequence-specific fashion. Proteins G and H both bind to site “A” but cannot bind to site “A” at the same time. Protein J binds to site “B” on the promoter. The promoter region is diagrammed in Figure 8-20.

Figure 8-20

You develop a cell-free transcriptional system to study the effects of proteins G, H, and J on the transcription of gene Q. Using this system, you can examine the effects of adding these proteins to the transcriptional system in equal amounts and measuring how much gene Q is produced. When you add these proteins to the system, you get the results shown in Table 8-20.

Table 8-20

Which proteins are likely to act as gene repressors? a. G b. H c. J d. both H and J ANS: A DIF: Moderate REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 22. You are interested in the regulation of gene Q. Proteins G, H, and J are proteins that are important for regulating gene Q, and bind to its promoter region in a sequence-specific fashion. Proteins G and H both bind to site “A” but cannot bind to site “A” at the same time. Protein J binds to site “B” on the promoter. The promoter region is diagrammed in Figure 820.

Figure 8-20

You develop a cell-free transcriptional system to study the effects of proteins G, H, and J on the transcription of gene Q. Using this system, you can examine the effects of adding these proteins to the transcriptional system in equal amounts and measuring how much gene Q is produced. When you add these proteins to the system, you get the results shown in Table 8-20.

Table 8-20

Your colleague looks at your data above and predicts that protein G will bind more strongly to the DNA at site A, compared to protein H. Which experiment above is critical for this prediction? a. #2 b. #3 c. #5 d. #6 ANS: C Experiment #5 examines binding in the presence of both G and H. Because no Q mRNA is produced, the repressor, G, is likely bound.

DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 23. You are interested in the regulation of gene Q. Proteins G, H, and J are proteins that are important for regulating gene Q, and bind to its promoter region in a sequence-specific fashion. Proteins G and H both bind to site “A” but cannot bind to site “A” at the same time. Protein J binds to site “B” on the promoter. The promoter region is diagrammed in Figure 8-20.

Figure 8-20

You develop a cell-free transcriptional system to study the effects of proteins G, H, and J on the transcription of gene Q. Using this system, you can examine the effects of adding these proteins to the transcriptional system in equal amounts and measuring how much gene Q is produced. When you add these proteins to the system, you get the results shown in Table 8-20.

Table 8-20

Which proteins do you predict are bound to the promoter in experiment #8? a. only H and J b. only G and H c. only G and J d. only J ANS: C From experiment #5, we predict that G is bound to site A. Because we see Q mRNA being produced, J must be bound at site B as well.

DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 24. You are interested in studying the transcriptional regulation of the Gip1 promoter. The Gip1 promoter contains a binding site for the Jk8 protein that overlaps with the binding site for the Pa5 protein. Jk8 and Pa5 cannot bind DNA at the same time, but both proteins are present at high levels in adult liver cells. The binding sites for Jk8 and Pa5 are shown in Figure 8-24.

Figure 8-24

Jk8 binds to site A while Pa5 binds to site B. You create mutations that remove the nonoverlapping sequences of either binding site A or B, and examine Gip1 mRNA production in adult liver cells that contain these mutations. The data you obtain from these experiments are shown in Table 8-24.

Table 8-24

You know that Gip1 is only expressed in adult liver cells and not in the liver of embryos. You also know that Jk8 and Pa5 behave similarly on other promoters in the embryo or in the adult, in terms of whether they act as repressors or gene activators. Given the data, use of which of the following mechanisms would make the most sense for regulating the Jk8 and Pa5 proteins: a. Jk8 is ubiquitylated and targeted for destruction in adult cells. b. Jk8, but not Pa5, is transcribed in embryonic liver cells. c. Jk8 binds to the promoter of the gene that encodes Jk8 in embryonic liver cells. d. Pa5 binds to the promoter of the gene that encodes Jk8 in embryonic liver cells.

ANS: D Jk8 must be present in adult cells to act as a gene activator and should not be produced in embryonic liver cells.

DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Evaluating 25. How are most eukaryotic transcription regulators able to affect transcription when their binding sites are far from the promoter? a. by binding to their binding site and sliding to the site of RNA polymerase assembly b. by looping out the intervening DNA between their binding site and the promoter c. by unwinding the DNA between their binding site and the promoter d. by attracting RNA polymerase and modifying it before it can bind to the promoter ANS: B Most eukaryotic transcription regulators that act at a distance are thought to do so by looping out the intervening DNA while at the same time binding, via the Mediator, to proteins that form the initiation complex at the promoter.

DIF: Easy REF: 8.2 OBJ: 8.2.g Explain how eukaryotic activator proteins can enhance transcription even when bound to sequences hundreds or thousands of nucleotide pairs away from a gene’s promoter.

MSC: Understanding 26. Which of the following statements about nucleosomes is TRUE? a. Nucleosomes activate transcription when bound to the promoter. b. Although RNA polymerase can access DNA packed within nucleosomes, the general transcription factors and transcriptional regulators cannot. c. Histone acetyltransferases affect transcription by both altering chromatin structure to allow accessibility to the DNA and by adding acetyl groups to histones that can bind proteins that promote transcription. d. Histone deacetylases remove lysines from histone tails.

ANS: C The binding of nucleosomes to the promoter inhibits transcription by blocking RNA polymerase, the general transcription factors, and transcriptional regulators from accessing the DNA. Histone deacetylases remove acetyl groups that are attached to lysines in the tail of histone proteins.

DIF: Easy REF: 8.2 OBJ: 8.2.i Compare how eukaryotic activator and repressor proteins exploit the mechanisms that regulate chromatin packaging to enhance or suppress transcription. MSC: Understanding 27. The expression of the BRF1 gene in mice is normally quite low, but mutations in a gene called BRF2 lead to increased expression of BRF1. You have a hunch that nucleosomes are involved in the regulation of BRF1 expression and so you investigate the position of nucleosomes over the TATA box of BRF1 in normal mice and in mice that lack either

the BRF2 protein (BRF2–) or part of histone H4 (HHF–) (histone H4 is encoded by the HHF gene). Table 8-27 summarizes your results. A normal functional gene is indicated by a plus sign (+).

Table 8-27

Which of the following conclusions CANNOT be drawn from your data? a. BRF2 is required for the repression of BRF1. b. BRF2 is required for the specific pattern of nucleosome positions over the BRF1 upstream region. c. The specific pattern of nucleosome positioning over the BRF1 upstream region is required for BRF1 repression. d. The part of histone H4 missing in HHF– mice is not required for the formation of nucleosomes.

ANS: C All the other conclusions can be drawn from the data. Because the BRF2+ HHF– mutant does not show the specific pattern of nucleosome positioning yet still has a low level of BRF1 expression, and because the BRF2– HHF– mutant has high levels of BRF1 expression (indicating that HHF is not required for BRF1 expression), it seems that repression of BRF1 could take place in the absence of nucleosome positioning. Because nucleosomes are formed in all cases, the missing portion of histone H4 is not required for their formation.

DIF: Difficult REF: OBJ: 8.2.i Compare how eukaryotic activator and repressor proteins exploit the mechanisms that regulate chromatin packaging to enhance or suppress transcription.

28. Enhancers can act over long stretches of DNA, but are specific about which genes they affect. How do eukaryotic cells prevent these transcription regulators from looping in the wrong direction and inappropriately turning on the transcription of a neighboring gene? a. The cell uses histone acetyltransferase to prevent enhancer binding to the DNA. b. The cell will group neighboring genes into operons so that they are transcribed as a single unit, regulated by a single enhancer. c. Chromosome loop-forming proteins arrange the DNA into topological associated domains such that individual genes and their associated enhancer binding regions are in proximity. d. There is no mechanism to prevent inappropriate enhancer action on neighboring genes. The cells will degrade inappropriately produced mRNA molecules.

ANS: C DIF: Easy REF: 8.2 OBJ: 8.2.j Describe how enhancers are prevented from inappropriately activating the transcription of nearby genes. MSC: Understanding 29. Combinatorial control of gene expression a. involves every gene using a different combination of transcriptional regulators for its proper expression. b. involves groups of transcription regulators working together to determine the expression of a gene. c. involves only the use of gene activators used together to regulate genes appropriately. d. is seen only when genes are arranged in operons.

ANS: B DIF: Easy REF: 8.3 OBJ: 8.3.b Define combinatorial control as it relates to the regulation of gene expres-

sion. MSC: Remembering 30. In principle, how many different cell types can an organism having four different types of transcription regulators and thousands of genes create? a. up to 4 b. up to 8 c. up to 16 d. thousands ANS: C The type of cell is determined by the particular combination of transcription factors active within it. With four different proteins available, there is one possibility with no proteins at all and one with all four proteins. There are four possibilities with one protein each, six possible combinations of two different proteins, and four possible combinations of three different proteins for a total of 16.

DIF: Moderate REF: 8.3 OBJ: 8.3.f Summarize how a limited number of transcription regulators can direct the differentiation of specific cell types. MSC: Applying 31. From the sequencing of the human genome, we believe that there are approximately 21,000 protein-coding genes in the genome, of which 1500–3000 are transcription factors. If every gene has a tissue-specific and signal-dependent transcription pattern, how can such a small number of transcriptional regulatory proteins generate a much larger set of transcriptional patterns?

ANS: Transcription regulators are generally used in combinations, thereby increasing the possible regulatory repertoire of gene expression with a limited number of proteins. DIF: REF: 8.3 OBJ: 8.3.f Summarize how a limited number of transcription regulators can direct the differentiation of specific cell types.

32. You are studying a set of mouse genes whose expression increases when cells are exposed to the hormone cortisol, and you believe that the same cortisol-responsive transcriptional activator regulates all of these genes. If your hypothesis is correct, which of the following statements below should be TRUE? a. The cortisol-responsive genes share a DNA sequence in their regulatory regions that binds the cortisol-responsive transcriptional activator. b. The cortisol-responsive genes must all be in an operon. c. The transcriptional regulators that bind to the regulatory regions of the cortisol-responsive genes must all be the same. d. The cortisol-responsive genes must not be transcribed in response to other hormones.

ANS: A DIF: Easy REF: 8.3 OBJ: 8.3.a Differentiate between general transcription factors and transcription regulators and compare how they participate in gene expression. MSC: Applying 33. Which of the following statements about how fruit flies can develop an eye in the middle of a leg is TRUE? a. When the Ey gene is expressed in adult leg cells, these cells dedifferentiate and become eye cells. b. The Ey gene encodes a transcription regulator that is the only transcription regulator used to produce a fruit fly eye. c. When the Ey gene is introduced into cells that would normally give rise to a leg, the transcription regulators used to control its expression in the leg are different from those that are normally used to control Ey expression in the eye. d. All the eye cells found in the adult leg are a single cell type and have identical characteristics.

ANS: C In order to express Ey in cells that would normally become a leg, scientists must use a different promoter (and thus, involve different transcription regulators) to force Ey expression in an inappropriate cell. In this experiment, Ey is expressed in the embryo in leg-cell precursors, and so the transformation into eye tissue does not involve dedifferentiation and redifferentiation. When Ey is expressed in the leg-cell precursors, the Ey transcriptional regulator controls the expression of many different genes, including other transcription regulators that assist in regulating genes important for differentiation into an eye. When Ey is expressed in the leg-cell precursors, an entire organ of several different cell types will develop.

DIF: Easy REF: 8.3 OBJ: 8.3.g Recall how a master transcription regulator can direct the formation of an entire organ.

MSC: Understanding 34. The modular nature of the Eve gene’s regulatory region means that a. there are seven regulatory elements and each element is sufficient for driving expression in a single stripe. b. all the regulatory elements for each stripe use the same transcriptional activators. c. the E. coli LacZ gene is normally only expressed in a single stripe—unlike Eve, which is expressed in seven stripes. d. transcription regulators only bind to the stripe 2 regulatory DNA segment in stripe 2.

ANS: A Different transcriptional activators are located in different regions of the Drosophila embryo and are differentially responsible for expression in different stripes. The E. coli LacZ gene is a reporter gene and is not normally expressed in flies; its expression depends on the specific regulatory element engineered into its promoter. In other stripes, transcriptional repressors may bind to the stripe 2 regulatory DNA segment while transcriptional activators will bind to this segment in stripe 2.

DIF: Easy REF: 8.3 OBJ: 8.3.e Explain how a combination of regulatory sequences could produce a well-defined “stripe” of gene expression in a consistent location along a developing fruit fly embryo. MSC: Understanding 35. Which of the following statements about the Ey transcriptional regulator is FALSE? a. Expression of Ey in cells that normally form legs in the fly will lead to the formation of an eye in the middle of the legs. b. The Ey transcription factor must bind to the promoter of every eye-specific gene in the fly. c. Positive feedback loops ensure that Ey expression remains switched on in the developing eye. d. A homolog of Ey is found in vertebrates; this homolog is also used during eye development.

ANS: B Ey turns on the transcription of other transcriptional regulators; it is the combined action of the genes directly regulated by Ey and the genes regulated by the regulators that Ey turns on that forms the eye.

DIF: Easy REF: 8.3 OBJ: 8.3.g Recall how a master transcription regulator can direct the formation of an entire o rgan. | 8.3.i Relate how positive feedback, DNA methylation, and histone modification allow differentiated cells to maintain their identity and to pass this information to daughter cells during cell division.

MSC: Understanding 36. Which of the following statements about iPS cells is FALSE? a. iPS cells can be made by adding a combination of transcription regulators to a fibroblast. b. iPS cells made from mouse cells can differentiate into almost any human cell type.

c. Stimulation by extracellular signal molecules causes iPS cells to differentiate. d. A cell that is dedifferentiated to become an iPS cell will undergo changes to its gene expression profile.

ANS: B When iPS cells differentiate, the cells follow a program dictated by their genome. Thus, mouse cells can only differentiate into mouse cells and cannot become human cells.

DIF: Easy REF: 8.3 OBJ: 8.3.h Outline how transcription regulators can be used to artificially reprogram cells in culture, including induced pluripotent stem cells and other specialized cell types. MSC: Understanding 37. The MyoD transcriptional regulator is normally found in differentiating muscle cells and participates in the transcription of genes that produce muscle-specific proteins, such as those needed in contractile tissue. Amazingly, expression of MyoD in fibroblasts causes these cells derived from skin connective tissue to produce proteins normally only seen in muscles. However, some other cell types do not transcribe muscle-specific genes when MyoD is expressed in them.

Which of the following statements below is the best explanation of why MyoD can cause fibroblasts to express muscle-specific genes? a. Unlike some other cell types, fibroblasts have not lost the muscle-specific genes from their genome. b. The muscle-specific genes must be in heterochromatin in fibroblasts. c. During their developmental history, fibroblasts have accumulated some transcriptional regulators in common with differentiating muscle cells. d. The presence of MyoD is sufficient to activate the transcription of muscle-specific genes in all cell types.

ANS: C In general, genes are not lost from the genome during differentiation. Heterochromatin is generally associated with transcriptionally silent regions of the chromosome. The presence of MyoD is not sufficient for the expression of muscle-specific genes, because some cell types do not transcribe muscle-specific genes even in the presence of MyoD. DIF: Moderate REF: 8.3 OBJ: 8.3.h Outline how transcription regulators can be used to artificially reprogram cells in culture, including induced pluripotent stem cells and other specialized cell types. MSC: Applying 38. In mammals, individuals with two X chromosomes are female, and individuals with an X and a Y chromosome are male. It had long been known that a gene located on the Y chromosome was sufficient to induce the gonads to form testes, which is the main male-determining factor in development, and researchers sought the product of this gene, the so-called testes-determining factor (TDF). For several years, the TDF was incorrectly thought to be a zinc finger protein encoded by a gene called BoY. Which of the following observations would most strongly suggest that BoY might NOT be the TDF? a. Some XY individuals that develop into females have mutations in a different gene, SRY, but are normal at BoY. b. BoY is not expressed in the adult male testes. c. Expression of BoY in adult females does not masculinize them. d. A few of the genes that are known to be expressed only in the testes have binding sites for the BoY protein in their upstream regulatory sequences, but most do not.

ANS: A XY individuals that develop as females presumably lack the testes-determining factor (TDF). If BoY is normal in these individuals, it would strongly suggest that BoY is not the TDF. Although expression of TDF is necessary for testes development, this does not mean that it must be expressed in adult males once the gonads have already formed. Similarly, even though TDF expression is sufficient to induce testes formation, once the structures have been formed, TDF

may not be able to exert any additional effect. TDF will not necessarily bind upstream of all genes whose expression it influences; some of the genes it regulates directly probably encode other transcription regulators that bind to regulatory sites different from the TDF site.

DIF: Easy REF: 8.3 OBJ: 8.3.a Differentiate between general transcription factors and transcription regulators and compare how they participate in gene expression. MSC: Applying 39. Which of the following is NOT a general mechanism that cells use to maintain stable patterns of gene expression as cells divide? a. a positive feedback loop, mediated by a transcriptional regulator that activates transcription of its own gene in addition to other cell-type-specific genes b. faithful propagation of condensed chromatin structures as cells divide c. inheritance of DNA methylation patterns when cells divide d. proper segregation of housekeeping proteins when cells divide ANS: D DIF: Easy REF: 8.3 OBJ: 8.3.i Relate how positive feedback, DNA methylation, and histone modification allow differentiated cells to maintain their identity and to pass this information to daughter cells during cell division.

MSC: Understanding 40. Which of the following statements about DNA methylation in eukaryotes is FALSE? a. Appropriate inheritance of DNA methylation patterns involves maintenance methyltransferase. b. DNA methylation involves a covalent modification of cytosine bases. c. Methylation of DNA attracts proteins that block gene expression. d. Immediately after DNA replication, each daughter helix contains one methylated DNA strand, which corresponds to the newly synthesized strand.

ANS: D DIF: Easy REF: 8.3 OBJ: 8.3.i Relate how positive feedback, DNA methylation, and histone modification allow differentiated cells to maintain their identity and to pass this information to daughter cells during cell division.

MSC: Understanding 41. Which of the following examples does NOT describe a mechanism of post-transcriptional control of gene expression? a. the alternative splicing of a gene, leading to the production of a muscle-specific protein b. the action of the RNA component of telomerase c. the activation of a protein when it is phosphorylated by a protein kinase d. translational inhibition of an mRNA by the binding of a protein at the 5′ untranslated region of the message.

ANS: B Telomerase is an enzyme important for maintaining the ends of chromosomes, and is not directly involved in posttranscriptional regulation of gene activity.

DIF: Easy REF: 8.4 OBJ: 8.4.a Define post-transcriptional control. MSC: Understanding 42. Using genetic engineering techniques, you remove the sequences that code for the ribosome-binding sequences of the bacterial LacZ gene. The removal of these sequences will lead to a. more LacZ protein being produced due to faster ribosome movement across the LacZ mRNA. b. transcriptional repression, resulting in fewer mRNA molecules being produced from this gene. c. a longer half-life for the LacZ mRNA. d. translational inhibition of the LacZ mRNA.

ANS: D DIF: Easy REF: 8.4 OBJ: 8.4.b Compare how sequences in bacterial and eukaryotic mRNAs regulate whether the message will be translated into protein. MSC: Understanding 43. miRNAs, tRNAs, and rRNAs all a. do not code for proteins. b. act in the nucleus. c. are packaged with other proteins to form RISC. d. form base pairs with mRNA molecules.

ANS: A DIF: Easy REF: 8.4 OBJ: 8.4.c Differentiate between noncoding RNAs and regulatory RNAs. MSC: Remembering 44. Which of the following is NOT involved in post-transcriptional control? a. the spliceosome b. Dicer c. Mediator d. RISC ANS: C DIF: Easy REF: 8.4 OBJ: 8.4.a Define post-transcriptional control. MSC: Remembering 45. MicroRNAs a. are produced from a precursor miRNA transcript. b. are found only in humans. c. control gene expression by base-pairing with DNA sequences. d. can degrade RNAs by using their intrinsic catalytic activity.

ANS: A DIF: East REF: 8.4 OBJ: 8.4.c Differentiate between noncoding RNAs and regulatory RNAs. MSC: Remembering 46. Which of the following statements about miRNAs is FALSE? a. One miRNA can regulate the expression of many genes. b. miRNAs are transcribed in the nucleus from genomic DNA. c. miRNAs are produced from rRNAs. d. miRNAs are made by RNA polymerase.

ANS: C DIF: Easy REF: 8.4 OBJ: 8.4.f Compare microRNAs (miRNAs) and small interfering RNAs (siRNAs) in terms of their generation, processing, structure, and mechanism of destroying target RNA molecules.

MSC: Understanding 47. Which of the following statements about RNAi is TRUE? a. The RNAi mechanism is found only in plants and animals. b. RNAi is induced when double-stranded, foreign RNA is present in the cell. c. RISC uses the siRNA duplex to locate complementary foreign RNA molecules. d. siRNAs bind to miRNAs to induce RNAi. ANS: B DIF: Easy REF: 8.4 OBJ: 8.4.e Summarize the role that RNAi plays in cell defense.

MSC: Understanding 48. The owners of a local bakery ask for your help in improving a special yeast strain they use to make bread. They would

like you to help them design experiments using RNA interference to turn off genes, to allow them to test their hypothesis that certain genes are important for the good flavors found in their bread. Of the components in the following list, which is the most important to check for in this yeast strain if you would like this project to succeed? a. foreign double-stranded RNA b. genes in the genome that code for RISC proteins c. miRNA genes in the genome d. single-stranded siRNAs within the cell ANS: B Without RISC proteins, the RNA degradation necessary to turn off genes would not occur. Foreign double-stranded RNA induces RNAi and cells with an intact RNAi response would not be expected to have foreign double-stranded RNA stably present. miRNA genes could be present in the genome of cells without an intact RNAi response if the ancestral cell did have RNAi capabilities that were lost though evolutionary time. Although the presence of singlestranded siRNAs within the cell may indicate the presence of Dicer, Dicer alone cannot carry out RNAi without RISC. In fact, if Dicer were absent but RISC were present, it might still be possible to carry out RNAi by providing the cell with short, 23-nucleotide-pair siRNAs. DIF: Moderate REF: 8.4 OBJ: 8.4.g Contrast the roles played by RISC and RITS in RNAi. MSC: Applying 49. The long noncoding RNA Xist a. is important for the formation of highly condensed heterochromatin on the X chromosome. b. binds to foreign RNAs in the cell to promote their destruction. c. catalyzes miRNA maturation. d. acts with the RTS complex to bind to complementary RNA molecules.

ANS: A DIF: Easy REF: 8.4 OBJ: 8.4.h Articulate how long noncoding (lnc) RNAs can silence genes on the X chromosome. MSC: Remembering

MULTIPLE SELECT 1. Which method or methods of controlling eukaryotic gene expression is NOT employed in prokaryotic cells? Select all that apply. a. controlling how often a gene is transcribed b. controlling how an RNA transcript is spliced c. controlling which mRNAs are exported from the nucleus to the cytosol d. controlling which mRNAs are translated into protein by the ribosomes ANS: B, C DIF: Easy REF: 8.3 OBJ: 8.3.c Compare how prokaryotes and eukaryotes coordinate the expression of multiple genes. MSC: Remembering

MATCHING 1. Match the following structures (1–4) in Figure 8-51 with the list below.

Figure 8-51

A. activator protein B. RNA polymerase C. general transcription factors D. Mediator 1. ANS: A DIF: Easy REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the role they play in gene expression and the proteins that bind to them. | 8.2.g Explain how eukaryotic activator proteins can enhance transcription even when bound to sequences hundreds or thousands of nucleotide pairs away from a gene’s promoter. MSC: Remembering 2. ANS: D DIF: Easy REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the role they play in gene expression and the proteins that bind to them. | 8.2.g Explain how eukaryotic activator proteins can enhance transcription even when bound to sequences hundreds or thousands of nucleotide pairs away from a gene’s promoter. MSC: Remembering 3. ANS: C DIF: Easy REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the role they play in gene expression and the proteins that bind to them. | 8.2.g Explain how eukaryotic activator proteins can enhance transcription even when bound to sequences hundreds or thousands of nucleotide pairs away from a gene’s promoter. MSC: Remembering 4. ANS: B DIF: Easy REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the role they play in gene expression and the proteins that bind to them. | 8.2.g Explain how eukaryotic activator proteins can enhance transcription even when bound to sequences hundreds or thousands of nucleotide pairs away from a gene’s promoter. MSC: Remembering

SHORT ANSWER 1. The gene for a hormone necessary for insect development contains binding sites for three transcription regulators called A, B, and C. Because the binding sites for A and B overlap, A and B cannot bind simultaneously. You make mutations in the binding sites for each of the proteins and measure hormone production in cells that contain equal

amounts of the A, B, and C proteins. Figure 8-52 summarizes your results. In each of the following sentences, fill in the blank one of the phrases within square brackets to make the statement consistent with the results.

Figure 8-52

1. Protein A binds to its DNA binding site __________ [more tightly/less tightly] than protein B binds to its DNA binding site.

ANS: more tightly DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 2. Protein A is a __________ [stronger/weaker] activator of transcription than protein B. ANS: weaker DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 3. Protein C is able to prevent activation by __________ [protein A only/protein B only/both protein A and protein B].

ANS: both protein A and protein B DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 2. The CAP activator protein and the Lac repressor both control the Lac operon (see Figure 8-53). You create cells that are mutant in the gene coding for the Lac repressor so that these cells lack the Lac repressor under all conditions. For these mutant cells, state whether the Lac operon will be switched on or off in the following situations, and explain why.

Figure 8-53

A. in the presence of glucose and lactose B. in the presence of glucose and the absence of lactose C. in the absence of glucose and the absence of lactose D. in the absence of glucose and the presence of lactose ANS: A. Operon off. CAP will not bind in the presence of glucose.

B. Operon off. Although normally the Lac repressor would bind in the absence of lactose, the lack of the Lac repressor in this case does not matter because the presence of glucose means that the CAP protein will not bind and activate transcription.

C. Operon on. Normally in the absence of both glucose and lactose, the operon would be off. However, because the cells lack the Lac repressor, the cells cannot sense the absence of lactose. Because the CAP protein will bind and activate transcription, the operon will be on.

D. Operon on. The CAP protein will bind and activate transcription because of the presence of glucose. It does not matter whether the Lac repressor gene is mutant, because there is lactose available.

DIF: Difficult REF: 8.2 OBJ: 8.2.f Outline how the Lac operon allows bacteria to utilize the alternative carbon sources lactose and glucose. MSC: Applying 3. You are interested in examining the regulation of the gene that encodes an enzyme, Tre-ase, important in metabolizing trehalose into glucose in bacteria. Trehalose is a disaccharide formed of two glucose units. It is known that two DNA-binding proteins, TreA and TreB, are important for binding to the promoter of the Tre-ase gene and are involved in regulating the transcription of the Tre-ase gene: TreA binds to the “A” site in the promoter region, and TreB binds to the “B” site. You make mutations in the TreA and TreB genes to create cells lacking these genes, observe what happens to transcription of the Tre-ase gene, and obtain the results in Table 8-54.

Table 8-54

A. What is the role for TreA in controlling Tre-ase expression? Explain.

B. What is the role for TreB in controlling Tre-ase expression? Explain.

C. From these data, what do you predict will happen to Tre-ase transcription (compared with that in normal cells) in the presence of trehalose if you were to create a version of the TreA protein that will constitutively bind to the “A” site in the Tre-ase promoter?

ANS: A. TreA is a gene repressor involved in turning off the Tre-ase gene when glucose is present. When cells lack TreB, the Tre-ase gene is not expressed when glucose and trehalose are present in the growth medium. Normally, Trease is not produced when glucose is present, suggesting that TreA is required to keep the Tre-ase gene switched off when glucose is present.

B. TreB is a gene activator involved in turning on the Tre-ase gene when trehalose is present. When cells lack TreB, the Tre-ase gene is not transcribed.

C. If TreA is constitutively bound to the “A” site, no transcription of the Tre-ase gene will occur. TreA is a gene repressor involved in turning off the Tre-ase gene when glucose is present. Transcription of the Tre-ase gene is seen inappropriately in cells lacking TreA when both glucose and trehalose are present in the growth medium, suggesting that the TreB gene activator can activate transcription of the Tre-ase gene in those circumstances and normally does not do so because of the repression mediated by TreA. Thus, if TreA were always bound, it would overcome the activation normally due to TreB, and the Tre-ase gene would not be transcribed.

DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 4. An allosteric transcription regulator called HisP regulates the enzymes for histidine biosynthesis in the bacterium E. coli. Histidine modulates HisP activity. On binding histidine, HisP alters its conformation, markedly changing its affinity for the regulatory sequences in the promoters of the genes for the histidine biosynthetic enzymes.

A. If HisP functions as a gene repressor, would you expect that HisP would bind more tightly or less tightly to the regulatory sequences when histidine is abundant? Explain your answer.

B. If HisP functions as a gene activator, would you expect that HisP would bind more tightly or less tightly to the regulatory sequences when histidine levels are low? Explain your answer.

ANS: A. If HisP functions as a gene repressor, it would bind more tightly to the regulatory sequences when histidine is abundant, because the histidine biosynthetic genes should be turned off when the cell has enough histidine.

B. If HisP functions as a gene activator, it should bind more tightly to the regulatory sequences when histidine levels are low. When histidine levels are low, the cell needs to synthesize more histidine.

DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 5. Bacterial cells can take up the amino acid tryptophan from their surroundings, or, if the external supply is insufficient, they can synthesize tryptophan by using enzymes in the cell. In some bacteria, the control of glutamine synthesis is similar to that of tryptophan synthesis, such that the glutamine repressor inhibits the transcription of the glutamine operon, which contains the genes that code for the enzymes required for glutamine synthesis. On binding to cellular glutamine, the glutamine repressor binds to a site in the promoter of the operon.

A. Why is glutamine-dependent binding to the operon a useful property for the glutamine repressor?

B. What would you expect to happen to the regulation of the enzymes that synthesize glutamine in cells expressing a mutant form of the glutamine repressor that cannot bind to DNA?

C. What would you expect to happen to the regulation of the enzymes that synthesize glutamine in cells expressing a mutant form of the glutamine repressor that binds to DNA even when no glutamine is bound to it?

ANS: A. If sufficient glutamine is present in cells, the glutamine repressor will block the synthesis of enzymes that would make more glutamine. Similarly, if cells are starved for glutamine, the unoccupied repressor would not bind to the DNA, and the enzymes that synthesize glutamine would be induced. These conditions permit a direct connection between the levels of glutamine and the expression of glutamine-synthesizing enzymes.

B. The glutamine synthesis enzymes would be permanently switched on, regardless of the level of glutamine in the cells.

C. The glutamine synthesis enzymes would be always switched off, regardless of the level of glutamine in the cells, because the repressor is always bound to the DNA. These cells will not be able to grow unless glutamine is added to the medium.

DIF: Difficult REF: 8.2 OBJ: 8.2.c Describe the structure of an operon and note the type of organism in which this arrangement is most common. | 8.2.d Review how, in bacteria, the amino acid tryptophan shuts down production of the enzymes responsible for its biosynthesis. MSC: Evaluating 6. In the absence of glucose, E. coli can proliferate by using the pentose sugar arabinose. As shown in Figure 8-57, the arabinose operon regulates the ability of E. coli to use arabinose. The araA, araB, and araD genes encode enzymes for the metabolism of arabinose. The araC gene encodes a transcription regulator that binds adjacent to the promoter of the arabinose operon. To understand the regulatory properties of the AraC protein, you engineer a mutant bacterium in which the araC gene has been deleted and look at the effect of the presence or absence of the AraC protein on the AraA enzyme.

Figure 8-57

A. If the AraC protein works as a gene repressor, would you expect araA RNA levels to be high or low in the presence of arabinose in the araC– mutant cells? What about in the araC– mutant cells in the absence of arabinose? Explain your answer.

B. Your findings from the experiment are summarized in Table 8-57.

Table 8-57

Do the results in Table 8-57 indicate that the AraC protein regulates arabinose metabolism by acting as a gene repressor or a gene activator? Explain your answer.

ANS: A. If the AraC protein acts as a gene repressor for the arabinose operon, araA RNA levels should be high in the pres-

ence or absence of arabinose when there is no AraC protein around. In fact, the araA RNA levels should be high all the time, regardless of the presence or absence of arabinose, because the AraA gene should be transcribed under all conditions in the absence of AraC. B. The results are consistent with AraC acting as a gene activator for the arabinose operon. A gene activator must bind to the promoter regions of the arabinose genes so as to stimulate their transcription. Thus, if the gene for the regulatory protein is deleted, the arabinose genes cannot be turned on.

DIF: Difficult REF: 8.2 OBJ: 8.2.c Describe the structure of an operon and note the type of organism in which this arrangement is most common. MSC: Applying 7. You have discovered an operon in a bacterium that is turned on only when sucrose is present and glucose is absent.

You have also isolated three mutants that have changes in the upstream regulatory sequences of the operon and whose behavior is summarized in the Table 8-58. You hypothesize that there are two gene regulatory sites, A and B, in the upstream regulatory sequence that are affected by the mutations. For this question, a plus (+) indicates a normal site and a minus (–) indicates a mutant site that no longer binds its transcription regulator.

Table 8-58

A. If mutant 1 has sites A– B+, which of these sites is regulated by sucrose and which by glucose?

B. Give the state (+ or—) of the A and B sites in mutants 2 and 3.

C. Which site is bound by a repressor and which by an activator?

ANS: A. Site A is regulated by sucrose, and site B by glucose.

B. Mutant 2 (A+ B–); mutant 3 (A– B–) or (A– B+).

C. Site A is bound by an activator, and site B by a repressor.

DIF: Difficult REF: 8.2 OBJ: 8.2.c Describe the structure of an operon and note the type of organism in which this arrangement is most common. MSC: Applying 8. You are interested in studying the transcriptional regulation of the Gip1 promoter. The Gip1 promoter contains a binding site for the Jk8 protein that overlaps with the binding site for the Pa5 protein. Jk8 and Pa5 cannot bind DNA at the same time, but both proteins are present at high levels in adult liver cells. The binding sites for Jk8 and Pa5 are shown in Figure 8-24.

Figure 8-24

Jk8 binds to site A while Pa5 binds to site B. You create mutations that remove the nonoverlapping sequences of

either binding site A or B, and examine Gip1 mRNA production in adult liver cells that contain these mutations. The data you obtain from these experiments are shown in Table 8-24.

Table 8-24

Which transcription regulatory protein is bound in experiment #1 (the normal situation)? Explain.

ANS: Protein Jk8 is bound to the Gip1 promoter because Gip1 mRNA is produced. Jk8 is a gene activator, as seen in experiment #2 when the A binding site is present (and thus Jk8 is bound to the promoter) and Gip1 mRNA is produced. Pa5 acts as a gene repressor, since experiment #3 shows that when Pa5 is bound, there is no Gip1 mRNA produced.

DIF: Moderate REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. MSC: Applying 9. The yeast GAL4 gene encodes a transcription regulator that can bind DNA upstream of genes required for the metabolism of the sugar galactose and turn them on. Gal4 has a DNA-binding domain and an activation domain. The DNA-binding domain allows it to bind to the appropriate sites in the promoters of the galactose metabolism genes.

The activation domain attracts histone-modifying enzymes and also binds to a component of the RNA polymerase II enzyme complex, attracting it to the promoter so that the regulated genes can be turned on when Gal4 is also bound to the DNA. When Gal4 is expressed normally, the genes can be maximally activated. You decide to try to produce more of the galactose metabolism genes by overexpressing the Gal4 protein at levels fiftyfold greater than normal.

You conduct experiments to show that you are overexpressing the Gal4 protein and that it is properly localized in the nucleus of the yeast cells. To your surprise, you find that too much Gal4 causes the galactose genes to be transcribed only at a low level. What is the most likely explanation for your findings?

ANS: For Gal4 to work properly, the DNA-bound Gal4 must attract histone-modifying enzymes and recruit RNA polymerase to the promoter. If there is too much Gal4 in the cell, the non-DNA-bound Gal4 (or free Gal4) will compete with the DNA-bound Gal4 for binding to histone-modifying enzymes and RNA polymerase. The excess amount of Gal4 forms nonproductive complexes with histone-modifying enzymes and RNA polymerase, preventing their recruitment to the promoter and lowering the level of transcription.

DIF: Difficult REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the roles they play in gene expression and the proteins that bind to them. | 8.2.i Compare how eukaryotic activator and repressor proteins exploit the mechanisms that regulate chromatin packaging to enhance or suppress transcription.

MSC: Evaluating 10. A virus produces a protein X that activates only a few of the virus’s own genes (V1, V2, and V3) when it infects cells.

The cellular proteins A (a zinc finger protein) and B (a homeodomain protein) are known to be repressors of the viral genes V1, V2, and V3. You examine the complete upstream gene regulatory sequences of these three viral genes and find the following: 1. V1 and V2 contain binding sites for the zinc finger protein, A, only.

2. V3 contains a binding site for the homeodomain protein, B, only.

3. The only sequence that all three genes have in common is the TATA box.

Label each of the choices below as LIKELY or UNLIKELY as an explanation for your findings. For each choice you label as UNLIKELY, explain why.

A. Protein X binds nonspecifically to the DNA upstream of V1, V2, and V3 and activates transcription.

B. Protein X binds to a repressor and prevents the repressor from binding upstream of V1, V2, and V3.

C. Protein X activates transcription by binding to the TATA box.

D. Protein X activates transcription by binding to and sequestering proteins A and B. E. Protein X represses transcription of the genes for proteins A and B. ANS: A. Unlikely. If protein X were to bind nonspecifically to DNA, it would not specifically regulate a particular subset of genes.

B. Unlikely. If a single repressor were to bind upstream of V1, V2, and V3, we would expect to have found a binding site common to all three genes, but there is none.

C. Unlikely. If a protein X that activated gene transcription were to bind to the TATA box, it would be likely to activate most genes transcribed by RNA polymerase II, because most genes contain TATA boxes.

D. Likely E. Likely DIF: Difficult REF: 8.2 OBJ: 8.2.h Summarize how eukaryotic repressor proteins decrease transcription.

MSC: Evaluating 11. Fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

The genes of a bacterial __________ are transcribed into a single mRNA. Many bacterial promoters contain a region known as a/an __________, to which a specific transcription regulator binds. Genes in which transcription is prevented are said to be __________. The interaction of small molecules, such as tryptophan, with __________ DNAbinding proteins, such as the tryptophan repressor, regulates bacterial genes. Genes that are being __________ expressed are being transcribed all the time. allosteric

negatively

positively

constitutively

operator

promoter

induced

operon

repressed

ANS: The genes of a bacterial operon are transcribed into a single mRNA. Many bacterial promoters contain a region known as an operator, to which a specific transcription regulator binds. Genes in which transcr iption is prevented are said to be repressed. The interaction of small molecules, such as tryptophan, with allosteric DNAbinding proteins, such as the tryptophan repressor, regulates bacterial genes. Genes that are being constitutively expressed are being transcribed all the time. DIF: Easy REF: 8.2 OBJ: 8.2.c Describe the structure of an operon and note the type of organism in which this arrangement is most common. | 8.2.d Review how, in bacteria, the amino acid tryptophan shuts down production of the enzymes responsible for its biosynthesis.

MSC: Understanding 12. For each of the following sentences, fill in the blanks with the best word or phrase in the list below. Not all words or

phrases will be used; use each word or phrase only once.

During transcription in __________ cells, transcription regulators that bind to DNA thousands of nucleotides away from a gene’s promoter can affect a gene’s transcription. The __________ is a complex of proteins that links distantly bound transcription regulators with the proteins bound closer to the transcription start site. Transcription activators can also interact with histone __________s, which alter chromatin by modifying lysines in the tail of histone proteins to allow greater accessibility to the underlying DNA. Gene repressor proteins can reduce the efficiency of transcription initiation by attracting histone __________s. Sometimes, many contiguous genes can become transcriptionally inactive as a result of chromatin remodeling, like the __________ found in interphase chromosomes. acetyltransferase

eukaryotic

operator

centrosome

helicase

peroxidase

deacetylase

heterochromatin

prokaryotic

deoxidase

leucine zipper

telomere

enhancer

Mediator

viral

ANS: During transcription in eukaryotic cells, transcription regulators that bind to DNA thousands of nucleotides away from a gene’s promoter can affect a gene’s transcription. The Mediator is a complex of proteins that links distantly bound transcription regulators with the proteins bound closer to the transcription start site. Transcription activators can also interact with histone acetyltransferases, which alter chromatin by modifying lysines in the tail of histone proteins to allow greater accessibility to the underlying DNA. Gene repressor proteins can reduce the efficiency of transcription initiation by attracting histone deacetylases. Sometimes, many contiguous genes can become transcriptionally inactive as a result of chromatin remodeling, like the heterochromatin found in interphase chromosomes.

DIF: Easy REF: 8.2 OBJ: 8.2.b Illustrate how transcription regulators recognize and bind to regulatory DNA s equences along a DNA double helix. | 8.2.i Compare how eukaryotic activator and repressor proteins exploit the mechanisms that regulate chromatin packaging to enhance or suppress transcription.

MSC: Understanding 13. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. combinatorial

feedback

phosphorylation

deacetylation

histone

pluripotency

differential

leucine zipper

proliferation

epigenetic

memory

receptor

expression

methylation

unwinding

The transmission of information important for gene regulation from parent to daughter cell, without altering the actual nucleotide sequence, is called __________ inheritance. This type of inheritance is seen with the inheritance of the covalent modifications on __________ proteins bound to DNA; these modifications are important for reestablishing the pattern of chromatin structure found on the parent chromosome. Another way to inherit chromatin structure involves DNA __________, a covalent modification that occurs on cytosine bases that typically turns off the transcription of a gene. Gene transcription patterns can also be transmitted across generations through positive __________ loops that can involve a transcription regulator activating its own transcription in addition to other genes.

These mechanisms all allow for cell __________, a property involving the maintenance of gene expression patterns

important for cell identity.

ANS: The transmission of information important for gene regulation from parent to daughter cell, without altering the actual nucleotide sequence, is called epigenetic inheritance. This type of inheritance is seen with the inheritance of the covalent modifications on histone proteins bound to DNA; these modifications are important for reestablishing the pattern of chromatin structure found on the parent chromosome. Another way to inherit chromatin structure involves DNA methylation, a covalent modification that occurs on cytosine bases that typically turns off the transcription of a gene.

Gene transcription patterns can also be transmitted across generations through positive feedback loops that can involve a transcription regulator activating its own transcription in addition to other genes. These mechanisms all allow for cell memory, a property involving the maintenance of gene expression patterns important for cell identity.

DIF: Easy REF: 8.3 OBJ: 8.3.i Relate how positive feedback, DNA methylation, and histone modification allow differentiated cells to maintain their identity and to pass this information to daughter cells during cell division. | 8.3.j Define epigenetic inheritance. MSC: Understanding 14. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. acetylation

methylation

riboswitch

destruction

mitochondria

RISC

Dicer

mRNA

rRNA

DNA

phosphorylation

single-stranded RNA

double-stranded RNA

prokaryotic

tRNA

MicroRNAs are noncoding RNAs that are incorporated into a protein complex called __________, which searches the __________s in the cytoplasm for sequence complementary to that of the miRNA. When such a molecule is found, it is then targeted for __________. RNAi is triggered by the presence of foreign __________ molecules, which are digested by the __________ enzyme into shorter fragments approximately 23 nucleotide pairs in length.

ANS: MicroRNAs are noncoding RNAs that are incorporated into a protein complex called RISC, which searches the mRNAs in the cytoplasm for sequence complementary to that of the miRNA. When such a molecule is foun d, it is then targeted for destruction. RNAi is triggered by the presence of foreign double-stranded RNA molecules, which are digested by the Dicer enzyme into shorter fragments approximately 23 nucleotide pairs in length.

DIF: Easy REF: 8.4 OBJ: 8.4.d Outline how microRNAs can block the production of a particular protein and specify how the mRNAs targeted by this mechanism can be destroyed. | 8.4.e Summarize the role that RNAi plays in cell defense. | 8.4.g Contrast the roles played by RISC and RITS in RNAi. MSC: Understanding 15. The Drosophila Eve gene has a complex promoter containing multiple binding sites for four transcription regulators: Bicoid, Hunchback, Giant, and Krüppel. Bicoid and Hunchback are activators of Eve transcription, whereas Giant and Krüppel repress Eve transcription. Figure 8-66A shows the patterns of expression of these regulators.

Figure 8-66

The Eve promoter contains modules that control expression in various stripes. You construct a reporter gene that contains the DNA 5 kb upstream of the Eve gene, so that this reporter contains the stripe 3 module, the stripe 2 module, the stripe 7 module, and the TATA box, all fused to the LacZ reporter gene (which encodes the βgalactosidase enzyme), as shown in Figure 8-66B. This construct results in expression of the β-galactosidase enzyme in three stripes, which correspond to the normal positions of stripes 3, 2, and 7.

A. By examining the overlap of sites on the stripe 2 module, as depicted in Figure 8-66B, what is the biological effect of having some of the transcription regulator binding sites overlap?

B. You make two mutant versions in which several of the binding sites in the Eve stripe 2 module have been deleted, as detailed in items (i) and (ii) below. Refer to Figure 8-66B for the positions of the binding sites. (Note, however, that because many of the binding sites overlap, it is not possible to delete all of one kind of site without affecting some of the other sites.) Match the appropriate mutant condition with the most likely pattern of Eve expression shown in Figure 8-66C. Explain your choices. i. deletion of the Krüppel-binding sites in stripe 2 ii. deletion of the two Bicoid-binding sites in the stripe 2 module that are marked with an asterisk (*) in Figure 8-66B ANS: A. Binding sites for the repressor proteins Krüppel and Giant do not seem to overlap. Binding sites for the activator

proteins Bicoid and Hunchback also do not seem to overlap. Instead, the binding sites for repressor proteins seem to overlap with the binding sites for activator proteins. These overlapping binding sites cause repressor and activator proteins to compete for binding to the DNA. It is thought that the binding of a repressor and an activator is mutually exclusive. The overlap between the repressor- and activator-binding sites allows Eve expression to be exquisitely sensitive to the levels of repressors and activators in the cell, and suggests that the repressors function by preventing activator binding. In fact, the repressors and activators can antagonize each other, allowing the creation of sharp stripes of transcription from smooth gradients of protein regulatory factors.

B. (i) Mutant embryo (b). When the Krüppel-binding sites are removed, the effects of the Krüppel repressor are eliminated. Stripe 2 expression now expands slightly in the posterior direction, which is to be expected because Hunchback and Bicoid expression extend slightly beyond the posterior end of stripe 2. (ii) Mutant embryo (c). When two of the Bicoid-binding sites are removed, expression from the promoter is less sensitive to the effects of the Bicoid activator. Thus, stripe 2 appears at its normal position, but the expression of β-galactosidase is decreased.

DIF: Difficult REF: 8.3 OBJ: 8.3.d Articulate how a reporter gene can be used to determine where or when a particular gene is activated during embryonic development. MSC: Analyzing 16. In principle, a eukaryotic cell can regulate gene expression at any step in the pathway from DNA to the active protein.

Place the types of control listed below at the appropriate places on the diagram in Figure 8-67.

Figure 8-67

A. translation control B. transcriptional control C. RNA splicing D. RNA degradation E. protein degradation ANS: See Figure 8-67A.

Figure 8-67A

DIF: Easy REF: 8.1 OBJ: 8.1.e Articulate the steps at which gene expression can be regulated and identify the step that, for most genes, is the main point of control. MSC: Remembering 17. Fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. allosteric

negatively

positively

constitutively

operator

promoter

induced

operon

repressed

The genes of a bacterial __________ are transcribed into a single mRNA. Many bacterial promoters contain a region known as a/an __________, to which a specific transcription regulator binds. Genes in which transcription is prevented are said to be __________. The interaction of small molecules, such as tryptophan, with __________ DNAbinding proteins, such as the tryptophan repressor, regulates bacterial genes. Genes that are being __________ expressed are being transcribed all the time.

ANS: The genes of a bacterial operon are transcribed into a single mRNA. Many bacterial promoters contain a region known as an operator, to which a specific transcription regulator binds. Genes in which transcription is prevented are said to be repressed. The interaction of small molecules, such as tryptophan, with allosteric DNA-binding proteins, such as the tryptophan repressor, regulates bacterial genes. Genes that are being constitutively expressed are being transcribed all the time.

DIF: Easy REF: 8.2 OBJ: 8.2.c Describe the structure of an operon and note the type of organism in which this arrangement is most common. | 8.2.d Review how, in bacteria, the amino acid tryptophan shuts down production of the enzymes responsible for its biosynthesis. MSC: Understanding 18. Label the following structures in Figure 8-51.

Figure 8-51

A. activator protein B. RNA polymerase C. general transcription factors D. Mediator ANS: A—1; B—4; C—3; D—2 Also see Figure 8-51A.

Figure 8-51A

DIF: Easy REF: 8.2 OBJ: 8.2.a Distinguish between promoters and regulatory DNA sequences in terms of the role they play in gene expression and the proteins that bind to them. | 8.2.g Explain how eukaryotic activator proteins can enhance transcription even when bound to sequences hundreds or thousands of nucleotide pairs away from a gene’s

promoter. MSC: Remembering 19. For each of the following sentences, fill in the blanks with the best word or phrase in the list below. Not all words or phrases will be used; use each word or phrase only once. acetyltransferase

eukaryotic

operator

centrosome

helicase

peroxidase

deacetylase

heterochromatin

prokaryotic

deoxidase

leucine zipper

telomere

enhancer

Mediator

viral

During transcription in __________ cells, transcription regulators that bind to DNA thousands of nucleotides away from a gene’s promoter can affect a gene’s transcription. The __________ is a complex of proteins that links distantly bound transcription regulators with the proteins bound closer to the transcription start site. Transcription activators can also interact with histone __________s, which alter chromatin by modifying lysines in the tail of histone proteins to allow greater accessibility to the underlying DNA. Gene repressor proteins can reduce the efficiency of transcription initiation by attracting histone __________s. Sometimes, many contiguous genes can become transcriptionally inactive as a result of chromatin remodeling, like the __________ found in interphase chromosomes.

ANS: During transcription in eukaryotic cells, transcription regulators that bind to DNA thousands of nucleotides away from a gene’s promoter can affect a gene’s transcription. The Mediator is a complex of proteins that links distantly bound transcription regulators with the proteins bound closer to the transcription start site. Transcription activators can also interact with histone acetyltransferases, which alter chromatin by modifying lysines in the tail of histone proteins to allow greater accessibility to the underlying DNA. Gene repressor proteins can reduce the efficiency of transcription initiation by attracting histone deacetylases. Sometimes, many contiguous genes can become transcriptionally inactive as a result of chromatin remodeling, like the heterochromatin found in interphase chromosomes.

DIF: Easy REF: 8.2 OBJ: 8.2.b Illustrate how transcription regulators recognize and bind to regulatory DNA sequences along a DNA double helix. | 8.2.i Compare how eukaryotic activator and repressor proteins exploit the mechanisms that regulate chromatin packaging to enhance or suppress transcription. MSC: Understanding

CHAPTER 9 How Genes and Genomes Evolve

GENERATING GENETIC VARIATION 9.1.a Summarize six basic mechanisms that generate genetic change.

9.1.b Contrast the fate of mutations that arise in somatic cells versus germ-line cells.

9.1.c Outline an experimental approach to determining the rate at which point mutations accumulate in bacteria.

9.1.d Compare the point mutation rates of E. coli and humans and assess the significance of these relative values.

9.1.e Recall how point mutations typically arise and evaluate how the locations of these mutations can dictate their effects on an organism’s appearance or fitness.

9.1.f Review how a point mutation affects our ability to digest lactose.

9.1.g Illustrate how homologous recombination can lead to gene duplication.

9.1.h Articulate how gene duplication can lead to the evolution of genes with new functions or to the generation of pseudogenes.

9.1.i Outline how gene duplication and divergence gave rise to the globin gene family.

9.1.j Present evidence that supports the occurrence of whole genome duplication events.

9.1.k Compare the mechanisms of exon shuffling and gene duplication.

9.1.l Express how exon shuffling can facilitate the evolution of new proteins.

9.1.m Explain how mobile genetic elements can alter the activity of regulation of a gene or promote gene duplication and exon shuffling.

9.1.n Describe the process of horizontal gene transfer and state how this form of genetic exchange has significantly impacted human health.

RECONSTRUCTING LIFE’S FAMILY TREE 9.2.a Define homologous genes and state the percentage of human genes that have clear homologs in species such as the fruit fly and nematode.

9.2.b Compare the fate of germ-line mutations that are deleterious, selectively neutral, or that provide a selective advantage.

9.2.c Contrast the types of DNA sequences that can accommodate mutations with those that cannot.

9.2.d Illustrate how selectively neutral mutations, combined with analysis of the fossil record, can be used to construct a phylogenetic tree.

9.2.e Outline how a comparison of the nucleotide sequences from two closely related organisms can be used to reconstruct the amino acid sequence of a protein from their extinct, common ancestor.

9.2.f Contrast the differences between the chimp and human genomes with those between the human and mouse genomes.

9.2.g Express what conserved synteny between two extant—or living—species indicates about the genome of their common ancestor.

9.2.h Explain how purifying selection leads to the conservation of functionally important DNA sequences and characterize the roles that these conserved sequences might play.

9.2.i Summarize how the Fugu genome differs from the human genome in terms of size and approximate number of genes.

9.2.j Recall why a ribosomal RNA gene was selected to construct a phylogenetic “tree of life.”

MOBILE GENETIC ELEMENTS AND VIRUSES 9.3.a Present the components of a typical transposon.

9.3.b Contrast cut-and-paste and replicative transposition for DNA-only transposons.

9.3.c Compare the movement of DNA-only transposons and retrotransposons.

9.3.d Differentiate between L1 elements and Alu sequences.

9.3.e Articulate how viruses differ from mobile genetic elements.

9.3.f Review how and why viruses use the host’s biochemical machinery to reproduce themselves.

9.3.g Outline the steps involved in the integration and replication of a retrovirus.

9.3.h Assess how retroviruses and retrotransposons are similar.

EXAMINING THE HUMAN GENOME 9.4.a State the total size of the human genome, the total number of genes, and the number of chromosomes used to carry this genetic information.

9.4.b Differentiate between the amount of the human genome that codes for proteins versus the amount that consists of mobile genetic elements.

9.4.c Review the structure of genes in the genomes of human, fly, and yeast, and compare how densely genes are distributed in the chromosomes of each organism.

9.4.d Compare the number of protein-coding genes present in humans versus Drosophila, C. elegans, and Arabidopsis, and analyze how these values relate to the complexity of the organism.

9.4.e Describe how the early estimate of 100,000 human genes was derived.

9.4.f Review how open reading frames are used in estimating gene number.

9.4.g Evaluate how RNA sequencing can provide a more accurate estimate of the number of genes in a genome than DNA sequencing.

9.4.h Express how changes in regulatory DNA sequences can contribute to the evolution of species.

9.4.i Summarize what has been revealed by the comparison of human and Neanderthal DNA. 9.4.j State the amount of variation that typically distinguishes one human genome from another and describe the most common form of genetic variation.

MULTIPLE CHOICE 1. Which of the following statements is FALSE? a. A mutation that arises in a mother’s somatic cell often causes a disease in her daughter. b. All mutations in an asexually reproducing single-celled organism are passed on to the progeny. c. In an evolutionary sense, somatic cells exist only to help propagate germ-line cells. d. A mutation is passed on to offspring only if it is present in the germ line.

ANS: A Mutations are carried in the genetic material, and the only genetic material passed along to the offspring of a sexually reproducing organism comes from a germ-line cell (not a somatic cell).

DIF: Easy REF: 9.1 OBJ: 9.1.b Contrast the fate of mutations that arise in somatic cells versus germ-line cells.

MSC: Understanding 2. Your friend works in a lab that is studying why a particular mutant strain of Drosophila grows an eye on its wing. Your friend discovers that this mutant strain of Drosophila is expressing a transcription factor incorrectly. In the mutant Drosophila, this transcription factor, which is normally expressed in the primordial eye tissue, is now misexpressed in the primordial wing tissue, thus turning on transcription of the set of genes required to produce an eye in the wing primordial tissue. If this hypothesis is true, which of the following types of genetic change would most likely lead to this situation? a. A mutation within the transcription factor gene that leads to a premature stop codon after the third amino acid. b. A mutation within the transcription factor gene that leads to a substitution of a positively charged amino acid for a negatively charged amino acid. c. A mutation within an upstream enhancer of the gene. d. A mutation in the TATA box of the gene.

ANS: C A mutation within an upstream enhancer of the gene will affect the regulation of gene expression. Mutations within the coding sequence will lead to a mutated protein being produced in the proper tissues at the proper time. A mutation in the TATA box of the gene will probably lead to no expression at all.

DIF: Moderate REF: 9.1 OBJ: 9.1.e Recall how point mutations typically arise and evaluate how the location of these mutations can dictate their effect on an organism’s appearance or fitness. MSC: Analyzing 3. You discover that the underlying cause of a disease is a protein that is now less stable than the non-disease-causing version of the protein. This change is most likely to be due to a. a mutation within a gene. b. a mutation within the regulatory DNA of a gene. c. gene duplication. d. horizontal gene transfer.

ANS: A DIF: Easy REF: 9.1 OBJ: 9.1.a Summarize the six basic mechanisms that generate genetic change.

MSC: Applying 4. You isolate a pathogenic strain of E. coli from a patient and discover that this E. coli strain is resistant to an antibiotic.

Common laboratory strains of E. coli are not resistant to this antibiotic, nor are any other previously isolated pathogenic E. coli strains. However, such resistance has been observed in other bacteria in the hospital in which the patient was treated. This newly discovered antibiotic resistance in E. coli is most likely due to a. a mutation within a gene. b. a mutation within the regulatory DNA of a gene. c. gene duplication. d. horizontal gene transfer.

ANS: D Horizontal gene transfer of antibiotic-resistance genes occurs commonly among bacterial strains, and human beings are hosts to a plethora of bacteria. Although it is possible that there is a gene in the normal E. coli genome that can lead to antibiotic resistance when mutated or duplicated, this is less likely, especially given that this type of antibiotic resistance has never been observed.

DIF: Moderate REF: 9.1 OBJ: 9.1.n Describe the process of horizontal gene transfer and state how this form of genetic exchange can ultimately affect human health. MSC: Applying 5. What is the most likely explanation of why the overall mutation rates in bacteria and in humans are roughly similar? a. Cell division needs to be fast. b. Most mutations are silent. c. There is a narrow range of mutation rates that offers an optimal balance between keeping the genome stable and generating sufficient diversity in a population. d. It benefits a multicellular organism to have some variability among its cells.

ANS: C Although it is true that most mutations are silent, this fact cannot explain the similar mutation rate.

DIF: Easy REF: 9.1 OBJ: 9.1.d Compare the point mutation rate of E. coli and humans and assess the significance of these relative values. MSC: Understanding 6. Two individuals are represented in Figure 9-6; individual 1 is one of the parents of individual 2. The asterisk indicates the occurrence of a single mutation.

Figure 9-6

What is the chance that individual 2 will inherit the mutation in individual 1? a. 100% b. 50% c. 1 in 100,000 d. none ANS: B Although the mutation in individual 1 arose before the differentiation of germ cells, these organisms are diploid and thus only half of individual 1’s germ cells will contain the mutation.

DIF: Easy REF: 9.1 OBJ: 9.1.b Contrast the fate of mutations that arise in somatic cells versus germ-line cells.

MSC: Applying 7. Figure 9-7 shows an experiment used to determine the spontaneous mutation rate in E. coli. If the spontaneous mutation rate in E. coli is 1 mistake in every 109 nucleotides copied, about how many colonies would you expect to see on the plates lacking histidine if you were to assay 1011 cells from the culture for their ability to form colonies?

Figure 9-7

a. 1 b. 2 c. 10 d. 100 ANS: D DIF: Easy REF: 9.1 OBJ: 9.1.c Outline an experimental approach to determining the rate at which point mutations accumulate in bacteria. MSC: Applying 8. Which of the following changes is least likely to arise from a point mutation in a regulatory region of a gene? a. A mutation that changes the time in an organism’s life during which a protein is expressed. b. A mutation that eliminates the production of a protein in a specific cell type. c. A mutation that changes the subcellular localization of a protein. d. A mutation that increases the level of protein production in a cell.

ANS: C Information for the subcellular localization of a protein is usually encoded within the translated portion of the gene.

DIF: Moderate REF: 9.1 OBJ: 9.1.e Recall how point mutations typically arise and evaluate how the location of these mutations can dictate their effect on an organism’s appearance or fitness. MSC: Understanding 9. Which of the following statements about gene families is FALSE? a. Because gene duplication can occur when crossover events occur, genes are always duplicated onto homologous chromosomes. b. Not all duplicated genes will become functional members of gene families.

c. Whole-genome duplication can contribute to the formation of gene families. d. Duplicated genes can diverge in both their regulatory regions and their coding regions.

ANS: A Regions of homology between nonhomologous chromosomes will cause gene duplications onto a different chromosome (as well as chromosome rearrangements).

DIF: Easy REF: 9.1 OBJ: 9.1.h Articulate how gene duplication can lead to the evolution of genes with new function or to the generation of pseudogenes. MSC: Understanding 10. Which of the following statements about the globin gene family is TRUE? a. The globin protein, which can carry oxygen molecules throughout an organism’s body, was first seen in ancient vertebrate species about 500 million years ago. b. The gene duplication that led to the expansion of the globin gene family led to the separation and distribution of globin on many chromosomes in mammals, such that no chromosome has more than a single functional member of the globin gene family. c. As globin gene family members diverged over the course of evolution, all the DNA sequence variations that have accumulated between family members are within the regulatory DNA sequences that affect when and how strongly each globin gene is expressed. d. Some of the duplicated globin genes that arose during vertebrate evolution acquired inactivating mutations and became pseudogenes in modern vertebrates.

ANS: D Some of the duplicated globin genes are now pseudogenes in modern vertebrates. Globin proteins have been found in insects, primitive fish, and marine worms in addition to vertebrates. Gene duplication does not necessarily mean distribution among chromosomes, and in humans, the β-globin genes are located in a cluster on Chromosome 11, while the α-globin genes are on Chromosome 16. Mutations have occurred in both the regulatory DNA sequences and within the protein-coding regions of the globin genes.

DIF: Easy REF: 9.1 OBJ: 9.1.h Articulate how gene duplication can lead to the evolution of genes with new function or to the generation of pseudogenes. | 9.1.i Outline how gene duplication and divergence gave rise to the globin gene family. MSC: Remembering 11. Which of the following statements about pseudogenes is FALSE? a. All pseudogenes code for microRNAs. b. Pseudogenes share significant nucleotide similarity with functional genes. c. Pseudogenes are no longer expressed as a protein product. d. There are approximately 11,000 pseudogenes in the human genome.

ANS: A DIF: Easy REF: 9.1 OBJ: 9.1.h Articulate how gene duplication can lead to the evolution of genes with new function or to the generation of pseudogenes. MSC: Remembering 12. Figure 9-12 shows the evolutionary history of the globin gene family members.

Figure 9-12

Given this information, which of the following statements is TRUE? a. The ancestral globin gene arose 500 million years ago. b. The α-globin gene is more closely related to the ε-globin gene than to the δ-globin gene. c. The nucleotide sequences of the two γ-globins will be most similar because they are the closest together on the chromosome. d. The fetal β-globins arose from a gene duplication that occurred 200 million years ago, which gave rise to a β-globin expressed in the fetus and a β-globin expressed in the adult.

ANS: D The α-globin and β-globin genes arose 500 million years ago, not the ancestral globin gene. The α-globin gene is as related to the ε-globin gene as it is to the δ-globin gene. The distance between the γ-globins on the chromosome cannot be used to predict sequence similarity.

DIF: Moderate REF: 9.1 OBJ: 9.1.h Articulate how gene duplication can lead to the evolution of genes with new function or to the generation of pseudogenes. | 9.1.i Outline how gene duplication and divergence gave rise to the glo bin gene family. MSC: Analyzing 13. Which of the following situations would facilitate the process of exon shuffling? a. shorter introns b. a haploid genome c. exons that code for more than one protein domain d. introns that contain regions of similarity to one another ANS: D Because exon shuffling can occur by recombination between introns, introns with regions of similarity to one another will facilitate shuffling. Exon shuffling is facilitated by long introns and by short exons that each code for one protein domain. A haploid genome will probably be LESS prone to exon shuffling than a diploid genome, because having two

copies of each gene allows an organism to keep one copy of the gene as a backup while it shuffles the other copy.

DIF: Easy REF: 9.1 OBJ: 9.1.a Summarize the six basic mechanisms that generate genetic change. | 9.1.k Compare the mechanisms of exon shuffling and gene duplication. | 9.1.l Express how exon shuffling can facilitate the evolution of new proteins. MSC: Applying 14. Which of the following is the least likely to be a selectively neutral mutation? (The codon table in Figure 9-14 will help you answer this question.)

Table 9-14

a. A mutation that deletes 50% of a pseudogene. b. A mutation that changes the CCC codon to the CCT codon in a protein-coding gene. c. A mutation that changes the TCC codon to the AGT codon in a protein-coding gene. d. A mutation that changes the TAT codon to the TAG codon in a protein-coding gene.

ANS: D TAT codes for tryptophan, while TAG codes for a stop codon, and this substitution is most likely to be detrimental to gene function. Although TCC and AGT look very different at the nucleotide level, both codons code for serine. Similarly, CCC and CCT will both code for proline. Pseudogenes do not make functional proteins, and thus removal of 50% of a pseudogene is likely to be neutral.

DIF: Moderate REF: 9.1 OBJ: 9.2.c Contrast the types of DNA sequences that can accommodate mutations with those that cannot. | 9.1.e Recall how point mutations typically arise and evaluate how the location of these mutations can dictate their effect on an organism’s appearance or fitness. MSC: Analyzing 15. Which of the following statements is TRUE? a. The position of introns in most genes is conserved among vertebrates. b. The more nucleotides there are in an organism’s genome, the more genes there will be in its genome. c. Because the fly Drosophila melanogaster and humans diverged from a common ancestor so long ago, any two fly genes will show more similarity to each other than it will to a human gene. d. Two closely related organisms are more likely to have a genome of the same size than a more evolutionarily

diverged animal.

ANS: A There is no necessary correlation between genome size and gene number or genome size and evolutionary relatedness. There are some fly genes, particularly those with a conserved function, that show much greater similarity to human genes than to another fly gene.

DIF: Moderate REF: 9.2 OBJ: 9.2.a Define homologous genes and state the percentage of human genes that have clear homologs in species such as the fruit fly and nematode. | 9.2.h Explain how purifying selection leads to the conservation of functionally important sequences and characterize the roles that these conserved sequences might play. | 9.4.d Compare the number of protein-coding genes present in humans versus Drosophila, C. elegans, and Arabidopsis, and analyze how these values relate to the complexity of the organism. | 9.4.c Review the structure of genes in the genomes of human, fly, and yeast, and compare how densely genes are distributed in the chromosomes of each organism. MSC: Understanding 16. Which of the following statements about homologous genes is TRUE? a. For protein-coding genes, homologous genes will show more similarity in their amino acid sequences than in their nucleotide sequences. b. Fewer than 1% of human genes have homologs in the nematode and the fruit fly. c. Most homologous genes arose by gene duplication. d. A gene in humans that has homologs in plants and prokaryotes will show the same level of similarity in nucleotide sequence when the human and prokaryotic sequences are compared as when the human and chimpanzee sequences are compared.

ANS: A Because of the degeneracy of the genetic code, nucleotide sequences can diverge but still code for identical amino acids.

DIF: Easy REF: 9.2 OBJ: 9.2.a Define homologous genes and state the percentage of human genes that have clear homologs in species such as fruit fly and nematode. | 9.2.c Contrast the types of DNA sequences that can accommodate mutations with those that cannot. | 9.2.e Outline how a comparison of the nucleotide sequences from two closely related organisms can be used to reconstruct the amino acid sequence of a protein from their extinct, common ancestor. MSC: Understanding 17. Figure 9-17 shows the nucleotide sequence from a protein-coding region of a gene in humans, chimpanzees, and gorillas and the protein sequence produced from this gene. The seventeen amino acids encoded by this DNA are numbered below. The two codons that are not conserved in all three species have been boxed. These two codons code for amino acids 3 and 15.

Figure 9-17

Which of these statements is consistent with these sequence-comparison data? a. The gorilla sequence is more similar to the chimp sequence than to the human sequence. b. Since these sequences are so similar, this protein must also be found in invertebrates. c. The chimp DNA sequence has likely diverged at the DNA coding for amino acid 15 from the sequence found in the last common ancestor of humans and chimps. d. The last common ancestor of chimps and gorillas most likely used AAA to code for amino acid number 3.

ANS: C Because humans and gorillas both use CCC to code for proline and because humans and chimps are more closely related than humans and gorillas, the last common ancestor of humans and chimps likely used CCC at amino acid 15 to code for proline.

DIF: Difficult REF: 9.2 OBJ: 9.2.d Illustrate how selectively neutral mutations, combined with analysis of the fossil record, can be used to construct a phylogenetic tree. | 9.2.e Outline how a comparison of the nucleotide sequences from two closely related organisms can be used to reconstruct the amino acid sequence of a protein from their extinct, common ancestor. MSC: Analyzing 18. Given the evolutionary relationship between higher primates shown in Figure 9-18, which of the following statements is FALSE?

Figure 9-18

a. The last common ancestor of humans, chimpanzees, gorillas, and orangutans lived about 15 million years ago. b. Chimpanzees are more closely related to gorillas than to humans. c. Humans and chimpanzees diverged about 6 million years ago. d. Orangutans are the most divergent of the four species shown in Figure 9-18.

ANS: B DIF: Easy REF: 9.2 OBJ: 9.2.d Illustrate how selectively neutral mutations, combined with analysis of the fossil record, can be used to construct a phylogenetic tree. MSC: Applying 19. You are interested in finding out how the budding yeast Saccharomyces cerevisiae is so good for making bread and have collected five new related species from the wild. You sequence the genomes of all of these new species and also consult with a fungal biologist to help you construct the phylogenetic tree shown in Figure 9-19. You find that species V, W, and X make pretty good bread whereas species Y and Z do not, suggesting that the last common

ancestor of species X and S. cerevisiae may have the genes necessary for making good bread. You compare the gene sequences of species X and S. cerevisiae and find many identical coding sequences, but you also identify nucleotides that differ between the two species. Which species would be the best to examine to determine what the sequence was in the last common ancestor of species X and S. cerevisiae?

Figure 9-19

a. species V b. species W c. species Y d. species Z ANS: C DIF: Moderate REF: 9.2 OBJ: 9.2.h Explain how purifying selection leads to the conservation of functionally important sequences and characterize the roles that these conserved sequences might play. MSC: Applying 20. Which of the following statements is FALSE? a. The human genome is more similar to the orangutan genome than it is to the mouse genome. b. A comparison of genomes shows that 90% of the human genome shares regions of conserved synteny with the mouse genome. c. Primates, dogs, mice, and chickens all have about the same number of genes. d. Genes that code for ribosomal RNA share significant similarity in all eukaryotes but are much more difficult to recognize in archaea.

ANS: D The gene that codes for the ribosomal RNA of the small ribosomal subunit is conserved in all living species.

DIF: Easy REF: 9.2 OBJ: 9.2.g Express what conserved synteny between two extant—or living—species indicates about the genome of their common ancestor. | 9.2.j Recall why a ribosomal RNA gene was selected to construct a phylogenetic “tree of life.” MSC: Understanding 21. The pufferfish, Fugu rubripes, has a genome that is one-tenth the size of mammalian genomes. Which of the following statements is NOT a possible reason for this size difference? a. Intron sequences in Fugu are shorter than those in mammals. b. Fugu lacks the repetitive DNA found in mammals.

c. The Fugu genome seems to have lost sequences faster than it has gained sequences over evolutionary time. d. Fugu has lost many genes that are part of gene families.

ANS: D DIF: Easy REF: 9.2 OBJ: 9.2.i Summarize how the Fugu genome differs from the human genome in terms of size and composition. MSC: Understanding 22. Which of the following regions of the genome is the LEAST likely to be conserved over evolutionary time? a. the upstream regulatory region of a gene that encodes the region conferring tissue specificity b. the upstream regulatory region of a gene that binds to RNA polymerase c. the portion of the genome that codes for proteins d. the portion of the genome that codes for RNAs that are not translated into protein ANS: A DIF: Easy REF: 9.2 OBJ: 9.2.c Contrast the types of genome sequence that can accommodate mutations with those that cannot. MSC: Understanding 23. Which of the following functions do you NOT expect to find in the set of genes found in all organisms on Earth? a. DNA replication b. DNA repair c. protein production d. RNA splicing ANS: D Not all organisms have introns and therefore not all organisms will have genes involved in RNA splicing.

DIF: Easy REF: 9.2 OBJ: 9.2.h Explain how purifying selection leads to the conservation of functionally important sequences and characterize the roles that these conserved sequences might play. MSC: Remembering 24. Which of the following generalities about genomes is TRUE? a. All vertebrate genomes contain roughly the same number of genes. b. All unicellular organisms contain roughly the same number of genes. c. The larger an organism, the more genes it has. d. The more types of cell an organism has, the more genes it has.

ANS: A DIF: Easy REF: 9.2 OBJ: 9.2.h Explain how purifying selection leads to the conservation of functionally important sequences and characterize the roles that these conserved sequences might play. MSC: Remembering 25. The evolutionary relationships between seven different species—G, H, J, K, L, M, and N—are diagrammed in Figure 9-25.

Figure 9-25

Given this information, which of the following statements is FALSE? a. These are all highly related species, because the sequence divergence between the most divergent species is 3%. b. Species M is as closely related to species G as it is to species J. c. Species N is more closely related to the last common ancestor of all of these species than to any of the other species shown in the diagram. d. Species G and H are as closely related to each other as species J and K are to each other.

ANS: C Species N has also diverged from the last common ancestor (just like all the other species in the diagram).

DIF: Easy REF: 9.2 OBJ: 9.2.h Explain how purifying selection leads to the conservation of functionally important sequences and characterize the roles that these conserved sequences might play. MSC: Applying 26. Which of the following statements about mobile genetic elements is TRUE? a. Mobile genetic elements can sometimes rearrange the DNA sequences of the genome in which they are embedded by accidentally excising neighboring chromosomal regions and reinserting these sequences into different places within the genome. b. DNA-only transposons do not code for proteins but instead rely on transposases found in cells that are infected by viruses. c. The two major families of transposable sequences found in the human genome are DNA-only transposons that move by replicative transposition. d. During replicative transposition, the donor DNA will no longer have the mobile genetic element embedded in its sequence when transposition is complete.

ANS: A DIF: Easy REF: 9.3 OBJ: 9.3.a Present the components of a typical transposon. MSC: Understanding 27. Which of the following is true of a retrovirus, but NOT of the Alu retrotransposon? a. It requires cellular enzymes to make copies. b. It can be inserted into the genome. c. It can be excised and moved to a new location in the genome. d. It encodes its own reverse transcriptase.

ANS: D DIF: Easy REF: 9.3 OBJ: 9.3.e Articulate how viruses differ from mobile genetic elements.

MSC: Remembering 28. Which of the following DNA sequences is NOT commonly carried on a DNA-only transposon? a. transposase gene b. reverse transcriptase gene c. recognition site for transposase d. antibiotic-resistance gene ANS: B The reverse transcriptase gene is found on retrotransposons that move via an RNA intermediate. The other choices are all sequences that can be found in mobile genetic elements.

DIF: Easy REF: 9.3 OBJ: 9.3.c Compare the movement of DNA-only transposons and retrotransposons.

MSC: Remembering 29. HIV is a human retrovirus that integrates into the host cell’s genome and will eventually replicate, produce viral proteins, and ultimately escape from the host cell. Which of the following proteins is not encoded in the HIV genome? a. reverse transcriptase b. envelope protein c. RNA polymerase d. capsid protein ANS: C Retroviruses like HIV will use host-cell RNA polymerase to make an RNA copy of the integrated viral DNA. DIF: Easy REF: 9.3 OBJ: 9.3.f Review how and why viruses use the host’s biochemical machinery to reproduce themselves. MSC: Remembering 30. In humans and in chimpanzees, 99% of the Alu retrotransposons are in corresponding positions. Which of the following statements below is the most likely explanation for this similarity? a. The Alu retrotransposon is not capable of transposition in humans. b. Most of the Alu sequences in the chimpanzee genome underwent duplication and divergence before humans and chimpanzees diverged. c. The Alu retrotransposons are in the most beneficial position in the genome for primates. d. The Alu retrotransposons must also be in the same position in flies.

ANS: B DIF: Easy REF: 9.1 OBJ: 9.1.g Illustrate how homologous recombination can lead to gene duplication. | 9.2.d Illustrate how selectively neutral mutations, combined with analysis of the fossil record, can be used to con struct a phylogenetic tree. MSC: Applying 31. Viral genomes a. can be made of DNA. b. can be made of RNA. c. can be either double-stranded or single-stranded. d. All of these answers are true.

ANS: D DIF: Easy REF: 9.3 OBJ: 9.3.f Review how and why viruses use the host’s biochemical machinery to reproduce themselves. MSC: Remembering

32. Viruses reproduce inside a host cell because a. viruses package DNA from the host-cell genome into the virus particle. b. viruses need host-cell reverse transcriptase to convert its RNA into DNA. c. viruses use host-cell ribosomes to produce viral coat proteins. d. all viruses must insert their genomes into the host-cell genome in order to be replicated.

ANS: C Viruses hijack the host cell’s machinery to replicate, which includes the host cell’s protein synthesis machinery. DNA from the host-cell genome is not typically packaged by viruses. Retroviruses provide their own reverse transcriptase, not the host cell. Although some viruses can integrate into the host-cell genome and become latent, this is not true of all viruses.

DIF: Easy REF: 9.3 OBJ: 9.3.f Review how and why viruses use the host’s biochemical machinery to reproduce themselves. MSC: Understanding 33. Which of the following statements about retroviruses is FALSE? a. Retroviruses are packaged with a few molecules of reverse transcriptase in each virus particle. b. Retroviruses use the host-genome integrase enzyme to create the provirus. c. The production of viral RNAs can occur long after the initial infection of the host cell by the retrovirus. d. Viral RNAs are translated by host-cell ribosomes to produce the proteins required for the production of viral particles.

ANS: B Integrase is typically encoded by the viral genome.

DIF: Easy REF: 9.3 OBJ: 9.3.g Outline the steps involved in the integration and replication of a retrovirus.

MSC: Remembering 34. A finished draft of the human genome was published in a. 1965. b. 1984. c. 2004. d. 2018.

ANS: C DIF: Easy REF: 9.4 OBJ: 9.4.e Describe how the early estimate of 100,000 human genes was derived.

MSC: Remembering 35. Which of the following statements about the human genome is FALSE? a. About 50% of the human genome is made up of mobile genetic elements. b. More of the human genome comprises intron sequences than exon sequences. c. About 1.5% of the human genome codes for exons. d. Only the exons are conserved between the genomes of humans and other mammals.

ANS: D About 5% of the human genome is highly conserved with other mammalian genomes, yet only about 1.5% of the human genome codes for exons.

DIF: Easy REF: 9.4 OBJ: 9.4.b Differentiate the amount of the human genome that codes for proteins versus the amount that consists of mobile genetic elements. MSC: Remembering 36. The nucleotide sequences between individuals differ by 0.1%, yet the human genome is made up of about 3 × 109

nucleotide pairs. Which of the following statements is FALSE? a. In most human cells, the homologous autosomes differ from each other by 0.1%. b. All changes between human individuals are single-nucleotide polymorphisms. c. Any two individuals (other than identical twins) will generally have more than 3 million genetic differences in their genomes. d. Much of the variation between human individuals was present 200,000 years ago, when the human population was small.

ANS: B DIF: Easy REF: 9.4 OBJ: 9.4.j State the amount of variation that typically distinguishes one human genome from another and describe the most common form of genetic variation. MSC: Remembering 37. Which of the following processes is NOT thought to contribute to the diversity in the genome seen between any two human individuals? a. exon shuffling b. single-nucleotide polymorphisms c. new mutations seen at birth that are not present in the genomes of either parent d. duplication and deletion of large blocks of sequence ANS: A DIF: Easy REF: 9.4 OBJ: 9.4.j State the amount of variation that typically distinguishes one human genome from another and describe the most common form of genetic variation. MSC: Understanding 38. Which of the following statements about what we have learned by comparing the modern-day human genome to other genomes is TRUE? a. Modern humans whose ancestors come from Europe or Asia share up to 2% of their genome with Neanderthals. b. Many of the 3 million genetic differences between the genome of two humans are the result of new mutations that have arisen within the last 2000 years. c. The human genome is far more gene-dense than the yeast genome. d. In syntenic regions of the human and mouse genomes, both gene order and the placements of more than 95% of the mobile genetic elements are conserved.

ANS: A DIF: Easy REF: 9.4 OBJ: 9.2.e Outline how a comparison of the nucleotide sequences from two closely related organisms can be used to reconstruct the amino acid sequence of a protein from their extinct, common ancestor. | 9.2.g Express what conserved synteny between two extant—or living—species indicates about the genome of their common ancestor. | 9.4.i Summarize what has been revealed by the comparison of human and Neanderthal DNA. MSC: Remembering 39. An SNP found in the conserved sequence of the regulatory region of a gene is likely to a. affect protein folding. b. affect when and where the gene is expressed. c. be a new mutation. d. be found in plants as well as humans.

ANS: B DIF: Easy REF: 9.4 OBJ: 9.4.h Express how changes in regulatory DNA sequences can fuel the evolution of species. MSC: Understanding 40. The yeast genome was sequenced more than 20 years ago, yet the total number of genes continues to be refined.

The sequencing of closely related yeast species was important for validating the identity of short (less than 100

nucleotides long) open reading frames (ORFs) that were otherwise difficult to predict. What is the main reason that these short ORFs are Difficult to find without the genomes of other yeast for comparison? a. Short ORFs are found only in yeast. b. The short ORFs code for RNAs. c. Many short stretches of DNA may lack a stop codon simply by chance, making it difficult to distinguish those DNA sequences that code for proteins from those that do not. d. Short ORFs occur mainly in gene-rich regions, making them difficult to identify by computer programs.

ANS: C DIF: Easy REF: 9.4 OBJ: 9.4.f Review how open reading frames can be used as an estimate for gene number. MSC: Remembering 41. Which of the following would NOT be useful when finding genes in a newly sequenced mammalian genome? a. searching for splicing sequences that signal an intron-exon boundary b. searching for sequences that code for proteins similar to those found in fruit flies c. matching sequences obtained from RNA-Seq back to the genome d. searching for long stretches of DNA sequence conservation with intron sequences from zebrafish ANS: D Intron sequences are typically not conserved when evolutionary distances are far (as would be the case between mammals and fish). On the other hand, mammals have proteins in common with flies, and thus protein sequences from flies can be used to help search for genes in mammals. Splicing sequences can be helpful for identifying intron-exon boundaries and RNA-Seq can be a useful tool for identifying transcribed regions of the genome.

DIF: Moderate REF: 9.4 OBJ: 9.2.a Define homologous genes and state the percentage of human genes that have clear homologs in species such as the fruit fly and nematode. | 9.2.e Outline how a comparison of the nucleotide sequences from two closely related organisms can be used to reconstruct the amino acid sequence of a protein from their extinct, common ancestor. | 9.4.c Contrast the structure and distribution of genes in the genomes of human, fly, and yeast. | 9.4.f Review how open reading frames are used in estimating gene number.

MSC: Understanding 42. Two individuals are represented in each choice in Figure 9-42; individual 1 is one of the parents of individual 2.

The asterisk seen in each choice indicates the occurrence of a single mutation during the cell division. Which of the choices in Figure 9-42 will lead to a mutation in every cell of the individual in which the original mutation occurred?

Figure 9-42

ANS: C DIF: Easy REF: 9.1 OBJ: 9.1.b Contrast the fate of mutations that arise in somatic cells versus germ-line cells. MSC: Applying

MATCHING 1. Match the type of phenotypic change below with the type of genetic change most likely to cause it. Each type of genetic change may be used more than once, or may not be used at all.

Types of genetic change: A. mutation within a gene B. gene duplication C. mutation in a regulatory region D. exon shuffling E. horizontal gene transfer Phenotypic changes: 1. A protein normally localized in the nucleus is now localized in the cytoplasm. _________ 2. A protein acquires a DNA-binding domain.

3. Tandem copies of a gene are found in the genome. _________ 4. A copy of a bacterial gene is now found integrated on a human chromosome. _________ 5. A protein becomes much more unstable.

6. A protein normally expressed only in the liver is now expressed in blood cells. ________ 1. ANS: A DIF: Moderate REF: 9.1 OBJ: 9.1.a Summarize the six basic mechanisms that generate genetic change. |

9.1.h Articulate how gene duplication can lead to the evolution of genes with new function or to the gener ation of pseudogenes. | 9.1.k Compare the mechanisms of exon shuffling and gene duplication. | 9.1.n Describe the process of horizontal gene transfer and state how this form of genetic exchange can ultimately affect human health. MSC: Applying 2. ANS: D DIF: Moderate REF: 9.1 OBJ: 9.1.a Summarize the six basic mechanisms that generate genetic change. | 9.1.h Articulate how gene duplication can lead to the evolution of genes with new function or to the gener ation of pseudogenes. | 9.1.k Compare the mechanisms of exon shuffling and gene duplication. | 9.1.n Describe the process of horizontal gene transfer and state how this form of genetic exchange can ultimately affect human health. MSC: Applying 3. ANS: B DIF: Moderate REF: 9.1 OBJ: 9.1.a Summarize the six basic mechanisms that generate genetic change. | 9.1.h Articulate how gene duplication can lead to the evolution of genes with new function or to the gener ation of pseudogenes. | 9.1.k Compare the mechanisms of exon shuffling and gene duplication. | 9.1.n Describe the process of horizontal gene transfer and state how this form of genetic exchange can ultimately affect human health. MSC: Applying 4. ANS: E DIF: Moderate REF: 9.1 OBJ: 9.1.a Summarize the six basic mechanisms that generate genetic change. | 9.1.h Articulate how gene duplication can lead to the evolution of genes with ne w function or to the generation of pseudogenes. | 9.1.k Compare the mechanisms of exon shuffling and gene duplication. | 9.1.n Describe the process of horizontal gene transfer and state how this form of genetic exchange can ultimately affect human health. MSC: Applying 5. ANS: A DIF: Moderate REF: 9.1 OBJ: 9.1.a Summarize the six basic mechanisms that generate genetic change. | 9.1.h Articulate how gene duplication can lead to the evolution of genes with new function or to the gener ation of pseudogenes. | 9.1.k Compare the mechanisms of exon shuffling and gene duplication. | 9.1.n Describe the process of horizontal gene transfer and state how this form of genetic exchange can ultimately affect human health. MSC: Applying 6. ANS: C DIF: Moderate REF: 9.1 OBJ: 9.1.a Summarize the six basic mechanisms that generate genetic change. | 9.1.h Articulate how gene duplication can lead to the evolution of genes with new function or to the gener ation of pseudogenes. | 9.1.k Compare the mechanisms of exon shuffling and gene duplication. | 9.1.n Describe the process of horizontal gene transfer and state how this form of genetic exchange can ultimately affect human health. MSC: Applying

SHORT ANSWER 1. For each of the following sentences, fill in the blanks with the best word or phrase in the list below. Not all words or phrases will be used; use each word or phrase only once. cellulose

intron

common

neutral

deleterious

somatic

gamete

unequal

homologous

zygote

Sexual reproduction in a multicellular organism involves specialized reproductive cells, called __________s, which come together to form a __________ that will divide to produce both reproductive and __________ cells. A point

mutation in the DNA is considered a __________ mutation if it changes a nucleotide that leads to no phenotypic consequence; a point mutation is considered __________ if it changes a nucleotide within a gene and causes the protein to be nonfunctional.

ANS: Sexual reproduction in a multicellular organism involves specialized reproductive cells, called gametes, which come together to form a zygote that will divide to produce both reproductive and somatic cells. A point mutation in the DNA is considered a neutral mutation if it changes a nucleotide that leads to no phenotypic consequence; a point mutation is considered deleterious if it changes a nucleotide within a gene and causes the protein to be nonfunctional.

DIF: Easy REF: 9.1 OBJ: 9.1.b Contrast the fate of mutations that arise in somatic cells versus germ-line cells. | 9.1.e Recall how point mutations typically arise and evaluate how the locations of these mutations can dictate their effects on an organism’s appearance or fitness. | 9.2.b Compare the fate of germ-line mutations that are deleterious, selectively neutral, or that provide a selective advantage. | 9.2.c Contrast the types of DNA sequences that can accommodate mutations with those that cannot. MSC: Understanding 2. Transposable elements litter the genomes of primates, and a few of them are still capable of moving to new regions of the genome. If a transposable element jumped into an important gene in one of your cells when you were a baby and caused a disease, is it likely that your child would also have the disease? Explain.

ANS: It is not likely that your child would have the disease, because it is unlikely that the mutation is carried in the germ line. Probably the mutation occurred in a cell that gave rise to somatic cells and not germ cells. Only mutations in germ cells are passed on to progeny.

DIF: Moderate REF: 9.3 OBJ: 9.3.a Present the components of a typical transposon. MSC: Evaluating 3. For each statement below, indicate whether it is TRUE or FALSE, and explain why.

A. To meet a challenge or develop a new function, evolution essentially builds from first principles, designing from scratch, to find the best possible solution.

B. Nearly every instance of DNA duplication leads to a new functional gene.

C. A pseudogene is very similar to a functional gene but cannot be expressed because of mutations.

D. Most genes in vertebrates are unique, and only a few genes are members of multigene families.

E. Horizontal gene transfer is very rare and thus has had little influence on the genomes of bacteria.

ANS: A. False. Evolution can work only by tinkering with the tools and materials on hand, not by starting from scratch to make completely new genes or pathways. New functions arise from the ancestral functions by a process of gradual mutational change, and thus may not represent the best possible solution to a problem.

B. False. Many duplications are subsequently lost or become pseudogenes, and only a few evolve into new genes.

C. True. Pseudogenes look very similar to normal genes but cannot produce a full-length protein, as a result of one or more disabling mutations.

D. False. A large proportion of the genes in vertebrates (and many other species) are members of multigene families.

E. False. By some estimates, 20% of the genomic DNA in some bacterial species arose by horizontal gene transfer.

DIF: Moderate REF: 9.1 OBJ: 9.1.h Articulate how gene duplication can lead to the evolution of genes with new function or to the generation of pseudogenes. | 9.1.n Describe the process of horizontal gene transfer and state how this form of genetic exchange can ultimately affect human health. MSC: Evaluating

4. For each of the following sentences, fill in the blanks with the best word or phrase in the list below. Not all words or phrases will be used; use each word or phrase only once. divergence

purifying selection

exon shuffling

single-nucleotide polymorphisms

gene duplication

synteny

horizontal gene transfer

unequal crossing-over

__________ may arise by recombination within introns and can create proteins with novel combinations of domains.

Scientists and government regulators must be very careful when introducing herbicide-resistant transgenic corn plants into the environment, because if resistant weeds arise from __________, then the herbicides could become useless.

Families of related genes can arise from a single ancestral copy by __________ and subsequent __________.

ANS: Exon shuffling may arise by recombination within introns and can create proteins with novel combinations of domains. Scientists and government regulators must be very careful when introducing herbicide-resistant transgenic corn plants into the environment, because if resistant weeds arise from horizontal gene transfer then the herbicides could become useless. Families of related genes can arise from a single ancestral copy by gene duplication and subsequent divergence.

DIF: Easy REF: 9.1 OBJ: 9.1.h Articulate how gene duplication can lead to the evolution of genes with new function or to the generation of pseudogenes. | 9.1.l Express how exon shuffling can facilitate the evolution of proteins with novel functions. | 9.1.n Describe the process of horizontal gene transfer and state how this form of genetic exchange can ultimately affect human health. MSC: Understanding 5. The spontaneous mutation rate in E. coli was determined to be 1 mistake for every 10 9 nucleotides copied. This was determined by measuring the frequency of a particular AT-to-GC change. This was accomplished using a strain of E. coli that started out unable to produce histidine (His –) because of an inserted UGA stop codon that disrupted the region coding for an enzyme required to produce histidine. When a spontaneous mutation arose that enabled the UGA stop codon to code for tryptophan, the E. coli cells were then able to produce the enzyme required for histidine production.

You discover that if the stop codon were to change to code for cysteine (instead of tryptophan), this change would also allow the bacteria to produce histidine. How would the previously calculated spontaneous mutation rate of 1 mistake every 109 nucleotides copied change, given this new information? Explain. (The codon table is shown in Figure 9-48 to help you answer this question.)

Figure 9-48

ANS: There would be a twofold increase. Two different mutations (UGA to either UGU or UGC) could arise to change the stop codon to a codon coding for cysteine. This mutation would lead to a doubling of the observed spontaneous mutation rate, from 1 in 109 to 2 in 109, in comparison with a single mutation that can change the stop codon to a codon coding for tryptophan (UGA to UGG). Because spontaneous mutations are rare, this would not be a particularly significant change.

DIF: Difficult REF: 9.1 OBJ: 9.1.c Outline an experimental approach to determining the rate at which point mutations accumulate in bacteria. MSC: Applying 6. Propose a reason to explain why highly repetitive regions of the genome are particularly susceptible to expansions and contractions in number.

ANS: Highly repetitive regions of the genome are particularly susceptible to unequal genetic exchange during h omologous recombination.

DIF: Moderate REF: 9.1 OBJ: 9.1.g Illustrate how homologous recombination can lead to gene duplication.

MSC: Evaluating 7. Consider a gene with a particular function. Mutation X and mutation Y each cause defects in the function of the encoded protein, yet a gene containing both mutations X and Y encodes a protein that works even better than the original protein. The odds are exceedingly small that a single mutational event will generate both mutations X and Y. Explain a simple way that an organism with a mutant gene containing both mutations X and Y could arise during evolution.

ANS: The simplest way to evolve the new gene is by duplication and divergence. If the gene is duplicated, then the cell or lineage can maintain one functional, intact old copy of the original gene and can thus tolerate the disabling mutations in the other copy. The other copy can first be modified by the X or Y mutation that impairs its function; second, at some later time, the gene with the single mutation can acquire the additional mutation to yield the doubly mutant X + Y gene with the new or improved function.

DIF: Moderate REF: 9.1 OBJ: 9.1.h Articulate how gene duplication can lead to the evolution of genes with new func-

tion or to the generation of pseudogenes. MSC: Evaluating 8. Some types of gene are more highly conserved than others. For each of the following pairs of gene functions, choose the one that is more likely to be highly conserved.

A. genes involved in sexual reproduction; genes involved in sugar metabolism B. DNA replication; developmental pathways C. hormone production; lipid synthesis ANS: A. sugar metabolism B. DNA replication C. lipid synthesis These pathways or phenomena are fundamental to the growth and proliferation of all cells, including bacteria, and thus are likely to be highly conserved from species to species.

DIF: Easy REF: 9.2 OBJ: 9.2.h Explain how purifying selection leads to the conservation of functionally important sequences and characterize the roles that these conserved sequences might play. MSC: Applying 9. Figure 9-52 shows a hypothetical phylogenetic tree. Use this tree to answer the following questions.

Figure 9-52

A. How many years ago did species M and N diverge from their last common ancestor?

B. How much nucleotide divergence is there on average between species M and N? C. Are species M and N more or less closely related to each other than species P and S are?

D. In looking for functionally important nucleotide sequences, is it more informative to compare the genome sequences of species M and N or those of species M and Q? ANS: A. M and N diverged 10 million years ago.

B. There is an average of 2.0% nucleotide substitution in species M compared with species N (follow the path connecting the two species, which is twice the distance between each one and their common ancestor).

C. Neither more nor less. They show roughly the same degree of relatedness. The sequence divergence between species M and N is about 2.0%, the same as that between species P and S. Both pairs of species diverged 10

million years ago.

D. It is more informative to compare species that are separated by a greater evolutionary distance; thus, comparing species M and Q, which diverged 20 million years ago, will be better able to identify sequences that are likely to be important for function. Closely related species share many sequences by chance, because there has been insufficient time for neutral mutations to accumulate.

DIF: Moderate REF: 9.2 OBJ: 9.2.d Illustrate how selectively neutral mutations, combined with analysis of the fossil record, can be used to construct a phylogenetic tree. MSC: Applying 10. For each statement below, indicate whether it is TRUE or FALSE, and explain why.

A. All highly conserved stretches of DNA in the genome are transcribed into RNA. B. To find functionally important regions of the genome, it is more useful to compare species whose last common ancestor lived 100 million years ago rather than 5 million years ago.

C. Most mutations and genome alterations have neutral consequences.

D. Proteins required for growth, metabolism, and cell division are more highly conserved than those involved in development and in response to the environment.

E. Introns and transposons tend to slow the evolution of new genes.

ANS: A. False. Many highly conserved stretches of DNA are not transcribed but instead contain information critical for regulating where and when genes are expressed.

B. True. Species that diverged recently have many identical stretches of DNA sequence by chance, whereas sequence similarity between species that diverged long ago is probably due to functional constraints. The sequences that are necessary to preserve the function of the gene will not be able to undergo changes and thus are more likely to be similar between species that diverged long ago.

C. True. Most genomic changes do not alter the amino acid sequence of proteins or the regulatory properties of genes. Even some mutations that cause minor alterations have little effect on protein function.

D. True. All organisms need to perform a similar basic set of fundamental functions, such as those for metabolism, protein synthesis, and DNA replication. Proteins involved in these functions are shared by descent, and their evolution is constrained. Different species and cells are likely to require different developmental paths and to encounter different environmental challenges, so the proteins involved in these processes will tend to be more variable. For example, bacteria do not undergo elaborate developmental programs and so lack many of the regulators of development found in eukaryotes.

E. False. Introns and transposons can act as sites where recombinational crossovers occur. Transposons can also catalyze genetic rearrangements. Rearrangements occurring within these sequences are less likely to be detrimental than those occurring elsewhere in the genome. In general, only the short intron sequences required for splicing are important to intron function; alterations in sequences outside the splicing sites may have no consequences for intron function and thus will not be subject to purifying selection.

DIF: Moderate REF: 9.2 OBJ: 9.1.e Recall how point mutations typically arise and evaluate how the locations of these mutations can dictate their effects on an organism’s appearance or fitness. | 9.2.c Contrast the types of DNA sequences that can accommodate mutations with those that cannot. | 9.2.e Outline how a comparison of the nucleotide sequences from two closely related organisms can be used to reconstruct the amino acid sequence of a protein from their extinct, common ancestor. | 9.2.h Explain how purifying selection leads to the conservation of functionally important sequences and characterize the roles that these conserved sequences might play. MSC: Evaluating

11. The genomes of some vertebrates are much smaller than those of others. For example, the genome of the pufferfish Fugu is much smaller than the human genome, and even much smaller than genomes of other fish, primarily because of the small size of its introns.

A. Describe a mechanism that might drive evolution toward small introns or loss of introns and could therefore account for the evolutionary loss of introns according to the “introns early” hypothesis.

B. Describe a mechanism that might drive evolution toward more or larger introns and could thereby account for the evolutionary appearance of introns according to the “introns late” hypothesis.

ANS: A. Spontaneous deletions or selection pressure to decrease the time or cost of DNA replication may cause a loss of introns.

B. Acquisition of intron sequences provides a selective advantage for those organisms that experience transposon insertions. According to this idea, introns became sinks for transposon and virus insertion to protect the rest of the genome. Alternatively, introns may provide another advantage to the host genome: by providing ample sites for crossing-over, larger introns could facilitate exon shuffling and thus the generation of new genes with novel functions.

DIF: Moderate REF: 9.2 OBJ: 9.2.i Summarize how the Fugu genome differs from the human genome in terms of size and composition. MSC: Understanding 12. There are about 700 eukaryotic genes that have obvious prokaryotic homologs. Would you predict that this set of genes would be enriched for genes that are used at distinct times during the life cycle of multicellular animals? Explain your answer.

ANS: No, these genes are likely to be involved in basic cellular functions such as glycolysis and protein production, processes important for both prokaryotes and eukaryotes. Genes involved in basic cellular functions are likely to be used all the time in an organism’s life and are less likely to be used at only a specific stage of life.

DIF: Easy REF: 9.2 OBJ: 9.2.h Explain how purifying selection leads to the conservation of functionally important sequences and characterize the roles that these conserved sequences might play. MSC: Understanding 13. You are working in a human genetics laboratory that studies causes and treatments for eye cataracts in newborns.

This disease is thought to be caused by a deficiency in the enzyme galactokinase, but the human gene that encodes this enzyme has not yet been identified. At a talk by a visiting scientist, you learn about a strain of baker’s yeast that contains a mutation called gal1– in its galactokinase gene. Because this gene is needed to metabolize galactose, the mutant strain cannot grow in galactose medium. Knowing that all living things evolved from a common ancestor and that distantly related organisms often have homologous genes that perform similar functions, you wonder whether the human galactokinase gene can function in yeast. Because you have an optimistic temperament, you decide to pursue this line of experimentation. You isolate mRNA gene transcripts from human cells, use reverse transcriptase to make complementary DNA (cDNA) copies of the mRNA molecules, and ligate the cDNAs into circular plasmid DNA molecules that can be stably propagated in yeast cells. You then transform the pool of plasmids into gal1– yeast cells so that each cell receives a single plasmid. What will happen when you spread the plasmid-containing cells on Petri dishes that contain galactose as a carbon source? How can this approach help you find the human gene encoding galactokinase?

ANS: On galactose medium, the original gal1– yeast cells cannot grow, nor can cells that received plasmids containing most human cDNA sequences. However, yeast cells that received a plasmid with the human galactokinase gene will

probably be able to grow on galactose medium and produce many progeny. This kind of “selection” procedure is very powerful, because even if only 1 in 100,000 cells has the ability to grow under particular conditions, it will be easy to find it. The other 99,999 cells will die in the Petri dish and will therefore be invisible to the investigator. Indeed, scientists have found that the human galactokinase gene can function perfectly well in yeast and thus can “rescue” the defect of the gal1– mutant. It was initially astonishing that genes from humans can function properly in yeast, but this phenomenon has now been observed for many genes.

DIF: Difficult REF: 9.2 OBJ: 9.2.a Define homologous genes and state the percentage of human genes that have clear homologs in species such as fruit fly and nematode. MSC: Applying 14. A. When a mutation arises, it can have three possible consequences: beneficial to the individual, selectively neutral, or detrimental. Order these from most likely to least likely.

B. The spread of a mutation in subsequent generations will, of course, depend on its consequences to individuals that inherit it. Order the three possibilities in part A to indicate which is most likely to spread and become overrepresented in subsequent generations, and which is most likely to become underrepresented or disappear from the population.

ANS: A. The order is selectively neutral; detrimental; beneficial. Most nucleotide changes in the genome, or mutations, will have little or no effect on the fitness of the individual because many changes are not located in regions that encode a protein or regulate the expression of a gene. Even changes within a coding region may not change the amino acid encoded or may cause a conservative amino acid change—for example, from one small nonpolar amino acid to another. Most changes that have a functional consequence will interfere with the regulation of a gene or the behavior of the encoded protein, usually rendering it useless and occasionally making it harmful or yielding a new function. Only very rarely will a mutation improve the performance of the gene or its encoded protein.

B. The order is beneficial; selectively neutral; detrimental. Individuals bearing beneficial mutations will be more likely to have more offspring than others in the population, and thus the beneficial mutations will become overrepresented in the population in subsequent generations. Individuals bearing detrimental mutations will be likely to have fewer children and grandchildren, and thus these mutations will be culled from the population, although perhaps not eliminated.

DIF: Easy REF: 9.2 OBJ: 9.1.b Contrast the fate of mutations that arise in somatic cells versus germ-line cells. | 9.1.e Recall how point mutations typically arise and evaluate how the locations of these mutations can dictate their effects on an organism’s appearance or fitness. | 9.2.c Contrast the types of DNA sequences that can accommodate mutations with those that cannot.

9.2.b Compare the fate of germ-line mutations that are deleterious, selectively neutral, or that provide a selective advantage. MSC: Understanding 15. Some retrotransposons and retroviruses integrate preferentially into regions of the chromosome that are packaged in euchromatin and are also located outside the coding regions of genes that contain information for making a protein.

Why might these mobile genetic elements have evolved this strategy?

ANS: The most evolutionarily successful mobile genetic elements are those that are best at reproducing themselves.

To increase the number of copies of a particular element, the element must meet two criteria: (1) it must not kill its host, and (2) it must maximize its ability to continue reproducing. If an element inserts into the coding region of a

gene, it might disable the gene and thereby confer a selective disadvantage in the reproduction or survival of its host.

Thus, elements that devised a way to avoid insertion into coding regions were probably better able to increase their copy number throughout the human population. If an element inserts into a heterochromatic region of a chromosome, its genes may not be expressed and therefore it may become immobile. Elements that devised a way to direct insertion into euchromatin would be more likely to maintain mobility and thereby increase their copy number over time.

DIF: Moderate REF: 9.3 OBJ: 9.3.h Appraise how retroviruses and retrotransposons are similar. MSC: Evaluating 16. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

0.5

1.5

3

4

9

23

30

50

75

The human genome has ~3 × 10__________ nucleotide pairs that make up the sequence of __________ sets of chromosomes. We estimate that this sequence encodes approximately 1.9 × 10 __________ protein-coding genes and about 5 × 10__________ non-protein-coding genes, which include structural catalytic, and regulatory RNAs. Approximately __________% of the human genome is made up of high-copy repetitive elements, while the percentage of DNA sequence in protein-coding exons is __________%.

ANS: The human genome has ~3 × 109 nucleotide pairs that make up the sequence of 23 sets of chromosomes. We estimate that this sequence encodes approximately 1.9 × 104 protein-coding genes and about 5 × 103 non-proteincoding genes, which include structural catalytic, and regulatory RNAs. Approximately 50% of the human genome is made up of high-copy repetitive elements, while the percentage of DNA sequence in protein-coding exons is 1.5%.

DIF: Easy REF: 9.4 OBJ: 9.4.a State the size of the human genome and recall how many chromosomes are used to carry this genetic information. | 9.4.b Differentiate the amount of the human genome that codes for proteins versus the amount that consists of mobile genetic elements. MSC: Remembering 17. The human genome has 3.2 × 109 nucleotide pairs. At its peak, the Human Genome Project was generating raw nucleotide sequences at a rate of 1000 nucleotides per second. At the rate of 1000 nucleotides per second, how long would it take to generate 3.2 × 109 nucleotides of sequence?

ANS: It would take approximately 37 days; as there are 60 seconds per minute, there are 86,400 seconds in a day. At the peak rate, you can obtain 86,400,000 nucleotides per day.

DIF: Easy REF: 9.4 OBJ: 9.4.a State the size of the human genome and recall how many chromosomes are used to carry this genetic information. MSC: Applying 18. The average size of a protein in a human cell is about 430 amino acids, yet the average gene in the human genome is 27,000 nucleotide pairs long. Explain.

ANS: In the human genome, the exons are relatively short, whereas the introns can be quite large. A protein 430 amino acid residues long will need fewer than 1300 nucleotides to code for it. Furthermore, many genes in the human genome undergo alternative splicing, so some of those 27,000 nucleotides may be coding for alternative exons that are not used every time the gene is transcribed.

DIF: Moderate REF: 9.4 OBJ: 9.4.a State the size of the human genome and recall how many chromosomes are used to carry this genetic information. | 9.4.c Review the structure of genes in the genomes of human, fly, and yeast, and compare how densely genes are distributed in the chromosomes of each organism. MSC: Evaluating

19. For each statement below, indicate whether it is TRUE or FALSE, and explain why.

A. The increased complexity of humans compared with flies and worms is largely due to the vastly larger number of genes in humans.

B. Most single-nucleotide polymorphisms cause no observable functional differences between individual humans.

C. There is little conserved synteny between human and mouse genes.

D. The differences between multicellular organisms are largely explained by the different kinds of genes carried on their chromosomes.

ANS: A. False. The number of genes differs only by about a factor of two. It is thought that the increased complexity of humans is due largely to differences in when and where the genes are expressed. Differences in the timing of splicing may also be a major contributor to the relative complexity of humans.

B. True. Nearly all single-nucleotide polymorphisms have no effect on the appearance or behavior of the individual, but a few cause important differences.

C. False. Human and mouse chromosomes show extensive synteny, with blocks of chromosomal DNA exhibiting homologous genes arranged in the same order between the two species.

D. False. Multicellular organisms are built from essentially the same toolbox of ge ne building blocks, but the parts are put together differently because of regulatory differences that dictate when and where and how much of each protein is made. Alternative splicing can also have an important role, as it can generate several proteins from a single gene in some species, yet the homologous gene in other species may produce only one protein.

DIF: Moderate REF: 9.2 OBJ: 9.2.a Define homologous genes and state the percentage of human genes that have clear homologs in species such as the fruit fly and nematode. | 9.2.c Contrast the types of DNA sequences that can accommodate mutations with those that cannot. | 9.2.g Express what conserved synteny indicates about the genome of a common ancestor. | 9.4.c Review the structure of genes in the genomes of human, fly, and yeast, and compare how densely genes are distributed in the chromosomes of each organism. MSC: Evaluating 20. Your friend discovered a new single-celled organism living under the polar ice caps, and brought it back to the laboratory, where it seems to be growing well in the 37°C incubator used for growing his other laboratory bacterial strains. Your friend is particularly interested in the proteins that allow this organism to survive in extreme cold.

Because he is specifically interested in proteins, he decides to perform RNA-Seq on cells grown in the lab incubator to determine which proteins are important for cold survival. What do you think the pitfalls of this approach might be?

Explain.

ANS: Although RNA-Seq can be very useful tool for identifying genes, the use of RNA-Seq in this case may not work for several reasons. Two of these are: 1. Sequences obtained through RNA-Seq represent actively transcribed genes. Your friend is studying the proteins that permit survival in extreme cold. Since the organism is no longer living in the extreme cold, the genes required for survival may no longer be expressed.

2. RNA-Seq determines the sequence from the RNAs in the cell. Because genes can be expressed at different levels, you will obtain sequence from the RNA molecules that represent genes that are abundantly transcribed more often than those that may be transcribed rarely. If the genes important for survival in extreme cold are transcribed at very low levels, their transcripts may be difficult to detect.

DIF: Moderate REF: 9.4 OBJ: 9.4.g Evaluate how RNA sequencing can provide a more accurate estimate of the number of genes in a genome. MSC: Understanding

21. Panels (A) and (B) of Figure 9-64 show substrates of exon shuffling and the outcome of exon shuffling after recombination. Horizontal lines and small filled circles represent chromosomes and centromeres, respectively.

Exons are labeled A, B, C, and D. Homologous recombination or shuffling may take place at short, repeated homologous DNA sequences in introns; because DNA sequences have a polarity, the repeated sequences can be considered to have a head and a tail and thus are drawn as arrows. A large X represents a recombinational crossover. Panel (A) shows that recombination between two direct repeats located on opposite sides of the centromere yields one circular product that contains a centromere and a second product that lacks a centromere and will therefore be lost when the cell divides. Panel (B) shows that recombination between inverted repeats flanking the centromere will keep the rearranged chromosome intact. Draw the products of recombination when the repeated sequences are located on different chromosomes, as shown in panels (C) and (D). Will these products be faithfully transmitted during cell division?

Figure 9-64

ANS: See Figure 9-64A. The products of panel (C) will be segregated to progeny cells reliably. In contrast, one product in panel (D) will have two centromeres and the other will lack a centromere. The chromosome without a centromere will be rapidly lost as cells divide. The chromosome containing two centromeres will probably be broken during mitosis and will subsequently be lost or severely damaged.

Figure 9-64A

DIF: Difficult REF: 9.1 OBJ: 9.1.b Contrast the fate of mutations that arise in somatic cells versus germ-line cells.

MSC: Applying

22. Your friend has sequenced the genome of her favorite experimental organism, a kind of yeast. She wants to identify the locations of all the genes in this genome. To aid her search, she collaborates with another researcher, one who has sequenced the genome of a distantly related yeast species. Luckily, the absence of introns simplifies the effort.

She and her collaborator use a computer program to align similar stretches of DNA sequence from the two genomes.

The program yields the graphical output shown in Figure 9-65, where the horizontal lines represent a portion of the two genomic sequences and vertical lines indicate where the sequences differ. (No vertical line means that the sequence is identical in the two yeasts.) Label both the functionally conserved regions and the divergent (nonconserved) sequences. Are all of the functionally conserved regions likely to be transcribed into RNA? If not, what might be the function of the nontranscribed conserved regions?

Figure 9-65

ANS: See Figure 9-65A. Not all of the functionally conserved regions will be transcribed into RNA. Some of the functionally conserved regions are likely to encode RNA and others are likely to be critical for regulating when the gene is transcribed and when it is turned off. These nontranscribed regulatory regions may be conserved nearly as much as the coding regions.

Figure 9-65A

DIF: Difficult REF: 9.4 OBJ: 9.4.h Express how changes in regulatory DNA sequences can fuel the evolution of species. MSC: Applying

CHAPTER 10 Analyzing the Structure and

Function of Genes ISOLATING AND CLONING DNA MOLECULES 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory.

10.1.b Express how the target sequence of a restriction enzyme relates to how often the sequence will appear in a long piece of DNA and the size of the DNA fragments it will generate.

10.1.c Outline how DNA fragments can be separated by size using gel electrophoresis.

10.1.d State how bands of DNA fragments on a gel can be visualized.

10.1.e Articulate why cloning in bacteria involves the generation of recombinant DNA. 10.1.f List the features typically present in a plasmid used as a cloning vector.

10.1.g Review how a DNA fragment is inserted into a plasmid.

10.1.h Summarize how a plasmid carrying a DNA fragment of interest can be inserted into bacteria and how the fragment, then greatly amplified, is subsequently isolated.

10.1.i Compare a genomic library and a cDNA library in terms of what each contains, how each is made, and how each might be used.

10.1.j Review how hybridization allows the detection of any DNA fragment of interest.

DNA CLONING BY PCR 10.2.a Summarize the main differences between cloning in bacteria and cloning by PCR. 10.2.b Review the purpose of each of the three basic steps involved in each cycle of PCR. 10.2.c Estimate the number of double-stranded DNA molecules produced after the 1st, 2nd, 3rd, and 30th PCR cycles.

10.2.d Recall why the DNA polymerase used in PCR is ideally suited for this application.

10.2.e Explain how PCR can be used to produce either genomic or cDNA clones.

10.2.f Outline how PCR can be used to detect an infectious agent in a sample of blood.

10.2.g Summarize how PCR can be used to produce a DNA fingerprint.

SEQUENCING DNA 10.3.a Review how chain-terminating dideoxynucleotides are used to determine the sequence of a piece of DNA. 10.3.b Distinguish between the shotgun method for assembling a genome sequence and the clone-by-clone approach,

which uses bacterial artificial chromosomes.

10.3.c Explain how the second-generation Illumina sequencing method uses PCR coupled with chain-terminating nucleotides to determine the sequence of a genome.

10.3.d State what distinguishes third-generation sequencing methods from older approaches and describe a thirdgeneration method.

10.3.e Review the types of information that can be gathered from comparative genome analyses.

EXPLORING GENE FUNCTION 10.4.a Describe the techniques of RNA-Seq and in situ hybridization and compare what each can reveal about gene expression in a particular cell type or tissue.

10.4.b Outline the construction of a reporter gene and review how such reporters can be used to assess patterns of gene expression.

10.4.c Review how reporter genes can be used to track the location of proteins within cells.

10.4.d Differentiate between the classical genetic approach to studying gene function and the methods referred to as reverse genetics.

10.4.e Review how RNA interference can be used to inhibit the activity of a target gene.

10.4.f Summarize the ways in which genetically modified animals can be used to explore the function of a given gene, and describe how these techniques can be adapted to study genes whose function is essential during development.

10.4.g Explain how the CRISPR system can be used to produce a transgenic organism in which a target gene is inactivated or replaced.

10.4.h Recall how the CRISPR system can be adapted to activate or repress a target gene.

10.4.i Summarize how investigators can produce mutant organisms that serve as models of human disease.

10.4.j Review the approach used to produce transgenic plants.

10.4.k State how expression vectors direct the production of large quantities of a selected protein.

10.4.l Outline how recombinant DNA techniques make it possible to analyze the activity of a protein starting from its gene or to assess gene function starting with its encoded protein.

MULTIPLE CHOICE 1. Recombinant DNA technologies involve techniques that permit the creation of custom-made DNA molecules that can be introduced back into living organisms. These technologies were first developed in the a. 1800s. b. 1950s. c. 1970s. d. 2000s.

ANS: C DIF: Easy REF: 10.1 OBJ: 10.1.e Articulate why the classical approach to cloning involves the generation of recombinant DNA. MSC: Remembering 2. Which of the following statements about restriction nucleases is FALSE? a. A reproducible set of DNA fragments will be produced every time a restriction nuclease digests a known piece of DNA. b. Restriction nucleases recognize specific sequences on single-stranded DNA. c. Some bacteria use restriction nucleases as protection from foreign DNA. d. Some restriction nucleases cut in a staggered fashion, leaving short, single-stranded regions of DNA at the ends of the cut molecule.

ANS: B Restriction nucleases cleave double-stranded DNA. DIF: Easy REF: 10.1 OBJ: 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory. MSC: Remembering 3. You have purified DNA from your recently deceased goldfish. Which of the following restriction nucleases would you use if you wanted to end up with DNA fragments with an average size of 70 kilobase pairs (kb) after complete digestion of the DNA? The recognition sequence for each enzyme is indicated in the right-hand column. a. Sau3AI GATC b. BamHI GGATCC d. NotI

GCGGCCGC

c. XzaI

GAAGGATCCTTC

ANS: C A restriction nuclease that has a 4-base-pair recognition sequence cuts on average once every 44 or 256 base pairs; one that has a 6-base-pair recognition sequence cuts once every 4 6 or 4096 base pairs; one that has an 8-base-pair recognition sequence cuts once every 48 or 65,536 base pairs; one that has a 12-base-pair recognition sequence cuts once every 412 or 16 million base pairs. Thus, to obtain fragments of about 70 kb in size, you would cut with a nuclease that recognizes an 8-base-pair site.

DIF: Moderate REF: 10.1 OBJ: 10.1.b Express how the target sequence of a restriction enzyme relates to how often the sequence will appear in a long piece of DNA and the size of the DNA fragments it will generate.

MSC: Understanding 4. You have a circular plasmid that can be cut by the restriction nuclease HindIII, as diagrammed in Figure 10-4.

Figure 10-4

If you were to cut this circular piece of DNA with HindIII, which of the answers below best predicts what you would

get? a. one linear piece of DNA b. two circular pieces of DNA c. two semicircular pieces of DNA d. two linear pieces of DNA ANS: A DIF: Easy REF: 10.1 OBJ: 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory. MSC: Applying 5. You have a piece of circular DNA that can be cut by the restriction nucleases XhoI and SmaI, as indicated in Figure 10-5.

Figure 10-5

If you were to cut this circular piece of DNA with both XhoI and SmaI, how many fragments of DNA would you end up with? a. 1 b. 2 c. 3 d. 4 ANS: B DIF: Easy REF: 10.1 OBJ: 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory. MSC: Applying 6. You have a piece of circular DNA that can be cut by the restriction nucleases EcoRI, HindIII, and NotI, as indicated in Figure 10-6.

Figure 10-6

Which of the following statements is FALSE? a. One piece of DNA will be obtained when this DNA is cut by NotI. b. A piece of DNA that cannot be cut by EcoRI will be obtained by cutting this DNA with both NotI and HindIII.

c. Two DNA fragments that cannot be cut by HindIII will be obtained when this DNA is cut by EcoRI and NotI. d. Two DNA fragments of unequal size will be created when this DNA is cut by both HindIII and EcoRI. ANS: D DIF: Moderate REF: 10.1 OBJ: 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory. MSC: Applying 7. You have a linear piece of DNA that can be cut by the restriction nucleases HindIII and EcoRI, as diagrammed in Figure 10-7.

Figure 10-7

If you were to cut this linear DNA with HindIII, what type of DNA fragments do you predict you would obtain? a. three linear pieces of DNA b. two linear pieces of DNA, only one of which can be cut by EcoRI c. two linear pieces of DNA, both of which can be cut by EcoRI d. two linear pieces of DNA, only one of which can be cut by HindIII ANS: B DIF: Easy REF: 10.1 OBJ: 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory. MSC: Applying 8. During gel electrophoresis, DNA fragments a. travel through a matrix containing a microscopic network of pores. b. migrate toward a negatively charged electrode. c. can be visualized without stains or labels. d. are separated on the basis of their sequence.

ANS: A DIF: Easy REF: 10.1 OBJ: 10.1.c Outline how DNA fragments can be separated by size using gel electrophoresis. MSC: Remembering 9. You are interested in a single-stranded DNA molecule that contains the following sequence: 5′- . . . . .GATTGCAT. . . . -3′ Which molecule will hybridize to your sequence of interest? a. 5′-GATTGCAT-3′ b. 5′-TACGTTAG-3′ c. 5′-CTAACGTA-3′ d. 5′-ATGCAATC-3′ ANS: D DIF: Moderate REF: 10.1 OBJ: 10.1.j Review how hybridization allows the detection of a DNA fragment of interest. MSC: Applying 10. During DNA renaturation, two DNA strands will a. break the covalent bonds that hold the nucleotides together while maintaining the hydrogen bonds that hold the two strands together. b. break the hydrogen bonds that hold the two strands together with no effect on the covalent bonds that hold the nucleotides together.

c. re-form a double helix if the two strands have complementary sequences. d. re-form a double helix if the two strands are identical in sequence.

ANS: C “DNA renaturation” refers to the process by which two DNA strands will re-form a double helix if the two strands are complementary, DNA strands should not renature if their sequences are identical. The breaking of the hydrogen bonds between two DNA strands is the process of denaturation.

DIF: Easy REF: 10.1 OBJ: 10.1.j Review how hybridization allows the detection of a DNA fragment of interest.

MSC: Understanding 11. DNA ligase is an enzyme used when making recombinant DNA molecules in the lab. In what normal cellular process is DNA ligase involved? a. none; it is only found in virally infected cells b. transcription c. transformation d. DNA replication ANS: D DIF: Easy REF: 10.1 OBJ: 10.1.g Review how a DNA fragment is inserted into a plasmid.

MSC: Remembering 12. Figure 10-12 shows the cleavage sites of several restriction nucleases.

Figure 10-12

You cut a vector using the PciI restriction nuclease. Which of the following restriction nucleases will generate a fragment that can be ligated into this cut vector with the addition of only ligase and ATP? a. HindIII b. NcoI c. MmeI d. NspV ANS: B However, you will not be able to excise the fragment after ligation, because you will destroy both the PciI site and the NcoI site, creating a new site with the sequence 5′-ACATGG-3′ 3′-TGTACC-5′ DIF: Difficult REF: 10.1 OBJ: 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory. | 10.1.b Express how the target sequence of a restriction enzyme relates to

how often the sequence will appear in a long piece of DNA and the size of the DNA fragments it will generate. MSC: Applying 13. Figure 10-13 depicts a strategy by which a DNA fragment produced by cutting with the EcoRI restriction nuclease can be joined to a DNA fragment produced by cutting DNA with the HaeIII restriction nuclease.

Figure 10-13

Note that cutting DNA with EcoRI produces a staggered end, whereas cutting DNA with HaeIII produces a blunt end.

Why must polymerase be added in this reaction? a. Polymerase will fill in the staggered end to create a blunt end. b. Polymerase is needed to seal nicks in the DNA backbone. c. Polymerase will add nucleotides to the end produced by the HaeIII restriction nuclease. d. Without polymerase, there will not be enough energy for the reaction to proceed.

ANS: A DIF: Moderate REF: 10.1 OBJ: 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory. | 10.1.b Express how the target sequence of a restriction enzyme relates to how often the sequence will appear in a long piece of DNA and the size of the DNA fragments it will generate. MSC: Understanding 14. DNA can be introduced into bacteria by a mechanism called a. transcription. b. ligation. c. replication. d. transformation.

ANS: D DIF: Easy REF: 10.1 OBJ: 10.1.h Summarize how a plasmid carrying a DNA fragment of interest can be inserted into bacteria and how the amplified fragment is subsequently isolated. MSC: Remembering 15. A plasmid a. can confer antibiotic resistance to a bacterium. b. is a single-stranded circular DNA molecule that can undergo horizontal transfer among bacteria. c. is a tool designed in the lab and never found in naturally occurring bacteria. d. always becomes part of the bacterial chromosome during transformation.

ANS: A Plasmids that are found in naturally occurring bacteria can confer antibiotic resistance to a bacterium; plasmids in the lab often carry an antibiotic-resistance gene so that scientists can select for bacteria that are carrying their recombinant DNA molecule. Plasmids are made of double-stranded DNA, and typically have their own replication origin, allowing for plasmids to replicate independently of the bacterial chromosome.

DIF: Easy REF: 10.1 OBJ: 10.1.f List the features typically present in a plasmid used as a cloning vector. | 10.1.h

Summarize how a plasmid carrying a DNA fragment of interest can be inserted into bacteria and how the amplified fragment is subsequently isolated. MSC: Remembering 16. Which of the following statements about DNA libraries is TRUE? a. Production of a DNA library involves the direct insertion of short DNA fragments into bacteria through transformation. b. By placing the library DNA into bacteria, the bacteria can be used to amplify the desired DNA fragments from the DNA library. c. Individual bacteria that have taken up most of the library DNA are selected for during the construction of a DNA library. d. The library DNA within the bacteria will only be replicated when it contains the plasmid with the gene of interest.

ANS: B To make a DNA library, the fragmented DNA must first be placed into an appropriate plasmid before it can be placed into the bacteria through transformation; these DNA fragments are not directly inserted into the bacterium. Any individual bacterium in a library should only carry a single DNA fragment from the library. The different plasmids within a DNA library will all be replicated within bacteria. Other techniques (such as hybridization to a DNA probe) can be used to identify and isolate the bacteria carrying the plasmid that has the gene of interest from the rest of the library.

DIF: Easy REF: 10.1 OBJ: 10.1.h Summarize how a plasmid carrying a DNA fragment of interest can be inserted into bacteria and how the amplified fragment is subsequently isolated. MSC: Understanding 17. Which of the following statements about genomic DNA libraries is FALSE? a. The larger the size of the fragments used to make the library, the fewer colonies you will have to examine to find a plasmid that contains the gene you are interested in studying. b. The larger the size of the fragments used to make the library, the more DNA you will need to examine to find your gene of interest once you have identified a plasmid that contains the gene you are interested in studying. c. The larger the genome of the organism from which a library is derived, the larger the fragments inserted into the vector will tend to be. d. The smaller the gene you are seeking, the more likely it is that the gene will be found on a single plasmid.

ANS: C The sizes of the fragments left after a restriction digest do not depend on the total size of the genome; they depend on the sequence of the genome and the frequency with which the restriction enzyme recognition site used to construct the library is found in the genome. The size of the fragments in the library will affect how many plasmids you will need to screen to identify your gene of interest as well as how much DNA you will need to examine to find your gene of interest. As a limiting case, think of what would happen if a fragment the size of the entire genome were inserted into the bacterial vector. You would have to screen only one colony to find the clone that hybridized to your probe, but you would need to scan through a lot more DNA to find out where on the insert your gene of interest lay. The larger the gene you are seeking, the more likely it is that there will be a restriction fragment in the gene (or that the gene will be broken if the DNA was fragmented by random shearing), and hence the less likely it is that the entire gene will be found in one clone.

DIF: Easy REF: 10.1 OBJ: 10.1.i Compare a genomic library and a cDNA library in terms of what each contains, how each is made, and how each might be used. MSC: Understanding 18. A DNA library has been constructed by purifying chromosomal DNA from mice, cutting the DNA with the restriction

enzyme NotI, and inserting the fragments into the NotI site of a plasmid vector. What information CANNOT be retrieved from this library? a. gene regulatory sequences b. intron sequences c. sequences of the telomeres (the ends of the chromosomes) d. amino acid sequences of proteins ANS: C The very ends of all of the chromosomes are unlikely to be NotI sites, meaning that the fragments containing the ends of the chromosomes will not be able to insert into the bacterial vector (because they have not been cut by NotI at both ends) and will be lost from the library. All sequences present in genomic DNA (which includes regulatory sequences and introns) should be present in a genomic library. The coding sequence of the gene (and hence the amino acid sequence of the encoded protein) is also present in a genomic clone, although it is interrupted by intron sequences and is therefore somewhat difficult (but not impossible) to determine.

DIF: Easy REF: 10.1 OBJ: 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory. | 10.1.i Compare a genomic library and a cDNA library in terms of what each contains, how each is made, and how each might be used. MSC: Understanding 19. PCR was invented in a. the 1800s. b. the 1950s. c. the 1980s. d. 2009.

ANS: C DIF: Easy REF: 10.2 OBJ: 10.2.a Summarize the main differences between the classical approach to cloning and cloning by PCR. MSC: Remembering 20. Starting with one double-stranded DNA molecule, how many cycles of PCR would you have to perform to produce about 100 double-stranded copies (assuming 100% efficiency per cycle)? a. 2 b. 7 c. 25 d. 100 ANS: B After seven cycles of PCR, you would have 128 molecules of DNA if the reaction were 100% efficient.

DIF: Easy REF: 10.2 OBJ: 10.2.c Estimate the number of double-stranded DNA molecules produced after the first, second, third, and 30th PCR cycles. MSC: Understanding 21. A double-stranded DNA molecule can be separated into single strands by heating it to 90°C because a. heat disrupts the hydrogen bonds holding the sugar–phosphate backbone together. b. DNA is negatively charged. c. heat disrupts hydrogen-bonding between complementary nucleotides. d. DNA is positively charged.

ANS: C DIF: Easy REF: 10.2 OBJ: 10.2.b Review the purpose of each of the three basic steps involved in each cycle

of PCR. MSC: Remembering 22. Which of the following statements about PCR is FALSE? a. PCR uses a DNA polymerase from a thermophilic bacterium. b. PCR is particularly powerful because after each cycle of replication, there is a linear increase in the amount of DNA available. c. For PCR, every round of replication is preceded by the denaturation of the double-stranded DNA molecules. d. The PCR will generate a pool of double-stranded DNA molecules, most of which will have DNA from primers at the 5′ ends.

ANS: B DIF: Easy REF: 10.2 OBJ: 10.2.e Explain how PCR can be used to produce genomic or cDNA clones.

MSC: Understanding 23. PCR involves a heating step, followed by a cooling step, and then DNA synthesis. What is the primary reason for why this cooling step is necessary? a. Cooling the reaction ensures the integrity of the covalent bonds holding the nucleotides together in the DNA strand. b. Cooling the reaction gives the DNA polymerase an opportunity to rest from the previous cycle so that it will be ready for the next round of synthesis. c. Transcription takes place during the cooling step. d. Cooling the reaction brings the temperature down to a level that is compatible with the short primers forming stable hydrogen bonds with the DNA to be amplified.

ANS: D The cooling step allows the short primers to anneal to the DNA template. During PCR, the reaction is heated to a temperature that will break the hydrogen bonds holding the strands together but should not be harmful to the covalent bonds holding the nucleotides together. The DNA polymerase used in PCR is a thermostable enzyme that can tolerate the high temperatures used in PCR. Transcription involves the production of RNA from DNA and does not occur during PCR. DIF: Easy REF: 10.2 OBJ: 10.2.b Review the purpose of each of the three basic steps involved in each cycle of PCR. MSC: Understanding 24. Which of the following limits the use of PCR to detect and isolate genes? a. The sequence at the beginning and end of the DNA to be amplified must be known. b. It also produces large numbers of copies of sequences beyond the 5′ or 3′ end of the desired sequence. c. It cannot be used to amplify cDNAs or mRNAs. d. It will amplify only sequences present in multiple copies in the DNA sample.

ANS: A To construct primers that will bracket the desired gene, you have to know the sequence at the beginning and end of the DNA to be copied.

DIF: Easy REF: 10.2 OBJ: 10.2.e Explain how PCR can be used to produce genomic or cDNA clones.

MSC: Understanding 25. Figure 10-25 shows a restriction map of a piece of DNA containing your favorite eukaryotic gene. The arrow indicates the position and orientation of the gene in the DNA. In part (B) of the figure are enlargements showing the portions of the DNA whose sequences have been used to make PCR primers A, B, C, and D. Which primer or primers can be

used to amplify the gene?

Figure 10-25

a. primers A and B b. primers A and C c. primers A and D d. primer A alone ANS: C In order for the PCR primers to amplify DNA, they must bind to complementary strands and synthesize the DNA toward the other primer. The use of primers A and B will lead to DNA synthesis away from each other. Use of primer A alone or primer A with primer C will lead to the synthesis of single-stranded DNA. DIF: Moderate REF: 10.2 OBJ: 10.2.e Explain how PCR can be used to produce genomic or cDNA clones.

MSC: Applying 26. You want to design a set of PCR primers to amplify a portion of a gene from a cDNA library. Which of the following concerns about PCR primer design is the most legitimate? a. You must be careful when designing your primers to take into account which DNA strand was transcribed in mRNA and make sure both primers are complementary to the mRNA. b. You must be certain not to include any DNA sequences in your primer that are upstream (5′) of the AUG start codon. c. You must make sure that all the DNA sequences in your primer lie within an exon, and do not span two exons. d. You must be certain that all the DNA sequences in your primer are not located downstream (3′) of the polyadenylation signal.

ANS: D Primers made from DNA sequences downstream of the polyadenylation signal will not be included on the mRNA molecule. mRNAs are converted into cDNAs, which are double-stranded, so it is not necessary to worry about which strand the primers will hybridize with. The transcriptional and translational start sites of an mRNA molecule differ, so including some sequences 5′ of the AUG start codon is not necessarily a problem. Because cDNAs are made from mRNAs, the exon-splicing events have already occurred so it shouldn’t matter if your primer sequences span two exons or lie within a single exon.

DIF: Easy REF: 10.2 OBJ: 10.2.e Explain how PCR can be used to produce genomic or cDNA clones.

MSC: Understanding 27. You want to design a set of PCR primers to specifically amplify a portion of a gene from genomic DNA. Which of the following statements about this set of primers is TRUE? a. One of the primers should contain sequences complementary to common repetitive sequences in the genome, to ensure that it will hybridize to the genomic DNA.

b. Each primer should span the junction of two adjacent exons and leave out intron sequences, to ensure that a portion of the coding portion of the gene is amplified and not noncoding regulatory sequence. c. If the primers include sequence from both an exon and the adjacent intron, this will help ensure that the amplified product is made from genomic DNA and not any contaminating mRNA. d. The primers should use “U”s instead of “T”s to ensure that the coding portion of the gene is amplified.

ANS: C Having sequences from the introns will help ensure the amplification of genomic DNA instead of amplification from contaminating RNA. Repetitive DNA sequence should be avoided, as using those sequences for your PCR primer may lead to regions other than the gene of interest being amplified. Primers that span two exons without the intervening intron will not hybridize efficiently to the genomic DNA and should not be used to amplify from genomic DNA. Both U (uracil) and T (thymine) will bind to A (adenine).

DIF: Easy REF: 10.2 OBJ: 10.2.e Explain how PCR can be used to produce genomic or cDNA clones.

MSC: Understanding 28. Your friend works at the Centers for Disease Control and Prevention and has discovered a brand-new virus that has recently been introduced into the human population. She has just developed a new assay that allows her to detect the virus by using PCR products made from the blood of infected patients. The assay uses primers in the PCR assay that hybridize to sequences in the viral genome.

Your friend is distraught because of the result she obtained (see Figure 10-28) when she looked at PCR products made using the blood of three patients suffering from the viral disease, using her own blood, and using a leaf from her petunia plant.

You advise your friend not to panic, as you believe she is missing an important control. Which one of the choices listed below is the best control for clarifying the results of her assay? Explain your answer.

Figure 10-28

a. a PCR assay using blood from a patient who is newly infected but does not yet show symptoms b. a PCR assay using blood from a dog c. a PCR assay using blood from an uninfected person d. repeating the experiments she has already done with a new tube of polymerase ANS: C Primers can sometimes hybridize to unintended sequences and produce unintended products. The appropriate con-

trol for your friend’s experiment would be DNA from an uninfected person; in that way, she would be able to determine whether the bands present in the PCR from her blood truly correspond to product generated from viral DNA rather than cross-hybridization to DNA sequences in the human genome, because the bands would be absent from a person uninfected by the virus in the former case only. Doing PCR from an infected but asymptomatic person would not be useful, because it would not allow your friend to distinguish whether she is infected. Although doing PCR from dog blood should not give any viral bands, any nonspecific products from a dog would probably be different from those in your friend. The absence of PCR fragments in the petunia lane suggests that there is no viral contaminant in any of your friend’s reagents, so using a new tube of polymerase is not the solution.

DIF: Moderate REF: 10.2 OBJ: 10.1.c Outline how DNA fragments can be separated by size using gel electrophoresis. |10.2.f Outline how PCR can be used to detect an infectious agent in a sample of blood. MSC: Applying 29. Why is an excess of normal deoxyribonucleoside triphosphate molecules (dNTPs) needed during dideoxy sequencing? a. DNA polymerase uses the dNTPs to synthesize a DNA molecule complementary to the molecule being sequenced. b. dNTPs are consumed as energy to fuel the sequencing reactions. c. When dNTP levels are too low, there will be very few chain-termination events. d. The dNTPs can hybridize to the fragment to be sequenced and serve as primers for DNA polymerase.

ANS: A DIF: Easy REF: 10.3 OBJ: 10.3.a Review how chain-terminating dideoxynucleotides are used to determine the sequence of a piece of DNA. MSC: Understanding 30. Why are dideoxyribonucleoside triphosphates used during DNA sequencing? a. They cannot be incorporated into DNA by DNA polymerase. b. They are incorporated into DNA particularly well by DNA polymerases from thermophilic bacteria. c. Incorporation of a dideoxyribonucleoside triphosphate leads to the termination of replication for that strand. d. Dideoxyribonucleoside triphosphates are more stable than deoxyribonucleoside triphosphates.

ANS: C DIF: Easy REF: 10.3 OBJ: 10.3.a Review how chain-terminating dideoxynucleotides are used to determine the sequence of a piece of DNA. MSC: Remembering 31. With fully automated Sanger sequencing, all four chain-terminating ddNTPs are added into a single reaction and loaded onto a capillary gel. The sequence can be determined even though all ddNTPs are mixed together because a. the Sanger sequencing reactions do not contain dNTPs and use the ddNTPS to synthesize the DNA. b. the Sanger sequencing reactions utilize ddNTPs, each labeled with a different fluorescent tag, which allows all four ddNTPs to be incorporated into a single molecule of DNA. c. the Sanger sequencing reactions generate a set of products, each of which carries a single fluorescent tag whose color reveals the identity of the base that is at the end of the product. d. the fully automated Sanger sequencing reactions do not require DNA polymerase because the bases are read as the DNA is pulled through a tiny pore at the end of the capillary gel.

ANS: C Sanger sequencing utilizes ddNTPs that carry a unique fluorescent tag. Sanger sequencing requires dNTPs as well as a small amount of the chain-terminating ddNTPs. Ideally, a single ddNTP is incorporated into each product; all four ddNTPs should not be incorporated into a single product. Sanger sequencing still requires DNA polymerase, and reads the fluorescent ddNTP at the end of each product.

DIF: Moderate REF: 10.3 OBJ: 10.3.a Review how chain-terminating dideoxynucleotides are used to determine the sequence of a piece of DNA. MSC: Understanding 32. Second-generation sequencing differs from Sanger sequencing because a. second-generation sequencing does not depend on chain-terminator nucleoside triphosphates. b. second-generation sequencing does not require DNA polymerase. c. for the cost per base sequenced, second-generation sequencing is much more expensive than Sanger sequencing. d. second-generation sequencing can sequence millions of pieces of DNA at the same time on a single glass slide.

ANS: D Second-generation sequencing is faster and cheaper because millions of reactions can be carried out in parallel on a solid support.

DIF: Easy REF: 10.3 OBJ: 10.3.c Explain how the second-generation Illumina sequencing method uses PCR coupled with chain-terminating nucleotides to determine the sequence of a genome. MSC: Remembering 33. Which of the following statements about Illumina sequencing is FALSE? a. The nucleoside triphosphates that are used have a fluorescent marker. b. The nucleoside triphosphates that are used have a 3′ chemical group that will terminate DNA synthesis. c. The termination of DNA synthesis by the nucleoside triphosphates that are used is reversible. d. Both normal and modified nucleoside triphosphates are used to synthesize DNA. ANS: D There are no normal dNTPs present in the reaction for Illumina sequencing.

DIF: Easy REF: 10.3 OBJ: 10.3.c Explain how the second-generation Illumina sequencing method uses PCR coupled with chain-terminating nucleotides to determine the sequence of a genome. MSC: Understanding 34. Which portion of an organism’s genome is most likely to diverge most rapidly over evolutionary time? a. the noncoding gene regulatory regions. b. the noncoding telomeric sequences. c. the exons. d. the sequences at the intron/exon border.

ANS: A DIF: Easy REF: 10.3 OBJ: 10.3.e Review the types of information that can be gathered from comparative genome analyses. MSC: Remembering 35. Which of the following techniques is not appropriate if you want to examine whether a gene is expressed in a specific tissue? a. in situ hybridization b. production of a cDNA library c. RNA-Seq d. next generation genome sequencing ANS: D Next generation genome sequencing will give you the sequence of the genomic DNA, which cannot quickly tell you about where a gene is expressed. All the other techniques will examine gene expression. In situ hybridization can be used to determine when and where a gene is expressed. cDNA libraries are produced from mRNA, and can tell you

about the genes expressed in that population of messages. RNA-seq can be used to analyze the RNAs expressed in a specific tissue or cell type.

DIF: Easy REF: 10.4 OBJ: 10.4.a Describe the techniques of RNA-Seq and in situ hybridization and compare what each reveals about gene expression in a particular cell type or tissue. MSC: Understanding 36. You create a recombinant DNA molecule that fuses the coding sequence of green fluorescent protein to the regulatory DNA sequences that control the expression of your favorite genes. Which of the following pieces of information can you NOT gain by examining the expression of this reporter gene? a. the tissue where the protein encoded by this gene is expressed b. the cell in which the protein encoded by this gene is expressed c. the specific location within the cell of the protein encoded by this gene d. when, during an organism’s development, this gene is expressed ANS: C The information for localizing proteins within a cell is found on the protein product and not in the regulatory DNA sequences. If you were to fuse your reporter gene to the DNA sequences that encode the protein to produce a GFP fusion protein, then you might determine the specific location of the protein within the cell.

DIF: Easy REF: 10.4 OBJ: 10.4.b Outline the construction of a reporter gene and review how such reporters can be used to assess patterns of gene expression. | 10.4.c Review how reporter genes can be used to track the location of proteins within cells. MSC: Understanding 37. Which of the following statements about RNA interference (or RNAi) is FALSE? a. RNAi is a natural mechanism used to regulate genes. b. During the process of RNAi, hybridization of a small RNA molecule with the mRNA degrades the mRNA. c. Because RNAi depends on the introduction of a double-stranded RNA into a cell or an organism, it is not a process that can cause heritable changes in gene expression. d. In C. elegans, RNAi can be introduced into the animals by feeding them with bacteria that produce the inhibitory RNA molecules.

ANS: C DIF: Easy REF: 10.4 OBJ: 10.4.e Review how RNA interference can be used to inhibit the activity of a target gene. MSC: Understanding 38. You have been hired to create a cat that will not cause allergic reactions in cat-lovers. Your coworkers have cloned the gene encoding a protein found in cat saliva, expressed the protein in bacteria, and shown that it causes violent allergic reactions in people. But you soon realize that even if you succeed in making a knockout cat lacking this gene, anyone who buys one will easily be able to make more hypoallergenic cats just by breeding them. Which of the following will ensure that people will always have to buy their hypoallergenic cats from you? a. Inject the modified embryonic stem (ES) cells into embryos that have a genetic defect to prevent the mature adult from reproducing. b. Implant the injected embryos into a female cat that is sterile as a result of a genetic defect. c. Sell only the offspring from the first litter of the female cat implanted with the injected embryos. d. Surgically remove the sexual organs of all the knockouts before you sell them.

ANS: D Neutering all the knockout animals that you sell is the only option of the four listed that will prevent happy pet owners

from becoming happy pet breeders. Injecting modified ES cells into embryos that will be sterile will not allow you to make any knockout cats because the first litter (which will at best have a few mosaics in which one copy of the gene has been knocked out in the germ cells) will be sterile and you will not be able to mate them. The genotype of the female cat in which you implant the embryos has no effect on the genotype of the embryos. The first litter will yield mosaic cats that still have one copy of the allergen-producing gene in their cells and are therefore not hypoallergenic.

DIF: Moderate REF: 10.4 OBJ: 10.4.f Summarize the ways in which genetically modified animals can be used to explore the function of a gene and describe how these techniques can be adapted to study genes whose function is essential during development. MSC: Applying 39. Insulin is a small protein that regulates blood sugar level and is given to patients who suffer from diabetes. Many years ago, diabetics were given insulin that had been purified from pig pancreas. Once recombinant DNA techniques became available, the DNA encoding insulin could be placed into an expression vector and insulin could be produced in bacteria. Which of the following is NOT a reason why purifying insulin from bacteria is a better way to produce insulin for diabetics than using insulin purified from a pig pancreas? a. Insulin can be easily produced in large quantities from cells carrying the cloned DNA sequence. b. The creation of transgenic pigs that expressed insulin was very expensive compared to the cost of creating bacteria that expressed insulin. c. Insulin made from a bacterial culture and then purified will be free of any possible contaminating viruses that pigs (and any other animals) harbor. Since pigs are more closely related to people than bacteria are, their viruses are more likely to be harmful to people than are viruses that might infect bacteria. d. The pig protein has slight amino acid differences compared to the human protein, so human insulin produced by bacteria will work better in people.

ANS: B Transgenic pigs were not created for the purification of insulin. Instead, scientists were purifying the endogenous insulin present in the pig pancreas that is made from the normal pig insulin gene.

DIF: Easy REF: 10.4 OBJ: 10.4.k State how expression vectors direct the production of large quantities of a selected protein. MSC: Remembering 40. Which of the following describes a feature found in bacterial expression vectors but not in cloning vectors? a. origin of replication b. cleavage sites for restriction nucleases c. promoter DNA sequences d. a polyadenylation signal ANS: C Without promoter sequences, there will be no gene expression. Origins of replication and cleavage sites for restriction nucleases are found in both cloning vectors and expression vectors. Bacterial mRNAs do not undergo polyadenylation.

DIF: Easy REF: 10.4 OBJ: 10.1.f List the features typically present in a plasmid used as a cloning vector. | 10.4.k State how expression vectors direct the production of large quantities of a selected protein. MSC: Remembering 41. The Cas9 enzyme used in the CRISPR system a. was created by scientists by modifying restriction enzymes. b. was discovered in bacteria as part of a mechanism to protect themselves from foreign DNA.

c. was found as part of a natural mechanism used by plants to protect themselves from viruses. d. was found in cultured mouse embryonic stem (ES) cells.

ANS: B DIF: Easy REF: 10.4 OBJ: 10.4.h Recall how the CRISPR system can be adapted to activate or repress a target gene. MSC: Remembering 42. How does the Cas9 system target where it produces a double-strand break in the DNA? a. A guide RNA molecule is associated with Cas9 and will direct Cas9 to bind at sequences complementary to the guide RNA. b. The Cas9 protein contains amino acids that can interact with specific sequences in the DNA, targeting Cas9 to those specific sites. c. The Cas9 protein binds to a recombinase, allowing it to disable the gene of interest. d. A guide RNA molecule associated with Cas9 provides the catalytic activity to cleave the DNA where Cas9 binds.

ANS: A The guide RNA associated with Cas9 directs it to a segment of DNA with a complementary sequence. Researchers take advantage of this property by supplying an appropriate guide RNA to target Cas9 to wherever in the genome researchers may wish to make a cut to alter a gene.

DIF: Easy REF: 10.4 OBJ: 10.4.h Recall how the CRISPR system can be adapted to activate or repress a target gene.

MSC: Remembering 43. To study gene function, scientists use various methods to disrupt the sequence of the gene. In some cases, this is not possible and conditional knockouts must be used instead. Which of the following statements about the Cre-Lox system for creating conditional knockouts is TRUE? a. Conditional knockouts can be produced because the gene of interest is flanked by LoxP sites only in the cell type of interest. b. Conditional knockouts can be produced by expressing the Cre recombinase gene only in the cell type of interest. c. The gene of interest is not expressed in the cell type of interest in a conditional mutant because Cre recombinase acts to inhibit RNA polymerase when it binds to the LoxP site. d. Nontarget tissues do not have the Cre recombinase gene, and thus do not produce Cre recombinase.

ANS: B The gene of interest is flanked by LoxP sites in all cells, but only gets removed in cells that express Cre recombinase.

Cre recombinase catalyzes recombination and does not directly inhibit the activity of RNA polymerase. All cells carry the Cre recombinase gene, but the Cre recombinase gene is only expressed in the cell type of interest in conditional knockouts.

DIF: Easy REF: 10.4 OBJ: 10.4.d Differentiate between the “classical genetic approach” to studying gene function and the methods referred to as “reverse genetics.” | 10.4.l Outline how recombinant DNA techniques make it possible to analyze the activity of a protein starting from its gene—or to assess gene function starting with its encoded protein.

MSC: Understanding 44. The sequencing of the human genome was a technological accomplishment. Which of the following did NOT contribute to the sequencing of the human genome in the early 2000s? a. shotgun sequencing b. restriction site mapping of BACs

c. automated Sanger sequencing d. Illumina sequencing ANS: D Illumina sequencing was not invented at the time the human genome was completed.

DIF: Easy REF: 10.3 OBJ: 10.3.b Distinguish between the shotgun method for assembling a genome sequence and the clone-by-clone, bacterial artificial chromosome approach. MSC: Remembering 45. Figure 10-45A depicts the restriction map of one segment of the human genome for four restriction nucleases W, X, Y, and Z. Figure 10-45B depicts the restriction maps of four individual BAC clones that contain segments of human DNA from the region depicted in Figure 10-45A.

Figure 10-45

From this information, how would you order these BAC clones, from left to right? a. 1, 2, 3, 4 b. 2, 1, 4, 3 c. 3, 4, 2, 1 d. 4, 1, 3, 2 ANS: B DIF: Difficult REF: 10.3 OBJ: 10.3.b Distinguish between the shotgun method for assembling a genome sequence and the clone-by-clone, bacterial artificial chromosome approach. MSC: Applying 46. You have a circular plasmid that has a single EcoRI site in it, as diagrammed in Figure 10-46, which also shows the cleavage site for EcoRI. Choose the answer below that best represents what the end of the DNA molecule will look like once you cut the plasmid with EcoRI. Note that only the very ends of the DNA molecule are shown in the answers.

Figure 10-46

ANS: C DIF: Difficult REF: 10.1 OBJ: 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory. MSC: Applying 47. Figure 10-45A depicts the restriction map of one segment of the human genome for four restriction nucleases W, X, Y, and Z. Figure 10-45B depicts the restriction maps of four individual BAC clones that contain segments of human DNA from the region depicted in Figure 10-45A.

Figure 10-45

From this information, how would you order these BAC clones, from left to right? a. 1, 2, 3, 4 b. 2, 1, 4, 3 c. 3, 4, 2, 1

d. 4, 1, 3, 2 ANS: B The order of the BAC clones relative to the segment of human DNA is shown in Figure 10-47A.

Figure 10-47A

DIF: Difficult REF: 10.3 OBJ: 10.3.b Distinguish between the shotgun method for assembling a genome sequence and the clone-by-clone, bacterial artificial chromosome approach. MSC: Applying

SHORT ANSWER 1. You have accidentally torn the labels off two tubes, each containing a different plasmid, and now do not know which plasmid is in which tube. Fortunately, you have restriction maps for both plasmids, shown in Figure 10-48. You have the opportunity to test just one sample from one of your tubes. You have equipment for agarose-gel electrophoresis, a standard set of DNA size markers, and the necessary restriction enzymes.

Figure 10-48

A. Outline briefly the experiment you would do to determine which plasmid is in which tube.

B. Which restriction enzyme or combination of restriction enzymes would you use in this experiment?

ANS: A. You would first digest your sample with a combination of restriction enzymes selected so that they give a set of fragment sizes that could have come from only one of the plasmids. Then you would run the resulting mixture of DNA fragments on a gel alongside a set of size markers and determine the size of each fragment. By looking at the restriction maps, you should then be able to match your results to one of the plasmids.

B. Digestion with any of the following combinations will enable you to distinguish which plasmid you have: HindIII + BglII; EcoRI + BglII; EcoRI + BglII + HindIII. The plasmids are the same size, so you cannot distinguish between them simply by making a single cut (with HindIII) and determining the size of the complete DNA by gel electr ophoresis. Nor can you distinguish them by cutting with all four restriction nucleases, because the set of fragment sizes produced from both plasmids will be the same. Cutting with BamHI or EcoRI on their own is not sufficient because you will get bands of the same size from both plasmid A and plasmid B. The only difference between the two plasmids is in the location of the BglII site relative to the two BamHI sites, so if you cut with an enzyme that cuts outside the BamHI fragment and with BglII, you will get different-sized fragments from the two plasmids.

DIF: Difficult REF: 10.1 OBJ: 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory. | 10.1.b Express how the target sequence of a restriction enzyme relates to how often the sequence will appear in a long piece of DNA and the size of the DNA fragments it will generate. | 10.1.c Outline how DNA fragments can be separated by size using gel electrophoresis. MSC: Creating

2. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. centrifugation

negative

digested

neutral

electrophoresis

positive

fluoresce

radioactive

gravity

sequencing

irradiate

smaller

larger

voltage

During gel __________, DNA fragments can be loaded into one end of an agarose slab to separate the fragments on the basis of charge. As __________ is applied across the agarose slab, the DNA molecules, which have a __________ charge, will migrate toward the __________ electrode. Because __________ DNA fragments will migrate more quickly, they will be found furthest away from the area of the gel where the DNA fragments were loaded. One method to visualize the DNA on the agarose slab involves staining the DNA with a dye that will __________ under ultraviolet light.

ANS: During gel electrophoresis, DNA fragments can be loaded into one end of an agarose slab to separate the fragments on the basis of charge. As voltage is applied across the agarose slab, the DNA molecules, which have a negative charge, will migrate toward the positive electrode. Because smaller DNA fragments will migrate more quickly, they will be found furthest away from the area of the slab where the DNA fragments were loaded. One method to visualize the DNA on the agarose slab involves staining the DNA with a dye that will fluoresce under ultraviolet light.

DIF: Easy REF: 10.1 OBJ: 10.1.c Outline how DNA fragments can be separated by size using gel electrophoresis. | 10.1.d State how bands of DNA fragments on a gel can be visualized. MSC: Understanding 3. Figure 10-50 shows the recognition sequences and sites of cleavage for the restriction enzymes SalI, XhoI, PstI, and SmaI, and also a plasmid with the sites of cleavage for these enzymes marked.

Figure 10-50

A. After which of the five treatments listed below can the plasmid shown in Figure 10-50 re-form into a circle simply by treating it with DNA ligase? Assume that after treatment any small pieces of DNA are removed, and it is the larger portion of plasmid that will reassemble into a circle.

After digestion with 1. SalI alone.

2. SalI and XhoI.

3. SalI and PstI. 4. SalI and SmaI. 5. SmaI and PstI. B. After which of the treatments can the plasmid re-form into a circle by the addition of DNA ligase after treating the cut DNA with DNA polymerase in a mixture containing the four deoxyribonucleotides? Again, assume that you are trying to get the larger portion of plasmid to reassemble into a circle.

ANS: A. 1 and 2. When SalI and XhoI cut DNA, the staggered ends left behind will match up by base-pairing and can therefore be joined by ligase alone.

B. 1, 2, and 4. SmaI cuts and leaves a blunt end. Addition of DNA polymerase and the four deoxyribonucleotides will fill in the 5′ overhangs generated by digestion with SalI and XhoI, leaving blunt ends. DNA ligase joins the blunt ends. However, 3′ overhangs (that is, those generated by PstI) will not be filled in, because DNA polymerase moves in a 5′-to-3′ direction. DNA ligase will not join 3′ overhangs to blunt-ended DNA, which are the situations presented in treatments 3 and 5.

DIF: Difficult REF: 10.1 OBJ: 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory. | 10.1.b Express how the target sequence of a restriction enzyme relates to how often the sequence will appear in a long piece of DNA and the size of the DNA fragments it will generate. MSC: Applying 4. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. blunt ends

DNA polymerase

RNA

cDNA

genomic

staggered ends

DNA ligase

probe

vector

Two fragments of DNA can be joined together by __________. Restriction enzymes that cut DNA straight across the double helix produce fragments of DNA with __________. A fragment of DNA is inserted into a __________ in order to be cloned in bacteria. A __________ library contains a collection of DNA clones derived from mRNAs. A __________ library contains a collection of DNA clones derived from chromosomal DNA. ANS: Two fragments of DNA can be joined together by DNA ligase. Restriction enzymes that cut DNA straight across the double helix produce fragments of DNA with blunt ends. A fragment of DNA is inserted into a vector in order to be cloned in bacteria. A cDNA library contains a collection of DNA clones derived from mRNAs. A genomic library contains a collection of DNA clones derived from chromosomal DNA. DIF: Easy REF: 10.1 OBJ: 10.1.h Summarize how a plasmid carrying a DNA fragment of interest can be inserted into bacteria and how the amplified fragment is subsequently isolated. | 10.1.i Compare a genomic library and a cDNA library in terms of what each contains, how each is made, and how each might be used.

MSC: Understanding 5. Name two features that a cloning vector for use in bacteria must contain. Explain your answers.

ANS: A cloning vector for use in bacteria must contain the following: 1. a bacterial replication origin (to allow the plasmid to be replicated) 2. an antibiotic-resistance gene or some other selectable marker gene (to allow selection for bacteria that have

taken up the recombinant plasmids) DIF: Easy REF: 10.1 OBJ: 10.1.f List the features typically present in a plasmid used as a cloning vector.

MSC: Understanding 6. What is the main reason for using a cDNA library rather than a genomic library to isolate a human gene from which you wish to make large quantities of the human protein in bacteria?

ANS: The gene isolated from a genomic library would still contain introns, and bacteria do not contain the biochemical machinery for removing introns by RNA splicing. The same gene isolated from the cDNA library will already have had its introns removed.

DIF: Easy REF: 10.1 OBJ: 10.1.i Compare a genomic library and a cDNA library in terms of what each contains, how each is made, and how each might be used. MSC: Understanding 7. Some plasmids from cDNA libraries can have defects because of the way in which a cDNA library is constructed. For each dilemma (A to D) listed below, indicate which of the outcomes (1 to 4) you might encounter, and explain why.

Outcomes may be used more than once.

Table 10-54

ANS: A. Outcome 1 would occur. If the mRNA is degraded from the 5′ end, it will still be reverse-transcribed and will end up in the library as a clone lacking its 5′ end.

B. Outcome 4 would occur. If the mRNA is degraded from the 3′ end, it will lack its 3′ poly-A tail. In the construction of a cDNA library, only molecules that still have their poly-A tail will be reverse-transcribed, so mRNAs lacking their 3′ end will not be represented in the library.

C. Outcome 1 would occur. If the 5′ end hybridizes to sequences in the middle of the gene, the “hairpin” formed when the single-stranded DNA loops back on itself to form the primer for DNA polymerase will be very large. After this loop has been digested, the remaining double-stranded DNA fragment will lack the 5′ end of the gene.

D. Outcome 2 would occur. If the gene has a long stretch of internal As, the poly T primer used in the reverse transcription step can hybridize to the internal poly-A stretch rather than to the poly-A tail, and the resulting cDNA will have lost its 3′ end.

DIF: Moderate REF: 10.1 OBJ: 10.1.i Compare a genomic library and a cDNA library in terms of what each contains, how each is made, and how each might be used. MSC: Analyzing 8. Indicate whether a cDNA library or a genomic DNA library will be more appropriate for use in the following applications.

A. You want to study the promoter of gene A.

B. Gene A encodes a tRNA and you wish to isolate a piece of DNA containing the full-length sequence of the tRNA. C. You discover that gene A is alternatively spliced and you want to see which predicted alternative splice products the cell actually produces.

D. You want to find both gene A and the genes located near gene A on the chromosome.

E. You want to express gene A in bacteria to produce lots of protein A. ANS: A. Genomic library. cDNAs are produced from mRNAs; therefore, the promoters will not be included in a cDNA library.

B. Genomic library. cDNAs are usually constructed by using an oligo dT primer; tRNAs do not have poly-A tails. If the cDNA library were made using random primers, it would be unlikely to contain the full-length version of the tRNA. C. cDNA library. Because cDNAs are produced from mRNAs, isolating cDNAs would tell you which splice variants are produced in a cell.

D. Genomic library. A genomic DNA fragment can contain the genes next to your gene of interest; a cDNA will not.

E. cDNA library. Bacteria do not have the ability to remove introns, which may exist in DNA isolated from a genomic library.

DIF: Moderate REF: 10.1 OBJ: 10.1.i Compare a genomic library and a cDNA library in terms of what each contains, how each is made, and how each might be used. MSC: Applying 9. Which of the restriction nucleases listed below can potentially cleave a segment of cDNA that encodes the peptide KIGDACF?

Table 10-56

ANS: The nucleotide sequences that can encode the peptide KIGDACF are shown below.

The enzyme NsiI cleaves at ATGCAT, which is underlined in the sequence above.

DIF: Difficult REF: 10.1 OBJ: 10.1.b Express how the target sequence of a restriction enzyme relates to how often the sequence will appear in a long piece of DNA and the size of the DNA fragments it will generate.

MSC: Applying 10. Why is a heat-stable DNA polymerase from a thermophilic bacterium (the Taq polymerase) used in the polymerase chain reaction rather than a DNA polymerase from E. coli or humans?

ANS: The PCR technique involves heating the reaction at the beginning of each cycle to separate the newly synthesized DNA into single strands so that they can act as templates for the next round of DNA synthesis. Using a heatstable polymerase avoids having to add it afresh for each round of DNA replication.

DIF: Easy REF: 10.2 OBJ: 10.2.d State the property possessed by the DNA polymerase used in PCR that makes it ideal for this method. MSC: Understanding 11. You want to amplify the DNA between the two stretches of sequence shown in Figure 10-58. Of the listed primers, choose the pair that will allow you to amplify the DNA by PCR.

Figure 10-58

ANS: The appropriate PCR primers are primer 1 (5′-GACCTGTGGAAGC-3′) and primer 8 (5′-TCAATCCCGTATG-3′).

The first primer will hybridize to the bottom strand and prime synthesis in the rightward direction. The second primer will hybridize to the top strand and prime synthesis in the leftward direction. (Remember that strands pair antiparallel.)

The middle two primers in each list (primers 2, 3, 6, and 7) would not hybridize to either strand. The remaining pair of primers (4 and 5) would hybridize, but would prime synthesis in the wrong direction—that is, outward, away from the central segment of DNA. Each wrong choice has been made at one time or another in most laboratories that use PCR. In most cases, the confusion arises because the conventions for writing nucleotide sequences have been ignored. By convention, nucleotide sequences are written 5′ to 3′, with the 5′ end on the left. For double-stranded DNA, the 5′ end of the top strand is on the left.

DIF: Difficult REF: 10.2 OBJ: 10.2.e Explain how PCR can be used to produce genomic or cDNA clones.

MSC: Applying

12. You have the amino acid sequence of a protein and wish to design PCR primers that will allow you to amplify this gene by PCR from genomic DNA. Using this protein sequence, you deduce a particular DNA sequence that can encode this protein. Why is it unwise to use only this DNA sequence you have deduced to design PCR primers for isolating the gene encoding your protein of interest from genomic DNA? ANS: Because most amino acids can be encoded by more than one codon, a given sequence of amino acids could be encoded by several different nucleotide sequences. PCR primers corresponding to all these possible sequences have to be synthesized, to be sure of including the one that corresponds to the actual nucleotide sequence of the gene and thus will hybridize with it.

DIF: Moderate REF: 10.2 OBJ: 10.2.e Explain how PCR can be used to produce genomic or cDNA clones.

MSC: Understanding 13. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.

DNA sequencing

polymerase chain reaction

DNA ligase

recombinant DNA

endonucleases

restriction enzymes

gel electrophoresis

ribonucleases

Nucleases that cut DNA only at specific short sequences are known as __________. DNA composed of sequences from different sources is known as __________. __________ can be used to separate DNA fragments of different sizes. Millions of copies of a DNA sequence can be made entirely in vitro by the __________ technique.

ANS: Nucleases that cut DNA only at specific short sequences are known as restriction enzymes. DNA composed of sequences from different sources is known as recombinant DNA. Gel electrophoresis can be used to separate DNA fragments of different sizes. Millions of copies of a DNA sequence can be made entirely in vitro by the polymerase chain reaction technique.

DIF: Easy REF: 10.1 OBJ: 10.1.b Express how the target sequence of a restriction enzyme relates to how often the sequence will appear in a long piece of DNA and the size of the DNA fragments it will generate. | 10.1.c Outline how DNA fragments can be separated by size using gel electrophoresis. MSC: Understanding 14. Assume that defects in a hypothetical gene X have been linked to a disease. Two copies of a defective gene X will predispose a child to the disease, while a single copy of the gene seems to produce no symptoms. Because early treatment can counteract the effects of the disease, a program of voluntary genetic testing is being performed with prospective parents. Caril Ann is pregnant with Charles’s child. You obtain DNA samples from Charles, Caril Ann, and the fetus. You conduct PCR from these DNA samples using two different primer sets that will detect two commonly found deletions in gene X that are associated with the disease. The gene and location of these primer sets are shown in Figure 10-61A. Your results are shown in Figure 10-61B.

Figure 10-61

A. Which of the three individuals have defects in gene X? B. Which individuals have a single defective gene and which have two defective copies of the gene?

C. Indicate the location of each individual’s defects on gene X. ANS: A. All three are affected.

B. The two parents have a single defective copy of the gene; the fetus has two defective copies.

C. As seen in the two gels diagrammed in Figure 10-61B, Caril Ann and Charles each have one copy of gene X without either deletion that can be detected by primer sets 1 & 2 and one copy of gene X that contains a deletion.

(The more quickly migrating band toward the bottom of the gel is the PCR product from the deleted region of the gene.) Caril Ann carries the copy of gene X that has the deletion that can be detected using primer set 1, while Charles carries the copy of gene X that has the deletion that can bet detected using primer set 2. The fetus has unfortunately inherited the copies of gene X with the deletions from both parents.

DIF: Difficult REF: 10.2 OBJ: 10.2.f Outline how PCR can be used to detect an infectious agent in a sample of blood.

MSC: Applying 15. You are studying a protein, and a small fragment of its sequence is shown below. You have decided that the glutamine in the protein segment has an important role in its function. You decide to change this glutamine to a lysine by changing only one base. Given the partial mRNA sequence that codes for this stretch of protein below, write the new sequence for this stretch of RNA if you were to make this change. Be sure to label the 5′ and 3′ ends.

Table 10-56

F

D

P

Q

G

S

H

5′-UUCGACCCGCAGGGCAGCCAC–3′ ANS: The RNA should now look like: 5′-UUCGACCCGAAGGGCAGCCAC–3′ The changed nucleotide is indicated in gray in the sequences above. The codon for glutamine (CAG) has been changed to AAG, which codes for lysine.

DIF: Moderate REF: 10.4 OBJ: 10.4.l Outline how recombinant DNA techniques make it possible to analyze the activity of a protein starting from its gene—or to assess gene function starting with its encoded protein.

MSC: Applying 16. You have been asked to consult for a biotech company that is seeking to understand why some fungi can live in very extreme environments, such as the high temperatures inside naturally occurring hot springs. The company has isolated two different fungal species, F. cattoriae and W. gravinius, both of which can grow at temperatures exceeding 95°C. The company has determined the following things about these fungal species:

By sequencing and examining their genomes, the biotech company hopes to understand why these species can live in extreme environments. However, the company only has the resources to sequence one genome, and would like your input as to which species should be sequenced and whether you believe a shotgun strategy will work in this case. (Be sure to explain your answer.)

ANS: Even though the genome of F. cattoriae is smaller, the W. gravinius genome is more attractive to sequence because it contains less repetitive DNA. Repetitive DNA makes the assembly of sequenced fragments difficult. Shotgun sequencing would not be an unrealistic approach for W. gravinius, because the genome of W. gravinius contains little repetitive DNA and is relatively small. The genome of H. influenzae is 1.83 megabases long and was successfully sequenced with the shotgun approach. (For comparison, the genome of S. cerevisiae is 14 megabases long.)

DIF: Difficult REF: 10.3 OBJ: 10.3.b Distinguish between the shotgun method for assembling a genome sequence and the clone-by-clone, bacterial artificial chromosome approach. | 10.3.e Review the types of information that can be gathered from comparative genome analyses. MSC: Analyzing 17. Which of the restriction nucleases listed below can potentially cleave a segment of cDNA that encodes the peptide KIGDACF?

Table 10-56

ANS: The enzyme NsiI cleaves at ATGCAT, which is underlined in the sequence below.

The nucleotide sequences that can encode the peptide KIGDACF are shown below.

DIF: Difficult REF: 10.1 OBJ: 10.1.b Express how the target sequence of a restriction enzyme relates to how often the sequence will appear in a long piece of DNA and the size of the DNA fragments it will generate.

MSC: Applying 18. You are studying a protein that contains the peptide sequence RDWKLVI. The part of the DNA encoding this peptide is included in the sequence shown below.

Table 10-56

5′-GGCGTGACTGGAAGCTAGTCATC-3′ 3′-CCGCACTGACCTTCGATCAGTAG-5′ This sequence does not contain any HindIII restriction enzyme sites; the target sequence for the HindIII restriction nuclease is shown in Figure 10-65.

Figure 10-65A

Your goal is to create a HindIII site on this plasmid without changing the coding sequence of the protein. Explain how you would do this.

ANS: The new DNA sequence is shown in Figure 10-65B.

Figure 10-65B

The change in the sequence is indicated by the rectangle; the HindIII recognition sequence has been underlined. The peptide encoded by this piece of DNA is indicated above the DNA sequence. Note that this DNA will still encode leucine, despite the change from a CUA codon to a CUU codon.

DIF: Difficult REF: 10.1 OBJ: 10.1.b Express how the target sequence of a restriction enzyme relates to how often the sequence will appear in a long piece of DNA and the size of the DNA fragments it will generate.

MSC: Applying 19. You have created a piece of recombinant DNA by placing a cDNA from a gene you believe is important for the

differentiation of liver cells (called LC1) onto an expression plasmid that contains all the sequences necessary for propagation of this DNA in bacteria and for the production of the LC1 protein in bacteria. A picture of this plasmid is shown in Figure 10-66A, with the segment of the DNA containing the LC1 gene depicted as a gray rectangle; the promoter sequence is depicted as a white rectangle. The LC1 protein is phosphorylated on serine 54; the nucleotide sequence of the portion of the DNA that encodes this region is shown below the diagram. All HindIII and SalI restriction sites have also been mapped on the plasmid; the recognition sequences for these restriction nucleases are shown in Figure 10-66B.

Table 10-56

Figure 10-66

A. Given the information above, write out the amino acids 52 to 57, encoded by the nucleotide sequence shown above. Be sure to number the amino acids appropriately. (Hint: Remember, serine is amino acid number 54.)

B. You want to create a mutant version of the LC1 gene that replaces the serine 54 found on this peptide with a glycine. You want to do this by changing only one nucleotide, and you also want to destroy the HindIII recognition sequence with this change. Write out a 21-nucleotide DNA sequence that can accommodate these changes. Be sure to (i) write out the DNA and label the 5′ and 3′ ends, (ii) underline the mutated HindIII recognition site, and (iii) circle any change made to the original sequence.

ANS: A. 52 alanine, 53 valine, 54 serine, 55 leucine, 56 tryptophan, 57 cysteine. See Figure 10-66A for how this answer was derived.

B. Either of the two nucleotide sequences shown in Figure A10-66B is correct. The peptide encoded is shown above the top sequence, for reference. The bottom nucleotide sequence is the reverse complement of the top sequence.

Figure 10-66A

DIF: Difficult REF: 10.1 OBJ: 10.1.a Recall the role played by restriction enzymes in bacteria and describe how these enzymes are employed in the laboratory. MSC: Applying

CHAPTER 11 Membrane Structure

THE LIPID BILAYER 11.1.a Estimate the thickness of the plasma membrane and recall its main constituents and functions.

11.1.b Assess the solubility of the lipids that constitute cell membranes.

11.1.c Articulate how the amphipathic nature of membrane lipids drives the assembly of a lipid bilayer, as well as the formation of closed vesicles, organelles, and even whole cells.

11.1.d Express why fats such as triacylglycerol coalesce into fat droplets in water.

11.1.e Recall the structures of phospholipids, glycolipids, triacylglycerol, and cholesterol.

11.1.f

Review how lipids move within the lipid bilayer.

11.1.g Summarize how the length and saturation of lipids’ hydrophobic tails affect the fluidity of a cell membrane.

11.1.h State how the inclusion of cholesterol affects the permeability of animal cell membranes.

11.1.i

Propose how single-celled organisms such as yeast and bacteria can maintain the fluidity of their membranes as

temperatures rise and fall.

11.1.j

Review the reasons why cells carefully regulate their membrane fluidity.

11.1.k State where new phospholipids are produced and explain how membranes are able to grow evenly.

11.1.l

Outline how the asymmetric distribution of phospholipids, characteristic of different membrane types, is estab-

lished and maintained.

11.1.m Describe how membranes retain their orientation during transfer between cell compartments.

11.1.n Contrast the actions of flippases and scramblases.

11.1.o Identify the lipids that show the most consistent or dramatic differences in distribution and indicate which monolayer they favor and why.

MEMBRANE PROTEINS 11.2.a Compare the relative mass and number of protein and lipid molecules in a typical animal cell membrane.

11.2.b List some functions of plasma membrane proteins.

11.2.c Categorize membrane proteins based on the way they interact with the lipid bilayer.

11.2.d Distinguish between integral membrane proteins and peripheral membrane proteins.

11.2.e Explain how the polypeptide chain of a transmembrane protein, with its hydrophilic backbone, is able to span the hydrophobic interior of the lipid bilayer.

11.2.f

Contrast the structure and potential function of single-pass and multipass transmembrane proteins.

11.2.g Describe the structure of a membrane-spanning β-barrel transmembrane protein.

11.2.h Recall how detergent molecules can extract proteins from a cell membrane.

11.2.i

Relate the structure and function of the membrane protein bacteriorhodopsin.

11.2.j

Contrast how the plasma membrane is mechanically stabilized in animal cells versus plant cells.

11.2.k Summarize how the cortex of red blood cells differs from that of other animal cell types.

11.2.l

Explain how hybrid cells allowed investigators to monitor the movement of membrane proteins within the plane of

the lipid bilayer.

11.2.m Describe how fluorescence recovery after photobleaching (FRAP) allows investigators to assess membrane fluidity.

11.2.n Outline the ways that cells restrict the lateral movement of their membrane proteins.

11.2.o Relate how single-particle tracking (SPT) microscopy allows investigators to track the movement of individual membrane proteins.

11.2.p Review the structure and function of the glycocalyx.

MULTIPLE CHOICE 1. Which of the following is most likely to occur after the lipid bilayer is pierced? a. The membrane reseals. b. The membrane collapses. c. A tear is formed. d. The membrane expands.

ANS: A DIF: Easy REF: 11.1 OBJ: 11.1.c Articulate how the amphipathic nature of membrane lipids drives the assembly of a lipid bilayer, as well as the formation of closed vesicles, organelles, and even whole cells.

MSC: Understanding 2. Which of the following functions of the plasma membrane is possible without membrane proteins? a. intercellular communication b. selective permeability c. cellular movement d. import/export of molecules ANS: B DIF: Easy REF: 11.1 OBJ: 11.1.a Estimate the thickness of the plasma membrane and recall its main constituents and functions. MSC: Understanding 3. Which type of lipids are the most abundant in the plasma membrane? a. phospholipids b. glycolipids c. sterols d. triacylglycerides ANS: A DIF: Easy REF: 11.1 OBJ: 11.1.a Estimate the thickness of the plasma membrane and recall its main constituents and functions. MSC: Remembering

4. Which of the following membrane lipids does not contain a fatty acid tail? a. phosphatidylcholine b. a glycolipid c. phosphatidylserine d. cholesterol ANS: D DIF: Easy REF: 11.1 OBJ: 11.1.e Recall the structures of phospholipids, glycolipids, triacylglycerol, and cholesterol. MSC: Remembering 5. Formation of a lipid bilayer is energetically favorable. How does this arrangement result in higher entropy for the system, and thus make bilayer formation energetically favorable? a. Polar head groups form a hydrogen-bonding network at the interface with water. b. Water molecules form cage-like structures around hydrophobic molecules. c. Hydrogen bonds form between neighboring polar head groups in the bilayer. d. Fatty acid tails are highly saturated and flexible.

ANS: B DIF: Easy REF: 11.1 OBJ: 11.1.c Articulate how the amphipathic nature of membrane lipids drives the assembly of a lipid bilayer, as well as the formation of closed vesicles, organelles, and even whole cells.

MSC: Understanding 6. Which of the following statements is TRUE? a. Phospholipids will spontaneously form liposomes in nonpolar solvents. b. In eukaryotes, all membrane-enclosed organelles are surrounded by one lipid bilayer. c. Membrane lipids diffuse within the plane of the membrane. d. Membrane lipids frequently flip-flop between one monolayer and the other.

ANS: C The remaining answers are false. Phospholipids form bilayers only in polar solvents. Nuclei and mitochondria are enclosed by two membranes. Membrane lipids rarely flip-flop between one monolayer and the other.

DIF: Easy REF: 11.1 OBJ: 11.1.f Review how lipids move within the lipid bilayer. MSC: Understanding 7. A bacterium is suddenly expelled from a warm human intestine into the cold world outside. Which of the following adjustments might the bacterium make to maintain the same level of membrane fluidity? a. produce lipids with hydrocarbon tails that are longer and have fewer double bonds b. produce lipids with hydrocarbon tails that are shorter and have more double bonds c. decrease the amount of cholesterol in the membrane d. decrease the amount of glycolipids in the membrane ANS: B DIF: Moderate REF: 11.1 OBJ: 11.1.i Propose how single-celled organisms such as yeast and bacteria can maintain the fluidity of their membranes as temperatures rise and fall. | 11.1.j Review the reasons why cells carefully regulate their membrane fluidity. MSC: Applying 8. Some lipases are able to cleave the covalent bonds between the glycerol backbone and the attached fatty acid. What final products do you expect to accumulate through the action of the enzyme monoacylglycerol lipase? a. phosphoglycerol and free fatty acid b. sterol and glycerol c. free phosphate and glycerol

d. glycerol and free fatty acid ANS: D DIF: Easy REF: 11.1 OBJ: 11.1.e Recall the structures of phospholipids, glycolipids, triacylglycerol, and cholesterol. MSC: Understanding 9. Which of the following phospholipid precursors is the most hydrophobic? a. triacylglycerol b. diacylglycerol c. phosphate d. glycerol ANS: A DIF: Easy REF: 11.1 OBJ: 11.1.e Recall the structures of phospholipids, glycolipids, triacylglycerol, and cholesterol. MSC: Remembering 10. Three phospholipids, X, Y, and Z, are distributed in the plasma membrane as indicated in Figure 11-10. For which of these phospholipids does a flippase probably exist?

Figure 11-10

a. X only b. Z only c. X and Y d. Y and Z ANS: C DIF: Moderate REF: 11.1 OBJ: 11.1.n Contrast the actions of flippases and scramblases.

MSC: Applying 11. Where does most new membrane synthesis take place in a eukaryotic cell? a. in the Golgi apparatus b. in the endoplasmic reticulum c. in the plasma membrane d. on ribosomes ANS: B DIF: Easy REF: 11.1 OBJ: 11.1.k State where new phospholipids are produced and explain how membranes are able to grow evenly. MSC: Remembering 12. Water molecules readily form hydrogen bonds with other polar molecules, and when they encounter nonpolar molecules they must form hydrogen-bonding networks with neighboring water molecules. Which of the following molecules will cause a “cage” of water to form? a. 2-methylpropane b. acetone

c. methanol d. urea ANS: A DIF: Easy REF: 11.1 OBJ: 11.1.b Assess the solubility of the lipids that constitute cell membranes.

MSC: Remembering 13. Membranes undergo spontaneous rearrangement if torn. Which of the following would happen if a cell membrane underwent a large tear?

Figure 11-13

ANS: C DIF: Moderate REF: 11.1 OBJ: 11.1.c Articulate how the amphipathic nature of membrane lipids drives the assembly of a lipid bilayer, as well as the formation of closed vesicles, organelles, and even whole cells.

MSC: Understanding 14. Membrane lipids are capable of many different types of movement. Which of these does not occur spontaneously in biological membranes? a. moving between lipid layers b. lateral movement c. rotation around the long axis of a fatty acid d. flexing of hydrocarbon chains ANS: A DIF: Easy REF: 11,1 OBJ: 11.1.f Review how lipids move within the lipid bilayer.

MSC: Remembering 15. Which of the following would yield the most highly mobile phospholipid (listed as number of carbons and number of double bonds, respectively)? a. 24 carbons with one double bond b. 15 carbons with two double bonds c. 20 carbons with two double bonds d. 16 carbons with no double bonds ANS: B DIF: Moderate REF: 11.1 OBJ: 11.1.f Review how lipids move within the lipid bilayer. MSC: Applying 16. Cholesterol serves several essential functions in mammalian cells. Which of the following is NOT influenced by cholesterol?

a. membrane permeability b. membrane fluidity c. membrane rigidity d. membrane thickness ANS: D DIF: Easy REF: 11.1 OBJ: 11.1.h State how the inclusion of cholesterol affects the permeability of animal cell membranes. MSC: Remembering 17. Most animal fats form a solid at room temperature, while plant fats remain liquid at room temperature. Which of the following is a feature of lipids in plant membranes that best explains this difference? a. unsaturated hydrocarbons b. longer hydrocarbon tails c. higher levels of sterols d. larger head groups ANS: A DIF: Easy REF: 11.1 OBJ: 11.1.g Summarize how the length and saturation of lipids’ hydrophobic tails affect the fluidity of a cell membrane. MSC: Understanding 18. New membrane phospholipids are synthesized by enzymes bound to the __________ side of the __________ membrane. a. cytosolic; mitochondrial b. luminal; Golgi c. cytosolic; endoplasmic reticulum d. extracellular; plasma ANS: C DIF: Easy REF: 11.1 OBJ: 11.1.k State where new phospholipids are produced and explain how membranes are able to grow evenly. MSC: Remembering 19. Membrane synthesis in the cell requires the regulation of growth for both halves of the bilayer and the selective retention of certain types of lipids on one side or the other. Which group of enzymes accomplishes both of these tasks? a. flippases b. phospholipases c. convertases d. glycosylases ANS: A DIF: Easy REF: 11.1 OBJ: 11.1.n Contrast the actions of flippases and scramblases. MSC: Remembering 20. Membrane curvature is influenced by the differential lipid composition of the two membrane monolayers. Which factor do you think has the largest impact on the curvature of biological membranes? a. amount of cholesterol b. charge of the lipid head group c. length of the hydrocarbon tails d. size of the lipid head group ANS: D DIF: Easy REF: 11.1 OBJ: 11.1.l Outline how the asymmetric distribution of phospholipids, characteristic of different membrane types, is established and maintained. MSC: Understanding

21. Membrane proteins, like membrane lipids, can move laterally by exchanging positions with other membrane components. Which type of membrane protein is expected to be the least mobile, based on their function? a. channels b. anchors c. receptors d. enzymes ANS: B DIF: Easy REF: 11.2 OBJ: 11.2.c Categorize membrane proteins based on the way they interact with the lipid bilayer. MSC: Remembering 22. A group of membrane proteins can be extracted from membranes only by using detergents. All the proteins in this group have a similar amino acid sequence at their C-terminus: -KKKKKXXC (where K stands for lysine, X stands for any amino acid, and C stands for cysteine). This sequence is essential for their attachment to the membrane. What is the most likely way in which the C-terminal sequence attaches these proteins to the membrane? a. The cysteine residue is covalently attached to a membrane lipid. b. The peptide spans the membrane as an α helix. c. The peptide spans the membrane as part of a β sheet. d. The positively charged lysine residues interact with an acidic integral membrane protein.

ANS: A DIF: Moderate REF: 11.2 OBJ: 11.2.h Recall how detergent molecules can extract proteins from a cell membrane. MSC: Understanding 23. Porin proteins form large, barrel-like channels in the membrane. Which of the following statements about these channels is FALSE? a. They are made primarily of α helices. b. They are made primarily of β sheets. c. They cannot form narrow channels. d. They have alternating hydrophobic and hydrophilic amino acids.

ANS: A DIF: Easy REF: 11.2 OBJ: 11.2.g Describe the structure of a membrane-spanning β-barrel transmembrane protein. MSC: Remembering 24. The amino acid sequences below represent the sequences of transmembrane helices. The characteristics of α helices that form a channel are different from those that form a single transmembrane domain. Select the helix that forms a single transmembrane domain. a. VGHSLSIFTLVISLGIFVFF b. IMIVLVMLLNIGLAILFVHF c. ILHFFHQYMMACNYFWMLCE d. VTLHKNMFLTYILNSMIIII ANS: B The amino acid sequence is entirely made up of hydrophobic side chains, indicating that it forms a helix in which all the side chains interact with the lipid bilayer.

DIF: Hard REF: 11.2 OBJ: 11.2.f Contrast the structure and potential function of single -pass and multipass transmembrane proteins. MSC: Applying 25. Which of the following substances is most commonly used to help purify a membrane protein?

a. high salt solution b. sucrose c. detergent d. ethanol ANS: C DIF: Easy REF: 11.2 OBJ: 11.2.h Recall how detergent molecules can extract proteins from a cell membrane.

MSC: Remembering 26. We know the detailed molecular structure and mechanism of action of the transmembrane protein bacteriorhodopsin.

This protein uses sunlight as the source of energy to pump __________ out of the cell. a. ATP b. H+ c. K+ d. Na+ ANS: B DIF: Easy REF: 11.2 OBJ: 11.2.i Relate the structure and function of the membrane protein bacteriorhodopsin. MSC: Remembering 27. In the photosynthetic archaean Halobacterium halobium, a membrane transport protein called bacteriorhodopsin captures energy from sunlight and uses it to pump protons out of the cell. The resulting proton gradient serves as an energy store that can later be tapped to generate ATP. Which statement best describes how bacteriorhodopsin operates? a. The absorption of sunlight triggers a contraction of the β barrel that acts as the protein’s central channel, squeezing a proton out of the cell. b. The absorption of sunlight triggers a shift in the conformation of the protein’s seven, membrane-spanning α helices, allowing a proton to leave the cell. c. The absorption of sunlight triggers a restructuring of bacteriorhodopsin’s otherwise unstructured core to form the channel through which a proton can exit the cell. d. The absorption of sunlight triggers the activation of an enzyme that generates ATP. ANS: B DIF: Easy REF: 11.2 OBJ: 11.2.i Relate the structure and function of the membrane protein bacteriorhodopsin. MSC: Understanding 28. Plasma membranes are extremely thin and fragile, requiring an extensive support network of fibrous proteins. This network is called the a. cortex. b. attachment complex. c. cytoskeleton. d. spectrin.

ANS: A DIF: Easy REF: 11.2 OBJ: 11.2.j Contrast how the plasma membrane is mechanically stabilized in animal cells versus plant cells. MSC: Remembering 29. Red blood cells have been very useful in the study of membranes and the protein components that provide structural support. Which of the following proteins is the principal fibrous protein in the cortex of the red blood cell? a. tubulin b. attachment proteins

c. actin d. spectrin ANS: D DIF: Easy REF: 11.2 OBJ: 11.2.k Summarize how the cortex of red blood cells differs from that of other animal cell types. MSC: Remembering 30. Which mechanism best describes the process by which a budding yeast cell designates the site of new bud formation during cell division? a. Proteins are tethered to the cell cortex. b. Proteins are tethered to the cell wall. c. Proteins are tethered to the proteins on the surface of another cell. d. Protein movement is limited by the presence of a diffusion barrier.

ANS: A DIF: Easy REF: 11.2 OBJ: 11.2.b List some functions of plasma membrane proteins. MSC: Remembering 31. Which mechanism best describes the process by which focal adhesions are formed to promote cell motility? a. Proteins are tethered to the cell cortex. b. Proteins are tethered to the extracellular matrix. c. Proteins are tethered to the proteins on the surface of another cell. d. Protein movement is limited by the presence of a diffusion barrier.

ANS: B DIF: Easy REF: 11.2 OBJ: 11.2.b List some functions of plasma membrane proteins. MSC: Remembering 32. Which mechanism best describes the process by which neutrophils are recruited by endothelial cells? a. Proteins are tethered to the cell cortex. b. Proteins are tethered to the extracellular matrix. c. Proteins interact with the proteins on the surface of another cell. d. Protein movement is limited by the presence of a diffusion barrier.

ANS: C DIF: Easy REF: 11.2 OBJ: 11.2.b List some functions of plasma membrane proteins. MSC: Remembering 33. Which mechanism best describes the process by which nutrients are taken up selectively at the apical surface of the epithelial cells that line the gut and released from their basal and lateral surfaces? a. Proteins are tethered to the cell cortex. b. Proteins are tethered to the extracellular matrix. c. Proteins are tethered to the proteins on the surface of another cell. d. Protein movement is limited by the presence of a diffusion barrier.

ANS: D DIF: Easy REF: 11.2 OBJ: 11.2.b List some functions of plasma membrane proteins. MSC: Remembering 34. Which mechanism best describes the process by which an antigen-presenting cell triggers an adaptive immune response? a. Proteins are tethered to the cell cortex. b. Proteins are tethered to the extracellular matrix. c. Proteins interact with the proteins on the surface of another cell. d. Protein movement is limited by the presence of a diffusion barrier.

ANS: C DIF: Easy REF: 11.2 OBJ: 11.2.b List some functions of plasma membrane proteins. MSC: Remembering

35. Consider the apical location of a particular protein expressed in epithelial cells, illustrated in part A of Figure 11-35.

Which type of defect described below is the most likely to cause the redistribution of that protein around the entire cell, shown in part B of Figure 11-35

Figure 11-35

a. a nonfunctional protein glycosylase b. the deletion of a junctional protein c. the truncation of a protein found in the extracellular matrix d. a nonfunctional flippase ANS: B DIF: Moderate REF: 11.2 OBJ: 11.2.b List some functions of plasma membrane proteins. MSC: Applying 36. Consider the apical location of a particular protein expressed in epithelial cells, illustrated in part A of Figure 11-36.

When a molecule that chelates calcium is added to the cell culture medium, you observe a redistribution of that protein around the entire cell, shown in part B of Figure 11-36. Which is most likely to be true about the role of calcium in maintaining an apical distribution of protein A?

Figure 11-36

a. Calcium is required to maintain the structural integrity of the junctional complex. b. Calcium is required for the binding of the junctional proteins to the cell cortex. c. Calcium is a structural component of protein A. d. Calcium inhibits the intracellular transport of protein A. ANS: A DIF: Moderate REF: 11.2 OBJ: 11.2.b List some functions of plasma membrane proteins. MSC: Applying 37. Diversity among the oligosaccharide chains found in the carbohydrate coating of the cell surface can be achieved in

which of the following ways? a. varying the types of sugar monomers used b. varying the types of linkages between sugars c. varying the number of branches in the chain d. all of these answers are correct ANS: D DIF: Easy REF: 11.2 OBJ: 11.2.b List some functions of plasma membrane proteins. MSC: Understanding 38. Which of the following statements about the carbohydrate coating of the cell surface is FALSE? a. It is not usually found on the cytosolic side of the membrane. b. It can play a role in cell–cell adhesion. c. The arrangement of the oligosaccharide side chains is highly ordered, much like the peptide bonds of a polypeptide chain. d. Specific oligosaccharides can be involved in cell–cell recognition.

ANS: C The sugars in an oligosaccharide side chain attached to the cell surface can be joined together in many different ways and in varied sequences.

DIF: Easy REF: 11.2 OBJ: 11.2.b List some functions of plasma membrane proteins. MSC: Understanding 39. Both glycoproteins and proteoglycans contribute to the carbohydrate layer on the surface of the cell. Which of the following statements about glycoproteins is FALSE? a. They can be secreted into the extracellular environment. b. They have only one transmembrane domain. c. They have long carbohydrate chains. d. They are recognized by lectins.

ANS: C DIF: Easy REF: 11.2 OBJ: 11.2.b List some functions of plasma membrane proteins. MSC: Remembering 40. The endothelial cells found closest to the site of an infection express proteins called lectins. Each lectin binds to a particular __________ that is presented on the surface of a target cell. a. oligosaccharide b. aminophospholipid c. polysaccharide d. sphingolipid ANS: A DIF: Easy REF: 11.2 OBJ: 11.2.b List some functions of plasma membrane proteins. MSC: Remembering 41. We can estimate the relative mobility of a population of molecules along the surface of a living cell by fluorescently labeling the molecules of interest, bleaching the label in one small area, and then measuring the speed of signal recovery as molecules migrate back into the bleached area. What is this method called? What does the abbreviation stand for? a. SDS b. SPT c. GFP d. FRAP

ANS: D, fluorescence recovery after photobleaching DIF: Easy REF: 11.2 OBJ: 11.2.m Describe how fluorescence recovery after photobleaching (FRAP) allows investigators to assess membrane fluidity. MSC: Remembering 42. It is possible to follow the movement of a single molecule or a small group of molecules. This requires the use of antibodies linked to small particles of gold, which appear as dark spots when tracked through video microscopy. What is this method called? What does the abbreviation stand for? a. SDS b. SPT c. GFP d. FRAP ANS: B, single-particle tracking DIF: Easy REF: 11.2 OBJ: 11.2.o Relate how single-particle tracking (SPT) microscopy allows investigators to track the movement of individual membrane proteins. MSC: Remembering

SHORT ANSWER 1. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Although cholesterol is a hydrophobic molecule, it has a hydrophilic head group like all other membrane lipids.

B. Phosphatidylserine is the most abundant type of phospholipid found in cell membranes.

C. Glycolipids lack the glycerol component found in phospholipids.

D. The highly ordered structure of the lipid bilayer makes its generation and maintenance energetically unfavorable.

ANS: A. True.

B. False. Phosphatidylcholine is the most abundant phospholipid found in cell membranes.

C. True.

D. False. The formation of a lipid bilayer is energetically favorable.

DIF: Easy REF: 11.1 OBJ: 11.1.e Recall the structures of phospholipids, glycolipids, triacylglycerol, and cholesterol.

MSC: Evaluating 2. Cholesterol can either make membranes more rigid or more fluid. Explain how this is possible.

ANS: There are two properties of phospholipids that affect how tightly they pack together: the length of the hydrocarbon chain and the number of double bonds. The degree of packing, in turn, influences the relative mobility of these molecules in the membrane. The effect of cholesterol is dependent upon both temperature and concentration. The presence of cholesterol at normal physiological concentrations helps maintain membrane fluidity at lower temperatures by solvating the long hydrocarbon tails of phospholipids. At higher temperatures, cholesterol stabilizes the membrane (observed as an increase in the melting temperature).

DIF: Moderate REF: 11.1 OBJ: 11.1.h State how the inclusion of cholesterol affects the permeability of animal cell membranes. MSC: Understanding 3. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. amphipathic

hydrophobic

phosphatidylserine

cholesterol

lipid bilayer

phospholipids

fatty acid tails

lipid monolayer

proteins

glycolipids

lipids

sterols

hydrophilic head groups

phosphatidylcholine

sugars

The specialized functions of different membranes are largely determined by the __________ they contain. Membrane lipids are __________ molecules, composed of a hydrophilic portion and a hydrophobic portion. All cell membranes have the same __________ structure, with the __________ of the phospholipids facing into the interior of the membrane and the __________ on the outside. The most common lipids in most cell membranes are the __________. The head group of a glycolipid is composed of __________.

ANS: The specialized functions of different membranes are largely determined by the proteins they contain. Membrane lipids are amphipathic molecules, composed of a hydrophilic portion and a hydrophobic portion. All cell membranes have the same lipid bilayer structure, with the fatty acid tails of the phospholipids facing into the interior of the membrane and the hydrophilic head groups on the outside. The most common lipids in most cell membranes are the phospholipids. The head group of a glycolipid is composed of sugars.

DIF: Easy REF: 11.1 OBJ: 11.1.a Estimate the thickness of the plasma membrane and recall its main constituents and functions. | 11.1.b Assess the solubility of the lipids that constitute cell membranes. MSC: Understanding 4. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Lipid-linked proteins are classified as peripheral membrane proteins because the polypeptide chain does not pass through the bilayer.

B. A protein can be embedded on the cytosolic side of the membrane bilayer by employing a hydrophobic α helix.

C. A protein that relies on protein–protein interactions to stabilize its membrane association is classified as a peripheral membrane protein because it can be dissociated without the use of detergents.

D. Membrane proteins that pump ions in and out of the cell are classified as enzymes.

ANS: A. False. Lipid-linked proteins are classified as integral membrane proteins because although they are not transmembrane proteins, they are covalently bound to membrane lipids and cannot be dissociated without disrupting the membrane’s integrity.

B. False. An embedded protein employs an amphipathic helix. The hydrophobic side interacts with the fatty acid tails of the membrane lipids, and the hydrophilic portion interacts with the aqueous components of the cytosol.

C. True.

D. False. Membrane proteins that pump ions in either direction across the membrane are in the functional class of transporters.

DIF: Easy REF: 11.2 OBJ: 11.2.c Categorize membrane proteins based on the way they interact with the lipid bilayer. | 11.2.d Distinguish between integral membrane proteins and peripheral membrane proteins.

MSC: Evaluating 5. Explain the reasons why membrane proteins are more difficult to isolate and purify for in vitro studies compared to cytosolic proteins.

ANS: Cytosolic proteins are soluble in aqueous solutions. They can be isolated fairly easily from the soluble fraction of a cell lysate. Membrane proteins contain both hydrophilic and hydrophobic segments and need to be isolated in association with micelles (vesicles that form when the bilayer is disrupted) to retain their active conformation and avoid

aggregation.

DIF: Hard REF: 11.2 OBJ: 11.2.h Recall how detergent molecules can extract proteins from a cell membrane.

MSC: Understanding 6. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. amphipathic

hydrophilic

noncovalently

cortical

hydrophobic

peripheral

covalently

integral

transmembrane

detergent

micelle

unfolded

There are several ways that membrane proteins can associate with the cell membrane. Membrane proteins that extend through the lipid bilayer are called __________ proteins and have __________ regions that are exposed to the interior of the bilayer. On the other hand, membrane-associated proteins do not span the bilayer and instead associate with the membrane through an α helix that is __________. Other proteins are __________ attached to lipid molecules that are inserted in the membrane. __________ membrane proteins are linked to the membrane through noncovalent interactions with other membrane-bound proteins.

ANS: There are several ways that membrane proteins can associate with the cell membrane. Membrane proteins that extend through the lipid bilayer are called transmembrane proteins and have hydrophobic regions that are exposed to the interior of the bilayer. On the other hand, membrane-associated proteins do not span the bilayer and instead associate with the membrane through an α helix that is amphipathic. Other proteins are covalently attached to lipid molecules that are inserted in the membrane. Peripheral membrane proteins are linked to the membrane through noncovalent interactions with other membrane-bound proteins.

DIF: Easy REF: 11.2 OBJ: 11.2.c Categorize membrane proteins based on the way they interact with the lipid bilayer. | 11.2.d Distinguish between integral membrane proteins and peripheral membrane proteins.

MSC: Understanding 7. Anemia, a condition that results in individuals with a low red blood cell count, can be caused by a number of factors.

Why do individuals with defects in the spectrin protein often have this condition?

ANS: Spectrin is the primary protein in the cortex of red blood cells. A defect in the spectrin protein directly affects the strength and shape of the cortex. Red blood cells that contain mutated spectrin molecules have an irregular shape and are prone to lysis as a result of cortical fragility, leading to a smaller population of red blood cells.

DIF: Easy REF: 11.2 OBJ: 11.2.k Summarize how the cortex of red blood cells differs from that of other animal cell types. MSC: Understanding 8. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. When a mouse cell is fused with a human cell, the movement of the respective membrane proteins is restricted to their original locations at the time of fusion.

B. Epithelial cell membranes are asymmetric, and proteins from the apical side of the cell membrane cannot diffuse into the basal side of the membrane.

C. The longest carbohydrates found on the surfaces of cells are linked to lipid molecules.

D. The only role of the carbohydrate layer on the cell surface is to absorb water, which creates a slimy surface and prevents cells from sticking to each other.

ANS: A. False. After about 1 hour, the mouse and human proteins present on the surface of the fused cell are found evenly dispersed throughout the plasma membrane.

B. True.

C. False. The very long, branched polysaccharides that are attached to integral membrane proteins are much longer than the oligosaccharides covalently attached to membrane lipids.

D. False. Although the absorption of water is an important role of the carbohydrates on the surface of the plasma membrane, a second critical role is that of cell–cell recognition, which is important in immune responses, wound healing, and other processes that rely on cell-type-specific interactions.

DIF: Easy REF: 11.2 OBJ: 11.2.b List some functions of plasma membrane proteins. MSC: Evaluating 9. Cell membranes are fluid, and thus proteins can diffuse laterally within the lipid bilayer. However, sometimes the cell needs to localize proteins to a particular membrane domain. Name three mechanisms that a cell can use to restrict a protein to a particular place in the cell membrane.

ANS: Any combination of the following four answers is acceptable.

1. The protein can be attached to the cell cortex inside the cell.

2. The protein can be attached to the extracellular matrix outside the cell.

3. The protein can be attached to other proteins on the surface of a different cell.

4. The protein can be restricted by a diffusion barrier, such as that set up by specialized junctional proteins at a tight junction.

DIF: Easy REF: 11.2 OBJ: 11.2.n Outline the ways that cells restrict the lateral movement of their membrane proteins.

MSC: Remembering 10. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. In order to study the activity of isolated transmembrane proteins, the membrane lipids must be completely stripped away.

B. FRAP is a method used to study the movement of individual proteins.

C. SDS is a mild detergent that is useful for the reconstitution of membrane components.

D. The speed of fluorescent signal recovery during a FRAP assay is a measure of lateral mobility for the molecule of interest.

ANS: A. False. The region of the protein that normally crosses the membrane must be stabilized by the presence of phospholipids for the purified protein to be active. For this reason, purified membrane proteins are often reconstituted into artificial lipid bilayers.

B. False. The FRAP method involves photobleaching of a small region of the membrane, which contains hundreds of target molecules, and follows the displacement of these molecules with neighboring molecules that have not been bleached.

C. False. SDS is a strong, ionic detergent that will break up membrane bilayers and also denature proteins.

D. True.

DIF: Easy REF: 11.2 OBJ: 11.2.m Describe how fluorescence recovery after photobleaching (FRAP) allows investigators to assess membrane fluidity. MSC: Evaluating

11. While many prokaryotic cells have a single membrane bilayer, all eukaryotic cells have a complex system of internal membrane-bound compartments. How might it be advantageous for the cell to have these additional compartments?

ANS: Compartmentalization using intracellular membranes allows eukaryotic cells to separate a variety of cell processes. Although this requires a higher degree of coordination, the cell also gains a more stringent degree of control over these processes (examples include: the separation of transcription and translation; the separation of enzymes involved in protein modifications for secreted versus cytosolic substrates; the separation of proteolytic events in the lysosomes versus the cytosol; the separation of anaerobic metabolism in the cytosol and aerobic metabolism in the mitochondria).

DIF: Easy REF: 11.1 OBJ: 11.1.a Estimate the thickness of the plasma membrane and recall its main constituents and functions. MSC: Understanding 12. Thermal motion promotes lateral position exchanges between lipid molecules within a monolayer. In an artificial bilayer, this movement has been estimated to be ~2 μm/second. This represents the entire length of a bacterial cell.

Do you expect the lateral movement of a lipid molecule within a biological membrane to be equally fast? Explain your answer.

ANS: No. Although the rate of movement may be similar, it will most likely be slower in a biological membrane. An artificial bilayer is primarily phospholipids. Biological membranes contain a large number of protein components and specialized membrane domains that could limit the rate of lateral diffusion.

DIF: Moderate REF: 11,2 OBJ: 11.2.n Outline the ways that cells restrict the lateral movement of their membrane proteins. MSC: Applying 13. Even though proteins can form channels across biological membranes using either α helices or β sheets, channels made of α helices are more versatile. Explain the physical constraints on β-barrel structures and why these constraints do not apply to channels made of α helices.

ANS: β-Barrel structures are composed of individual β strands that form a β sheet that needs to be curved to make the structure of a pore in the membrane. The physical constraints are due to very specific positioning of each strand to maintain the necessary hydrogen-bonding network within the sheet. The relative positions of α helices can vary and still form strong interactions with other helices in the transmembrane region of a protein.

DIF: Moderate REF: 11.2 OBJ: 11.2.g Describe the structure of a membrane-spanning β-barrel transmembrane protein. MSC: Understanding 14. Sodium dodecyl sulfate (SDS) and Triton X-100 are both detergents that can be used to lyse cells to study individual components. Your first project as a Ph.D. student is to create an in vitro model that allows you to study the mechanism of action of a novel glucose transporter protein. You use SDS to lyse the cells. Later experiments demonstrate that the transporter is not functional. Provide a possible explanation for what happened, propose a solution for your next experiment, and articulate the reasons you think the new method will be more successful.

ANS: SDS is a strong ionic detergent. When cells are exposed to SDS, membrane proteins are not only extracted from the membrane, they are completely unfolded. After denaturation, they cannot be studied as functional molecules.

Therefore, it is likely that your use of SDS for cell lysis denatured and inactivated the transporter protein. Using a more mild detergent like Triton X-100 would be a single change that might yield better results. Triton X-100 is amphipathic, having a smaller nonpolar portion and a polar but uncharged end, which allows it to mimic more closely the type of

solvation effect of the membrane lipids. Triton X-100 forms a shell around the hydrophobic portion of the protein without disrupting the existing structure. This makes it possible to then place the active transporter protein into a new, synthetic membrane bilayer for study in vitro.

DIF: Hard REF: 11.2 OBJ: 11.2.h Recall how detergent molecules can extract proteins from a cell membrane.

MSC: Applying 15. You have isolated two mutants of a normally pear-shaped microorganism that have lost their distinctive shape and are now round. One of the mutants has a defect in a protein that you call A and the other has a defect in a protein that you call B. First, you grind up each type of mutant cell and normal cells separately and separate the plasma membranes from the cytoplasm, forming the first cell extract. Then you set aside a portion of each fraction for later testing. Next, you wash the remaining portion of the membrane fractions with a low concentration of urea (which will unfold proteins and disrupt their ability to interact with other proteins) and centrifuge the mixture. The membranes and their constituent proteins form a pellet, and the proteins liberated from the membranes by the urea wash remain in the supernatant. When you check each of the fractions for the presence of A or B, you obta in the results given below.

Figure 11-15

Use these initial observations to classify membrane proteins A and B. What additional information can be inferred from this data set that helps you understand the mutant phenotype?

ANS: The results from the extracts of normal cells show that protein A is an integral membrane protein that remains in the membrane through all the treatments, whereas protein B is a peripheral membrane protein that can be removed from the membrane by urea. In the cell extracts from the mutants with a defect in A, the A protein still remains in the membrane, but the B protein does not. This is consistent with the mutation in A affecting its interaction with B. The same results are obtained when the B protein is mutant, which is consistent with the idea that A and B interact. The loss of an interaction between an integral membrane protein and a protein in the cortex would be more likely to result in a change in cell shape than the loss of an interaction between an integral membrane protein and a protein on the exterior of the cell.

DIF: Hard REF: 11.2 OBJ: 11.2.c Categorize membrane proteins based on the way they interact with the lipid bilayer. | 11.2.d Distinguish between integral membrane proteins and peripheral membrane proteins.

MSC: Applying 16. Three different membrane components are shown in Figure 11-16. Using the list below, identify the three components, and label the chemical groups indicated.

A. glycerol B. sugar C. phospholipid D. glycolipid E. sterol

F. unsaturated hydrocarbon G. saturated hydrocarbon H. sterol polar head group

Figure 11-16

ANS: See Figure 11-16A.

Figure 11-16A

DIF: Easy REF: 11.1 OBJ: 11.1.e Recall the structures of phospholipids, glycolipids, triacylglycerol, and cholesterol.

MSC: Remembering 17. Explain the importance of glycolipids in the plasma membrane, identify where they are produced inside the cell, and describe the mechanism by which they are transported to the plasma membrane and presented to the extracellular environment. Draw a diagram to support your answer.

ANS: Glycolipids are found on the surface of healthy cells and contribute to the cell’s defense against chemical damage and infectious agents. Glycolipids are produced by enzymes inside the Golgi apparatus. They are then transported to the plasma membrane through a process of vesicle budding. These secretory vesicles then fuse with the plasma membrane. The glycolipids that were facing the lumen of the Golgi apparatus will now face the extracellular environment (Figure 11-17).

Figure 11-17

DIF: Hard REF: 11.1 OBJ: 11.1.e Recall the structures of phospholipids, glycolipids, triacylglycerol, and cholesterol.

MSC: Creating

CHAPTER 12 Transport Across Cell Membranes

PRINCIPLES OF TRANSMEMBRANE TRANSPORT 12.1.a Recall why water-soluble molecules and ions have difficulty crossing a lipid bilayer.

12.1.b Distinguish between simple diffusion and facilitated transport.

12.1.c Review the properties that govern the rate at which a given solute can cross a protein-free lipid bilayer.

12.1.d State the function of membrane transport proteins and differentiate between transporters and channels.

12.1.e Indicate how the concentration of ions in the cell differs from that outside the cell.

12.1.f

Define “resting membrane potential.”

12.1.g Compare how transporters and channels discriminate among solutes, moving only a select subset across the membrane.

12.1.h Distinguish between active and passive transport, and indicate which type of membrane transport protein carries out each.

12.1.i

Review the forces that govern the passive transport of charged and uncharged solutes across a cell membrane.

12.1.j

Compare the electrochemical gradients for sodium and potassium ions, and recall how the membrane potential and concentration gradient for each ion combine to influence its movement across the plasma membrane.

12.1.k Describe the ways in which water can move across cell membranes, and articulate what governs whether water will enter or exit a cell.

12.1.l

Contrast how plant cells, animal cells, and protozoa maintain their osmotic equilibrium.

TRANSPORTERS AND THEIR FUNCTIONS 12.2.a Outline how the glucose transporter in the plasma membrane of mammalian cells imports glucose after a meal and exports glucose to provide fuel for other tissues in the body.

12.2.b List three sources of energy used by transmembrane pumps to actively transport a solute against its concentration gradient.

12.2.c Review how the sodium pump in animal cells uses the energy supplied by ATP hydrolysis to maintain the concentration gradients of sodium and potassium ions.

12.2.d Recall how the toxin ouabain affects the operation of the sodium pump.

12.2.e Explain why—and how—cells keep the cytosolic concentration of calcium ions low.

12.2.f

Differentiate between a symport, an antiport, and a uniport.

12.2.g Compare the uniport glucose transporter and the glucose-sodium symport in terms of their activities and roles in glucose transport in intestinal epithelial cells.

12.2.h Express how the sodium−proton exchanger in the plasma membrane allows animal cells to control the pH of their cytosol.

12.2.i

Review how proton pumps provide the energy for the transport of solutes in plants, fungi, and bacteria.

ION CHANNELS AND THE MEMBRANE POTENTIAL 12.3.a Explain how the structure of ion channels leads to their ion selectivity.

12.3.b Compare how conformational changes of ion channels and transporters with respect to the passage of individual solutes across the membrane.

12.3.c Recall how potassium leak channels participate in establishing the cell’s resting membrane potential.

12.3.d Estimate the resting membrane potential in animal cells.

12.3.e Review how the Nernst equation can be used to calculate the resting membrane potential across a membrane.

12.3.f

Outline how patch-clamp recording can be used to study the activity of ion channels.

12.3.g Review the mechanisms that can regulate ion flow through a channel.

12.3.h Define “gating” and list the types of conditions that can alter the opening and closing of ion channels.

12.3.i

Summarize the process by which ion channels in auditory hair cells allow the detection of sound.

12.3.j

List several cell types that use voltage-gated ion channels.

12.3.k Explain how voltage sensors allow voltage-gated ion channels to respond to changes in membrane potential.

ION CHANNELS AND NERVE CELL SIGNALING 12.4.a Review the anatomy of a nerve cell and discuss the direction in which electrical signals travel from one neuron to another.

12.4.b Contrast the passive spread of electrical excitation within a nerve cell membrane with the traveling wave of an action potential.

12.4.c Outline how membrane depolarization triggers an action potential and how the action potential spreads along the membrane.

12.4.d Recall the conformations that Na+-gated ion channels cycle through as an action potential travels along the axonal membrane.

12.4.e Compare the roles that Na+-gated and K+-gated ion channels play in an action potential.

12.4.f

Explain how investigators determined which ions are involved in an action potential.

12.4.g Outline how studies using the squid giant axon revealed the different roles of potassium and sodium ions in establishing the resting membrane potential and propagating an action potential.

12.4.h Review how an electrical signal is converted to a chemical signal at a nerve terminal.

12.4.i

Summarize how transmitter-gated ion channels convert the chemical signal carried by a neurotransmitter back

into an electrical signal.

12.4.j

Contrast how the receptors for excitatory and inhibitory neurotransmitters alter the activity of a postsynaptic cell.

12.4.k Describe the events that take place when acetylcholine binds to its receptor on a vertebrate muscle cell.

12.4.l

Compare the mechanisms of action of curare and strychnine.

12.4.m Recall how the drugs Prozac and Ambien exert their neurological effects.

12.4.n Propose how nerve cells can integrate and interpret a variety of incoming signals to generate an appropriate response.

12.4.o Explain why channelrhodopsin can be used to artificially manipulate the activity of neurons and neural circuits.

12.4.p Outline how optogenetics can be used to study the role that particular neurons play in the behavior of an an imal.

MULTIPLE CHOICE 1. Which of the following channels would not be expected to generate a change in voltage by movement of its substrate across the membrane where it is found? a. an aquaporin b. a sodium channel c. a calcium channel d. a proton channel ANS: A Aquaporin channels are found in the plasma membrane of some cells, where they facilitate the diffusion of water across the membrane. Because water is an uncharged molecule, its movement would not be expected to alter the voltage across the membrane.

DIF: Easy REF: 12.1 OBJ: 12.1.k Describe the ways in which water can move across cell membranes and articulate what governs whether water will enter or exit a cell. MSC: Understanding 2. Although the extracellular environment has a high sodium ion concentration and the intracellular environment has a high potassium ion concentration, both must be neutralized by negatively charged molecules. In the extracellular case, what is the principal anion? a. HCO3− b. Cl− c. PO43− d. OH− ANS: B DIF: Easy REF: 12.1 OBJ: 12.1.e Indicate how the concentration of ions in the cell differs from that outside the cell. MSC: Remembering 3. Below is a list of molecules with different chemical characteristics. Knowing that all molecules will eventually diffuse across a phospholipid bilayer, select the option below that most accurately predicts the relative rates of diffusion of these molecules (fastest to slowest). alanine estrogen propanol sodium a. alanine > propanol > sodium > estrogen

b. sodium > propanol > alanine > estrogen c. estrogen > propanol > sodium > alanine d. estrogen > propanol > alanine > sodium ANS: D Estrogen is a steroid hormone and will diffuse the fastest across the membrane. Propanol is a small, uncharged molecule with a polar group. Alanine is an amino acid, and although it has a small, nonpolar side group, amino acids are charged molecules. Sodium is an ion and will move the slowest across the bilayer.

DIF: Easy REF: 12.1 OBJ: 12.1.c Review the properties that govern the rate at which a given solute can cross a protein-free lipid bilayer. MSC: Understanding 4. Cells use membranes to help maintain set ranges of ion concentrations inside and outside the cell. Which of the following ions is the most abundant inside a typical mammalian cell? a. Na+ b. K+ c. Ca2+ d. Cl− ANS: B DIF: Easy REF: 12.1 OBJ: 12.1.e Indicate how the concentration of ions in the cell differs from that outside the cell. MSC: Remembering 5. Cells use membranes to help maintain set ranges of ion concentrations inside and outside the cell. Which of the following ions is the most abundant outside a typical mammalian cell? a. Na+ b. K+ c. Ca2+ d. Cl− ANS: A DIF: Easy REF: 12.1 OBJ: 12.1.e Indicate how the concentration of ions in the cell differs from that outside the cell. MSC: Remembering 6. Cells use membranes to help maintain set ranges of ion concentrations inside and outside the cell. Which of the following negatively charged ions is NOT primarily used to buffer positive charges inside the cell? a. PO43− b. OH− c. Cl− d. HCO3− ANS: C DIF: Easy REF: 12.1 OBJ: 12.1.e Indicate how the concentration of ions in the cell differs from that outside the cell. MSC: Remembering 7. Negatively charged ions are required to balance the net positive charge from metal ions such as K +, Na+, and Ca2+.

Which of the following negatively charged ions is the most abundant outside the cell and which ion most often neutralize (written in parentheses)? a. Cl− (Ca2+) b. PO43− (K+) c. PO43− (Ca2+)

d. Cl− (Na+) ANS: D DIF: Easy REF: 12.1 OBJ: 12.1.e Indicate how the concentration of ions in the cell differs from that outside the cell. MSC: Remembering 8. Which of the following statements about resting membrane potential is FALSE? a. The resting membrane potential for most animal cells is 0 mV, because the positive and negative ions are in balance. b. The resting membrane potential for most animal cells is positive, because Na + ions are so plentiful inside cells. c. The resting membrane potential for most animal cells is negative, because the inside of the cell is more negatively charged than the outside of the cell. d. At the resting membrane potential, no ions enter or exit the cell.

ANS: C DIF: Easy REF: 12.1 OBJ: 12.1.f Define “resting membrane potential.” MSC: Understanding 9. A hungry yeast cell lands in a vat of grape juice and begins to feast on the sugars there, producing carbon dioxide and ethanol in the process: C6H12O6 + 2ADP + 2Pi + H+ → 2CO2 + 2CH3CH2OH + 2ATP + 2H2O Unfortunately, the grape juice is contaminated with proteases that attack some of the transport proteins in the yeast cell membrane, and the yeast cell dies. Which of the following is the most likely cause of the yeast cell’s demise? a. toxic buildup of carbon dioxide inside the cell b. toxic buildup of ethanol inside the cell c. diffusion of ATP out of the cell d. inability to import sugar into the cell ANS: D DIF: Moderate REF: 12.1 OBJ: 12.1.g Compare how transporters and c hannels discriminate among solutes, moving only a select subset across the membrane. MSC: Applying 10. Ion channels are classified as membrane transport proteins. They discriminate between ions based on size and charge. In addition to Na+, which one of the following ions would you expect to be able to freely diffuse through a Na + channel? Explain your answer. a. Mg2+ b. H+ c. K+ d. Cl− ANS: B If an ion channel is open, it will allow any ion that is under a certain size and that has the correct charge to pass through. H+ is the only ion listed that is both smaller and has the same charge of +1.

DIF: Moderate REF: 12.1 OBJ: 12.1.g Compare how transporters and channels discriminate among solutes, moving only a select subset across the membrane. MSC: Applying 11. Some cells express aquaporin proteins—they are channel proteins that facilitate the flow of water molecules through the plasma membrane. What regulates the rate and direction of water diffusion across the membrane? a. aquaporin conformation b. resting membrane potential

c. solute concentrations on either side of the membrane d. availability of ATP ANS: C DIF: Easy REF: 12.1 OBJ: 12.1.k Describe the ways in which water can move across cell membranes and articulate what governs whether water will enter or exit a cell. MSC: Remembering 12. Transporters, in contrast to channels, work by a. specific recognition of transport substrates. b. a gating mechanism. c. filtering solutes by charge. d. filtering solutes by size.

ANS: A DIF: Easy REF: 12.1 OBJ: 12.1.g Compare how transporters and channels discriminate among solutes, moving only a select subset across the membrane. MSC: Remembering 13. Pumps are transporters that are able to harness energy provided by other components in the cells to drive the movement of solutes across membranes, against their concentration gradient. This type of transport is called a. active transport. b. free diffusion. c. facilitated diffusion. d. passive transport.

ANS: A DIF: Easy REF: 12.1 OBJ: 12.1.h Distinguish between active and passive transport and indicate which type of membrane transport protein carries out each. MSC: Remembering 14. Active transport requires the input of energy into a system so as to move solutes against their electrochemical and concentration gradients. Which of the following is NOT one of the common ways to perform active transport? a. Na+-coupled b. K+-coupled c. ATP-driven d. light-driven ANS: B Because K+ is a positively charged ion and the outside of the plasma membrane is positively charged, K + has a very small electrochemical gradient across the membrane even though its concentration gradient is large. Because there is little net movement across the membrane for K+, it would not make a good source of energy to drive the transport of other molecules against their respective gradients.

DIF: Moderate REF: 12.1 OBJ: 12.1.e Indicate how the concentration of ions in the cell differs from that outside the cell. | 12.1.h Distinguish between active and passive transport and indicate which type of membrane transport protein carries out each. MSC: Understanding 15. The Na+-K+ ATPase is also known as the Na+-K+ pump. It is responsible for maintaining the high extracellular sodium ion concentration and the high intracellular potassium ion concentration. What happens immediately after the pump hydrolyzes ATP? a. Na+ is bound b. ADP is bound c. The pump is phosphorylated.

d. The pump changes conformation.

ANS: C The phosphorylation of the pump causes the conformational change, and it occurs after the binding of Na +.

DIF: Easy REF: 12.2 OBJ: 12.2.c Review how the sodium pump in animal cells uses the energy supplied by ATP hydrolysis to maintain the concentration gradients of sodium and potassium ions. MSC: Remembering 16. You have generated antibodies that recognize the extracellular domain of the Ca 2+-pump. Adding these antibodies to animal cells blocks the active transport of Ca2+ from the cytosol into the extracellular environment. What do you expect to observe with respect to intracellular Ca2+? a. Ca2+-pumps in vesicle membranes keep cytosolic calcium levels low. b. Ca2+-pumps in the endoplasmic reticulum membrane keep cytosolic calcium levels low. c. Ca2+-pumps in the Golgi apparatus keep cytosolic calcium levels low. d. Ca2+ concentrations in the cytosol increase at a steady rate.

ANS: B In addition to the Ca2+-pumps in the plasma membrane, Ca2+-pumps are also found in the membrane of the endoplasmic reticulum (ER). Those in the ER membrane will continue to remove calcium ions from the cytosol, keeping calcium levels low.

DIF: Moderate REF: 12.2 OBJ: 12.2.e Explain why—and how—cells keep the cytosolic concentration of calcium ions low. MSC: Applying 17. Cells make use of H+ electrochemical gradients in many ways. Which of the following proton transporters is used to regulate pH in animal cells? a. light-driven pump b. H+ ATPase c. H+ symporter d. Na+-H+ exchanger ANS: D The high extracellular concentration of Na+ is employed by a transporter that pumps protons out of animal cells as Na + is brought into the cell. The other transporters are found only in bacterial cells.

DIF: Easy REF: 12.2 OBJ: 12.2.h Express how the sodium–proton exchanger in the plasma membrane allows animal cells to control the pH of their cytosol. MSC: Remembering 18. Which of the following statements is TRUE? a. Amoebae have transporter proteins that actively pump water molecules from the cytoplasm to the cell exterior. b. Bacteria and animal cells rely on the Na+-K+ pump in the plasma membrane to prevent lysis resulting from osmotic imbalances. c. The Na+-K+ pump allows animal cells to thrive under conditions of very low ionic strength. d. The Na+-K+ pump helps to keep both Na+ and Cl− ions out of the cell.

ANS: D The Na+-K+ pump keeps Na+ out directly by pumping it out and keeps Cl − out indirectly by helping to maintain the negative membrane potential.

DIF: Easy REF: 12.2 OBJ: 12.2.c Review how the sodium pump in animal cells uses the energy supplied by ATP hy-

drolysis to maintain the concentration gradients of sodium and potassium ions. MSC: Understanding 19. Ca2+-pumps in the plasma membrane and endoplasmic reticulum are important for a. maintaining osmotic balance. b. preventing Ca2+ from altering the activity of molecules in the cytosol. c. providing enzymes in the endoplasmic reticulum with Ca2+ ions that are necessary for their catalytic activity. d. maintaining a negative membrane potential.

ANS: B The major purpose of the Ca2+-pumps is to keep the cytosolic concentration of Ca2+ low. When Ca2+ does move into the cytosol, it alters the activity of many proteins; hence, Ca2+ is a powerful signaling molecule.

DIF: Easy REF: 12;2 OBJ: 12.2.e Explain why—and how—cells keep the cytosolic concentration of calcium ions low.

MSC: Remembering 20. Which of the following occur WITHOUT coupling transport of the solute to the movement of a second solute? a. import of glucose into gut epithelial cells b. export of Ca2+ from the cytosol c. export of H+ from animal cells for pH regulation d. the export of Na+ from cells to maintain resting membrane potential ANS: B Ca2+ is exported using ATP-powered pumps. There are no other solutes that are being moved by these pumps.

DIF: Easy REF: 12.2 OBJ: 12.2.a Outline how the glucose transporter in the plasma membrane of mammalian cells imports glucose after a meal and exports glucose to provide fuel for other tissues in the body. | 12.2.e Explain why— and how—cells keep the cytosolic concentration of calcium ions low. | 12.2.h Express how the sodium–proton exchanger in the plasma membrane allows animal cells to control the pH of their cytosol.

MSC: Understanding 21. Which of the following best describes the behavior of a gated channel? a. It stays open continuously when stimulated. b. It opens more frequently in response to a given stimulus. c. It opens more widely as the stimulus becomes stronger. d. It requires a stimulus to change from closed to open.

ANS: B DIF: Moderate REF: 12.3 OBJ: 12.3.h Define “gating” and list the types of conditions that can alter the opening and closing of ion channels. MSC: Understanding 22. The stimulation of auditory nerves depends on the opening and closing of channels in the auditory hair cells. Which type of gating mechanism do these cells use? a. voltage-gated b. extracellular ligand-gated c. intracellular ligand-gated d. stress-gated ANS: D Sound waves cause vibrations of the tectorial membrane. These vibrations cause the bundles of stereocilia to tilt. This tilting physically pulls the filament that links a cilium to the ion channel in neighboring cilia, which then pulls the gate

on that ion channel open.

DIF: Easy REF: 12.3 OBJ: 12.3.i Summarize the process by which ion channels in auditory hair cells allow the detection of sound. MSC: Remembering 23. Voltage-gated channels contain charged protein domains, which are sensitive to changes in membrane potential. By responding to a threshold in the membrane potential, these voltage sensors trigger the opening of the channels.

Which of the following best describes the behavior of a population of channels exposed to such a threshold? a. Some channels remain closed and some open completely. b. All channels open completely. c. All channels open partly, to the same degree. d. All channels open partly, each to a different degree.

ANS: A Individual channels are either completely open or completely closed. However, in a given population, there will be a mixture of open and closed channels.

DIF: Moderate REF: 12.3 OBJ: 12.3.k Explain how voltage sensors allow voltage-gated ion channels to respond to changes in membrane potential. MSC: Understanding 24. When the net charge on either side of the plasma membrane is zero, what else is true? a. There is an equal number of K+ ions on each side of the plasma membrane. b. The K+ leak channels are open. c. The electrochemical potential across the membrane is zero. d. The resting membrane potential is between −20 mV and −200 mV. ANS: C DIF: Easy REF: 12.3 OBJ: 12.3.d Estimate the resting membrane potential in animal cells.

MSC: Remembering 25. K+ leak channels are found in the plasma membrane. These channels open and close in an unregulated, random fashion. What do they accomplish in a resting cell? a. They set the K+ concentration gradient to zero. b. They set the membrane potential to zero. c. They disrupt the resting membrane potential. d. They keep the electrochemical gradient for K+ at zero.

ANS: D DIF: Easy REF: 12.3 OBJ: 12.3.c Recall how potassium leak channels participate in establishing the cell’s resting membrane potential. MSC: Remembering 26. The Nernst equation can be used to calculate the membrane potential based on the ratio of the outer and inner ion concentration. In a resting cell, membrane potential is calculated taking only K + ions into account. What is V when Co = 15 mM and Ci = 106 mM? a. 438.1 mV b. −52.7 mV c. 52.7 mV d. −5.3 mV ANS: B Knowing that V = 62 × log(Co/Ci), substitute the outer and inner concentration values:

V = 62 × log(15/106) V = 62 × −0.849 V = −52.7 mV DIF: Easy REF: 12.3 OBJ: 12.3.e Review how the Nernst equation can be used to calculate the resting membrane potential across a membrane. MSC: Applying 27. When using the Nernst equation to calculate membrane potential, we are making several assumptions about conditions in the cell. Which of the following is NOT a good assumption? a. The temperature is 37°C. b. The plasma membrane is primarily permeable to Na+. c. At rest, the interior of the cell is more negatively charged than the exterior. d. K+ is the principal positive ion in the cell.

ANS: B The cell has K+ leak channels. At rest, the cell is mostly permeable to K+ because of the presence of these channels.

DIF: Easy REF: 12.3 OBJ: 12.3.e Review how the Nernst equation can be used to calculate the resting membrane potential across a membrane. MSC: Analyzing 28. If Na+ channels are opened in a cell that was previously at rest, how will the resting membrane potential be affected? a. The membrane potential is not affected by Na+. b. It becomes more negative. c. It becomes more positive. d. It is permanently reset.

ANS: C As Na+ ions move into the cell, the net charge becomes more positive (less negative) and the membrane potential changes to reflect the Co/Ci for both Na+ and K+ ions.

DIF: Easy REF: 12.3 OBJ: 12.3.e Review how the Nernst equation can be used to calculate the resting membrane potential across a membrane. MSC: Applying 29. In a method called patch-clamping, a glass capillary can be converted into a microelectrode that measures the electrical currents across biological membranes. Which of the following statements about the patch-clamp method is FALSE? a. The glass capillary adheres to a “patch” of membrane through the application of suction. b. The aperture in the glass capillary used to make a microelectrode is about 1 μm in diameter. c. If the experimental conditions are held constant, fluctuations in electrical currents across the patch of membrane are still observed. d. Single-channel patch-clamp recordings have demonstrated that gated membrane channels will only open and close in response to specific stimuli.

ANS: D DIF: Easy REF: 12.3 OBJ: 12.3.f Outline how patch-clamp recording can be used to study the activity of ion channels. MSC: Remembering 30. Which of the following is required for the secretion of neurotransmitters in response to an action potential? a. neurotransmitter receptors b. Na+-K+ pumps

c. voltage-gated K+ channels d. voltage-gated Ca2+ channels ANS: D Voltage-gated Ca2+ channels open in response to the depolarization caused as the action potential moves toward the nerve terminal. The influx of calcium from outside the cell causes the synaptic vesicles to fuse with the plasma membrane and release large amounts of neurotransmitter into the synaptic cleft.

DIF: Easy REF: 12.4 OBJ: 12.4.h Review how an electrical signal is converted to a chemical signal at a nerve terminal. MSC: Remembering 31. Figure 12-31 illustrates changes in membrane potential during the formation of an action potential. What membrane characteristic or measurement used to study action potentials is indicated by the arrow?

Figure 12-31

a. effect of a depolarizing stimulus b. resting membrane potential c. threshold potential d. action potential ANS: C DIF: Easy REF: 12.4 OBJ: 12.4.c Outline how membrane depolarization triggers an action potential and how the action potential spreads along the membrane. MSC: Remembering 32. Figure 12-32 illustrates changes in membrane potential during the formation of an action potential. What membrane characteristic or measurement used to study action potentials is indicated by the arrow?

Figure 12-32

a. effect of a depolarizing stimulus b. resting membrane potential c. threshold potential d. action potential ANS: B DIF: Easy REF: 12.4 OBJ: 12.4.c Outline how membrane depolarization triggers an action potential and how the action potential spreads along the membrane. MSC: Remembering 33. Figure 12-33 illustrates changes in membrane potential during the formation of an action potential. What membrane characteristic or measurement used to study action potentials is indicated by the arrow?

Figure 12-33

a. effect of a depolarizing stimulus b. resting membrane potential c. threshold potential d. action potential ANS: D DIF: Easy REF: 12.4 OBJ: 12.4.c Outline how membrane depolarization triggers an action potential and how the action potential spreads along the membrane. MSC: Remembering

34. Figure 12-34 illustrates changes in membrane potential during the formation of an action potential. What membrane characteristic or measurement used to study action potentials is indicated by the arrow?

Figure 12-34

a. effect of a depolarizing stimulus b. resting membrane potential c. threshold potential d. action potential ANS: A DIF: Easy REF: 12.4 OBJ: 12.4.c Outline how membrane depolarization triggers an action potential and how the action potential spreads along the membrane. MSC: Remembering 35. The stimulation of a motor neuron ultimately results in the release of a neurotransmitter at the synapse between the neuron and a muscle cell. What type of neurotransmitter is used at these neuromuscular junctions? a. acetylcholine b. glutamate c. GABA d. glycine ANS: A DIF: Easy REF: 12.4 OBJ: 12.4.h Review how an electrical signal is converted to a chemical signal at a nerve terminal. | 12.4.k Describe the events that take place when acetylcholine binds to its receptor on a vertebrate muscle cell. MSC: Remembering 36. Both excitatory and inhibitory neurons form junctions with muscles. By what mechanism do inhibitory neurotransmitters prevent the postsynaptic cell from firing an action potential? a. by closing Na+ channels b. by preventing the secretion of excitatory neurotransmitters c. by opening K+ channels d. by opening Cl− channels ANS: D Inhibitory neurons release inhibitory neurotransmitters such as GABA and glycine. They bind to and open ligandgated Cl− channels. If Na+ channels are open, Cl− ions will rush into the cell as well, neutralizing the positive charges car-

ried by Na+.

DIF: Easy REF: 12.4 OBJ: 12.4.j Contrast how the receptors for excitatory and inhibitory neurotransmitters alter the activity of a postsynaptic cell. MSC: Remembering 37. Which of the following statements best reflects the nature of synaptic plasticity? a. New synapses are created due to the postnatal generation of neurons. b. Synaptic response changes in magnitude depending on frequency of stimulation. c. There is a change in the type of neurotransmitter used at the synapse. d. Neuronal connections are pruned during normal development.

ANS: B DIF: Easy REF: 12.4 OBJ: 12.4.a Review the anatomy of a nerve cell and discuss the direction in which electrical signals travel from one neuron to another. MSC: Understanding 38. Approximately, how many distinct synapses are established on the dendrites and cell body of a motor neuron in the spinal cord? a. tens b. hundreds c. thousands d. millions ANS: C DIF: Easy REF: 12.4 OBJ: 12.4.n Propose how nerve cells can integrate and interpret a variety of incoming signals to generate an appropriate response. MSC: Remembering 39. Which of the following statements about GABA receptors is FALSE? a. They are located on postsynaptic membranes. b. They are ligand-gated channels. c. They inhibit synaptic signaling. d. They promote neuronal uptake of Na+.

ANS: D DIF: Easy REF: 12.4 OBJ: 12.4.i Summarize how transmitter-gated ion channels convert the chemical signal carried by a neurotransmitter back into an electrical signal. | 12.4.j Contrast how the receptors for excitatory and inhibitory neurotransmitters alter the activity of a postsynaptic cell. MSC: Analyzing 40. Which of the following gated ion channels are involved in inhibitory synaptic signaling? a. voltage-gated Na+ channels b. voltage-gated Ca2+ channels c. glycine-gated Cl− channels d. glutamate-gated cation channels ANS: C DIF: Easy REF: 12.4 OBJ: 12.4.j Contrast how the receptors for excitatory and inhibitory neurotransmitters alter the activity of a postsynaptic cell. MSC: Remembering

MULTIPLE SELECT 1. Eukaryotic cell membranes are highly specialized and have characteristic sets of transporters and channels. Below, match the transporter with the appropriate membrane type. More than one transporter can be matched to a membrane type.

Membrane

1. plasma 2. lysosomal 3. endoplasmic reticulum 4. inner mitochondrial Transporter A. nucleotide B. Proton C. Glucose D. ADP/ATP E. Na+/K+ F. Ca2+ ANS: 1. ANS: A, C, E, F DIF: Easy REF: 12.1 OBJ: 12.1.d State the function of membrane transport proteins and differentiate between transporters and channels. MSC: Remembering 2. ANS: B DIF: Easy REF: 12.1 OBJ: 12.1.d State the function of membrane transport proteins and differentiate between transporters and channels. MSC: Remembering 3. ANS: F DIF: Easy REF: 12.1 OBJ: 12.1.d State the function of membrane transport proteins and differentiate between transporters and channels. MSC: Remembering 4. ANS: D DIF: Easy REF: 12.1 OBJ: 12.1.d State the function of membrane transport proteins and differentiate between transporters and channels. MSC: Remembering

MATCHING 1. Match the numbered lines in the diagram with the following structures.

A. nerve terminal B. cell body C. axon D. dendrite

Figure 12-1

1. ANS: B DIF: Easy REF: 12.4 OBJ: 12.4.a Review the anatomy of a nerve cell and relate the direction in which electrical signals travel from one neuron to another. MSC: Remembering 2. ANS: D DIF: Easy REF: 12.4 OBJ: 12.4.a Review the anatomy of a nerve cell and relate the direction in which

electrical signals travel from one neuron to another. MSC: Remembering 3. ANS: C DIF: Easy REF: 12.4 OBJ: 12.4.a Review the anatomy of a nerve cell and relate the direction in which electrical signals travel from one neuron to another. MSC: Remembering 4. ANS: A DIF: Easy REF: 12.4 OBJ: 12.4.a Review the anatomy of a nerve cell and relate the direction in which electrical signals travel from one neuron to another. MSC: Remembering

SHORT ANSWER 1. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. CO2 and O2 are water-soluble molecules that diffuse freely across cell membranes.

B. The differences in permeability between artificial lipid bilayers and cell membranes arise from variations in phospholipid content.

C. Transporters are similar to channels, except that they are larger, allowing folded proteins as well as smaller organic molecules to pass through them.

D. Cells expend energy in the form of ATP hydrolysis so as to maintain ion concentrations that differ from those found outside the cell.

ANS: A. True.

B. False. The primary difference between cell membranes and artificial membranes is that cell membranes have proteins responsible for creating a selective permeability, which varies with the location and function of the membrane.

C. False. Transporters work by changing conformation after specific binding of the solute to be transported. Channels exclude molecules on the basis of size and charge, but do not depend on specific recognition of the molecules moving through.

D. True.

DIF: Easy REF: 12.1 OBJ: 12.1.b Distinguish between simple diffusion and facilitated transport. | 12.1.c Review the properties that govern the rate at which a given solute can cross a protein-free lipid bilayer. | 12.1.d State the function of membrane transport proteins and differentiate between transporters and channels. MSC: Evaluating 2. Identify the molecule in each pair that is more likely to diffuse through the lipid bilayer.

A. amino acids or

benzene

B. Cl−

or

ethanol

C. glycerol

or

RNA

D. H2O

or

O2

E. adenosine

or

ATP

ANS: The two basic properties governing the likelihood of whether a molecule will diffuse through a lipid bilayer are the size of the molecule and the charge of the molecule. A smaller mol ecule will be more likely to diffuse through the lipid bilayer than a larger molecule. A nonpolar (hydrophobic) molecule will be more likely to diffuse through the l ipid bilayer than a polar molecule, which is more likely to diffuse through the lipid bilay er than a charged molecule.

A. benzene (small nonpolar versus larger uncharged)

B. ethanol (polar versus charged) C. glycerol (small polar versus very large, highly charged) D. O2 (nonpolar versus polar) E. adenosine (polar versus highly charged) DIF: Moderate REF: 12.1 OBJ: 12.1.c Review the properties that govern the rate at which a given solute can cross a protein-free lipid bilayer. MSC: Applying 3. We can test the relative permeability of a phospholipid bilayer by using a synthetic membrane that does not contain any protein components. Some uncharged, polar molecules are found to diffuse freely across these membranes, to varying degrees. Which of the following has the lowest rate of diffusion across an artificial membrane? Why? A. glucose B. water C. glycerol D. ethanol ANS: A; Glucose will have a negligible degree of diffusion across the synthetic bilayer. Not only is it polar, it is also larger than other molecules that are able to diffuse across the membrane.

DIF: Easy REF: 12.1 OBJ: 12.1.c Review the properties that govern the rate at which a given solute can cross a protein-free lipid bilayer. MSC: Understanding 4. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. active

hydrophilic

noncovalent

amino acid

hydrophobic

passive

amphipathic

ion channels

transporter proteins

A molecule moves down its concentration gradient by __________ transport, but requires __________ transport to move up its concentration gradient. Transporter proteins and ion channels function in membrane transport by providing a __________ pathway through the membrane for specific polar solutes or inorganic ions. __________ are highly selective in the solutes they transport, binding the solute at a specific site and changing conformation so as to transport the solute across the membrane. On the other hand, __________ discriminate between solutes mainly on the basis of size and electrical charge.

ANS: A molecule moves down its concentration gradient by passive transport, but requires active transport to move up its concentration gradient. Transporter proteins and ion channels function in membrane transport by providing a hydrophilic pathway through the membrane for specific polar solutes or inorganic ions. Transporter proteins are highly selective in the solutes they transport, binding the solute at a specific site and changing conformation so as to transport the solute across the membrane. On the other hand, ion channels discriminate between solutes mainly on the basis of size and electrical charge.

DIF: Easy REF: 12.1 OBJ: 12.1.d State the function of membrane transport proteins and differentiate between transporters and channels. MSC: Understanding 5. What chemical principles explain the observation that a protein-free lipid bilayer is a billion times more permeable to water than to a sodium ion?

ANS: Water is a small, uncharged molecule that diffuses directly across the membrane, despite needing to pass

through a hydrophobic lipid core. In cells, ions pass through lipid bilayers via ion channels or transporters. Because ions carry a charge, they have strong electrostatic interactions with water molecules. There are no compensating interactions in the hydrophobic lipid core of the bilayer, which prevents ions from entering.

DIF: Moderate REF: 12.1 OBJ: 12.1.c Review the properties that govern the rate at which a given solute can cross a protein-free lipid bilayer. | 12.1.k Describe the ways in which water can move across cell membranes and articulate what governs whether water will enter or exit a cell. MSC: Analyzing 6. The toxicity of mercury depends greatly upon the formulation of the metal. Which of them (elemental, metal ion, or methyl-mercury) is more likely to be absorbed through the skin? Explain your answer.

ANS: Organo-mercury compounds such as methyl-mercury are absorbed rapidly through the cellular membranes of skin cells and reach the bloodstream. Elemental mercury is a relatively large, uncharged atom, which is not absorbed to an appreciable degree. Mercury salts (the ionic form of the metal) can pass through membranes only of those cells that express transport channels with large enough pores.

DIF: Hard REF: 12.1 OBJ: 12.1.c Review the properties that govern the rate at which a given solute can cross a protein-free lipid bilayer. MSC: Applying 7. Indicate whether the statements below are TRUE or FALSE. If a statement is false, explain why it is false.

A. Facilitated diffusion can be described as the favorable movement of one solute down its concentration gradient being coupled with the unfavorable movement of a second solute up its concentration gradient.

B. Transporters undergo transitions between different conformations, depending on whether the substrate-binding pocket is empty or occupied.

C. The electrochemical gradient for K+ across the plasma membrane is small. Therefore, any movement of K+ from the inside to the outside of the cell is driven solely by its concentration gradient.

D. The net negative charge on the cytosolic side of the membrane enhances the rate of glucose import into the cell by a uniporter.

ANS: A. False. This describes coupled transport, which is one type of active transport. Facilitated diffusion can also be called passive transport, in which a solute always moves down its concentration gradient.

B. True.

C. True.

D. False. Glucose is an uncharged molecule, and its import is not directly affected by the voltage difference across the membrane if glucose is being transported alone. If the example given were the Na +/glucose symporter, we would have to consider the charge difference across the membrane.

DIF: Easy REF: 12.1 OBJ: 12.1.b Distinguish between simple diffusion and facilitated transport. | 12.1.g Compare how transporters and channels discriminate among solutes, moving only a select subset across the membrane. | 12.1.i Review the forces that govern the passive transport of charged and uncharged solutes across a cell membrane.

MSC: Evaluating 8. Describe the two components of an electrochemical gradient. Use your description to suggest which of the following is influenced by a larger electrochemical gradient: (1) Na + moving into the cell; (2) K+ moving out of the cell. Explain your reasoning.

ANS: An electrochemical gradient is composed of two forces: the energy from the concentration gradient, the force derived from the charge differential across the membrane, referred to as the membrane potential. There is a larger elec-

trochemical gradient driving the movement of Na+ into the cell compared to the net force moving K+ out of the cell. In the case of Na+ the concentration gradient and the membrane potential both drive movement of Na+ into the cell. In the case of K+ these two forces work in opposition, reducing the net magnitude of the electrochemical gradient.

DIF: Easy REF: 12.1 OBJ: 12.1.j Compare the electrochemical gradients for sodium and potassium ions and recall how the membrane potential and concentration gradient for each ion combine to influence its movement across the plasma membrane. MSC: Evaluating 9. It is thought that the glucose transporter switches between two conformational states in a completely random fashion.

How is it possible for such a system to move glucose across the membrane efficiently in a single direction?

ANS: Although the opening of the glucose transporter on one side of the membrane or the other is random, the binding of glucose into the binding site of the transporter is not a random event. The affinity between the glucose molecule and the transporter governs the binding event: transporter + glucose ↔ transporter − glucose At high glucose concentrations, the complex formation is favored; at low glucose concentrations, dissociation of glucose from the transporter is favored. Therefore, as long as there is a large concentration gradient, efficient transport can occur by the simple rules of binding equilibria.

DIF: Moderate REF: 12.1 OBJ: 12.2.a Outline how the glucose transporter in the plasma membrane of mammalian cells imports glucose after a meal and exports glucose to provide fuel for other tissues in the body.

MSC: Understanding 10. If ATP production is blocked in an animal cell, the cell will swell up. Explain this observation.

ANS: ATP is required to power the Na +-K+ pump, which is necessary for maintaining osmotic balance. The pump requires ATP hydrolysis to drive its pumping cycle. So, in the absence of ATP production, the Na + concentration inside the cell will increase. This is followed by passive diffusion of water across the membrane, causing the cell to swell.

DIF: Moderate REF: 12.2 OBJ: 12.2.c Review how the sodium pump in animal cells uses the energy supplied by ATP hydrolysis to maintain the concentration gradients of sodium and potassium ions.

MSC: Understanding 11. Indicate whether the statements below are TRUE or FALSE. If a statement is false, explain why it is false.

A. The extracellular concentration of Ca2+ is approximately 104-fold higher than the concentration of Ca2+ in the cytosol.

B. The low cytosolic Ca2+ concentration sensitizes the cell to an influx of Ca2+, ensuring a rapid response to environmental stimuli.

C. Cytosolic Ca2+ concentration is kept low by the use of chelators such as EDTA. D. The primary mechanism by which Ca 2+ acts as a signaling molecule is by increasing the net charge in the cytosol.

ANS: A. True.

B. True.

C. False. Ca2+ concentrations in the cytosol are kept low by the action of ATP-driven calcium pumps in the endoplasmic reticulum membrane and the plasma membrane.

D. False. Ca2+ binds tightly to many proteins in the cell, which in turn changes their activity. This interaction is the primary mechanism by which Ca2+ signaling occurs.

DIF: Easy REF: 12.2 OBJ: 12.2.e Explain why—and how—cells keep the cytosolic concentration of calcium ions low.

MSC: Evaluating 12. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. antiport

coupled

membrane potential

ATP hydrolysis

electrochemical

symport

concentration

light-driven

uniport

For an uncharged molecule, the direction of passive transport across a membrane is determined solely by its __________ gradient. On the other hand, for a charged molecule, the __________ must also be considered. The net driving force for a charged molecule across a membrane therefore has two components and is referred to as the __________ gradient. Active transport allows the movement of solutes against this gradient. The transporter proteins called __________ transporters use the movement of one solute down its gradient to provide the energy to drive the uphill transport of a second solute. When this transporter moves both ions in the same direction across the membrane, it is considered a/an __________; if the ions move in opposite directions, the transporter is considered a/an __________.

ANS: For an uncharged molecule, the direction of passive transport across a membrane is determined solely by its concentration gradient. On the other hand, for a charged molecule, the membrane potential must also be considered. The net driving force for a charged molecule across a membrane therefore has two components and is r eferred to as the electrochemical gradient. Active transport allows the movement of solutes against this gradient.

The transporter proteins called coupled transporters use the movement of one solute down its gradient to provide the energy to drive the uphill transport of a second solute. When this transporter moves both ions in the same direction across the membrane, it is considered a symport; if the ions move in opposite directions, the transporter is considered an antiport.

DIF: Easy REF: 12.2 OBJ: 12.2.f Differentiate between a symport, an antiport, and a uniport.

MSC: Understanding 13. The movement of glucose into the cell, against its concentration gradient, can be powered by the co-transport of Na+ into the cell. Explain the thermodynamic factors governing this transporter. Under what conditions would transport of glucose slow or come to a halt?

ANS: The movement of Na+ ions from an area that has a high Na+ concentration to a new area of low Na+ concentration is energetically favorable because the net entropy in the system is increasing. As long as the difference in Na + ion concentration across the membrane is large enough, the entropic factor will drive the import of glucose into the cell.

This effect is amplified by the electrical component of the electrochemical gradient, the membrane potential, that drive Na+ in the same direction. If the magnitude of the Na+ electrochemical gradient were reduced (by reducing the membrane potential or concentration gradient) below that required to drive glucose against its concentration gradient, glucose import would cease.

DIF: Hard REF: 12.2 OBJ: 12.2.g Compare the uniport glucose transporter and the glucose-sodium symport in terms of their activity and role in glucose transport in intestinal epithelial cells. MSC: Applying

14. Describe the process by which gut epithelial cells use transporters to take up ingested glucose (against the concentration gradient) and to distribute glucose to other tissues by moving it back out of the cell (down the concentration gradient).

ANS: Gut epithelial cells use two different transporters to take glucose up from the gut and distribute it into the bloodstream and to other tissues. These transporters are located at opposite sides of the cell: the apical side of the cell (which faces the gut) contains a Na +-glucose symporter. This symporter couples the entry of Na + down its electrochemical gradient to the active import of glucose against its concentration gradient. The Na +-glucose symporter is restricted to the apical side of the cell by tight-junction complexes in the plasma membrane, which link neighboring epithelial cells together. On the basolateral side of the cell, there is another transporter that facilitates movement of glucose down its concentration gradient, out of the cell. This transporter is a uniporter that only transports glucose in one direction: from the cytosol to the extracellular matrix. T he location of this uniporter is also restricted by the presence of the tight junctions, so that the epithelial cell will not transport glucose back into the lumen of the gut.

DIF: Hard REF: 12.2 OBJ: 12.2.g Compare the uniport glucose transporter and the glucose-sodium symport in terms of their activity and role in glucose transport in intestinal epithelial cells. MSC: Understanding 15. Define homeostasis and illustrate your understanding of this concept by describing the mechanism of one transmembrane transport protein and how it contributes to homeostasis.

ANS: Homeostasis is the tendency of any living system to maintain an internal equilibrium (temperature, osmolarity, pH, glucose levels, iron levels, etc.), independent of external conditions. Any exam ple can be used to illustrate how transmembrane transport proteins contribute to the homeostasis in a cell. In muscle cells, ATP -driven Ca 2+ pumps keep cytosolic Ca 2+ levels low until the Ca 2+ is released from the sarcoplasmic reticulum to trigger muscle contraction. After the cytosol is flooded with Ca 2+, the pump works to move the Ca 2+ ions back across the membrane of the sarcoplasmic reticulum to restore homeostasis.

DIF: Moderate REF: 12.2 OBJ: 12.2.e Explain why—and how—cells keep the cytosolic concentration of calcium ions low. MSC: Understanding 16. The flow of ions through a gated channel can be studied using a method called “patch-clamp recording.”

A. How is a detached patch-clamp experiment set up, and what exactly does it mean to “clamp” an ion channel?

B. How is it possible to collect the recordings shown in Figures Q12-16A and Q12-16B from a single ion channel?

Figure 12-16

ANS: A. A detached patch-clamp experiment requires the removal of a portion of the cell membrane by sealing the microelectrode to the membrane surface. After lifting the patch of membrane stuck to the microelectrode, it is placed into a solution of controlled medium. The voltage applied to the membrane patch can be fixed (clamped) while other parameters are studied.

B. By manipulating ion concentrations in the two chambers or simply reversing the direction of the current in the system, the ion flow through the channel can be reversed, resulting in the recording of negative values for current when the channel opens.

DIF: Hard REF: 12.3 OBJ: 12.3.f Outline how patch-clamp recording can be used to study the activity of ion channels.

MSC: Applying 17. Indicate whether the statements below are TRUE or FALSE. If a statement is false, explain why it is false.

A. Gap junctions are large pores that connect the cytosol to the extracellular space.

B. Aquaporin channels are found in the plasma membrane, and allow the rapid passage of water molecules and small ions in and out of cells.

C. The ion selectivity of a channel depends solely on the charge of the amino acids lining the pore inside the channel.

D. Most ion channels are gated, which allows them to open and close in response to a specific stimulus, rather than allowing the constant, unregulated flow of ions.

ANS: A. False. Gap junctions are used to connect the cytosol of adjacent cells, allowing the sharing of ions and small metabolites. Because gap junctions are large channels, if they were open while facing the extracellular environment, the ability of the plasma membrane to serve as a permeability barrier would be greatly reduced.

B. False. Charged molecules (even protons, which are very small) are not able to pass through aquaporins.

C. False. Selectivity depends on three parameters: the diameter, shape, and charge of the ion trying to pass through the pore of the channel.

D. True.

DIF: Easy REF: 12.3 OBJ: 12.3.a Explain how the structure of ion channels leads to their ion selectivity. | 12.3.h Define “gating” and list the types of conditions that can alter the opening and closing of ion channels.

MSC: Evaluating 18. Because the Na+/K+ pump moves these ions at the same rate in opposing directions across the plasma membrane, it creates a net balance of charges on each side of the membrane (i.e., membrane potential = 0). Given this fact, explain how cells generate a membrane potential across the plasma membrane and suggest the consequences for cell functions if this capacity were handicapped.

ANS: The presence of K+ channels in the plasma membrane allows the free diffusion of K+ out of the cell. These K+ leak channels create a voltage difference across the membrane (membrane potential ≠ 0). Without the “leak” capability in the membrane, the transport of glucose and other small molecules that depend on the electrochemical gradient would be reduced.

DIF: Moderate REF: 12.3 OBJ: 12.3.c Recall how potassium leak channels participate in establishing the cell’s resting membrane potential. MSC: Understanding 19. Describe the two forces that drive an ion across the plasma membrane and explain how the Nernst equation takes into account both of these forces. Use the components of the equation to support your explanation and be sure to

specify the assumptions being made when using the Nernst equation to calculate membrane potential.

ANS: The forces that drive the movement of an ion across the plasma membrane include a concentration gradient (that is, there is a negative change in free energy associated with an increase in entropy for ions in solution) and an electrical component (the force resulting from the attraction between molecules of opposite charges). The Nernst equation expresses the change in voltage across the membrane as it relates to a change in the ratio of ions on either side of the plasma membrane. As written below, the voltage changes by 62 millivolts with every tenfold change in the ion concentration ratio across the membrane.

V = 62 log10(Co/Ci) As written in this simplified form, the equation assumes that the flow of ions has reached an equilibrium at 37°C and that the ions that are moving are positive ions. Because the ions that are moving across the membrane are typically K+ ions (resting membrane potential) or Na+ ions (action potentials) these assumptions hold true for most of the biological systems being examined.

DIF: Moderate REF: 12.3 OBJ: 12.3.e Review how the Nernst equation can be used to calculate the resting membrane potential across a membrane. MSC: Understanding 20. Indicate whether the statements below are TRUE or FALSE. If a statement is false, explain why it is false.

A. Neurotransmitters are small molecules released into the synaptic cleft after the fusion of synaptic vesicles with the presynaptic membrane.

B. Action potentials are usually mediated by voltage-gated Ca2+ channels.

C. Voltage-gated Na+ channels become automatically inactivated shortly after opening, which ensures that the action potential cannot move backward along the axon.

D. Voltage-gated K+ channels also open immediately in response to local depolarization, reducing the magnitude of the action potential.

ANS: A. True.

B. False. Action potentials are usually mediated by voltage-gated Na+ channels.

C. True.

D. False. Voltage-gated K+ channels respond more slowly than the voltage-gated Na+ channels. Because voltagegated K+ channels do not open until the action potential reaches its peak, they do not affect its magnitude. Instead, they help to restore the local membrane potential quickly while the voltage-gated Na+ channels are in the inactivated conformation.

DIF: Easy REF: 12.4 OBJ: 12.4.c Outline how membrane depolarization triggers an action potential and how the action potential spreads along the membrane. | 12.4.d Recall the conformations Na +-gated ion channels cycle through as an action potential travels along the axonal membrane. | 12.4.e Compare the roles that Na+-gated and K+-gated ion channels play in an action potential. | 12.4.h Review how an electrical signal is converted to a chemical signal at a nerve terminal. MSC: Evaluating 21. Order and describe all the molecular events required for the propagation of an action potential. Is propagation along the axon bidirectional? Why or why not?

ANS: An initial event depolarizes the plasma membrane in the neuron, this change in membrane potential activates the voltage-gated Na+ channels, causing them to open. Na+ ions flood into the cytosol, locally depolarizing the mem-

brane, which then activates resting Na+ channels further down the neuron. The action potential is only propagated in one direction because after the Na+ channels close, they cannot be immediately reactivated. Only after the membrane is repolarized through the action of the Na+/K+ pump, the channels flip back into their resting state, ready for the next depolarization event.

DIF: Hard REF: 12.4 OBJ: 12.4.c Outline how membrane depolarization triggers an action potential and how the action potential spreads along the membrane. | 12.4.d Recall the conformations Na +-gated ion channels cycle through as an action potential travels along the axonal membrane. MSC: Evaluating 22. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. anions

hyperpolarization

neutral

axon

less

positive

cytoskeleton

ligand

pressure

dendrites

more

synaptic vesicle

depolarization

negative

voltage

The action potential is a wave of __________ that spreads rapidly along the neuronal plasma membrane. This wave is triggered by a local change in the membrane potential to a value that is __________ negative than the resting membrane potential. The action potential is propagated by the opening of __________-gated channels. During an action potential, the membrane potential changes from __________ to __________. The action potential travels along the neuron’s __________ to the nerve terminals. Neurons chiefly receive signals at their highly branched __________.

ANS: The action potential is a wave of depolarization that spreads rapidly along the neuronal plasma membrane.

This wave is triggered by a local change in the membrane potential to a value that is less negative than the resting membrane potential. The action potential is propagated by the opening of voltage-gated channels. During an action potential, the membrane potential changes from negative to positive. The action potential travels along the neuron’s axon to the nerve terminals. Neurons chiefly receive signals at their highly branched dendrites.

DIF: Easy REF: 12.4 OBJ: 12.4.a Review the anatomy of a nerve cell and relate the direction in which electrical signals travel from one neuron to another. | 12.4.c Outline how membrane depolarization triggers an action potential and how the action potential spreads along the membrane. MSC: Understanding 23. The stimulation of a motor neuron ultimately results in the release of a neurotransmitter at the synapse between the neuron and a muscle cell. How is the chemical signal converted into an electrical signal in the postsynaptic muscle cell?

ANS: Most neurotransmitter receptors function as ligand-gated ion channels. These ion channels are similar to voltage-gated channels, except that they do not open in response to a change in voltage across the membrane, but to the binding of a neurotransmitter. In the neuromuscular junction, the neurotransmitter acetylcholine binds to the acetylcholine receptor, which allows Na+ to enter the muscle cell, altering its membrane potential. In this way, a chemical signal (acetylcholine) is converted back into an electrical signal (change in membrane potential).

DIF: Moderate REF: 12.4 OBJ: 12.4.i Summarize how transmitter-gated ion channels convert the chemical signal carried by a neurotransmitter back into an electrical signal. MSC: Understanding 24. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all

words or phrases will be used; each word or phrase should be used only once. acetylcholine receptor 2+

Ca

channels

chemical

GABA receptor +

K channels +

Na channels

postsynaptic presynaptic synapses

electrical Neurons communicate with each other through specialized sites called __________. Many neurotransmitter receptors are ligand-gated ion channels that open transiently in the __________ cell membrane in response to neurotransmitters released by the __________ cell. Ligand-gated ion channels in nerve cell membranes convert __________ signals into __________ ones. Neurotransmitter release is stimulated by the opening of voltage-gated __________ in the nerve-terminal membrane.

ANS: Neurons communicate with each other through specialized sites called synapses. Many neurotransmitter receptors are ligand-gated ion channels that open transiently in the postsynaptic cell membrane in response to neurotransmitters released by the presynaptic cell. Ligand-gated ion channels in nerve cell membranes convert chemical signals into electrical ones. Neurotransmitter release is stimulated by the opening of voltage-gated Ca2+ channels in the nerve-terminal membrane. DIF: Easy REF: 12.4 OBJ: 12.4.h Review how an electrical signal is converted to a chemical signal at a nerve terminal. | 12.4.i Summarize how transmitter-gated ion channels convert the chemical signal carried by a neurotransmitter back into an electrical signal. MSC: Understanding 25. Studies on the squid giant axon were instrumental to our current understanding of how action potentials are generated. You decide to do some experiments on the squid giant axon yourself.

A. You remove the cytoplasm in an axon and replace it with an artificial cytoplasm that contains twice the normal concentration of K+ by adding KOAc, where OAc− is an anion to which the membrane is impermeable. In this way, you double the internal concentration of K+ while maintaining the bulk electrical balance of the cytoplasmic solution. Will this make the resting potential of the membrane more or less negative?

B. You add NaCl to the extracellular fluid and effectively double the amount of extracellular Na + ions. How does this affect the action potential?

C. You replace half of the NaCl in the extracellular fluid with choline chloride. (Choline is a monovalent cation much larger than Na+. Note that the presence of choline will not impede the flow of Na+ through its channels.) How will this affect the action potential?

ANS: A. Increasing the concentration of K+ in the cytoplasm of the squid axon will make the membrane potential more negative. Doubling the amount of K+ increases the driving force for K+ to move out of the cell, leaving the inside of the cell more negative and thus decreasing the membrane potential. (Remember, from the Nernst equation, the driving force for an ion across a membrane is proportional to the ratio of the concentration of the ion on the outside to the concentration of the ion on the inside.)

B. Doubling the amount of Na+ in the extracellular fluid will increase the height of the peak of the action potential.

Again, this is because now the driving force for Na + to enter the cell is greater than it was before. Thus, when Na+ channels open, the flux of Na + ions is now greater. (Remember that flux is the number of ions entering per second.)

C. The action potential in this case will reach a height that is less than that normally achieved. (Choline is added in this case to maintain bulk electrical neutrality. Because Na+ channels are not permeable to choline, choline does

not contribute to the electrochemical gradient.) You have now halved the concentration of Na + and thus decreased the driving force for Na+ to enter the cell.

DIF: Hard REF: 12.4 OBJ: 12.4.g Outline how studies using the squid giant axon revealed the different roles of potassium and sodium ions in establishing the resting membrane potential and propagating an action potential.

MSC: Applying 26. The field of neurobiology is seeing rapid advances in our understanding of neural circuitry in the brain. Part of this work involves the physical mapping of all synapses (a project dubbed the connectome); another critical advance is in the area of optogenetics, which allows scientists to dissect neural circuits that determine specific behaviors in a range of organisms from fruit flies to monkeys.

A. Describe the method of optogenetics.

B. Optogenetics has great potential for deepening our understanding of behavior, learning, memory, and cognitive development. However, it is not as likely to be used directly for treatment of problems such as depression or anxiety. Why not? Consider the limitations of the method when answering this question.

ANS: A. Optogenetics is a method in which genetic engineering techniques are used to introduce light-gated channels into a selected set of target neurons. Light of a specific wavelength is then used to open the channel, which allows the investigator to directly control the activity of these neurons in the living organism. When the channels used are light-gated Na+ channels, stimulation will allow Na+ to enter the neurons, triggering an action potential. Light-gated Cl- channels can similarly be used to inhibit neural activity.

B. Optogenetics currently uses viral vectors to introduce the genes encoding the light-gated channels into the neurons of interest. This approach can be potentially problematic, as insertion of this engineered DNA at off-target sites in the genome could cause mutations that would alter or destroy the activity of critical genes. In addition, the stimulation of neurons by optogenetics requires a light source. In the case of a human patient, this would mean inserting a fiber optic light source into the region of the brain that contains the foreign channels. This would represent a highly experimental treatment, hence it is currently only used in animal models to explore circuitry we do not yet fully understand.

DIF: Hard REF: 12.4 OBJ: 12.4.p Outline how optogenetics can be used to study the role that particular neurons play in the behavior of an animal. MSC: Understanding 27. You have prepared lipid vesicles (spherical lipid bilayers) that contain Na +-K+ pumps as the sole membrane protein.

All of the Na+-K+ pumps are oriented in such a way that the portion of the molecule that normally faces the cytosol is on the inside of the vesicle and the portion of the molecule that normally faces the extracellular space is on the outside of the vesicle. Assume that each pump transports one Na+ ion in one direction and one K+ ion in the other direction during each pumping cycle (see Figure 12-27 for how the Na+-K+ pump normally functions in the plasma membrane).

Figure 12-27

Predict what would happen in each of the following conditions: A. The solutions inside and outside the vesicles contain both Na+ and K+ ions but no ATP. B. The solution outside the vesicles contains both Na+ and K+ ions; the solution inside contains both Na+ and K+ ions and ATP. C. The solution outside contains Na+; the solution inside contains Na+ and ATP. ANS: A. Without any ATP to provide energy for the Na+-K+ pumps, no ions will be pumped.

B. The pumps will use the energy from ATP hydrolysis to transport Na+ out of the vesicles and K+ into the vesicles.

(The pumps will stop working either when the amount of ATP inside the vesicle is depleted or when the K + outside the vesicles is depleted.)

C. The pump will bind a molecule of Na+, causing the ATPase activity to hydrolyze ATP and transfer the phosphate group onto the pump. A conformational change will occur, leading to the release of Na + from the vesicle. However, because there is no K+ outside the vesicle, the pump will get stuck at that step and subsequent steps of the cycle will not occur.

DIF: Hard REF: 12.2 OBJ: 12.2.c Review how the sodium pump in animal cells uses the energy supplied by ATP hydrolysis to maintain the concentration gradients of sodium and potassium ions. MSC: Applying 28. The Aeroschmidt weed contains an ATP-driven ion pump in its vacuolar membrane that pumps potentially toxic heavy metal ions such as Zn2+ and Pb2+ into the vacuole. The pump protein cycles through a phosphorylated and an unphosphorylated state and works in a similar way to the Na+-K+ pump of animal cells. To study its action, you

incorporate the unphosphorylated form of the protein into phospholipid vesicles containing K + in their interiors. (You ensure that all of the protein molecules are oriented such that their cytosolic domains face the outside of your vesicles.) When you add Zn2+ and ATP to the solution outside such vesicles, you find that Zn 2+ is pumped into the vesicle lumen. You then expose vesicles containing the pump protein to the solutes shown in Table 12-28.

Table 12-28

You then determine the amount of phosphorylated and unphosphorylated ATP-driven ion pump protein in each sample. Your results are summarized in Table 12-28A, where a minus sign indicates an absence of a type of protein and a plus sign indicates its presence.

Table 12-28A

You treat vesicles as in lane F, but before determining the phosphorylation state of the protein, you wash away the outside buffer and replace it with a buffer containing only Zn 2+. What do you expect to happen in this sample, and what are the results you expect to observe with respect to phosphorylation and movement of K +?

ANS: Because the pump is mechanistically similar to the Na+-K+ pump, the transport of ions is driven by ATP hydrolysis and the pump is transiently phosphorylated; phosphorylation is stimulated by one ion and dephosphorylation is stimulated by the other ion. Because all of the protein is in the phosphorylated form in the absence of Zn2+ (lane F), Zn2+ must be required for dephosphorylation. K+, then, must bind to the dephosphorylated form and stimulate its ATPase/autophosphorylation. Therefore, if Zn2+ is added to the phosphorylated pump, Zn2+ will stimulate dephosphorylation, trigger a conformational change, and be injected into the vesicle. K + would stimulate phosphorylation of the pump, but if there is no ATP in the solution, no phosphorylation and hence no K + movement will occur.

DIF: Hard REF: 12.2 OBJ: 12.2.c Review how the sodium pump in animal cells uses the energy supplied by ATP hydrolysis to maintain the concentration gradients of sodium and potassium ions. MSC: Applying 29. Describe how synaptic signaling is influenced by the action of tranquilizers (such as Valium) compared to the effects of antidepressants (such as Prozac).

ANS: Tranquilizers promote inhibitory signaling at the synapse. They do this by making GABA-gated channels open more easily in response to the inhibitory neurotransmitter GABA. This channel, when open, allows Cl− to flow into the neuron, making the cell more difficult to depolarize. Antidepressants are used to enhance the neuronal signaling in neurons that have serotonin receptors. Prozac specifically blocks the reuptake of serotonin in the synaptic cleft. This results in a net increase in serotonin available for excitatory signaling to the postsynaptic neuron.

DIF: Moderate REF: 12.4 OBJ: 12.4.m Recall how the drugs Prozac and Ambien exert their neurological effects.

MSC: Understanding 30. For each of the following sentences, fill in the blank with the appropriate type of gating for the ion channel described.

You can use the same type of gating mechanism more than once.

1. The acetylcholine receptor in skeletal muscle cells is a/an __________ ion channel.

ANS: ligand-gated DIF: Easy REF: 12.3 OBJ: 12.3.h Define “gating” and list the types of conditions that can alter the opening and closing of ion channels. | 12.3.i Summarize the process by which ion channels in auditory hair cells allow the detection of sound. MSC: Understanding 2. __________ ion channels are found in the hair cells of the mammalian cochlea.

ANS: stress-gated DIF: Easy REF: 12.3 OBJ: 12.3.h Define “gating” and list the types of conditions that can alter the opening and closing of ion channels. | 12.3.i Summarize the process by which ion channels in auditory hair cells allow the detection of sound. MSC: Understanding 3. __________ ion channels in the mimosa plant propagate the leaf-closing response.

ANS: voltage-gated DIF: Easy REF: 12.3 OBJ: 12.3.h Define “gating” and list the types of conditions that can alter the opening and closing of ion channels. | 12.3.i Summarize the process by which ion channels in auditory hair cells allow the detection of sound. MSC: Understanding 4. __________ ion channels respond to changes in membrane potential.

ANS: voltage-gated DIF: Easy REF: 12.3 OBJ: 12.3.h Define “gating” and list the types of conditions that can alter the opening and closing of ion channels. | 12.3.i Summarize the process by which ion channels in auditory hair cells allow the detection of sound. MSC: Understanding 5. Many receptors for neurotransmitters are __________ ion channels.

ANS: ligand-gated DIF: Easy REF: 12.3 OBJ: 12.3.h Define “gating” and list the types of conditions that can alter the opening and closing of ion channels. | 12.3.i Summarize the process by which ion channels in auditory hair cells allow the detection of sound. MSC: Understanding 31. Fill in Table 12-31. In the “Type of transport” column, designate whether the transporter works by uniport, symport, or antiport mechanisms.

Table 12-31

ANS:

Table 12-31A

DIF: Easy REF: 12.2 OBJ: 12.2.f Differentiate between a symport, an antiport, and a uniport.

MSC: Remembering

CHAPTER 13 How Cells Obtain Energy from Food

THE BREAKDOWN AND UTILIZATION OF SUGARS AND FATS 13.1.a Define cell respiration.

13.1.b List three categories of food molecules that can serve as energy sources for cells, and identify the sugar whose breakdown generates most of the energy produced by the majority of animal cells.

13.1.c Summarize why cells need enzymes to maximize the energy that can be harvested from the oxidation of a fuel molecule such as glucose.

13.1.d Outline the three stages of catabolism, indicating where each takes place and identifying the stage’s major products.

13.1.e Recall the efficiency with which energy released from the oxidative breakdown of sugars and fats is captured by cells.

13.1.f

Summarize the amount of ATP energy invested and the amount recouped during the breakdown of a glucose

molecule during glycolysis.

13.1.g List the end products of glycolysis.

13.1.h Explain why energy is released by the splitting of glucose into pyruvate during glycolysis.

13.1.i

Present the reactions by which ATP is generated by substrate-level phosphorylation in steps 7 and 10 of glycoly-

sis.

13.1.j

List four enzyme types involved in glycolysis and indicate their functions.

13.1.k Recall how the generation of NADH in step 6 of glycolysis is linked to an oxidation reaction.

13.1.l

Describe the consequence of phosphorylating glucose in the first step of glycolysis.

13.1.m Contrast the fermentation pathway in an oxygen-starved muscle cell with the pathway in a yeast cell that is growing anaerobically.

13.1.n Identify the products that would build up in the cytosol if glycolysis were to take place in the absence of oxygen in cells that cannot carry out fermentation.

13.1.o Recall why—and in which step—glycolysis would halt in the absence of oxygen in cells that cannot carry out fermentation.

13.1.p Contrast anaerobic respiration with fermentation.

13.1.q Explain what it means for a bond to be described as having “high energy.”

13.1.r

Explain how the generation of 1,3-bisphosphoglycerate in step 6 of glycolysis drives the production of ATP in

step 7.

13.1.s Summarize how the pyruvate produced by glycolysis is converted into acetyl CoA, and state where the process takes place.

13.1.t

Review how and where the fatty acids derived from fat are converted into acetyl CoA.

13.1.u Recall where glycolysis and acetyl CoA production take place in aerobic prokaryotes.

13.1.v Explain why oxygen is required for the citric acid cycle to continue.

13.1.w Recount where the oxygen used during the oxidation of glucose to carbon dioxide ultimately goes.

13.1.x Identify the origin of the oxygen atoms used during the citric acid cycle to produce carbon dioxide.

13.1.y State where the citric acid cycle takes place in animal cells, in plant cells, and in prokaryotes.

13.1.z Outline the fate of the acetyl group carbons that enter the citric acid cycle.

13.1.aa List the activated carriers produced for each molecule of acetyl CoA that enters the citric acid cycle.

13.1.ab Recall in which steps of the citric acid cycle carbon dioxide is released.

13.1.ac Describe the role of oxaloacetate in the citric acid cycle.

13.1.ad Compare NADH with reduced flavin adenine dinucleotide.

13.1.ae Review how intermediates of glycolysis and the citric acid cycle can be used to synthesize other molecules needed by the cell.

13.1.af Summarize how oxygen is used by the electron transport chain.

13.1.ag Compare the number of ATP molecules generated by glycolysis with the number produced by the complete oxidation of glucose to water and carbon dioxide.

13.1.ah Outline briefly how the energy captured during glycolysis and the citric acid cycle is used to generate ATP during oxidative phosphorylation.

13.1.ai Review how the use of malonate, which inhibits the enzyme succinate dehydrogenase, revealed the cyclical nature of the citric acid cycle.

REGULATION OF METABOLISM 13.2.a Name several uses for the metabolite pyruvate.

13.2.b Summarize the ramifications of metabolites being substrates for a number of different enzymes.

13.2.c Define gluconeogenesis and identify the type of cell in mammals in which the reaction is likely to occur.

13.2.d State the amounts of ATP and GTP that must be consumed to produce a single molecule of glucose by gluconeogenesis.

13.2.e Identify the three reactions in glycolysis that are so energetically favorable that they are effectively irreversible, and review how gluconeogenesis circumvents these steps to convert pyruvate into glucose.

13.2.f

Summarize how the regulation of phosphofructokinase and fructose 1,6-bisphosphatase help control whether glu-

cose will be synthesized or oxidized.

13.2.g Outline how, when food is scarce, cells can break down glycogen to produce energy via glycolysis.

13.2.h Review how glycogen synthesis and breakdown is coordinated by feedback regulation.

13.2.i

Contrast the types of molecules broken down for energy after a meal versus after an overnight fast.

13.2.j

Compare the amount of energy provided by stored fats and stored glycogen.

13.2.k Recall where—and in what form—plant cells store their sugars and fats.

MULTIPLE CHOICE 1. Glycolysis is an anaerobic process used to catabolize glucose. What does it mean for this process to be anaerobic? a. No oxygen is required. b. No oxidation occurs. c. It takes place in the lysosome. d. Glucose is broken down by the addition of electrons.

ANS: A Glycolysis takes place in the cytosol, and although oxidation of glucose is taking place, no molecular oxygen is used.

DIF: Easy REF: 13.1 OBJ: 13.1.a Define cell respiration. MSC: Remembering 2. Which of the following stages in the breakdown of the piece of toast you had for breakfast generates the most ATP? a. the digestion of starch to glucose b. glycolysis c. the citric acid cycle d. oxidative phosphorylation ANS: D Oxidative phosphorylation produces about 28 ATP molecules.

DIF: Easy REF: 13.1 OBJ: 13.1.d Outline the three stages of catabolism, indicating where each takes place and identifying the major products of each stage. MSC: Understanding 3. The advantage to the cell of the gradual oxidation of glucose during cellular respiration compared with its combustion to CO2 and H2O in a single step is that a. more free energy is released for a given amount of glucose oxidized. b. no energy is lost as heat. c. energy can be extracted in usable amounts. d. more CO2 is produced for a given amount of glucose oxidized.

ANS: C The amount of free energy released by glucose oxidation is the same as combustion, and the amount of CO 2 released is the same.

DIF: Easy REF: 13.1 OBJ: 13.1.ag Compare the number of ATP molecules generated by glycolysis with the number produced by the complete oxidation of glucose to water and carbon dioxide. MSC: Understanding 4. Foods are broken down into simple molecular subunits for distribution and use throughout the body. Which type of simple subunits, listed below, is used preferentially as an energy source? a. simple sugars b. proteins c. free fatty acids d. glycerol ANS: A DIF: Easy REF: 13.1 OBJ: 13.1.b List three categories of food molecules that can serve as energy sources for

cells and identify the sugar whose breakdown generates most of the energy produced by a majority of animal cells.

MSC: Remembering 5. The final metabolite produced by glycolysis is a. acetyl CoA. b. pyruvate. c. 3-phosphoglycerate. d. glyceraldehyde 3-phosphate.

ANS: B DIF: Easy REF: Remembering OBJ: 13.1.g List the end products of glycolysis. MSC: Remembering 6. Glycolysis generates more stored energy than it expends. What is the net number of activated carrier molecules produced in this process (number and type of molecules produced minus the number of those molecules used as input)? a. 6 ATP, 2 NADH b. 4 ATP, 4 NADH c. 2 ATP, 2 NADH d. 4 ATP, 2 NADH ANS: C Although 4 ATP and 2 NADH are produced, 2 ATP are converted to ADP in the early steps of glycolysis. In contrast, NADH is not converted to NAD+ in any of the reactions.

DIF: Easy REF: 13.1 OBJ: 13.1.f Summarize the amount of ATP energy invested and the amount recouped during the breakdown of a molecule of glucose during glycolysis. MSC: Remembering 7. Which of the following steps or processes in aerobic respiration include the production of carbon dioxide? a. breakdown of glycogen b. glycolysis c. conversion of pyruvate to acetyl CoA d. oxidative phosphorylation ANS: C One carbon is oxidized and released as carbon dioxide when pyruvate is converted to acetyl CoA. DIF: Easy REF: 13.1 OBJ: 13.1.s Summarize how the pyruvate produced by glycolysis is converted into acetyl CoA and state where the process takes place. MSC: Remembering 8. In step 4 of glycolysis, a six-carbon sugar (fructose 1,6-bisphosphate) is cleaved to produce two three-carbon molecules (dihydroxyacetone phosphate and glyceraldehyde 3-phosphate). Which enzyme catalyzes this reaction? a. aldolase b. phosphoglucose isomerase c. enolase d. triose phosphate isomerase ANS: A DIF: Easy REF: 13.1 OBJ: 13.1.j List four enzyme types involved in glycolysis and indicate their functions.

MSC: Remembering 9. The conversion of glyceraldehyde 3-phosphate to 1,3 bisphosphoglycerate in step 6 of glycolysis generates a “highenergy” phosphoanhydride bond. Which of the following BEST describes what happens to that bond in step 7?

a. It is hydrolyzed to drive the formation of ATP. b. It is hydrolyzed to drive the formation of NADH. c. It is hydrolyzed to generate pyruvate. d. It is oxidized to CO2.

ANS: A DIF: Easy REF: 13.1 OBJ: 13.1.r Explain how the generation of 1,3-bisphosphoglycerate in step 6 of glycolysis drives the production of ATP in step 7. MSC: Understanding 10. Steps 7 and 10 of glycolysis result in substrate-level phosphorylation. Which of the following best describes this process? a. ATP is being hydrolyzed to phosphorylate the substrate. b. The energy derived from substrate oxidation is coupled to the conversion of ADP to ATP. c. Two successive phosphates are transferred, first to AMP, then to ADP, finally forming ATP. d. The substrate is hydrolyzed using ATP as an energy source.

ANS: B DIF: Easy REF: 13.1 OBJ: 13.1.i Present the reactions by which ATP is generated by substrate-level phosphorylation in steps 7 and 10 of glycolysis. MSC: Understanding 11. Which of the following descriptions best matches the function of a kinase? a. an enzyme that catalyzes the rearrangement of bonds within a single molecule b. an enzyme that catalyzes a change in the position of a specific chemical group within a single molecule c. an enzyme that catalyzes the oxidation of a molecule by removing a hydride ion d. an enzyme that catalyzes the addition of phosphate groups to other molecules ANS: D DIF: Easy REF: 13.1 OBJ: 13.1.j List four enzyme types involved in glycolysis and indicate their functions.

MSC: Remembering 12. Which of the following descriptions best matches the function of a mutase? a. an enzyme that catalyzes the rearrangement of bonds within a single molecule b. an enzyme that catalyzes a change in the position of a specific chemical group within a single molecule c. an enzyme that catalyzes the oxidation of a molecule by removing a hydride ion d. an enzyme that catalyzes the addition of phosphate groups to other molecules ANS: B DIF: Easy REF: 13.1 OBJ: 13.1.j List four enzyme types involved in glycolysis and indicate their functions.

MSC: Remembering 13. Which of the following descriptions best matches the function of a dehydrogenase? a. an enzyme that catalyzes the rearrangement of bonds within a single molecule b. an enzyme that catalyzes a change in the position of a specific chemical group within a single molecule c. an enzyme that catalyzes the oxidation of a molecule by removing a hydride ion d. an enzyme that catalyzes the addition of phosphate groups to other molecules ANS: C DIF: Easy REF: 13.1 OBJ: 13.1.j List four enzyme types involved in glycolysis and indicate their functions.

MSC: Remembering 14. On a diet consisting of nothing but protein, which of the following is the most likely outcome? a. loss of weight because amino acids cannot be used for the synthesis of fat b. muscle gain because the amino acids will go directly into building muscle

c. tiredness because amino acids cannot be used to generate energy d. excretion of more nitrogenous (ammonia-derived) wastes than with a more balanced diet ANS: D Ammonia is an end product of amino acid metabolism, but not when sugars and fats are metabolized, when you would expect more nitrogenous waste to be excreted.

DIF: Moderate REF: 13.1 OBJ: 13.1.b List three categories of food molecules that can serve as energy sources for cells and identify the sugar whose breakdown generates most of the energy produced by a majority of animal cells.

MSC: Applying 15. Which of the following processes do NOT take place in the mitochondria? a. citric acid cycle b. conversion of pyruvate to activated acetyl groups c. oxidation of fatty acids to acetyl CoA d. glycogen breakdown ANS: D DIF: Easy REF: 13.1 OBJ: 13.1.s Summarize how the pyruvate produced by glycolysis is converted into acetyl CoA and state where the process takes place. | 13.1.t Review how and where the fatty acids derived from fat are converted into acetyl CoA. | 13.1.y State where the citric acid cycle takes place in animal cells, plant cells, and in prokaryotes. MSC: Remembering 16. Which reaction does the enzyme phosphoglucose isomerase catalyze? a. glucose → glucose 6-phosphate b. fructose 6-phosphate → fructose 1,6-bisphosphate c. glucose 6-phosphate → fructose 6-phosphate d. glucose → glucose 1-phosphate ANS: C The isomerase part of the enzyme name indicates that it catalyzes an isomerization reaction, and the phosphoglucose part of the name indicates the type of substrate used.

DIF: Easy REF: 13.1 OBJ: 13.1.j List four enzyme types involved in glycolysis and indicate their functions.

MSC: Remembering 17. What purpose does the phosphorylation of glucose to glucose 6-phosphate by the enzyme hexokinase serve as the first step in glycolysis? a. It helps drive the uptake of glucose from outside the cell. b. It generates a high-energy phosphate bond. c. It converts ATP to a more useful form. d. It enables the glucose 6-phosphate to be recognized by phosphofructokinase, the next enzyme in the glycolytic pathway.

ANS: A DIF: Easy REF: 13.1 OBJ: 13.1.j List four enzyme types involved in glycolysis and indicate their functions.

MSC: Remembering 18. Which of the following cells rely exclusively on glycolysis to supply them with ATP? a. anaerobically growing yeast b. aerobic bacteria

c. skeletal muscle cells d. plant cells ANS: A All the other cells can perform oxidative phosphorylation to generate additional ATP. DIF: Easy REF: 13.1 OBJ: 13.1.p Contrast anaerobic respiration and fermentation. MSC: Understanding 19. In anaerobic conditions, skeletal muscle produces a. lactate and CO2. b. ethanol and CO2. c. lactate only. d. ethanol only.

ANS: C DIF: Easy REF: 13.1 OBJ: 13.1.m Contrast the fermentation pathways in an oxygen-starved muscle cell and in a yeast cell that is growing anaerobically. MSC: Remembering 20. Select the best option to fill in the blanks of the following statement: Fermentation is a/an __________ process that converts __________into carbon dioxide and __________. a. anaerobic, pyruvate, ethanol b. anaerobic, lactate, ethanol c. eukaryotic, glyceraldehyde 3-phosphate, ethanol d. prokaryotic, lactate, propanol ANS: A DIF: Easy REF: 13.1 OBJ: 13.1.p Contrast anaerobic respiration and fermentation.

MSC: Understanding 21. Glyceraldehyde 3-phosphate dehydrogenase operates by stripping a hydride ion from its substrate. Which molecule is the recipient of the proton and two electrons during this transfer? a. oxygen b. acetyl CoA c. NAD+ d. FADH ANS: C DIF: Easy REF: 13.1 OBJ: 13.1.k Recall how the generation of NADH in step 6 of glycolysis is linked to an oxidation reaction. MSC: Remembering 22. The first energy-generating steps in glycolysis begin when glyceraldehyde 3-phosphate undergoes an energetically favorable reaction in which it is simultaneously oxidized and phosphorylated by the enzyme glyceraldehyde 3phosphate dehydrogenase to form 1,3-bisphosphoglycerate, with the accompanying conversion of NAD + to NADH. In a second energetically favorable reaction catalyzed by a second enzyme, the 1,3-bisphosphoglycerate is then converted to 3-phosphoglycerate, with the accompanying conversion of ADP to ATP. Which of the following statements is TRUE about this reaction? a. The reaction glyceraldehyde 3-phosphate → 1,3-bisphosphoglycerate should be inhibited when levels of NADH fall. b. The ΔG° for the oxidation of the aldehyde group on glyceraldehyde 3-phosphate to form a carboxylic acid is more negative than the ΔG° for ATP hydrolysis. c. The energy stored in the phosphate bond of glyceraldehyde 3-phosphate contributes to driving the reaction

forward. d. The cysteine side chain on the enzyme is oxidized by NAD +.

ANS: B This is another way of stating that the energetically favorable oxidation of glyceraldehyde 3-phosphate provides sufficient energy to ultimately drive the energy-requiring step of ATP synthesis from ADP. DIF: Moderate REF: 13.1 OBJ: 13.1.r Explain how the generation of 1,3-bisphosphoglycerate in step 6 of glycolysis drives the production of ATP in step 7. MSC: Understanding 23. The simultaneous oxidation and phosphorylation of glyceraldehyde 3-phosphate forms a highly reactive covalent thioester bond between a cysteine side chain (reactive group −SH) on the enzyme (glyceraldehyde 3-phosphate dehydrogenase) and the oxidized intermediate (see the arrow in Figure 13-23A). If the enzyme had a serine (reactive group −OH) instead of a cysteine at this position, which could form only a much lower energy bond to the oxidized substrate (see the arrow in Figure 13-23B), how might this new enzyme act?

Figure 13-23

a. It would oxidize the substrate and phosphorylate it without releasing it. b. It would oxidize the substrate but not release it. c. It would phosphorylate the substrate on the 2 position instead of the 1 position. d. It would behave just like the normal enzyme.

ANS: B The phosphorylation and release of the product from the normal enzyme is possible because a phosphate molecule can attack the high-energy thioester bond formed between the oxidized substrate and enzyme. If the bond between the oxidized substrate and enzyme is of much lower energy, the enzyme will not be able to transfer the oxidized substrate to a phosphate group, and substrate and enzyme will remain covalently bound.

DIF: Hard REF: 13.1 OBJ: 13.1.q Explain what it means for a bond to be described as having “high energy.”

MSC: Applying 24. In the absence of oxygen, yeast cells can switch to a completely anaerobic metabolism called fermentation. Which of the following is a final product of fermentation in yeast?

Figure 13-24

a. Figure (a) b. Figure (b) c. Figure (c) d. Figure (d) ANS: B Ethanol and CO2 are the products of fermentation in yeast when grown anaerobically. Figure (a) is lactate, (c) is acetaldehyde, and (d) is pyruvate.

DIF: Easy REF: 13.1 OBJ: 13.1.p Contrast anaerobic respiration and fermentation. MSC: Remembering 25. Pyruvate must move from the cytosol into the mitochondria, where it is oxidized to form CO 2 and acetyl CoA by the pyruvate dehydrogenase complex. How many different enzymes and what total number of polypeptides, respectively, are required to perform this oxidation process in the mitochondrion? a. 1; 60 b. 3; 3 c. 3; 26 d. 3; 60 ANS: D DIF: Moderate REF: 13.1 OBJ: 13.1.s Summarize how the pyruvate produced by glycolysis is converted into acetyl CoA and state where the process takes place. MSC: Remembering 26. The citric acid cycle is a critical sequence of reactions for energy production, which take place in the matrix of the mitochondria. The reaction cycle requires materials from the cytosol to be converted into acetyl CoA, which represents the starting point of a new cycle. Which of the following statements about acetyl CoA is TRUE? a. Amino acids can be converted into acetyl CoA. b. Pyruvate is converted into acetyl CoA in the cytosol. c. Triacylglycerol molecules are transported into the mitochondrial matrix and cleaved by lipases to produce acetyl CoA. d. Oxaloacetate is converted directly into acetyl CoA to feed the citric acid cycle.

ANS: A DIF: Easy REF: 13.1 OBJ: 13.1.s Summarize how the pyruvate produced by glycolysis is converted into acetyl CoA and state where the process takes place. | 13.1.t Review how and where the fatty acids derived from fat are converted into acetyl CoA. MSC: Understanding 27. The citric acid cycle is a series of oxidation reactions that removes carbon atoms from substrates in the form of CO 2.

Where do the oxygen atoms in the carbon dioxide molecules come from? a. water

b. phosphates c. molecular oxygen d. acetyl CoA ANS: A DIF: Easy REF: 13.1 OBJ: 13.1.x Identify the origin of the oxygen atoms used during the citric acid cycle to produce carbon dioxide. MSC: Remembering 28. Fatty acids can easily be used to generate energy for the cell. Which of the following fatty acids will yield more energy? Explain your answer. a. CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH = CH-COOH b. CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-COOH c. CH3-CH = CH-CH2-CH2-CH2-CH2-CH = CH-COOH d. CH3-CH2-CH2-CH2-CH2-CH2-CH2-COOH ANS: B This fatty acid has 10 carbons and has a completely saturated hydrocarbon tail. The total number of carbons determines how many acetyl CoA molecules can be derived from it; in this case it is 5. The fact that all hydrocarbon bonds are saturated means that an FADH2 molecule will be generated in the first step of the fatty acid cycle, converting a carbon–carbon single bond into a carbon–carbon double bond. If that bond is already reduced, the cycle begins at the second step. As a result, FADH2 is not generated in that cycle, lowering the total number of cofactors that can be used by the electron-transport chain to generate a proton gradient.

DIF: Hard REF: 13.1.t Review how and where the fatty acids derived from fat are converted into acetyl CoA. OBJ: 13.1 MSC: Analyzing 29. The citric acid cycle is a series of oxidation reactions that removes carbon atoms from substrates in the form of CO 2.

Once a molecule of acetyl CoA enters the citric acid cycle, how many complete cycles are required for both of the carbon atoms in its acetyl group to be oxidized to CO2? a. 1 b. 2 c. 3 d. 4 ANS: D The carbons from acetyl CoA are carried through the entire first reaction cycle. In the second cycle, one of these carbons is oxidized to CO2. Two more cycles later, the second carbon molecule from that same acetyl CoA is oxidized to CO2.

DIF: Easy REF: 13.1 OBJ: 13.1.z Outline the fate of the acetyl group carbons that enter the citric acid cycle.

MSC: Understanding 30. In step 1 of the citric acid cycle, citrate is generated by the enzyme citrate synthase. The enzyme combines the twocarbon acetyl group from acetyl CoA and the four-carbon oxaloacetate. What is the source of energy that drives this reaction forward? a. a high-energy phosphodiester bond b. a transfer of high-energy electrons c. a high-energy thioester bond d. the heat of molecular collision

ANS: C DIF: Moderate REF: 13.1 OBJ: 13.1.q Explain what it means for a bond to be described as having “high energy.” MSC: Understanding 31. In step 2 of the citric acid cycle, the enzyme aconitase generates isocitrate from citrate. Which of the following statements about this reaction is TRUE? a. There is a substantial free-energy difference between the reactants and products of this reaction. b. The unbonded electrons from hydroxide ions provide energy for this reaction. c. The aconitase enzyme functions as a mutase in this reaction. d. The reaction sequence first generates one molecule of water and then consumes one molecule of water.

ANS: D DIF: Easy REF: 13.1 OBJ: 13.1.z Outline the fate of the acetyl group carbons that enter the citric acid cycle.

MSC: Remembering 32. In step 3 of the citric acid cycle, the oxidation of isocitrate and the production of CO 2 are coupled to the reduction of NAD+, generating NADH and an α-ketoglutarate molecule. In the isocitrate molecule shown in Figure 13-32, which carbon is lost as CO2 and which is converted to a carbonyl carbon?

Figure 13-32

a. 4 and 6 b. 6 and 5 c. 5 and 4 d. 6 and 4 ANS: D DIF: Moderate REF: 13.1 OBJ: 13.1.ab Recall in which steps of the citric acid cycle carbon dioxide is released. MSC: Understanding 33. In step 4 of the citric acid cycle, the reduction of NAD + to NADH is coupled to the generation of CO 2 and the formation of a high-energy thioester bond. Which molecule provides the sulfhydryl group necessary to form the thioester bond? a. pyruvate b. acetyl CoA c. CoA d. cysteine side chain in the catalytic pocket ANS: C DIF: Moderate REF: 13.1 OBJ: 13.1.z Outline the fate of the acetyl group carbons that enter the citric acid cycle. | 13.1.aa List the activated carriers produced for each molecule of acetyl CoA that enters the citric acid c ycle. | 13.1.ab Recall in which steps of the citric acid cycle carbon dioxide is released.

MSC: Understanding 34. In step 4 of the citric acid cycle, the reduction of NAD + to NADH is coupled to the generation of CO 2 and the

formation of a high-energy thioester bond. The energy of the thioester bond is harnessed in step 5. What is the energy used for? a. to generate a molecule of GTP b. to generate a molecule of ATP c. to generate a proton gradient d. to generate a molecule of NADH ANS: A DIF: Moderate REF: 13.1 OBJ: 13.1.q Explain what it means for a bond to be described as having “high energy.” MSC: Understanding 35. Step 6 of the citric acid cycle is catalyzed by succinate dehydrogenase. Keeping in mind that dehydrogenases catalyze redox reactions, which are the products of the reaction in which succinate is oxidized? a. fumarate, NADH b. fumarate, FADH2 c. fumarate, FADH2 d. succinyl CoA, NADH ANS: B DIF: Easy REF: 13.1 OBJ: 13.1.z Outline the fate of the acetyl group carbons that enter the citric acid cycle.

MSC: Remembering 36. In the final step of the citric acid cycle, oxaloacetate is regenerated through the oxidation of malate and this is coupled with the production of which other molecule? a. FADH b. NADH c. GTP d. CO2 ANS: B DIF: Easy REF: 13.1 OBJ: 13.1.ai Review how the use of malonate, which inhibits the enzyme succinate dehydrogenase, revealed the cyclical nature of the citric acid cycle. [how we know] MSC: Remembering 37. The oxygen-dependent reactions required for cellular respiration were originally thought to occur in a linear pathway.

By using a competitive inhibitor for one enzyme in the pathway, investigators discovered that these reactions occur in a cycle. What compound served as the inhibitor? a. malonate b. malate c. fumarate d. succinate ANS: A DIF: Easy REF: 13.1 OBJ: 13.1.ai Review how the use of malonate, which inhibits the enzyme succinate dehydrogenase, revealed the cyclical nature of the citric acid cycle. MSC: Remembering 38. The oxygen-dependent reactions required for cellular respiration were originally thought to occur in a linear pathway.

By using a competitive inhibitor for one enzyme in the pathway, investigators discovered that these reactions occur in a cycle. Which enzyme was inhibited? a. aconitase b. isocitrate dehydrogenase c. malate dehydrogenase

d. succinate dehydrogenase ANS: D DIF: Easy REF: 13.1 OBJ: 13.1.ai Review how the use of malonate, which inhibits the enzyme succinate dehydrogenase, revealed the cyclical nature of the citric acid cycle. MSC: Remembering 39. The oxygen-dependent reactions required for cellular respiration were originally thought to occur in a linear pathway.

By using a competitive inhibitor for one enzyme in the pathway, investigators discovered that these reactions occur in a cycle. Which product in the reaction pathway builds up when the inhibitor is added? a. citrate b. succinate c. fumarate d. malate ANS: B DIF: Easy REF: 13.1 OBJ: 13.1.ai Review how the use of malonate, which inhibits the enzyme succinate dehydrogenase, revealed the cyclical nature of the citric acid cycle. MSC: Remembering 40. In the final stage of the oxidation of food molecules, a gradient of protons is formed across the inner mitochondrial membrane, which is normally impermeable to protons. If cells were exposed to an agent that causes the membrane to become freely permeable to protons, which of the following effects would you expect to observe? a. The ratio of ATP to ADP in the cytoplasm would fall. b. NADH would build up. c. Carbon dioxide production would cease. d. The consumption of oxygen would fall.

ANS: A If the inner mitochondrial membrane became permeable to protons, the electron-transport chain would continue to oxidize NADH to NAD+, transport electrons, and pump protons, so the consumption of oxygen would not fall. However, without the energy stored in a proton gradient, there is no way of driving the synthesis of ATP. DIF: Moderate REF: 13.1 OBJ: 13.1.ah Outline briefly how the energy captured during glycolysis and the citric acid cycle is used to generate ATP during oxidative phosphorylation. MSC: Applying 41. Pyruvate is an important metabolic intermediate that can be converted into several other compounds, depending on which enzyme is catalyzing the reaction. Which of the following CANNOT be produced from pyruvate in a single enzyme-catalyzed reaction? a. lactate b. oxaloacetate c. citrate d. alanine ANS: C DIF: Moderate REF: 13.2 OBJ: 13.2.a Name several uses for the metabolite pyruvate.

MSC: Understanding 42. When glucose is being used up and not replaced from food intake, the blood sugar level can be maintained by synthesizing glucose from smaller molecules such as pyruvate or lactate. This process is called gluconeogenesis.

Which organ is principally responsible for supplying glucose to the rest of the body when glucose reserves are low? a. liver

b. pancreas c. spleen d. gall bladder ANS: A DIF: Easy REF: 13.2 OBJ: 13.2.c Define gluconeogenesis and identify the type of cell in mammals in which the reaction is likely to occur. MSC: Remembering 43. Step 3 in glycolysis requires the activity of phosphofructokinase to convert fructose 6-phosphate into fructose 1,6bisphosphate. Which of the following molecules is an allosteric inhibitor of this enzyme? a. Pi b. AMP c. ADP d. ATP ANS: D When ATP levels are high, the cell does not need to break down more glucose to generate ATP. Thus, with ATP acting as an allosteric inhibitor of a key glycolytic step, there is a rapid on/off switch for this pathway.

DIF: Easy REF: 13.2 OBJ: 13.2.f Summarize how the regulation of phosphofructokinase and fructose 1,6bisphosphatase help control whether glucose will be synthesized or oxidized. MSC: Remembering 44. The conversion of fructose 1,6-bisphosphate to fructose 6-phosphate is catalyzed by a fructose 1,6-bisphosphatase and is one of the final steps in gluconeogenesis. Which of the following molecules is an allosteric activator of this enzyme? a. Pi b. AMP c. ADP d. ATP ANS: D DIF: Easy REF: 13.2 OBJ: 13.2.f Summarize how the regulation of phosphofructokinase and fructose 1,6bisphosphatase help control whether glucose will be synthesized or oxidized. MSC: Remembering 45. Which of the following polymers of glucose is used as a vehicle to store energy reserves in animal cells? a. glucagon b. glycogen c. starch d. glycerol ANS: B DIF: Easy REF: 13.2 OBJ: 13.2.g Outline how, when food is scarce, cells can break down glycogen to produce energy via glycolysis. MSC: Remembering 46. The intermediates of the citric acid cycle are constantly being depleted because they are used to produce many of the amino acids needed to make proteins. The enzyme pyruvate carboxylase converts pyruvate to oxaloacetate to replenish these intermediates. Bacteria, but not animal cells, have additional enzymes that can carry out the reaction acetyl CoA + isocitrate → oxaloacetate + succinate. Which of the following compounds will NOT support the growth of animal cells when used as the major source of carbon in food, but will support the growth of nonphotosynthetic bacteria? a. pyruvate

b. glucose c. fatty acids d. fructose ANS: C In oxidative metabolism, fatty acids can only be converted to acetyl CoA, which is completely oxidized to carbon dioxide through the citric acid cycle. In addition, bacteria can use some of this acetyl CoA as a source of carbon atoms to replenish the components of the citric acid cycle, whereas animals cannot.

DIF: Moderate REF: 13.1 OBJ: 13.1.t Review how and where the fatty acids derived from fat are converted into acetyl CoA. | 13.1.ac Describe the role of oxaloacetate in the citric acid cycle. MSC: Applying 47. Pyruvate can be converted into many other molecules by various biosynthetic and metabolic pathways, which makes it a central hub in the regulation of cellular metabolism. Which of the following molecules is NOT made from pyruvate? a. oxaloacetate b. ethanol c. lactate d. NADH ANS: D Pyruvate cannot be converted into NADH, but it can be converted into the other metabolites in one or two steps.

DIF: Easy REF: 13.2 OBJ: 13.2.a Name several uses for the metabolite pyruvate. MSC: Understanding 48. In humans, glycogen is a more useful food-storage molecule than fat because a. a gram of glycogen produces more energy than a gram of fat. b. it can be utilized to produce ATP under anaerobic conditions, whereas fat cannot. c. it binds water and is therefore useful in keeping the body hydrated. d. for the same amount of energy storage, glycogen occupies less space in a cell than does fat.

ANS: B The breakdown of glycogen to glucose does not require oxygen; the glucose can then enter glycolysis and generate ATP by a fermentation process that produces lactic acid. In contrast, fats are broken down to acetyl CoA; this must enter the citric acid cycle, which requires oxygen to keep turning.

DIF: Moderate REF: 13.2 OBJ: 13.2.j Compare the amount of energy provided by stored fats and stored glycogen.

MSC: Analyzing 49. The concentration of H+ ions inside the mitochondrial matrix is lower than it is in the cytosol or the mitochondrial intermembrane space. What would be the IMMEDIATE effect of a membrane-permeable compound that carries and releases protons into the mitochondrial matrix? a. inhibition of the electron-transport chain b. inhibition of ATP synthesis c. increased import of ADP into the matrix d. inhibition of the citric acid cycle ANS: B DIF: Moderate REF: 13.2 OBJ: 13.2.h Review how glycogen synthesis and breakdown is coordinated by feedback regulation. MSC: Applying

50. Which of the following descriptions best matches the function of an isomerase? a. an enzyme that catalyzes the rearrangement of bonds within a single molecule b. an enzyme that removes a specific chemical group from a molecule c. an enzyme that catalyzes the oxidation of a molecule by removing a hydride ion d. an enzyme that catalyzes the addition of phosphate groups to other molecules ANS: A and B DIF: Easy REF: 13.1 OBJ: 13.1.j List four enzyme types involved in glycolysis and indicate their functions. MSC: Remembering

SHORT ANSWER 1. Provide a brief description of the process by which the body derives energy from food. Include in your description the three stages of catalysis, and delineate the input and output that connect each stage.

ANS: In the first stage of catabolism, polymers are broken down into smaller subunits in the digestive system, which stretches from the mouth to the gut. Proteins are converted to amino acids, fats to fatty acids and glycerol, and carbohydrates simple sugars, including the monosaccharide glucose. In the second stage of catabolism, these simple subunits are further broken down to generate the activated carrier acetyl CoA. Acetyl CoA is the molecular input for the third stage of metabolism. In the last stage of catabolism, the acetyl CoA is oxidized to CO2, coupled to the production of large amounts of ATP, which is used as chemical energy for the cell. DIF: Easy REF: 13.1 OBJ: 13.1.d Outline the three stages of catabolism, indicating where each takes place and identifying the major products of each stage. MSC: Understanding 2. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. CO2 and H2O are generated during the oxidation of food molecules.

B. Activated carrier molecules store heat energy for the cell to use later.

C. Catabolism is a general term that refers to the processes by which large molecules are synthesized from smaller molecules.

D. The oxidation of sugar is an energetically favorable process.

ANS: A. True.

B. False. Activated carriers have high-energy bonds that can drive other reactions when broken. Heat may be released during these reactions and may increase the reaction rates, but is not a form of energy that is stored in biological systems.

C. False. Catabolism comprises the metabolic reactions that are involved in breaking large molecules into smaller molecules. Anabolism encompasses the reverse types of reactions: synthesizing larger molecules from smaller molecules.

D. True.

DIF: Easy REF: 13.1 OBJ: 13.1.q Explain what it means for a bond to be described as having “high energy.”

MSC: Evaluating 3. Although ATP and NADH are both important activated carrier molecules, ATP hydrolysis provides the direct molecular energy for most biochemical reactions. Why do the mitochondria also need to generate high levels of NADH? ANS: NADH is an activated carrier molecule used as a cofactor for many enzymes that catalyze redox reactions.

NADH also donates electrons to the electron-transport chain, which is essential for the production of ATP. DIF: Moderate REF: 13.1 OBJ: 13.1.ah Outline briefly how the energy captured during glycolysis and the citric acid cycle is used to generate ATP during oxidative phosphorylation. MSC: Evaluating 4. Compare and contrast the catabolism of carbohydrates, proteins and fats. Specifically, highlight the similarities and differences in how they are broken down (extracellular, and intracellular) for conversion into acetyl-CoA, the entry point into the citric acid cycle.

ANS:

The breakdown of all carbohydrates, proteins, and fats into smaller subunits occurs first in the stomach and small intestines.

Stage 1

Carbohydrates are converted into simple sugars.

Proteins are broken down into amino acids.

Fats are cleaved into glycerol and fatty acid chains.

DIF: Hard

Once inside the cell, each of the simpler subunits are broken down further to produce 2-carbon acetyl groups REF: joined to coenzyme A (via its SH group) to become acetyl CoA, which is a direct substrate for oxidation in the 13.1 citric acid cycle.

OBJ: Stage Glucose must be converted to Both amino acids and fatty acids are transported directly into the 13.1.b 2 pyruvate for transport into the mitochondrial matrix for conversion to acetyl CoA. mitochondrial matrix. This hapList pens in the cytosol during a prothree cess called glycolysis, which produces two pyruvate molecategocules per glucose. ries of Two pyruvate molecules are Amino acids require a deamina- Fatty acids are combined direct-food transported into the mitochontion step prior to combination ly with coenzyme A, before undrial matrix and converted into with coenzyme A. dergoing a series of stepwise moleacetyl CoA. oxidation reactions that convert cules the entire fatty acid chain to the that can serve as energy sources for cells and identify the sugar whose breakdown generates most of the energy 2-carbon segments in acetyl produced by a majority of animal cells. | 13.1.t Review how and where the fatty CoA.acids d erived from fat are converted into acetyl CoA.

MSC: Analyzing

5. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

Biological systems __________ hydrocarbons in stepwise __________ reactions. Glucose is the preferred energy source in eukaryotes, and must be converted into __________ before it can be transported across the __________ mitochondrial membrane for use in the citric acid cycle. The citric acid cycle is critical for the mitochondrion’s subsequent production of __________.

ATP

catabolic

inner

outer

permeable

acetyl CoA

electrons

isomerize

oxidize

protons

anabolic

glycolysis

kinase

peptides

pyruvate

anaerobic

GTP

NADH

ANS: Biological systems oxidize hydrocarbons in stepwise catabolic reactions. Glucose is the preferred energy

source in eukaryotes, and must be converted into pyruvate before it can be transported across the inner mitochondrial membrane for further oxidation in the citric acid cycle. The citric acid cycle is critical for the mitochondrion’s subsequent production of ATP. DIF: Easy REF: 13.1 OBJ: 13.1.g List the end products of glycolysis. MSC: Understanding 6. Explain how the generation of ATP by oxidative phosphorylation differs from ATP generation by substrate-level phosphorylation.

ANS: During the oxidation of acetyl CoA that occurs in the citric acid cycle, NADH and FADH 2 are generated. In oxidative phosphorylation, the high-energy electrons from these two cofactors are passed along the electron-transport chain in the inner mitochondrial membrane, finally being used to convert oxygen gas to two water molecules. Much of the energy released by this chain of electron transfers is harnessed to pump protons out of the mitochondrial matrix. It is the resulting chemiosmotic gradient across the inner mitochondria membrane that powers the production of ATP from ADP by ATP synthase.

In the case of substrate-level phosphorylation, high-energy phosphate groups produced during oxidation of a substrate are transferred from the enzyme substrate directly to ADP, producing ATP. DIF: Moderate REF: 13.1 OBJ: 13.1.i Present the reactions by which ATP is generated by substrate-level phosphorylation in steps 7 and 10 of glycolysis. | 13.1.ah Outline briefly how the energy captured during glycolysis and the citric acid cycle is used to generate ATP during oxidative phosphorylation. MSC: Analyzing 7. Name the catabolic process that use substrate-level phosphorylation, identify where it takes place and how it is connected to oxidative phosphorylation.

ANS: Glycolysis employs substrate-level phosphorylation. The process occurs in the cytosol and generates a net total of two ATP molecules for each glucose molecule. The connection to oxidative phosphorylation is through the product of glycolysis: 2 pyruvate molecules for each glucose molecule. Pyruvate is shuttled into the mitochondria and converted into acetyl CoA. This acetyl CoA is oxidized in the citric acid cycle to generate the high-energy electrons in NADH and FADH2 that feed the electron-transport chain in the mitochondrial inner membrane that drives oxidative phosphorylation.

DIF: Moderate REF: 13.1 OBJ: 13.1.i Present the reactions by which ATP is generated by substrate-level phosphorylation in steps 7 and 10 of glycolysis. | 13.1.ah Outline briefly how the energy captured during glycolysis and the citric acid cycle is used to generate ATP during oxidative phosphorylation. MSC: Analyzing 8. Where does oxidative phosphorylation take place and what materials are the reactants and the products in this process?

ANS: Oxidative phosphorylation takes place in mitochondria as part of a multistage oxidation process. First, the citric acid cycle generates NADH and FADH2, which donate their high-energy electrons to the electron-transport chain.

These electrons ultimately reduce molecular oxygen to water, and the energy of their oxidation is used to pump protons across a membrane. The resulting proton gradient is harnessed by the enzyme ATP synthase to drive the production of ATP from ADP and Pi. DIF: Moderate REF: 13.1 OBJ: 13.1.ah Outline briefly how the energy captured during glycolysis and the citric acid cycle is used to generate ATP during oxidative phosphorylation.

MSC: Analyzing 9. Anaerobically growing yeast further metabolizes the pyruvate produced by glycolysis to CO 2 and ethanol as part of a

series of fermentation reactions.

A. What other important reaction occurs during this fermentation step?

B. Why is this reaction (that is, the answer to part A) essential for the anaerobically growing cell?

ANS: A. NADH → NAD+.

B. Under anaerobic conditions, it is the only means of regenerating the NAD + required for glycolysis, the main energygenerating pathway of an anaerobically growing yeast cell.

DIF: Easy REF: 13.1 OBJ: 13.1.p Contrast anaerobic respiration and fermentation. MSC: Understanding 10. It can be useful to analyze the steps of glycolysis with respect to the four basic types of enzymes required by this central catabolic pathway and to consider whether each enzyme produces or harvests the energy of an activated carrier. For each step of glycolysis (see Figure 13-60 or Panel 13-1), indicate which type of enzyme (of the four listed below and in Table 13-60) is required—or if none apply. Also, indicate whether an activated energy carrier is involved, and, if so, how.

Step 1 __________ Step 2 __________ Step 3 __________ Step 4 __________ Step 5 __________ Step 6 __________ Step 7 __________ Step 8 __________ Step 9 __________ Step 10 __________ Enzyme types: kinase, isomerase, mutase, dehydrogenase ANS: Step 1 kinase, energy in the form of ATP consumed Step 2 isomerase Step 3 kinase, energy in the form of ATP consumed Step 4 none of the above Step 5 isomerase Step 6 dehydrogenase, energy in the form of NADH produced Step 7 kinase (catalyzing its reverse reaction)*; energy in the form of ATP produced Step 8 mutase Step 9 none of the above Step 10 kinase (catalyzing its reverse reaction)*; energy in the form of ATP produced *During glycolysis, both phosphoglycerate kinase and pyruvate kinase remove rather than add phosphates. However, if the appropriate conditions are established in vitro, these enzymes can also catalyze the backward reaction that adds a phosphate—thus functioning like a typical kinase.

DIF: Easy REF: 13.1 OBJ: 13.1.j List four enzyme types involved in glycolysis and indicate their functions.

MSC: Remembering

11. In the reaction cycle involved in the oxidation of pyruvate, what are the advantages of having three enzyme activities contained in a single large complex instead of having three smaller and physically independent enzymes?

ANS: By co-localizing three enzyme activities in a large, layered complex, the substrates are already bound and properly positioned for rapid enzyme catalysis, and the free energy released by one reaction can be readily harnessed for the next.

DIF: Moderate REF: 13.1 OBJ: 13.1.s Summarize how the pyruvate produced by glycolysis is converted into acetyl CoA and state where the process takes place. MSC: Understanding 12. Glycolysis and the citric acid cycle comprise two different sets of oxidation reactions. The reaction sequence for glycolysis is linear, whereas the reaction sequence for the citric acid cycle forms a circle. How does this difference in the arrangement of reactions influence the rate of these processes when an excess amount of a single intermediate is added?

ANS: Primarily, what is seen is that the citric acid cycle occurs more rapidly after the addition of any one of the intermediates. This means that if one intermediate is added, levels of all of them increase. In glycolysis, the intermediates downstream of the intermediate being added will be affected.

DIF: Moderate REF: 13.1 OBJ: 13.1.d Outline the three stages of catabolism, indicating where each takes place and identifying the major products of each stage. MSC: Applying 13. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. During glycolysis, glucose molecules are broken down to yield CO 2 and H2O. B. The cleavage of fructose 1,6-bisphosphate yields two molecules of glyceraldehyde 3-phosphate.

C. Anaerobic respiration is not the same as fermentation, as only the former requires an electron-transport chain.

D. When subjected to anaerobic conditions, glycolysis in mammalian cells continues and causes a buildup of pyruvate in the cytosol.

E. The pyruvate dehydrogenase complex catalyzes three different, but linked, enzymatic reactions.

F. Amino acids can be transported into the mitochondria and converted into acetyl CoA. ANS: A. False. At the end of a series of the 10 different reactions involved in glycolysis, the final products are two molecules of pyruvate. Pyruvate will later be broken down into CO2 and H2O in the citric acid cycle.

B. False. When fructose 1,6-bisphosphate is cleaved, the products are dihydroxyacetone phosphate and glyceraldehyde 3-phosphate. Only after the subsequent isomerization of dihydroxyacetone phosphate is the second molecule of glyceraldehyde 3-phosphate produced.

C. True.

D. False. Under anaerobic conditions, mammalian cells convert pyruvate to lactate in a fermentation process. The lactate is subsequently excreted from the cell.

E. True.

F. True.

DIF: Easy REF: 13.1 OBJ: 13.1.g List the end products of glycolysis. | 13.1.p Contrast anaerobic respiration and fermentation. MSC: Evaluating 14. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase may be used more than once.

ADP

FADH2

H+

moderately

protons

ATP

favorable

higher

NADH

severely

cytosol

glucose

lower

NAD+

slightly

electrons

GTP

matrix

Pi

unfavorable

High-energy electrons are transferred through a series of molecules, and the energy released during these transfers is used to generate a gradient of __________, or __________. Because their concentration is __________ outside than inside the mitochondria, the flow of __________, or __________, down the concentration gradient is energetically very __________ and can thus be coupled to the production of ATP from ADP. Thus, oxidative phosphorylation refers to the oxidation of __________ and __________ molecules and the phosphorylation of __________. Without this process, the yield of ATP from each glucose molecule would be __________ decreased.

ANS: High-energy electrons are transferred through a series of molecules, and the energy released during these transfers is used to generate a gradient of protons, or H+. Because their concentration is higher outside than inside the mitochondria, the flow of protons, or H+, down the concentration gradient is energetically very favorable and can thus be coupled to the production of ATP from ADP. Thus, oxidative phosphorylation refers to the oxidation of NADH and FADH2 molecules and the phosphorylation of ADP. Without this process, the yield of ATP from each glucose molecule would be severely decreased.

DIF: Easy REF: 13.1 OBJ: 13.1.ah Outline briefly how the energy captured during glycolysis and the citric acid cycle is used to generate ATP during oxidative phosphorylation. MSC: Understanding 15. In step 7 of the citric acid cycle, fumarase catalyzes the addition of a water molecule to a carbon–carbon double bond (see Panel 13-2). Can this be considered an oxidation reaction? Explain your answer.

ANS: Yes. In the citric acid cycle, the overall process oxidizes carbon molecules to produce carbon dioxide. Although, fumarase does not directly remove electrons from its substrate (as do the enzymes that catalyze steps 3, 4, 6, and 8), the addition of water across the double bond in fumarate leaves one of its carbon atoms in a more oxidized state. This is because one of the carbons becomes bonded to an oxygen atom. In this arrangement, the carbon atom shares its electrons unequally across the new bond (see Figure 3-65). DIF: Moderate REF: 13.1 OBJ: 13.1.z Outline the fate of the acetyl group carbons that enter the citric acid cycle. MSC: Understanding 16. The oxidative reactions of cellular respiration were the focus of intense study in the 1930s. These reactions are represented in a linear pathway, as they were thought to occur. Each product is designated as a lettered compound (A through H) in Figure 13-16.

Figure 13-16

A. What was the first observation that Krebs made when he added malonic acid to the minced muscle samples, and what was his conclusion about how and where it was acting in the reactions he was studying?

B. What happens when the malonate block is introduced and subsequently compound A is added in excess? What is the result if compound G is added after the block, instead of A? How did Krebs attempt to reconcile these two results?

C. What additional observation led Krebs to hypothesize that what was previously thought to be a linear sequence of

reactions is actually a cyclic sequence of reactions? How did this idea further explain the earliest observations that the addition of any single compound in the pathway greatly increases oxygen uptake by the muscle tissue?

ANS: A. When malonic acid was added to the minced muscle samples, compound E accumulated and the concentrations of compounds F, G, and H were severely diminished. Krebs interpreted this to mean that malonic acid blocked one of the reactions by inhibiting the enzyme involved. The inhibitory effect of malonic acid on the reaction cycle became known as the malonate block.

B. After the malonate block is in place, if compound A is added in excess, compound E will accumulate, inhibiting the pathway at this point and preventing the conversion of E to F. If the experiment is repeated and compo und G is added in excess instead of compound A, the result is the same. The first explanation for this surprising r esult was that the F → G → H pathway somehow also led to an intermediate that could be converted into co mpound E. C. When Krebs combined pyruvate and oxaloacetate (compound H) with the muscle suspensions, he noticed an immediate production of citrate (compound A). This result made it logical to conclude that H is recycled and used in the first reaction. The model of a reaction cycle explained all the other previous results, including the observation that any one compound can be added to increase O2 uptake. A large amount of any intermediate will drive the production of the others, as long as pyruvate is available. The cofactors produced are used in the electrontransport chain, which terminates with the splitting of oxygen and its reduction to form water.

DIF: Hard REF: 13.1 OBJ: 13.1.ai Review how the use of malonate, which inhibits the enzyme succinate dehydrogenase, revealed the cyclical nature of the citric acid cycle. MSC: Understanding 17. Do you expect the cell to produce more ATP from one glucose molecule or from one fatty acid molecule? Explain your answer.

ANS: More ATP is produced from fat catabolism than from glucose catabolism. Most ATP is generated in the mitochondria, and the amount depends on the production of the NADH and FADH 2 cofactors in the Krebs cycle. The Krebs cycle relies on the input of acetyl CoA. Each glucose molecule can be converted into two acetyl CoA molecules.

A molecule of fat will have three fatty acid chains, with an average length of 12–16 carbons. Even if we assume very short fatty acid chains of six carbons each (the length of a glucose molecule), this would mean the production of three acetyl CoA molecules per chain, and nine total for the triacylglycerol.

DIF: Moderate REF: 13.1OBJ: 13.1.t Review how and where the fatty acids derived from fat are converted into acetyl CoA. MSC: Applying 18. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. The proteins of the electron-transport chain remove a pair of high-energy electrons from the cofactors NADH and FADH2, after which the electrons move across the inner mitochondrial membrane to maintain the voltage gradient.

B. Gluconeogenesis is a linear reaction pathway that the cell employs to generate glucose from pyruvate and is exactly the reverse of the reactions in the glycolytic pathway.

C. With respect to the amount of energy stored in molecules of the body, 6 g of glycogen is the equivalent of 1 g of fat.

D. Glycogen phosphorylase cleaves glucose monomers from the glycogen polymer, phosphorylating them at the same time so that they can be fed unchanged into the glycolytic pathway.

ANS:

A. False. Although the proteins of the electron-transport chain collect electrons from the NADH and FADH 2 cofactors, these high-energy electrons go through a series of transfers along the electron-transport chain. The energy released with each transfer moves protons across the inner mitochondrial membrane. It is this proton gradient that provides the energy to synthesize ATP. B. False. Gluconeogenesis can begin with pyruvate as a building block to make glucose, but there are three reactions in glycolysis that are irreversible because of a large free-energy barrier. Alternative enzymes and reaction pathways are used to bypass this problem, and they require the input of energy in the form of ATP and GTP. C. True.

D. False. When glycogen phosphorylase cleaves a glucose monomer from glycogen, the product is glucose 1phosphate. Before it can be used in glycolysis, it needs to be isomerized to glucose 6-phosphate.

DIF: 13.2 REF: 13.2 OBJ: 13.2.c Define gluconeogenesis and identify the type of cell in mammals in which the reaction is likely to occur. | 13.2.g Outline how, when food is scarce, cells can break down glycogen to produce energy via glycolysis. MSC: Evaluating 19. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. amino acids

galactose

nucleotides

carbon fixation

glucose

oxidative phosphorylation

citric acid cycle

glycogen

pyruvate

fatty acid

insulin

starch

fermentation

lactate

triacylglycerol

A carbon atom in a CO2 molecule in the atmosphere eventually becomes a part of one of the enzymes that catalyzes glycolysis in one of your cells. The CO2 first enters a cell in a corn leaf, where photosynthesis fixes the carbon to make it part of a sugar molecule; this travels from the leaf to an ear of corn, where it is stored as part of a polysaccharide __________ molecule in the corn seed. You then eat a corn chip made from the corn seed. You digest the corn seed, and the free __________ travels in your bloodstream, eventually being taken up by a liver cell and stored as __________. When required, this storage molecule breaks down into glucose 1-phosphate, which enters the glycolytic pathway. Glycolysis produces __________, which is converted into acetyl CoA, which enters the __________. Several intermediates in this process can provide the carbon skeleton for the production of __________, which are then incorporated into the enzymes that catalyze steps in glycolysis.

ANS: A carbon atom in a CO2 molecule in the atmosphere eventually becomes a part of one of the enzymes that catalyzes glycolysis in one of your cells. The CO2 first enters a cell in a corn leaf, where photosynthesis fixes the carbon to make it part of a sugar molecule; this travels from the leaf to an ear of corn, where it is stored as part of a polysaccharide starch molecule in the corn seed. You then eat a corn chip made from the corn seed. You digest the corn seed, and the free glucose travels in your bloodstream, eventually being taken up by a liver cell and stored as glycogen.

When required, this storage molecule breaks down into glucose 1-phosphate, which enters the glycolytic pathway.

Glycolysis produces pyruvate, which is converted into acetyl CoA, which enters the citric acid cycle. Several intermediates in this process can provide the carbon skeleton for the production of amino acids, which are then incorporated into the enzymes that catalyze steps in glycolysis.

DIF: Easy REF: 13.2 OBJ: 13.2.k Recall where—and in what form—plant cells store their sugars and fats.

MSC: Understanding

20. Living systems run on a glucose economy even though the metabolism of fatty acids is a more efficient catabolic process. Explain this observation and suggest why no-carb diets are ill-advised.

ANS: Unlike glucose, fatty acids are transported directly into the mitochondrial matrix and readily converted into acetyl CoA molecules. A simple condensation reaction shortens the fatty acid by two carbons and generates acetyl CoA, NADH, and FADH2 in each reaction cycle.

There are multiple consequences of using lipids as your only energy source: 1. Several critical organs preferentially store or use glucose. Skeletal muscles store glycogen, the brain does not store energy, and it needs glucose. Without glucose in your system, you will feel weakness, fatigue, and dizziness.

2. There will be an accumulation of excess of free fatty acids in the absence of glucose, and this excess is broken down into ketones. Ketones in the bloodstream cause an acidification of the blood. Ketoacidosis can be induced by extremely low-carb diets, lack of insulin in diabetic patients, or the consumption of extreme levels of alcohol without other sources of energy.

DIF: Moderate REF: 13.1 OBJ: 13.1.b List three categories of food molecules that can serve as energy sources for cells and identify the sugar whose breakdown generates most of the energy produced by a majority of animal cells.

MSC: Understanding 21. Although the outer mitochondrial membrane is permeable to all small molecules, the inner mitochondrial membrane is essentially impermeable in the absence of specific transport proteins. Consider this information and what you have learned about the citric acid cycle to address the following questions.

A. The ATP generated by oxidative respiration is used throughout the cell. The majority of ATP production occurs in the mitochondrial matrix. How do you think ATP is made accessible to enzymes in the cytosol and other organelles?

B. If the inner mitochondrial membrane were rendered as permeable as the outer membrane, how would that affect oxidative phosphorylation? Which specific processes would stop and which remain?

C. Present two types of benefits derived from separating the reactions of glycolysis in the cytosol from those that occur during the citric acid cycle in the mitochondrion.

ANS: A. The ATP must be transported across the inner mitochondrial membrane, after which it freely diffuses into the cytosol through the permeable outer membrane. Embedded in the inner membrane are dedicated ADP/ATP antiporters that serve the dual purpose of exporting ATP and bringing in new ADP, which can then be converted into ATP during oxidative phosphorylation.

B. During oxidative phosphorylation, the NADH and FADH2 generated by the citric acid cycle donate their electrons to an electron-transport chain in the inner mitochondrial membrane. As the electrons move along this chain, the energy released is used to drive protons across the inner mitochondrial membrane. This movement of protons produces a proton gradient across the membrane, which then serves as a source of energy for the generation of ATP. If the inner mitochondrial membrane were made “leaky,” the proton gradient would dissipate. Thus, although acetyl CoA would continue to be oxidized by the citric acid cycle, and electrons donated to the electron-transport chain, these processes could no longer promote the production of ATP. C. Compartmentalization provides a basic mechanism for the regulation of independent sets of reactions, including the citric acid cycle, fatty acid oxidation, glycolysis, and gluconeogenesis. In some cases, metabolic reactions are physically compartmentalized to separate anabolic from catabolic reactions. For example, in the mitochondrial ma-

trix, oxaloacetate is used in the citric acid cycle to help oxidize the acetyl carbons of acetyl CoA. However, oxaloacetate in the cytosol tends to be consumed by biosynthetic enzymes that use the molecule as a precursor for the production of amino acids such as aspartate. By keeping these reactions separate, the cell can control whether a molecule is used in an anabolic or catabolic reaction. A second advantage of compartmentalization is the colocalization and concentration of enzymes with their substrates, which can enhance reaction rates.

DIF: Hard REF: 13.1 OBJ: 13.1.ah Outline briefly how the energy captured during glycolysis and the citric acid cycle is used to generate ATP during oxidative phosphorylation. MSC: Understanding 22. You are stranded on an island with no way to communicate or signal for help. Your only hope is that somebody has tracked your last known location and will begin a search from there. Nevertheless, it will be a while before you have a chance of seeing another person. You have found a fresh stream of water at the center of the island, but other than green leafy plants, food options appear nonexistent.

A. What substances are used as energy stores within your body? In which organs?

B. Which will be used first in the absence of ingesting external sources of energy? How long will this source last?

C. What other substance will your body then turn to? On average, how long before you run out of that fuel source?

Explain why this second fuel source is a better choice than the first-used source as your long-term energy store.

D. What enzyme is required for the breakdown and usage of the energy source you listed in part B? If you are a person with a deficiency in that enzyme, how does this affect your other estimates?

ANS: A. Glycogen (a polymer of glucose, stored in the liver and skeletal muscle) and triacylglycerol (storage molecule for fatty acids in adipose) are the fuel molecules you have stored. The protein in your skeletal muscle can be broken down as the last fuel option your body has available.

B. Glycogen in the liver will be the first fuel source consumed. This will be burned within the first 12 –24 hours. Glycogen stored in your skeletal muscle tissue will be used as the liver glycogen levels go down, as your system attempts to maintain normal blood glucose levels.

C. The triacylglycerol (TAG) molecules in adipose cells will be used as glycogen levels wane. These lipids are cleaved into glycerol and free fatty acids, and the fatty acids are released into the bloodstream for use by the rest of your body. TAG is a better storage molecule because fatty acid metabolism is more efficient (yields more ATP per molecule than glucose) and can be stored in a smaller volume. Glycogen is highly hydrated, and takes up more intracellular space than lipid droplets. Lipid stores could fuel your body as long as one month, depending on the net volume of lipids stored and your level of activity.

D. Glycogen is broken in to glucose subunits by glycogen phosphorylase. A deficiency in this enzyme would mean that your ability to break down glycogen is impaired. In order to supplement the fuel to meet your body’s needs, you may burn fatty acids more quickly. You should also expect your glycogen stores to last longer in both the liver and skeletal muscle.

DIF: Moderate REF: 13.2 OBJ: 13.2.g Outline how, when food is scarce, cells can break down glycogen to produce energy via glycolysis. MSC: Applying 23. Figure 13-23 represents a cell lining the gut. Draw numbered, labeled lines to indicate exactly where inside a cell the following processes take place.

Figure 13-23

1. glycolysis 2. citric acid cycle 3. conversion of pyruvate to activated acetyl groups 4. oxidation of fatty acids to acetyl CoA 5. glycogen breakdown 6. release of fatty acids from triacylglycerols 7. oxidative phosphorylation ANS: See Figure 13-23A.

Figure 13-23A

DIF: Easy REF: 13.1 OBJ: 13.1.d Outline the three stages of catabolism, indicating where each takes place and identifying the major products of each stage. | 13.1.s Summarize how the pyruvate produced by glycolysis is converted into acetyl CoA and state where the process takes place. | 13.1.t Review how and where the fatty acids derived from fat are converted into acetyl CoA. MSC: Remembering 24. Fill in the spaces in the table below. For steps 1, 4, 5, and 8, name the correct substrates, enzymes, or products. For all the other steps, name the enzyme and draw the missing structure.

ANS: DIF: Easy REF: 13.1 OBJ: 13.1.j List four enzyme types involved in glycolysis and indicate their functions.

MSC: Remembering 25. The citric acid cycle is outlined in Figure 13-25. Some of these reactions produce small molecules that are used in the electron-transport chain or as energy for other reactions. Select from the list below to fill in the empty boxes. Keep in mind that some choices may be used more than once and others not used at all.

Figure 13-25

A. ATP B. ADP C. GTP D. GDP E. NAD+ F. NADH G. FADH H. FADH2 ANS:

Figure 13-25A

DIF: Easy REF: 13.1 OBJ: 13.1.aa List the activated carriers produced for each molecule of acetyl CoA that enters the citric acid cycle. MSC: Remembering

CHAPTER 14

Energy Generation in Mitochondria

and Chloroplasts ENERGY GENERATION IN MITOCHONDRIA AND CHLOROPLASTS 14.0.a Identify the molecule that serves as the main source of chemical energy in a cell.

14.0.b Differentiate between the mechanisms of ATP production by glycolysis and by oxidative phosphorylation.

14.0.c Compare where the production of ATP by oxidative phosphorylation takes place in plant cells, animal cells, bacteria, and archaea.

14.0.d Summarize the membrane-based processes involved in oxidative phosphorylation and photosynthesis.

14.0.e Summarize the stages involved in generating ATP by oxidative phosphorylation.

14.0.f

List the evidence suggesting that both mitochondria and chloroplasts evolved from bacteria that were engulfed by

ancestral cells.

MITOCHONDRIA AND OXIDATIVE PHOSPHORYLATION 14.1.a State the percentage of the free energy available in a molecule of glucose that is captured during glycolysis, and relate that efficiency with that of a gasoline-powered engine.

14.1.b Review the consequences of disrupting electron transport in mitochondria.

14.1.c Explain how mitochondrial replacement therapy can be used to prevent the transmission of mitochondrial d efects.

14.1.d Compare where mitochondria are located in a heart muscle cell, sperm, and fibroblast.

14.1.e Recall how many mitochondria are present in a typical cell and how their numbers can change with the energy needs of the cell.

14.1.f

Describe the structure of a mitochondrion and distinguish the functions and compositions of its different mem-

branes and compartments.

14.1.g Review how pyruvate and fatty acids move from the cytosol to the mitochondrial matrix, and identify the metabolic intermediate into which both are converted before entering the citric acid cycle.

14.1.h Recall the activated carriers, generated by the citric acid cycle, that will transfer high-energy electrons to the electron transport chain.

14.1.i

Outline the process that allows much of the energy contained in the high-energy electrons of activated carriers to

be stored in the high-energy phosphate bonds of ATP. 14.1.j

Distinguish the source of the high-energy electrons that power ATP production during cell respiration and photosynthesis.

14.1.k List the components of the electron transport chain in their order of operation, including mobile electron carriers, and describe the functions of each.

14.1.l

Compare the electron affinities of the components of the electron transport chain, and describe the change in en-

ergy of electrons as they move along the chain.

14.1.m Recall the direction in which protons are pumped across the inner mitochondrial membrane and describe the resulting pH difference in the mitochondrial matrix and intermembrane space.

14.1.n Review the membrane potential and pH gradients across the inner mitochondrial membrane, and state in which direction it is energetically favorable for protons to flow.

14.1.o Explain how ATP synthase acts as a motor to convert the energy of protons flowing down an electrochemical gradient into the chemical bond energy in ATP. 14.1.p Describe the conditions under which ATP synthase will act as a proton pump and hydrolyze ATP. 14.1.q Outline how the electrochemical proton gradient is used to drive the transport of metabolites across the inner mitochondrial membrane in eukaryotic cells and to power the rotation of flagella in motile bacteria.

14.1.r

Express the ratio at which cells maintain the concentrations of ATP and ADP, and state how cyanide works to

upset this balance.

14.1.s Compare where electrons donated by NADH and FADH2 enter the respiratory chain, and relate how much ATP is ultimately produced by each activated carrier.

MOLECULAR MECHANISMS OF ELECTRON TRANSPORT AND PROTON PUMPING 14.2.a Identify the main source of the protons that are pumped across the inner mitochondrial membrane by the electron transport chain.

14.2.b Summarize how electron carriers are able to transfer a proton from one side of the membrane to the other.

14.2.c Relate redox potential to electron affinity and describe how the redox potentials of reduced/oxidized nicotinamide adenine dinucleotide and oxygen/water align with their functions in the respiratory chain.

14.2.d Estimate the number of ATP molecules that could be synthesized from the energy released by the transfer of two electrons from NADH to molecular oxygen.

14.2.e Explain why NADH does not donate its electrons directly to molecular oxygen in living systems.

14.2.f

Contrast the redox potentials of iron-sulfur clusters and iron atoms held in heme groups, and relate where pro-

teins containing these metals are positioned along the electron transport chain.

14.2.g Compare the redox potentials of components in the electron transport chain and state which way electrons will flow.

14.2.h Identify the metals that help give cytochrome c oxidase its high redox potential.

14.2.i

Name the atoms or molecules that are oxidized or reduced by cytochrome c oxidase.

14.2.j

Summarize why cytochrome c oxidase must bind oxygen tightly.

14.2.k Outline how conformational change in cytochrome c oxidase pumps protons across the inner mitochondrial membrane.

14.2.l

Explain how specialized carrier proteins in the inner mitochondrial membrane of brown fat cells allow those cells

to oxidize fats to produce heat.

14.2.m Summarize the effects that uncoupling agents such as dinitrophenol have on the proton motive force in intact mitochondria.

14.2.n Outline how investigators used an artificial system including bacteriorhodopsin and ATP synthase to demonstrate the role that a proton gradient plays in producing ATP.

CHLOROPLASTS AND PHOTOSYNTHESIS 14.3.a Describe the structure of a chloroplast and indicate the functions of its membranes and compartments, including where chlorophyll and the photosynthetic machinery are contained.

14.3.b Outline the events that take place during stage 1 of photosynthesis and compare this process to the oxidative phosphorylation that occurs in mitochondria.

14.3.c Review the events that take place in stage 2 of photosynthesis and indicate where these reactions occur.

14.3.d Identify one mechanism by which plant cells avoid synthesizing sugars in the absence of sunlight.

14.3.e Recall the wavelengths of light that are absorbed or reflected by chlorophyll.

14.3.f

Delineate the functions of chlorophyll’s porphyrin ring and hydrophobic tail.

14.3.g List the components of a photosystem.

14.3.h Summarize how light energy, captured by a chlorophyll molecule in an antenna complex, gets transferred to the chlorophyll special pair in the reaction center.

14.3.i

Outline how the transfer of an electron from the chlorophyll special pair to a mobile electron carrier creates a

charge separation that effectively converts light energy into chemical energy.

14.3.j

Differentiate between photosystems I and II, indicate the electron carriers to which they transfer their high-energy electrons, and state the source of the electrons that replace those donated by their chlorophyll special pairs.

14.3.k Name the components of the photosynthetic electron transport chains and describe their functions.

14.3.l

State what drives the production of ATP by ATP synthase during photosynthesis.

14.3.m Identify the component of the photosynthetic machinery that generates all of the oxygen we breathe.

14.3.n Explain how the photosynthetic machinery produces enough energy to remove electrons from water and uses them to produce NADPH. 14.3.o Compare the redox potentials of oxygen/water, oxidized/reduced nicotimade adenine dinucleotide phosphate, plastoquinone, plastocyanin, and ferredoxin, and indicate which way electrons will flow.

14.3.p Identify the molecules that provide the energy to convert carbon dioxide into sugars.

14.3.q Recall the role that ribulose 1,5-bisphosphate plays in the carbon fixation cycle.

14.3.r

Name the three-carbon product of carbon fixation and state how many molecules of ATP and NADPH are re-

quired for its synthesis.

14.3.s Summarize the possible fates of glyceraldehyde 3-phosphate generated by the carbon fixation cycle.

14.3.t

Express how plant cells produce the ATP they need to fuel their metabolic needs.

14.3.u Review how the products of photosynthesis feed into oxidative phosphorylation—and vice versa.

THE EVOLUTION OF ENERGY-GENERATING SYSTEMS 14.4.a Propose how the first living cells on Earth, whether prokaryotic or eukaryotic, likely generated their ATP. 14.4.b Explain why the buildup of organic acids, produced by fermentation, might have favored the survival of cells that evolved a mechanism for pumping protons.

14.4.c Summarize how oxygen began accumulating in the atmosphere.

14.4.d Review the impact that the liberation of oxygen into the atmosphere had on the efficiency of energy production.

14.4.e Compare nitrogen fixation with carbon fixation.

14.4.f

Express what the use of proton gradients by Methanococcus suggests about the evolution of chemiosmotic cou-

pling.

MULTIPLE CHOICE 1. The overall relationship that links bond-forming reactions to membrane transport processes in the mitochondria is called a. chemiosmotic coupling. b. proton pumping. c. electron transfer. d. ATP synthesis.

ANS: A DIF: Easy REF: 14.0 OBJ: 14.0.d Summarize the membrane-based processes involved in oxidative phosphorylation and photosynthesis. MSC: Remembering 2. Which of the following is NOT part of the process known as “oxidative phosphorylation”? a. Molecular oxygen serves as a final electron acceptor. b. FADH2 and NADH become oxidized as they transfer a pair of electrons to the electron-transport chain. c. The electron carriers in the electron-transport chain toggle between reduced and oxidized states as electrons are passed along. d. ATP molecules are produced in the cytosol as glucose is converted into pyruvate.

ANS: D During glycolysis, ATP generation occurs simultaneously with two of the key reactions that convert glucose to pyruvate in the cytosol. This production of ATP in the absence of oxygen is not part of oxidative phosphorylation; it is referred to as “substrate-level phosphorylation.”

DIF: Easy REF: 14.0 OBJ: 14.0.b Differentiate between the mechanism of ATP production by glycolysis and by oxi-

dative phosphorylation. MSC: Understanding 3. Which of the following statements describes the phosphorylation event that occurs during the process known as “oxidative phosphorylation”? a. A phosphate group is added to ADP. b. ATP is hydrolyzed in order to add phosphate groups to protein substrates. c. A phosphate group is added to molecular oxygen. d. Inorganic phosphate is transported into the mitochondrial matrix, increasing the local phosphate concentration.

ANS: A DIF: Easy REF: 14.0 OBJ: 14.0.e Summarize the stages involved in generating ATP by oxidative phosphorylation. MSC: Remembering 4. Modern eukaryotes depend on mitochondria to generate most of the cell’s ATP. How many molecules of ATP can a single molecule of glucose generate? a. 30 b. 2 c. 20 d. 36 ANS: A Glycolysis of a single glucose molecule generates 2 ATP molecules. Oxidative phosphorylation in the mitochondria generates an additional 28 ATP molecules, making a total of 30 ATP molecules for each glucose molecule.

DIF: Easy REF: 14.0 OBJ: 14.0.a Identify the molecule that serves as the main source of chemical energy in a cell.

MSC: Remembering 5. Which of the following statements about mitochondrial division is TRUE? a. Mitochondria divide in synchrony with the cell. b. The rate of mitochondrial division is the same in all cell types. c. Mitochondrial division is mechanistically similar to prokaryotic cell division. d. Mitochondria cannot divide and produce energy for the cell at the same time.

ANS: C DIF: Easy REF: 14.0 OBJ: 14.0.f List the evidence that both mitochondria and chloroplasts evolved from bacteria that were engulfed by ancestral cells. MSC: Understanding 6. Which of the following statements describes the mitochondrial outer membrane? a. It is permeable to molecules with molecular mass as high as 5000 daltons. b. It contains transporters for ATP molecules. c. It contains proteins that are released during apoptosis. d. It contains enzymes required for the oxidation of fatty acids.

ANS: A DIF: Easy REF: 14.1 OBJ: 14.1.f Describe the structure of a mitochondrion and distinguish the functions and compositions of its different membranes and compartments. MSC: Remembering 7. Which of the following statements describes the mitochondrial inner membrane? a. It is permeable to molecules with molecular mass as high as 5000 daltons. b. It contains transporters for ATP molecules. c. It contains proteins that are released during apoptosis.

d. It contains enzymes required for the oxidation of fatty acids.

ANS: B DIF: Easy REF: 14.1 OBJ: 14.1.f Describe the structure of a mitochondrion and distinguish the functions and compositions of its different membranes and compartments. MSC: Remembering 8. Which of the following statements describes the mitochondrial intermembrane space? a. It is permeable to molecules with molecular mass as high as 5000 daltons. b. It contains transporters for ATP molecules. c. It contains proteins that are released during apoptosis. d. It contains enzymes required for the oxidation of fatty acids.

ANS: C DIF: Easy REF: 14.1 OBJ: 14.1.f Describe the structure of a mitochondrion and distinguish the functions and compositions of its different membranes and compartments. MSC: Remembering 9. Which of the following statements describes the mitochondrial matrix? a. It is permeable to molecules with molecular mass as high as 5000 daltons. b. It contains transporters for ATP molecules. c. It contains proteins that are released during apoptosis. d. It contains enzymes required for the oxidation of fatty acids.

ANS: D DIF: Easy REF: 14.1 OBJ: 14.1.f Describe the structure of a mitochondrion and distinguish the functions and compositions of its different membranes and compartments. MSC: Remembering 10. NADH contains a high-energy bond that, when cleaved, donates a pair of electrons to the electron-transport chain.

What are the immediate products of this bond cleavage? a. NAD+ + OH− b. NAD+ + H− c. NAD− + H+ d. NAD + H ANS: B DIF: Easy REF: 14.1 OBJ: 14.1.h Recall the activated carriers, generated by the citric acid cycle, that will transfer high-energy electrons to the electron-transport-chain. MSC: Remembering 11. Electron transport is coupled to ATP synthesis in mitochondria, in chloroplasts, and in the thermophilic bacterium Methanococcus. Which of the following is likely to affect the coupling of electron transport to ATP synthesis in ALL of these systems? a. a potent inhibitor of cytochrome c oxidase b. the removal of oxygen c. the absence of light d. an ADP analog that inhibits ATP synthase ANS: D All chemiosmotic coupling systems involve a proton gradient that ATP synthase uses to bind to ADP and phosphorylate it. Hence, agents that prevent ADP from binding the synthase or that dissipate the proton gradient affect all chemiosmotic systems. Cytochrome c oxidase and oxygen are required only for mitochondria and aerobic bacteria (not Methanococcus); light is required only for chloroplasts and photosynthetic bacteria (not Methanococcus).

DIF: Easy REF: 14.1 OBJ: 14.1.o Explain how ATP synthase acts as a motor to convert the energy of protons flowing

down their electrochemical gradient into the chemical bond energy in ATP. MSC: Applying 12. What is the source of protons that are pumped out of the mitochondrial matrix in Stage 1 of oxidative phosphorylation? a. NADH b. H2O c. FADH d. H2S ANS: B DIF: Easy REF: 14.1 OBJ: 14.1.i Outline the process that allows much of the energy contained in the highenergy electrons of activated carriers to be stored in the high-energy phosphate bonds of ATP. MSC: Remembering 13. What is the final result of the electron transfers in Stage 1 of the membrane-based processes that drive ATP synthesis in mitochondria? a. OH− is oxidized to O2 b. pyruvate is oxidized to CO2 c. O2 is reduced to H2O d. H− is converted to H2 ANS: C Contrary to what the term “oxidative phosphorylation” may imply, the phosphorylation event depends on the reduction of molecular oxygen, converting it to water.

DIF: Easy REF: 14.1 OBJ: 14.1.i Outline the process that allows much of the energy contained in the high-energy electrons of activated carriers to be stored in the high-energy phosphate bonds of ATP. MSC: Remembering 14. Osmosis describes the movement of water across a biological membrane and down its concentration gradient. In chemiosmosis, useful energy is harnessed by the cell from the movement of __________ across the inner mitochondrial membrane into the matrix __________ a concentration gradient. a. ATP, against b. protons, down c. electrons, down d. ADP, against ANS: B DIF: Easy REF: 14.1 OBJ: 14.1.o Explain how ATP synthase acts as a motor to convert the energy of protons flowing down their electrochemical gradient into the chemical bond energy in ATP. MSC: Remembering 15. Which of the following components of the electron-transport chain does NOT act as a proton pump? a. NADH dehydrogenase b. cytochrome c c. cytochrome c reductase d. cytochrome c oxidase ANS: B DIF: Easy REF: 14.1 OBJ: 14.1.k List the components of the electron-transport-chain in their order of operation, including mobile electron carriers, and describe the functions of each. MSC: Remembering 16. Which component of the electron-transport chain is required to combine the pair of electrons with molecular oxygen?

a. cytochrome c b. cytochrome b-c1 complex c. ubiquinone d. cytochrome c oxidase ANS: D DIF: Easy REF: 14.1 OBJ: 14.1.k List the components of the electron-transport-chain in their order of operation, including mobile electron carriers, and describe the functions of each. MSC: Remembering 17. During Stage 2 of oxidative phosphorylation, ATP synthesis is powered by movement of __________ ions through the __________. a. H+; H+ pump b. OH-; porin complex c. H+; ATP synthase d. elections; electron-transport chain ANS: C DIF: Easy REF: 14.1 OBJ: 14.1.o Explain how ATP synthase acts as a motor to convert the energy of pr otons flowing down their electrochemical gradient into the chemical bond energy in ATP. MSC: Remembering 18. Which of the following statements is TRUE? a. The NADH dehydrogenase complex does not pump protons across the membrane. b. The pH in the mitochondrial matrix is higher than the pH in the intermembrane space. c. The proton concentration gradient and the membrane potential across the inner mitochondrial membrane tend to work against each other in driving protons from the intermembrane space into the matrix. d. The difference in proton concentration across the inner mitochondrial membrane has a much larger effect than the membrane potential on the total proton-motive force.

ANS: B The pumping of protons out of the matrix into the intermembrane space creates a difference in proton concentration between the two sides of the membrane, with the matrix at a higher pH (i.e., more alkaline) than the intermembrane space, which tends to equilibrate with the cytosol (which has a neutral pH). The electrons in NADH are at a higher energy than the electrons in reduced ubiquinone, and proton pumping occurs linked to its electron transfers. Instead, the number of protons that can be pumped by each complex is determined by the difference in energy between the electrons in each substrate–product pair (i.e., the difference between the electrons in NADH and reduced ubiquinone, compared with that between reduced ubiquinone and reduced cytochrome c).

DIF: Easy REF: 14.1 OBJ: 14.1.m Recall the direction in which protons are pumped across the inner mitochondrial membrane and describe the resulting pH difference in the mitochondrial matrix and intermembrane space.

MSC: Understanding 19. The mitochondrial ATP synthase consists of several different protein subunits. Which subunit binds to ADP + Pi and catalyzes the synthesis of ATP as a result of a conformational change? a. transmembrane H+ carrier b. F1 ATPase head c. peripheral stalk d. central stalk

ANS: B DIF: Easy REF: 14.1 OBJ: 14.1.o Explain how ATP synthase acts as a motor to convert the energy of pr otons flowing down their electrochemical gradient into the chemical bond energy in ATP. MSC: Remembering 20. Bongkrekic acid is an antibiotic that inhibits the ATP/ADP transport protein in the inner mitochondrial membrane.

Which of the following will allow electron transport to occur in mitochondria treated with bongkrekic acid? a. placing the mitochondria in anaerobic conditions b. adding FADH2 c. making the inner membrane permeable to protons d. inhibiting the ATP synthase ANS: C Inhibition of the ATP/ADP translocase prevents the export of ATP generated by oxidative phosphorylation in exchange for an import of ADP into the matrix. The ensuing buildup of ATP at the expense of ADP inhibits the ATP synthase. Because protons are no longer being used to power the ATP synthase, the proton gradient is not dissipated; the increasingly steep proton gradient makes it increasingly difficult for the electron-transport proteins to pump protons out of the matrix, and electron transport quickly stops. Hence, the inner membrane becomes permeable to protons, allowing electron transport to resume (although no ATP will be synthesized).

DIF: Moderate REF: 14.1 OBJ: 14.1.o Explain how ATP synthase acts as a motor to convert the energy of protons flowing down their electrochemical gradient into the chemical bond energy in ATP. MSC: Applying 21. The relationship of free-energy change (ΔG) to the concentrations of reactants and products is important because it predicts the direction of spontaneous chemical reactions. In the hydrolysis of ATP to ADP and inorganic phosphate (Pi), the standard free-energy change (ΔG°) is −7.3 kcal/mole. The free-energy change depends on concentrations according to the following equation: ΔG = ΔG° + 1.42 log10 ([ADP] [Pi]/[ATP]) In a resting muscle, the concentrations of ATP, ADP, and Pi are approximately 0.005 M, 0.001 M, and 0.010 M, respectively. What is the ΔG for ATP hydrolysis in resting muscle? a. −11.1 kcal/mole b. −8.72 kcal/mole c. 6.01 kcal/mole d. −5.88 kcal/mole ANS: A The ΔG for ATP hydrolysis is −11.1 kcal/mole. This result is calculated by substituting values into the equation given: ΔG = −7.3 kcal/mole + 1.42 log10 ([0.001 M] [0.010 M]/[0.005 M]) = −7.3 kcal/mole + 1.42 log10 (0.002) = −11.1 kcal/mole.

DIF: Moderate REF: 3.2 OBJ: 3.2.d Review the relationship between /delta G, equilibrium, and the concentrations of a reaction’s substrates and products. MSC: Applying 22. The relationship of free-energy change (ΔG) to the concentrations of reactants and products is important because it predicts the direction of spontaneous chemical reactions. In the hydrolysis of ATP to ADP and inorganic phosphate (Pi), the standard free-energy change (ΔG°) is −7.3 kcal/mole. The free-energy change depends on concentrations according to the following equation:

ΔG = ΔG° + 1.42 log10 ([ADP] [Pi]/[ATP]) In a resting muscle, the concentrations of ATP, ADP, and Pi are approximately 0.005 M, 0.001 M, and 0.010 M, respectively. What is the ΔG for ATP synthesis in resting muscle? a. −6.01 kcal/mole b. 5.88 kcal/mole c. 8.72 kcal/mole d. 11.1 kcal/mole ANS: D The ΔG for ATP hydrolysis is −11.1 kcal/mole. This result is calculated by substituting values into the equation given: ΔG = −7.3 kcal/mole + 1.42 log10 ([0.001 M] [0.010 M]/[0.005 M]) = −7.3 kcal/mole + 1.42 log10 (0.002) = −11.1 kcal/mole. The ΔG for synthesis is +11.1 kcal/mole because the forward and reverse reactions always have the same numerical value for ΔG but with the sign reversed.

DIF: Moderate REF: 3.2 OBJ: 3.2.d Review the relationship between /delta G, equilibrium, and the concentrations of a reaction’s substrates and products. MSC: Applying 23. The relationship of free-energy change (ΔG) to the concentrations of reactants and products is important because it predicts the direction of spontaneous chemical reactions. Consider, for example, the hydrolysis of ATP to ADP and inorganic phosphate (Pi). The standard free-energy change (ΔG°) for this reaction is −7.3 kcal/mole. The free-energy change depends on concentrations according to the following equation: ΔG = ΔG° + 1.42 log10 ([ADP] [Pi]/[ATP]) In a resting muscle, the concentrations of ATP, ADP, and Pi are approximately 0.005 M, 0.001 M, and 0.010 M, respectively. At [Pi] = 0.010 M, what will be the ratio of [ATP] to [ADP] at equilibrium? a. 1.38 × 106 b. 1 c. 7.2 × 10−8 d. 5.14 ANS: C At equilibrium, the ΔG is equal to zero by definition. The ratio of [ATP] to [ADP] at equilibrium is less than 1:10 7. This result is calculated by setting ΔG = 0, so that 1.42 log10 ([ADP] [Pi] / [ATP]) = −ΔG° = 7.3 kcal/mole. Solving for [ADP] / [ATP], the equation becomes log10 ([ADP] [0.010] / [ATP]) = 7.3/1.42 = 5.14; then [ADP] / [ATP] = (105.14) /(0.010) = 13.8 × 106. Thus, the reciprocal [ATP] / [ADP] is 7.2 × 10−8.

DIF: Moderate REF: 3.2 OBJ: 3.2.d Review the relationship between /delta G, equilibrium, and the concentrations of a reaction’s substrates and products. MSC: Applying 24. NADH and FADH2 carry high-energy electrons that are used to power the production of ATP in the mitochondria.

These cofactors are generated during glycolysis, the citric acid cycle, and the fatty acid oxidation cycle. Which molecule below can produce the most ATP? Explain your answer. a. NADH from glycolysis b. FADH2 from the fatty acid cycle c. NADH from the citric acid cycle d. FADH2 from the citric acid cycle

ANS: C NADH produced in glycolysis does not contribute directly to ATP production in the mitochondria because it cannot be imported into the matrix. If the energy is transferred to a different carrier, some of the stored energy is lost. FADH 2, from either the fatty acid cycle or the citric acid cycle, contributes less energy than NADH from the citric acid cycle because the electrons are donated further down the chain. Fewer electron transfers means that fewer protons are pumped across the membrane.

DIF: Moderate REF: 14.1 OBJ: 14.1.s Compare where electrons donated by NADH and FADH2 enter the respiratory chain and relate how much ATP is ultimately produced by each activated carrier. MSC: Understanding 25. Which of the following statements about “redox potential” is TRUE? a. Redox potential is a measure of a molecule’s capacity to strip electrons from oxygen. b. For molecules that have a strong tendency to pass along their electrons, the standard redox potential is negative. c. The transfer of electrons from cytochrome c oxidase to oxygen has a negative redox potential. d. A molecule’s redox potential is a measure of the molecule’s capacity to pass along electrons to oxygen.

ANS: B DIF: Easy REF: 14.2 OBJ: 14.2.g Compare the redox potentials of components in the electron-transport-chain and state which way electrons will flow. MSC: Understanding 26. Which of the following statements is TRUE? a. Only compounds with negative redox potentials can donate electrons to other compounds under standard conditions. b. Compounds that donate one electron have higher redox potentials than those compounds that donate two electrons. c. The ΔE0′ of a redox pair does not depend on the concentration of each member of the pair. d. The free-energy change, ΔG, for an electron-transfer reaction does not depend on the concentration of each member of a redox pair.

ANS: C By definition, E0′ refers to the standard state of equal concentrations of each member of the redox pair. Therefore, ΔE0′ does not vary with the actual concentrations. Compounds with positive redox potentials can donate electrons to other compounds under standard conditions, so long as the electron acceptor has a higher (more positive) redox potential. Compounds that are able to donate only one electron do not necessarily have higher redox potentials than compounds that are able to donate two electrons. Although the ΔE0′ of a reaction is directly proportional to the ΔG° of a reaction and both are independent of the concentrations of substrates and products, the ΔG depends on these concentrations.

DIF: Easy REF: 14.2 OBJ: 14.2.g Compare the redox potentials of components in the electron-transport-chain and state which way electrons will flow. MSC: Understanding 27. Which ratio of NADH to NAD+ in solution will generate the largest positive redox potential? a. 1:10 b. 10:1 c. 1:1 d. 5:1 ANS: A Choice (a) is correct. NAD+ is the electron acceptor; NADH is the electron donor. If there is an excess of NAD + in so-

lution, there is less capacity to donate electrons and more capacity to accept electrons. This is reflected by a redox potential that is more positive than the alternative conditions.

DIF: Moderate REF: 14.2 OBJ: 14.2.g Compare the redox potentials of components in the electron-transport-chain and state which way electrons will flow. MSC: Applying 28. Ubiquinone is one of two mobile electron carriers in the electron-transport chain. Where does the additional pair of electrons reside in the reduced ubiquinone molecule? a. The electrons are added directly to the aromatic ring. b. The electrons are added to each of two ketone oxygens on the aromatic ring. c. The electrons are added to the hydrocarbon tail, which hides them inside the membrane bilayer. d. Both electrons, and one proton, are added to a single ketone oxygen bound to the aromatic ring.

ANS: B DIF: Easy REF: 14.2 OBJ: 14.2.g Compare the redox potentials of components in the electron-transport-chain and state which way electrons will flow. MSC: Remembering 29. Which of the following reactions has a sufficiently large free-energy change to enable it to be used, in principle, to provide the energy needed to synthesize one molecule of ATP from ADP and Pi under standard conditions? See Table 14-29. Recall that ΔG° = —n (0.023) ΔE0′, and ΔE0′ = E0′ (acceptor) − E0′ (donor). a. the reduction of a molecule of pyruvate by NADH b. the reduction of a molecule of cytochrome b by NADH c. the reduction of a molecule of cytochrome b by reduced ubiquinone d. the oxidation of a molecule of reduced ubiquinone by cytochrome c

Table 14-29

ANS: B For a reaction to drive ATP synthesis under standard conditions, the ΔG° of the reaction must be less than −7.3 kcal/mole. Because ΔG° = −n (0.023) ΔE0′, the value of ΔE0′ must be greater than 317 mV/n, where n is the number of electrons transferred. ΔE0′ is 130 mV for the reduction of a molecule of pyruvate by NADH, 390 mV for the reduction of a molecule of cytochrome b by NADH, 40 mV for the reduction of a molecule of cytochrome b by ubiquinone, 200 mV for the oxidation of a molecule of ubiquinone by cytochrome c, and 590 mV for the oxidation of cytochrome c by oxygen. The numbers of electrons transferred in each of the above reactions are two, one, one, one, and one, respectively. Thus, only the second and fifth of these reactions are sufficient to drive ATP synthesis.

DIF: Easy REF: 14.2 OBJ: 14.2.g Compare the redox potentials of components in the electron-transport-chain and state which way electrons will flow. MSC: Remembering 30. Cytochrome c oxidase is an enzyme complex that uses metal ions to help coordinate the transfer of four electrons to

O2. Which metal atoms are found in the active site of this complex? a. two iron atoms b. one iron atom and one copper atom c. one iron atom and one zinc atom d. one zinc atom and one copper atom ANS: B DIF: Easy REF: 14.2 OBJ: 14.2.h Identify the metals that help give cytochrome c oxidase its high redox potential. MSC: Remembering 31. Which of the following statements is TRUE? a. Ubiquinone is a small, hydrophobic protein containing a metal group that acts as an electron carrier. b. A 2Fe2S iron–sulfur center carries one electron, whereas a 4Fe4S iron-sulfur center carries two electrons. c. Iron–sulfur centers generally have a higher redox potential than do cytochromes. d. Mitochondrial electron carriers with the highest redox potential generally contain copper ions and/or heme groups.

ANS: D Cytochrome c oxidase, which is the last carrier in the mitochondrial electron-transport chain and therefore has the highest redox potential, contains copper ions and a heme group. Ubiquinone is not a protein and does not contain a metal group. Both 2Fe2S and 4Fe4S centers carry one electron. Iron–sulfur centers generally have a lower redox potential than do cytochromes. The heme group in cytochrome c contains a charged iron ion. The interiors of proteins are often hydrophobic, favoring a relatively high redox potential, because reduction of the iron ion decreases its charge, and charges are energetically unfavorable in a hydrophobic environment.

DIF: Easy REF: 14.2 OBJ: 14.2.f Contrast the redox potentials of iron-sulfur clusters and iron atoms held in heme groups and relate where proteins containing these metals are positioned along the electron-transport-chain.

MSC: Understanding 32. Which of the following is not an electron carrier that participates in the electron-transport chain? a. cytochrome b. quinone c. rhodopsin d. copper ion ANS: C DIF: Easy REF: 14.2 OBJ: 14.2.g Compare the redox potentials of components in the electron-transport-chain and state which way electrons will flow. MSC: Remembering 33. Which of the following statements about cytochrome c is TRUE? a. Cytochrome c shuttles electrons between the NADH dehydrogenase complex and cytochrome c reductase complex. b. When cytochrome c becomes reduced, two cysteines (sulfur-containing amino acids) become covalently bound to a heme group. c. The pair of electrons accepted by cytochrome c are added to the porphyrin ring of the bound heme group. d. Cytochrome c is the last protein in the electron-transport chain, passing its electrons directly to molecular oxygen, a process that reduces O2 to H2O. ANS: C DIF: Easy REF: 14.2 OBJ: 14.2.f Contrast the redox potentials of iron-sulfur clusters and iron atoms held in

heme groups and relate where proteins containing these metals are positioned along the electron-transport-chain.

MSC: Understanding 34. Experimental evidence supporting the chemiosmotic hypothesis was gathered by using artificial vesicles containing a protein that can pump protons in one direction across the vesicle membrane to create a proton gradient. Which protein was used to generate the gradient in a highly controlled manner? a. cytochrome c oxidase b. NADH dehydrogenase c. cytochrome c d. bacteriorhodopsin ANS: D Bacteriorhodopsin is a transmembrane protein that pumps protons across the membrane when exposed to light. The other proteins pump protons but are part of the electron-transport chain, so they would not be good options to show that it is only the proton gradient that is required in this system rather than any specific intermediate in the electrontransport chain.

DIF: Easy REF: 14.2 OBJ: 14.2.n Outline how investigators used an artificial system including bacteriorhodopsin and ATP synthase to demonstrate the role that a proton gradient plays in producing ATP. MSC: Remembering 35. Photosynthesis is a process that takes place in chloroplasts and uses light energy to generate high-energy electrons, which are passed along an electron-transport chain. Where are the proteins of the electron-transport chain located in chloroplasts? a. thylakoid space b. stroma c. inner membrane d. thylakoid membrane ANS: D DIF: Easy REF: 14.3 OBJ: 14.3.a Describe the structure of a chloroplast and indicate the functions of its membranes and compartments, including where chlorophyll and the photosynthetic machinery are contained.

MSC: Remembering 36. In stage 1 of photosynthesis, a proton gradient is generated and ATP is synthesized. Where do protons become concentrated in the chloroplast? a. thylakoid space b. stroma c. inner membrane d. thylakoid membrane ANS: A DIF: Easy REF: 14.3 OBJ: 14.3.a Describe the structure of a chloroplast and indicate the functions of its membranes and compartments, including where chlorophyll and the photosynthetic machinery are contained.

MSC: Remembering 37. The ATP synthase found in chloroplasts is structurally similar to the ATP synthase in mitochondria. Given that ATP is being synthesized in the stroma, where will the F 0 portion of the ATP synthase be located? a. thylakoid space b. stroma

c. inner membrane d. thylakoid membrane ANS: D DIF: Easy REF: 14.3 OBJ: 14.3.a Describe the structure of a chloroplast and indicate the functions of its membranes and compartments, including where chlorophyll and the photosynthetic machinery are contained.

MSC: Remembering 38. Stage 2 of photosynthesis, sometimes referred to as the dark reactions, involves the reduction of CO 2 to produce organic compounds such as sucrose. What cofactor is the electron donor for carbon fixation? a. H2O b. NADH c. FADH2 d. NADPH ANS: D DIF: Easy REF: 14.3 OBJ: 14.3.c Review the events that take place in stage 2 of photosynthesis and indicate where these reactions occur. MSC: Remembering 39. In the electron-transport chain in chloroplasts, __________-energy electrons are taken from __________. a. high; H2O. b. low; H2O. c. high; NADPH. d. low; NADPH. ANS: B DIF: Easy REF: 14.3 OBJ: 14.3.b Outline the events that take place during stage 1 of photosynthesis and compare this process to the oxidative phosphorylation that occurs in mitochondria. MSC: Remembering 40. The photosystems in chloroplasts contain hundreds of chlorophyll molecules, most of which are part of a. plastoquinone. b. the antenna complex. c. the reaction center. d. the ferredoxin complex.

ANS: B DIF: Easy REF: 14.3 OBJ: 14.3.g List the components of a photosystem. MSC: Remembering 41. If you shine light on chloroplasts and measure the rate of photosynthesis as a function of light intensity, you get a curve that reaches a plateau at a fixed rate of photosynthesis, x, as shown in Figure 14-41.

Figure 14-41

Which of the following conditions will increase the value of x? a. increasing the number of chlorophyll molecules in the antenna complexes b. increasing the number of reaction centers c. adding a powerful oxidizing agent d. decreasing the wavelength of light used ANS: B The rate of photosynthesis will increase with increasing light intensity until photons hit all of the reaction centers directly. At saturating levels of light, the number of reaction centers that are still capable of being excited limits the rate of photosynthesis, which can be increased only by increasing the number of reaction centers or by increasing the rate at which the reaction centers are restored to their low-energy state. Increasing the number of chlorophyll molecules in the antenna complexes, the energy per photon of light, or the rate at which chlorophyll molecules are able to transfer energy electrons to one another will have no effect on either of these parameters. Adding a powerful oxidizing agent might, if anything, interfere with the reduction of the reaction center back to its resting state.

DIF: Moderate REF: 14.3 OBJ: 14.3.h Summarize how light energy, captured by a chlorophyll molecule in an antenna complex, gets transferred to the chlorophyll special pair in the reaction center. MSC: Understanding 42. If you add a compound to illuminated chloroplasts that inhibits the NADP + reductase, NADPH generation ceases, as expected. However, ferredoxin does not accumulate in the reduced form because it is able to donate its electrons not only to NADP+ (via NADP+ reductase) but also back to the cytochrome b6-f complex. Thus, in the presence of the compound, a “cyclic” form of photosynthesis occurs in which electrons flow in a circle from ferredoxin, to the cytochrome b6-f complex, to plastocyanin, to photosystem I, to ferredoxin. What will happen if you now also inhibit photosystem II? a. Less ATP will be generated per photon absorbed. b. ATP synthesis will cease. c. Plastoquinone will accumulate in the oxidized form. d. Plastocyanin will accumulate in the oxidized form.

ANS: C If you now inhibit photosystem II, you will deprive plastoquinone, which can still donate its electrons to the cytochrome b6-f complex, of an electron source. Hence, plastoquinone will accumulate in its oxidized form. In contrast, all of the other components downstream of plastoquinone will be able to cycle between their oxidized and reduced states. ATP synthesis will continue, because electrons are still being fed through the cytochrome b6-f complex, and the same amount of ATP will be generated.

DIF: Hard REF: 14.3 OBJ: 14.3.l State what drives the production of ATP by ATP synthase during photosynthesis.

MSC: Applying 43. The enzyme ribulose bisphosphate carboxylase (Rubisco) normally adds carbon dioxide to ribulose 1,5-bisphosphate.

However, it will also catalyze a competing reaction in which O 2 is added to ribulose 1,5-bisphosphate to form 3phosphoglycerate and phosphoglycolate. Assume that phosphoglycolate is a compound that cannot be used in any further reactions. If O2 and CO2 have the same affinity for Rubisco, which of the following is the lowest ratio of CO 2 to O2 at which a net synthesis of sugar can occur? a. 1:3 b. 1:2

c. 3:1 d. 2:1 ANS: C Three molecules of O2 are required to form three molecules of 3-phosphoglycerate and three molecules of phosphoglycolate. To break even (that is, simply to keep the Calvin cycle going with no net sugar produced), you need to have enough 3-phosphoglycerate to synthesize ribulose 1,5-bisphosphate again. Therefore, for every three molecules of O2 that react with ribulose 1,5-bisphosphate, you need to generate two additional molecules of 3-phosphoglycerate. For every three molecules of CO2 that go into the Calvin cycle, one molecule of 3-phosphoglycerate is formed. So if you have at least six molecules of CO2 per three molecules of O2 going through the Calvin cycle, you will break even. Only if you have a ratio of CO2 to O2 higher than 6:3 (2:1) can you have a net synthesis of carbohydrate.

DIF: Hard REF: 14.3 OBJ: 14.3.q Recall the role that ribulose 1,5-bisphosphate plays in the carbon fixation cycle.

MSC: Applying 44. Which of the following statements about the possible fates of glyceraldehyde 3-phosphate is FALSE? a. It can be exported from the chloroplast to the cytosol for conversion into sucrose. b. It can be used to make starch, which is stored inside the stroma of the chloroplast. c. It can be used as a precursor for fatty acid synthesis and stored as fat droplets in the stroma. d. It can be transported into the thylakoid space for use as a secondary electron acceptor downstream of the electron-transport chain.

ANS: D DIF: Easy REF: 14.3 OBJ: 14.3.s Summarize the possible fates of glyceraldehyde 3-phosphate generated by the carbon-fixation-cycle. MSC: Understanding 45. Oxidative phosphorylation, as it occurs in modern eukaryotes, is a complex process that probably arose in simple stages in primitive bacteria. Which mechanism is proposed to have arisen first as this complex system evolved? a. electron transfers coupled to a proton pump b. the reaction of oxygen with an ancestor of cytochrome c oxidase c. ATP-driven proton pumps d. the generation of ATP from the energy of a proton gradient ANS: C DIF: Easy REF: 14.4 OBJ: 14.4.a Propose how the first living cells on Earth, whether prokaryotic or eukaryotic, likely generated their ATP. MSC: Remembering 46. Below is a list of breakthroughs in energy metabolism in living systems. Which is the correct order in which they are thought to have evolved?

A. H2O-splitting enzyme activity B. light-dependent transfer of electrons from H2S to NADPH C. the consumption of fermentable organic acids D. oxygen-dependent ATP synthesis a. A, C, D, B b. C, A, B, D c. B, C, A, D d. C, B, A, D ANS: D DIF: Moderate REF: 14.4 OBJ: 14.4.c Summarize how oxygen began accumulating in the atmosphere.

MSC: Understanding 47. Which of the phylogenetic trees in Figure 14-47 is the most accurate? Explain your answer. Note: the mitochondria and chloroplasts are from maize, but they are treated as independent “organisms” for the purposes of this question.

Figure 14-47

a. Tree (a) b. Tree (b) c. Tree (c) d. Tree (d) e. Tree (e) ANS: C Mitochondria are most closely related to Bacillus, and chloroplasts to cyanobacteria. Maize (a eukaryote) is more closely related to Giardia (a simple eukaryote) than it is to bacteria (prokaryotes).

DIF: Moderate REF: 14.4 OBJ: 14.4.a Propose how the first living cells on Earth, whether prokaryotic or eukaryotic, likely generated their ATP. MSC: Analyzing

SHORT ANSWER 1. Based upon what you know about metabolism, explain how electrons are stripped from food molecules and used to drive the electron-transport chain.

ANS: Food molecules are ultimately converted into acetyl CoA. Electrons removed during the generation of acetyl CoA are added to the cofactors NAD+ and FAD to generate the reduced cofactors NADH and FADH2, respectively.

The two carbon atoms in the acetyl group of acetyl CoA are then fed into the citric acid cycle, where they are oxidized

to two molecules of CO2. The electrons removed during this oxidation are also captured by the activated carriers NADH and FADH2. The high-energy electrons in all of these activated carriers, which derived from carbons that were formerly part of food molecules are now transferred to the proteins in the electron-transport chain.

DIF: REF: 14.1 OBJ: 14.1.h Recall the activated carriers, generated by the citric acid cycle, that will transfer highenergy electrons to the electron-transport-chain. MSC: Understanding 2. The citric acid cycle generates NADH and FADH2, which are then used in the process of oxidative phosphorylation to make ATP. The reactions in the citric acid cycle do not utilize oxygen. Yet the citric acid cycle stops almost immediately when O2 is removed. Explain this observation.

ANS: The citric acid cycle stops almost immediately when oxygen is removed because several steps in the cycle require the oxidized forms of NAD+ and FAD. In the absence of oxygen, these electron carriers can be reduced by the reactions of the citric acid cycle but cannot be reoxidized by the electron-transport chain that participates in oxidative phosphorylation.

DIF: Moderate REF: 14.1 OBJ: 14.1.h Recall the activated carriers, generated by the citric acid cycle, that will transfer high-energy electrons to the electron-transport-chain. MSC: Understanding 3. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. The number and location of mitochondria within a cell can change, depending on both the cell type and the amount of energy required.

B. The inner mitochondrial membrane contains porins, which allow pyruvate to enter for use in the citric acid cycle.

C. The inner mitochondrial membrane is actually a series of discrete, flattened, membrane-enclosed compartments called cristae, similar to what is seen in the Golgi apparatus.

D. The intermembrane space of the mitochondria is chemically equivalent to the cytosol with respect to pH and the small molecules present.

ANS: A. True.

B. False. The outer mitochondrial membrane contains porins, allowing the passage of all molecules with a mass of less than 5000 daltons. Although pyruvate must pass through the inner membrane, it does so in a highly regulated manner via specific transporter channels.

C. False. Although the cristae do look like individual compartments on the basis of the images of the inner structure of the mitochondria, the inner membrane is a single, albeit highly convoluted, membrane.

D. True.

DIF: Easy REF: 14.1 OBJ: 14.1.f Describe the structure of a mitochondrion and distinguish the functions and compositions of its different membranes and compartments. MSC: Evaluating 4. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. The dark reactions of photosynthesis occur only in the absence of light.

B. Much of the glyceraldehyde 3-phosphate made in the chloroplast ends up producing the molecules needed by the mitochondria to produce ATP. C. Ribulose 1,5-bisphosphate is similar to oxaloacetate in the Krebs cycle in that they are both regenerated at the end of their respective cycles.

D. Each round of the Calvin cycle uses five molecules of CO 2 to produce one molecule of glyceraldehyde 3phosphate and one of pyruvate.

ANS: A. False. The dark reactions are those involved in carbon fixation and are named as such because they do not require light.

B. True.

C. True.

D. False. Three molecules of CO2 are required for each round of the Calvin cycle, and the product is one molecule of glyceraldehyde 3-phosphate and the recycling of the ribulose 1,5-bisphosphate molecule.

DIF: Easy REF: 14.3 OBJ: 14.3.q Recall the role that ribulose 1,5-bisphosphate plays in the carbon-fixation-cycle. | 14.3.s Summarize the possible fates of glyceraldehyde 3-phosphate generated by the carbon-fixation-cycle.

MSC: Evaluating 5. Name the mitochondrial localization of each of the following macromolecules.

A. porin proteins B. mitochondrial chromosome C. citric acid cycle enzymes D. electron-transport chain proteins E. ATP synthase F. pyruvate transporter ANS: A. Porin is in the outer membrane.

B. The mitochondrial genome is in the matrix.

C. The citric acid cycle enzymes are in the matrix.

D. The proteins of the electron-transport chain are in the inner membrane.

E. ATP synthase is in the inner membrane.

F. The transport protein for pyruvate is in the inner membrane.

DIF: Easy REF: 14.1 OBJ: 14.1.f Describe the structure of a mitochondrion and distinguish the functions and compositions of its different membranes and compartments. MSC: Remembering 6. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. acetyl groups

glucose +

carbon dioxide

NAD

chemiosmosis

NADH

fatty acids

NADPH oxidative phosphorylation

+

NADP

oxygen pyruvate

Mitochondria can use both __________ and __________ directly as fuel. __________ produced in the citric acid cycle donates electrons to the electron-transport chain. The citric acid cycle oxidizes __________ and produces __________ as a waste product. __________ acts as the final electron acceptor in the electron-transport chain. The synthesis of ATP in mitochondria is also known as __________.

ANS: Mitochondria can use both pyruvate and fatty acids directly as fuel. NADH produced in the citric acid cycle donates electrons to the electron-transport chain. The citric acid cycle oxidizes acetyl groups and produces carbon dioxide as a waste product. Oxygen acts as the final electron acceptor in the electron-transport chain. The synthesis of ATP in mitochondria is also known as oxidative phosphorylation.

DIF: Easy REF: 14.1 OBJ: 14.1.g Review how pyruvate and fatty acids move from the cytosol to the mitochondrial matrix and identify the metabolic intermediate into which both are converted before entering the citric acid cycle. | 14.1.i Outline the process that allows much of the energy contained in the high-energy electrons of activated carriers to be stored in the high-energy phosphate bonds of ATP. MSC: Understanding 7. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. cytochrome b-c1 complex

oxidation

second

cytochrome c

oxidation–reduction

the

cytochrome c oxidase

phosphorylation

third

first

plastoquinone

uniquinone

NADH dehydrogenase

reduction

NADH donates electrons to the __________ of the three respiratory enzyme complexes in the mitochondrial electrontransport-chain. __________ is a small protein that acts as a mobile electron carrier in the respiratory chain. __________ transfers electrons to oxygen. Electron transfer in the chain occurs in a series of __________ reactions.

The first mobile electron carrier in the respiratory chain is __________.

ANS: NADH donates electrons to the first of the three respiratory enzyme complexes in the mitochondrial electrontransport chain. Cytochrome c is a small protein that acts as a mobile electron carrier in the respiratory chain. Cytochrome c oxidase transfers electrons to oxygen. Electron transfer in the chain occurs in a series of oxidation– reduction reactions. The first mobile electron carrier in the respiratory chain is ubiquinone.

DIF: Easy REF: 14.1 OBJ: 14.1.k List the components of the electron-transport-chain in their order of operation, including mobile electron carriers, and describe the functions of each. MSC: Understanding 8. Some bacteria can live both aerobically and anaerobically. How does the ATP synthase in the plasma membrane of the bacterium help such bacteria to keep functioning in the absence of oxygen?

ANS: In the absence of oxygen, the respiratory chain no longer pumps protons, and thus no proton electrochemical gradient is generated across the bacterial membrane. In these conditions, the ATP synthase uses some of the ATP generated by glycolysis in the cytosol to pump protons out of the bacterium, thus forming the proton gradient across the membrane that the bacterium requires for importing vital nutrients by coupled transport. DIF: Moderate REF: 14.1 OBJ: 14.1.p Describe the conditions under which ATP synthase will act as a proton pump and hydrolyze ATP. MSC: Understanding 9. What are the major molecules that need to be transported into and out of the mitochondrion to keep the citric acid cycle, electron-transport chain, and ATP synthesis going?

ANS: ADP/ATP, inorganic phosphate, and pyruvate. Because the inner mitochondrial membrane is relatively impermeable, there are many other transport substrates and respective protein transporters needed to keep these processes running. DIF: Moderate REF: 14.1 OBJ: 14.1.h Recall the activated carriers, generated by the citric acid cycle, that will transfer high-energy electrons to the electron-transport-chain. MSC: Understanding 10. Describe the steps by which the F0 portion of the ATP synthase harnesses the proton-motive force to help synthesize ATP. What would you expect to observe if the proton gradient were reversed? Explain your answer.

ANS: Protons flow through a channel that exists between the subunits of the transmembrane H + carrier, which forms a ring (the rotor). The flow of protons through this carrier makes the rotor and its attached stalk rotate. As the stalk in F 0 rotates, it rubs against proteins in the stationary F 1 portion of the ATP synthase. The resulting mechanical deformation produces a conformational change in the subunits of the F1 ATPase that causes them to produce ATP. When the proton gradient is reversed, the F1 portion of the ATP synthase catalyzes the hydrolysis of ATP to ADP and Pi, rather than the reverse reaction of ATP synthesis; this causes protons to be pumped out of the matrix against their electrochemical gradient, as the rotor and its stalk rotate in the direction opposite to that involved in ATP synthesis.

DIF: Moderate REF: 14.1 OBJ: 14.1.p Describe the conditions under which ATP synthase will act as a proton pump and hydrolyze ATP. MSC: Applying 11. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. The driving force that pulls protons into the matrix is called the proton-motive force, which is a combination of the large force due to the pH gradient and the smaller force that results from the voltage gradient across the inner mitochondrial membrane.

B. Under anaerobic conditions, the ATP synthase can hydrolyze ATP instead of synthesizing it.

C. ATP is moved out of the matrix, across the inner mitochondrial membrane, in a co-transporter that also brings ADP into the matrix.

D. Brown fat cells make less ATP because they have an inefficient ATP synthase.

ANS: A. False. Although it is true that both the pH gradient and the voltage gradient are components of the proton-motive force, it is the voltage gradient (also referred to as the membrane potential) that is the greater of the two.

B. True.

C. True.

D. False. The inner mitochondrial membranes in brown fat cells contain a transport protein that allows protons to move down their gradient without passing through the ATP synthase. As a result, less ATP is made and most of the energy from the proton gradient is released as heat.

DIF: Easy REF: 14.1 OBJ: 14.1.n Review the membrane potential and pH gradients across the inner mitochondrial membrane and state in which direction it is energetically favorable for protons to flow. | 14.1.p Describe the conditions under which ATP synthase will act as a proton pump and hydrolyze ATP. | MSC: Evaluating 12. Mitochondrial structure and the reaction products generated inside the matrix are critical for generating stores of energy. Answer the following questions based on what you know about mitochondrial structure and processes.

A. The gradients used to generate ATP are maintained across the inner mitochondrial membrane. Why don’t we observe a similar gradient generation across the outer mitochondrial membrane?

B. The proton-motive force created by the electrochemical proton gradient is the source of free energy utilized in ATP formation. Describe the two components contributing to the total proton-motive force.

ANS: A. The outer mitochondrial membrane contains large, channel-forming proteins called porins, which makes this membrane permeable to small molecules. Protons, other ions, nucleotides, and many other small molecules flow freely across this membrane, making it impossible to establish a gradient of any of these molecules on either side of this membrane.

B. One component of the proton-motive force is the concentration gradient of protons (or pH gradient) across the

membrane. This pH gradient makes it energetically favorable for protons to flow back into the matrix. A second component of the proton-motive force is a charge differential across the membrane, referred to as the membrane potential. Because the matrix side of the membrane has a net negative charge, and protons have a positive charge, this component of the proton-motive force also drives the movement of protons back into the matrix.

DIF: Moderate REF: 14.1 OBJ: 14.1.f Describe the structure of a mitochondrion and distinguish the functions and compositions of its different membranes and compartments. | 14.1.n Review the membrane potential and pH gradients across the inner mitochondrial membrane and state in which direction it is energetically favorable for protons to flow.

MSC: Understanding 13. A. Match each equation in column A with the corresponding standard redox potential in column B. Column A

Column B −

1. H2O ↔ ½O2 + 2H+ + 2 e

A. +30 mV

2. reduced ubiquinone ↔ oxidized ubiquinone + 2H+ + 2 e−

B. +820 mV

−

C. +230 mV

3. NADH ↔ NAD + H + 2 e +

+

−

4. reduced cytochrome c ↔ oxidized cytochrome c + e

D. −320 mV

B. How do these standard redox potentials support our understanding of the stepwise electron transfers that occur in the electron-transport chain?

C. Why would it not be advantageous for living systems to evolve a mechanism for the direct transfer of electrons from NADH to O2?

ANS: A. 1—B; 2—A; 3—D; 4—C B. Each successive member of the electron-transport chain is a better electron acceptor, which permits a unidirectional series of electron transfers until reaching O2, which is the best electron acceptor and the final destination of the electrons, forming water as oxygen is consumed.

C. If NADH directly donated electrons to O2, a large amount of energy would be released as heat and lost, rather than used as a way for the cell to generate chemical energy in the form of ATP. DIF: Moderate REF: 14.2 OBJ: 14.2.g Compare the redox potentials of components in the electron-transport-chain and state which way electrons will flow. | 14.2.e Explain why NADH does not donate its electrons directly to molecular oxygen in living systems. MSC: Understanding 14. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Ubiquinone is associated with the inner mitochondrial membrane as a protein-bound electron carrier molecule.

B. Ubiquinone can transfer only one electron in each cycle.

C. The iron–sulfur centers in NADH dehydrogenase are relatively poor electron acceptors.

D. Cytochrome c oxidase binds O2 using an iron–heme group, where four electrons are shuttled one at a time.

ANS: A. False. Ubiquinone is an aromatic compound that uses its long hydrocarbon tail to associate with the inner mitochondrial membrane.

B. False. Ubiquinone can transfer one or two electrons. In the case in which only one electron is transferred, the molecule contains an unpaired electron, which is highly reactive.

C. True.

D. True.

DIF: Easy REF: 14.2 OBJ: 14.2.f Contrast the redox potentials of iron-sulfur clusters and iron atoms held in heme groups and relate where proteins containing these metals are positioned along the electron-transport-chain. | 14.2.g Compare the redox potentials of components in the electron-transport-chain and state which way electrons will flow. | 14.2.h Identify the metals that help give cytochrome c oxidase its high redox potential.

MSC: Evaluating 15. Explain how the diffusion of ubiquinone in three dimensions is restricted in order to ensure its proximity to the other enzyme complexes for the rapid transfer of electrons?

ANS: Ubiquinone is a membrane-anchored protein, localized to the inner mitochondrial membrane via its hydrocarbon tail. This limits diffusion to lateral movement along the plane of the membrane, and ensures relative proximity to ETC complexes.

DIF: Easy REF: 14.1 OBJ: 14.1.f Describe the structure of a mitochondrion and distinguish the funct ions and compositions of its different membranes and compartments. MSC: Understanding 16. Consider a redox reaction between molecules A and B. Molecule A has a redox potential of −100 mV and molecule B has a redox potential of +100 mV. For the transfer of electrons from A to B, is the ΔG° positive or negative or zero?

Under what conditions will the reverse reaction, transfer of electrons from B to A, occur?

ANS: The ΔG° is negative. The sign of ΔG° is the opposite of that of ΔE0′ = E0′ (acceptor) − E0′ (donor). The acceptance of electrons by B from A has a ΔE0′ = 100 − (−100) = 200. The reverse reaction, the donation of electrons from B to A, has a positive ΔG° and is therefore unfavorable under standard conditions. Remember that, by definition, the concentrations of A and its redox pair A′ are equal under standard conditions; similarly, the concentration of B is equal to the concentration of its redox pair B′. B will be able to donate electrons to A only when [B] > [B′] and/or [A] < [A′] to such an extent that the ΔG for electron transfer becomes negative.

DIF: Hard REF: 14.2 OBJ: 14.2.g Compare the redox potentials of components in the electron-transport-chain and state which way electrons will flow. MSC: Applying 17. For each of the following sentences, choose one of the options enclosed in square brackets to make a correct statement.

An electron bound to a molecule with low affinity for electrons is a [high/low]-energy electron. Transfer of an electron from a molecule with low affinity to one with higher affinity has a [positive/negative] ΔG° and is thus [favorable/unfavorable] under standard conditions. If the reduced form of a redox pair is a strong electron donor with a [high/low] affinity for electrons, it is easily oxidized; the oxidized member of such a redox pair is a [weak/strong] electron acceptor.

ANS: An electron bound to a molecule with low affinity for electrons is a high-energy electron. Transfer of an electron from a molecule with low affinity to one with higher affinity has a negative ΔG° and is thus favorable under standard conditions. If the reduced form of a redox pair is a strong electron donor with a low affinity for electrons, it is easily oxidized; the oxidized member of such a redox pair is a weak electron acceptor.

DIF: Easy REF: 14.2 OBJ: 14.2.g Compare the redox potentials of components in the electron-transport-chain and state which way electrons will flow. MSC: Understanding 18. Explain how scientists used artificial vesicles to prove that the generation of ATP by the ATP synthase was not powered by a single high-energy intermediate but rather by a proton gradient. Be sure to describe the two

experiments that were negative controls (no ATP generated), the positive control (ATP generated as expected), and a fourth experiment proving that the gradient is the required energy source.

ANS: In all the experiments, artificial liposomes were generated and exposed to light, and the surrounding solution was checked for an increase in ATP. In the first experiment, the liposomal membranes contained only bacteriorhodopsin, a bacterial protein that pumps protons and is activated by light. In this negative control, ATP was not expected to be produced, and it was not. In the second experiment, also a negative control, the liposomes contained only ATP synthase. Again, if the chemiosmotic hypothesis was correct, ATP should not have been generated, which was what was observed. In the third experiment, both bacteriorhodopsin and ATP synthase were present in the liposomal membrane. When exposed to light, protons were pumped into the vesicle and ATP was generated. In the fourth experiment, to show that the ATP production was solely a result of the proton gradient, an uncoupling agent was added to the solution containing liposomes with bacteriorhodopsin and ATP synthase. In this case, even though the protons were being pumped into the liposomes, a gradient did not build up; this was because of the presence of the uncoupling agent, which made the membrane permeable to protons. No ATP was generated, proving that it was the proton gradient that was the energy source for ATP synthesis.

DIF: Hard REF: 14.2 OBJ: 14.2.n Outline how investigators used an artificial system including bacteriorhodopsin and ATP synthase to demonstrate the role that a proton gradient plays in producing ATP. MSC: Understanding 19. The respiratory chain is relatively inaccessible in the experimental manipulation of intact mitochondria. After disrupting mitochondria with ultrasound, however, it is possible to isolate functional submitochondrial particles, which consist of broken cristae that have resealed inside-out into small, closed vesicles. In these vesicles, the components that originally faced the matrix are now exposed to the surrounding medium.

A. How might such an arrangement aid in the study of electron transport and ATP synthesis?

B. Consider an anaerobic preparation of such submitochondrial particles. If a small amount of oxygen is added, do you predict that the preparation will consume oxygen in respiration reactions? Will the medium outside the particles become more acidic or more basic? What, if anything, will change if the flow of protons through ATP synthase is blocked by an inhibitor? Explain your answer.

ANS: A. This arrangement of components within the vesicles allows the experimental manipulation of the medium surrounding the vesicles, which permits the consequences of different conditions in the mitochondrial matrix to be examined. The medium can be altered by changing pH, adding electron carriers and oxygen, and providing ADP and Pi, for example. The oxidation of electron carriers, the consumption of oxygen, and the production of ATP can be measured in the medium. By changing the composition of the medium, it should be possible, for example, to identify the electron carriers that can donate electrons from the matrix to the transport chain (the side of the membrane that normally faces the matrix is now on the outside), to assess the redox potentials of various components of the transport chain, and to determine the dependence of ATP synthesis on the pH gradient across the membrane and on the ATP/ADP ratio.

B. Respiration reactions will rapidly consume at least some of the added oxygen. During the anaerobic conditions, the electron carriers in the electron-transport chain were reduced; on the addition of oxygen, electrons will be transferred to oxygen, thereby reducing the oxygen and oxidizing the carriers. Concomitantly with the electron flow, protons will be pumped from the medium into the vesicles, thereby making the medium slightly more basic and the inside of the vesicles acidic. Inhibition of the ATP synthase will not have an immediate effect on oxygen consumption or proton pumping. However, the proton concentration inside the vesicles will quickly become too high to

continue the activity of the electron-transport-coupled proton pumping, and thus electron transport and oxygen consumption will cease.

DIF: Hard REF: 14.2 OBJ: 14.2.g Compare the redox potentials of components in the electron-transport-chain and state which way electrons will flow. MSC: Analyzing 20. Porphyrin ring molecules are critical both for oxidative phosphorylation in the mitochondria and for photosynthesis in chloroplasts. Compare and contrast the role of this type of molecule in each process and explain the roles of O 2 and H2O in each.

ANS: Porphyrin rings are large, stable, aromatic molecules with catalytic metal ions in their centers. They can a ccept and donate electrons, making them critical for the redox reactions required for the electron -transport chain. In oxidative phosphorylation, a tightly bound heme in the cytochrome c oxidase complex is the porphyrin molecule that binds molecular oxygen, allowing this enzyme to coordinate the stepwise reduction of O 2 to two OH- ions (which requires four electrons) and the pumping of protons out of the mitochondrial matrix. The porphyrin molecule critical for photosynthesis is chlorophyll. For example, the reaction center of photosystem II contains antenna co mplexes (sets of protein-bound chlorophyll molecules) that harvest light energy and transfer the energy to excite the special pair of chlorophyll molecules at the reaction center. Chlorophyll molecules contain Mg 2+ rather than the Fe 2+ of cytochromes. The special pair of chlorophyll molecules in the reaction center transfers high-energy electrons to plastoquinone, a mobile electron carrier that then shuttles electrons to the electron -transport chain in the thylakoid membrane. Finally, the special pair that has lost its excited electron (and is thereby positively charged) regains an electron from water, in a reaction that is catalyzed by a tightly bound manganese-containing, water-splitting enzyme. Once four electrons have been removed from two water molecules in this way, a molecule of oxygen gas (O2) is released as a product.

DIF: Hard REF: 14.3 OBJ: 14.3.f Delineate the functions of chlorophyll’s porphyrin ring and hydrophobic tail.

MSC: Analyzing 21. Indicate whether the following statements are TRUE or FALSE. If a statement is false, explain why it is false.

A. Carbon fixation can be described as a process by which gaseous carbon-containing molecules are captured and incorporated into biological hydrocarbon molecules.

B. The electron-transport proteins, utilized in stage 1 of photosynthesis, reside in the inner membrane of the chloroplast.

C. Similar to oxidative phosphorylation, the electrons passed along the chloroplast electron-transport chain are ultimately passed on to a molecule of O2, to produce H2O. D. Stage 2 of photosynthesis involves a cycle of reactions that does not directly depend on energy derived from sunlight.

ANS: A. True.

B. False. The electron-transport system in chloroplasts resides in the thylakoid membrane.

C. False. The recipient of electrons in the chloroplast electron-transport chain is the NADP+ cofactor, which becomes reduced to NADPH. D. True.

DIF: Easy REF: 14.3 OBJ: 14.3.b Outline the events that take place during stage 1 of photosynthesis and compare this process to the oxidative phosphorylation that occurs in mitochondria. | 14.3.c Review the events that take place in

stage 2 of photosynthesis and indicate where these reactions occur. MSC: Evaluating 22. Use the terms provided below to fill in the blanks. Not all words or phrases will be used; each word or phrase may be used more than once. benzene

longer

porphyrin

blue

orange

red

electrons

photons

shorter

heme Photons from sunlight that are in the __________ wavelength range are preferentially absorbed by chlorophyll molecules to raise the energy levels of electrons in the __________ ring. The __________ reflected are lower in energy, which is indicated in the ________, green wavelengths detected by the human eye.

ANS: Photons from sunlight that are in the red wavelength range are preferentially absorbed by chlorophyll molecules to raise the energy levels of electrons in the porphyrin ring. The photons reflected are lower in energy, which is indicated in the longer, green wavelengths detected by the human eye.

DIF: Easy REF: 14.3 OBJ: 14.3.e Recall the wavelengths of light that are absorbed or reflected by chlorophyll.

MSC: Understanding 23. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

3-phosphoglycerate

NADPH

starch

ATP

pyruvate

sucrose

glyceraldehyde 3-phosphate

ribose 1,5-bisphosphate

NADH

ribulose 1,5-bisphosphate

In the carbon-fixation process in chloroplasts, carbon dioxide is initially added to the sugar __________. The final product of carbon fixation in chloroplasts is the three-carbon compound __________. This is converted into __________ (which can be used directly by the mitochondria), into __________ (which is exported to other cells), and into __________ (which is stored in the stroma). The carbon-fixation cycle requires energy in the form of __________ and reducing power in the form of __________.

ANS: In the carbon-fixation process in chloroplasts, carbon dioxide is initially added to the sugar ribulose 1,5bisphosphate. The final product of carbon fixation in chloroplasts is the three-carbon compound glyceraldehyde 3phosphate. This is converted into pyruvate (which can be used directly by the mitochondria), into sucrose (which is exported to other cells), and into starch (which is stored in the stroma). The carbon-fixation cycle requires energy in the form of ATP and reducing power in the form of NADPH. DIF: Easy REF: 14.3 OBJ: 14.3.p Identify the molecules that provide the energy to convert carbon dioxide into sugars. | 14.3.q Recall the role that ribulose 1,5-bisphosphate plays in the carbon-fixation-cycle. | 14.3.s Summarize the possible fates of glyceraldehyde 3-phosphate generated by the carbon-fixation-cycle. MSC: Understanding 24. Compare and contrast the process of reducing N2 to NH3 in Methanococcus jannaschii to the reduction of O2 to H20 in the mitochondrial matrix. What inferences can you draw with respect to the early evolution of energy-generating systems.

ANS: Methanococcus jannaschii fixes nitrogen gas, reducing it to ammonia, which is necessary to make nitrogen-

containing organic molecules (amino acids and nucleic acids). The reduction of N 2 requires a large investment of energy and the transfer of electrons. The electron transfer occurs inside membrane protein c omplexes and pumps protons to produce an electrochemical gradient. The proton gradient powers an ATP synthase to generate ATP, supplying chemical energy other reactions for the cell. This chemiosmotic coupling is very similar to what happens across the inner mitochondrial membrane where O2 is reduced by the cytochrome c oxidase complex. Electrons are moved through the ETC and used to reduce molecular oxygen. Simultaneously, protons are pumped across the membrane generating a proton gradient, and the flow of protons back through the ATP synthase is coupled to the generation of ATP. The similarities indicate that the use of chemical/charge gradients across membranes to power the production of ATP evolved almost 4 billion years ago.

DIF: Moderate REF: 14.4 OBJ: 14.4.f Express what the use of proton gradients by Methanococcus suggests about the evolution of chemiosmotic coupling. MSC: Analyzing 25. Describe how a standard flashlight battery can convert energy into useful work and explain how this is similar to the energy conversions in the mitochondria.

ANS: A battery contains chemicals that generate negatively charged ions at one pole, and it is able to cause the continuous transfer of electrons along a metal wire if that pole is connected to the other end of the battery. The energy released by the electron-transfer process driven by the battery can be harnessed to do useful work, as when it is used to run an electric motor. Likewise, the energy released by the electron transfers that occur between the protein complexes in the electron-transport chain does useful work when it drives the movement of protons to one side of the membrane, since the resulting proton gradient is then used to generate chemical energy in the form of ATP. DIF: Easy REF: 14.0 OBJ: 14.0.e Summarize the stages involved in generating ATP by oxidative phosphorylation.

MSC: Understanding 26. Human infants have a much larger portion of brown adipose tissue than adult humans. It was found that the mitochondria in brown adipocytes (brown fat cells) have a novel protein in the inner mitochondrial membrane. This protein, called the uncoupling protein (UCP), was found to transport protons from the intermembrane space into the matrix.

A. What is the impact of UCP on oxidative phosphorylation in the mitochondria of brown fat?

B. Propose an explanation for the higher proportion of brown fat cells in infants compared to adults.

ANS: A. A protein that transports protons into the mitochondrial matrix would diminish the proton gradient. Without the proton gradient, ATP will not be generated. However, the electron-transport chain can still work, as long as oxygen is present. The UCP, therefore, is a biological uncoupler of the oxidative phosphorylation process. The electrontransport chain will run in a futile cycle that does not convert the energy from redox reactions into chemical energy (ATP), but instead releases this energy as heat.

B. The thermogenesis resulting from the action of UCP is important for helping infants maintain a constant body temperature. As our body mass increases with age, our body temperatures are probably less susceptible to fluctuations, and therefore adults do not require the same amount of brown fat as infants.

DIF: Moderate REF: 14.2 OBJ: 14.2.l Explain how specialized carrier proteins in the inner mitochondrial membrane of brown fat cells allow those cells to oxidize fats to produce heat. MSC: Applying 27. In 1925, David Keilin used a simple spectroscope to observe the characteristic absorption bands of the cytochromes that participate in the electron-transport chain in mitochondria. A spectroscope passes a very bright light through the

sample of interest and then through a prism to display the spectrum from red to blue. If molecules in the sample absorb light of particular wavelengths, dark bands will interrupt the colors of the rainbow. His key discovery was that the absorption bands disappeared when oxygen was introduced and then reappeared when the samples became anoxic. Subsequent findings demonstrated that different cytochromes absorb light of different frequencies. When light of a characteristic wavelength shines on a mitochondrial sample, the amount of light absorbed is proportional to the amount of a particular cytochrome present in its reduced form. Thus, spectrophotometric methods can be used to measure how the amounts of reduced cytochromes change over time in response to various treatments. If isolated mitochondria are incubated with a source of electrons such as succinate, but without oxygen, electrons enter the respiratory chain, reducing each of the electron carriers almost completely. When oxygen is then introduced, the carriers oxidize at different rates, as can be seen from the decline in the amount of reduced cytochrome (see Figure 14-27). Note that cytochromes a1 and a3 cannot be distinguished and thus are listed as cytochrome (a1 + a3). How does this result allow you to order the electron carriers in the respiratory chain? What is their order?

Figure 14-27

ANS: This result allows you to order the electron carriers in the respiratory chain because when oxygen is added, the last carrier in the chain will be oxidized first. This is because oxygen is the final sink for the electrons that flow through the chain, and it participates directly in a redox reaction with the last electron carrier. The wave of oxidation will then proceed backward through the chain toward the first electron carrier in the chain; this is because the oxidation of each carrier will convert it to a form that can accept electrons from the “upstream” carrier in the chain, thereby oxidizing each upstream carrier sequentially. The order of cytochromes in the respiratory chain is the reverse of the order in which they are oxidized (that is, the order in which the reduced form is lost). Listed from first to last, the cytochromes in the chain are b, c1, c, and (a1 + a3).

DIF: Hard REF: 14.2 OBJ: 14.2.g Compare the redox potentials of components in the electron-transport-chain and state which way electrons will flow. MSC: Applying

CHAPTER 15

Intracellular Compartments and

Protein Transport MEMBRANE-ENCLOSED ORGANELLES 15.1.a Present two strategies cells use to isolate and organize their various chemical reactions, and identify which approach is more highly developed in eukaryotes than in prokaryotes.

15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each.

15.1.c Identify the eukaryotic organelles that are surrounded by double membranes.

15.1.d Recall what percentage of the volume of a typical eukaryotic cell is occupied by its membrane-enclosed organelles, and compare the area occupied by the plasma membrane versus its endoplasmic reticulum.

15.1.e Describe the relationship between the ER and the nuclear membrane.

15.1.f

Compare the functions of rough ER and smooth ER, and describe the types of cell in which each can be found.

15.1.g List the organelles that form the endomembrane system and review how the interiors of these organelles communicate with one another and with the cell exterior.

15.1.h Contrast the evolution of the nucleus with that of mitochondria and chloroplasts.

PROTEIN SORTING 15.2.a Outline the mechanisms by which proteins can enter membrane-enclosed organelles, and identify the organelles that use them.

15.2.b List the membrane-bound organelles that can receive proteins directly from the cytosol.

15.2.c Describe the fate of proteins that lack a sorting signal.

15.2.d Review where the proteins found in mitochondria and chloroplasts are synthesized.

15.2.e Relate what would happen if an ER signal sequence were removed from an ER protein and attached to a cytosolic protein.

15.2.f

Predict what would happen to a protein that bears both a nuclear localization signal and a nuclear export signal.

15.2.g Articulate what would happen to a protein that bears both an ER sorting signal and a nuclear localization signal.

15.2.h Describe the structure of the nuclear envelope.

15.2.i

Contrast the conformation adopted by proteins during nuclear transport and that of proteins transported into mito-

chondria and chloroplasts.

15.2.j

Explain how nuclear pores restrict the passage of larger molecules while allowing small, water-soluble molecules to pass freely between the nucleus and cytosol.

15.2.k Review how nuclear import receptors escort proteins bearing a nuclear localization signal from the cytosol into the

nucleus.

15.2.l

Describe the types of molecules transported by nuclear export receptors.

15.2.m Summarize how the energy supplied by GTP is used to drive nuclear transport.

15.2.n Articulate how mitochondrial proteins are recognized and transported into the mitochondrial matrix, and describe the role played by chaperones inside the organelle.

15.2.o Outline the two mechanisms through which proteins are transported into peroxisomes.

15.2.p Name the organelle that serves as an entry point for proteins destined for all organelles or for the cell surface.

15.2.q Contrast the destinations of the transmembrane and water-soluble proteins that are transferred from the cytosol to the ER. 15.2.r

Distinguish between free ribosomes and membrane-bound ribosomes.

15.2.s Identify the source of energy for the transport of proteins into the ER. 15.2.t

Compare a polyribosome and the rough ER.

15.2.u Review how the signal-recognition particle (SRP) and SRP receptor guide proteins containing an ER signal sequence to the ER membrane.

15.2.v Recall the location, functions, and ultimate fate of the ER signal sequence on soluble proteins.

15.2.w Compare the locations and fates of the ER signal and stop-transfer sequences of single-pass transmembrane proteins to the start-transfer and stop-transfer sequences of multipass transmembrane proteins.

VESICULAR TRANSPORT 15.3.a Contrast the endocytic and exocytic pathways in terms of directionality, purpose, and participation of certain organelles and membranes.

15.3.b Recall the function and fate of the protein coat that surrounds transport vesicles.

15.3.c Summarize how clathrin-coated vesicles select their cargo molecules and then bud from their parent membranes.

15.3.d Compare the processes of vesicle tethering, docking, and fusion, and list the proteins involved in each.

SECRETORY PATHWAYS 15.4.a List the types of covalent modifications that take place in the ER and describe the functions these modifications serve.

15.4.b Review how proteins are selected for glycosylation in the ER and how these sugars are attached to the protein.

15.4.c Recall how proteins destined to function in the ER are kept in the ER—or returned to the ER if they accidentally enter the Golgi apparatus.

15.4.d Summarize how misfolded or not fully assembled proteins are retained in the ER. 15.4.e Explain the unfolded protein response and outline the consequences of triggering this program.

15.4.f

Describe the structure and location of the Golgi apparatus.

15.4.g Articulate the ways in which proteins move from one cisterna to another within the Golgi apparatus.

15.4.h Compare how proteins are sorted in the cis and trans Golgi networks.

15.4.i

Recall how and where additional oligosaccharides are added to glycoproteins within the Golgi stack.

15.4.j

Contrast the constitutive and regulated exocytosis pathways, and describe the behavior of the proteins secreted by each pathway.

15.4.k Explain how lipids are delivered to—and removed from—the plasma membrane.

15.4.l

Outline how treatment with a protease can be used to determine whether a protein is transported into an orga-

nelle in vitro.

15.4.m Outline how temperature-sensitive mutants have been used to dissect the protein secretory pathway in yeast.

15.4.n Explain how GFP can be used to monitor the location and movement of proteins in living cells.

ENDOCYTIC PATHWAYS 15.5.a Distinguish between pinocytosis and phagocytosis, and indicate the types of cells that engage in each.

15.5.b Outline how microorganisms can be recognized, engulfed, and digested by phagocytic cells.

15.5.c Explain why pinocytosis does not decrease the surface area or volume of a cell.

15.5.d Compare pinocytosis and receptor-mediated endocytosis.

15.5.e Describe how cholesterol is transported into the cell cytosol by receptor-mediated endocytosis.

15.5.f

Summarize how viruses take advantage of receptor-mediated endocytosis.

15.5.g Outline the possible fates of receptor proteins and cargo molecules following endocytosis into endosomes.

15.5.h Recall how endosomes maintain their pH and explain how pH plays a crucial role in the endosomal sorting process.

15.5.i

Recall how lysosomes maintain their pH and describe the role that pH plays in lysosomal function.

15.5.j

List the types of enzymes present in lysosomes and summarize how these enzymes are transported into lysosomes.

15.5.k Outline the pathways that different materials follow to arrive in lysosomes.

15.5.l

Explain what happens to the final products of macromolecule digestion by lysosomes.

15.5.m Describe autophagy and explain when or why cells might use this pathway.

MULTIPLE CHOICE 1. Which of the following statements about the endoplasmic reticulum (ER) is FALSE? a. The ER is the major site for new membrane synthesis in the cell. b. Proteins to be delivered to the ER lumen are synthesized on the smooth ER.

c. Steroid hormones are synthesized on the smooth ER. d. The ER membrane is contiguous with the outer nuclear membrane.

ANS: B Proteins to be delivered to the ER lumen are synthesized on rough ER; these areas appear “rough” because ribosomes are attached to the cytosolic surface of these ER regions.

DIF: Easy REF:15.1 OBJ: 15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each. | 15.1.e Describe the relationship between the ER and the nuclear membrane. | 15.1.f Compare the functions of rough ER and smooth ER, and describe the types of cell in which each can be found.

MSC: Remembering 2. Which of the following statements about membrane-enclosed organelles is TRUE? a. In a typical cell, the area of the endoplasmic reticulum membrane far exceeds the area of plasma membrane. b. The nucleus is the only organelle that is surrounded by a double membrane. c. Other than the nucleus, most organelles are small and thus, in a typical cell, only about 10% of a cell’s volume is occupied by membrane-enclosed organelles; the other 90% of the cell volume is the cytosol. d. The nucleus is the only organelle that contains DNA. ANS: A The area of the endoplasmic reticulum membrane is 20–30 times that of the plasma membrane in a typical cell. Like the nucleus, chloroplasts and mitochondria are also surrounded by a double membrane. The cytosol is about half the volume of a typical eukaryotic cell, with membrane-enclosed organelles making up the other half of the volume.

Chloroplasts and mitochondria also carry their own genome, whereas the nucleus carries the genome of the organism.

DIF: Easy REF: 15.1 OBJ: 15.1.c Identify the eukaryotic organelles that are surrounded by a double membrane. | 15.1.d Recall what percentage of the volume of a typical eukaryotic cell is occupied by its membrane-enclosed organelles, and compare the area occupied by the plasma membrane versus its endoplasmic reticulum.

MSC: Remembering 3. Which of the following organelles is not part of the endomembrane system? a. Golgi apparatus b. the endosome c. mitochondria d. lysosomes ANS: C Mitochondria are not part of the endomembrane system, which is thought to have arisen initially through invagination of the plasma membrane. Instead, mitochondria (and chloroplasts) are thought to have evolved from a bacterium that was engulfed by a primitive eukaryotic cell.

DIF: Easy REF: 15.1 OBJ: 15.1.g List the organelles that form the endomembrane system and review how the interiors of these organelles communicate with one another and with the cell exterior. MSC: Remembering 4. Which of the following statements is TRUE? a. Lysosomes are believed to have originated from the engulfment of bacteria specialized for digestion. b. The nuclear membrane is thought to have arisen from the plasma membrane invaginating around the DNA.

c. Because bacteria do not have mitochondria, they cannot produce ATP in a membrane-dependent fashion. d. Chloroplasts and mitochondria share their DNA. ANS: B Lysosomes are part of the endomembrane system and are not thought to have come from the engulfment of an ancient prokaryotic cell. Bacteria use their plasma membrane for ATP production. Chloroplasts and mitochondria have their own DNA and do not share.

DIF: Easy REF: 15.1 OBJ: 15.1.c Identify the eukaryotic organelles that are surrounded by a double membrane. | 15.1.a Present two strategies cells use to isolate and organize their various chemical reactions, and identify which approach is more highly developed in eukaryotes than in prokaryotes. | 15.1.h Contrast the evolution of the nucleus with that of mitochondria and chloroplasts. MSC: Remembering 5. Where are chloroplast proteins translated? a. in the cytosol b. in the chloroplast c. on the endoplasmic reticulum d. in both the cytosol and the chloroplast ANS: D Proteins in the chloroplast are translated in the cytosol and in the chloroplast. The chloroplast proteins that are encoded by the nuclear DNA are translated in the cytosol, and the sorting signals on the protein direct them to the chl oroplast. The chloroplast proteins encoded by the chloroplast DNA are translated on ribosomes inside the chloroplast.

DIF: Moderate REF: 15.2 OBJ: 15.2.d Review where the proteins found in mitochondria and chloroplasts are synthesized. | 15.2.a Outline the mechanisms by which proteins can enter membrane-enclosed organelles, and identify the organelles that use them. MSC: Understanding 6. Proteins that are fully translated in the cytosol do not end up in a. the cytosol. b. the mitochondria. c. the interior of the nucleus. d. transport vesicles.

ANS: D Proteins destined for transport vesicles will be translated on ribosomes associated with the endoplasmic reticulum.

DIF: Moderate REF: 15.2 OBJ: 15.2.b List the membrane-bound organelles that can receive proteins directly from the cytosol. | 15.2.p Name the organelle that serves as an entry point for proteins destined for all organelles or for the cell surface. MSC: Understanding 7. Proteins that are fully translated in the cytosol and lack a sorting signal will end up in the a. cytosol. b. mitochondria. c. interior of the nucleus. d. nuclear membrane.

ANS: A Proteins produced in the cytosol that lack sorting signals remain in the cytosol. Proteins produced in the cytosol and

destined for the mitochondria or the interior of the nucleus will have a sorting signal to direct the protein to its proper location. Proteins destined for the nuclear membrane are not translated in the cytosol.

DIF: Moderate REF: 15.2 OBJ: 15.2.c Describe the fate of proteins that lack a sorting signal.

MSC: Understanding 8. Signal sequences that direct proteins to the correct compartment are a. added to proteins through post-translational modification. b. added to a protein by a protein translocator. c. encoded in the amino acid sequence and sufficient for targeting a protein to its correct destination. d. always removed once a protein is at the correct destination.

ANS: C Signal sequences are found within the amino acid sequence of proteins. They are sometimes removed when the protein is at the correct destination, but not all are removed. For example, nuclear import signals are not rem oved once a protein is inside the nucleus. A protein translocator resides in the membrane and helps transport soluble proteins across the membrane, but does not add signal sequences to proteins.

DIF: Moderate REF: 15.2 OBJ: 15.2.a Outline the mechanisms by which proteins can enter membrane-enclosed organelles and identify the organelles that use them. | 15.2.v, Recall the location, functions, and ultimate fate of the ER signal sequence on soluble proteins. | 15.2.c Describe the fate of proteins that lack a sorting signal.

MSC: Understanding 9. What is the role of the nuclear localization sequence in a nuclear protein? a. It is bound by cytoplasmic proteins that direct the nuclear protein to the nuclear pore. b. It is a hydrophobic sequence that enables the protein to enter the nuclear membranes. c. It aids in protein unfolding so that the protein can thread through nuclear pores. d. It prevents the protein from diffusing out of the nucleus through nuclear pores.

ANS: A The nuclear localization signal binds to a nuclear import receptor, found in the cytosol. The nuclear localization signal typically contains positively charged amino acids, not hydrophobic ones. Proteins are not unfolded as they enter the nucleus, do not diffuse through the nuclear pores, and are actively transported in and out of the nucleus.

DIF: Easy REF: 15.2 OBJ: 15.2.k Review how nuclear import receptors escort proteins bearing a nuclear localization signal from the cytosol into the nucleus. MSC: Remembering 10. Which of the following statements about nuclear transport is TRUE? a. mRNAs and proteins transit the nucleus through different types of nuclear pores. b. Nuclear import receptors bind to proteins in the cytosol and bring the proteins to the nuclear pores, where the proteins are released from the receptors into the pores for transit into the nucleus. c. Nuclear pores contain proteins with disordered segments that fill the channel and allow small water-soluble molecules to pass through in a non-selective fashion. d. Nuclear pores are made up of many copies of a single protein.

ANS: C Many of the proteins that line the nuclear pore have largely disordered segments. mRNAs and proteins can move through the same nuclear pore. Nuclear import receptors bind to proteins in the cytosol and transit with them across

the nuclear pore into the nucleus. Nuclear pores are made up of many copies of multiple proteins.

DIF: Easy REF: 15.2 OBJ: 15.2.j Explain how nuclear pores restrict the passage of larger molecules while allo wing small, water-soluble molecules to pass freely between the nucleus and cytosol. | 15.2.k Review how nuclear import receptors escort proteins bearing a nuclear localization signal from the cytosol into the nucleus.

MSC: Understanding 11. A large protein that passes through the nuclear pore must have an appropriate a. sorting sequence, which typically contains the positively charged amino acids lysine and arginine. b. sorting sequence, which typically contains the hydrophobic amino acids leucine and isoleucine. c. sequence to interact with the nuclear fibrils. d. Ran-interacting protein domain.

ANS: A The nuclear localization signal, which typically contains several positively charged lysines or arginines, is recognized by the nuclear import receptors. The nuclear import receptor interacts with the fibrils of the nuclear pore and Ran, which hydrolyses GTP. DIF: Easy REF: 15.2 OBJ: 15.2.k Review how nuclear import receptors escort proteins bearing a nuclear localization signal from the cytosol into the nucleus. MSC: Remembering 12. Your friend works in a biotechnology company and has discovered a drug that blocks the ability of Ran to exchange GDP for GTP. What is the most likely effect of this drug on nuclear transport? a. Nuclear transport receptors would be unable to bind cargo. b. Nuclear transport receptors would be unable to enter the nucleus. c. Nuclear transport receptors would be unable to release their cargo in the nucleus. d. Nuclear transport receptors would interact irreversibly with the nuclear pore fibrils.

ANS: C When Ran-GTP binds to the nuclear transport receptor, cargo is released. If Ran could not exchange its GDP for GTP, this would not happen. Ran-GTP is not needed for cargo binding, for nuclear entry, or for interactions with the nuclear pore fibrils during nuclear import.

DIF: Moderate REF: 15.2 OBJ: 15.2.m Summarize how the energy supplied by GTP is used to drive nuclear transport. MSC: Applying 13. Which of the following statements is TRUE? a. The signal sequences on mitochondrial proteins are usually at the C-terminus. b. Most mitochondrial proteins are not imported from the cytosol but are synthesized inside the mitochondria. c. Chaperone proteins in the mitochondria facilitate the movement of proteins across the outer and inner mitochondrial membranes. d. Mitochondrial proteins cross the membrane in their native, folded state.

ANS: C The signal sequences on a protein destined for the mitochondria are on its N-terminus. Although some mitochondrial proteins are synthesized inside the mitochondria from the mitochondrial genome, most mitochondrial proteins are encoded by genes in the nucleus and imported into the mitochondria after synthesis in the cytosol. Mitochondrial proteins are unfolded as they enter the mitochondria through protein translocators.

DIF: Easy REF: 15.2 OBJ: 15.2.n Articulate how mitochondrial proteins are recognized and transported into the mito-

chondrial matrix, and describe the role played by chaperones inside the organelle. MSC: Understanding 14. Which of the following statements about transport into mitochondria and chloroplasts is FALSE? a. The signal sequence on proteins destined for these organelles is recognized by a receptor protein in the outer membrane of these organelles. b. After a protein moves through the protein translocator in the outer membrane of these organelles, the protein diffuses in the lumen until it encounters a protein translocator in the inner membrane. c. Proteins that are transported into these organelles are unfolded as they are being transported. d. Signal peptidase will remove the signal sequence once the protein has been imported into these organelles.

ANS: B Once a protein is bound to the import receptor, the protein—in a complex that includes the protein translocator—will diffuse along the outer membrane until it reaches a specialized site where the inner and outer membranes contact each other, and will then be translocated simultaneously across the inner and outer membranes.

DIF: Easy REF: 15.2 OBJ: 15.2.i Contrast the conformation adopted by proteins during nuclear transport and that of proteins transported into mitochondria and chloroplasts. MSC: Understanding 15. Which of the following statements about peroxisomes is FALSE? a. Most peroxisomal proteins are synthesized in the ER. b. Peroxisomes synthesize phospholipids for the myelin sheath. c. Peroxisomes contain enzymes that help inactivate toxins. d. Proteins do not need to unfold to enter the peroxisome.

ANS: A Although peroxisomes can get some membrane-embedded proteins from the ER, most peroxisomal proteins are imported from the cytosol.

DIF: Easy REF: 15.2 OBJ: 15.2.o Outline the two mechanisms through which proteins are transported into peroxisomes. MSC: Understanding 16. Most proteins destined to enter the endoplasmic reticulum a. are transported across the membrane after their synthesis is complete. b. are completely translated on free ribosomes in the cytosol. c. begin to cross the membrane while still being synthesized. d. remain within the endoplasmic reticulum.

ANS: C Proteins destined to enter the endoplasmic reticulum have an N-terminal signal sequence that leads to the docking of the ribosome synthesizing the protein onto the ER and the entry of the protein across the ER membrane as the pol ypeptide chain is being synthesized.

DIF: Easy REF: 15.2 OBJ: 15.2.q Contrast the destinations of the transmembrane and water-soluble proteins that are transferred from the cytosol to the ER. | 15.2.t Compare a polyribosome and the rough ER. MSC: Understanding 17. After isolating the rough endoplasmic reticulum from the rest of the cytoplasm, you purify the RNAs attached to it.

Which of the following proteins do you expect the RNA from the rough endoplasmic reticulum to encode? a. soluble secreted proteins

b. ER membrane proteins c. plasma membrane proteins d. all of these answers are correct ANS: D The rough ER consists of ER membranes and polyribosomes that are in the process of translating and translocating proteins into the ER membrane and lumen. Thus, all proteins that end up in the lysosome, Golgi apparatus, or plasma membrane or that are secreted will be encoded by the RNAs associated with the rough ER. DIF: Moderate REF: 15.2 OBJ: 15.2.r Distinguish between free ribosomes and membrane-bound ribosomes. | 15.2.p Name the organelle that serves as an entry point for proteins destined for all organelles or for the cell surface. | 15.2.q Contrast the destinations of the transmembrane and water-soluble proteins that are transferred from the cytosol to the ER. MSC: Applying 18. In which cellular location would you expect to find ribosomes translating mRNAs that encode ribosomal proteins? a. the nucleus b. on the rough ER c. in the cytosol d. in the lumen of the ER ANS: C Ribosomes are cytoplasmic proteins and thus their protein components are translated in the cytosol.

DIF: Moderate REF: 15.2 OBJ: 15.2.b List the membrane-bound organelles that can receive proteins directly from the cytosol. | 15.2.q. Contrast the destinations of the transmembrane and water-soluble proteins that are transferred from the cytosol to the ER. MSC: Applying 19. You are interested in Fuzzy, a soluble protein that functions within the ER lumen. Given that information, which of the following statements must be TRUE? a. Fuzzy has a C-terminal signal sequence that binds to SRP. b. Only one ribosome can be bound to the mRNA encoding Fuzzy during translation. c. Fuzzy must contain a hydrophobic stop-transfer sequence. d. Once the signal sequence from Fuzzy has been cleaved, the signal peptide will be ejected into the ER membrane and degraded.

ANS: D ER signal sequences are removed from ER luminal proteins once they are translocated into the cytosol. ER signal sequences are typically at the N-terminus (and not the C-terminus) and bind to SRP. Hydrophobic stop-transfer sequences are found on membrane-inserted proteins and not on soluble proteins. More than one ribosome can bind to an mRNA molecule.

DIF: Moderate REF: 15.2 OBJ: 15.2.w Compare the locations and fates of the ER signal sequence and stop-transfer sequence of single-pass transmembrane proteins and the start-transfer and stop-transfer sequences of multipass transmembrane proteins. | 15.2.v Recall the location, functions, and ultimate fate of the ER signal sequence on soluble proteins. | 15.2.u Review how the signal-recognition particle (SRP) and SRP receptor guide proteins containing an ER signal sequence to the ER membrane. MSC: Applying 20. Which of the following statements about a protein in the lumen of the ER is FALSE?

a. A protein in the lumen of the ER is synthesized by ribosomes on the ER membrane. b. Some of the proteins in the lumen of the ER can end up in the extracellular space. c. Some of the proteins in the lumen of the ER can end up in the lumen of an organelle in the endomembra ne system. d. Some of the proteins in the lumen of the ER can end up in the plasma membrane.

ANS: D Plasma membrane proteins come from proteins in the ER membrane, not from the ER lumen.

DIF: Moderate REF: 15.2 OBJ: 15.2.q Contrast the destinations of the transmembrane and water-soluble proteins that are transferred from the cytosol to the ER. MSC: Understanding 21. Which of the following statements is TRUE? a. Proteins destined for the ER are translated by a special pool of ribosomes whose subunits are always associated with the outer ER membrane. b. Proteins destined for the ER translocate their associated mRNAs into the ER lumen where they are translated. c. Proteins destined for the ER are translated by cytosolic ribosomes and are targeted to the ER when a signal sequence emerges during translation. d. Proteins destined for the ER are translated by a pool of cytosolic ribosomes that contain ER-targeting sequences that interact with ER-associated protein translocators.

ANS: C DIF: Easy REF: 15.2 OBJ: 15.2.v Recall the location, functions, and ultimate fate of the ER signal sequence on soluble proteins. MSC: Understanding 22. Figure 15-22 shows the organization of a protein that normally resides in the plasma membrane. The boxes labeled 1 and 2 represent membrane-spanning sequences and the arrow represents a site of action of signal peptidase. Given this diagram, which of the following statements must be TRUE?

Figure 15-22

a. The N-terminus of this protein is cytoplasmic. b. The C-terminus of this protein is cytoplasmic. c. The mature version of this protein will span the membrane twice, with both the N- and C-terminus in the cytoplasm. d. none of these answers are correct.

ANS: B The mature version of this protein will span the membrane once, with membrane-spanning segment 2 in the membrane and the C-terminus facing the cytoplasm.

DIF: Hard REF: 15.2 OBJ: 15.2.w Compare the locations and fates of the ER signal sequence and stop-transfer sequence of single-pass transmembrane proteins and the start-transfer and stop-transfer sequences of multipass transmembrane proteins. MSC: Applying 23. Which of the following choices reflects the appropriate order of locations through which a protein destined for the plasma membrane travels?

a. lysosome → endosome → plasma membrane b. ER → lysosome → plasma membrane c. Golgi → lysosome → plasma membrane d. ER → Golgi → plasma membrane ANS: D DIF: Easy REF: 15.3 OBJ: 15.3.a Contrast the endocytic and exocytic pathways in terms of directionality, purpose, and the participating organelles and membranes. MSC: Understanding 24. Which of the following statements about vesicle budding from the Golgi is FALSE? a. Clathrin molecules are important for binding to and selecting cargoes for transport. b. Adaptins interact with clathrin. c. Once vesicle budding occurs, clathrin molecules are released from the vesicle. d. Clathrin molecules act at the cytosolic surface of the Golgi membrane.

ANS: A Cargo binds to cargo receptors. Adaptin molecules capture cargo receptors, which bind to the appropriate cargo molecules for incorporation into the vesicle.

DIF: Easy REF: 15.3 OBJ: 15.3.c Summarize how clathrin-coated vesicles select their cargo molecules and then bud from their parent membrane. | 15.3.b Recall the function and fate of the protein coat that surrounds transport vesicles.

MSC: Understanding 25. Molecules to be packaged into vesicles for transport are selected by a. clathrin. b. adaptins. c. dynamin. d. SNAREs. ANS: B DIF: Easy REF: 15.3 OBJ: 15.3.c Summarize how clathrin-coated vesicles select their cargo molecules and then bud from their parent membrane. 15.3.b | Recall the function and fate of the protein coat that surrounds transport vesicles. MSC: Remembering 26. Which of the following protein families are NOT involved in directing transport vesicles to the target membrane? a. SNAREs b. Rabs c. tethering proteins d. adaptins ANS: D Adaptins are involved in vesicle budding and are removed during the uncoating process, and thus should not be present when the vesicle reaches its target.

DIF: Easy REF: 15.3 OBJ: 15.3.d Compare the processes of vesicle tethering, docking, and fusion, and list the proteins involved in each. MSC: Analyzing 27. Your friend has just joined a lab that studies vesicle budding from the Golgi and has been given a cell line that does not form mature vesicles. He wants to start designing some experiments but was not listening carefully when he was told about the molecular defect of this cell line. He is too embarrassed to ask and comes to you for help. He does

recall that this cell line forms coated pits but vesicle budding and the removal of coat proteins don’t happen. Which of the following proteins might be lacking in this cell line? a. clathrin b. Rab c. dynamin d. adaptin ANS: C Given that coated pits can form but no vesicle budding is seen, dynamin is the most likely answer. Since coated pits are formed, clathrin and adaptin are unlikely to be the answer, because they are involved in the initial shaping of the vesicle into the pit. Rab proteins are involved in the recognition of the transport vesicle with its target membrane and not with vesicle budding.

DIF: Moderate REF: 15.3 OBJ: 15.3.c Summarize how clathrin-coated vesicles select their cargo molecules and then bud from their parent membrane. MSC: Applying 28. An individual transport vesicle a. contains only one type of protein in its lumen. b. will fuse with only one type of membrane. c. is endocytic if it is traveling toward the plasma membrane. d. is enclosed by a membrane with the same lipid and protein composition as the membrane of the donor organelle.

ANS: B An individual transport vesicle only fuses with one type of membrane. An individual vesicle may contain more than one type of protein in its lumen, all of which will contain the same sorting signal (or will lack specific sorting signals).

Endocytic vesicles generally move away from the plasma membrane. The vesicle membrane will not necessarily contain the same lipid and protein composition as the donor organelle, because the vesicle is formed from a selected subsection of the organelle membrane from which it budded.

DIF: Easy REF: 15.3 OBJ: 15.3.c Summarize how clathrin-coated vesicles select their cargo molecules and then bud from their parent membrane. MSC: Understanding 29. Which of the following statements about vesicular membrane fusion is FALSE? a. Membrane fusion does not always immediately follow vesicle docking. b. The hydrophilic surfaces of membranes have water molecules associated with them that must be displaced before vesicle fusion can occur. c. The GTP hydrolysis of the Rab proteins provides the energy for membrane fusion. d. The interactions of the v-SNAREs and the t-SNAREs pull the vesicle membrane and the target organelle membrane together so that their lipids can intermix.

ANS: C Rab proteins are important for docking, but are not involved in the catalysis of membrane fusion.

DIF: Moderate REF: 15.3 OBJ: 15.3.d Compare the processes of vesicle tethering, docking, and fusion, and list the proteins involved in each. MSC: Understanding 30. N-linked oligosaccharides on secreted glycoproteins are attached to a. nitrogen atoms in the polypeptide backbone. b. the serine or threonine in the sequence Asn-X-Ser/Thr.

c. the N-terminus of the protein. d. the asparagine in the sequence Asn-X-Ser/Thr. ANS: D DIF: Easy REF: 15.4 OBJ: 15.4.a List the types of covalent modifications that take place in the ER and describe the functions these modifications serve. MSC: Remembering 31. Which of the following statements about disulfide bond formation is FALSE? a. Disulfide bonds do not form under reducing environments. b. Disulfide bonding occurs by the oxidation of pairs of cysteine side chains on the protein. c. Disulfide bonding stabilizes the structure of proteins. d. Disulfide bonds form spontaneously within the ER because the lumen of the ER is oxidizing.

ANS: D An enzyme in the ER lumen catalyzes disulfide bond formation.

DIF: Easy REF: 15.4 OBJ: 15.4.a List the types of covalent modifications that take place in the ER and describe the functions these modifications serve. MSC: Understanding 32. Cells have oligosaccharides displayed on their cell surface that are important for cell–cell recognition. Your friend discovered a transmembrane glycoprotein, GP1, on a pathogenic yeast cell that is recognized by human immune cells. He decides to purify large amounts of GP1 by expressing it in bacteria. To his purified protein he then adds a branched 14-sugar oligosaccharide to the asparagine of the only Asn-X-Ser sequence found on GP1 (Figure 15-32).

Unfortunately, immune cells do not seem to recognize this synthesized glycoprotein. Which of the following statements is a likely explanation for this problem?

Figure 15-32

a. The oligosaccharide should have been added to the serine instead of the asparagine. b. The oligosaccharide should have been added one sugar at a time. c. The oligosaccharide needs to be further modified before it is mature.

d. The oligosaccharide needs a disulfide bond.

ANS: C Oligosaccharides are usually further modified by enzymes in the ER and the Golgi before the glycoprotein is inserted into the plasma membrane. The other choices are untrue, and thus are not good explanations. Oligosaccharides are added to the Asn and not the Ser and are added as a branched 14-sugar oligosaccharide. Disulfide bonding occurs between cysteines of proteins and not in sugars.

DIF: Hard REF: 15.4 OBJ: 15.4.a List the types of covalent modifications that take place in the ER and describe the functions these modifications serve. | 15.4.b Review how proteins are selected for glycosylation in the ER and how these sugars are attached to the protein. MSC: Applying 33. Different glycoproteins can have a diverse array of oligosaccharides. Which of the statements below about this diversity is TRUE? a. Extensive modification of oligosaccharides occurs in the extracellular space. b. Different oligosaccharides are covalently linked to proteins in the ER and the Golgi. c. A diversity of oligosaccharyl transferases recognizes specific protein sequences, resulting in the linkage of a variety of oligosaccharides to proteins. d. Oligosaccharide diversity comes from modifications that occur in the ER and the Golgi of the 14-sugar oligosaccharide added to the protein in the ER. ANS: D DIF: Easy REF: 15.4 OBJ: 15.4.a List the types of covalent modifications that take place in the ER and describe the functions these modifications serve. MSC: Understanding 34. Which of the following statements about the protein quality control system in the ER is FALSE? a. Chaperone proteins help misfolded proteins fold properly. b. Proteins that are misfolded are degraded in the ER lumen. c. Protein complexes are checked for proper assembly before they can exit the ER. d. A chaperone protein will bind to a misfolded protein to retain it in the ER. ANS: B Proteins that are misfolded are exported from the ER into the cytosol, where they are degraded.

DIF: Easy REF: 15.4 OBJ: 15.4.d Summarize how misfolded proteins or proteins that have not fully assembled are retained in the ER. MSC: Understanding 35. Which of the following statements about the unfolded protein response (UPR) is FALSE? a. Activation of the UPR results in the production of more ER membrane. b. Activation of the UPR results in the production of more chaperone proteins. c. Activation of the UPR occurs when receptors in the cytoplasm sense misfolded proteins. d. Activation of the UPR results in the cytoplasmic activation of gene regulatory proteins.

ANS: C The receptors for the unfolded proteins are on the ER membrane, and they sense the misfolded proteins using their luminal domains.

DIF: Easy REF: 15.4 OBJ: 15.4.e Explain the unfolded protein response and outline the consequences of triggering this program. MSC: Analyzing 36. Vesicles from the ER enter the Golgi at the

a. medial cisternae. b. trans Golgi network. c. cis Golgi network. d. trans cisternae.

ANS: C DIF: Easy REF: 15.4 OBJ: 15.4.h Compare how proteins are sorted in the cis and trans Golgi networks.

MSC: Remembering 37. Which of the following statements about secretion is TRUE? a. The membrane of a secretory vesicle will fuse with the plasma membrane when it discharges its contents to the cell’s exterior. b. Vesicles for regulated exocytosis will not bud off the trans Golgi network until the appropriate signal has been received by the cell. c. The signal sequences of proteins destined for constitutive exocytosis ensure their packaging into the correct vesicles. d. Proteins destined for constitutive exocytosis aggregate as a result of the acidic pH of the trans Golgi network.

ANS: A When secretory vesicles fuse with the plasma membrane, the membrane of the secretory vesicle becomes part of the plasma membrane. Vesicles for regulated exocytosis bud from the trans Golgi network and accumulate at the plasma membrane until the appropriate signal has been received. There are no signal sequences for proteins destined for exocytosis. Those proteins that are to be secreted by regulated exocytosis aggregate in the trans Golgi network as a result of the acidic pH and high Ca 2+ concentrations; those proteins that do not aggregate are packed into transport vesicles for constitutive exocytosis.

DIF: Easy REF: 15.4 OBJ: 15.4.j Contrast the constitutive and regulated exocytosis pathways, and describe the behavior of the proteins secreted by each pathway. | 15.4.k Explain how lipids are delivered to—and removed from— the plasma membrane. MSC: Understanding 38. Which of the following statements about phagocytic cells in animals is FALSE? a. Phagocytic cells are important in the gut to take up large particles of food. b. Phagocytic cells scavenge dead and damaged cells and cell debris. c. Phagocytic cells can engulf invading microorganisms and deliver them to their lysosomes for destruction. d. Phagocytic cells extend pseudopods that surround the material to be ingested.

ANS: A Although some unicellular eukaryotes ingest food particles by phagocytosis, phagocytosis is not involved in digestion in the animal gut.

DIF: Easy REF: 15.5 OBJ: 15.5.a Distinguish between pinocytosis and phagocytosis, and indicate the types of cells that engage in each. MSC: Understanding 39. Which of the following is NOT a process that delivers material to the lysosome? a. pinocytosis b. phagocytosis c. transcytosis d. autophagy

ANS: C Transcytosis is the process by which the endosome delivers cargo across the cell to a different plasma membrane domain.

DIF: Easy REF: 15.5 OBJ: 15.5.k Outline the pathways that different materials follow to arrive in lysosomes.

MSC: Remembering 40. You are working in a biotech company that has discovered a small-molecule drug called H5434. H5434 binds to LDL receptors when they are bound to cholesterol. H5434 binding does not alter the conformation of the LDL receptor’s intracellular domain. Interestingly, in vitro experiments demonstrate that addition of H5434 increases the affinity of LDL for cholesterol and prevents cholesterol from dissociating from the LDL receptor even in acidic conditions. Which of the following is a reasonable prediction of what may happen when you add H5434 to cells? a. Cytosolic cholesterol levels will remain unchanged relative to normal cells. b. Cytosolic cholesterol levels will decrease relative to normal cells. c. The LDL receptor will remain on the plasma membrane. d. The uncoating of vesicles will not occur.

ANS: B Normally, cholesterol dissociates from the LDL receptor in the acidic environment of the endosomes and is released into the cytosol. If the drug prevents cholesterol from dissociating from the LDL receptor in acidic conditions, cholesterol may not become released into the cytosol, and thus cytosolic cholesterol levels are likely to decrease relative to those in normal cells. There is no reason to believe that the LDL receptor will remain on the plasma membrane, because the cytosolic region of the receptor is not directly altered by the drug. Vesicle uncoating is also unlikely to be altered, because this occurs after vesicles are pinched off from the membrane.

DIF: Moderate REF: 15.5 OBJ: 15.5.e Describe how cholesterol is transported into the cell cytosol by receptormediated endocytosis. MSC: Applying

MATCHING 1. Match each cell part with the correct label in Figure 15-1.

Figure 15-1

1. nucleus 2. peroxisome 3. rough endoplasmic reticulum 4. Golgi apparatus 5. cytosol 6. endosome 7. plasma membrane 8. lysosome 9. mitochondrion 10. free ribosomes 1. ANS: D DIF: Easy REF: 15.1 OBJ: 15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each. MSC: Remembering 2. ANS: G DIF: Easy REF: 15.1 OBJ: 15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each. MSC: Remembering 3. ANS: C DIF: Easy REF: 15.1 OBJ: 15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each. MSC: Remembering 4. ANS: B DIF: Easy REF: 15.1 OBJ: 15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each. MSC: Remembering 5. ANS: H DIF: Easy REF: 15.1 OBJ: 15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each. MSC: Remembering 6. ANS: J DIF: Easy REF: 15.1 OBJ: 15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each. MSC: Remembering 7. ANS: E DIF: Easy REF: 15.1 OBJ: 15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each. MSC: Remembering 8. ANS: I DIF: Easy REF: 15.1 OBJ: 15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each. MSC: Remembering 9. ANS: A DIF: Easy REF: 15.1 OBJ: 15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each. MSC: Remembering 10. ANS: F DIF: Easy REF: 15.1 OBJ: 15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each. MSC: Remembering 2. Match the components involved with ER transport with the appropriate cellular location. Locations can be used more than once, or not at all.

Location A. cytosol B. ER lumen C. ER membrane Components

1. signal-recognition particle 2. protein translocator 3. mRNA 4. SRP receptor 5. active site of signal peptidase 1. ANS: A DIF: Easy REF: 15.2 OBJ: 15.2.u Review how the signal-recognition particle (SRP) and SRP receptor guide proteins containing an ER signal sequence to the ER membrane. MSC: Remembering 2. ANS: C DIF: Easy REF: 15.2 OBJ: 15.2.u Review how the signal-recognition particle (SRP) and SRP receptor guide proteins containing an ER signal sequence to the ER membrane. MSC: Remembering 3. ANS: A DIF: Easy REF: 15.2 OBJ: 15.2.u Review how the signal-recognition particle (SRP) and SRP receptor guide proteins containing an ER signal sequence to the ER membrane. MSC: Remembering 4. ANS: C DIF: Easy REF: 15.2 OBJ: 15.2.u Review how the signal-recognition particle (SRP) and SRP receptor guide proteins containing an ER signal sequence to the ER membrane. MSC: Remembering 5. ANS: B DIF: Easy REF: 15.2 OBJ: 15.2.u Review how the signal-recognition particle (SRP) and SRP receptor guide proteins containing an ER signal sequence to the ER membrane. MSC: Remembering 3. Match each cellular comparment with the correct label in Figure 15-3.

Figure 15-3

A. cisterna B. Golgi stack C. secretory vesicle D. trans Golgi network E. cis Golgi network 1. Label Line 1 ANS: B DIF: Easy REF: 15.4 OBJ: 15.4.f Describe the structure and location of the Golgi apparatus.

MSC: Remembering 2. Label Line 2

ANS: E DIF: Easy REF: 15.4 OBJ: 15.4.f Describe the structure and location of the Golgi apparatus.

MSC: Remembering 3. Label Line 3 ANS: A DIF: Easy REF: 15.4 OBJ: 15.4.f Describe the structure and location of the Golgi apparatus.

MSC: Remembering 4. Label Line 4 ANS: D DIF: Easy REF: 15.4 OBJ: 15.4.f Describe the structure and location of the Golgi apparatus.

MSC: Remembering 5. Label Line 5 ANS: C DIF: Easy REF: 15.4 OBJ: 15.4.f Describe the structure and location of the Golgi apparatus.

MSC: Remembering

SHORT ANSWER 1. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. cytosol

Golgi apparatus

nucleus

endoplasmic reticulum

lysosome

peroxisomes

endosomes

mitochondria

plasma membrane

The __________ makes up about half of the total cell volume of a typical eukaryotic cell. Ingested materials within the cell will pass through a series of compartments called __________ on their way to the __________, which contains digestive enzymes and will ultimately degrade the particles and macromolecules taken into the cell and will also degrade worn-out organelles. The __________ has a cis and trans face and receives proteins and lipids from the __________, a system of interconnected sacs and tubes of membranes that typically extends throughout the cell.

ANS: cytosol, endosomes, lysosome, Golgi apparatus, endoplasmic reticulum The cytosol makes up about half of the total cell volume of a typical eukaryotic cell. Ingested materials within the cell will pass through a series of compartments called endosomes on their way to the lysosome, which contains digestive enzymes and will ultimately degrade the particles and macromolecules taken into the cell and will also degrade worn-out organelles. The Golgi apparatus has a cis and trans face and receives proteins and lipids from the endoplasmic reticulum, a system of interconnected sacs and tubes of membranes that typically extends throughout the cell.

DIF: Easy REF: 15.1 OBJ: 15.1.b List the major membrane-enclosed organelles of the eukaryotic cell and briefly describe the function of each. MSC: Understanding 2. You discover a fungus that contains a strange star-shaped organelle not found in any other eukaryotic cell you have seen. On further investigation, you find the following.

1. The organelle possesses a small genome in its interior.

2. The organelle is surrounded by two membranes.

3. Vesicles do not pinch off from the organelle membrane.

4. The interior of the organelle contains proteins similar to those of many bacteria.

5. The interior of the organelle contains ribosomes.

How might this organelle have arisen?

ANS: A genome, a double membrane, ribosomes, and proteins similar to those found in bacteria are evidence for an organelle having evolved from an engulfed bacterium.

DIF: Moderate REF: 15.1 OBJ: 15.1.h Contrast the evolution of the nucleus with that of mitochondria and chlor oplasts. MSC: Applying 3. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. amino acid sequence

Golgi apparatus

sorting signal

endoplasmic reticulum

plasma membrane

transport vesicles

folded

protein translocators

unfolded

Plasma membrane proteins are inserted into the membrane in the __________. The address information for protein sorting in a eukaryotic cell is contained in the __________ of the proteins. Proteins enter the nucleus in their __________ form. Proteins that remain in the cytosol do not contain a __________. Proteins are transported into the Golgi apparatus via __________. The proteins transported into the endoplasmic reticulum by __________ are in their __________form.

ANS: Plasma membrane proteins are inserted into the membrane in the endoplasmic reticulum. The address information for protein sorting in a eukaryotic cell is contained in the amino acid sequence of the proteins. Proteins enter the nucleus in their folded form. Proteins that remain in the cytosol do not contain a signal. Proteins are transported into the Golgi apparatus via vesicles. The proteins transported into the endoplasmic reticulum by protein translocators are in their unfolded form.

DIF: Easy REF: 15.2 OBJ: 15.2.a Outline the mechanisms by which proteins can enter membrane-enclosed organelles and identify the organelles that use them. | 15.2.b List the membrane-bound organelles that can receive proteins directly from the cytosol. | 15.2.c Describe the fate of proteins that lack a sorting signal. MSC: Understanding 4. A gene regulatory protein, A, contains a typical nuclear localization signal but surprisingly is usually found in the cytosol. When the cell is exposed to hormones, protein A moves from the cytosol into the nucleus, where it turns on genes involved in cell division. When you purify protein A from cells that have not been treated with hormones, you find that protein B is always complexed with it. To determine the function of protein B, you engineer cells lacking the gene for protein B. You compare normal and defective cells by using differential centrifugation to separate the nuclear fraction from the cytoplasmic fraction, and then separating the proteins in these fractions by gel electrophoresis. You identify the presence of protein A and protein B by looking for their characteristic bands on the gel. The gel you run is shown in Figure 15-4.

Figure 15-4

On the basis of these results, what is the function of protein B? Explain your conclusion and propose a mechanism for how protein B works.

ANS: The data on the gel show that protein A is always found in the nucleus in the absence of protein B. Therefore, any mechanism that is proposed must explain this result. One possible answer is that protein B binds protein A and masks the nuclear localization signal. In the presence of hormone, protein B interacts with the hormone, which changes its conformation so that it can no longer bind protein A. When protein B no longer binds to protein A, the nuclear localization signal on protein A is now exposed and protein A can enter the nucleus. Therefore, in the absence of protein B, the nuclear localization signal on protein A is always exposed and protein A resides in the nucleus. Another possible answer is that protein B binds protein A and sequesters it by keeping protein A in some subcellular compartment, away from the nucleus. In the presence of hormone, protein B interacts with the hormone, changing its conformation so that it can no longer bind to protein A. When protein B is not present, protein A can enter the nucleus in the presence or absence of hormone.

DIF: Hard REF: 15.2 OBJ: 15.2.j Explain how nuclear pores restrict the passage of larger molecules while allowing small, water-soluble molecules to pass freely between the nucleus and cytosol. | 15.2.k Review how nuclear import receptors escort proteins bearing a nuclear localization signal from the cytosol into the nucleus.

MSC: Evaluating 5. You are trying to identify the peroxisome-targeting sequence in the thiolase enzyme in yeast. The thiolase enzyme normally resides in the peroxisome and therefore must contain amino acid sequences that are used to target the enzyme for import into the peroxisome. To identify the targeting sequences, you create a set of hybrid genes that encode fusion proteins containing part of the thiolase protein fused to another protein, histidinol dehydrogenase (HDH). HDH is a cytosolic enzyme required for the synthesis of the amino acid histidine and cannot function if it is localized in the peroxisome. You genetically engineer a series of yeast cells to express these fusion proteins instead of their own versions of these enzymes. If the fusion proteins are imported into the peroxisome, the HDH portion of the protein cannot function and the yeast cells cannot grow on a medium lacking histidine. You obtain the results shown in Figure 15-5.

Figure 15-5

What region of the thiolase protein contains the peroxisomal targeting sequence? Explain your answer.

ANS: The peroxisomal targeting sequence lies between amino acids number 100 and number 125. Any fusion protein containing this sequence can be targeted for import into the peroxisome (because the yeast cannot grow on a medium lacking histidine), whereas the fusion proteins lacking this region do not target the fusion protein for import into the peroxisome (because the yeast do grow on medium lacking histidine). The most important pieces of data are from the fusion protein containing amino acids 100–200 of the thiolase protein fused to HDH and the fusion protein containing amino acids 1–125 of the thiolase protein fused to HDH. Neither of these fusion proteins allow growth on medium lacking histidine and can be used to define the minimal region necessary for targeting thiolase for import into the peroxisome. (Note that although these experiments show that amino acids 100–125 are necessary, these experiments do not show that this region is sufficient for peroxisomal targeting. It is possible that the region consisting of amino acids 100–125 is sufficient, or it could be that this region collaborates with redundant signals between amino acids 1 and 100 or between amino acids 125 and 200.)

DIF: Hard REF: 15.2 OBJ: 15.2.o Outline the two mechanisms through which proteins are transported into peroxisomes. MSC: Applying 6. What would happen in each of the following cases? Assume in each case that the protein involved is a soluble protein, not a membrane protein.

A. You add a signal sequence (for the ER) to the N-terminal end of a normally cytosolic protein.

B. You change the hydrophobic amino acids in an ER signal sequence into charged amino acids.

C. You change the hydrophobic amino acids in an ER signal sequence into other hydrophobic amino acids.

D. You move the N-terminal ER signal sequence to the C-terminal end of the protein.

ANS: A. The protein will now be transported into the ER lumen.

B. The altered signal sequence will not be recognized and the protein will remain in the cytosol.

C. The protein will still be delivered into the ER. It is the distribution of hydrophobic amino acids that is important, not

the actual sequence.

D. The protein will not enter the ER. Because the C-terminus of the protein is the last part to be made, the ribosomes synthesizing this protein will not be recognized by the signal-recognition particle (SRP) and hence not carried to the ER. DIF: Hard REF: 15.2 OBJ: 15.2.e Relate what would happen if an ER signal sequence were removed from an ER protein and attached to a cytosolic protein. | 15.2.v Recall the location, functions, and ultimate fate of the ER signal sequence on soluble proteins. MSC: Applying 7. Briefly describe the mechanism by which an internal stop-transfer sequence in a protein causes the protein to become embedded in the lipid bilayer as a transmembrane protein with a single membrane-spanning region. Assume that the protein has an N-terminal signal sequence and just one internal hydrophobic stop-transfer sequence.

ANS: The N-terminal signal sequence initiates translocation and the protein chain starts to thread through the translocation channel. When the stop-transfer sequence enters the translocation channel, the channel discharges both the signal sequence and the stop-transfer sequence sideways into the lipid bilayer. The signal sequence is then cleaved, so that the protein remains held in the membrane by the hydrophobic stop-transfer sequence. DIF: Moderate REF: 15.2 OBJ: 15.2.w Compare the locations and fates of the ER signal sequence and stop-transfer sequence of single-pass transmembrane proteins and the start-transfer and stop-transfer sequences of multipass transmembrane proteins. MSC: Applying 8. Using genetic engineering techniques, you have created a set of proteins that contain two (and only two) conflicting signal sequences that specify different compartments. Predict which signal would win out for the following combinations. Explain your answers.

A. Signals for import into the nucleus and import into the ER. B. Signals for export from the nucleus and import into the mitochondria.

C. Signals for import into mitochondria and retention in the ER. ANS: A. The protein would enter the ER. The signal for a protein to enter the ER is recognized as the protein is being synthesized and the protein will end up either in the ER or on the ER membrane. Cytosolic nuclear transport proteins recognize proteins destined for the nucleus once those proteins are fully synthesized and fully folded.

B. The protein would enter the mitochondria. For a nuclear export signal to work, the protein would have to end up in the nucleus first and thus would need a nuclear import signal for the nuclear export signal to be used.

C. The protein would enter the mitochondria. To be retained in the ER, the protein needs to enter the ER. Because there is no signal for ER import, the ER retention signal would not function.

DIF: Hard REF: 15.2 OBJ: 15.2.g Articulate what would happen to a protein that bears both an ER sorting signal and a nuclear localization signal. | 15.2.f Predict what would happen to a protein that bears both a nuclear localization signal and a nuclear export signal. | 15.2.k Review how nuclear import receptors escort proteins bearing a nuclear localization signal from the cytosol into the nucleus. | 15.2.n Articulate how mitochondrial proteins are recognized and transported into the mitochondrial matrix, and describe the role played by chaperones inside the organelle. | 15.2.l Describe the types of molecules transported by nuclear export receptors. | 15.4.c Recall how proteins destined to function in the ER are kept in the ER—or returned to the ER if they accidentally enter the Golgi apparatus.

MSC: Applying 9. v-SNAREs and t-SNAREs mediate the recognition of a vesicle at its target membrane so that a vesicle displaying a

particular type of v-SNARE will only fuse with a target membrane containing a complementary type of t-SNARE. In some cases, v-SNAREs and t-SNAREs may also mediate the fusion of identical membranes. In yeast cells, right before the formation of a new cell, vesicles derived from the vacuole will come together and fuse to form a new vacuole destined for the new cell. Unlike the situation we have discussed in class, the vacuolar vesicles contain both v-SNAREs and t-SNAREs. Your friend is trying to understand the role of these SNAREs in the formation of the new vacuole and consults with you regarding the interpretation of his data.

Your friend has designed an ingenious assay for the fusion of vacuolar vesicles by using alkaline phosphatase. The protein alkaline phosphatase is made in a “pro” form that must be cleaved for the protein to be active. Your frien d has designed two different strains of yeast: strain A produces the “pro” form of alkaline phosphatase (pro -Pase), whereas strain B produces the protease that can cleave pro-Pase into the active form (Pase). Neither strain has the active form of the alkaline phosphatase, but when vacuolar vesicles from the strains A and B are mixed, fusion of vesicles generates active alkaline phosphatase, whose activity can be measured and quantified (Figure 15 -9A).

Figure 15-9

Your friend has taken each of these yeast strains and further engineered them so that they express only the vSNAREs, only the t-SNAREs, both SNAREs (the normal situation), or neither SNARE. He then isolates vacuolar vesicles from all strains and tests the ability of each variant form of strain A to fuse with each variant form of strain B, by using the alkaline phosphatase assay. The data are shown in the graph in Figure 15-9B. On this graph, the SNARE present on the vesicle of the particular yeast strain is indicated as “v” (for the presence of the v-SNARE) and “t” (for the presence of the t-SNARE).

What do his data say about the requirements for v-SNAREs and t-SNAREs in the vacuolar vesicles? Is it important to have a specific type of SNARE (that is, v-SNARE or t-SNARE) on each vesicle?

ANS: To get maximal levels of vacuolar vesicle fusion, vesicles from each strain must carry both v-SNAREs and tSNAREs. Experiment 1, which represents the normal scenario, is the only experiment in which 100% alkaline phosphatase activity is measured. However, as long as complementary SNAREs are present on the vesicles, some vesicle fusion does occur (see experiments 3, 4, 6, 7, 8, and 9). If both vesicles are missing v-SNAREs (experiment 2) or tSNAREs (experiment 5) or both SNAREs (experiments 10 and 11), the level of fusion is very low. It does not matter

whether a t-SNARE or a v-SNARE is on the vesicle of a particular strain, as long as the vesicle from the other strain contains a complementary SNARE (compare experiments 3 and 4, 6 and 7, and 8 and 9).

DIF: Hard REF: 15.3 OBJ: 15.3.d Compare the processes of vesicle tethering, docking, and fusion, and list the proteins involved in each. MSC: Applying 10. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. carbohydrate

endosome

lysosome

disulfide bonds

exocytic

protein

endocytic

Golgi apparatus

secretory

endomembrane

hydrogen bonds

endoplasmic reticulum

ionic bonds

Proteins are transported out of a cell via the __________ or __________ pathway. Fluids and macromolecules are transported into the cell via the __________ pathway. All proteins being transported out of the cell pass through the __________ and the __________. Transport vesicles link organelles of the __________ system. The formation of __________ in the endoplasmic reticulum stabilizes protein structure.

ANS: Proteins are transported out of a cell via the secretory or exocytic pathway. Fluid and macromolecules are transported into the cell via the endocytic pathway. All proteins being transported out of the cell pass through the endoplasmic reticulum and the Golgi apparatus. Transport vesicles link organelles of the endomembrane system. The formation of disulfide bonds in the endoplasmic reticulum stabilizes protein structure.

DIF: Easy REF: 15.3 OBJ: 15.3.a Contrast the endocytic and exocytic pathways in terms of directionality, purpose, and the participating organelles and membranes. MSC: Understanding 11. Name a type of protein modification that can occur in the ER but not in the cytosol.

ANS: Proteins in the ER can undergo disulfide bond formation. (This does not occur in the cytosol because of its reducing environment.)

(Signal-sequence cleavage and N-linked glycosylation are also acceptable answers.)

DIF: Easy REF: 15.4 OBJ: 15.4.a List the types of covalent modifications that take place in the ER and describe the functions these modifications serve. MSC: Remembering 12. If you remove the ER retention signal from a protein that normally resides in the ER lumen, where do you predict the protein will ultimately end up? Explain your reasoning.

ANS: The protein would end up in the extracellular space. Normally, the protein would go from the ER to the Golgi apparatus, get captured because of its ER retention signal, and return to the ER. However, without the ER retention signal, the protein would evade capture, ultimately leave the Golgi via the default pathway, and become secreted into the extracellular space. The protein would not be retained anywhere else along the secretory pathway: it presumably has no signals to promote such localization because it normally resides in the ER lumen.

DIF: Hard REF: 15.2 OBJ: 15.2.a Outline the mechanisms by which proteins can enter membrane-enclosed organelles and identify the organelles that use them. | 15.4.c Recall how proteins destined to function in the ER are kept in the ER—or returned to the ER if they accidentally enter the Golgi apparatus MSC: Evaluating 13. In a cell capable of regulated secretion, what are the three main classes of proteins that must be separated before

they leave the trans Golgi network?

ANS: The three main classes of protein that must be sorted before they leave the trans Golgi network in a cell capable of regulated secretion are (1) those destined for lysosomes, (2) those destined for secretory vesicles, and (3) those destined for immediate delivery to the cell surface.

DIF: Easy REF: 15.4 OBJ: 15.4.h Compare how proteins are sorted in the cis and trans Golgi networks.

MSC: Remembering 14. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. chaperone

Golgi apparatus

pseudopods

cholesterol

mycobacterium

rough ER

clathrin

phagocytosis

SNARE

endosome

pinocytosis

transcytosis

Eukaryotic cells are continually taking up materials from the extracellular space by the process of endocytosis. One type of endocytosis is __________, which uses __________ proteins to form small vesicles containing fluids and molecules. After these vesicles have pinched off from the plasma membrane, they will fuse with the __________, where materials that are taken into the vesicle are sorted. A second type of endocytosis is __________, which is used to take up large vesicles that can contain microorganisms and cellular debris. Macrophages are especially suited for this process, as they extend __________ (sheetlike projections of their plasma membrane) to surround the invading microorganisms.

ANS: Eukaryotic cells are continually taking up materials from the extracellular space by the process of endocytosis.

One type of endocytosis is pinocytosis, which uses clathrin proteins to form small vesicles containing fluids and molecules. After these vesicles have pinched off from the plasma membrane, they will fuse with the endosome, where materials that are taken into the vesicle are sorted. A second type of endocytosis is phagocytosis, which is used to take up large vesicles that can contain microorganisms and cellular debris. Macrophages are especially sui ted for this process, as they extend pseudopods (sheetlike projections of their plasma membrane) to surround the invading microorganisms.

DIF: Easy REF: 15.5 OBJ: 15.5.a Distinguish between pinocytosis and phagocytosis, and indicate the types of cells that engage in each. MSC: Understanding 15. Name three possible fates for an endocytosed molecule that has reached the endosome.

ANS: 1. recycled to the original membrane 2. destroyed in the lysosome 3. transcytosed across the cell to a different membrane DIF: Easy REF: 15.5 OBJ: 15.5.g Outline the possible fates of receptor proteins and cargo molecules following endocytosis into endosomes. MSC: Remembering 16. If a lysosome breaks, what protects the rest of the cell from lysosomal enzymes?

ANS: The lysosomal enzymes are all acid hydrolases, which have optimal activity at the low pH (about 5.0) found in the interior of lysosomes. If a lysosome were to break, the acid hydrolases would find themselves at pH 7.2, the pH of

the cytosol, and would therefore do little damage to cellular constituents.

DIF: Easy REF: 15.5 OBJ: 15.5.i Recall how lysosomes maintain their pH and describe the role that pH plays in lysosomal function. MSC: Remembering 17. You have created a green fluorescent protein (GFP) fusion to a protein that is normally secreted from yeast cells.

Because you have learned about the use of temperature-sensitive mutations in yeast to study protein and vesicle transport, you obtain three mutant yeast strains, each defective in some aspect of the protein secretory process.

Being a good scientist, you of course also obtain a wild-type control strain. You decide to examine the fate of your GFP fusion protein in these various yeast strains and engineer the mutant strains to express your GFP fusion protein.

However, in your excitement to do the experiment, you realize that you did not label any of the mutant yeast strains and no longer know which strain is defective in what process. You end up numbering your strains with the numbers 1 to 4, and then you carry out the experiment anyway, obtaining the results shown in Figure 15-17 (the black dots represent your GFP fusion protein).

Figure 15-17

Name the process that is defective in each of these strains. Remember that one of these strains is your wild-type control.

ANS: Strain A has protein accumulating in the ER, which means that this cell has a mutation that blocks transport from the ER to the Golgi apparatus. Strain B has secreted protein, and therefore is the wild-type control. Strain C has protein accumulating in the Golgi apparatus, and thus has a mutation that blocks exit of proteins from the Golgi apparatus. Strain D has protein accumulating in the cis Golgi network, and thus has a mutation that blocks the travel of proteins through the Golgi cisternae.

DIF: Hard REF: 15.4 OBJ: 15.4.m Outline how temperature-sensitive mutants have been used to dissect the protein secretory pathway in yeast. | 15.4.n Explain how GFP can be used to monitor the location and movement of proteins in living cells. MSC: Applying 18. Fill in the blank with the membrane-enclosed compartments in a eukaryotic cell where each of the functions listed below takes place.

1. Photosynthesis occurs in the ANS: chloroplast. DIF: Easy REF: 15.2 OBJ: 15.2.b List the membrane-bound organelles that can receive proteins directly from the cytosol. | 15.2.d Review where the proteins found in mitochondria and chloroplasts are synthesized. MSC: Remembering 2. Transcription occurs in the ANS: nucleus. DIF: Easy REF: 15.2 OBJ: 15.2.b List the membrane-bound organelles that can receive proteins directly from the cytosol. | 15.2.d Review where the proteins found in mitochondria and chloroplasts are synthesized.

MSC: Remembering

3. Oxidative phosphorylation occurs in the ANS: mitochondrion. DIF: Easy REF: 15.2 OBJ: 15.2.b List the membrane-bound organelles that can receive proteins directly from the cytosol. | 15.2.d Review where the proteins found in mitochondria and chloroplasts are synthesized. MSC: Remembering 4. Modification of secreted proteins occurs in the ANS: Golgi apparatus.; rough endoplasmic reticulum.; rough ER. DIF: Easy REF: 15.2 OBJ: 15.2.b List the membrane-bound organelles that can receive proteins directly from the cytosol. | 15.2.d Review where the proteins found in mitochondria and chloroplasts are synthesized. MSC: Remembering 5. Steroid hormone synthesis occurs in the ANS: smooth ER.; smooth endoplasmic reticulum. DIF: Easy REF: 15.2 OBJ: 15.2.b List the membrane-bound organelles that can receive proteins directly from the cytosol. | 15.2.d Review where the proteins found in mitochondria and chloroplasts are synthesized. MSC: Remembering 6. Degradation of worn-out organelles occurs in the ANS: lysosome. DIF: Easy REF: 15.2 OBJ: 15.2.b List the membrane-bound organelles that can receive proteins directly from the cytosol. | 15.2.d Review where the proteins found in mitochondria and chloroplasts are synthesized. MSC: Remembering 7. New membrane synthesis occurs in the ANS: ER.; endoplasmic reticulum. DIF: Easy REF: 15.2 OBJ: 15.2.b List the membrane-bound organelles that can receive proteins directly from the cytosol. | 15.2.d Review where the proteins found in mitochondria and chloroplasts are synthesized. MSC: Remembering 8. Breakdown of lipids and toxic molecules occurs in the ANS: peroxisome. DIF: Easy REF: 15.2 OBJ: 15.2.b List the membrane-bound organelles that can receive proteins directly from the cytosol. | 15.2.d Review where the proteins found in mitochondria and chloroplasts are synthesized. MSC: Remembering 19. For each of the following sentences, fill in the blank with one of the two options enclosed in square brackets to make a correct statement.

1. When a secretory vesicle fuses with the plasma membrane by __________ [exocytosis/endocytosis], the vesicle’s membrane becomes part of the plasma membrane.

ANS: exocytosis DIF: Easy REF: 15.3 OBJ: 15.3.a Contrast the endocytic and exocytic pathways in terms of directionality, purpose, and the participating organelles and membranes. MSC: Understanding 2. The surface areas of the plasma membrane is kept relatively constant because __________ [exocytosis/endocytosis] returns lipids and proteins to the interior of the cell.

ANS: endocytosis DIF: Easy REF: 15.3 OBJ: 15.3.a Contrast the endocytic and exocytic pathways in terms of directionality, purpose, and the participating organelles and membranes. MSC: Understanding 3. Insulin is secreted from pancreatic cells by the __________ [regulated/constitutive] exocytosis pathway.

ANS: regulated DIF: Easy REF: 15.3 OBJ: 15.3.a Contrast the endocytic and exocytic pathways in terms of directionality, purpose, and the participating organelles and membranes. MSC: Understanding 4. The interior of the trans Golgi network is __________ [acidic/alkaline].

ANS: acidic DIF: Easy REF: 15.3 OBJ: 15.3.a Contrast the endocytic and exocytic pathways in terms of directionality, purpose, and the participating organelles and membranes. MSC: Understanding 5. Proteins that are constitutively secreted __________ [aggregate/do not aggregate] in the trans Golgi network.

ANS: do not aggregate DIF: Easy REF: 15.3 OBJ: 15.3.a Contrast the endocytic and exocytic pathways in terms of directionality, purpose, and the participating organelles and membranes. MSC: Understanding 20. Figure 15-20 shows the organization of a protein that resides on the ER membrane. The N- and C-termini of the protein are labeled. Boxes 1, 2, and 3 represent membrane-spanning sequences. Non-membrane-spanning regions of the protein are labeled “X,” “Y,” and “Z.”

Figure 15-20

Once this protein is fully translocated, where will region Y be? a. in the cytoplasm b. in the ER lumen c. inserted into the ER membrane d. degraded by signal peptidase ANS: A The final topology of the protein on the ER membrane is diagrammed in Figure 15-20A.

Figure 15-20A

DIF: Hard REF: 15.2 OBJ: 15.2.w Compare the locations and fates of the ER signal sequence and stop-transfer sequence of single-pass transmembrane proteins and the start-transfer and stop-transfer sequences of multipass transmembrane proteins. MSC: Applying 21. Figure 15-21 shows the orientation of a multipass transmembrane protein after it has completed its entry into the ER membrane (part A) and after it gets delivered to the plasma membrane (part B). This protein has an N-terminal signal sequence (depicted as the dark gray membrane-spanning box), which signal peptidase cleaves off in the endoplasmic reticulum. The other membrane-spanning domains in the protein are represented as open boxes. Given that any hydrophobic membrane-spanning domain can act as either a start-transfer region or a stop-transfer region, draw the final consequences of the actions described below on the orientation of the protein in the plasma membrane. Indicate on your drawing the extracellular space, the cytosolic face, and the plasma membrane, as well as the N- and Cterminus of the protein.

Figure 15-21

A. deleting the first signal sequence B. changing the hydrophobic amino acids in the first, cleaved sequence to charged amino acids C. changing the hydrophobic residues in every other transmembrane sequence to charged residues, starting with the first, cleaved signal sequence ANS: A. Deleting the first signal sequence completely would convert the next membrane-spanning domain into an internal start-transfer signal and would invert the orientation of the protein (see Figure 15A-21A).

B. Changing the hydrophobic amino acids to charged amino acids destroys the ability of the sequence both to act as a signal sequence and to become a membrane-spanning sequence. Therefore, the adjacent membrane-spanning domain will now become an internal start-transfer sequence and the protein will be inverted, as seen above in part A. The mutated signal sequence would not get cleaved off, because it would remain on the cytoplasmic side of the membrane and signal peptidase is found only inside the ER (see Figure 15A-21B).

C. Mutating every other membrane-spanning region so that they are now charged (and thus cannot span the membrane) would decrease the number of transmembrane regions and increase the size of the loops between membrane-spanning regions (see Figure 15A-21C).

Figure 15-21A

DIF: Hard REF: 15.2 OBJ: 15.2.w Compare the locations and fates of the ER signal sequence and stop-transfer sequence of single-pass transmembrane proteins and the start-transfer and stop-transfer sequences of multipass transmembrane proteins. MSC: Applying 22. A plasma membrane protein carries an oligosaccharide containing mannose (Man), galactose (Gal), sialic acid (SA), and N-acetylglucosamine (GlcNAc). These sugars are added to the protein as it proceeds through the secretory pathway. First, a core oligosaccharide containing Man and GlcNAc is added, followed by Gal, Man, SA, and GlcNAc in a particular order. Each addition is catalyzed by a different transferase acting at a different stage as the protein

proceeds through the secretory pathway. You have isolated mutants defective for each of the transferases, purified the membrane protein from each of the mutants, and identified which sugars are present in each mutant protein.

Table 15-22 summarizes the results.

Table 15-22

From these results, match each of the transferases (A, B, C, D) to its subcellular location selected from the list below.

(Assume that each location contains only one enzyme.)

1. central Golgi cisternae 2. cis Golgi network 3. ER 4. trans Golgi network 1. ANS: B 2.

ANS: D

3.

ANS: A

4.

ANS: C

Proteins are modified in a stepwise fashion in the Golgi apparatus, with early steps taking place in the cis Golgi network, intermediate steps taking place in the central Golgi cisternae, and late steps occurring in the trans Golgi network. If each enzyme produces the substrate for the next step, then a mutant lacking the enzyme that catalyzes the addition of the first sugar will be missing all of the sugars, a mutant lacking the enzyme that catalyzes the addition of the second sugar will contain the first sugar but will lack th e other three, and so on. By this logic, mannose and GlcNAc must be the first sugars added, additional GlcNAc is the second added, galactose the third, and SA the last. Hence, the oligosaccharide protein transferase must be in the ER, the GlcNAc transferas e in the cis Golgi network, the galactose transferase in the central Golgi cisternae, and the SA transferase in the trans Golgi network.

DIF: Hard REF: 15.4 OBJ: 15.4.g Articulate the ways in which proteins move from one cisterna to another within the Golgi apparatus. MSC: Applying 23. Figure 15-23 shows the orientation of the Krt1 protein on the membrane of a Golgi-derived vesicle that will fuse with the plasma membrane.

Figure 15-23

Given this diagram, which of the following statements is TRUE? a. When this vesicle fuses with the plasma membrane, the entire Krt1 protein will be secreted into the extracellular space. b. When this vesicle fuses with the plasma membrane, the C-terminus of Krt1 will be inserted into the plasma membrane. c. When this vesicle fuses with the plasma membrane, the N-terminus of Krt1 will be in the extracellular space. d. When this vesicle fuses with the plasma membrane, the N-terminus of Krt1 will be cytoplasmic.

ANS: C The orientation of Krt1 as the vesicle fuses with the plasma membrane is shown in Figure A15-23. The darker-colored lines in the membrane represent the membranes contributed by the vesicle during fusion.

Figure 15-23A

DIF: Moderate REF: 15.3 OBJ: 15.3.a Contrast the endocytic and exocytic pathways in term s of directionality, purpose, and the participating organelles and membranes. MSC: Applying 24. Fibroblast cells from patients W, X, Y, and Z, each of whom has a different inherited defect, all contain “inclusion bodies,” which are lysosomes filled with undigested material. You wish to identify the cellular basis of these defects.

The possibilities are: 1. a defect in one of the lysosomal hydrolases 2. a defect in the phosphotransferase that is required for mannose-6-phosphate tagging of the lysosomal hydrolases 3. a defect in the mannose-6-phosphate receptor, which binds mannose-6-phosphate-tagged lysosomal proteins in the trans Golgi network and delivers them to lysosomes When you incubate some of these mutant fibroblasts in a medium in which normal cells have been grown, you find that the inclusion bodies disappear. Because of these results, you suspect that the constitutive exocytic pathway in normal cells is secreting lysosomal hydrolases that are being taken up by the mutant cells. (It is known that some mannose-6-phosphate receptor molecules are found in the plasma membrane and can take up and deliver lysosomal

proteins via the endocytic pathway.) You incubate cells from each patient with medium from normal cells and medium from each of the other mutant cell cultures, and get the results summarized in Table 15-24.

Table 15-24

Indicate which defect (1, 2, 3) each patient (W, X, Y, Z) is most likely to have.

ANS: W—3 (defect in mannose-6-phosphate receptor) X—2 (defect in phosphotransferase) Y—1; Z—1 (defect in lysosomal hydrolases); these will be defects in two different lysosomal acid hydrolases A cell that has no mannose-6-phosphate receptor will be able to make all the lysosomal hydrolases properly but will not be able to send them to the lysosome and will also not be able to scavenge hydrolases from the external media.

Hence, this cell line cannot be rescued by a culture medium that has had lysosomal hydrolases secreted into it and thus will not be rescued by any of the media tested here. A cell line that has no phosphotransferase will be able to scavenge hydrolases from the external medium, but because all of the cell’s own hydrolases will lack the mannose-6phosphate tag, it will be rescued only by medium from a cell line that is able to make all of the hydrolases. Cell lines lacking one hydrolase will be rescued by medium from any cell line that is able to secrete that hydrolase in a mannose-6-phosphate-tagged form; in addition, media from cultures of cells lacking a hydrolase will rescue any cell line with another type of defect.

DIF: Hard REF: 15.5 OBJ: 15.5.k Outline the pathways that different materials follow to arrive in lysosomes.

MSC: Applying

CHAPTER 16 Cell Signaling

GENERAL PRINCIPLES OF CELL SIGNALING 16.1.a Define signal transduction and list the basic components involved in this process in cells.

16.1.b Distinguish the main types of signal-mediated cell−cell communication and identify the type of extracellular signal molecule involved in each.

16.1.c Outline the main classes of extracellular signal molecules, and describe the types of receptors to which they bind.

16.1.d Explain how the same signal molecule can induce different responses in different target cells.

16.1.e Recall how a combination of signals can evoke a response that is different from the sum of the effects that each signal can trigger on its own.

16.1.f

Differentiate the types of cell responses that occur rapidly with those that take minutes or hours to execute.

16.1.g Name the basic components needed for an extracellular signal molecule to change the behavior of a target cell, and identify the site at which the primary step in signal transduction takes place.

16.1.h Review the main functions of an intracellular signaling pathway and identify the steps at which each can take place.

16.1.i

Compare positive and negative feedback and contrast the types of responses produced by each.

16.1.j

Summarize how phosphorylation can act as a molecular switch, and identify the types of proteins that add and remove this chemical modification.

16.1.k Distinguish the two main types of GTP-binding proteins.

16.1.l

Describe how monomeric GTPases toggle between active and inactive forms.

16.1.m Differentiate the three main classes of cell-surface receptors and provide an example of each.

16.1.n List some foreign substances that alter physiology by interacting with cell-surface receptors.

16.1.o Describe the type of signal transduction carried out by ion-channel-coupled receptors.

G-PROTEIN-COUPLED RECEPTORS 16.2.a Review the structure of G-protein-coupled receptors (GPCRs) and describe the types of extracellular signal molecules that bind to them.

16.2.b Recall the general structure of a G protein and describe how the protein responds when activated by a GPCR. 16.2.c Summarize the factors that determine the duration of a GPCR-stimulated response.

16.2.d Contrast how cholera toxin and pertussis toxin exert their effects.

16.2.e Relate the speeds of the responses produced by G proteins activating an ion channel versus activating a membrane-bound enzyme.

16.2.f

Name the classes of enzymes that are the most frequent targets of G proteins, and list the second messenger

molecules they produce.

16.2.g Outline how cyclic AMP is produced in response to G protein activation, and recall how caffeine can potentiate this response.

16.2.h Compare a signaling pathway in which cyclic AMP produces a response within seconds to one in which the response takes minutes or hours to develop.

16.2.i

Recall the location and action of the second messenger molecules produced by activated phospholipase C.

16.2.j

List several biological processes triggered by calcium ions.

16.2.k Explain how cells keep the concentration of calcium ions in the cytosol low and how they terminate a calcium ion signal.

16.2.l

Review how calcium-responsive proteins such as calmodulin propagate a calcium ion signal.

16.2.m Outline how the gas nitric oxide (NO) can act as a signaling molecule to trigger the relaxation of smooth mu scle cells.

16.2.n Recall why nitric oxide acts as a paracrine signal only on cells near its site of synthesis.

16.2.o Outline how GPCRs in the photoreceptors of the retina transmit an extremely rapid signal in response to stimulation by light.

16.2.p Summarize how adaptation in the intracellular signaling cascade of photoreceptors allows the eye to respond to dim or bright light.

ENZYME-COUPLED RECEPTORS 16.3.a Compare the general structures of GPCRs and enzyme-coupled receptors such as receptor-tyrosine kinases (RTKs).

16.3.b Review how the binding of a signal molecule activates RTKs to trigger the assembly of an intracellular signaling complex.

16.3.c Recall how signals transmitted by RTKs can be terminated.

16.3.d List several intracellular signaling proteins activated by RTKs. 16.3.e Outline how RTKs activate the MAP kinase signaling module.

16.3.f

Indicate how Ras can fuel uncontrolled proliferation in cancer.

16.3.g Review how extracellular signals that promote cell growth and survival activate PI-3-kinase signaling pathways.

16.3.h Compare how Akt promotes cell survival via Bad and stimulates cell growth via Tor. 16.3.i

Review how different types of receptors can trigger a rise in the cytosolic concentration of calcium ions.

16.3.j

Describe a method to identify proteins that interact in response to stimulation by an extracellular signal.

16.3.k Outline how a set of mutant RTKs can be used to determine which tyrosines serve as docking sites for the intracellular signaling proteins that propagate the signal.

16.3.l

Review how a technology such as RNA interference or CRISPR can be used to assess the importance of a particu-

lar protein in a signaling pathway.

16.3.m Explain how mutant proteins can be used to determine the order in which proteins participate in a signaling pathway.

16.3.n Review how the Notch receptor activates target genes in response to activation by Delta, and explain how this pathway controls the specialization of nerve cells in developing Drosophila embryos.

16.3.o Outline how steroid hormones trigger the transcription of different sets of target genes.

16.3.p Contrast the cell signaling systems used by plants and animals.

16.3.q Describe how the ethylene signaling pathway regulates ripening of fruits.

16.3.r

Explain how multiple signaling pathways can integrate information to produce a coordinated cell response.

MULTIPLE CHOICE 1. When a signal needs to be sent to most cells throughout a multicellular organism, the signal most suited for this is a a. neurotransmitter. b. hormone. c. dissolved gas. d. scaffold.

ANS: B DIF: Easy REF: 16.1 OBJ: 16.1.b Distinguish the main types of signal-mediated cell–cell communication and identify the type of extracellular signal molecules involved in each. MSC: Understanding 2. During nervous-system development in Drosophila, the membrane-bound protein Delta acts as an inhibitory signal to prevent neighboring cells from developing into neuronal cells. Delta is involved in __________ signaling. a. endocrine b. paracrine c. neuronal d. contact-dependent ANS: D DIF: Easy REF: 16.1 OBJ: 16.1.b Distinguish the main types of signal-mediated cell–cell communication and identify the type of extracellular signal molecules involved in each. MSC: Remembering 3. Which of the following statements is TRUE? a. Because endocrine signals are broadcast throughout the body, all cells will respond to the hormonal signal. b. The regulation of inflammatory responses at the site of an infection is an example of paracrine signaling. c. Paracrine signaling involves the secretion of signals into the bloodstream for distribution throughout the organism. d. The axons of neurons typically signal target cells using membrane-bound signaling molecules that act on receptors in the target cells.

ANS: B Only the cells with a receptor for the hormone will respond to the signal. Paracrine signaling involves signaling in a more local fashion, unlike endocrine signaling, where signals are sent through the bloodstream. Axons typically signal using diffusible neurotransmitters that are released at the synapse.

DIF: Easy REF: 16.1 OBJ: 16.1.b Distinguish the main types of signal-mediated cell–cell communication and identify the

type of extracellular signal molecules involved in each. MSC: Understanding 4. The lab you work in has discovered a previously unidentified extracellular signal molecule called QGF, a 75,000 dalton protein. You add purified QGF to different types of cells to determine its effect on these cells. When you add QGF to heart muscle cells, you observe an increase in cell contraction. When you add it to fibroblasts, they undergo cell division. When you add it to nerve cells, they die. When you add it to glial cells, you do not see any effect on cell division or survival. Given these observations, which of the following statements is most lik ely to be TRUE? a. Because it acts on so many diverse cell types, QGF probably diffuses across the plasma membrane into the cytoplasm of these cells. b. Glial cells do not have a receptor for QGF. c. QGF activates different intracellular signaling pathways in heart muscle cells, fibroblasts, and nerve cells to produce the different responses observed. d. Heart muscle cells, fibroblasts, and nerve cells must all have the same receptor for QGF. ANS: C Because heart muscle cells, fibroblasts, and nerve cells all respond to QGF with different outcomes, it is likely that QGF activates different effector proteins in these different cell types, leading to the diversity of outcomes observed in the experiment. QGF is unlikely to diffuse across the cell membrane, given that it is a large protein. Although glial cells do not die or divide in response to QGF, they could have a receptor for QGF, as receptor activation could lead to some other response. Often, a signal molecule can bind to different types of receptor on different cell types, so heart muscle, fibroblasts, and nerve cells may or may not have the same receptor.

DIF: Moderate REF: 16.1 OBJ: 16.1.d Explain how the same signal molecule can induce different responses in different target cells. MSC: Applying 5. Acetylcholine is a signaling molecule that elicits responses from heart muscle cells, salivary gland cells, and skeletal muscle cells. Which of the following statements is FALSE? a. Heart muscle cells decrease their rate and force of contraction when they receive acetylcholine, whereas skeletal muscle cells contract. b. Heart muscle cells, salivary gland cells, and skeletal muscle cells all express an acetylcholine receptor that belongs to the transmitter-gated ion channel family. c. Active acetylcholine receptors on salivary gland cells and heart muscle cells activate different intracellular signaling pathways. d. Heart muscle cells, salivary gland cells, and skeletal muscle cells all respond to acetylcholine within minutes of receiving the signal.

ANS: B Only skeletal muscle cells express an acetylcholine receptor that belongs to the transmitter-gated ion channel family; salivary gland cells and heart muscle cells express a different receptor. The other choices are all true.

DIF: Easy REF: 16.1 OBJ: 16.1.c Outline the main classes of extracellular signal molecules and describe the type of receptor to which they bind. | 16.1.d Explain how the same signal molecule can induce different responses in different target cells. MSC: Remembering 6. Which of the following statements is TRUE?

a. Extracellular signal molecules that are hydrophilic must bind to a cell-surface receptor so as to signal a target cell to change its behavior. b. To function, all extracellular signal molecules must be transported by their receptor across the plasma membrane into the cytosol. c. A cell-surface receptor capable of binding only one type of signal molecule can mediate only one kind of cell response. d. Any foreign substance that binds to a receptor for a normal signal molecule will always induce the same response that is produced by that signal molecule on the same cell type.

ANS: A A hydrophilic molecule cannot diffuse across the membrane and it can therefore only affect a cell if it binds to a cell-surface receptor. Most signal molecules remain bound to the extracellular domain of the receptor, whereas the intracellular domain mediates signal transduction; although many signal molecules are endocytosed with their r eceptor, they remain inside membrane-bounded compartments and are therefore not transported into the cytosol. A cell-surface receptor capable of binding only one type of signal molecule can stimulate more than one kind of cell response, depending on the types of intracellular signaling pathway it activates. Foreig n substances that bind to a receptor for a normal signal molecule can sometimes induce the same response as the natural signal molecule, but in other cases they can block the binding of the natural signal molecule wit hout activating the receptor.

DIF: Easy REF: 16.1 OBJ: 16.1.g Name the basic components needed for an extracellular signal molecule to change the behavior of a target cell and identify the site at which the primary step in signal transduction takes place.

MSC: Understanding 7. Which of the following statements about molecular switches is FALSE? a. Phosphatases remove the phosphate from GTP on GTP-binding proteins, turning them off. b. Protein kinases transfer the terminal phosphate from ATP onto a protein. c. Serine/threonine kinases are the most common types of protein kinase. d. A GTP-binding protein exchanges its bound GDP for GTP to become activated.

ANS: A GTP-binding proteins themselves hydrolyze their bound GTP to GDP, using their own intrinsic GTPase activity.

DIF: Easy REF: 16.1 OBJ: 16.1.l Describe how monomeric GTPases toggle between active and inactive forms. | 16.1.j Summarize how phosphorylation can act as a molecular switch and identify the types of proteins that add and remove this chemical modification. MSC: Understanding 8. Foreign substances like nicotine, morphine, and menthol exert their initial effects by a. killing cells immediately, exerting their physiological effects by causing cell death. b. diffusing through cell plasma membranes and binding to transcription factors to change gene expression. c. interacting with cell-surface receptors, causing the receptors to transduce signal inappropriately in the absence of the normal stimulus. d. removing cell-surface receptors from the plasma membrane.

ANS: C Foreign substances can alter physiology by interacting with cell-surface receptors. Although some foreign substances will remove cell-surface receptors from the plasma membrane (choice D), this is a long-term response and not part of the initial response.

DIF: Easy REF: 16.1 OBJ: 16.1.n List some foreign substances that alter physiology by interacting with cell-surface receptors. MSC: Remembering 9. Cell lines A and B both survive in tissue culture containing serum but do not proliferate. Factor F is known to stimulate proliferation in cell line A. Cell line A produces a receptor protein (R) that cell line B does not produce. To test the role of receptor R, you introduce this receptor protein into cell line B, using recombinant DNA techniques.

You then test all of your various cell lines in the presence of serum for their response to factor F, with the results summarized in Table 16-9.

Table 16-9

Which of the following cannot be concluded from your results above? a. Binding of factor F to its receptor is required for proliferation of cell line A. b. Receptor R binds to factor F to induce cell proliferation in cell line A. c. Cell line A expresses a receptor for factor F. d. Factor F is not required for proliferation in cell line B. ANS: B Expressing receptor R in cell line B causes proliferation in the absence of factor F, suggesting that something else in the serum can bind to receptor R in cell line B. We do not know the effect of eliminating receptor R from cell line A, and thus we cannot say whether binding of receptor R to its ligand is required for cell line A to proliferate. The data support all the other answers. Because cell line A proliferates only in the presence of factor F, binding of factor F must be required for cell line A to proliferate. Because cell line A proliferates in response to factor F, it must express a receptor for factor F since it responds to factor F. Because cell line B can proliferate in the absence of factor F once R is expressed, factor F is not required for cell line B to proliferate.

DIF: Hard REF: 16.1 OBJ: 16.1.a Define signal transduction and list the basic components involved in this process in cells. MSC: Applying 10. The following happens when a G-protein-coupled receptor activates a G protein. a. The β subunit exchanges its bound GDP for GTP. b. The GDP bound to the α subunit is phosphorylated to form bound GTP. c. The α subunit exchanges its bound GDP for GTP. d. It activates the α subunit and inactivates the βγ complex.

ANS: C When a G protein is active, the α subunit will exchange its bound GDP for GTP. Both the α subunit and the βγ complex can activate downstream targets.

DIF: Easy REF: 16.2 OBJ: 16.2.b Recall the general structure of a G protein and describe how the protein responds when activated by a GPCR. MSC: Remembering

11. Which of the following statements about G-protein-coupled receptors (GPCRs) is FALSE? a. GPCRs are the largest family of cell-surface receptors in humans. b. GPCRs are used in endocrine, paracrine, and neuronal signaling. c. GPCRs are found in yeast, mice, and humans. d. The different classes of GPCR ligands (proteins, amino acid derivatives, or fatty acids) bind to receptors with different numbers of transmembrane domains.

ANS: D Although it is true that many types of ligands can bind to and activate GPCRs, all GPCRs have a similar structure with seven transmembrane domains.

DIF: Easy REF: 16.2 OBJ: 16.2.a Review the structure of G-protein-coupled receptors (GPCRs) and describe the types of extracellular signal molecules that bind to them. MSC: Understanding 12. The length of time a G protein will signal is determined by the a. activity of phosphatases that turn off G proteins by dephosphorylating Gα. b. activity of phosphatases that turn GTP into GDP. c. degradation of the G protein after Gα separates from Gβγ. d. GTPase activity of Gα.

ANS: D DIF: Easy REF: 16.2 OBJ: 16.2.c Summarize the factors that determine the duration of a GPCR-stimulated response. MSC: Understanding 13. Acetylcholine binds to a GPCR on heart muscle, making the heart beat more slowly. The activated receptor stimulates a G protein, which opens a K+ channel in the plasma membrane, as shown in Figure 16-13. Which of the following would enhance this effect of the acetylcholine?

Figure 16-13

a. addition of a high concentration of a nonhydrolyzable analog of GTP b. addition of a drug that prevents the α subunit from exchanging GDP for GTP c. mutations in the acetylcholine receptor that weaken the interaction between the receptor and acetylcholine d. mutations in the acetylcholine receptor that weaken the interaction between the receptor and the G prot ein ANS: A The heart is induced to beat more slowly by the binding of acetylcholine to a GPCR, activating a G protein whose βγ complex binds to and opens K + channels. The addition of high concentrations of a nonhydrolyzable analog of GTP will increase the length of time that the G protein βγ complex remains free of the α subunit and able to activate the K+ channel; this will therefore enhance the effect of acetylcholine. All the other choices will make it more difficult for the signal to proceed from the GPCR to the K+ channel.

DIF: Hard REF: 16.2 OBJ: 16.2.b Recall the general structure of a G protein and describe how the protein responds when activated by a GPCR. | 16.2.c Summarize the factors that determine the duration of a GPCR-stimulated response. MSC: Applying 14. During the mating process, yeast cells respond to pheromones secreted by other yeast cells. These pheromones bind GPCRs on the surface of the responding cell and lead to the activation of G proteins inside the cell. When a wild-type yeast cell senses the pheromone, its physiology changes in preparation for mating: the cell stops growing until it finds a mating partner. If yeast cells do not undergo the appropriate response after sensing a pheromone, they are considered sterile. Yeast cells that are defective in one or more components of the G protein have characteristic

phenotypes in the absence and presence of the pheromone, which are listed in Table 16-14.

Table 16-14

Which of the following models is consistent with the data from the analysis of these mutants? Explain your answer. a. The α subunit activates the mating response but is inhibited when bound to βγ. b. The βγ subunit activates the mating response but is inhibited when bound to α. c. The G protein is inactive; either the free α or free βγ complex is capable of activating the mating response. d. The G protein is active; both free α and free βγ complex are required to inhibit the mating response.

ANS: B Single mutations in the β and γ subunits of the G protein permit growth in the absence of pheromone and display a sterile phenotype in the presence of pheromone, whereas loss of the α subunit causes growth arrest in either the presence or the absence of pheromone. Because arrested growth is a normal response that occurs when the cells sense a pheromone, it must be the βγ complex that normally activates the cellular mating response when released from the α subunit. In cells lacking α, the βγ complex causes growth arrest inappropriately because α normally inhibits the action of the complex until it is activated by the binding of pheromone to the yeast cell. This interpret ation is consistent with the analysis of the double mutants in which α is deleted together with either β or γ. In these double mutants, normal growth is seen in the absence of pheromone because β and γ act together as a complex to activate target proteins; a lack of either β or γ leads to no response to pheromone, even when α is not present to inhibit the response.

DIF: Hard REF: 16.2 OBJ: 16.2.b Recall the general structure of a G protein and describe how the protein responds when activated by a GPCR. MSC: Applying 15. You are interested in how cyclic-AMP-dependent protein kinase A (PKA) functions to affect learning and memory, and you decide to study its function in the brain. It is known that, in the cells you are studying, PKA works via a signal transduction pathway like the one depicted in Figure 16-15. Furthermore, it is also known that activated PKA phosphorylates the transcriptional regulator called Nerd that then activates transcription of the gene Brainy. Which situation described below will lead to an increase in Brainy transcription?

Figure 16-15

a. a mutation in the Nerd gene that produces a protein that cannot be phosphorylated by PKA b. a mutation in the nuclear import sequence of PKA from PPKKKRKV to PPAAAAAV c. a mutation in the gene that encodes cAMP phosphodiesterase that makes the enzyme inactive d. a mutation in the gene that encodes adenylyl cyclase that renders the enzyme unable to interact with the α subunit of the G protein ANS: C cAMP phosphodiesterase is important for converting cAMP into AMP and thus down-regulating the activity of PKA. Without cAMP phosphodiesterase, transcription of Brainy will be increased. All the other choices will lead to inactivation of the signaling pathway and a decrease in Brainy transcription. A mutant form of Nerd that cannot be phosphorylated will not be active. If PKA cannot be imported into the nucleus, it will be unable to phosphorylate Nerd. Adenylyl cyclase interaction with the α subunit of the G protein is important for the G protein’s activation.

DIF: Hard REF: 16.2 OBJ: 16.2.c Summarize the factors that determine the duration of a GPCR-stimulated response. | 16.2.f Name the classes of enzymes that are the most frequent targets of G proteins, and list the second messenger molecules they produce. | 16.2.g Outline how cyclic AMP is produced in response to G protein activation and recall how caffeine can potentiate this response. MSC: Applying 16. Adrenaline stimulates glycogen breakdown in skeletal muscle cells by ultimately activating glycogen phosphorylase,

the enzyme that breaks down glycogen, as depicted in Figure 16-16.

Figure 16-16

Which of the following statements is FALSE? a. A constitutively active mutant form of PKA in skeletal muscle cells would lead to a decrease in the amount of unphosphorylated phosphorylase kinase. b. A constitutively active mutant form of PKA in skeletal muscle cells would not increase the affinity of adrenaline for the adrenergic receptor. c. A constitutively active mutant form of PKA in skeletal muscle cells would lead to an excess in the amount of glucose available. d. A constitutively active mutant form of PKA in skeletal muscle cells would lead to an excess in the amount of glycogen available.

ANS: D A constitutively active mutant form of PKA in skeletal muscle cells would lead to a decrease in the amount of glycogen available, because active PKA stimulates enzymes that are responsible for the breakdown of glycogen so that glucose can be produced. All the other statements are true.

DIF: Hard REF: 16.2 OBJ: 16.2.f Name the classes of enzymes that are the most frequent targets of G proteins, and list the second messenger molecules they produce. | 16.2.g Outline how cyclic AMP is produced in response to G protein activation and recall how caffeine can potentiate this response. MSC: Applying 17. Activated protein kinase C (PKC) can lead to the modification of the membrane lipids in the vicinity of the active PKC. Figure 16-17 shows how G proteins can indirectly activate PKC. You have discovered the enzyme activated by PKC that mediates the lipid modification. You call the enzyme Rafty and demonstrate that activated PKC directly

phosphorylates Rafty, activating it to modify the plasma membrane lipids in the vicinity of the cell where PKC is active; these lipid modifications can be detected by dyes that bind to the modified lipids. Cells lacking Rafty do not have these modifications, even when PKC is active. Which of the following conditions would lead to signal-independent modification of the membrane lipids by Rafty?

Figure 16-17

a. the expression of a constitutively active phospholipase C b. a mutation in the GPCR that binds the signal more tightly c. a Ca2+ channel in the endoplasmic reticulum with an increased affinity for IP3 d. a mutation in the gene that encodes Rafty such that the enzyme can no longer be phosphorylated by PKC ANS: A A constitutively active phospholipase C will lead to the constitutive production of IP 3 and diacylglycerol, leading to activation of PKC in a signal-independent manner; thus, Rafty activation and the lipid modification will be signal independent. A mutation in the GPCR that bind the signal more tightly and a Ca 2+ channel with an increased affinity for IP will increase activity of the signal transduction pathway in a signal-dependent manner. A mutation that renders Rafty such that it can no longer by phosphorylated by PKC will prevent PKC from activating Rafty and will thus prevent the lipid modifications.

DIF: Moderate REF: 16.2 OBJ: 16.2.a Review the structure of G-protein-coupled receptors (GPCRs) and describe the types of extracellular signal molecules that bind to them. | 16.2.i Recall the location and action of the second messenger molecules produced by activated phospholipase C. | 16.2.j List several biological processes triggered by calcium ions. MSC: Applying 18. You are interested in cell-size regulation and discover that signaling through a GPCR called ERC1 is important in controlling cell size in embryonic rat cells. The G protein downstream of ERC1 activates adenylyl cyclase, which ultimately leads to the activation of PKA. You discover that cells that lack ERC1 are 15% smaller than normal cells, while cells that express a mutant, constitutively activated version of PKA are 15% larger than normal cells. Given these results, which of the following treatments to embryonic rat cells should lead to smaller cells? a. addition of a drug that causes cyclic AMP phosphodiesterase to be hyperactive

b. addition of a drug that prevents GTP hydrolysis by Gα c. addition of a drug that activates adenylyl cyclase d. addition of a drug that mimics the ligand of ERC1 ANS: A Hyperactivating cyclic AMP phosphodiesterase will degrade the cAMP, terminating the signal more quickly than usual.

All other answers will lead to larger cells.

DIF: Hard REF: 16.2 OBJ: 16.2.b Recall the general structure of a G protein and describe how the protein responds when activated by a GPCR. | 16.2.f Name the classes of enzymes that are the most frequent targets of G proteins, and list the second messenger molecules they produce. | 16.2.g Outline how cyclic AMP is produced in response to G protein activation and recall how caffeine can potentiate this response. MSC: Applying 19. The local mediator nitric oxide stimulates the intracellular enzyme guanylyl cyclase by a. activating a G protein. b. activating a receptor tyrosine kinase. c. diffusing into cells and stimulating the cyclase directly. d. activating an intracellular protein kinase.

ANS: C DIF: Easy REF: 16.2 OBJ: 16.2.m Outline how the gas nitric oxide (NO) can act as a signaling molecule to trigger the relaxation of smooth muscle cells. MSC: Remembering 20. Figure 16-20 shows the pathway through which nitric oxide (NO) triggers smooth muscle relaxation in a blood vessel wall. Which of the following situations would lead to relaxation of the smooth muscle cells in the absence of acetylcholine?

Figure 16-20

a. a smooth muscle cell that has a defect in guanylyl cyclase such that it cannot bind NO b. a muscle cell that has a defect in guanylyl cyclase such that it constitutively converts GTP to cyclic GMP c. a muscle cell that has cyclic GMP phosphodiesterase constitutively active d. a drug that blocks an enzyme involved in the metabolic pathway from arginine to NO ANS: B A constitutively active guanylyl cyclase will produce cyclic GMP (cGMP) even in the absence of a signal and thus will lead to relaxation of smooth muscle cells in the absence of acetylcholine. A defect in guanylyl cyclase such that it cannot bind NO would lead to a block in the production of cGMP such that even if NO were to reach the smooth muscle cells, relaxation would not occur. Constitutively active cyclic GMP phosphodiesterase would not lead to muscle-cell re-

laxation independently of acetylcholine, because cyclic GMP phosphodiesterase is involved in the degradation of cGMP. A drug that blocks an enzyme important for NO synthesis will lead to a block in the production of NO. DIF: Moderate REF: 16.2 OBJ: 16.2.m Outline how the gas nitric oxide (NO) can act as a signaling molecule to trigger the relaxation of smooth muscle cells. MSC: Applying 21. The growth factor Superchick stimulates the proliferation of cultured chicken cells. The receptor that binds Superchick is a receptor tyrosine kinase (RTK), and many chicken tumor cell lines have mutations in the gene that encodes this receptor. Which of the following types of mutation would be expected to promote uncontrolled cell proliferation? a. a mutation that prevents dimerization of the receptor b. a mutation that destroys the kinase activity of the receptor c. a mutation that inactivates the protein tyrosine phosphatase that normally removes the phosphates from tyrosines on the activated receptor d. a mutation that prevents the binding of the normal extracellular signal to the receptor ANS: C RTKs are usually activated by signal-induced dimerization, which allows the receptors to phosphorylate themselves and activate intracellular signaling proteins that are stimulated by the phosphorylated receptor. A fter it is activated, the receptor is dephosphorylated, and thereby inactivated, by a protein tyrosine phosphatase. Therefore, a mut ation in the gene that encodes the protein tyrosine phosphatase will inappropriately increase the activity of the r eceptor and promote uncontrolled cell proliferation. Mutations that prevent dimerization of the receptor (including mutations that prevent ligand binding) or autophosphorylation (which requires the kinase activity of the receptor) will inactivate the receptor.

DIF: Moderate REF: 16.3 OBJ: 16.3.b Review how the binding of a signal molecule activates RTKs to trigger the assembly of an intracellular signaling complex. MSC: Applying 22. The growth factor RGF stimulates proliferation of cultured rat cells. The receptor that binds RGF is a receptor tyrosine kinase called RGFR. Which of the following types of alteration would be most likely to prevent receptor dimerization? a. a mutation that increases the affinity of RGFR for RGF b. a mutation that prevents RGFR from binding to RGF c. changing the tyrosines that are normally phosphorylated on RGFR dimerization to alanines d. changing the tyrosines that are normally phosphorylated on RGFR dimerization to glutamic acid ANS: B Binding of a ligand to RTKs leads to their dimerization, and thus a mutation that prevents RGFR from binding to RGF will prevent dimerization. A mutation that increases the affinity of RGFR for RGF will increase dimerization in the presence of ligand. RTKs become phosphorylated on dimerization. However, changing the relevant tyrosines to alanine will block receptor activation but should not cause or prevent dimerization. Because glutamic acid is neg atively charged, it can mimic the addition of a phosphate to an amino acid; thus, changing the relevant tyrosines to glutamic acid may mimic receptor activation, but it should not cause or prevent receptor dimerization.

DIF: Moderate REF: 16.3 OBJ: 16.3.b Review how the binding of a signal molecule activates RTKs to trigger the assembly of an intracellular signaling complex. MSC: Applying 23. A protein kinase can act as an integrating device in signaling if it

a. phosphorylates more than one substrate. b. catalyzes its own phosphorylation. c. is activated by two or more proteins in different signaling pathways. d. initiates a phosphorylation cascade involving two or more protein kinases.

ANS: C Integrating devices are able to relay signals from more than one signaling pathway. Being activated by two or more proteins in different signaling pathways allows a kinase (or any other signaling molecule) to be affected by more than one upstream signal. The other choices affect the output signal that a kinase is able to produce, not its ability to integrate upstream signals from more than one signaling pathway.

DIF: Easy REF: 16.2 OBJ: 16.3.b Review how the binding of a signal molecule activates RTKs to trigger the assembly of an intracellular signaling complex. | 16.3.r Explain how multiple signaling pathways can integrate information to produce a coordinated cell response. MSC: Understanding 24. Which of the following mechanisms is NOT directly involved in inactivating an activated RTK? a. dephosphorylation by serine/threonine phosphatases b. dephosphorylation by protein tyrosine phosphatases c. removal of the RTK from the plasma membrane by endocytosis d. digestion of the RTK in lysosomes ANS: A RTKs are phosphorylated on tyrosines by their dimerization partner, which is also a tyrosine kinase, and thus the r eversal of these phosphorylations involves protein tyrosine phosphatases, and not protein serine/threonine phosphatases. Endocytosis of the receptor and its ultimate digestion in the lysosome are other methods that the cell uses to down-regulate active receptors.

DIF: Easy REF: 16.3 OBJ: 16.3.c Recall how signals transmitted by RTKs can be terminated. MSC: Understanding 25. You are interested in further understanding the signal transduction pathway that controls the production of Pig1, a protein important for regulating cell size. Activation of the TRK receptor leads to activation of the GTP-binding protein, Ras, which then activates a protein kinase that phosphorylates the SZE transcription factor. SZE only interacts with the nuclear transport receptor when it is phosphorylated. SZE is a gene activator for the Pig1 gene. This pathway is diagrammed in Figure 16-25.

Figure 16-25

Normal cells grown under standard conditions (without ligand) are 14 μm in diameter while normal cells exposed to TRK ligand are 10.5 μm in diameter. Given this situation, which of the following conditions do you predict will more likely lead to smaller cells? a. addition of TRK ligand and a drug that stimulates the GTPase activity of Ras b. addition of TRK ligand and a drug that inhibits the activity of the phosphatase that acts on SZE c. addition of TRK ligand and a drug that stimulates the degradation of Pig1 d. addition of TRK ligand and a drug that inhibits Pig1 binding to DNA ANS: B The activation of the TRK receptor and its downstream signaling pathway leads to smaller cells. Thus, by inhibiting the dephosphorylation of SZE, SZE will likely activate the expression of Pig1 longer than it would in normal cells, leading to smaller cells. All other scenarios interfere with TRK receptor signaling, and should lead to cells that are not as small.

DIF: Hard REF: 16.3 OBJ: 16.3.b Review how the binding of a signal molecule activates RTKs to trigger the assembly of an intracellular signaling complex. | 16.3.c Recall how signals transmitted by RTKs can be terminated. | 16.3.d List several intracellular signaling proteins activated by RTKs. MSC: Applying 26. Which of the following statements is TRUE? a. MAP kinase is important for phosphorylating MAP kinase kinase. b. PI 3-kinase phosphorylates a lipid in the plasma membrane.

c. Ras becomes activated when an RTK phosphorylates its bound GDP to create GTP. d. Dimerization of GPCRs leads to Gα activation.

ANS: B MAP kinases are phosphorylated by MAP kinase kinases. Ras exchanges its GDP for GTP when activated. GPCRs do not dimerize upon ligand binding.

DIF: Easy REF: 16.3 OBJ: 16.3.d List several intracellular signaling proteins activated by RTKs. | 16.3.e Outline how RTKs activate the MAP kinase signaling module. | 16.3.f Indicate how Ras can fuel uncontrolled proliferation in cancer. | 16.2.a Review the structure of G-protein-coupled receptors (GPCRs) and describe the types of extracellular signal molecules that bind to them. | 16.2.b Recall the general structure of a G protein and describe how the protein responds when activated by a GPCR. MSC: Understanding 27. The activation of the serine/threonine protein kinase Akt requires phosphoinositide 3-kinase (PI 3-kinase) to a. activate the RTK. b. create phosphorylated lipids that serve as docking sites that localize Akt to the plasma membrane. c. directly phosphorylate Akt. d. create DAG. ANS: B PI 3-kinase activity causes the localization of Akt to the plasma membrane, where it is phosphorylated by another protein kinase. The RTK is activated before PI 3-kinase is activated. PI 3-kinase phosphorylates lipids, not proteins such as Akt. DAG is created by phospholipase C and is not involved in this particular signaling pathway.

DIF: Easy REF: 16.3 OBJ: 16.3.g Review how extracellular signals that promote cell growth and survival activate PI-3kinase signaling pathways. MSC: Remembering 28. Akt promotes the survival of many cells by affecting the activity of Bad and Bcl2, as diagrammed in Figure 16-28.

Figure 16-28

Which of the following statements is FALSE? a. In the presence of a survival signal, Akt is phosphorylated. b. In the absence of a survival signal, Bad inhibits the cell-death inhibitor protein Bcl2. c. In the presence of a survival signal, the cell-death inhibitory protein Bcl2 is active. d. In the absence of a survival signal, Bad is phosphorylated.

ANS: D Bad is phosphorylated in the presence of a survival signal. When the survival signal is not present, Bad binds to the

cell-death inhibitor protein Bcl2, promoting cell death. All the other statements are correct.

DIF: Easy REF: 16.3 OBJ: 16.3.h Compare how Akt promotes cell survival via Bad and stimulates cell growth via Tor. MSC: Applying 29. When the cytosolic tail of the __________ receptor is cleaved, it migrates to the nucleus and affects gene regulation. a. nuclear b. Notch c. growth factor d. G-protein coupled ANS: B DIF: Easy REF: 16.3 OBJ: 16.3.n Review how the Notch receptor activates target genes in response to activation by Delta and explain how this pathway controls the specialization of nerve cells in developing Drosophila embryos. MSC: Remembering 30. All members of the nuclear receptor family a. are cell-surface receptors. b. do not undergo conformational changes. c. are found only in the cytoplasm. d. interact with signal molecules that diffuse through the plasma membrane.

ANS: D All members of the nuclear receptor family are intracellular proteins tha t interact with signal molecules that can diffuse through the plasma membrane. Once activated, nuclear receptors regulate gene transcription in the nucleus. The binding of the signal molecule induces a large conformational change in the receptor protein. T his conformational change activates the nuclear receptors, allowing them to promote or inhibit the transcription of the appropriate genes.

DIF: Easy REF: 16.3 OBJ: 16.3.o Outline how steroid hormones trigger the transcription of different sets of target genes. MSC: Understanding 31. Which of the following statements is FALSE? a. Nucleotides and amino acids can act as extracellular signal molecules. b. Some signal molecules can bind directly to intracellular proteins that bind DNA and regulate gene transcription. c. Some signal molecules are transmembrane proteins. d. Dissolved gases such as nitric oxide (NO) can act as signal molecules, but because they cannot interact with proteins they must act by affecting membrane lipids.

ANS: D NO can diffuse across the plasma membrane and directly activate intracellular proteins such as the enzyme guanylyl cyclase.

DIF: Easy REF: 16.2 OBJ: 16.3.a Compare the general structures of GPCRs and enzyme-coupled receptors such as receptor tyrosine kinases (RTKs). | 16.3.b Review how the binding of a signal molecule activates RTKs to trigger the assembly of an intracellular signaling complex. | 16.2.m Outline how the gas nitric oxide (NO) can act as a signaling molecule to trigger the relaxation of smooth muscle cells. | 16.3.n Review how the Notch receptor activates target genes in response to activation by Delta, and explain how this pathway controls the specialization of nerve cells in developing Drosophila embryos. | 16.3.o Outline how steroid hormones trigger the transcription of different sets of target genes. MSC: Understanding

32. The last common ancestor to plants and animals was a unicellular eukaryote. Thus, it is thought that multicellularity and the attendant demands for cell communication arose independently in these two lineages. This evolutionary viewpoint accounts nicely for the vastly different mechanisms that plants and animals use for cell communication.

Fungi use signaling mechanisms and components that are very similar to those used in animals. Which of the phylogenetic trees shown in Figure 16-32 does this observation support?

Figure 16-32

ANS: B The similarities in signaling mechanisms between animals and fungi support the phylogenetic tree in which fungi branched from the animal lineage after plants and animals separated. This branching order is supported by a wide variety of other data, including genomic sequence comparisons.

DIF: Easy REF: 16.3 OBJ: 16.3.p Contrast the cell signaling systems used by plants and animals. MSC: Applying 33. The ethylene response in plants involves a dimeric transmembrane receptor. When the receptor is not bound to ethylene, the receptor binds to and activates a protein kinase, which activates an intracellular signaling pathway that leads to the degradation of a transcriptional regulator important for transcribing the ethylene-responsive genes (see Figure 16-33). You discover a phosphatase that is important for ethylene signaling, and you name it PtpE. Plants lacking PtpE never turn on ethylene-responsive genes, even in the presence of ethylene. You find that PtpE dephosphorylates serine 121 on the transcriptional regulator. Furthermore, plants lacking PtpE degrade the transcriptional regulator in the presence of ethylene.

Figure 16-33

Which of the following statements is inconsistent with your data? a. When the transcriptional regulator is phosphorylated, it activates transcription of the ethylene-responsive genes. b. When the transcriptional regulator is not phosphorylated, it binds to DNA. c. Activation of the protein kinase that binds to the ethylene receptor leads to inactivation of PtpE. d. Binding of ethylene to its receptor leads to the activation of PtpE. ANS: A Cells lacking PtpE do not transcribe the ethylene-responsive genes, suggesting that dephosphorylation of the transcriptional regulator is required before it can activate transcription. From your data, PtpE is either constitutively active or inactivated by the active protein kinase downstream of the ethylene receptor. The latter scenario would permit maximum repression in the absence of ethylene. All the other choices are consistent with your data.

DIF: Moderate REF: 16.3 OBJ: 16.3.q Describe how the ethylene signaling pathway regulates ripening of fruits.

MSC: Applying 34. Figure 16-34 shows that intracellular signaling pathways can be highly interconnected.

Figure 16-34

From the information in Figure 16-34, which of the following statements is FALSE? a. The GPCR and the RTK both activate phospholipase C. b. Activation of either the GPCR or the RTK will lead to activation of transcriptional regulators. c. CaM-kinase is only activated when the GPCR is active and not when the RTK is active. d. Ras is activated only when the RTK is active and not when the GPCR is active.

ANS: C CaM-kinase is activated by calmodulin, which is ultimately activated by phospholipase C. Either the GPCR or the RTK activates phospholipase C. All the other statements are correct.

DIF: Hard REF: 16.3 OBJ: 16.3.r Explain how multiple signaling pathways can integrate information to produce a coordinated cell response. MSC: Applying 35. When Ras is activated, cells will divide. A dominant-negative form of Ras clings too tightly to GDP. You introduce a dominant-negative form of Ras into cells that also have a normal version of Ras. Which of the following statements is TRUE? a. The cells you create will divide less frequently than normal cells in response to the extracellular signals that typically activate Ras. b. The cells you create will run out of the GTP necessary to activate Ras. c. The cells you create will divide more frequently compared to normal cells in response to the extracellular signals that typically activate Ras. d. The normal Ras in the cells you create will not be able to bind GDP because the dominant-negative Ras binds to GDP too tightly.

ANS: A Dominant-negative Ras is considered “dominant” because it prevents normal Ras from doing its job. This mutant protein, when overexpressed in a cell, binds to—and essentially monopolizes—other signaling partners in the pathway, but it cannot activate the target proteins that lie downstream. In this way, dominant -negative Ras mutants block Ras signaling and inhibit cell proliferation. The creation of GTP in the cell is independent of Ras and its associated signaling pathways. An activated Ras mutant will cause cells to divide more frequently. In this situ ation, the Ras pathway does not work and cells should not divide more frequently. The binding of GDP to dominant negative Ras should not affect the ability of normal Ras to bind GDP, as the concentration of GTP in the cell is quite high, and the normal Ras is inactive and bound to the GDP it created from GTP the last time it was active.

DIF: Hard REF: 16.3 OBJ: 16.3.f Indicate how Ras can fuel uncontrolled proliferation in cancer. | 16.3.m Explain how mutant proteins can be used to determine the order in which proteins participate in a signaling pathway.

MSC: Applying 36. Figure 16-36 shows how normal signaling works with a Ras protein acting downstream of an RTK. You examine a cell line with a constitutively active Ras protein that is always signaling. Which of the following conditions will turn off signaling in this cell line?

Figure 16-36

a. addition of a drug that prevents protein X from activating Ras b. addition of a drug that increases the affinity of protein Y and Ras c. addition of a drug that blocks protein Y from interacting with its target d. addition of a drug that increases the activity of protein Y ANS: C If protein Y cannot interact with its target, signaling will not occur. Increasing the activity of protein Y or increasing the affinity of protein Y and Ras would not turn off signaling. Preventing protein X from activating Ras would have no effect in a cell line with a constitutively active Ras protein, because Ras is already active.

DIF: Moderate REF: 16.3 OBJ: 16.3.m Explain how mutant proteins can be used to determine the order in which proteins participate in a signaling pathway. MSC: Applying

MATCHING 1. Given the generic signaling pathway in Figure 16-37, match the letter with its corresponding descriptor below.

Figure 16-1

1. receptor protein 2. effector proteins 3. intracellular signaling proteins 4. ligand 1. ANS: B DIF: Easy REF: 16.1 OBJ: 16.1.a Define signal transduction and list the basic components involved in this process in cells. MSC: Remembering 2. ANS: D DIF: Easy REF: 16.1 OBJ: 16.1.a Define signal transduction and list the basic components involved in this process in cells. MSC: Remembering 3. ANS: C DIF: Easy REF: 16.1 OBJ: 16.1.a Define signal transduction and list the basic components involved in this process in cells. MSC: Remembering 4. ANS: A DIF: Easy REF: 16.1 OBJ: 16.1.a Define signal transduction and list the basic components involved in this process in cells. MSC: Remembering 2. Match the target of the G protein with the appropriate signaling outcome (A–C).

A. cleavage of inositol phospholipids B. increase in cAMP levels C. changes in membrane potential 1. adenylyl cyclase ANS: B DIF: Easy REF: 16.2 OBJ: 16.2.e Relate the speed of the response produced by G proteins that activate an ion channel versus a membrane-bound enzyme. | 16.2.f Name the classes of enzymes that are the most frequent target of G proteins and list the second messenger molecules they produce. MSC: Remembering

2. ion channels ANS: C DIF: Easy REF: 16.2 OBJ: 16.2.e Relate the speed of the response produced by G proteins that activate an ion channel versus a membrane-bound enzyme. | 16.2.f Name the classes of enzymes that are the most frequent target of G proteins and list the second messenger molecules they produce. MSC: Remembering 3. phospholipase C ANS: A DIF: Easy REF: 16.2 OBJ: 16.2.e Relate the speed of the response produced by G proteins that activate an ion channel versus a membrane-bound enzyme. | 16.2.f Name the classes of enzymes that are the most frequent target of G proteins and list the second messenger molecules they produce. MSC: Remembering 3. Match the class of cell-surface receptor (1–3) with the best description of its function (A–E). Not all descriptors will be used.

A. alter the membrane potential directly by changing the permeability of the plasma membrane B. signal by opening and closing in a ligand-independent manner C. must be coupled with intracellular monomeric GTP-binding proteins D. all receptors of this class are polypeptides with seven transmembrane domains E. discovered for their role in responding to growth factors in animal cells 1. G-protein-coupled receptors ANS: D DIF: Easy REF: 16.1 OBJ: 16.1.m Differentiate the three main classes of cell-surface receptors and provide an example of each. MSC: Remembering 2. Ion-channel-coupled receptors ANS: A DIF: Easy REF: 16.1 OBJ: 16.1.m Differentiate the three main classes of cell-surface receptors and provide an example of each. MSC: Remembering 3. Enzyme-coupled receptors ANS: E DIF: Easy REF: 16.1 OBJ: 16.1.m Differentiate the three main classes of cell-surface receptors and provide an example of each. MSC: Remembering

SHORT ANSWER 1. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. amplification

G protein

phosphorylation

contact-dependent

K+ channel

receptor

endocrine

neuronal

target

epithelial

paracrine

Cells can signal to each other in various ways. A signal that must be relayed to the entire body is most efficiently sent by __________ cells, which produce hormones that are carried throughout the body through the bloodstream. On the other hand, __________ methods of cell signaling do not require the release of a secreted molecule and are used for very localized signaling events. During __________ signaling, the signal remains in the neighborhood of the secreting cell and thus acts as a local mediator on nearby cells. Finally, __________ signaling involves the conversion of electrical impulses into a chemical signal. Cells receive signals through a __________, which can be an integral mem-

brane protein or can reside inside the cell.

ANS: Cells can signal to each other in various ways. A signal that must be relayed to the entire body is most efficiently sent by endocrine cells, which produce hormones that are carried throughout the body through the bloodstream. On the other hand, contact-dependent methods of cell signaling do not require the release of a secreted molecule and are used for very localized signaling events. During paracrine signaling, the signal remains in the neighborhood of the secreting cell and thus acts as a local mediator on nearby cells. Finally, neuronal signaling involves the conversion of electrical impulses into a chemical signal. Cells receive signals through a receptor, which can be an integral membrane protein or can reside inside the cell.

DIF: Easy REF: 16.1 OBJ: 16.1.b Distinguish the main types of signal-mediated cell–cell communication and identify the type of extracellular signal molecules involved in each.

MSC: Understanding 2. List in order the following types of cell signaling from the type of signaling in which the signal molecule travels the least distance to the type of signaling in which the signal molecule travels the largest distance. contact-dependent signaling

neuronal signaling

endocrine signaling

paracrine signaling

ANS: contact-dependent signaling, neuronal signaling, paracrine signaling, endocrine signaling DIF: Easy REF: 16.1 OBJ: 16.1.b Distinguish the main types of signal-mediated cell–cell communication and identify the type of extracellular signal molecules involved in each. MSC: Understanding 3. Explain why the signal molecules used in neuronal signaling work at a longer range than those used in contactdependent signaling.

ANS: The neurotransmitter released from a neuron in neuronal signaling must diffuse across the synaptic cleft to reach receptors on the target cell. In contrast, in contact-dependent signaling, the signal molecule is attached to the plasma membrane of the signaling cell and interacts with receptors located on the plasma membrane of the receiving cell; thus, the cells must be in direct contact for this type of signaling to occur.

DIF: Easy REF: 16.1 OBJ: 16.1.b Distinguish the main types of signal-mediated cell–cell communication and identify the type of extracellular signal molecules involved in each. MSC: Understanding 4. Choose the phrase in each pair that is likely to occur more rapidly in response to an extracellular signal.

A. changes in cell secretion/increased cell division B. changes in protein phosphorylation/changes in proteins being synthesized C. changes in mRNA levels/changes in membrane potential ANS: A. changes in cell secretion B. changes in protein phosphorylation C. changes in membrane potential DIF: Easy REF: 16.1 OBJ: 16.1.f Differentiate the types of cell responses that occur rapidly with those that take minutes or hours to execute. MSC: Applying 5. Receipt of extracellular signals can change cell behavior quickly (for example, in seconds or less) or much more slowly (for example, in hours).

A. What kind of molecular changes could cause quick changes in cell behavior?

B. What kind of molecular changes could cause slow changes in cell behavior?

C. Explain why the response you named in A results in a quick change, whereas the response you named in B results in a slow change.

ANS: A. Any answer that involves the modification of existing cell components is correct. Protein phosphorylation, protein dephosphorylation, protein ubiquitylation, lipid phosphorylation, and lipid cleavage are all examples of correct answers.

B. Responses that involve alterations in gene expression occur slowly.

C. Modification of existing cell components can happen quickly, whereas responses that depend on changes in gene expression take much longer, because the genes will need to be transcribed, the mRNAs will need to be translated, and the proteins need to accumulate to high-enough levels to instigate change.

DIF: Easy REF: 16.1 OBJ: 16.1.f Differentiate the types of cell responses that occur rapidly with those that take minutes or hours to execute. MSC: Applying 6. Can signaling via a steroid hormone receptor lead to amplification of the original signal? If so, how?

ANS: Because the interactions of the signal molecule with its receptor and of the activated receptor with its gene are both one-to-one, there is no amplification in this part of the signaling pathway. The signal can, however, be amplified when the target genes are transcribed, because each activated gene produces multiple copies of mRNA, each of which is used to make multiple copies of the protein that the gene encodes.

DIF: Easy REF: 16.3 OBJ: 16.3.o Outline how steroid hormones trigger the transcription of different sets of target genes. MSC: Understanding 7. When the neurotransmitter acetylcholine is applied to skeletal muscle cells, it binds the acetylcholine receptor and causes the muscle cells to contract. Succinylcholine, which is a chemical analog of acetylcholine, binds to the acetylcholine receptor on skeletal muscle cells but causes the muscle cells to relax; it is therefore often used by surgeons as a muscle relaxant. Propose a model for why succinylcholine causes muscle relaxation. What might be the mechanism to explain the different activities of acetylcholine and succinylcholine on the acetylcholine receptor?

ANS: Although succinylcholine can bind to the acetylcholine receptor, it does not activate the receptor and therefore does not cause the muscle cell to contract. Instead, succinylcholine blocks the ability of acetylcholine to bind to the receptor and thereby prevents acetylcholine from stimulating muscle contraction.

DIF: Moderate REF: 16.1 OBJ: 16.1.g Name the basic components needed for an extracellular signal molecule to change the behavior of a target cell and identify the site at which the primary step in signal transduction takes place.

MSC: Applying 8. For each of the following sentences, select the best word or phrase from the list below to fill in the blanks. Not all words or phrases will be used; each word or phrase should be used only once. acetylase

decouple

GTP-binding

AMP-binding

decrease

neurotransmitter

amplify

effector

protein kinases

autocrine

esterases

protein phosphatases

cleavage

integrate

receptors

convolute

GMP-binding

sterols

An extracellular signal molecule can act to change a cell’s behavior by acting through cell-surface __________ that control intracellular signaling proteins. These intracellular signaling proteins ultimately change the activity of __________ proteins that bring about cell responses. Intracellular signaling proteins can __________ the signal received to evoke a strong response from just a few extracellular signal molecules. A cell that receives more than one extracellular signal at the same time can __________ this information using intracellular signaling proteins. __________ proteins can act as molecular switches, letting a cell know that a signal has been received. Enzymes that phosphorylate proteins, termed __________, can also serve as molecular switches; the actions of these enzymes are countered by the activity of __________.

ANS: An extracellular signal molecule can act to change a cell’s behavior by acting through cell-surface receptors that control intracellular signaling proteins. These intracellular signaling proteins ultimately change the activity of effector proteins that bring about cell responses. Intracellular signaling proteins can amplify the signal received to evoke a strong response from just a few extracellular signal molecules. A cell that receives more than one extracellular signal at the same time can integrate this information using intracellular signaling proteins. GTP-binding proteins can act as molecular switches, letting a cell know that a signal has been received. Enzymes that phosphorylate proteins, termed protein kinases, can also serve as molecular switches; the actions of these enzymes are countered by the activity of protein phosphatases.

DIF: Easy REF: 16.1 OBJ: 16.1.a Define signal transduction and list the basic components involved in this process in cells. | 16.1.j Summarize how phosphorylation can act as a molecular switch and identify the types of proteins that add and remove this chemical modification. MSC: Understanding 9. Name the three main classes of cell-surface receptor.

ANS: ion-channel-coupled receptors; G-protein-coupled receptors; enzyme-coupled receptors DIF: Easy REF: 16.1 OBJ: 16.1.m Differentiate the three main classes of cell-surface receptors and provide an example of each. MSC: Remembering 10. For each of the following sentences, select the best word or phrase from the list below to fill in the blanks. Not all words or phrases will be used; each word or phrase should be used only once. adenylyl cyclase

cholera toxin

GTPase

AMP

diacylglycerol

phosphodiesterase

ATP

five

seven

ATPase

four

three

2+

GDP

twelve

cAMP

GTP

two

Ca

G-protein-coupled receptors (GPCRs) all have a similar structure with __________ transmembrane domains. When a GPCR binds an extracellular signal, an intracellular G protein, composed of __________ subunits, becomes activated. __________ of the G-protein subunits are tethered to the plasma membrane by short lipid tails. When unstimulated, the α subunit is bound to __________ which is exchanged for __________ on stimulation. The __________ activity of the α subunit is important for inactivating the G protein. __________ inhibits this activity of the α subunit, thereby keeping the subunit in an active state.

ANS: G-protein-coupled receptors (GPCRs) all have a similar structure with seven transmembrane domains. When a

GPCR binds an extracellular signal, an intracellular G protein, composed of three subunits, becomes activated. Two of the G-protein subunits are tethered to the plasma membrane by short lipid tails. When unstimulated, the α subunit is bound to GDP, which is exchanged for GTP on stimulation. The intrinsic GTPase activity of the α subunit is important for inactivating the G protein. Cholera toxin inhibits this activity of the α subunit, thereby keeping the subunit in an active state. DIF: Easy REF: 16.2 OBJ: 16.2.b Recall the general structure of a G protein and describe how the protein responds when activated by a GPCR. | 16.2.d Contrast how cholera toxin and pertussis toxin exert their effects.

MSC: Understanding 11. Indicate by writing “yes” or “no” whether amplification of a signal could occur at the particular steps described below.

Explain your answers.

A. An extracellular signaling molecule binds and activates a GPCR. B. The activated GPCRs cause Gα to separate from Gβ and Gγ.

C. Adenylyl cyclase produces cyclic AMP. D. cAMP activates protein kinase A. E. Protein kinase A phosphorylates target proteins.

ANS: A. No. Each signaling molecule activates only one receptor molecule.

B. Yes. Each activated GPCR activates many G-protein molecules.

C. Yes. Each activated adenylyl cyclase molecule can generate many molecules of cAMP. D. No. In unstimulated cells, protein kinase A is held inactive in a protein complex. Binding of cAMP to the complex induces a conformational change, releasing the active protein kinase A. Therefore, one cAMP cannot activate more than one molecule of protein kinase A. E. Yes. Each activated protein kinase A molecule can phosphorylate many molecules of each type of target protein.

DIF: Moderate REF: 16.2 OBJ: 16.2.b Recall the general structure of a G protein and describe how the protein responds when activated by a GPCR. | 16.2.g Outline how cyclic AMP is produced in response to G protein activation and recall how caffeine can potentiate this response. MSC: Evaluating 12. Acetylcholine acts at a GPCR on heart muscle to make the heart beat more slowly. It does so by ultimately opening K + channels in the plasma membrane (as diagrammed in Figure 16-51), which decreases the cell’s excitability by making it harder to depolarize the plasma membrane.

Indicate whether each of the following conditions would increase or decrease the effect of acetylcholine.

A. addition of a drug that stimulates the GTPase activity of the Gα subunit B. mutations in the K+ channel that keep it closed all the time C. modification of the Gα subunit by cholera toxin D. a mutation that decreases the affinity of the βγ complex of the G protein for the K+ channel E. a mutation in the acetylcholine receptor that prevents its localization on the cell surface F. adding acetylcholinesterase to the external environment of the cell ANS: A. Decrease. An increase in the GTPase activity of the Gα subunit will decrease the length of time that the G protein is active.

B. Decrease. If the K+ channel remains closed, acetylcholine will not slow the heart.

C. Increase. Cholera toxin inhibits the GTPase activity of the Gα subunit, keeping the subunit in an active state for a

longer time.

D. Decrease. The activated βγ complex binds to and activates the K+ channel; decreasing their affinity for each other will decrease the time that the K+ channel is open, effectively decreasing the effect of acetylcholine.

E. Decrease. If there is no receptor on the cell surface, cells will be unable to respond to acetylcholine.

F. Decrease. Acetylcholinesterase degrades acetylcholine and thus will decrease the effect of acetylcholine.

DIF: Hard REF: 16.2 OBJ: 16.2.b Recall the general structure of a G protein and describe how the protein responds when activated by a GPCR. | 16.2.d Contrast how cholera toxin and pertussis toxin exert their effects. | 16.2.g Outline how cyclic AMP is produced in response to G protein activation and recall how caffeine can potentiate this response.

MSC: Applying 13. When adrenaline binds to adrenergic receptors on the surface of a muscle cell, it activates a G protein, initiating an intracellular signaling pathway in which the activated α subunit activates adenylyl cyclase, thereby increasing cAMP levels in the cell. The cAMP molecules then activate a cAMP-dependent kinase (PKA) that, in turn, activates enzymes that result in the breakdown of muscle glycogen, thus lowering glycogen levels. You obtain muscle cells that are defective in various components of the signaling pathway. Referring to Figure 16-20, indicate how glycogen levels would be affected in the presence of adrenaline in the following cells. Would they be higher or lower than in normal cells treated with adrenaline?

A. cells that lack adenylyl cyclase B. cells that lack the GPCR C. cells that lack cAMP phosphodiesterase D. cells that have an α subunit that cannot hydrolyze GTP but can interact properly with the β and γ subunits ANS: A. higher B. higher C. lower D. lower DIF: Hard REF: 16.2 OBJ: 16.2.b Recall the general structure of a G protein and describe how the protein responds when activated by a GPCR. | 16.2.g Outline how cyclic AMP is produced in response to G protein activation and recall how caffeine can potentiate this response. MSC: Applying 14. A calmodulin-regulated kinase (CaM-kinase) is involved in spatial learning and memory. This kinase is able to phosphorylate itself such that its kinase activity is now independent of the intracellular concentration of Ca 2+. Thus, the kinase stays active after Ca2+ levels have dropped. Mice completely lacking this CaM-kinase have severe spatial learning defects but are otherwise normal.

A. Each of the following mutations also leads to similar learning defects. For each case explain why.

(1) a mutation that prevents the kinase from binding ATP (2) a mutation that deletes the calmodulin-binding part of the kinase (3) a mutation that destroys the site of autophosphorylation B. What would be the effect on the activity of CaM-kinase if there were a mutation that reduced its interaction with the protein phosphatase responsible for inactivating the kinase?

ANS: A. Because a complete lack of the CaM-kinase causes a learning defect, we can assume that mutations leading to

inactivation of the kinase would also have a similar effect.

(1) Protein kinases have a binding site for ATP, which is the source of the phosphate used for phosphorylating their target proteins; if the kinase cannot bind ATP, it will be inactive.

(2) Because binding to calmodulin in the presence of Ca2+ activates CaM-kinases, deletion of the calmodulinbinding portion would inactivate the kinase.

(3) A mutation that destroys the site of autophosphorylation will also impair the normal function of the kinase, because the kinase will become inactive as soon as Ca 2+ levels decrease.

B. The kinase would stay active for longer after a transient increase in intracellular Ca2+ concentration.

DIF: Hard REF: 16.2 OBJ: 16.2.l Review how calcium-responsive proteins such as calmodulin propagate a calcium ion signal. MSC: Applying 15. Activated GPCRs activate G proteins by reducing the strength of binding of GDP to the α subunit of the G protein, allowing GDP to dissociate and GTP (which is present at much higher concentrations in the cell than GDP) to bind in its place. How would the activity of a G protein be affected by a mutation that reduces the affinity of the α subunit for GDP without significantly changing its affinity for GTP? ANS: The mutant G protein would be constantly active. Each time the α subunit hydrolyzed GTP to GDP, the GDP would dissociate spontaneously, allowing GTP to bind and reactivate the α subunit, especially because the intracellular concentration of GTP is higher than that of GDP. Normally, GDP is tightly bound by the α subunit, which keeps the G protein in its inactive state until interaction with an appropriate activated GPCR stimulates the release of GDP. DIF: Moderate REF: 16.2 OBJ: 16.2.c Summarize the factors that determine the duration of a GPCR -stimulated response. MSC: Applying 16. The rod photoreceptors in the eye are extremely sensitive to light. The cells sense light through a signal transduction cascade involving light activation of a GPCR that activates a G protein that activates cyclic GMP phosphodiesterase.

How would you expect the addition of the following drugs to affect the light-sensing ability of the rod cells? Explain your answers.

A. a drug that inhibits cyclic GMP phosphodiesterase B. a drug that is a nonhydrolyzable analog of GTP ANS: A. A drug that inhibits cyclic GMP phosphodiesterase would decrease any light response in the rod cell. Normally, cyclic GMP is continuously being produced in the eye. The perception of light by a rod cell normally leads to the activation of cyclic GMP phosphodiesterase, which then hydrolyzes cyclic GMP molecules. This causes Na + channels to close, which changes the membrane potential and alters the signal sent to the brain. If cyclic GMP phosphodiesterase were blocked, levels of cyclic GMP would remain high and there would be no cellular response to light.

B. A drug that is a nonhydrolyzable analog of GTP would lead to a prolonged response to light. This is because a nonhydrolyzable analog of GTP would prevent the G protein from turning itself off by hydrolyzing its bound GTP to GDP. Continued activation of the G protein would keep cyclic GMP phosphodiesterase levels higher than normal, leading to a prolonged period of lowered levels of cyclic GMP. This in turn would cause Na + channels to be closed for longer than normal, leading to a prolonged change in the membrane potential and an extended light response.

DIF: Hard REF: 16.2 OBJ: 16.2.o Outline how GPCRs in the photoreceptors of the retina transmit an extremely rapid signal in response to stimulation by light. MSC: Applying

17. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. adenylyl cyclase

endoplasmic reticulum

nuclear

average

extracellular

peroxisome

Ca2+

high

phospholipase C

calmodulin

intracellular

protein kinase A

colorful

low

protein kinase C

Ca2+ can trigger biological effects in cells because an unstimulated cell has an extremely __________ concentration of free Ca2+ in the cytosol, compared with its concentration in the __________ space and in the __________ creating a steep electrochemical gradient. When Ca2+ enters the cytosol, it interacts with Ca2+-responsive proteins such as __________ which also binds diacylglycerol, and __________ which activates CaM-kinases.

ANS: Ca2+ can trigger biological effects in cells because an unstimulated cell has an extremely low concentration of free Ca2+ in the cytosol, compared with its concentration in the extracellular space and in the endoplasmic reticulum, creating a steep electrochemical gradient. When Ca2+ enters the cytosol, it interacts with Ca2+-responsive proteins such as protein kinase C, which also binds diacylglycerol, and calmodulin, which activates CaM-kinases.

DIF: Easy REF: 16.2 OBJ: 16.2.l Review how calcium-responsive proteins such as calmodulin propagate a calcium ion signal. MSC: Understanding 18. Antibodies are Y-shaped molecules that have two identical binding sites. Suppose that you have obtained an antibody that is specific for the extracellular domain of an RTK. When the antibody binds to the RTK, it brings together two RTK molecules. If cells containing the RTK were exposed to the antibody, would you expect the kinase to be activated, inactivated, or unaffected? Explain your reasoning.

ANS: The RTK will probably become activated on binding of the antibody molecule. This is because signal-induced dimerization usually activates RTKs. When RTK molecules are brought together, their cytoplasmic kinase domains become activated and each receptor phosphorylates the other. DIF: Moderate REF: 16.3 OBJ: 16.3.b Review how the binding of a signal molecule activates RTKs to trigger the assembly of an intracellular signaling complex.

MSC: Evaluating 19. You are interested in cell-size regulation and discover that signaling through an enzyme-coupled receptor is important for the growth (enlargement) of mouse liver cells. Activation of the receptor activates adenylyl cyclase, which ultimately leads to the activation of PKA, which then phosphorylates a transcription factor called TFS on threonine 42. This phosphorylation is necessary for the binding of TFS to its specific sites on DNA, where it then activates the transcription of Sze2, a gene that encodes a protein important for liver cell growth. You find that liver cells lacking the receptor are 15% smaller than normal cells, whereas cells that express a constitutively activated version of PKA are 15% larger than normal liver cells. Given these results, predict whether you would expect the cell’s size to be bigger or smaller than normal cells if cells were treated in the following fashion.

A. You change threonine 42 on TFS to an alanine residue.

B. You create a version of the receptor that is constitutively active.

C. You add a drug that inhibits adenylyl cyclase.

D. You add a drug that increases the activity of cyclic AMP phosphodiesterase.

E. You mutate the cAMP-binding sites in the regulatory subunits of PKA, so that the complex binds cAMP more tightly.

ANS: A. Smaller. This mutation will make a TFS that cannot be phosphorylated by PKA. B. Bigger.

C. Smaller.

D. Smaller. cAMP phosphodiesterase is involved in converting cAMP to AMP and will down-regulate this signaling pathway.

E. Bigger. Higher affinity of the PKA complex for cAMP will increase its activity, and thus cells will be bigger.

DIF: Hard REF: 16.3 OBJ: 16.3.d List several intracellular signaling proteins activated by RTKs. MSC: Applying 20. Male cockroaches with mutations that strongly decrease the function of an RTK called RTKX are oblivious to the charms of their female comrades. This particular RTK binds to a small molecule secreted by sexually mature females.

Most males carrying loss-of-function mutations in the gene for Ras protein are also unable to respond to females. You have just read a paper in which the authors describe how they have screened cockroaches that are mutant in RTKX for additional mutations that partly restore the ability of males to respond to females. These mutations decrease the function of a protein that the authors call Z. Which of the following types of protein could Z be? Explain your answer.

A. a protein that activates the Ras protein by causing Ras to exchange GDP for GTP B. a protein that stimulates hydrolysis of GTP by the Ras protein C. an adaptor protein that mediates the binding of the RTKX to the Ras protein D. a transcriptional regulator required for the expression of the Ras gene ANS: B; Mutations that increase the activity of Ras should mimic the effect of stimulating RTK X in a receptorindependent fashion. Because the intracellular concentration of GTP is higher than that of GDP, some proportion of the Ras molecules is expected to be GTP-bound and active; ridding the cells of a protein that stimulates GTP hydrolysis will increase this pool of active Ras. Mutants that cannot stimulate Ras to exchange GDP for GTP will have the same phenotype as mutants lacking Ras, as will mutants lacking a transcriptional regulator required for expression of the Ras gene. Defects in an adaptor protein that mediates the binding of receptor X to Ras will have no further effect on a mutant already lacking the receptor.

DIF: Hard REF: 16.3 OBJ: 16.3.b Review how the binding of a signal molecule activates RTKs to trigger the assembly of an intracellular signaling complex. 16.3.f Indicate how Ras can fuel uncontrolled proliferation in cancer. 16.3.m Explain how mutant proteins can be used to determine the order in which proteins participate in a signaling pathway.

MSC: Applying 21. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. cyclic GMP

IGF

NO

Delta

IP3

phosphodiesterase

diacylglycerol

MAP kinase

Ras

Cells signal to one another in various ways. Some use extracellular signal molecules that are dissolved gases, such as __________ which can diffuse easily into cells. Most receptor tyrosine kinases activate __________ , a small GTPbinding protein found at the cytosolic face of the plasma membrane. Some intracellular signaling pathways involve a series of protein kinases that phosphorylate each other, as seen in the __________ signaling module. Lipids can also

relay signals in the cell, as we observe when phospholipase C cleaves the sugar-phosphate head off a lipid molecule to generate the two small messenger molecules __________ (which remains embedded in the plasma membrane) and __________ (which diffuses into the cytosol).

ANS: Cells signal to one another in various ways. Some use extracellular signal molecules that are dissolved gases, such as NO, which can diffuse easily into cells. Most receptor tyrosine kinases activate Ras, a small GTP-binding protein found at the cytosolic face of the plasma membrane. Some intracellular signaling pathways involve a series of protein kinases that phosphorylate each other, as seen in the MAP kinase signaling module. Lipids can also relay signals in the cell, as we observe when phospholipase C cleaves the sugar-phosphate head off a lipid molecule to generate the two small messenger molecules diacylglycerol (which remains embedded in the plasma membrane) and IP3 (which diffuses into the cytosol). DIF: Easy REF: 16.3 OBJ: 16.3.g Review how extracellular signals that promote cell growth and survival activate PI-3-kinase signaling pathways. | 16.3.e Outline how RTKs activate the MAP kinase signaling module. MSC: Understanding 22. When activated by the extracellular signal protein platelet-derived growth factor (PDGF), the PDGF receptor phosphorylates itself on multiple tyrosines (as indicated in Figure 16-61A by the circled Ps; the numbers next to these Ps indicate the amino acid number of the tyrosine). These phosphorylated tyrosines serve as docking sites for proteins that interact with the activated PDGF receptor. These proteins are indicat ed in the figure, and include the proteins A, B, C, and D. One of the cell’s responses to PDGF is an increase in DNA synthesis, which can be measured by the incorporation of radioactive thymidine into the DNA. To determine which protein or proteins—A, B, C, or D—are responsible for the activation of DNA synthesis, you construct mutant versions of the PDGF receptor that retain one or more tyrosine phosphorylation sites. You express these mutant versions in cells that do not make their own PDGF receptor. In these cells, the various mutant versions of the PDGF receptor are expressed normally, and, in response to PDGF binding, become phosphorylated on whichever tyrosines remain. You measure the level of DNA synthesis in cells that express the various mutant receptors and obtain the data shown in Figure 16-22B.

Figure 16-22

A. From these data, which, if any, of proteins A, B, C, and D are involved in the stimulation of DNA synthesis by PDGF? Explain your answer.

B. Which, if any, of these proteins inhibit DNA synthesis? Explain your answer.

C. Which, if any, of these proteins seem to have no detectable role in DNA synthesis? Explain your answer.

ANS: A. Proteins A and D stimulate DNA synthesis. PDGF receptors that can bind to only A or D (see experiments 2 and 5) can stimulate DNA synthesis to about 50% of normal amounts (which is represented by experiment 1).

Proteins A and D are both needed and are used in an additive fashion; this is evident from experiment 6: when a PDGF receptor can bind both A and D, the DNA synthesis level is close to that obtained with the normal rece ptor.

B. Protein B is an inhibitor of DNA synthesis. Consequently, receptors with binding sites for B and D (see experiment 7) stimulate a lower DNA synthesis rate than do receptors that bind only D (experiment 5).

C. Protein C has no detectable role in DNA synthesis. Receptors that can bind only C (experiment 4) activate DNA synthesis about as much as receptors that do not bind any of the four proteins (experiment 9; the negative control).

Furthermore, the binding of protein C does not affect the response mediated by protein D when the receptor can bind both C and D (experiment 8).

DIF: Hard REF: 16.3 OBJ: 16.3.b Review how the binding of a signal molecule activates RTKs to trigger the assembly of an intracellular signaling complex. 16.3.k Outline how a set of mutant RTKs can be used to determine which tyrosines serve as docking sites for the intracellular signaling proteins that propagate the signal. MSC: Applying 23. Two protein kinases, PK1 and PK2, work sequentially in an intracellular signaling pathway. You create cells that contain inactivating mutations in the genes that encode either PK1 or PK2 and find that these cells no longer respond to a particular extracellular signal. You also create cells containing a version of PK1 that is permanently active and find that the cells behave as though they are receiving the signal even when the signal is not present. When you introduce the permanently active version of PK1 into cells that have an inactivating mutation in PK2, you find that these cells also behave as though they are receiving the signal even when no signal is present.

A. From these results, does PK1 activate PK2, or does PK2 activate PK1? Explain your answer.

B. You now create a permanently active version of PK2 and find that cells containing this version behave as though they are receiving the signal even when the signal is not present. What do you predict will happen if you introduce the permanently active version of PK2 into cells that have an inactivating mutation in PK1?

ANS: A. Normally, PK2 activates PK1. We are told that PK1 and PK2 normally work sequentially in an intracellular signaling pathway. If PK1 is permanently activated, a response is seen independently of whether or not PK2 is present. If PK1 activated PK2, no response should be seen if PK1 were activated in the absence of PK2.

B. You would predict that no response to the signal would be observed. This is because PK2 normally needs to activate PK1 for the cells to respond to the signal. When PK2 is permanently activated in the absence of PK1, PK1 is not there to relay the signal.

DIF: Hard REF: 16.3 OBJ: 16.3.m Explain how mutant proteins can be used to determine the order in which proteins participate in a signaling pathway. MSC: Applying 24. Nuclear receptors have binding sites for a signaling molecule and a DNA sequence. How is it that the same nuclear receptor, which binds to a specific DNA sequence, can regulate different genes in different cell types?

ANS: The specific genes regulated in response to an activated steroid hormone receptor depends not only on the genes having the appropriate DNA sequence for binding the receptor but also on a variety of other nuclear proteins that influence gene expression, some of which vary between different cell types. DIF: Easy REF: 16.3 OBJ: 16.3.o Outline how steroid hormones trigger the transcription of different sets of target genes. MSC: Understanding

25. Your friend is studying mouse fur color and has isolated the GPCR responsible for determining its color, as well as the extracellular signal that activates the receptor. She finds that, on addition of the signal to pigment cells (cells that produce the pigment determining fur color), cAMP levels rise in the cell. She starts a biotech company, and the company isolates more components of the signaling pathway responsible for fur color. Using transgenic mouse technology, the company genetically engineers mice that are defective in various proteins involved in determining fur color. The company obtains the following results.

• Normal mice have beige (very light brown) fur color.

• Mice lacking the extracellular signal have white fur.

• Mice lacking the GPCR have white fur.

• Mice lacking cAMP phosphodiesterase have dark brown fur.

Your friend has also made mice that are defective in the α subunit of the G protein in this signaling pathway. The defective α subunit works normally except that, once it binds GTP, it cannot hydrolyze GTP to GDP. What color do you predict that the fur of these mice will be? Why? ANS: These mice will have dark brown fur. The inability to hydrolyze GTP to GDP will lead to inappropriate activation of the signaling pathway that makes pigment. Too much pigment will be produced, as seen in the mice lacking cAMP phosphodiesterase (which lack the ability to damp the signal), and the mice will end up with dark brown fur.

DIF: Hard REF: 16.2 OBJ: 16.2.c Summarize the factors that determine the duration of a GPCR-stimulated response.

MSC: Applying 26. Bacteria undergo chemotaxis toward amino acids, which usually indicates the presence of a food source. Chemotaxis receptors bind a particular amino acid and cause changes in the bacterial cell that induce the cell to move toward the source of the amino acid. Four types of chemotaxis receptor that mediate responses to different amino acids have been identified in a bacterium. The receptors are called ChrA, ChrB, ChrC, and ChrD. Each receptor specifically senses serine, aspartate, glutamate, or glycine, although you do not know which receptor senses which amino acid. You have been given a wild-type bacterial strain that contains all four receptors, as well as various mutant bacterial strains that are lacking one or more of the receptors. To figure out which receptor senses which amino acid, you conduct experiments in which you fill a capillary tube with an amino acid to attract the bacteria, dip the capillary tube into a solution containing bacteria, remove the capillary tube after 5 minutes, and count the number of bacteria in the capillary tube.

Your results are shown in Table 16-26.

Table 16-26

From these results, indicate which receptor is used for which amino acid.

ANS: ChrA senses glycine, ChrB senses aspartate, ChrC senses glutamate, and ChrD senses serine. To figure this out, you must match the pattern of intact receptors with the pattern of responses to the various amino acids. For example, ChrD is missing in strain 2, which does not sense serine. Therefore, ChrD is the receptor used to sense serine. Since we know that one of the receptors senses glutamate, but all strains respond to glutamate, ChrC must be the sensor for glu-

tamate because it is present in all strains. DIF: Moderate REF: 16.1 OBJ: 16.1.a Define signal transduction and list the basic components involved in this process in cells. | 16.1.g Name the basic components needed for an extracellular signal molecule to change the behavior of a target cell and identify the site at which the primary step in signal transduction takes place. MSC: Applying

CHAPTER 17 Cytoskeleton

INTERMEDIATE FILAMENTS 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules.

17.1.b Describe the location and main function of intermediate filaments.

17.1.c Recall how the structure of intermediate filaments relates to their strength and durability.

17.1.d Summarize how intermediate filaments are assembled and describe their polarity.

17.1.e Review how intermediate filament proteins can differ from one another and how these differences relate to the function of the intermediate filament.

17.1.f

Describe three disorders that involve defects in intermediate filaments.

17.1.g Compare the structure of the nuclear lamina with that of cytoplasmic intermediate filaments.

17.1.h Review how the nuclear lamina disassembles and re-forms during each cell division.

17.1.i

Recall how intermediate filaments are stabilized by cross-linking accessory proteins, and explain how these pro-

teins help to position the nucleus within the cell interior.

MICROTUBULES 17.2.a Review the general location and functions of microtubules.

17.2.b List several examples of organizing centers from which microtubules grow.

17.2.c Describe the structure of microtubules and recall how microtubules are assembled from tubulin dimers.

17.2.d State the polarity of a microtubule and summarize how this polarity affects its assembly and function.

17.2.e Describe the structure of a centrosome and review how the centrosome nucleates the growth of microtubules.

17.2.f

Explain why microtubules require organizing centers such as centrosomes to nucleate their growth.

17.2.g Describe dynamic instability and indicate how this behavior relates to microtubule function.

17.2.h Summarize how dynamic instability is controlled by GTP hydrolysis.

17.2.i

Contrast the actions of colchicine and Taxol and explain why both are used to treat human cancers.

17.2.j

Recall how and why cells modify the dynamic instability of their microtubules.

17.2.k Compare the movement of cell components—including organelles, membrane vesicles, and macromolecules—by free diffusion and by microtubule-guided transport.

17.2.l

Outline how microtubules participate in cell polarization.

17.2.m Compare kinesins and cytoplasmic dyneins in terms of their structure, their movement along the microtubule, and how they interact with cargo.

17.2.n Compare the roles that kinesins and cytoplasmic dyneins have in positioning the organelles in a eukaryotic cell, and describe the effect that colchicine treatment has on organelle placement.

17.2.o Summarize how fluorescent marker proteins and non-hydrolyzable ATP analogs can be used to study the activity of motor proteins such as kinesin or myosin.

17.2.p Compare the functions and movements of cilia and flagella.

17.2.q Describe the arrangement of microtubules inside a cilium or flagellum.

17.2.r

Outline how ciliary dynein allows a cilium to bend.

ACTIN FILAMENTS 17.3.a List several cell structures formed by actin filaments.

17.3.b Compare actin filaments and microtubules in terms of width, length, polarity, and cross-linking.

17.3.c Compare the polymerization of actin filaments to that of microtubules.

17.3.d Explain treadmilling and identify the conditions under which this behavior takes place.

17.3.e Compare the actions of cytochalasin and phalloidin and describe their effects on cell behavior.

17.3.f

Outline the functions of common actin-binding proteins, including thymosin, profilin, formins, actin-related proteins

(ARPs), and myosin.

17.3.g Describe the structure and function of the cell cortex.

17.3.h Differentiate between lamellipodia and filopodia.

17.3.i

Distinguish the roles played by actin in the protrusion, attachment, and contraction involved in cell movement.

17.3.j

Explain how ARPs and formins aid in the assembly and extension of protrusions at the cell’s leading edge.

17.3.k Summarize how different members of the Rho family of GTPases alter the organization of actin filaments.

17.3.l

Compare the structure, binding properties, and general function of myosins I and II.

MUSCLE CONTRACTION 17.4.a List tissues in which actin and myosin filaments are organized in contractile bundles.

17.4.b Distinguish the structures of a skeletal muscle fiber and a myofibril.

17.4.c Describe the structure of a sarcomere.

17.4.d Outline the sliding-filament mechanism of muscle contraction and relaxation, and describe how ATP binding and hydrolysis drive the conformational changes that underlie this movement.

17.4.e Explain how excitation of the muscle cell membrane triggers a rise in the cytosolic concentration of calcium ions.

17.4.f

Summarize how calcium ions trigger muscle contraction and how contraction is subsequently reversed.

17.4.g Explain how calcium ions stimulate contraction in nonmuscle cells or in smooth muscle, and compare this mode of activation with that of skeletal muscle.

MULTIPLE CHOICE 1. Which of the following statements about the cytoskeleton is FALSE? a. The cytoskeleton is made up of three types of protein filaments. b. The cytoskeleton controls the location of organelles in eukaryotic cells. c. Covalent bonds between protein monomers hold together cytoskeletal filaments. d. The cytoskeleton of a cell can change in response to the environment.

ANS: C The protein monomers of the cytoskeleton are held together by noncovalent interactions between the protein monomers. All the other statements are true.

DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin fi laments, and microtubules. MSC: Understanding 2. Which of the following statements about the cytoskeleton is TRUE? a. All eukaryotic cells have actin, microtubules, and intermediate filaments in their cytoplasm. b. The cytoskeleton provides a rigid and unchangeable structure important for the shape of the cell. c. The three cytoskeletal filaments perform distinct tasks in the cell and act completely independently of one another. d. Actin filaments and microtubules have an inherent polarity, with a plus end that grows more quickly than the minus end.

ANS: D Not all eukaryotic cells have cytoplasmic intermediate filaments. The cytoskeleton is not rigid and unchangeable; in fact, it can be quite dynamic. Each of the three cytoskeletal systems is not completely independent. For example, proteins such as plectin are known to link intermediate filaments to the actin and microtubule cytoskeleton.

DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin fi laments, and microtubules. MSC: Understanding 3. Which of the statements below about intermediate filaments is FALSE? a. They can stay intact in cells treated with concentrated salt solutions. b. They can be found in the cytoplasm and the nucleus. c. They can be anchored to the plasma membrane at a cell–cell junction. d. Each filament is about 10 μm in diameter.

ANS: D Intermediate filaments are about 10 nm (not μm) in diameter. All the other statements are true.

DIF: Easy REF: 17.1 OBJ: 17.1.b Describe the location and main function of intermediate filaments. | 17.1.c Recall how the structure of intermediate filaments relates to their strength and durability. MSC: Understanding 4. All intermediate filaments are of similar diameter because a. the central rod domains are similar in size and amino acid sequence. b. the globular domains are similar in size and amino acid sequence. c. covalent bonds among tetramers allow them to pack together in a similar fashion. d. there is only a single type of intermediate filament in every organism.

ANS: A

Globular domains vary among intermediate filaments in size and have different types of amino acids. The interactions among all intermediate filament subunits involve noncovalent bonding. There are several classes of intermediate filaments and an organism can have more than one class (and sometimes, more than one member of each class).

DIF: Easy REF: 17.1 OBJ: 17.1.c Recall how the structure of intermediate filaments relates to their strength and durability. | 17.1.d Summarize how intermediate filaments are assembled and describe their polarity. | 17.1.e Review how intermediate filament proteins can differ from one another and how these differences relate to the function of the intermediate filament. MSC: Understanding 5. Intermediate filaments help protect animal cells from mechanical stress because filaments a. directly extend from the interior of the cell to the extracellular space and into the next cell, linking one cell to the next, helping to distribute locally applied forces. b. in each cell are indirectly connected to the filaments of a neighboring cell through the desmosome, creating a continuous mechanical link between cells. c. remain independent of other cytoskeletal elements and keep the mechanical stress away from other cellular components. d. make up the desmosome junctions that connect cells; these junctions are more important than the internal network of filaments for protecting cells against mechanical stress.

ANS: B Intermediate filaments do not directly extend from cell to cell. The linking of intermediate filaments to other cytoskeletal elements (like actin) is thought to help protect cells from mechanical stress. Desmosome junctions are made up of many different kinds of proteins, including cadherins in the extracellular space (which mediate cell–cell adhesion) as well as proteins within the cytoplasm that mediate the attachment of desomosomes to intermediate filaments. These junctions alone are not sufficient for protection against mechanical stress and need the interaction with the intermediate filament network in the cell.

DIF: Easy REF: 17.1 OBJ: 17.1.b Describe the location and main function of intermediate filaments. | 17.1.c Recall how the structure of intermediate filaments relates to their strength and durability. | 17.1.i Recall how intermediate filaments are stabilized by cross-linking accessory proteins, and explain how these proteins help to position the nucleus within the cell interior. MSC: Understanding 6. Intermediate filaments are made from elongated fibrous proteins that are assembled into a ropelike structure. Figure 176 shows the structure of an intermediate filament subunit. You are interested in how intermediate filaments are formed, and you create an intermediate filament subunit whose α-helical region is twice as long as that of a normal intermediate filament by duplicating the normal α-helical region while keeping a globular head at the N-terminus and a globular tail at the C-terminus; you call this subunit IFαd. If you were to assemble intermediate filaments using IFαd as the subunit, which of the following predictions describes the most likely outcome?

Figure 17-6

a. Filaments assembled using IFαd will interact with different cytoskeletal components. b. Filaments assembled using IFαd will form dimers that are twice as long as dimers assembled from normal intermediate

filaments. c. Sixteen tetramers assembled from IFαd will be needed for a ropelike structure to form. d. Dimers of IFαd will form by interactions with the N-terminal globular head and the C-terminal globular tail.

ANS: B Because the α-helical region is twice as long, you would predict that a coiled-coil dimer made up of two IFαd subunits would be about twice as long as a dimer assembled from a normal intermediate filament subunit. Because the globular head and tail regions usually interact with other cellular components, doubling the size of the α-helical region without changes in the globular regions is unlikely to cause changes in protein interactions. Eight tetramers are usually needed to form a ropelike filament, and it is unlikely that 16 will be needed with IFαd, because it is the length of the coiled-coil region that is most affected by a doubling in size of the α-helical region. Interactions in the α-helical region are important for dimerization.

DIF: Moderate REF: 17.1 OBJ: 17.1.d Summarize how intermediate filaments are assembled and describe their polarity. | 17.1.c Recall how the structure of intermediate filaments relates to their strength and durability. MSC: Applying 7. Keratins, neurofilaments, and vimentins are all categories of intermediate filaments. Which of the following properties is not true of these types of intermediate filaments? a. They strengthen cells against mechanical stress. b. Dimers associate by noncovalent bonding to form a tetramer. c. They are found in the cytoplasm. d. Phosphorylation causes disassembly during every mitotic cycle.

ANS: D Keratins, neurofilaments, and vimentins are cytoplasmic intermediate filaments, which tend to be very stable once formed. The nuclear intermediate filaments are disassembled and r eformed during mitosis; this process is regulated by phosphorylation.

DIF: Easy REF: 17.1 OBJ: 17.1.b Describe the location and main function of intermediate filaments. | 17.1.e Review how intermediate filament proteins can differ from one another and how these differences relate to the function of the intermediate filament. | 17.1.h Review how the nuclear lamina disassembles and re-forms during each cell division.

MSC: Remembering 8. You are studying nuclear lamins and use recombinant DNA technology to alter the coding sequence of a nuclear lamin gene. The alteration you make creates a situation such that the gene now codes for a nuclear lamin protein that can no longer be phosphorylated when the nuclear envelope is broken down during mitosis. What do you predict would happen if the cell only had the altered nuclear lamin gene (and not the unaltered version)? a. Mitosis should proceed as usual because the dephosphorylation of the lamin is what is important for nuclear lamina assembly during mitosis, so phosphorylation will not be necessary. b. Disassembly of the nuclear lamins will occur prematurely because the lamins cannot be phosphorylated. c. Nuclear lamins will no longer disassemble properly during mitosis. d. Nuclear lamins will be unable to produce dimers, as the coiled-coil formation will be disrupted.

ANS: C The lack of the phosphorylation site will prevent disassembly, as lamins should be more stable if they cannot be phosphorylated. Although it is true that dephosphorylation of the lamin is necessary for the reassembly of the nuclear lamins

at the end of mitosis, the cycle of phosphorylation and dephosphorylation is important for mitosis. Disassembly of the lamins occurs when they are phosphorylated, as this weakens the bonds between the lamin tetramers. Dimer formation depends on the α-helical rods; the binding of the lamin tetramers is what is affected by phosphorylation.

DIF: Easy REF: 17.1 OBJ: 17.1.h Review how the nuclear lamina disassembles and re-forms during each cell division.

MSC: Applying 9. You are interested in understanding the regulation of nuclear lamina assembly. To create an in vitro system for studying this process you start with partly purified nuclear lamina subunits to which you will add back purified cellular components to drive nuclear lamina assembly. Before you start doing experiments, your instructor suggests that you consider what type of conditions would be most amenable to the assembly of the nuclear lamina from its individual subunits in vitro. Which of the following additions do you predict would be most likely to enhance the assembly of the nuclear lamina? a. addition of phosphatase inhibitors b. addition of ATP c. addition of a concentrated salt solution that is 10 times the concentration normally found in the nucleoplasm d. addition of protein kinase inhibitors ANS: D The phosphorylation of nuclear lamins by protein kinases induces conformational changes that weaken the binding between nuclear lamin tetramers; thus, inhibiting protein kinases may enhance assembly of the nuclear lamina. Adding phosphatase inhibitors or ATP will enhance the activity of any copurifying protein kinases and enhance disassembly.

Because noncovalent protein–protein interactions hold the nuclear lamina together, the addition of a very concentrated salt solution will inhibit proper nuclear lamina assembly.

DIF: Moderate REF: 17.1 OBJ: 17.1.h Review how the nuclear lamina disassembles and re-forms during each cell division. MSC: Applying 10. Which of the following statements about the structure of microtubules is FALSE? a. Microtubules are built from protofilaments that come together to make a hollow structure. b. The two ends of a protofilament are chemically distinct, with α-tubulin exposed at one end and β-tubulin exposed at the other end. c. Within a microtubule, all protofilaments are arranged in the same orientation, giving the microtubule structural polarity. d. α-Tubulin and β-tubulin are covalently bound to make the tubulin dimer that then assembles into protofilaments.

ANS: D α-Tubulin and β-tubulin bind with each other through noncovalent interactions.

DIF: Easy REF: 17.2 OBJ: 17.2.c Describe the structure of microtubules and recall how microtubules are assembled from tubulin dimers. | 17.2.d State the polarity of a microtubule and summarize how this polarity affects its assembly and function. MSC: Understanding 11. Which of the following statements about the function of the centrosome is FALSE? a. Microtubules emanating from the centrosome have alternating polarity such that some have their plus end attached to the centrosome while others have their minus end attached to the centrosome. b. Centrosomes contain hundreds of copies of the γ-tubulin ring complex important for microtubule nucleation. c. Centrosomes typically contain a pair of centrioles, which is made up of a cylindrical array of short microtubules. d. Centrosomes are the major microtubule-organizing center in animal cells.

ANS: A Microtubules emanating from the centrosome are all arranged with their minus ends at the centrosomes and the plus ends extending into the cytoplasm.

DIF: Easy REF: 17.2 OBJ: 17.2.e Describe the structure of a centrosome and review how the centrosome nucleates the growth of microtubules. | 17.2.b List several examples of organizing centers from which microtubules grow.

MSC: Understanding 12. Which of the following statements about microtubules is TRUE? a. Motor proteins move in a directional fashion along microtubules by using the inherent structural polarity of a protofilament. b. The centromere nucleates the microtubules of the mitotic spindle. c. Because microtubules are subject to dynamic instability, they are used only for transient structures in a cell. d. ATP hydrolysis by a tubulin heterodimer is important for controlling the growth of a microtubule.

ANS: A Microtubules are nucleated by the centrosome (not the centromere). Although microtubules are subject to dynamic instability, their interaction with microtubule-binding proteins can stabilize them so that they can be used to form stable structures such as cilia and flagella. GTP (not ATP) hydrolysis is important for controlling the growth of a microtubule.

DIF: Easy REF: 17.2 OBJ: 17.2.d State the polarity of a microtubule and summarize how this polarity affects its assembly and function. | 17.2.g Describe dynamic instability and indicate how this behavior relates to microtubule function. | 17.2.h Summarize how dynamic instability is controlled by GTP hydrolysis. | 17.2.m Compare kinesins and cytoplasmic dyneins in terms of their structure, their movement along the microtubule, and how they interact with cargo.

MSC: Understanding 13. The hydrolysis of GTP to GDP carried out by tubulin molecules a. provides the energy needed for tubulin to polymerize. b. occurs because the pool of free GDP has run out. c. tips the balance in favor of microtubule assembly. d. allows the behavior of microtubules called dynamic instability.

ANS: D The hydrolysis of GTP to GDP occurs after a GTP-bound tubulin molecule is incorporated into a microtubule, and it makes the microtubule more susceptible to disassembly. It is the resulting switch in microtubule stability that gives rise to the phenomenon known as dynamic instability.

DIF: Easy REF: 17.2 OBJ: 17.2.h Summarize how dynamic instability is controlled by GTP hydrolysis.

MSC: Understanding 14. The microtubules in a cell form a structural framework that can have all the following functions except which one? a. holding internal organelles such as the Golgi apparatus in particular positions in the cell b. creating long, thin cytoplasmic extensions that protrude from one side of the cell c. strengthening the plasma membrane d. moving materials from one place to another inside a cell ANS: C One function of actin filaments, but not microtubules, is to provide a meshwork beneath the plasma membrane that

helps to form and strengthen this membrane. Microtubules have all of the other functions that are listed.

DIF: Easy REF: 17.2 OBJ: 17.2.a Review the general location and functions of microtubules. MSC: Understanding 15. You discover a protein, MtA, and find that it binds to the plus ends of microtubules in cells. The hypothesis that best explains this localization is that MtA a. is involved in stabilizing microtubules. b. binds to GTP-bound tubulin on microtubules. c. is important for the interaction of microtubules with the centrosome. d. will not bind to purified microtubules in a test tube.

ANS: B GTP-bound tubulin molecules are found at the growing end of a microtubule, which is its plus end. A function of MtA in stabilizing microtubules cannot be inferred simply from its localization, as protein factors that stabilize and d estabilize microtubules can bind to plus ends. The localization of MtA at the plus end of the microtubule means that MtA is at the end of the microtubule furthest from the centrosome. Whether MtA will bind to purified microtubules in a test tube cannot be inferred from the discovery of its localization at the plus end of microtubules in cells.

DIF: Easy REF: 17.2 OBJ: 17.2.d State the polarity of a microtubule and summarize how this polarity affects its assembly and function. | 17.2.h Summarize how dynamic instability is controlled by GTP hydrolysis.

MSC: Understanding 16. Which of the following statements regarding dynamic instability is FALSE? a. Each microtubule filament grows and shrinks independently of its neighbors. b. The GTP cap helps protect a growing microtubule from depolymerization. c. GTP hydrolysis by the tubulin dimer promotes microtubule shrinking. d. The newly freed tubulin dimers from a shrinking microtubule can be immediately captured by growing microtubules and added to their plus end.

ANS: D A newly dissociated tubulin dimer will be bound to GDP; this GDP will need to be exchanged for GTP before it can be added to a newly growing microtubule.

DIF: Easy REF: 17.2 OBJ: 17.2.g Describe dynamic instability and indicate how this behavior relates to microt ubule function. | 17.2.h Summarize how dynamic instability is controlled by GTP hydrolysis. MSC: Understanding 17. Which of the situations below will enhance microtubule shrinkage? a. addition of a drug that inhibits GTP exchange on free tubulin dimers b. addition of a drug that inhibits hydrolysis of the GTP carried by tubulin dimers c. addition of a drug that increases the affinity of tubulin molecules carrying GDP for other tubulin molecules d. addition of a drug that blocks the ability of a tubulin dimer to bind to γ-tubulin ANS: A A drug that inhibits GTP exchange on free tubulin dimers will effectively decrease the available pool of GTP -bound tubulin dimers available for addition to microtubule ends, thus tipping the balance in favor of microtubule disassembly. A drug that inhibits hydrolysis of GTP carried by tubulin dimers or that increases the affinity of GDP-bound tubulin dimers for each other will stabilize growing microtubules. Blocking the ability of a tubulin dimer to bind to γ-tubulin will decrease the rate of new microtubule formation but should not enhance microtubule shrinkage.

DIF: Moderate REF: 17.2 OBJ: 17.2.g Describe dynamic instability and indicate how this behavior relates to micr otubule function. | 17.2.h Summarize how dynamic instability is controlled by GTP hydrolysis. MSC: Applying 18. The graph in Figure 17-18 shows the time course of the polymerization of pure tubulin in vitro. Assume that the starting concentration of free tubulin is higher than it is in cells.

Figure 17-18

Three parts of the curve are labeled above it as A, B, and C. You conduct a similar in vitro tubulin-polymerization experiment, only this time you include purified centrosomes in your preparation. When you plot your data, which part of your graph should be most dissimilar to the curve shown in Figure 17-18? a. A b. B c. C d. None. The shape of my graph should be identical to the graph produced when tubulin is polymerized in the absence of purified centrosomes.

ANS: A Purified centrosomes should enhance the nucleation of microtubules, and thus decrease the lag time (seen in part A of the graph) for microtubule polymerization that occurs when microtubules are polymerized from only pure tubulin.

DIF: Moderate REF: 17.2 OBJ: 17.2.e Describe the structure of a centrosome and review how the centrosome nucleates the growth of microtubules. | 17.2.g Describe dynamic instability and indicate how this behavior relates to m icrotubule function. MSC: Applying 19. Which of the following statements about organellar movement in the cell is FALSE? a. Organelles undergo saltatory movement in the cell. b. Only the microtubule cytoskeleton is involved in organellar movement. c. Motor proteins involved in organellar movement use ATP hydrolysis for energy. d. Organelles are attached to the tail domain of motor proteins.

ANS: B Both the actin cytoskeleton and the microtubule cytoskeleton are involved in organellar movement.

DIF: Easy REF: 17.2 OBJ: 17.2.j Recall how and why cells modify the dynamic instability of their microtubules. Com-

pare the movement of cell components—including organelles, membrane vesicles, and macromolecules—by free diffusion and by microtubule-guided transport. | 17.2.n Compare the roles that kinesins and cytoplasmic dyneins have in positioning the organelles in a eukaryotic cell, and describe the effect that colchicine treatment has on organelle placement. | 17.3.f Outline the functions of common actin-binding proteins, including thymosin, profilin, formins, actinrelated proteins (ARPs), and myosin. MSC: Understanding 20. Microtubules are important for transporting cargo in nerve cell axons, as diagrammed in Figure 17 -20. Notice that the two types of cargo are traveling in opposite directions. Which of the following statements is likely to be FALSE?

Figure 17-20

a. The gray cargo is attached to dynein. b. The black cargo and the gray cargo require ATP hydrolysis for their motion. c. The black cargo moving toward the axon terminal contains a domain that specifically interacts with the tail domain of a particular kind of motor. d. The black cargo and the gray cargo are moving along microtubules of opposite polarity.

ANS: D Microtubules in nerve cell axons are generally organized such that their plus ends are facing the axon terminal while the minus ends reside in the cell body. Thus, the gray cargo is likely to be attached to a dynein motor because it is moving toward the cell body. Because cargo attaches to the tail domains of both dynein and kinesin motors, the attachment of a cargo to either tail is unlikely to affect directionality. Both dynein and kinesin require ATP hydrolysis for their movement.

DIF: Easy REF: 17.2 OBJ: 17.2.m Compare kinesins and cytoplasmic dyneins in terms of their structure, their movement along the microtubule, and how they interact with cargo. MSC: Understanding 21. Kinesins and dyneins a. have tails that bind to the filaments. b. move along both microtubules and actin filaments. c. often move in opposite directions to each other. d. derive their energy from GTP hydrolysis.

ANS: C All other answers are false. The motor heads bind to the filaments (choice A). Both motors move along microtubules (choice B) and use ATP hydrolysis for energy (choice D).

DIF: Easy REF: 17.2 OBJ: 17.2.m Compare kinesins and cytoplasmic dyneins in terms of their structure, their movement along the microtubule, and how they interact with cargo. MSC: Remembering

22. Which of the following statements about the movement of materials in a nerve axon is TRUE? a. Movement along microtubules in the axon is slower than free diffusion, but necessary due to the importance of directional transport. b. The small jerky steps seen when vesicles move along microtubules are due to the shrinkage of microtubules that occurs when axonal microtubules undergo dynamic instability. c. Microtubules within an axon are arranged such that all microtubules point in the same direction with their minus ends toward the nerve cell body. d. Microtubules within the axon support the unidirectional motion of materials from the nerve cell body to the axon terminal, while materials transported back from the axon terminal to the cell body move along intermediate filaments.

ANS: C Movement along microtubules is faster than free diffusion. The microtubules within the axon are typically stabilized and not undergoing dynamic instability. The saltatory movement of the cargo is due to the properties of the motor proteins that transport the cargo along microtubule tracks. Movement in axons is bidirectional.

DIF: Easy REF: 17.2 OBJ: 17.2.k Outline how microtubules participate in cell polarization. MSC: Remembering 23. Which of the following items is not important for flagellar movement? a. sarcoplasmic reticulum b. ATP c. dynein d. microtubules ANS: A The sarcoplasmic reticulum is important for muscle contraction. All other items are important for flagellar movement.

DIF: Easy REF: 17.2 OBJ: 17.2.p Describe the arrangement of microtubules inside a cilium or flagella. | 17.4.f Summarize how calcium ions trigger muscle contraction and how contraction is subsequently reversed.

MSC: Remembering 24. Figure 17-24A shows how the movement of dynein causes the flagellum to bend. If instead of the normal situation, the polarity of the adjacent doublet of microtubules were to be reversed (see Figure 17-24B), what do you predict would happen?

Figure 17-24

a. No bending would occur. b. Bending would occur exactly as diagrammed in Figure 17-24A. c. Bending would occur, except that the right microtubule doublet would move down relative to the left one. d. The two microtubule doublets would slide away from each other.

ANS: C Because the polarity of the microtubule bundle is reversed, the dynein motors should walk in the opposite direction from the normal situation diagrammed in Figure 17-40A. Microtubule sliding would occur if the linking proteins were absent, which is not true here.

DIF: Moderate REF: 17.2 OBJ: 17.2.p Compare the functions and movements of cilia and flagella. | 17.2.q Describe the arrangement of microtubules inside a cilium or flagellum. | 17.2.r Outline how ciliary dynein allows a cilium to bend. MSC: Applying 25. Which of the following statements about actin is FALSE? a. ATP hydrolysis decreases actin filament stability. b. Actin at the cell cortex helps govern the shape of the plasma membrane. c. Actin filaments are nucleated at the side of existing actin filaments in lamellipodia. d. The dynamic instability of actin filaments is important for cell movement.

ANS: D Dynamic instability is a phenomenon associated with microtubules and not actin. Actin disassembly and assembly are both important for cell movement. However, this differs from dynamic instability in that the growth of actin filaments occurs at the leading edge; this growth occurs in a directed fashion because of actin -binding proteins that promote the formation of new filaments at the leading edge. Actin-binding proteins that destabilize actin filaments promote actin disassembly away from the leading edge. The actin assembly and disassembly in moving cells di ffers from the stochastic growth and disassembly of the microtubules.

DIF: Easy REF: 17.3 OBJ: 17.3.c Compare the polymerization of actin filaments to that of microtubules. | 17.3.g Describe the structure and function of the cell cortex. MSC: Understanding 26. Consider the mechanism by which actin and tubulin polymerize. Which of the items below does not describe something similar about the polymerization mechanisms of actin and microtubules? a. Although both filaments can grow from both ends, the growth rate is faster at the plus ends. b. Depolymerization initiates at the plus ends of filaments. c. Nucleotide hydrolysis promotes depolymerization of filaments. d. Free subunits (actin and tubulin) carry nucleoside triphosphates.

ANS: B The shrinkage of microtubules that occurs involves a switch from growth to shrinkage only at the plus end of microtubules.

However, actin loses subunits from its minus end during actin treadmilling.

DIF: Easy REF: 17.3 OBJ: 17.3.c Compare the polymerization of actin filaments to that of microtubules.

MSC: Understanding 27. For both actin and microtubule polymerization, nucleotide hydrolysis is important for a. stabilizing the filaments once they are formed. b. increasing the rate at which subunits are added to the filaments.

c. promoting nucleation of filaments. d. decreasing the binding strength between subunits on filaments.

ANS: D ATP hydrolysis in actin polymerization decreases the binding strength between monomers in the actin filaments; GTP hydrolysis during tubulin polymerization decreases the binding strength between the tubulin subunits in the microtubule.

DIF: Easy REF: 17.3 OBJ: 17.3.c Compare the polymerization of actin filaments to that of microtubules. | 17.3.d Explain treadmilling and identify the conditions under which this behavior takes place. MSC: Remembering 28. Compared to the normal situation, in which actin monomers carry ATP, what do you predict would happen if actin monomers that bind a nonhydrolyzable form of ATP were incorporated into actin filaments? a. Actin filaments would grow longer. b. Actin filaments would grow shorter because depolymerization would be enhanced. c. Actin filaments would grow shorter because new monomers could not be added to the filaments. d. No change, as the addition of monomers binding nonhydrolyzable ATP would not affect actin filament length.

ANS: A Addition of monomers carrying a nonhydrolyzable form of ATP would stabilize the interactions between the monomers of a filament, stabilizing the filament and inhibiting depolymerization, resulting in longer actin filaments.

DIF: Easy REF: 17.3 OBJ: 17.3.c Compare the polymerization of actin filaments to that of microtubules. | 17.3.d Explain treadmilling and identify the conditions under which this behavior takes place. MSC: Applying 29. Which of the following statements is FALSE? a. Formins promote the formation of unbranched actin filaments. b. Actin filaments are usually excluded from the cell cortex. c. Integrins are transmembrane proteins that can bind to the extracellular matrix. d. ARPs can promote the formation of branched actin filaments.

ANS: B Much of the actin in the cell is concentrated in the cell cortex, the region of the cell just beneath the plasma membrane.

DIF: Easy REF: 17.3 OBJ: 17.3.j Explain how ARPs and formins aid in the assembly and extension of protrusions at the cell’s leading edge. MSC: Understanding 30. Cell movement involves the coordination of many events in the cell. Which of the following phenomena is not required for cell motility? a. myosin-mediated contraction at the rear of the moving cell b. integrin association with the extracellular environment c. nucleation of new actin filaments d. release of Ca2+ from the sarcoplasmic reticulum ANS: D The release of Ca2+ from the sarcoplasmic reticulum is important for muscle contraction, not cell motility.

DIF: Easy REF: 17.3 OBJ: 17.3.i Distinguish the roles played by actin in the protrusion, attachment, and contraction involved in cell movement. | 17.4.f Summarize how calcium ions trigger muscle contraction and how contraction is

subsequently reversed. MSC: Understanding 31. Figure 17-31 shows the leading edge of a lamellipodium. Which of the following statements is FALSE?

Figure 17-31

a. Nucleation of new filaments near the leading edge pushes the plasma membrane forward. b. ARP proteins nucleate the branched actin filaments in the lamellipodium. c. Capping proteins bind to the minus end of actin filaments. d. There is more ATP-bound actin at the leading edge than in the actin filaments away from the leading edge.

ANS: C Capping protein binds to the plus end of actin filaments, preventing further assembly or disassembly from the growing end.

DIF: Easy REF: 17.3 OBJ: 17.3.i Distinguish the roles played by actin in the protrusion, attachment, and contraction involved in cell movement. | 17.3.f Outline the functions of common actin-binding proteins, including thymosin, profilin, formins, actin-related proteins (ARPs), and myosin. | 17.3.j Explain how ARPs and formins aid in the assembly and extension of protrusions at the cell’s leading edge. MSC: Understanding 32. You are examining a cell line in which activation of the Rho family member Rac promotes lamellipodia formation.

Which of the following statements is most likely to be TRUE? a. Cells carrying a Rac mutation that makes Rac act as if it is always bound to GTP will polymerize more unbranched actin filaments than normal cells. b. Cells carrying a Rac mutation that makes Rac unable to exchange GDP for GTP will polymerize more unbranched actin filaments than normal cells. c. Cells carrying a Rac mutation that makes Rac act as if it is always bound to GTP will polymerize more branched actin filaments than normal cells. d. Cells carrying a Rac mutation that makes Rac unable to exchange GDP for GTP will polymerize more branched actin filaments than normal cells.

ANS: C Activation of Rac promotes lamellipodia formation by enhancing actin nucleation using the ARP complex, which promotes the formation of branched actin filaments. Because lamellipodia formation involves branched actin filaments, a

mutation that creates a constitutively active form of Rac (a GTP-bound form of Rac) will promote the formation of a greater number of branched actin filaments. Rac that is mutated and unable to exchange GDP for GTP will not be active.

DIF: Moderate REF: 17.3 OBJ: 17.3.k Summarize how different members of the Rho family of GTPases alter the organization of actin filaments. MSC: Understanding 33. Your friend works in a biotech company that has just discovered a drug that seems to promote lamellipodia formation in cells. Which of the following molecules is unlikely to be directly involved in the pathway that this drug affects? a. Rac b. ARP c. actin d. myosin ANS: D Myosins are not directly involved in lamellipodia formation. Lamellipodium formation involves branched actin structures that use ARP for their formation, and thus ARP and actin are likely to be involved. The Rho family member Rac triggers lamellipodia formation when activated, and thus may be involved.

DIF: Easy REF: 17.3 OBJ: 17.3.f Outline the functions of common actin-binding proteins, including thymosin, profilin, formins, actin-related proteins (ARPs), and myosin. | 17.3.j Explain how ARPs and formins aid in the assembly and extension of protrusions at the cell’s leading edge. | 17.3.k Summarize how different members of the Rho family of GTPases alter the organization of actin filaments. MSC: Understanding 34. Which of the following structures shorten during muscle contraction? a. myosin filaments b. flagella c. sarcomeres d. actin filaments ANS: C Sarcomeres contain actin filaments and myosin filaments that slide past each other during muscle contraction, leading to shortening of the sarcomere; the actin filaments and myosin filaments do not change in length. Flagella are microtubule-based structures that are not present on muscle cells.

DIF: Easy REF: 17.4 OBJ: 17.4.c Describe the structure of a sarcomere. MSC: Remembering 35. Which of the following conditions is likely to decrease the likelihood of skeletal muscle contraction? a. partial depolarization of the T-tubule membrane, such that the resting potential is closer to zero b. addition of a drug that blocks Ca2+ binding to troponin c. an increase in the amount of ATP in the cell d. a mutation in tropomyosin that decreases its affinity for the actin filament ANS: B Ca2+ binding to troponin leads to a conformational change that causes a movement in tropomyosin so that myosin can bind to actin to initiate contraction. Thus, if troponin cannot bind Ca2+, the likelihood of contraction decreases. Partial depolarization of the T-tubule membrane will make it easier to depolarize the membrane, increasing the likelihood of muscle contraction. ATP is required for myosin movement, so increasing the amount of ATP in the cell will not de-

crease contraction. Tropomyosin normally binds to actin and blocks myosin binding, so a mutation in tropomyosin that decreases its affinity for actin should not decrease the likelihood of muscle contraction.

DIF: Easy REF: 17.4 OBJ: 17.4.f Summarize how calcium ions trigger muscle contraction, and how contraction is subsequently reversed. MSC: Applying 36. Which of the following statements about skeletal muscle contraction is FALSE? a. When a muscle cell receives a signal from the nervous system, voltage-gated channels open in the T-tubule membrane. b. The changes in voltage across the plasma membrane that occur when a muscle cell receives a signal from the nervous system cause an influx of Ca2+ into the sarcoplasmic reticulum, triggering a muscle contraction. c. A change in the conformation of troponin leads to changes in tropomyosin such that it no longer blocks the binding of myosin heads to the actin filament. d. During muscle contraction, the Z discs move closer together as the myosin heads walk toward the plus ends of the actin filaments.

ANS: B Muscle contraction is triggered by an efflux of Ca2+ from the sarcoplasmic reticulum into the cytosol.

DIF: Easy REF: 17.4 OBJ: 17.4.d Outline the sliding-filament mechanism of muscle contraction and relaxation, and describe how ATP binding and hydrolysis drive the conformational changes that underlie this movement. | 17.4.e Explain how excitation of the muscle cell membrane triggers a rise in the cytosolic concentration of calcium ions. | 17.4.f Summarize how calcium ions trigger muscle contraction and how contraction is subsequently reversed.

MSC: Understanding 37. Consider the in vitro motility assay using purified kinesin and purified polymerized microtubules shown in Figure 1737. The three panels are images taken at 1-second intervals. In this figure, three microtubules have been numbered to make it easy to identify them. Which of the following statements about this assay is FALSE?

Figure 17-37

a. Kinesin molecules are attached by their tails to a glass slide. b. The microtubules used in this assay must be polymerized using conditions that stabilize tubule formation or else they would undergo dynamic instability. c. ATP must be added for this assay to work. d. Addition of the nonhydrolyzable ATP analog (AMP-PNP) would cause the microtubules to move faster.

ANS: D

Addition of AMP-PNP would block movement, because ATP hydrolysis is required for the kinesin to step along a microtubule. The addition of AMP-PNP would cause the microtubules to attach to the kinesin heads without being released. Kinesin molecules are attached to the slide by their tails (the cargo-binding domain) so that the heads are available to move the microtubules along the slides. If they were in solution with the microtubules, there would be no force and thus no movement. The microtubules used in this assay are stabilized with a nonhydrolyzable form of GTP, because otherwise they might shrink during the course of the assay. ATP is required for kinesin movement, and thus must be added for this assay to work.

DIF: Easy REF: 17.2 OBJ: 17.2.m Compare kinesins and cytoplasmic dyneins in terms of their structure, their movement along the microtubule, and how they interact with cargo. MSC: Understanding 38. Figure 17-38 shows an electron micrograph of a skeletal muscle fiber, where various points along a fiber and various regions have been labeled.

Figure 17-38

Which of the following statements is TRUE about muscle contraction? a. Point A will move closer to point B. b. Point B will move closer to point C. c. Region D will become smaller. d. Region E will shrink in size.

ANS: B The dark region in the center of the micrograph corresponds to the thick filaments of a myofibril and is composed of many myosin molecules. The light regions (labeled A, B, and C) correspond to actin filaments, which are attached to the Z discs (also see Figure 17-38A). During muscle contraction, the myosin filaments will travel along the actin filaments, bringing points B and C closer together. Points A and B will not move relative to each other, as contraction occurs within each sarcomere. The absolute size of region D will not change during contraction, and the width of the myofibril will not shrink.

Figure 17-38A

DIF: Easy REF: 17.4 OBJ: 17.4.b Distinguish the structures of a skeletal muscle fiber and a myofibril.

MSC: Understanding

MATCHING 1. Match one of the three major classes of cytoskeletal elements (A–C) to each statement below.

A. actin B. intermediate filaments C. microtubules 1. monomer that binds ATP ANS: A DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate fil aments, actin filaments, and microtubules. MSC: Remembering 2. includes keratin and neurofilaments ANS: B DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 3. important for formation of the contractile ring during cytokinesis ANS: A DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate fil aments, actin filaments, and microtubules. MSC: Remembering 4. supports and strengthens the nuclear envelope ANS: B DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 5. Their stability involves a GTP cap.

ANS: C DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 6. used in the eukaryotic flagellum ANS: C DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 7. a component of the mitotic spindle ANS: C DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 8. can be connected through desmosomes ANS: B DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 9. directly involved in muscle contraction ANS: A DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 10. abundant in filopodia

ANS: A DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 2. Match the type of intermediate filament with its appropriate location.

A. nerve cells B. epithelia C. nucleus D. connective tissue 1. lamins ANS: C DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 2. neurofilaments ANS: A DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 3. vimentins ANS: D DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 4. keratins ANS: B DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 3. Match each statement with the type of microtubule (A–D) it refers to.

A. ciliary microtubule B. microtubule of the mitotic spindle C. both types of microtubule D. neither type of microtubule 1. The basal body is the organizing center.

ANS: A DIF: Easy REF: 17.2 OBJ: 17.2.o Compare the functions and movement of cilia and flagella.

MSC: Remembering 2. The monomer is sequestered by profilin.

ANS: D DIF: Easy REF: 17.2 OBJ: 17.2.o Compare the functions and movement of cilia and flagella.

MSC: Remembering 3. It is arranged in a “9 + 2” array.

ANS: A DIF: Easy REF: 17.2 OBJ: 17.2.o Compare the functions and movement of cilia and flagella.

MSC: Remembering 4. It is nucleated at the centrosome.

ANS: B DIF: Easy REF: 17.2 OBJ: 17.2.o Compare the functions and movement of cilia and flagella.

MSC: Remembering 5. It uses dynein motors.

ANS: C DIF: Easy REF: 17.2 OBJ: 17.2.o Compare the functions and movement of cilia and flagella.

MSC: Remembering 6. It is involved in sperm motility.

ANS: D DIF: Easy REF: 17.2 OBJ: 17.2.o Compare the functions and movement of cilia and flagella.

MSC: Remembering 7. It is involved in moving fluid over the surface of cells.

ANS: A DIF: Easy REF: 17.2 OBJ: 17.2.o Compare the functions and movement of cilia and flagella.

MSC: Remembering 4. Actin can adopt a variety of shapes. Match each actin form with the correct structure in Figure 17-4.

Figure 17-4

1. lamellipodia ANS: C DIF: Easy REF: 17.3 OBJ: 17.3.a List several cell structures formed by actin filaments.

MSC: Remembering 2. contractile bundles ANS: B DIF: Easy REF: 17.3 OBJ: 17.3.a List several cell structures formed by actin filaments.

MSC: Remembering 3. contractile ring ANS: D DIF: Easy REF: 17.3 OBJ: 17.3.a List several cell structures formed by actin filaments.

MSC: Remembering 4. microvilli ANS: A DIF: Easy REF: 17.3 OBJ: 17.3.a List several cell structures formed by actin filaments.

MSC: Remembering 5. The following proteins are important for cell movement. Match the following proteins with their function.

A. nucleation of new actin filaments at the side of an existing filament B. regulation of the availability of actin monomers C. important for the growth of straight, unbranched actin filaments D. contracting the rear of the cell E. involvement in focal contacts 1. myosin

ANS: D DIF: Easy REF: 17.3 OBJ: 17.3.f Outline the functions of common actin-binding proteins, including thymosin, profilin, formins, actin-related proteins (ARPs), and myosin. MSC: Remembering 2. ARP proteins ANS: A DIF: Easy REF: 17.3 OBJ: 17.3.f Outline the functions of common actin-binding proteins, including thymosin, profilin, formins, actin-related proteins (ARPs), and myosin. MSC: Remembering 3. profilin ANS: B DIF: Easy REF: 17.3 OBJ: 17.3.f Outline the functions of common actin-binding proteins, including thymosin, profilin, formins, actin-related proteins (ARPs), and myosin. MSC: Remembering 4. integrins ANS: E DIF: Easy REF: 17.3 OBJ: 17.3.f Outline the functions of common actin-binding proteins, including thymosin, profilin, formins, actin-related proteins (ARPs), and myosin. MSC: Remembering 5. formins ANS: C DIF: Easy REF: 17.3 OBJ: 17.3.f Outline the functions of common actin-binding proteins, including thymosin, profilin, formins, actin-related proteins (ARPs), and myosin. MSC: Remembering

SHORT ANSWER 1. Identify the cytoskeletal structures (black lines) depicted in the epithelial cells shown in Figure 17-1.

Figure 17-1

ANS: (A) microtubules (B) intermediate filaments (C) actin DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 2. Rank the following cytoskeletal filaments from smallest to largest in diameter (1 = smallest in diameter, 4 = largest) _____ intermediate filaments _____ microtubules _____ actin filament _____ myofibril

ANS: __2___ intermediate filaments (10 nm diameter) __3___ microtubules (25 nm) __1___ actin filament (5–9 nm) __4___ myofibril (1–2 μm) DIF: Easy REF: 17.1 OBJ: 17.1.a Contrast the structures of the subunits that form intermediate filaments, actin filaments, and microtubules. MSC: Remembering 3. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. antiparallel

four

tail

β barrel

globular

ten

coiled-coil

head

trimeric

covalent

rod

twenty-five

eight

seven

two

Intermediate filaments are elongated fibrous proteins with an N-terminal globular __________ region and a C-terminal globular __________ region; these regions flank the elongated rod domain. The α-helical region of the rod interacts with the α-helical region of another monomer in a __________ configuration to form a dimer. __________ dimers will line up to form a staggered tetramer. __________ strands of tetramers come together and twist together to form the __________ nm filament. The __________ domains are exposed on the surface of the intermediate filament, allowing for interaction with cytoplasmic components.

ANS: Intermediate filaments are elongated fibrous proteins with an N-terminal globular head region and a C-terminal globular tail region; these regions flank the elongated rod domain. The α-helical region of the rod interacts with the αhelical region of another monomer in a coiled-coil configuration to form a dimer. Two dimers will line up to form a staggered tetramer. Eight strands of tetramers come together and twist together to form the ten nm filament. The globular domains are exposed on the surface of the intermediate filament, allowing for interaction with cytoplasmic components.

DIF: Easy REF: 17.1 OBJ: 17.1.c Recall how the structure of intermediate filaments relates to their strength and durability. MSC: Understanding 4. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. desmosome

lamin

synapse

keratin

neurofilament

vimentin

kinase

plectin

Intermediate filaments are found mainly in cells that are subject to mechanical stress. Gene mutations that disrupt intermediate filaments cause some rare human genetic diseases. For example, the skin of people with epidermolysis bullosa simplex is very susceptible to mechanical injury; people with this disorder have mutations in their __________ genes, which code for the intermediate filament found in epithelial cells. These filaments are usually connected from cell to cell through junctions called __________s. The main filaments found in muscle cells belong to the __________ family; people with disruptions in these intermediate filaments can have muscular dystrophy. In the nervous system,

__________s help strengthen the extremely long extensions often present in nerve cell axons; disruptions in these intermediate filaments can lead to neurodegeneration. People who carry mutations in the gene for __________, an important protein for cross-linking intermediate filaments, have a disease that combines symptoms of epidermolysis bullosa simplex, muscular dystrophy, and neurodegeneration. Humans with progeria, a disease that causes premature aging, carry mutations in a nuclear __________.

ANS: Intermediate filaments are found mainly in cells that are subject to mechanical stress. Gene mutations that disrupt intermediate filaments cause some rare human genetic diseases. For example, the skin of people with epidermolysis bullosa simplex is very susceptible to mechanical injury; people with this disorder have mutations in their keratin genes, the intermediate filament found in epithelial cells. These filaments are usually connected from cell to cell through junctions called desmosomes. The main filaments found in muscle cells belong to the vimentin family; people with disruptions in these intermediate filaments can have muscular dystrophy. In the nervous system, neurofilaments help strengthen the extremely long extensions often present in nerve cell axons; disruptions in these intermediate filaments can lead to neurodegeneration. People who carry mutations in the gene for plectin, an important protein for cross-linking intermediate filaments, have a disease that combines symptoms of epidermolysis bullosa simplex, muscular dystrophy, and neurodegeneration. Humans with progeria, a disease that causes premature aging, carry mutations in a nuclear lamin.

DIF: Easy REF: 17.1 OBJ: 17.1.c Recall how the structure of intermediate filaments relates to their strength and durability. MSC: Understanding 5. Phosphorylation of nuclear lamins regulates their assembly and disassembly during mitosis. You add a drug to cells undergoing mitosis that inhibits the activity of an enzyme that dephosphorylates nuclear lamins. What do you predict will happen to these cells? Why? ANS: Cells should become arrested in mitosis. Normally, the lamins are phosphorylated during mitosis, causing disassembly of the nuclear envelope. At the end of mitosis, the nuclear lamins are dephosphorylated, causing the lamins to reassemble. Inhibition of this last step should therefore prevent the nuclear lamins from reassembling a fter mitosis.

DIF: Moderate REF: 17.1 OBJ: 17.1.h Review how the nuclear lamina disassembles and re-forms during each cell division. MSC: Applying 6. Place the following in order of size, from the smallest to the largest.

A. protofilament B. microtubule C. α-tubulin D. tubulin dimer E. mitotic spindle ANS: C, D, A, B, E DIF: Easy REF: 17.2 OBJ: 17.2.c Describe the structure of microtubules and recall how microtubules are assembled from tubulin dimers. MSC: Remembering 7. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. α-tubulin

dynein

nine

ATP

four

thirteen

basal body

γ-tubulin

twenty-one

β-tubulin

GTP

UTP

centrosome

kinesin two

vimentin

δ-tubulin

myosin

Microtubules are formed from the tubulin heterodimer, which is composed of the nucleotide-binding __________ protein and the __________ protein. Tubulin dimers are stacked together into protofilaments; __________ parallel protofilaments form the tube-like structure of a microtubule. __________ rings are important for microtubule nucleation and are found in the __________ , which is usually found near the cell’s nucleus in cells that are not undergoing mitosis. A microtubule that is quickly growing will have a __________ cap that helps prevent the loss of subunits from its growing end. Stable microtubules are used in cilia and flagella; these microtubules are nucleated from a ___________ and involve a “__________ plus two” array of microtubules. The motor protein __________ generates the bending motion in cilia; the lack of this protein can cause Kartagener’s syndrome in humans.

ANS: Microtubules are formed from the tubulin heterodimer, which is composed of the nucleotide-binding β-tubulin protein and the α-tubulin protein. Tubulin dimers are stacked together into protofilaments; thirteen parallel protofilaments form the tube-like structure of a microtubule. γ-Tubulin rings are important for microtubule nucleation and are found in the centrosome, which is usually found near the cell’s nucleus in cells that are not undergoing mitosis. A microtubule that is quickly growing will have a GTP cap that helps prevent the loss of subunits from its growing end.

Stable microtubules are used in cilia and flagella; these microtubules are nucleated from a basal body and involve a “nine plus two” array of microtubules. The motor protein dynein generates the bending motion in cilia; the lack of this protein can cause Kartagener’s syndrome in humans.

DIF: Easy REF: 17.2 OBJ: 17.2.c Describe the structure of microtubules and recall how microtubules are assembled from tubulin dimers. | 17.2.e Describe the structure of a centrosome and review how the centrosome nucleates the growth of microtubules. | 17.2.p Describe the arrangement of microtubules inside a cilium or flagella. | 17.2.q Outline how ciliary dynein allows a cilium to bend. MSC: Understanding 8. Your friend discovers a protein that she names EBP. EBP binds to microtubule plus ends, and she hypothesizes a role for EBP in increasing dynamic instability. To determine the function of EBP, she examines its effect on microtubules. She polymerizes microtubules from purified centrosomes in a Petri plate and determines the number of shrinking microtubules over a three-minute time interval for different concentrations of EBP. The data she obtained are shown in Figure 17-8.

Figure 17-8

Is this result consistent with her hypothesis? Explain.

ANS: No, the data graphed in Figure 17-8 are not consistent with her hypothesis. If EBP were to increase the dynamic instability of microtubules, you would expect an increase in both the number of shrinking microtubules and the number of growing microtubules. Since the graph shows a reduction in the number of shrinking microtubules, dynamic instability has evidently decreased. Further experiments would be needed to determine the net effect of these changes on microtubules and whether they become longer or shorter on average following EBP treatment.

DIF: Moderate REF: 17.2 OBJ: 17.2.g Describe dynamic instability and indicate how this behavior relates to microtubule function. MSC: Applying 9. You are curious about the dynamic instability of microtubules and decide to join a lab that works on microtubule polymerization. The people in the lab help you grow some microtubules in culture using conditions that allow you to watch individual microtubules under a microscope. You can see the microtubules growing and shrinking, as you expect. The professor who runs the lab gets in a new piece of equipment, a very fine laser beam that can be used to sever microtubules. She is very excited and wants to sever growing microtubules at their middle, using the laser beam.

A. Do you predict that the newly exposed microtubule plus ends will grow or shrink? Explain your answer.

B. What do you expect would happen to the newly exposed plus ends if you were to grow the microtubules in the presence of an analog of GTP that cannot be hydrolyzed, and you then severed the microtubules in the middle with a laser beam?

ANS: A. The newly exposed microtubule plus ends will most probably shrink if you sever the microtubules in the middle.

This is because a microtubule grows by adding GTP-carrying subunits to the plus end. The GTP is hydrolyzed over time, leaving only a cap of GTP-carrying subunits at the plus end, with the remainder of the tubulin protofilament containing GDP-carrying subunits. Therefore, if you sever a growing microtubule in its middle, you will most probably create a plus end that contains GDP-carrying subunits. The GDP-carrying subunits are less tightly bound than the GTP-carrying subunits and will peel away from each other, causing depolymerization of the microtubule and shrinkage.

B. If you were to polymerize the microtubules in the presence of a nonhydrolyzable analog of GTP and you then severed the microtubules with a laser, the newly exposed plus end would contain a GTP cap and so would probably continue to grow.

DIF: Hard REF: 17.2 OBJ: 17.2.g Describe dynamic instability and indicate how this behavior relates to microtubule function. | 17.2.h Summarize how dynamic instability is controlled by GTP hydrolysis. MSC: Applying 10. The graph in Figure 17-10 shows the time course of the polymerization of pure tubulin in vitro. You can assume that the starting concentration of free tubulin is much higher than it is in cells.

Figure 17-10

A. Explain the reason for the initial lag in the rate of microtubule formation.

B. Why does the curve level out after point C? ANS: A. Before they can polymerize to form microtubules, tubulin molecules have to form small aggregates that act as nucleation centers. This aggregation step is slow because the molecules have to come together in the right configuration. This is why there is a lag phase before microtubules start to be formed.

B. After point C, an equilibrium point has been reached, where the rates of polymerization and depolymerization are exactly balanced.

DIF: Moderate REF: 17.2 OBJ: 17.2.f Explain why microtubules require organizing ce nters such as centrosomes to nucleate their growth. | 17.2.g Describe dynamic instability and indicate how this behavior relates to microtubule function. MSC: Applying 11. Do you agree or disagree with this statement? Explain your answer.

Minus-end directed microtubule motors (like dyneins) deliver their cargo to the periphery of the cell, whereas plus-end directed microtubule motors (like kinesins) deliver their cargo to the interior of the cell.

ANS: Disagree. The plus ends of microtubules usually point toward the cell periphery, whereas the minus ends point toward the cell center. This is because the γ-tubulin in the centrosome serves to nucleate microtubule growth. Because the centrosomes are near the center of the cell, the minus ends of microtubules are located there. Therefore, a minus-end directed microtubule motor would direct its cargo toward the center of the cell, and a plus-end directed microtubule motor would direct its cargo toward the cell periphery.

DIF: Moderate REF: 17.2 OBJ: 17.2.k Outline how microtubules participate in cell polarization. | 17.2.l Compare kinesins and cytoplasmic dyneins in terms of their structure, movement along the microtubule, and how they interact with cargo. MSC: Understanding 12. Match the following labels to the numbered lines on Figure 17-12.

Figure 17-12

A. minus end of microtubule B. tail of motor protein C. cargo of motor protein D. head of motor protein Which of the two motors in Figure 17-12 is most probably a kinesin? Explain your answer.

ANS: A—4; B—2; C—1; D—3 The top motor is more likely to be kinesin, because kinesins usually move toward the plus end of the microtubules.

DIF: Easy REF: 17.2 OBJ: 17.2.k Outline how microtubules participate in cell polarization. | 17.2.l Compare kinesins and cytoplasmic dyneins in terms of their structure, movement along the microtubule, and how they interact with cargo.

MSC: Understanding 13. Some lower vertebrates such as fish and amphibians can control their color by regulating specialized pigment cells called melanophores. These cells contain small, pigmented organelles, termed melanosomes, that can be dispersed throughout the cell, making the cell darker, or aggregated in the center of the cell to make the cell lighter. You purify the melanosomes from melanophores that have either aggregated or dispersed melanosomes and find that: 1. aggregated melanosomes co-purify with dynein; 2. dispersed melanosomes co-purify with kinesin.

Given this set of data, propose a mechanism for how the aggregation and dispersal of melanosomes occur.

ANS: The melanosomes are transported in the cell on microtubules. When it is advantageous for the animal to become lighter, a signal is sent to the pigment cell that causes the melanosomes to associate with dynein. Because dynein is a minus-end directed motor, it will transport the melanosomes toward the center of the cell, causing the melanosomes to aggregate in the center and the cell to take on a lighter appearance. When the animal wants to become darker, a signal is sent to the pigment cell that causes the melanosomes to associate with kinesin. Kinesin is usually a plus-end directed motor and will move the melanosomes away from the center of the cell so that they are more dispersed, making the cell look darker.

DIF: Hard REF: 17.2 OBJ: 17.2.k Outline how microtubules participate in cell polarization. | 17.2.l Compare kinesins and cytoplasmic dyneins in terms of their structure, movement along the microtubule, and how they interact with cargo.

MSC: Applying 14. Cytochalasin is a drug that caps actin filament plus ends, thus preventing actin polymerization. Phalloidin is a drug that binds to and stabilizes actin filaments, preventing actin depolymerization. Even though these drugs have opposite

effects on actin polymerization, the addition of either of these drugs instantaneously freezes the cell movements that depends on actin filaments. Explain why drugs that have opposite effects on actin filaments can have a similar effect on cell movements.

ANS: These drugs both stop cell movements because actin polymerization and depolymerization are both required for this process. As cells move forward, the growth of actin filaments near the plasma membrane helps push out the membrane. As this occurs, continuous depolymerization of actin filaments occurs at the actin filaments away from the plasma membrane.

DIF: Hard REF: 17.2 OBJ: 17.3.e Compare the actions of cytochalasin and phalloidin and describe their effects on cell behavior. MSC: Applying 15. In the budding yeast, activation of the GTP-binding protein Cdc42 occurs on binding of an external signal (pheromone) to a G-protein-coupled receptor. Activation of Cdc42 promotes actin polymerization. Predict what would happen to actin polymerization, in comparison with pheromone-treated cells, in the following cases.

A. You add pheromone to an inhibitor of G-protein-coupled receptors.

B. You add pheromone to a nonhydrolyzable analog of GTP. ANS: A. Less actin polymerization. Cdc42 will not be able to be activated by the G-protein-coupled receptor.

B. More actin polymerization. Cdc42 will be more active, because it will bind the nonhydrolyzable form of GTP and will not be able to be turned off.

DIF: Moderate REF: 17.3 OBJ: 17.3.k Summarize how different members of the Rho family of GTPases alter the organization of actin filaments. MSC: Applying 16. Do you agree or disagree with the following statement? Explain your answer.

When skeletal muscle receives a signal from the nervous system to contract, the signal from the motor neuron triggers the opening of a voltage-sensitive Ca2+ channel in the muscle cells’ plasma membrane, allowing Ca 2+ to flow into the cell.

ANS: Disagree. The increase in intracellular Ca2+ during muscle contraction comes from an intracellular source. The Ca2+ is released from the lumen of the sarcoplasmic reticulum, which is a specialized region of endoplasmic reticulum inside a muscle cell. The signal from the nerve terminal triggers an action potential in the muscle cell plasma membrane, which causes a voltage-sensitive transmembrane protein in the membranous transverse tubules to open a Ca2+release channel in the membrane of the sarcoplasmic reticulum.

DIF: Moderate REF: 17.4 OBJ: 17.4.e Explain how excitation of the muscle cell membrane triggers a rise in the cytosolic concentration of calcium ions. | 17.4.f Summarize how calcium ions trigger muscle contraction and how contraction is subsequently reversed. MSC: Evaluating 17. Kinesins were purified by adding the nonhydrolyzable analog AMP-PNP to cytoplasmic extracts containing microtubules, purifying the microtubules, and then releasing the kinesin proteins, which were still attached to the microtubules, by adding ATP. Would this trick have worked to purify myosin motors attached to actin filaments?

Explain.

ANS: No, addition of a nonhydrolyzable form of ATP would not increase the affinity of myosin for actin. This is because when the myosin motor binds to ATP, the myosin head undergoes a conformational change that reduces its affinity for actin.

DIF: Moderate REF: 17.3 OBJ: 17.2.m Compare kinesins and cytoplasmic dyneins in terms of their structure, their movement along the microtubule, and how they interact with cargo. | 17.3.l Compare the structure, binding properties, and general function of myosins I and II. MSC: Applying 18. You are interested in studying kinesin movements. You therefore prepare silica beads and coat them with kinesin molecules so that each bead, on average, has only one kinesin molecule attached to it. You add these kinesin-coated beads to a preparation of microtubules you have polymerized. Using video microscopy, you watch the kinesin [labeled with green fluorescent protein (GFP)] move down the microtubules.

A. Kinesin-GFP has been measured to move along microtubules at a rate of 0.3 μm/sec, and single-molecule studies have revealed that kinesin moves along microtubules progressively, with each step being 8 nm. How many steps can the kinesin molecule take in 4 seconds, assuming that the kinesin stays attached to the microtubule for the entire 4 seconds?

B. Because each kinesin molecule is thought to take approximately 100 steps before falling off the microtubule, will you see your silica beads detach from the microtubule during your 4 seconds of observation?

C. What would you predict would happen to the kinesin-coated silica beads if you were to add AMP-PNP (a nonhydrolyzable ATP analog)?

ANS: A. You would expect the kinesin molecule to travel 150 steps. The calculation is as follows: 0.3 μm = 300 nm. Therefore, in 4 seconds, the kinesin molecule could travel 1200 nm if it were to move at a rate of 0.3 μm/sec. Because the step size is 8 nm, 1200 nm/(8 nm per step) = 150 steps.

B. Yes, you should see silica beads detach some time during the 4 seconds of observation, because each kinesin will take more than 100 steps in that 4-second time frame (see part A above).

C. If you were to add AMP-PNP, you would no longer see the silica beads moving down the microtubule. It is thought that one molecule of ATP is hydrolyzed per step that kinesin takes; without ATP hydrolysis, translocation of the beads will be inhibited. However, you may still see the beads associated with the microtubules, because AMPPNP does not inhibit the association of kinesin with the microtubule.

DIF: Hard REF: 17.2 OBJ: 17.2.l Compare kinesins and cytoplasmic dyneins in terms of their structure, movement along the microtubule, and how they interact with cargo. MSC: Applying 19. In the three cell outlines in Figure 17-62, indicate the arrangement of the microtubules, showing clearly their free and attached ends. On each figure, indicate the plus end for one of the microtubules.

Figure 17-19

ANS: See Figure 17-19A.

Figure 17-19A

DIF: Easy REF: 17.2 OBJ: 17.2.d State the polarity of a microtubule and summarize how this polarity affects its assembly and function. MSC: Understanding 20. Indicate whether each of the following statements refers to a ciliary microtubule, a microtubule of the mitotic spindle, both types of microtubule, or neither type of microtubule.

A. The basal body is the organizing center.

B. The monomer is sequestered by profilin.

C. It is arranged in a “9 + 2” array.

D. It is nucleated at the centrosome.

E. It uses dynein motors.

F. It is involved in sperm motility.

G. It is involved in moving fluid over the surface of cells.

ANS: A. ciliary microtubules B. neither C. ciliary microtubules D. microtubules of the mitotic spindle E. both F. neither (this involves flagellar microtubules) G. ciliary microtubules DIF: Easy REF: 17.2 OBJ: 17.2.o Compare the functions and movement of cilia and flagella. MSC: Remembering 21. Figure 17-21 shows two isolated outer-doublet microtubules from a eukaryotic flagellum with their associated dynein molecules.

Figure 17-21

A. Sketch what will happen to this structure if it is supplied with ATP. B. Sketch what will happen to this structure if the linking proteins are removed and it is supplied with ATP. C. In a complete flagellum, what would happen if all the dynein molecules were active at the same time?

ANS: A. See Figure A17-21A. B. See Figure A17-21B. (Note to instructor: this question should be marked as correct only if the microtubule is shown bending in the correct direction [in A] and the correct microtubule is shown pushed forward [in B]).

Figure 17-21A

C. The flagellum will not bend because there is no significant relative motion of one microtubule doublet to another: each is trying to push its neighbor forward at the same time. For the flagellum to bend, sets of dynein molecules on one side of the flagellum must be selectively activated.

DIF: Hard REF: 17.2 OBJ: 17.2.p Describe the arrangement of microtubules inside a cilium or flagella.

MSC: Creating 22. You isolate some muscle fibers to examine what regulates muscle contraction. When you bathe the muscle fibers in a solution containing ATP and Ca 2+, you see muscle contraction (experiment 3 in Table 17-22). Ca2+ is necessary, as solutions containing ATP alone or nothing do not stimulate contraction and thus the muscle remains in a relaxed state (experiments 1 and 2 in Table 17-22). From what you know about the mechanism of muscle contraction, fill in your predictions of whether the muscle will be contracted or relaxed for experiments 4, 5, and 6.

Explain your answers.

Table 17-22

Extra credit: In what state would the muscle be if you added Ca 2+ but no ATP?

ANS: See Table 17-22A.

Table 17-22A

In experiment 4, the muscle will be relaxed because troponin will not be able to bind Ca 2+. By preventing troponin from binding to Ca2+, troponin will not be able to undergo the conformational change that causes tropomyosin to alter its association with actin. This altered association is normally required for myosin to bind actin. In the absence of troponin regulation by Ca2+, myosin cannot bind actin and the muscle cannot contract.

In experiment 5, the muscle will contract because tropomyosin cannot bind to actin. If tropomyosin cannot bind actin, myosin can. The presence of ATP means that ATP will be available for myosin to hydrolyze, causing muscle contraction.

In experiment 6, the muscle will remain relaxed. The presence of Ca2+ will induce a conformational change in troponin that causes tropomyosin to shift, exposing actin for myosin to bind. However, when myosin binds, the myosin molecule will attach and then release because the myosin will bind the nonhydrolyzable analog of ATP. Because no ATP hydrolysis can occur, the muscle will remain in the relaxed state.

Extra credit: The muscle will be in a rigor state. If Ca2+ is added without ATP, troponin can bind Ca2+ and undergo a conformational change, which causes tropomyosin to shift and exposes the actin for the myosin to bind. However, myosin will bind actin and remain in the attached state, because a myosin head lacking a bound nucleotide is locked onto the actin until nucleotide binding occurs.

DIF: Hard REF: 17.4 OBJ: 17.4.d Outline the sliding-filament mechanism of muscle contraction and relaxation, and describe how ATP binding and hydrolysis drive the conformational changes that underlie this movement. | 17.4.f Summarize how calcium ions trigger muscle contraction, and how contraction is subsequently reversed.

MSC: Applying

CHAPTER 18 The Cell-Division Cycle

OVERVIEW OF THE CELL CYCLE 18.1.a List the four phases of the eukaryotic cell cycle and summarize what takes place in each.

18.1.b Identify the phases that are shortened during the cleavage divisions of early embryos, and explain the effects of these divisions on cell size.

18.1.c Summarize the function of the cell-cycle control system and describe the transition points at which progression through the cycle is regulated.

18.1.d Explain why studies of yeast can lead to insights into the biology of human cancers.

THE CELL-CYCLE CONTROL SYSTEM 18.2.a Compare how cyclins and Cdks vary in concentration and activity during the cell cycle.

18.2.b Recall why different cyclin-Cdk complexes trigger different events in the cell cycle.

18.2.c Explain how extracts prepared from cells in different phases of the cell cycle have been used to identify components of the cell-cycle control system.

18.2.d Explain how studies of mutant yeast were used to dissect the cell-cycle control system.

18.2.e Describe how the synthesis and destruction of cyclins regulate progression from one phase of the cell cycle to the next.

18.2.f

Recall how dephosphorylation helps trigger the abrupt activation of cyclin-Cdk complexes.

18.2.g Indicate how Cdk inhibitors can help regulate progression through the cell cycle.

18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle.

G1 PHASE 18.3.a Explain why the G1-to-S transition in yeast cells is sometimes called “Start.”

18.3.b Review how cyclin-Cdk complexes are inhibited as cells enter G1.

18.3.c Describe how cells behave when deprived of mitogens.

18.3.d Explain how Rb blocks cell proliferation and how mitogens reverse this inhibition.

18.3.e Review how DNA damage can arrest cells in G1.

18.3.f

State what occurs when the DNA damage detected in G1 is too extensive to be repaired.

18.3.g Contrast the arrested state G0 with the cell-cycle withdrawal that occurs during terminal differentiation.

18.3.h Indicate which phases of the cell cycle vary the most in length to influence the rates of cell division in the adult body.

S PHASE 18.4.a Contrast the origin recognition complex (ORC) and the prereplicative complex in terms of composition and when each assembles on the DNA. 18.4.b Detail how S-Cdk initiates replication and prevents re-replication during the same cell cycle.

18.4.c Summarize how incomplete or incorrect replication can arrest the cell cycle in G 2.

M PHASE 18.5.a Explain how M-Cdk is activated at the end of G2 and indicate why this activation is sudden and explosive.

18.5.b Compare cohesins and condensins in terms of structure, function, and how and when they assemble onto chromosomal DNA. 18.5.c Compare the mitotic spindle and contractile ring in terms of composition, location, and the role each plays in cell division.

18.5.d List the stages of M phase.

MITOSIS 18.6.a Outline the centrosome cycle, indicating how and when it is initiated.

18.6.b Describe the structure of the mitotic spindle and explain how and when it begins to form.

18.6.c Summarize how and when the nuclear envelope breaks down.

18.6.d Review how chromosomes are captured by spindle microtubules, and describe the structure of the point of attachment.

18.6.e Define bi-orientation and explain its importance for chromosome segregation.

18.6.f

Recall the type of cells that lack centrosomes.

18.6.g Describe the role of chromosomes in assembly of the mitotic spindle.

18.6.h Explain why chromosomes align at the spindle equator during metaphase, and present experimental evidence for this mechanism.

18.6.i

Summarize the molecular events that trigger the separation of sister chromatids at the start of anaphase.

18.6.j

Compare the changes in the mitotic spindle that underlie chromosome segregation during anaphase A and anaphase B, and delineate the driving forces responsible for each process.

18.6.k Explain how the spindle assembly checkpoint ensures that all chromosomes are attached to the spindle and why this checkpoint can delay the onset of anaphase and the exit from mitosis.

18.6.l

Describe how the nuclear envelope reassembles during telophase.

CYTOKINESIS 18.7.a Recall when cytokinesis takes place with respect to mitosis.

18.7.b Define the cleavage furrow and explain how its position is determined.

18.7.c Describe how asymmetric divisions are set up during embryonic development.

18.7.d Review the contractile ring in terms of its composition, assembly, and mechanism of action.

18.7.e Summarize how cells change in shape and attachment throughout the cell cycle.

18.7.f

Outline the process of cytokinesis in plant cells.

18.7.g Explain how membrane-enclosed organelles are distributed to daughter cells during cell division.

CONTROL OF CELL NUMBERS AND CELL SIZE 18.8.a List the three fundamental cell processes that determine the size of an an animal’s organs or body.

18.8.b Explain how apoptosis participates in the development of a mouse or a human—or in the metamorphosis of a frog.

18.8.c Recall how apoptosis balances cell division in an adult tissue such as liver.

18.8.d Distinguish the appearance of cell necrosis from that of apoptosis.

18.8.e Contrast the causes of necrosis and apoptosis and describe the consequences of each on nearby cells and tissues.

18.8.f

Summarize how apoptosis is mediated by a proteolytic caspase cascade.

18.8.g Outline how apoptosis is regulated and initiated by the Bcl2 family of proteins.

18.8.h Review how death receptors can stimulate apoptosis.

18.8.i

List the classes of signal proteins that influence cell fate and summarize their functions.

18.8.j

Explain how survival factors suppress apoptosis, and describe how they regulate the number of neurons in the developing nervous system.

18.8.k Review how mitogens stimulate cell division, and explain how they aid in wound healing.

18.8.l

Contrast cell growth and cell division in terms of their dependence on cell-cycle control.

18.8.m Outline how growth factors function to increase cell size.

18.8.n Summarize the benefits of a single extracellular signal protein acting as both a growth factor and a mitogen.

18.8.o Describe an extracellular signal protein that inhibits tissue growth.

18.8.p Recall how dysregulation of cell growth, survival, and division can lead to cancer.

MULTIPLE CHOICE 1. What would be the most obvious outcome of repeated cell cycles consisting of S phase and M phase only?

a. The cells would not be able to replicate their DNA. b. The mitotic spindle could not assemble. c. The cells would get larger and larger. d. The cells produced would get smaller and smaller.

ANS: D The cells produced would get smaller and smaller, as they would not have sufficient time to double their mass before dividing.

DIF: Easy REF: 18.1 OBJ: 18.1.b Identify the phases that are shortened during the cleavage divisions of early embryos, and explain the effects of these divisions on cell size. MSC: Understanding 2. A mutant yeast strain stops proliferating when shifted from 25°C to 37°C. When these cells are analyzed at the two different temperatures, using a machine that sorts cells according to the amount of DNA they contain, the graphs in Figure 18-2 are obtained.

Figure 18-2

Which of the following would NOT explain the results with the mutant? a. inability to initiate DNA replication b. inability to begin M phase c. inability to activate proteins needed to enter S phase d. inappropriate production of a signal that causes the cells to remain in G 1 ANS: B At 37°C, the cells all have one genome-worth of DNA, meaning that they have not replicated their DNA and therefore

have not entered S phase. Cells that are unable to begin M phase should have two genomes-worth of DNA, as they would have completed DNA replication and arrested in G2.

DIF: Hard REF: 18.1 OBJ: 18.1.c Summarize the function of the cell-cycle control system and describe the transition points at which progression through the cycle is regulated. MSC: Applying 3. Which of the following events does NOT usually occur during interphase? a. Cells grow in size. b. The nuclear envelope breaks down. c. DNA is replicated. d. The centrosomes are duplicated.

ANS: B DIF: Easy REF: 18.1 OBJ: 18.1.a List the four phases of the eukaryotic cell cycle and summarize what takes place in each. MSC: Remembering 4. In which phase of the cell cycle do cells check to determine whether the DNA is fully and correctly replicated? a. at the transition between G1 and S b. when cells enter G0 c. during M d. at the end of G2 ANS: D Cells will check whether the DNA is fully and correctly replicated at the end of G. It does not make sense to monitor DNA replication before S phase because DNA replication has not yet occurred. When cells enter G 0, they do not replicate their DNA. During M phase, chromosomes are condensed for chromosome segregation, so it would be difficult for the cell to examine the replicated DNA for errors at that point.

DIF: Easy REF: 18.1 OBJ: 18.1.c Summarize the function of the cell-cycle control system and describe the transition points at which progression through the cycle is regulated. MSC: Remembering 5. Which of the following statements about the cell cycle is FALSE? a. Once a cell decides to enter the cell cycle, the time from start to finish is the same in all eukaryotic cells. b. An unfavorable environment can cause cells to arrest in G 1. c. A cell has more DNA during G2 than it did in G1. d. The cleavage divisions that occur in an early embryo have short G 1 and G2 phases.

ANS: A DIF: Easy REF: 18.1 OBJ: 18.1.a List the four phases of the eukaryotic cell cycle and summarize what takes place in each. | 18.1.b Identify the phases that are shortened during the cleavage divisions of early embr yos, and explain the effects of these divisions on cell size. | 18.1.c Summarize the function of the cell -cycle control system and describe the transition points at which progression through the cycle is regulated.

MSC: Understanding 6. Which of the following descriptions is consistent with the behavior of a cell that lacks a protein required for a checkpoint mechanism that operates in G2? a. The cell would be unable to enter M phase. b. The cell would be unable to enter G2. c. The cell would enter M phase under conditions when normal cells would not. d. The cell would pass through M phase more slowly than normal cells.

ANS: C Normal cells arrest at the G2 checkpoint if DNA replication is incomplete or DNA is damaged. Cells without this mechanism may enter M phase with unreplicated or damaged DNA, whereas normal cells would not.

DIF: Easy REF: 18.1 OBJ: 18.1.c Summarize the function of the cell-cycle control system and describe the transition points at which progression through the cycle is regulated. MSC: Applying 7. Progression through the cell cycle requires a cyclin to bind to a Cdk because a. the cyclins are the molecules with the enzymatic activity in the complex. b. the binding of a cyclin to Cdk is required for Cdk enzymatic activity. c. cyclin binding inhibits Cdk activity until the appropriate time in the cell cycle. d. without cyclin binding, a cell-cycle checkpoint will be activated.

ANS: B Cdks require cyclins for enzymatic activity. Cyclins have no enzymatic activities themselves, and cyclin binding to Cdk activates the Cdk. As far as we know, cyclin–Cdk binding is not directly monitored by checkpoints.

DIF: Easy REF: 18.2 OBJ: 18.2.a Compare how cyclins and Cdks vary in concentration and activity during the cell cycle. MSC: Understanding 8. Levels of Cdk activity change during the cell cycle, in part because a. the Cdks phosphorylate each other. b. the Cdks activate the cyclins. c. Cdk degradation precedes entry into the next phase of the cell cycle. d. cyclin activity change during the cycle.

ANS: D Cyclin activity fluctuates throughout the cell cycle and are required for Cdk activity. M-cyclins are degraded in a cellcycle-dependent fashion (not Cdks), and they are required for Cdk activity. Cdks do not phosphorylate each other.

The Cdks do not activate the cyclins.

DIF: Easy REF: 18.2 OBJ: 18.2.a Compare how cyclins and Cdks vary in concentration and activity during the cell cycle. MSC: Understanding 9. The concentration of mitotic cyclin (M cyclin) a. rises markedly during M phase. b. is activated by phosphorylation. c. falls toward the end of M phase as a result of ubiquitylation and degradation. d. is highest in G1 phase.

ANS: C The concentration of mitotic cyclin rises gradually during G 2 and it is ubiquitylated and degraded during late M phase.

DIF: Easy REF: 18.2 OBJ: 18.2.e Describe how the synthesis and destruction of cyclins regulate progression from one phase of the cell cycle to the next. MSC: Remembering 10. You have isolated a strain of mutant yeast cells that divides normally at 30°C but cannot enter M phase at 37°C. You have isolated its mitotic cyclin and mitotic Cdk and find that both proteins are produced and can form a normal M-Cdk complex at both temperatures. Which of the following temperature-sensitive mutations could NOT be responsible for

the behavior of this strain of yeast? a. inactivation of a protein kinase that acts on the mitotic Cdk kinase b. inactivation of an enzyme that ubiquitylates M cyclin c. inactivation of a phosphatase that acts on the mitotic Cdk kinase d. a decrease in the levels of a transcriptional regulator required for producing sufficient amounts of M cyclin ANS: B A cell with a mutation that prevents ubiquitylation of M cyclin could enter mitosis but could not exit mitosis properly.

DIF: Hard REF: 18.2 OBJ: 18.2.d Explain how studies of mutant yeast were used to dissect the cell-cycle control system. | 18.2.e Describe how the synthesis and destruction of cyclins regulate progression from one phase of the cell cycle to the next. MSC: Applying 11. You engineer yeast cells that express the M cyclin during S phase by replacing the gene regulatory sequences of the M cyclin gene with those of the S cyclin gene. Keeping in mind that yeast cells have one common Cdk that binds to all cyclins, which of the following outcomes is LEAST likely during this experiment? a. There will be both M cyclin–Cdk and S cyclin–Cdk complexes in the cell during S phase. b. Some substrates that are normally phosphorylated in M phase will now be phosphorylated in S phase. c. G1 cyclin–Cdks will be activated earlier in G1. d. S-Cdk targets will be phosphorylated during S phase.

ANS: C The activity of G 1 cyclin–Cdks should not be affected by S-Cdk (which normally occurs in S phase) or M-Cdk (which likely now exists within the cell due to the inappropriate expression of M cyclin from the S cyclin promoter).

DIF: Hard REF: 18.2 OBJ: 18.2.b Recall why different cyclin-Cdk complexes trigger different events in the cell cycle. | 18.2.d Explain how studies of mutant yeast were used to dissect the cell-cycle control system. | 18.2.e Describe how the synthesis and destruction of cyclins regulate progression from one phase of the cell cycle to the next.

MSC: Analyzing 12. Which of the following statements is FALSE? a. Cdc25 dephosphorylation of Wee1 activates the kinase, promoting the G2/M transition. b. Phosphorylation of mitotic Cdk by the inhibitory kinase (Wee1) makes the Cdk inactive. c. Inhibiting the Cdc25 phosphatase will delay the G2/M transition. d. The activating phosphatase (Cdc25) removes the phosphates from mitotic Cdk that were added by Wee1, so that M-Cdk will be active.

ANS: A The Cdc25 phosphatase acts to dephosphorylate M-Cdk and does not dephosphorylate Wee1. Furthermore, Wee1 activation would inhibit M-Cdk activity, which would inhibit (and not promote) the G2/M transition.

DIF: Easy REF: 18.2 OBJ: 18.2.f Recall how dephosphorylation helps trigger the abrupt activation of cyclin-Cdk complexes. MSC: Understanding 13. MPF activity was discovered when cytoplasm from a Xenopus M-phase cell was injected into Xenopus oocytes, inducing the oocytes to form a mitotic spindle. In a control experiment, Xenopus interphase cytoplasm was injected into oocytes and shown not to induce the formation of a mitotic spindle. Which of the following statements is NOT a legitimate conclusion from the control experiment?

a. The piercing of the oocyte membrane by a needle is insufficient to cause mitotic spindle formation. b. An increased volume of cytoplasm is insufficient to cause mitotic spindle formation. c. Injection of extra RNA molecules is insufficient to cause mitotic spindle formation. d. Components of an interphase nucleus are insufficient to cause mitotic spindle formation.

ANS: D The interphase cytoplasmic extract used as the control would not have contained nuclear components (because the nuclear membrane was intact) and so one cannot conclude that components of an interphase nucleus are insufficient to cause mitotic spindle formation. The other statements are all true.

DIF: Moderate REF: 18.2 OBJ: 18.2.c Explain how extracts prepared from cells in different phases of the cell cycle have been used to identify components of the cell-cycle control system. MSC: Understanding 14. Which of the following is NOT good direct evidence that the cell-cycle control system is conserved through billions of years of divergent evolution? a. A yeast cell lacking a Cdk function can use the human Cdk to substitute for its missing Cdk during the cell cycle. b. The amino acid sequences of cyclins in plants are similar to the amino acid sequences of cyclins in humans. c. The Cdk proteins in humans share conserved phosphorylation sites with the Cdk proteins in yeast. d. Yeast cells have only one Cdk, whereas humans have many Cdks.

ANS: D Although it is true that yeast cells have one Cdk and human cells have many, this statement does not provide any evidence for the conservation of function across evolutionary time. All the other statements do provide such evidence.

DIF: Easy REF: 18. OBJ: 18.1.d Explain why studies of yeast can lead to insights into the biology of human cancers. | 18.2.c Explain how extracts prepared from cells in different phases of the cell cycle have been used to identify components of the cell-cycle control system. | 18.2.d Explain how studies of mutant yeast were used to dissect the cell-cycle control system. MSC: Understanding 15. Mitogens are a. extracellular signals that stimulate cell division. b. transcription factors important for cyclin production. c. kinases that cause cells to grow in size. d. produced by mitotic cells to keep nearby neighboring cells from dividing.

ANS: A DIF: Easy REF: 18.3 OBJ: 18.3.c Describe how cells behave when deprived of mitogens.

MSC: Remembering 16. The Retinoblastoma (Rb) protein blocks cells from entering the cell cycle by a. phosphorylating Cdk. b. marking cyclins for destruction by proteolysis. c. inhibiting cyclin transcription. d. activating apoptosis.

ANS: C DIF: Easy REF: 18.3 OBJ: 18.3.d Explain how Rb blocks cell proliferation and how mitogens reverse this inhibition. MSC: Remembering 17. The G1 DNA damage checkpoint a. causes cells to proceed through S phase more quickly.

b. involves the degradation of p53. c. is activated by errors caused during DNA replication. d. involves the inhibition of cyclin–Cdk complexes by p21.

ANS: D DIF: Easy REF: 18.3 OBJ: 18.3.e Review how DNA damage can arrest cells in G1. MSC: Understanding 18. Cells in the G0 state a. do not divide. b. cannot reenter the cell cycle. c. have entered this arrest state from either G1 or G2. d. have duplicated their DNA. ANS: A DIF: Easy REF: 18.3 OBJ: 18.3.g Contrast the arrested state G0 with the cell-cycle withdrawal that occurs during terminal differentiation. MSC: Remembering 19. A cell that is terminally differentiated will a. replicate its DNA. b. reenter the cell cycle only once a year. c. dismantle the cell-cycle control system. d. arrest after S phase.

ANS: C A terminally differentiated cell cannot reenter the cell cycle. DIF: Easy REF: 18.3 OBJ: 18.3.g Contrast the arrested state G0 with the cell-cycle withdrawal that occurs during terminal differentiation. MSC: Remembering 20. Which of the following statements is FALSE? a. DNA synthesis begins at origins of replication. b. The loading of the origin recognition complexes (ORCs) is triggered by S-Cdk. c. The phosphorylation and degradation of Cdc6 help to ensure that DNA is replicated only once in each cell cycle. d. DNA synthesis can only begin after prereplicative complexes assemble on the ORCs. ANS: B DIF: Easy REF: 18.4 OBJ: 18.4.a Contrast the origin recognition complex (ORC) and the prereplicative complex in terms of composition and when each assembles on the DNA. | 18.4.b Detail how S-Cdk initiates replication and prevents re-replication during the same cell cycle. MSC: Understanding 21. How does S-Cdk help guarantee that replication occurs only once during each cell cycle? a. It blocks the rise of Cdc6 concentrations early in G1. b. It phosphorylates and inactivates DNA helicase. c. It phosphorylates and inactivates Cdc6. d. It promotes the assembly of a prereplicative complex.

ANS: C The phosphorylation and inactivation of Cdc6 ensures that replication occurs once and only once during each cell cycle. The concentration of Cdc6 rises early in G1, independent of S-Cdk. Cdc6 guides the assembly of the prereplicative complex at an origin. By phosphorylating Cdc6, S-Cdk inactivates it, so that once an origin fires and replicates, Cdc6 cannot reinitiate replication. S-Cdk phosphorylates and activates DNA helicases.

DIF: Easy REF: 18.4 OBJ: 18.4.b Detail how S-Cdk initiates replication and prevents re-replication during the same

cell cycle. MSC: Understanding 22. You create cells with a version of the origin recognition complex, ORC, that cannot be phosphorylated by S-Cdk and thus cannot be inactivated. Which of the following statements describes the likely consequence of this change in ORC? a. Cells will enter S phase prematurely. b. Cells will replicate some regions of the genome more than once in a cell cycle. c. ORC will be unable to bind to DNA. d. DNA helicases will not be able to open up the double helix at the replication origin.

ANS: B If ORC is not properly inactivated (and thus remains active), ORC could bind to DNA and reassemble a prereplicative complex on a DNA molecule that has already undergone replication. The assembly of an inappropriate prereplicative complex would lead to an inappropriate reinitiation of replication.

DIF: Moderate REF: 18.4 OBJ: 18.4.a Contrast the origin recognition complex (ORC) and the prereplicative complex in terms of composition and when each assembles on the DNA. | 18.4.b Detail how S-Cdk initiates replication and prevents re-replication during the same cell cycle. MSC: Applying 23. Which of the following does not occur during M phase in animal cells? a. growth of the cell b. condensation of chromosomes c. breakdown of nuclear envelope d. attachment of chromosomes to microtubules ANS: A Cell size increases throughout interphase and not during M phase. All of the other phenomena are observed in M phase.

DIF: Easy REF: 18.5 OBJ: 18.5.d List the stages of M phase. MSC: Understanding 24. Condensins a. are degraded when cells enter M phase. b. assemble into complexes on the DNA when phosphorylated by M-Cdk. c. are involved in holding sister chromatids together. d. bind to DNA before DNA replication begins.

ANS: B DIF: Easy REF: 18.5 OBJ: 18.5.b Compare cohesins and condensins in terms of structure, function, and how and when they assemble onto chromosomal DNA. MSC: Remembering 25. At the end of DNA replication, the sister chromatids are held together by the a. kinetochores. b. securins. c. cohesins. d. histones.

ANS: C DIF: Remembering REF: 18.5 OBJ: 18.5.b Compare cohesins and condensins in terms of structure, function, and how and when they assemble onto chromosomal DNA. MSC: Remembering

26. Which of the following statements is TRUE? a. The mitotic spindle is largely made of intermediate filaments. b. The contractile ring is made largely of microtubules and actin filaments. c. The contractile ring divides the nucleus in two. d. The mitotic spindle helps segregate the chromosomes to the two daughter cells.

ANS: D DIF: Easy REF: 18.5 OBJ: 18.5.c Compare the mitotic spindle and contractile ring in terms of composition, location, and the role each plays in cell division. MSC: Understanding 27. Sister chromatid separation occurs because __________ are destroyed by the APC/C. a. securins b. cohesins c. kinetochores d. condensins ANS: A The APC/C initiates sister chromatid separation by triggering destruction of securin, which results in activation of separase. Separase then cleaves the cohesins that hold the sister chromatids together.

DIF: Easy REF: 18.6 OBJ: 18.6.i Summarize the molecular events that trigger separation of sister chromatids at the start of anaphase. MSC: Remembering 28. The principal microtubule-organizing center in animal cells is the a. centrosome. b. centromere. c. kinetochore. d. cell cortex.

ANS: A DIF: Easy REF: 18.6 OBJ: 18.6.b Describe the structure of the mitotic spindle and explain how and when it begins to form. MSC: Remembering 29. Which word or phrase below best describes the phase in mitosis depicted in Figure 18-29?

Figure 18-29

a. anaphase

b. prometaphase c. S-phase checkpoint d. metaphase ANS: D DIF: Easy REF: 18.6 OBJ: 18.6.h Explain why chromosomes align at the spindle equator during metaphase, and present experimental evidence for this mechanism. MSC: Remembering 30. Which letter is associated with the line that is pointing to the interpolar microtubules in Figure 18-29?

Figure 18-29

a. A b. B c. C d. D e. E ANS: E DIF: Easy REF: 18.6 OBJ: 18.6.b Describe the structure of the mitotic spindle and explain how and when it begins to form. MSC: Remembering 31. Disassembly of the nuclear envelope a. causes the inner nuclear membrane to separate from the outer nuclear membrane. b. results in the conversion of the nuclear envelope into protein-free membrane vesicles. c. is triggered by the phosphorylation of integrins. d. must occur for kinetochore microtubules to form in animal cells.

ANS: D In animal cells, kinetochore microtubules cannot form if the chromosomes in the nucleus are separated from the microtubules in the cytoplasm because of the presence of the nuclear envelope. (But in some other cells, such as the yeast S. cerevisiae, the nuclear envelope never breaks down and yet chromosomes can attach to microtubules emanating from the spindle poles within the nucleus.) The nuclear envelope disassembles by breaking up into vesicles containing lipids from both the outer and inner envelopes. Integral membrane proteins of the nuclear envelope and some of the nuclear lamins remain associated with the vesicles. Phosphorylation of lamins (not integrins) triggers the breakdown of the nuclear lamina.

DIF: Easy REF: 18.6 OBJ: 18.6.c Summarize how and when the nuclear envelope breaks down.

MSC: Understanding 32. Which of the following statements about kinetochores is TRUE? a. Kinetochores assemble onto chromosomes during late prophase. b. Kinetochores contain DNA-binding proteins that recognize sequences at the telomere of the chromosome. c. Kinetochore proteins bind to the tubulin molecules at the minus end of microtubules. d. Kinetochores assemble on chromosomes that lack centromeres.

ANS: A DIF: Easy REF: 18.6 OBJ: 18.6.d Review how chromosomes are captured by spindle microtubules and describe the structure of the point of attachment. MSC: Remembering 33. A friend declares that chromosomes are held at the metaphase plate by microtubules that push on each chromosome from opposite sides. Which of the following observations does not support your belief that the microtubules are pulling on the chromosomes? a. the jiggling movement of chromosomes at the metaphase plate b. the way in which chromosomes behave when the attachment between sister chromatids is severed c. the way in which chromosomes behave when the attachment to one kinetochore is severed d. the shape of chromosomes as they move toward the spindle poles at anaphase ANS: A The jiggling movement is simply a sign that the chromosomes are subject to forces from both sides, which could be the microtubules pulling, pushing, or both. All the other observations suggest pulling. When the attachment between sister chromatids is severed, both daughter chromosomes move toward their respective poles, which suggests that they are being pulled. When the attachment to one kinetochore is severed, the whole chromosome moves to the opposite pole, showing that the kinetochore microtubules are pulling on their attached chromatid, not pushing it. Similarly, the shape of the chromosomes as they move toward the pole suggests that the chromosomes are being pulled.

DIF: Moderate REF: 18.6 OBJ: 18.6.j Compare the changes in the mitotic spindle that underlie chromosome segregation during anaphase A and anaphase B, and delineate the driving forces responsible for each process.

MSC: Understanding 34. Which of the following statements about the anaphase-promoting complex (APC) is FALSE? a. It promotes the degradation of proteins that regulate M phase. b. It inhibits M-Cdk activity. c. It is continuously active throughout the cell cycle. d. M-Cdk stimulates its activity.

ANS: C The APC becomes activated in mid- to late M phase.

DIF: Easy REF: 18.6 OBJ: 18.6.i Summarize the molecular events that trigger separation of sister chromatids at the start of anaphase. MSC: Understanding 35. Which of the following statements is TRUE? a. Anaphase A must be completed before anaphase B can take place. b. In cells in which anaphase B predominates, the spindle will elongate much less than in cells in which anaphase A dominates. c. In anaphase A, both kinetochore and interpolar microtubules shorten.

d. In anaphase B, microtubules associated with the cell cortex shorten.

ANS: D Anaphase A and anaphase B generally occur at the same time. In cells in which anaphase B predominates, the spindle will elongate more than in cells in which anaphase A predominates. In anaphase A, only the kinetochore microtubules shorten.

DIF: Easy REF: 18.6 OBJ: 18.6.j Compare the changes in the mitotic spindle that underlie chromosome segregation during anaphase A and anaphase B, and delineate the driving forces responsible for each process.

MSC: Understanding 36. Which of the following precede the re-formation of the nuclear envelope during M phase in animal cells? a. assembly of the contractile ring b. decondensation of chromosomes c. reassembly of the nuclear lamina d. transcription of nuclear genes ANS: A The contractile ring in an animal cell begins to assemble in anaphase. The chromosomes do not decondense, the lamina does not re-form, and transcription does not begin until the formation of the nuclear envelope is complete and nuclear proteins have been imported through the nuclear pores.

DIF: Easy REF: 18.6 OBJ: 18.6.l Describe how the nuclear envelope reassembles during telophase.

MSC: Remembering 37. A cell with nuclear lamins that cannot be phosphorylated in M phase will be unable to a. reassemble its nuclear envelope at telophase. b. disassemble its nuclear lamina at prometaphase. c. begin to assemble a mitotic spindle. d. condense its chromosomes at prophase.

ANS: B If the lamins cannot be phosphorylated during mitosis, the cells will be unable to disassemble their nuclear lamina, preventing the breakdown of the nuclear envelope at prometaphase. The mitotic spindle begins to form before the nuclear envelope breaks down, forming a sort of cage around the nucleus. Lamins are not involved in chromosome condensation.

DIF: Easy REF: 18.6 OBJ: 18.6.l Describe how the nuclear envelope reassembles during telophase.

MSC: Understanding 38. Cytokinesis in animal cells a. requires ATP. b. leaves a small circular “scar” of actin filaments on the inner surface of the plasma membrane. c. is often followed by phosphorylation of integrins in the plasma membrane. d. is assisted by motor proteins that pull on microtubules attached to the cell cortex.

ANS: A All cell movement requires ATP, and in cytokinesis actin and myosin molecules are moving relative to one another to cause contraction of the contractile ring. The myosin is an ATPase that hydrolyzes ATP to power this movement. The

assembly of the contractile ring requires a general rearrangement of the filaments in the cell cortex. The contractile ring in animal cells disassembles completely after mitosis, leaving no trace. Phosphorylation of integrins (which weakens the hold of these transmembrane proteins on the extracellular matrix, thereby allowing cells to round up) generally precedes cytokinesis and is part of the general rearrangement of cell structure that accompanies cell division. Microtubules do not have an important role in animal cytokinesis.

DIF: Easy REF: 18.6 OBJ: 18.7.d Review the contractile ring in terms of its composition, assembly, and mechanism of action. MSC: Understanding 39. Which of the following statements is FALSE? a. The cleavage furrow is a puckering of the plasma membrane caused by the constriction of a ring of filaments attached to the plasma membrane. b. The cleavage furrow will not begin to form in the absence of a mitotic spindle. c. The cleavage furrow always forms perpendicular to the interpolar microtubules. d. The cleavage furrow always forms in the middle of the cell.

ANS: D Although the furrow always forms perpendicular to the interpolar microtubules about midway between the spindle poles, if the spindle were in an asymmetrical position (which can occur normally during development), cell division would not always occur in the middle of the cell.

DIF: Easy REF: 18.7 OBJ: 18.7.b Define the cleavage furrow and explain how its position is determined.

MSC: Understanding 40. Which of the following statements is FALSE? a. Cytokinesis in plant cells is mediated by the microtubule cytoskeleton. b. Small membrane vesicles derived from the Golgi apparatus deliver new cell-wall material for the new wall of the dividing cell. c. The phragmoplast forms from the remains of interpolar microtubules of the mitotic spindle. d. Motor proteins walking along the cytoskeleton are important for the contractile ring that guides formation of the new cell wall.

ANS: D No contractile ring is formed during plant cytokinesis.

DIF: Easy REF: 18.7 OBJ: 18.7.f Outline the process of cytokinesis in plant cells. MSC: Understanding 41. Which organelle fragments during mitosis? a. endoplasmic reticulum b. Golgi apparatus c. mitochondrion d. chloroplast ANS: B DIF: Easy REF: 18.7 OBJ: 18.7.g Explain how membrane-enclosed organelles are distributed to daughter cells during cell division. MSC: Remembering 42. Programmed cell death occurs a. by means of an intracellular suicide program. b. rarely and selectively only during animal development. c. only in unhealthy or abnormal cells.

d. only during embryonic development.

ANS: A Programmed cell death results from an intracellular suicide program that eliminates unneeded, unwanted, or damaged cells. It occurs frequently and happens even in healthy cells throughout the lifetime of an individual.

DIF: Easy REF: 18.8 OBJ: 18.8.f Summarize how apoptosis is mediated by a proteolytic caspase cascade.

MSC: Remembering 43. Apoptosis differs from necrosis in that necrosis a. requires the reception of an extracellular signal. b. causes DNA to fragment. c. causes cells to swell and burst, whereas apoptotic cells shrink and condense. d. involves a caspase cascade.

ANS: C DIF: Easy REF: 18.8 OBJ: 18.8.d Distinguish the appearance of cells that die by necrosis and apoptosis.

MSC: Understanding 44. Which of the following statements about apoptosis is TRUE? a. Cells that constitutively express Bcl2 will be more prone to undergo apoptosis. b. The prodomain of procaspases contains the catalytic activity necessary for procaspase activation. c. Bax and Bak promote apoptosis by binding to procaspases in the apoptosome. d. Apoptosis can be promoted by the release of cytochrome c into the cytosol from mitochondria.

ANS: D Release of cytochrome c into the cytosol can promote apoptosis. Bcl2 tends to inhibit rather than promote apoptosis.

The prodomain of procaspases does not contain the catalytic activity needed for procaspase activation and is usually discarded from the active caspase. When activated, Bax and Bak promote apoptosis by stimulating the release of cytochrome c from mitochondria into the cytosol, not by binding to procaspases in the apoptosome.

DIF: Easy REF: 18.8 OBJ: 18.8.f Summarize how apoptosis is mediated by a proteolytic caspase cascade.

MSC: Understanding

MATCHING 1. Match the following labels to the label lines (A–H) on Figure 18-1.

Figure 18-1

1. G1 phase DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle. MSC: Remembering

2. mitotic cyclin ANS: F DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle. MSC: Remembering 3. mitotic Cdk ANS: C DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle. MSC: Remembering 4. S phase ANS: B DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle. MSC: Remembering 5. S-phase cyclin ANS: D DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle. MSC: Remembering 6. S-phase Cdk ANS: A DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle. MSC: Remembering 7. MPF ANS: G DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle. MSC: Remembering 8. G2 phase ANS: H DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle. MSC: Remembering 2. Cells can pause in G1 when DNA is damaged, and can pause in S when there are replication errors. Match each mechanism below (1–4) with the type of arrest it applies to (A–D).

A. G1 arrest, an B. S-phase arrest, C. both types of arrest D. neither type of arrest 1. p53 activates the transcription of a Cdk inhibitor.

ANS: A DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle. MSC: Understanding 2. Cyclins are phosphorylated and destroyed.

ANS: D DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle. MSC: Understanding 3. Cdk is unable to phosphorylate its substrates.

ANS: C DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle. MSC: Understanding 4. The Cdc25 phosphatase is inhibited.

ANS: B DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either p ause or contin-

ue through different transition points in the cell cycle. MSC: Understanding 3. Match each of the main classes of spindle microtubules with their functions and features from the options below.

A. stabilized by interactions with each other via motor proteins B. interact with the cell cortex C. link chromosomes to a spindle pole 1. interpolar microtubules ANS: A DIF: Easy REF: 18.6 OBJ: 18.6.b Describe the structure of the mitotic spindle and explain how and when it begins to form. MSC: Remembering 2. aster microtubules ANS: B DIF: Easy REF: 18.6 OBJ: 18.6.b Describe the structure of the mitotic spindle and explain how and when it begins to form. MSC: Remembering 3. kinetochore microtubules ANS: C DIF: Easy REF: 18.6 OBJ: 18.6.b Describe the structure of the mitotic spindle and explain how and when it begins to form. MSC: Remembering

TRUE/FALSE 1. Is the following statement TRUE or FALSE? After the nuclear envelope breaks down, microtubules gain access to the chromosomes and, every so often, a randomly probing microtubule captures a chromosome and ultimately connects to the kinetochore to become a kinetochore microtubule of the spindle.

ANS: T DIF: Easy REF: 18.6 OBJ: 18.6.d Review how chromosomes are captured by spindle microtubules and describe the structure of the point of attachment. MSC: Understanding

SHORT ANSWER 1. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase may be used once. assemble

G1

phosphorylation

condense

G2

polymerization

cytokinesis

M

S

degrade

replicate

segregate

A cell contains the most DNA after __________ phase of the cell cycle, the period when a cell __________s its DNA. A cell is smallest in size after __________ phase of the cell cycle, which includes __________, the process by which the cytoplasm is cleaved in two. S and M phases are separated by gap phases, with the first one called the __________ phase and the second one, which follows S phase, called the __________ phase of the cell cycle. During mitosis, the mitotic spindle is built to __________ chromosomes.

ANS: A cell contains the most DNA after S phase of the cell cycle, the period when a cell replicates its DNA. A cell is smallest in size after M phase of the cell cycle, which includes cytokinesis, the process by which the cytoplasm is

cleaved in two. S and M phases are separated by gap phases, with the first one called the G1 phase and the second one, which follows S phase, called the G2 phase of the cell cycle. During mitosis, the mitotic spindle is built to segregate chromosomes.

DIF: Easy REF: 18.1 OBJ: 18.1.a List the four phases of the eukaryotic cell cycle and summarize what takes place in each. MSC: Understanding 2. What would happen to the progeny of a cell that proceeded to mitosis and cell division after entering S phase but had not completed S phase? Keep in mind that highly condensed chromatin, including the centromere region, is replicated late in S phase. Explain your answer.

ANS: The daughter cells would probably die. Those chromosomes that had not completed replication in S phase would have only one centromere, because the centromere is the last part of the chromosome to be replicated; the chromosome would therefore be segregated to only one of the two daughter cells at random. At least one, and probably both, of the daughter cells would thus receive an incomplete set of chromosomes and would be unlikely to be viable. Even if one daughter cell, by chance, received a full set of chromosomes, some of these chromosomes would be incompletely replicated and the cell would probably still not be viable.

DIF: Hard REF: 18.6 OBJ: 18.6.d Review how chromosomes are captured by spindle microtubules and describe the structure of the point of attachment. MSC: Applying 3. Are the statements below TRUE or FALSE? Explain your answer.

A. Statement 1: Generally, in a given organism, the S, G2, and M phases of the cell cycle take a defined and stereotyped amount of time in most cells.

B. Statement 2: Therefore, the cell-cycle control system operates primarily by a timing mechanism, in which the entry into one phase starts a timer set for sufficient time to complete the required tasks. After a given amount of time has elapsed, a molecular “alarm” triggers movement to the next phase.

ANS: A. True. In nearly all cells in an organism, the S, G 2, and M phases of the cell cycle take the same amount of time.

The different timing of cell division in different cell types is due to variable lengths of the G1 phase or to withdrawal into the G0 state.

B. False. The cell-cycle control system does use a timing mechanism of sorts, but it also employs various surveillance and feedback mechanisms (checkpoints). Cells will not embark on later events or phases until the earlier events or phases have been completed successfully. In response to a defect or a delay in a cell-cycle event, cells engage molecular brakes to arrest the progression of the cell cycle at various checkpoints, to allow time for completion or repair.

DIF: Easy REF: 18.2 OBJ: 18.2.h Summarize the mechanisms that allow cells to either pause or continue through different transition points in the cell cycle. MSC: Evaluating 4. You have isolated a mutant in which a fraction of the new cells die soon after cell division and a fraction of the living cells have an extra copy of one or more chromosomes. When you grow the cells under conditions in which they transit the cell cycle more slowly, the defect disappears, suggesting that the mitotic spindle and segregation machinery are normal. Propose a basis for the defect.

ANS: This mutant may be lacking the checkpoint mechanism that delays the onset of anaphase and chromosome segregation until all chromosomes have attached properly to the mitotic spindle. If cells attempt chromosome segre-

gation before all chromosomes have attached properly, some of the daughter cells will receive too few chromosomes (and thus will probably die) and other cells will receive additional chromosomes. Normally, cells use such a surveillance system to monitor the spindle attachment of each chromosome and engage molecular brakes until all chromosomes have attached properly. The molecular brakes will be dispensable in some dividing cells if all of the chromosomes rapidly become properly attached to the spindle. In other dividing cells, it will take longer for some of the chromosomes to find their appropriate attachments, and thus the molecular brakes will be essential to ensure faithful segregation of the duplicated copies of each chromosome to the two daughter cells. Slowing the cycle will allow more cells to segregate their chromosomes properly even in the absence of this “spindle attachment” checkpoint.

DIF: Hard REF: 18.6 OBJ: 18.6.k Explain how the spindle assembly checkpoint ensures that all chromosomes are attached to the spindle and why this checkpoint can delay the onset of anaphase and the exit from mitosis.

MSC: Evaluating 5. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. able

hexokinase

short

asynchronously

inhibitory

sperm

Cdk

long

steady

conserved

M

stimulatory

cyclin

maturation promoting factor

substrate

divergent

oscillate

synchronously

egg

PI 3-kinase

ubiquitin

fibroblast

regulin

unable

G1

S

uniform

G2 Many features of __________ cells make them suitable for biochemical studies of the cell-cycle control system. For example, the cells are unusually large and are arrested in a __________-like phase. When the cells are triggered to resume cycling, the cell divisions have especially __________ G1 and G2 phases and occur __________. Studies with Xenopus eggs identified a partly purified activity called __________ that drives a resting Xenopus oocyte into M phase. MPF activity was found to __________ during the cell cycle, although the amount of its kinase component, called __________, remained constant. The regulatory component of MPF, called __________, has a __________ effect on MPF activity and plays a part in regulating interactions with its __________s. The components of MPF are evolutionarily __________ from yeast to humans, so that the corresponding human genes are __________ to function in yeast.

ANS: Many features of egg cells make them suitable for biochemical studies of the cell-cycle control system. For example, the cells are unusually large and are arrested in a G2-like phase. When the cells are triggered to resume cycling, the cell divisions have especially short G1 and G2 phases and occur synchronously. Studies with Xenopus eggs identified a partly purified activity called maturation promoting factor that drives a resting Xenopus oocyte into M phase. MPF activity was found to be oscillate during the cell cycle, although the amount of its kinase component, called Cdk, remained constant. The regulatory component of MPF, called cyclin, has a stimulatory effect on MPF activity and plays a part in regulating interactions with its substrates. The components of MPF are evolutionarily conserved from yeast to humans, so that the corresponding human genes are able to function in yeast.

DIF: Easy REF: 18.2 OBJ: 18.2.c Explain how extracts prepared from cells in different phases of the cell cycle have been used to identify components of the cell-cycle control system. MSC: Understanding 6. Irradiated mammalian cells usually stop dividing and arrest at a G 1 checkpoint. Place the following events in the order in which they occur.

A. production of p21 B. DNA damage C. inhibition of cyclin–Cdk complexes D. accumulation and activation of p53 ANS: B, D, A, C DIF: Easy REF: 18.3 OBJ: 18.3.e Review how DNA damage can arrest cells in G1.

MSC: Understanding 7. What is the main molecular difference between cells in a G0 state and cells that have simply paused in G1?

ANS: In G0, the cell-cycle control system is partly dismantled, so that some of the Cdks and cyclins are not present.

Cells paused in G1, by contrast, still contain all the components of the cell-cycle control system. Whereas the latter cells can rapidly progress through the cycle when conditions are right, G 0 cells need to synthesize the missing cellcycle control proteins so as to reenter the cycle, which usually takes 8 hours or more.

DIF: Moderate REF: 18.3 OBJ: 18.3.b Review how cyclin-Cdk complexes are inhibited as cells enter G1.

MSC: Analyzing 8. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

APC

G2 phase

metaphase

Cdks

interphase

microtubules

condensation

intraphase

mitosis

cytokinesis

kinesins

myosins

meiosis

M phase

S phase

G1 phase

M-Cdk

S-Cdk

G1-Cdk The cell cycle consists of an alternation between __________, which appears as a period of dramatic activity under the microscope, and a preparative period called __________, which consists of three phases called __________, __________, and __________. During M phase, the nucleus divides in a process called __________, and the cytoplasm splits in two in a process called __________. The cell-cycle control system relies an increase in the activity of __________ to trigger DNA replication. Inactivation of __________ is required to exit from M phase after chromosome segregation.

ANS: The cell cycle consists of an alternation between M phase, which appears as a period of dramatic activity under the microscope, and a preparative period called interphase, which consists of three phases called G1 phase, S phase, and G2 phase. During M phase, the nucleus divides in a process called mitosis, and the cytoplasm splits in two in a process called cytokinesis. The cell-cycle control system relies on an increase in the activity of S-Cdk to trigger DNA replication. Inactivation of M-Cdk is required to exit from M phase after chromosome segregation.

DIF: Easy REF: 18.2 OBJ: 18.1.a List the four phases of the eukaryotic cell cycle and summarize what takes place in each. MSC: Understanding

9. Which stage of mitosis in an animal cell does each part of Figure 18-9 represent?

Figure 18-9

ANS: A—telophase; B—prophase; C—anaphase; D—prometaphase DIF: Easy REF: 18.5 OBJ: 18.5.d List the stages of M phase. MSC: Remembering 10. Figure 18-10 shows a living cell from the lung epithelium of a newt at different stages in M phase. Order these light micrographs into the correct sequence and identify the stage in M phase that each represents.

Figure 18-10

ANS: The correct order is listed as follows, with stages in parentheses: E (prophase), D (prometaphase), C (metaphase), A (anaphase), F (telophase), and B (cytokinesis). DIF: Easy REF: 18.5 OBJ: 18.5.d List the stages of M phase.

MSC: Applying 11. Name the stage of M phase in which the following events occur. Place the numbers 1–8 next to the letter headings to

indicate the normal order of events.

A. alignment of the chromosomes at the spindle equator B. attachment of spindle microtubules to chromosomes C. breakdown of nuclear envelope D. pinching of cell in two E. separation of two centrosomes and initiation of mitotic spindle assembly F. re-formation of the nuclear envelope G. condensation of the chromosomes H. separation of sister chromatids ANS: A. 5, metaphase B. 4, prometaphase C. 3, prometaphase D. 8, cytokinesis E. 2, prophase F. 7, telophase G. 1, prophase H. 6, anaphase DIF: Easy REF: 18.5 OBJ: 18.5.d List the stages of M phase. MSC: Remembering 12. Before a cell can enter M phase, two structures from Figure 18-12 must be duplicated. Write the letters corresponding to the lines pointing to these structures, and write the names of the structures next to the letters.

Figure 18-12

ANS: A. centrosome; B. chromosome DIF: Easy REF: 18.6 OBJ: 18.6.b Describe the structure of the mitotic spindle and explain how and when it begins to form. MSC: Understanding 13. Examine the schematic representation of centrosome duplication in Figure 18-13. By analogy with DNA replication, would you classify centrosome duplication as conservative or semiconservative? Explain your answer.

Figure 18-13

ANS: Centrosome duplication is semiconservative. The paired centrioles in the centrosome separate, and each serves to nucleate the assembly of a new centriole. As a consequence, each new centrosome consists of one old and one new centriole. Thus, centrosome duplication is analogous to DNA replication, in which the new double helix consists of one old DNA strand and one newly replicated DNA strand.

DIF: Hard REF: 18.6 OBJ: 18.6.a Outline the centrosome cycle, indicating how and when it is initiated.

MSC: Applying 14. Before chromosomes segregate in M phase, they and the segregation machinery must be appropriately prepared.

Indicate whether the following statements are TRUE or FALSE. If false, change a single noun to make the statement true.

A. Sister chromatids are held together by condensins from the time they arise by DNA replication until the time they separate at anaphase.

B. Cohesins are required to make the chromosomes more compact and thus to prevent tangling between different chromosomes.

C. Microtubule-dependent motor proteins and microtubule polymerization and depolymerization are mainly responsible for the organized movements of chromosomes during mitosis.

D. The centromere nucleates a radial array of microtubules called an aster, and its duplication is triggered by SCdk. E. Each centrosome contains a pair of centrioles and hundreds of γ-tubulin rings that nucleate the growth of microtubules.

ANS: A. False. Sister chromatids are held together by cohesins from the time they arise by DNA replication until the time they separate at anaphase.

B. False. Condensins are required to make the chromosomes more compact and thus to prevent tangling between different chromosomes.

C. True.

D. False. The centrosome nucleates a radial array of microtubules called an aster, and its duplication is triggered by S-Cdk.

E. True.

DIF: Moderate REF: 18.5 OBJ: 18.5.b Compare cohesins and condensins in terms of structure, function, and how and when they assemble onto chromosomal DNA. | 18.5.c Compare the mitotic spindle and contractile ring in terms of composition, location, and the role each plays in cell division. MSC: Analyzing 15. The cytoskeleton of an animal cell changes markedly between G 1 and early M phase (prophase) of the cell cycle. For each of the following sentences, choose one of the options enclosed in square brackets that best describes the changes to the cytoskeleton and its components.

Before mitosis, the number of centrosomes must [increase/decrease]. At the beginning of [anaphase/prophase] in animal cells, the centrosomes separate in a process driven partly by interactions between the [plus/minus] ends of microtubules arising from the two centrosomes. Centrosome separation initiates the assembly of the bipolar mitotic spindle and is associated with a sudden [increase/decrease] in the dynamic instability of microtubules.

[Interpolar/astral/kinetochore] microtubules are formed in an overlap zone where two microtubules from opposite centrosomes interact. During anaphase [A/B], kinetochore microtubules are shortened, dragging chromosomes toward their spindle pole.

ANS: Before mitosis, the number of centrosomes must increase. At the beginning of prophase in animal cells, the centrosomes separate in a process driven partly by interactions between the plus ends of microtubules arising from the two centrosomes. Centrosome separation initiates the assembly of the bipolar mitotic spindle and is associated with a sudden increase in the dynamic instability of microtubules. Interpolar microtubules are formed in an overlap zone where two microtubules from opposite centrosomes interact. During anaphase A, kinetochore microtubules are shortened, dragging chromosomes toward their spindle pole.

DIF: Easy REF: 18.6 OBJ: 18.6.a Outline the centrosome cycle, indicating how and when it is initiated. | 18.6.b Describe the structure of the mitotic spindle and explain how and when it begins to form. | 18.6.j Compare the changes in the mitotic spindle that underlie chromosome segregation during anaphase A and anaphase B, and delineate the driving forces responsible for each process. MSC: Understanding 16. Consider an animal cell that has eight chromosomes (four pairs of homologous chromosomes) in G 1 phase. How many of each of the following structures will the cell have at mitotic prophase?

A. sister chromatids B. centromeres C. kinetochores D. centrosomes E. centrioles ANS: A—16; B—16; C—16; D—2; E—4 DIF: Moderate REF: 18.6 OBJ: 18.6.a Outline the centrosome cycle, indicating how and when it is initiated. | 18.6.d Review how chromosomes are captured by spindle microtubules and describe the structure of the point of attachment.

MSC: Applying 17. Why should it be that drugs such as colchicine, which inhibit microtubule polymerization, and drugs such as Taxol®, which stabilize microtubules, both inhibit mitosis?

ANS: Mitosis requires that the spindle microtubules behave dynamically—continuously polymerizing and depolymerizing—to probe the cell cortex, to seek attachments to kinetochores, and to segregate the chromosomes. Static micro-

tubules are unable to do any of these things.

DIF: Hard REF: 18.6 OBJ: 18.6.d Review how chromosomes are captured by spindle microtubules and describe the structure of the point of attachment. | 18.6.i Summarize the molecular events that trigger separation of sister chromatids at the start of anaphase. MSC: Applying 18. The lengths of microtubules in various stages of mitosis depend on the balance between the activities of proteins that destabilize microtubules, and microtubule-associated proteins that stabilize them. If you created cells with an increased number of proteins that destabilize molecules, do you predict the length of the mitotic spindle will be longer, shorter, or unchanged, relative to the corresponding stage of mitosis in wild-type cells? What do you predict for a cell with increased numbers of microtubule stabilizing proteins? Explain your reasoning.

ANS: In a cell with excessive microtubule-destabilizing molecules, the balance between these destabilizing proteins and the proteins that stabilize microtubules will be disrupted. Increased microtubule-destabilizing activity relative to stabilizing activity will lead to an increased frequency of microtubule catastrophes, the sudden shift from growth to shrinkage. Thus, microtubules will be shorter on average because they spend less time growing slowly and more time shrinking rapidly. Shorter microtubules would probably result in a shorter mitotic spindle. Conversely, increased microtubule stabilizing activity relative to destabilizing activity will stabilize microtubules by enhancing polymerization or inhibiting depolymerization, thereby promoting the formation of longer microtubules and probably a longer mitotic spindle. The normal balance between destabilizing and stabilizing microtubules has been shown to be necessary for the formation of a bipolar spindle, likely because an imbalance may create microtubules that are so long or so short that they cannot form a spindle at all.

DIF: Hard REF: 18.6 OBJ: 18.6.b Describe the structure of the mitotic spindle and explain how and when it begins to form. MSC: Applying 19. You have discovered a new protein that regulates microtubule dynamics. First, you isolated proteins from a cellular extract that bound to a tubulin affinity column. You then separated the proteins from each other by loading the mixture of proteins on an ion-exchange column, eluting the column with increasing salt concentration, and collecting small “fractions” of protein as they dripped from the column. To test whether each fraction contained microtubule regulators, you mixed it with fluorescent tubulin and purified centrosomes, and then analyzed the reaction microscopically to measure the size of the astral microtubules formed. You found that fractions 8, 9, and 10 promoted the formation of unusually long astral microtubules. Because electrophoretic separation of the fractions on a gel revealed a plentiful protein with an apparent molecular mass of 98 kD, you named the protein p98.

A. Propose two ways in which p98 might change the dynamic behavior of microtubules to account for the observed change in microtubule length. (Hint: There are four simple possible mechanisms.)

B. Video microscopy of fluorescent tubulin in reactions with purified centrosomes allowed you to follow the behavior of individual microtubules over time. You graphed the changes in microtubule length in the absence (Figure 18-19A) and presence (Figure 18-19B) of p98. Five representative microtubules are shown for each condition. Does p98 alter the rate of microtubule growth or shrinkage? Does p98 alter the frequency of catastrophes (a sudden and rapid decline in microtubule length) or rescues (when a microtubule switches from shrinking to growing)? Explain your answers.

C. After demonstrating the consequences of p98 addition on microtubule dynamics in vitro with the use of purified components, you want to determine whether the protein has the same effects in a complex cellular extract that naturally contains p98. You remove the p98 protein from an extract of Xenopous eggs in mitosis by using antibodies that specifically recognize p98. The p98-depleted extract is then mixed with sperm nuclei, centrosomes,

and fluorescent tubulin. How would you expect the microtubules to behave?

Figure 18-19

ANS: A. The four possible mechanisms by which a protein can promote microtubule growth are (1) increasing the rate of polymerization, (2) decreasing the rate of depolymerization, (3) inhibiting the occurrence of catastrophes (when a microtubule shifts abruptly from growing to shrinking), and (4) stimulating the occurrence of rescues (when a microtubule switches from shrinking to growing).

B. p98 increased the rate of microtubule growth, as seen by the steeper slopes of the lines when p98 was added. The data do not allow a conclusion to be made about the rate of microtubule shrinkage. The protein decreased the rate of catastrophes: two of the five microtubules in the absence of p98 underwent catastrophe, whereas none did so in the presence of p98. No rescues were observed in either the presence or the absence of p98, so it is uncertain whether the protein alters the rescue rate.

C. On the basis of the previous experiments, the p98-depleted extract would be expected to have shorter and more dynamic microtubules. (Ideally, this kind of experiment would be supplemented with the production and examination of intact cells in which p98 function has been abolished by mutation or another means, such as RNA-mediated interference.)

DIF: Hard REF: 18.6 OBJ: 18.6.b Describe the structure of the mitotic spindle and explain how and when it begins to form. MSC: Applying 20. Imagine that you could microinject cytochrome c into the cytosol of both wild-type cells and cells that were lacking both Bax and Bak, which are apoptosis-promoting members of the Bcl2 family of proteins. Would you expect one, both, or neither of the cell lines to undergo apoptosis? Explain your reasoning.

ANS: Both. The presence or absence of Bak and Bax would not affect whether a microinjection of cytochrome c would promote apoptosis, because Bax and Bak act upstream of cytochrome c by promoting its release from mitochondria.

By promoting the formation of the apoptosome and the activation of procaspases, microinjection of cytochrome c bypasses the need for Bax or Bak in promoting apoptosis.

DIF: Hard REF: 18.8 OBJ: 18.8.g Outline how apoptosis is regulated and initiated by the Bcl2 family of proteins.

MSC: Analyzing 21. The number of cells in an adult tissue or animal depends on cell proliferation. What else does it depend on?

ANS: Programmed cell death also influences cell numbers. Most animal cells require survival signals from other cells to

avoid programmed cell death, so that the levels of such signals can help determine how many cells live and how many die.

DIF: Easy REF: 18.8 OBJ: 18.8.c Recall how apoptosis balances cell division in an adult tissue such as liver.

MSC: Understanding 22. What is the cause of the massive amount of programmed cell death of nerve cells (neurons) that occurs in the developing vertebrate nervous system, and what purpose does it serve?

ANS: Immature neurons are produced in excess of the number that will eventually be required. They compete for the limited amount of survival factors secreted by the target cells they contact. Those cells that fail to get enough survival factor undergo programmed cell death. Up to half or more of the original nerve cells die in this way. This competitive mechanism helps match the number of developing nerve cells to the number of target cells they contact.

DIF: Easy REF: 18.8 OBJ: 18.8.j Explain how survival factors suppress apoptosis, and describe how they regulate the number of neurons in the developing nervous system. MSC: Understanding 23. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. anaphase

differentiation

myostatin

annihilation

G0

nourishment

arrestase

G1

nutrition

Bcl2

G2

phosphatases

biosynthetic

growth factors

proliferation

cyclin

interphase

receptor

cascades

ligand

S

caspase

M

survival factors

Cdk

mitogens

transcription

The survival, __________, and size of each cell in an animal are controlled by extracellular signal molecules secreted by neighboring and distant cells. Many of these signal molecules bind to a cell-surface __________ and trigger various intracellular signaling pathways. One class of signal molecules, called __________, stimulates cell division by releasing the molecular brakes that keep cells in the __________or __________ phase of the cell cycle.

Members of a second class of signal molecules are called __________, because they stimulate cell growth and an increase in cell mass. The third class of signal molecules, called __________, inhibits apoptosis by regulating members of the __________ family of proteins. In addition to such stimulatory factors, some signal proteins, such as __________ act negatively on other cells, inhibiting their survival, growth, or proliferation.

ANS: The survival, proliferation, and size of each cell in an animal are controlled by extracellular signal molecules secreted by neighboring and distant cells. Many of these signal molecules bind to a cell-surface receptor and trigger various intracellular signaling pathways. One class of signal molecules, called mitogens, stimulates cell division by releasing the molecular brakes that keep cells in the G0 or G1 phase of the cell cycle. Members of a second class of signal molecules are called growth factors, because they stimulate cell growth and an increase in cell mass. The third class of signal molecules, called survival factors, inhibits apoptosis by regulating members of the Bcl2 family of proteins. In addition to such stimulatory factors, some signal proteins, such as myostatin act negatively on other cells, inhibiting their survival, growth, or proliferation.

DIF: Easy REF: 18.8 OBJ: 18.8.i List the classes of signal proteins that influence cell fate and summarize their functions. | 18.8.j Explain how survival factors suppress apoptosis, and describe how they regulate the number of neurons in the developing nervous system. | 18.8.m Outline how growth factors function to increase cell size.

MSC: Understanding 24. Of the following mutations, which are likely to cause cell-cycle arrest? If you predict a cell-cycle arrest, indicate whether the cell will arrest in early G1, late G1, or G2. Explain your answers.

A. a mutation in a gene encoding a cell-surface mitogen receptor that makes the receptor active even in the absence of the mitogen B. a mutation that destroyed the kinase activity of S-Cdk C. a mutation that allowed G1-Cdk to be active independently of its phosphorylation status D. a mutation that removed the phosphorylation sites on the Rb protein E. a mutation that inhibited the activity of Rb ANS: B and D are likely to cause cell-cycle arrest.

A. Because ligand-independent activation of a mitogen receptor will probably make the cell divide when it otherwise might not, this mutation is unlikely to cause a cell-cycle arrest.

B. This mutation is likely to cause cell-cycle arrest in late G1. Without S-Cdk activity, the cells will probably be unable to enter S phase.

C. Phosphorylation-independent activity of G1-Cdk is unlikely to lead to cell-cycle arrest; the cells should progress through the cycle, although the kinetics and fine regulation of the cycle may be altered.

D. This mutation is likely to cause cell-cycle arrest in early G1. Unphosphorylated Rb will inhibit the transcription of genes required for progression through G1 and into S phase. This inhibition is normally released on phosphorylation of Rb. If Rb cannot be phosphorylated, it will always inhibit transcription of these genes, leading to arrest in early G1.

E. If Rb is inactivated by a mutation, cells will be more likely to divide in the absence of extracellular mitogens, which is the opposite of a cell-cycle arrest.

DIF: Hard REF: 18.3 OBJ: 18.3.b Review how cyclin-Cdk complexes are inhibited as cells enter G1. | 18.3.d Explain how Rb blocks cell proliferation and how mitogens reverse this inhibition. MSC: Applying 25. Your friend is intrigued by Cdks and purifies Cdk from the daisy plant. She is able to determine the sequence of 18 amino acids from the daisy Cdk protein. Using these data, she aligns the daisy Cdk sequence with one human Cdk protein and the Cdk proteins from two different kinds of yeast, as shown in Figure 18-25. Sequences identical between the human and fungal proteins have been boxed.

Figure 18-25

Such conserved amino acid sequences are often involved in protein–protein interactions. Indeed, the threonine (T) in the central “YTHE” block is known to be phosphorylated by a kinase that activates Cdks in human and yeast. The

surrounding conserved sequences could thus be important for the interaction of this Cdk-activating kinase with Cdk. In the daisy Cdk, however, not all of the amino acids in these conserved blocks match the sequences from yeast and human, as indicated by the arrows marked #1 and #2.

A. When you replace the S. cerevisiae (budding yeast) Cdk with the daisy Cdk, you discover the daisy Cdk does not interact with the Cdk-activating kinase inside the yeast cell. Circle the amino acid in the daisy Cdk that is likely to be the most disruptive to its interaction with Cdk-activating kinase. Explain.

B. Based on the information in Figure 18-25 and what you know about the Cdk1–Cdk-activating kinase interaction, would you predict that the human Cdk will interact with yeast Cdk-activating kinase? Explain.

ANS: A. The lysine (K) at the position of arrow #2 in the daisy Cdk is likely the culprit for the disrupted interaction between Cdk and Cdk-activating kinase (see Figure 18-25A). This changes a conserved glutamic acid (E) to a lysine, which results in a change from a negatively charged amino acid to a positively charged amino acid. Although there is also a difference in the daisy sequence at the position of arrow #1, this change from an arginine (R) to a lysine is less dramatic, as both are positively charged amino acids.

Figure 18-25A

B. You would predict that the human Cdk will interact with the yeast Cdk-activating kinase, because the amino acids at the positions of arrows #1 and #2 are identical between human and yeast.

DIF: Hard REF: 18.2 OBJ: 18.1.d Explain why studies of yeast can lead to insights into the biology of human cancers. | 18.2.d Explain how studies of mutant yeast were used to dissect the cell-cycle control system. MSC: Analyzing

CHAPTER 19

Sexual Reproduction and the Power

of Genetics THE BENEFITS OF SEX 19.1.a Provide examples of asexual reproduction and indicate how the offspring relate, genetically, to the parent organism.

19.1.b Distinguish among somatic cells, germline cells, and gametes in terms of function and whether the cells are diploid or haploid.

19.1.c Compare sperm, egg, and zygote in terms of appearance and chromosome complement.

19.1.d Compare genes and alleles.

19.1.e Explain how sexual reproduction generates genetic diversity.

19.1.f

Summarize the potential advantages conferred by sexual reproduction.

MEIOSIS AND FERTILIZATION 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell.

19.2.b Describe the products of meiosis in terms of chromosomal complement and genetic similarity with the parental germline cell.

19.2.c Review the process of homolog pairing and explain its importance in meiosis.

19.2.d Describe the chromosome composition of a bivalent.

19.2.e Contrast how duplicated chromosomes align at the metaphase plate in mitosis to their alignment in meiosis I. 19.2.f

Outline the main events that take place during meiotic homologous recombination.

19.2.g Review the function of the synaptonemal complex.

19.2.h Describe a chiasma and indicate how many are typically found in each bivalent.

19.2.i

Explain how crossing-over helps ensure the proper segregation of homologs.

19.2.j

Contrast the behaviors of kinetochores during meiosis I and II.

19.2.k Differentiate how cohesins are degraded during anaphase in meiosis I from their degradation during anaphase in meiosis II. 19.2.l

Describe the events that take place between the first and second meiotic divisions.

19.2.m Outline the forms of genetic reassortment that give rise to new chromosome combinations during meiosis.

19.2.n Explain how nondisjunction gives rise to aneuploid gametes, and recall the consequences of this type of genetic

error.

19.2.o Describe where and how sperm and egg unite during fertilization, and outline the mechanisms that prevent multiple sperm from entering the egg.

MENDEL AND THE LAWS OF INHERITANCE 19.3.a Explain why Mendel chose to study inheritance using the traits and organism he selected.

19.3.b Define what it means for an organism to be true-breeding, and describe the appearance of the offspring Mendel observed when cross-fertilizing two true-breeding plants with each other.

19.3.c Distinguish genotype from phenotype.

19.3.d Describe the observations that led Mendel to postulate that traits are specified by pairs of alleles, one of which is dominant and the other recessive.

19.3.e Define heterozygosity and homozygosity and describe how they relate to phenotype.

19.3.f

State Mendel’s law of segregation and explain how it gives rise to the phenotypic ratios he observed in the F2

generation.

19.3.g Explain how pedigrees can be used to study the inheritance of recessive traits in human populations.

19.3.h Outline the type of breeding experiment that led to Mendel’s law of independent assortment.

19.3.i

Summarize how the behavior of chromosomes during meiosis underlies Mendel’s laws of inheritance.

19.3.j

Review how genes present on the same chromosome can segregate independently.

19.3.k Explain why loss-of-function mutations can be recessive or dominant.

19.3.l

Describe the effect a dominant mutation has on an organism’s phenotype.

19.3.m Review how gain-of-function mutations in the Ras gene promote the development of cancer.

19.3.n Distinguish the fates of deleterious mutations that are dominant versus recessive.

GENETICS AS AN EXPERIMENTAL TOOL 19.4.a Outline the steps involved in the classical genetic approach to studying gene activity, and explain how mutagens can be used to expedite the process.

19.4.b Describe how genetic screens are used to identify mutants with a phenotype of interest.

19.4.c Compare how mutations that have lethal effects can be studied in diploid and haploid organisms.

19.4.d Review how a complementation test can be used to determine whether two mutations affect the same gene.

EXPLORING HUMAN GENETICS 19.5.a Define polymorphism and list the most common types of polymorphisms in the human genome.

19.5.b Recall how much variation exists between the genome of any two individuals.

19.5.c Explain why haplotype blocks are present in the human genome.

19.5.d Review how an analysis of haplotype blocks can be used to estimate when a particular mutation or allele arose in the human population.

19.5.e Describe how genetic and environmental factors can interact to give rise to a “disease phenotype.”

19.5.f

Explain why monogenic disorders tend to be rare, yet they persist in the human population.

19.5.g Recall how consanguineous marriages affect the likelihood of inheriting a disease-causing allele.

19.5.h Review why some monogenic diseases might be more common in certain populations.

19.5.i

Provide an example of a loss-of-function mutation that is beneficial rather than deleterious.

19.5.j

Explain why the risk-enhancing alleles associated with multigenic disorders are relatively common in the population.

19.5.k Outline how genome-wide association studies can be used to search for genes that predispose individuals to common diseases.

19.5.l

Review how SNPs can be used to produce a genetic linkage map and to facilitate the search for alleles that pre-

dispose to disease.

19.5.m Distinguish between polymorphisms and mutations in terms of prevalence and when they arose in our evolutionary history.

19.5.n Explain how genome sequencing can be used to identify rare variants and more recent mutations.

19.5.o Recall what genome sequencing studies reveal about the prevalence of mutations in apparently healthy individuals.

MULTIPLE CHOICE 1. Organisms that reproduce sexually a. must be haploid, unlike organisms that reproduce asexually. b. can reproduce only with a partner that carries the same alleles. c. create zygotes that are genetically identical to each other. d. undergo a sexual reproductive cycle that involves an alternation of haploid cells with the generation of diploid cells.

ANS: D DIF: Easy REF: 19.1 OBJ: 19.1.e Explain how sexual reproduction generates genetic diversity.

MSC: Remembering 2. Which of the following statements is TRUE? a. Another name for the fertilized egg cell is the zygote. b. Diploid organisms reproduce only sexually. c. All sexually reproducing organisms must have two copies of every chromosome. d. Gametes have only one chromosome.

ANS: A Some diploid organisms (for example, many plants) are capable of asexual reproduction. Many organisms have sex chromosomes that are only present in one copy in the diploid organism. Gametes have only one member of each pair

of homologous chromosomes, but because most organisms have more than one pair of homologous chromosomes, most gametes have more than one chromosome.

DIF: Easy REF: 19.1 OBJ: 19.1.b Distinguish among somatic cells, germline cells, and gametes in terms of function and whether the cells are diploid or haploid. | 19.1.c Compare sperm, egg, and zygote in terms of appearance and chromosome complement. MSC: Remembering 3. Which of the following statements is FALSE? a. Asexual reproduction typically gives rise to offspring that are genetically identical. b. Mutations in somatic cells are passed on to individuals of the next generation. c. Sexual reproduction allows for a wide variety of gene combinations. d. Gametes are specialized sex cells.

ANS: B DIF: Easy REF: 19.1 OBJ: 19.1.b Distinguish between somatic cells, germline cells, and gametes in terms of function and whether the cells are diploid or haploid. | 19.1.e Explain how sexual reproduction generates genetic diversity. MSC: Understanding 4. Somatic cells __________. a. are not necessary for sexual reproduction in all eukaryotic organisms. b. are used to produce germ-line cells when organisms reach sexual maturity c. leave no progeny. d. do not contain sex chromosomes.

ANS: C Somatic cells are used to form the rest of the animal’s body, which is required to support sexual reproduction even if these cells are not used to generate the germ cells. Germ-line cells are generally specified early in development, before organisms reach sexual maturity. Somatic cells do contain sex chromosomes.

DIF: Easy REF: 19.1 OBJ: 19.1.b Distinguish between somatic cells, germline cells, and gametes in terms of function and whether the cells are diploid or haploid. MSC: Understanding 5. Which of the following statements about the benefits of sexual reproduction is FALSE? a. Sexual reproduction permits enhanced survival because the gametes that carry alleles enhancing survival in harsh environments are used preferentially during fertilization. b. Unicellular organisms that can undergo sexual reproduction have an increased ability to adapt to harsh environments. c. Sexual reproduction reshuffles genes, which is thought to help species survive in novel or varying environments. d. Sexual reproduction can speed the elimination of deleterious alleles.

ANS: A Alleles enhancing survival in harsh environments will not be selected for until the organism encounters the harsh environment.

DIF: Easy REF: 19.1 OBJ: 19.1.f Summarize the potential advantages conferred by sexual reproduction.

MSC: Understanding 6. During sexual reproduction, novel mixtures of alleles are generated. This is because a. in all diploid species, two alleles exist for every gene. b. a diploid individual has two different alleles for every gene.

c. every gamete produced by a diploid individual has several different alleles of a single gene. d. during meiosis, the segregation of homologs is random such that different gametes end up with different alleles of each gene.

ANS: D Many alleles can exist for any gene, and a diploid individual can have either two different or two identical alleles of any given gene. Every gamete only has a single allele of any gene.

DIF: Easy REF: 19.1 OBJ: 19.1.e Explain how sexual reproduction generates genetic diversity.

MSC: Understanding 7. Which of the following does not describe a situation of asexual reproduction? a. a bacterium multiplying by simple cell division b. using a part of a plant to create a new independent plant c. using in vitro fertilization to combine a sperm and an egg to create an embryo d. the parthenogenetic development of eggs produced by some species of lizards ANS: C Asexual reproduction gives rise to offspring that are genetically identical to the parent. The in vitro fertilization process requires gametes (sperm and egg), which, when combined, will produce an offspring that is genetically distinct from either parent.

DIF: Easy REF: 19.1 OBJ: 19.1.a Provide examples of asexual reproduction and indicate how the offspring relate, genetically, to the parent organism. MSC: Understanding 8. Both budding yeast and the bacteria E. coli are unicellular life forms. Which of the following statements explains why budding yeast can undergo sexual reproduction while E. coli cannot? a. Unlike E. coli, budding yeast can alternate between a diploid state and a haploid state. b. Unlike E. coli, budding yeast cannot multiply by undergoing cell division. c. Unlike E. coli, haploid budding yeast cells can undergo meiosis to produce the gametes necessary for sexual reproduction. d. E. coli DNA is unable to undergo homologous recombination, making it incapable of producing gametes.

ANS: A Diploid budding yeast cells can either reproduce by cell division or undergo meiosis to produce haploid gametes that can either fuse with another haploid gamete to form a diploid cell, or can become a free-living haploid cell that multiplies by cell division. A budding yeast cell must be diploid to undergo meiosis; haploid cells cannot undergo meiosis.

E. coli can undergo homologous recombination and uses this process for DNA repair. In fact, E. coli can shuffle their genomes through a process of conjugation, where DNA is transferred from one bacterium to another and may get incorporated in the E. coli genetic material by recombination. Although this conjugation process can lead to the transfer of genetic material, it is not typically considered sexual reproduction.

DIF: Easy REF: 19.1 OBJ: 19.1.a Provide examples of asexual reproduction and indicate how the offspring relate, genetically, to the parent organism. | 19.1.b Distinguish between somatic cells, germline cells, and gametes in terms of function and whether the cells are diploid or haploid. MSC: Understanding 9. The formation of a bivalent during meiosis ensures that a. one chromatid from the mother and one chromatid from the father will segregate together during meiosis I.

b. all four sister chromatids remain together until the cell is ready to divide. c. recombination will occur between identical sister chromatids. d. the sex chromosomes, which are not identical, will line up separately at the metaphase plate during meiosis I. ANS: B Both chromatids from a single parent will segregate together during meiosis I. Recombination occurs between nonidentical sister chromatids. The sex chromosomes will come together to form a bivalent, despite not being identical.

DIF: Easy REF: 19.2 OBJ: 19.2.d Describe the chromosome composition of a bivalent. MSC: Understanding 10. Imagine meiosis in a diploid organism that only has a single chromosome. Like most diploid organisms, it received one copy of this chromosome from each of its parents and the two homologs are genetically distinct. If only a single homologous recombination event occurs during meiosis, which of the following choices below correctly describes the four gametes formed? a. None of the gametes will contain chromosomes identical to the chromosomes found in the original diploid cell. b. All four of the gametes will have chromosomes identical to the chromosomes found in the original diploid cell. c. Three of the gametes will have chromosomes identical to the chromosomes found in the original diploid cell, while one of the gametes will have chromosomes that are different. d. Two of the gametes will have chromosomes identical to the chromosomes found in the original diploid cell, while two of the gametes will have chromosomes that are different.

ANS: D A single recombination event will lead to an exchange of genetic information between two of the chromatids—one from each parent (for example, see Figure 19-7).

DIF: Easy REF: 19.2 OBJ: 19.2.b Describe the products of meiosis in terms of chromosomal complement and genetic similarity with the parental germline cell. | 19.2.f Outline the main events that take place during meiotic homologous recombination. MSC: Understanding 11. There are organisms that go through meiosis but do not undergo recombination when forming haploid gametes.

Which of the following statements correctly describes the gametes produced by such an organism? (Assume that these organisms are diploid, that each of the two homologous chromosomes are genetically distinct as typically found in the wild, and that these organisms have more than one chromosome.) a. All gametes formed during a single meiosis will be identical. b. Due to the random assortment of homologs, each of the gametes formed during a single meiosis will be different. c. This organism could potentially produce 2n genetically distinct gametes, where n is its haploid number of chromosomes. d. The fusion of any two gametes produced by such an organism that does not undergo recombination during meiosis will create a cell that is genetically identical to that individual.

ANS: C Although it is true that the homologs will randomly assort in meiosis I, this will only create two different types of genetic combinations in a single meiosis. Thus, four gametes will be produced, and the two products of meiosis II will be genetically identical to each other. Because the two homologous chromosomes are genetically distinct, the four gametes formed during meiosis cannot be identical. Because the two homologs are genetically distinct and segregate randomly in meiosis I, the fusion of any two gametes produced by an individual will lead to unique combinations of maternal and paternal chromosomes, and thus the resulting cell will not be genetically identical to the individual.

DIF: Moderate REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. | 19.2.b Describe the products of meiosis in terms of chromosomal complement and genetic similarity with the parental germline cell. MSC: Understanding 12. In the absence of recombination, how many genetically different types of gametes can an organism with five homologous chromosome pairs produce? a. 5 b. 10 c. 32 d. 64 ANS: C Because homologous chromosomes assort randomly at meiosis and a gamete has two choices for each chromosome (because sexual organisms are diploid), there are 25, or 32, possible genetically different gametes.

DIF: Moderate REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. | 19.2.b Describe the products of meiosis in terms of chromosomal complement and genetic similarity with the parental germline cell. MSC: Applying 13. Which of the following statements most correctly describes meiosis? a. Meiosis involves two rounds of DNA replication followed by a single cell division. b. Meiosis involves a single round of DNA replication followed by four successive cell divisions. c. Meiosis involves four rounds of DNA replication followed by two successive cell divisions. d. Meiosis involves a single round of DNA replication followed by two successive cell divisions.

ANS: D DIF: Easy REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. MSC: Understanding 14. A diploid cell containing 32 chromosomes will make a haploid cell containing __________ chromosomes. a. 8 b. 16 c. 30 d. 64 ANS: B DIF: Easy REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. | 19.2.b Describe the products of meiosis in terms of chromosomal complement and genetic similarity with the parental germline cell. MSC: Applying 15. You examine a worm that has two genders: males that produce sperm and hermaphrodites that produce both sperm and eggs. The diploid adult has four homologous pairs of chromosomes that undergo very little recombination. Given a choice, the hermaphrodites prefer to mate with males, but just to annoy the worm, you pluck a hermaphrodite out of the wild and fertilize its eggs with its own sperm. Assuming that all the resulting offspring are viable, what fraction do you expect to be genetically identical to the parent worm? Assume that each chromosome in the original hermaphrodite is genetically distinct from its homolog. a. all b. none c. 1/16

d. 1/256 ANS: C Because each chromosome is genetically distinct from its homolog, the parent is heterozygous for each chromosome and thus can produce 24 = 16 types of egg and 24 = 16 types of sperm. Any of the eggs produced will be able to give rise to an adult that is identical to the parent, but to do so it must be fertilized by the right type of sperm. For each type of egg, only 1 of the 16 possible sperm will produce a diploid that is identical to the parent. Therefore, 1 out of 16 of the offspring should be identical to the parent. In other words, a sexually reproducing organism with several heterozygous chromosomes has a relatively high probability of producing genetically distinct offspring even when the parent mates to itself.

DIF: Hard REF: 19.2 OBJ: 19.1.c Compare sperm, egg, and zygote in terms of appearance and chromosome complement. | 19.2.b Describe the products of meiosis in terms of chromosomal complement and genetic similarity with the parental germline cell. | 19.2.b Describe the products of meiosis in terms of chromosomal complement and genetic similarity with the parental germline cell. MSC: Applying 16. Which of the following statements about meiosis is TRUE? a. During meiosis, the paternal chromosomes pair with the maternal chromosomes before lining up at the metaphase plate. b. Unicellular organisms that have a haploid state undergo meiosis instead of mitosis during cell division. c. Meiosis produces four genetically identical cells. d. In general, meiosis is faster than mitosis.

ANS: A DIF: Easy REF: 19.2 OBJ: 19.2.e Contrast how duplicated chromosomes align at the metaphase plate in mitosis and in meiosis I. MSC: Understanding 17. During recombination a. sister chromatids undergo crossing-over with each other. b. chiasmata hold chromosomes together. c. one crossover event occurs for each pair of human chromosomes. d. the synaptonemal complex keeps the sister chromatids together until anaphase II. ANS: B Non-sister chromatids undergo crossing-over; because sister chromatids are identical, there would be no exchange of genetic material if sister chromatids underwent recombination. The number of crossover events can vary for each meiosis; on average, two to three crossover events occur between each pair of human chromosomes. Although the sister chromatids do not separate until anaphase II, their arms become unglued because the cohesins holding them together are degraded during anaphase I. The synaptonemal complex is important for holding together and aligning the duplicated homologs and is not involved directly in sister-chromatid cohesion.

DIF: Easy REF: 19.2 OBJ: 19.2.h Describe a chiasma and indicate how many are typically found in each bivalent.

MSC: Remembering 18. After the first meiotic cell division, a. two haploid gametes are produced. b. cells are produced that contain the same number of chromosomes as somatic cells. c. the number of chromosomes will vary depending on how the paternal and maternal chromosomes align at the

metaphase plate. d. DNA replication occurs.

ANS: B DIF: Easy REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. MSC: Understanding 19. Which of the following would NOT lead to aneuploidy during meiosis? a. sister chromatids segregating inappropriately b. nonsister chromatids segregating inappropriately c. a reciprocal rearrangement of parts between nonhomologous chromosomes (for example, the left arm of Chromosome 2 exchanging places with the right arm of Chromosome 3) d. an extra set of chromosomes produced during S phase (for example, if paternal Chromosome 3 were replicated twice) ANS: C Aneuploidy describes cells with an incorrect chromosome number. A reciprocal rearrangement of parts between nonhomologous chromosomes will not lead to aneuploidy—it will simply change the assortment of genes attached to each centromere, while leaving the total amount of DNA and the total number of chromosomes unchanged.

DIF: Moderate REF: 19.2 OBJ: 19.2.n Explain how nondisjunction gives rise to aneuploid gametes, and recall the consequences of this type of genetic error. MSC: Understanding 20. A single nondisjunction event during meiosis a. will block recombination. b. will occur only during meiosis II. c. cannot occur with sex chromosomes. d. will result in the production of two normal gametes if it occurs during meiosis II. ANS: D A single nondisjunction event will lead to the segregation of one homologous pair of chromosomes, resulting in two aneuploid gametes and two normal gametes.

DIF: Moderate REF: 19.2 OBJ: 19.2.n Explain how nondisjunction gives rise to aneuploid gametes, and recall the consequences of this type of genetic error. MSC: Understanding 21. During fertilization in humans, a. a wave of Ca2+ ions is released in the fertilized egg’s cytoplasm. b. only one sperm binds to the unfertilized egg. c. a sperm moves in a random fashion until it encounters an egg. d. several sperm pronuclei compete in the cytoplasm to fuse with the egg nucleus.

ANS: A Many sperm can bind to an egg. Cells surrounding the egg release a chemoattractant signal, attracting the sperm to the correct place; sperm are moving toward the chemoattractant signal and not moving randomly. Mechanisms exist to ensure that only one sperm fertilizes each egg, and thus only one sperm pronucleus will reach the cytoplasm of the egg.

DIF: Easy REF: 19.2 OBJ: 19.2.o Describe where and how sperm and egg unite during fertilization, and outline the mechanisms that prevent multiple sperm from entering the egg. MSC: Understanding

22. Do you agree or disagree with the following statement? Explain your answer.

If a diploid organism has 16 chromosomes (and thus 8 pairs of homologous chromosomes), that organism can produce only 28 genetically different gametes.

ANS: Disagree.

There are two distinct mechanisms for genetic variation in gametes. One method, the reassortment of chromosomes during meiosis, would result in a diploid organism with 16 chromosomes producing 2 8 genetically different gametes.

However, a much greater number of genetically different gametes can be produced as a result of the recombination between chromosomes that occurs during every meiosis.

DIF: Moderate REF: 19.2 OBJ: 19.2.m Outline the forms of genetic reassortment that give rise to new chromosome combinations during meiosis. MSC: Evaluating 23. Which of the following statements about Mendel’s experiments is FALSE? a. The pea plants could undergo both cross-fertilization and self-fertilization. b. The true-breeding strains were homozygous for the traits that Mendel examined. c. The egg can carry either the allele from the maternal or the paternal chromosome. d. All traits that Mendel studied were recessive.

ANS: D The traits that Mendel studied were inherited in a discrete fashion. For this to occur, for each pair of alleles, one allele is necessarily dominant and the other is recessive.

DIF: Easy REF: 19.3 OBJ: 19.3.a Explain why Mendel chose to study inheritance using the traits and organism he selected. | 19.3.b Define what it means for an organism to be true-breeding, and describe the appearance of the offspring Mendel observed when crossbreeding two true-breeding plants. MSC: Understanding 24. Figure 19-24 diagrams one of Mendel’s experiments using the round and wrinkled seed traits.

Figure 19-24

Which of the following could be considered a true-breeding strain for the seed-shape phenotype? a. all of the round-seeded plants produced in the F2 generation b. all of the wrinkle-seeded plants produced in the F2 generation c. all of the round-seeded plants produced in the F1 generation d. half of the round-seeded plants produced in the F2 generation ANS: B DIF: Moderate REF: 19.3 OBJ: 19.3.b Define what it means for an organism to be true-breeding, and describe the appearance of the offspring Mendel observed when crossbreeding two true-breeding plants.

MSC: Understanding 25. Figure 19-24 diagrams one of Mendel’s experiments using the round and wrinkled seed traits.

Figure 19-24

If you crossed the round-seeded plants obtained in the F1 generation with a true-breeding strain of round-seeded plants, how many wrinkle-seeded plants would you expect to obtain in the next generation? a. none b. 25% c. 75% d. all ANS: A If R is the allele for round seeds and r is the allele for wrinkled seeds, Rr × RR would yield all round-seeded plants, because R is dominant.

DIF: Moderate REF: 19.3 OBJ: 19.3.b Define what it means for an organism to be true-breeding, and describe the appearance of the offspring Mendel observed when crossbreeding two true-breeding plants.

MSC: Understanding 26. Figure 19-24 diagrams one of Mendel’s experiments using the round and wrinkled seed traits.

Figure 19-24

If you crossed the round-seeded plants obtained in the F1 generation with a true-breeding strain of wrinkle-seeded plants, how many round-seeded plants would you expect to obtain in the next generation? a. 25% b. 50% c. 75% d. 100% ANS: B If R is the allele for round seeds and r is the allele for wrinkled seeds, Rr × rr would be expected to yield 50% roundseeded plants and 50% wrinkle-seeded plants.

DIF: Moderate REF: 19.3 OBJ: 19.3.b Define what it means for an org anism to be true-breeding, and describe the appearance of the offspring Mendel observed when crossbreeding two true -breeding plants.

MSC: Understanding 27. Figure 19-24 diagrams one of Mendel’s experiments using the round and wrinkled seed traits.

Figure 19-24

Which of the following statements about the round-seeded pea plants obtained in the F2 generation is FALSE? a. These plants are phenotypically identical for seed shape. b. Two-thirds of these plants are expected to be heterozygous for the seed-shape allele. c. We expect 25% of these plants to be homozygous for the seed-shape allele. d. If these plants were crossed to wrinkle-seeded plants, some of these plants would produce only round-seeded plants.

ANS: C One-third of these plants are expected to be homozygous for the seed-shape allele.

DIF: Moderate REF: 19.3 OBJ: 19.3.f State Mendel’s law of segregation and explain how it gives rise to the phenotypic ratios he observed in the F2 generation. MSC: Understanding 28. Figure 19-24 diagrams one of Mendel’s experiments using the round and wrinkled seed traits.

Figure 19-24

Which of the following statements about the experiment diagrammed in Figure 19-28 is TRUE? a. If you crossed all the round-seeded pea plants from the F2 generation with true-breeding wrinkle-seeded pea plants, you would get more round-seeded pea plants in the next generation than if you crossed all the roundseeded pea plants from the F1 generation with true-breeding wrinkle-seeded pea plants. b. The reason you do not see wrinkle-seeded pea plants in the F1 generation is because the round-seeded pea plants used to create the F1 generation were not true-breeding strains. c. The gene for round-seeded pea plants is on a different chromosome from the gene for wrinkle-seeded pea plants, which is why you get 25% wrinkle-seeded pea plants in the F2 generation. d. If you crossed the round-seeded pea plants from the F2 generation with the wrinkle-seeded pea plants from the F2 generation, you should get 100% round-seeded pea plants.

ANS: A If you crossed the round-seeded pea plants from the F2 generation with true-breeding wrinkle-seeded pea plants, you would get two-thirds round-seeded pea plants. On the other hand, if you crossed the round-seeded pea plants from the F1 generation with true-breeding wrinkle-seeded pea plants, you would get one-half round-seeded pea plants.

DIF: Moderate REF: 19.3 OBJ: 19.3.f State Mendel’s law of segregation and explain how it gives rise to the phenotypic ratios he observed in the F2 generation. MSC: Understanding 29. Which of the following reasons was essential for Mendel to disprove the theory of blended inheritance? a. The traits that Mendel examined all involved genes that did not display linkage. b. The traits that Mendel examined all involved the reproductive structures of the pea plant. c. Mendel pioneered techniques permitting the fusion of male and female gametes from the same plant to produce a zygote.

d. The traits that Mendel examined involved an allele that was dominant and an allele that was recessive.

ANS: D To see the traits disappear in the F1 generation and reappear in the F2 generation, one of the alleles for each trait needed to be dominant while the other was recessive. Although it is true that the traits Mendel examined involved unlinked genes that were involved in reproductive structures, these characteristics were not necessary for Mendel to disprove the theory of blended inheritance. The plants that Mendel worked with had both male and female reproductive structures and could self-fertilize naturally.

DIF: Easy REF: 19.3 OBJ: 19.3.d Describe the observations that led Mendel to postulate that traits are specified by pairs of alleles, one of which is dominant, the other recessive. MSC: Understanding 30. Which of the following reasons was essential for Mendel’s law of independent assortment? a. All the traits that Mendel examined involved genes that did not display linkage. b. Several of the phenotypes that Mendel examined involved color. c. Mendel observed chromosomal segregation in pea-plant cells. d. Mendel carried out his experiments on plants and not on fungi.

ANS: A To see the 9:3:3:1 segregation from a dihybrid cross, the two traits cannot be linked. Although it is true that several phenotypes Mendel examined involved color, this characteristic was not necessary for Mendel to determine the law of independent assortment. Mendel did not observe chromosomal segregation. The law of independent assortment holds true in animal, fungal, and plant cells.

DIF: Easy REF: 19.3 OBJ: 19.3.h Outline the type of breeding experiment that led to Mendel’s law of independent assortment. MSC: Understanding 31. Loss-of-function mutations a. cause the production of proteins that are active in inappropriate circumstances. b. will usually show a phenotype when heterozygous. c. are only present in a population at barely detectable levels. d. are usually recessive.

ANS: D DIF: Easy REF: 19.3 OBJ: 19.3.k Explain why loss-of-function mutations can be recessive or dominant.

MSC: Remembering 32. Which of the following statements about conditional alleles is FALSE? a. Conditional alleles allow for the study of lethal mutations in haploid organisms. b. Conditional alleles are only defective under high temperature conditions. c. Conditional alleles can be used to study tissue-specific gene function in diploid organisms. d. Conditional alleles behave like the wild-type allele under permissive conditions.

ANS: B Although some conditional alleles are defective at high temperature, conditional alleles can be produced that are defective under different types of conditional. For example, there are cold-sensitive alleles that are defective at lower temperatures.

DIF: Easy REF: 19.4 OBJ: 19.4.a Outline the steps involved in the classical genetic approach to studying gene activity, and explain how mutagens can be used to speed the process. MSC: Understanding

33. You conduct a genetic screen on peas and isolate four mutant strains, each carrying a recessive mutation that causes the production of red peas (instead of the wild-type green peas). To test whether these mutations are in the same gene, you perform complementation tests between the four different true-breeding strains. The results from these complementation tests are shown in Table 19-33.

Table 19-33

Given this data, how many genes do these four alleles represent? a. 1 b. 2 c. 3 d. 4 ANS: B Strains 1 and 4 are mutated in the same gene. Strains 2 and 3 are mutated in the same gene; this gene is different from the gene that is defective in strains 1 and 4.

DIF: Moderate REF: 19.4 OBJ: 19.4.d Review how a complementation test can be used to determine whether two mutations affect the same gene. MSC: Applying 34. Haplotype blocks can be seen in humans because a. disease genes are found in haplotype blocks. b. modern humans descended from a relatively small population of about 10,000 individuals that existed about 2,000 generations ago. c. human germ cells do not undergo recombination. d. new mutations cannot be introduced into existing haplotype blocks.

ANS: B The relatively small population of human ancestors means that the chromosomes in modern-day humans are a shuffled set of a relatively small number of chromosome sets. Because this small population existed only about 2,000 generations ago, there has not been enough time for recombination to scramble these haplotype blocks. Disease genes can certainly be found in haplotype blocks, although this is not why haplotype blocks exist. Human germ cells do undergo recombination during meiosis. New mutations can be introduced into an existing haplotype block.

DIF: Easy REF: 19.5 OBJ: 19.5.c Explain why haplotype blocks are present in the human genome.

MSC: Understanding 35. The single-nucleotide polymorphisms found in the human population a. are important for genetic mapping because they represent mutations in genes important for human disease.

b. are rarely found among blood relatives. c. can be linked into haplotype blocks. d. arose mainly during the past 10,000 years.

ANS: C Haplotype blocks are segments of chromosomes spanning a set of single-nucleotide polymorphisms that are linked and tend to be inherited as a unit.

DIF: Easy REF: 19.5 OBJ: 19.5.a Define polymorphism and list the most common types of polymorphisms in the human genome. MSC: Remembering 36. Which of the following statements about genome-wide association studies (GWAS) is FALSE? a. GWAS use SNPs to compare populations of people with disease and people without disease to look for SNPs more likely to be present in those with disease. b. GWAS can be used even if more than one gene can cause the disease of interest. c. Sometimes GWAS will identify SNPs that are associated with a disease, but these SNPs do not affect the gene product of the gene that causes the disease. d. Studies using GWAS only examine SNPs that occur very rarely (<0.001%) in the population, as those SNPs are most likely to cause disease.

ANS: D GWAS are best at detecting SNPs that are common in the human population. However, because they are so common, these variants are likely to alter disease susceptibility slightly, as otherwise they would have been strongly selected against in the population during evolution (and therefore rare). Significant differences in the inheritance of a very rare SNP will be difficult to determine in a GWAS unless the number of people used in the study is extremely large.

DIF: Easy REF: 19.5 OBJ: 19.5.k Outline how genome-wide association studies can be used to search for genes that predispose individuals to common diseases. MSC: Understanding 37. Finding co-inheritance of an SNP variant and a disease tells scientists that a. everybody who carries this SNP will get the disease. b. sequences within the SNP cause the disease. c. a gene important for causing the disease is linked to the SNP. d. SNPs on other chromosomes will not be co-inherited with the disease.

ANS: C DIF: Easy REF: 19.5 OBJ: 19.5.l Review how SNPs can be used to produce a genetic linkage map and to facilitate the search for alleles that predispose to disease. MSC: Understanding

MATCHING 1. Indicate whether each of the following is TRUE for (A) meiosis, (B) mitosis, (C) both, or (D) neither.

1. formation of a bivalent ANS: A DIF: Easy REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. MSC: Understanding 2. genetically identical products

ANS: B DIF: Easy REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. MSC: Understanding 3. condensation of chromosomes ANS: C DIF: Easy REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. MSC: Understanding 4. segregation of all paternal chromosomes to one cell ANS: D DIF: Easy REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. MSC: Understanding 5. involvement of DNA replication ANS: C DIF: Easy REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. MSC: Understanding

SHORT ANSWER 1. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase can be used only once. allele

germ

pollen

bivalent

meiosis

somatic

eggs

mitosis

sperm

gametes

pedigree

zygote

To reproduce sexually, an organism must create haploid __________ cells, or __________, from diploid cells via a specialized nuclear division called __________. During mating, the father’s haploid cells, called __________ in animals, fuse with the mother’s haploid cells, called __________. Cell fusion produces a diploid cell called a __________, which undergoes many rounds of cell division to create the entire body of the new individual. The cells produced from the initial fusion event include __________ cells that form most of the tissues of the body as well as the __________-line cells that give rise to the next generation of progeny.

ANS: To reproduce sexually, an organism must create haploid germ cells, or gametes, from diploid cells via a specialized nuclear division called meiosis. During mating, the father’s haploid cells, called sperm in animals, fuse with the mother’s haploid cells, called eggs. Cell fusion produces a diploid cell called a zygote, which undergoes many rounds of cell division to create the entire body of the new individual. The cells produced from the initial fusion event include somatic cells that form most of the tissues of the body as well as the germ-line cells that give rise to the next generation of progeny.

DIF: Easy REF: 19.1 OBJ: 19.1.b Distinguish between somatic cells, germline cells, and gametes in terms of function and whether the cells are diploid or haploid. | 19.1.c Compare sperm, egg, and zygote in terms of appearance and chromosome complement. MSC: Understanding 2. Why is sexual reproduction more beneficial to a species living in an unpredictable environment than to one living in a constant environment?

ANS: The real benefit in sexual reproduction seems to be that parents produce children that are genetically unlike either parent and that are not genetically identical to each other. Sexual reproduction provides more variation in the

population than asexual reproduction could provide, and is an advantage if the environment is variable, because any one combination of the parents’ characteristics, however well adapted to the prevailing conditions, may or may not be the best in a new situation.

DIF: Easy REF: 19.1 OBJ: 19.1.f Summarize the potential advantages conferred by sexual reproduction.

MSC: Understanding 3. Is the following statement TRUE or FALSE? Explain.

Somatic cells leave no progeny and thus, in an evolutionary sense, exist only to help create, sustain, and propagate the germ cells.

ANS: True. Only germ cells contribute genetic material to the next generation of organisms, and thus only germ cells leave an evolutionary legacy in the gene pool of the species. The only contribution of somatic cells to subsequent generations arises through the assistance they provide to dissemination of the genetic material in the germ cells. In accordance with this, mutations that arise in somatic cells are not passed along to offspring.

DIF: Easy REF: 19.1 OBJ: 19.1.b Distinguish between somatic cells, germline cells, and gametes in terms of function and whether the cells are diploid or haploid. MSC: Understanding 4. Sexual reproduction is a large drain on the limited resources of an individual. Nonetheless, sexual reproduction is common. In fact, to allow sexual reproduction, organisms have evolved many elaborate anatomical structures, cellular processes, and chemical signals. For example, flowers exist entirely to further the goal of sexual reproduction, and many plants have enlisted the help of bees and birds to aid in the dissemination of their germ cells. Describe one reason why most multicellular organisms have evolved to reproduce sexually instead of relying solely on asexual reproduction.

ANS: A definitive explanation of the evolutionary advantages of sexual reproduction over asexual reproduction is elusive, but several benefits seem clear.

1. The reshuffling of genes that occurs during sexual reproduction generates most of the diversity between organisms within a species. Having a large variety of different genetic combinations in a population may help guarantee that at least a few individuals will survive after a sudden and unpredictable change in the environment. This is why some ecologists are concerned about the global trend toward monoculture; namely, limiting cultivation to only a small number of varieties of each species of plant, each variety being inbred and genetically uniform.

2. The competition between males for the available females may increase the frequency of advantageous alleles in the gene pool while efficiently eliminating deleterious combinations of alleles. This is because, in many species, only the strongest and healthiest males succeed in mating, whereas other males, carrying less successful combinations of alleles, leave no progeny.

3. Sexual reproduction may help to eliminate deleterious recessive genes from a population in another way also: a recessive deleterious mutation in a gene may be “unmasked” in a haploid germ cell because, in contrast with a heterozygous diploid cell, the haploid cell carrying the mutation contains no normal allele to provide the gene function. Because there is competition between germ cells for fertilizing partner cells, gametes with recessive deleterious mutations are less likely to form zygotes.

DIF: Easy REF: 19.1 OBJ: 19.1.f Summarize the potential advantages conferred by sexual reproduction.

MSC: Understanding 5. Meiosis is a highly specialized nuclear division in which several events occur in a precisely defined order. Please

order the meiotic events listed below.

1. loss of cohesins near centromeres 2. chromosome replication 3. degradation of cohesins bound to chromosome arms 4. formation of chiasmata (chiasmata = plural of chiasma) 5. homolog pairing 6. alignment of chromosomes at the metaphase plate ANS: 2, 5, 4, 6, 3, 1 Chromosome replication (2) occurs during meiotic S phase. During meiotic G2, or prophase, the two homologous chromosomes pair (5) and undergo recombination to produce reciprocal exchanges of genetic material that are visible as chiasmata (4). Meiotic division I begins with chromosome condensation and proceeds to metaphase, when chromosomes align at a central plate (6). Anaphase of meiotic division I is triggered by loss of the cohesin glue on the chromosome arms (3), which allows the homologs to be segregated. Anaphase of meiotic division II is triggered by degradation of cohesins near the centromere (1), which allows sister chromatids to be segregated.

DIF: Easy REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. | 19.2.c Review the process of homolog pairing and explain its importance in meiosis. | 19.2.h Describe a chiasma and indicate how many are typically found in each bivalent. | 19.2.k Differentiate how cohesins are degraded during anaphase in meiosis I and II. MSC: Understanding 6. For each of the following sentences, choose one of the options enclosed in square brackets to make a correct statement.

Starting with a single diploid cell, mitosis produces [two/four] [identical/different] [haploid/diploid] cells, whereas meiosis yields [two/four] [identical/different] [haploid/diploid] cells. This is accomplished in meiosis because a single round of chromosome [replication/segregation] is followed by two sequential rounds of [replication/segregation]. Mitosis is more like meiosis [I/II] than meiosis [I/II]. In meiosis I, the kinetochores on sister chromatids behave [independently/coordinately] and thus attach to microtubules from the [same/opposite] spindle.

The cohesin-mediated glue between [chromatids/homologs] is regulated differently near the centromeres than along the chromosome arms. Cohesion is lost first at the [centromeres/arms] to allow segregation of [chromatids/homologs] and is lost later at the [centromeres/arms] to trigger segregation of [chromatids/homologs].

ANS: Starting with a single diploid cell, mitosis produces two identical diploid cells, whereas meiosis yields four different haploid cells. This is accomplished in meiosis because a single round of chromosome replication is followed by two sequential rounds of segregation. Mitosis is more like meiosis II than meiosis I. In meiosis I, the kinetochores on sister chromatids behave coordinately and thus attach to microtubules from the same spindle. The cohesin-mediated glue between chromatids is regulated differently near the centromeres than along the chromosome arms. Cohesion is lost first at the arms to allow segregation of homologs and is lost later at the centromeres to trigger segregation of chromatids.

DIF: Easy REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. MSC: Understanding 7. Imagine a diploid sexually reproducing organism, Diploidus sexualis, that contains three pairs of chromosomes. This organism is unusual in that no recombination between homologous chromosomes occurs during meiosis.

A. Assuming that the chromosomes are distributed independently during meiosis, how many different types of sperm or egg cells can a single individual of this species produce?

B. What is the likelihood that two siblings of this species will be genetically identical? You can assume that the homologous chromosomes of each parent are different from one another and from their counterparts in the other parent.

ANS: A. Eight different types With respect to each of the three chromosomes, an individual can produce two kinds of gamete. The gamete can receive the copy that the individual received from “Mom” or the copy from “Dad.” Thus, with three chromosomes, there are 2 × 2 × 2 = 8 possible gametes.

B. 1/64 The mother and father together can produce 8 × 8 = 64 different genetic combinations.

DIF: Hard REF: 19.2 OBJ: 19.2.b Describe the products of meiosis in terms of chromosomal complement and genetic similarity with the parental germline cell. MSC: Applying 8. You have received exactly half of your genetic material from your mother, who received exactly half of her genetic material from her mother (your grandmother).

A. Explain why it is unlikely that you share EXACTLY one-quarter of your genetic material with your grandmother, and instead it is more accurate to say that in general, people receive an AVERAGE of one-quarter of their genetic endowment from each grandparent.

B. Consider a gene on Chromosome 3 that you received from your grandmother. Is it likely you received an entire Chromosome 3 from your grandmother? Why or why not?

C. What portion of your genetic material do you share with your sibling? Your aunt? Your cousin?

ANS: A. It is unlikely that you share exactly one-quarter of your genetic material with your grandmother, but it is true that organisms receive an average of one-quarter of their genes from each grandparent. When cells in your mother’s germ line were undergoing meiosis, the chromosomes that she received from your grandmother and your grandfather were shuffled by recombination and then randomly assorted into the gametes. Each of these gametes probably received slightly less or slightly more than half of its genetic material from your grandmother. Thus, fusion of your mother’s gamete with your father’s created a zygote that shared approximately, but not exactly, one-quarter of its genes with your grandmother; this zygote divided repeatedly to form all the cells in your body.

B. It is unlikely that you received the entire Chromosome 3 from your grandmother, because each chromosome undergoes at least one recombinational crossover with its homolog to ensure proper chromosome segregation in meiotic division I. C. You share an average of one-half of your genetic material with your sibling, one-quarter with your aunt, and oneeighth with your cousin. Because you and your sibling each get half of each parent’s DNA, you have approximately half of your genetic material in common with your sibling. As your mother and her sister share approximately half of their genetic material, you and your aunt share 1/2 × 1/2 = 1/4 of your genetic material. And as your cousin has half of your aunt’s genetic material, you and your cousin share 1/2 × 1/4 = 1/8 of your genetic material.

DIF: Hard REF: 19.2 OBJ: 19.2.b Describe the products of meiosis in terms of chromosomal complement and genetic similarity with the parental germline cell. MSC: Analyzing

9. Meiosis includes a recombination checkpoint that is analogous to the checkpoints in cell-cycle progression. Doublestrand breaks in the DNA initiate recombination in meiosis. The broken end of a DNA molecule finds the corresponding sequence on a homologous chromosome and exchanges a chromosomal segment with its homolog, thereby repairing the break. Ongoing recombination sends a negative regulatory signal that prevents cells from entering meiotic division I. A. Mutations in several genes inactivate the recombination checkpoint. What do you predict will happen if a cell proceeds through meiotic division I before completing recombination?

B. What will happen if a cell fails to initiate recombination and proceeds through meiotic division I? Meiotic division II? ANS: A. If a cell proceeds through meiotic division I when its chromosomes are broken and incompletely repaired, segregation will be disastrous. Some recombination intermediates will be pulled to opposite spindle poles, thus breaking the DNA. Other chromosome fragments lacking centromeres will not be attached to microtubules and thus will segregate randomly, causing some meiotic products to have too little DNA and others to have too much.

B. In the absence of recombination, the homologs will not segregate from each other properly in meiotic division I but the sister chromatids will segregate normally in meiotic division II. In meiotic division I, the unrecombined homologous chromosomes will not be held together by chiasmata. Thus, the homologous chromosomes will line up independently on the metaphase plate and segregate randomly, causing chromosome nondisjunction. Some products will have both homologs of a given chromosome and others will have none. In meiotic division II, the events will proceed normally and sister chromatids will be properly segregated to opposite poles. Nonethele ss, because the chromosome sets at the start of meiotic division II were distributed unevenly, the gametes produced after meiotic division II will be aneuploid (that is, they will have an incorrect number of chromosomes).

DIF: Hard REF: 19.2 OBJ: 19.2.f Outline the main events that take place during meiotic homologous recombination.

MSC: Applying 10. In some fungi, cell division during meiosis gives rise to an ordered spore sac containing a row of four haploid spores, as shown in Figure 19-10A. The position of each spore within the sac reflects its relation to its neighbors; in other words, spores that result from the same meiosis II division are positioned next to each other. You notice that a strain of the fungus produced by crossing a dark-colored strain with a light-colored strain gives rise mostly to spore sacs as shown in Figure 19-10B, with a few spore sacs like those in Figure 19-10C. Indicate whether the following statements are correct. Explain your reasoning.

Figure 19-10

A. Meiosis I and meiosis II in the fungus occur in the reverse order from that which occurs in humans.

B. Recombination in the fungus can occur during prophase I. C. Recombination in the fungus cannot occur during prophase II. D. The spore sacs in Figure 19-10C result from recombination between the centromere and the gene responsible for spore color.

E. Recombination has occurred between the gene responsible for spore color and the end of the chromosome arm.

ANS: During normal meiosis, the homologs are separated during meiotic division I and the sister chromatids are separated during meiotic division II. This pattern of chromosome segregation, in the absence of recombination, always gives the pattern of spores that was shown in the question (Figure 19-10B). To yield the unusual spore sacs, there must be recombination between the two homologous chromosomes. Such recombination always occurs when both homologs are in the same cell (that is, in prophase I of meiosis).

A. Incorrect. If meiosis I and meiosis II were reversed, we would expect a completely random pattern of spores in spore sacs, because the two homologs could go to either cell in the second division, as shown in Figure 19-10C. A reversal of meiosis I and II cannot explain the observation that most of the spore sacs appear like those in Figure 19-10B. B. Correct. The unusual spore sacs must result from recombination between the homologs carrying different alleles of the color gene. Recombination between the homologs occurs during prophase I. C. Not a valid deduction. The results shown cannot distinguish whether recombination has occurred during prophase II. During prophase II, there is only one homolog in each cell; if recombination took place between the two sister chromatids of this homolog we would not be able to detect it: if there had been no recombination in meiosis I, the sister chromatids would be identical and crossing-over between them would cause no genetic change; and if recombination had occurred in meiosis I, further recombination in meiosis II would still lead only to the same three types of spore sac shown in Figure 19-10C. D. Correct. Segregation of chromosomes is mediated by microtubules that pull on the centromeres of chromosomes.

Thus, unusual segregation patterns arise from recombination (and thus the loss of a covalent connection) between

the centromere and the gene of interest.

E. Incorrect. If recombination had taken place only between the gene for spore color and the end of that chromosome arm, it would look as though no recombination had taken place (Figure 19-10B) because the centromere would still be covalently connected to the gene of interest. (See also the answer to part D.) DIF: Hard REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. | 19.2.l Describe the events that take place between the first and second meiotic divisions. | 19.2.m Outline the forms of genetic reassortment that give rise to new chromosome combinations during meiosis. MSC: Analyzing 11. In mammals, there are two sex chromosomes, X and Y, which behave like homologous chromosomes during meiosis.

Normal males have one X chromosome and one Y chromosome, and normal females have two X chromosomes.

Males with an extra Y chromosome (XYY) are found occasionally. Which of the following could give rise to such an XYY male? Explain your answer.

A. nondisjunction in the first meiotic division of spermatogenesis; normal meiosis in the mother B. nondisjunction in the second meiotic division of spermatogenesis; normal meiosis in the mother C. nondisjunction in the first meiotic division of oogenesis; normal meiosis in the father D. nondisjunction in the second meiotic division of oogenesis; normal meiosis in the father ANS: B Nondisjunction in one or the other meiotic division will give rise to gametes with two sex chromosomes instead of one.

Because the only source of Y chromosomes is the father, the gametes that produce an XYY male must be X (a normal egg) and YY (an abnormal sperm). Nondisjunction in the second meiotic division of spermatogenesis could give rise to a YY gamete that fertilizes a normal egg and thereby creates an individual with the karyotype XYY. Nondisjunction in the first meiotic division of spermatogenesis would result in two XY sperm and two sperm with no sex chromosome.

Nondisjunction during oogenesis cannot yield an individual with two Y chromosomes.

DIF: Moderate REF: 19.2 OBJ: 19.2.n Explain how nondisjunction gives rise to aneuploid gametes, and recall the consequences of this type of genetic error. MSC: Applying 12. When a reciprocal translocation occurs, part of one chromosome is exchanged with a part of another chromosome.

For example, one-half of Chromosome 3 may now be found fused to Chromosome 10, and part of Chromosome 10 is now found fused to Chromosome 3. In a balanced reciprocal translocation, an even exchange of material occurs such that no genetic information is extra or missing. Individuals can carry balanced reciprocal translocations and be quite healthy. Consider the case where a gamete containing a balanced reciprocal translocation of a single chromosome is used to fertilize a genetically normal egg. Explain why individuals carrying a single balanced reciprocal translocation might have problems with chromosome segregation during meiosis but not in mitosis.

ANS: During meiosis, homologous chromosomes pair to form bivalents. The chromosomes with the translocations will have difficulty pairing with the homologous chromosomes of normal structure during meiosis. However, during mitosis, there is no need for the homologous chromosomes to interact. DIF: Moderate REF: 19.2 OBJ: 19.2.c Review the process of homolog pairing and explain its importance in meiosis. MSC: Applying 13. Is the following statement TRUE or FALSE? Explain.

The phenotype of an organism reflects all of the alleles carried by that individual.

ANS: False. The phenotype, or the observable traits, of an organism often does not fully reflect its genotype, or the

catalog of all alleles in the chromosomes. The reason is that some alleles are dominant (call these A) and other alleles are recessive (call these a). An individual who is heterozygous (Aa) for a dominant allele will have the same phenotype as one who is homozygous (AA). DIF: Easy REF: 19.6 OBJ: 19.3.c Distinguish genotype and phenotype.

MSC: Understanding 14. With respect to gene E on the chromosome drawn in Figure 19-14, which gene is least likely to behave according to Mendel’s law of independent assortment? Explain your answer.

Figure 19-14

ANS: Gene F is least likely to behave according to Mendel’s law of independent assortment because it lies the closest to gene E and is least likely to be separated from gene E by recombination. DIF: Moderate REF: 19.3 OBJ: 19.3.j Review how genes present on the same chromosome can segregate independently. MSC: Applying 15. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase may be used more than once. allele

genotype

monohybrid

chromosome

heterozygous

pedigree

dependent

homozygous

phenotype

dihybrid

independent

segregation

Gregor Mendel studied pea plants and developed some very important ideas about how genes are inherited. These studies used plant strains that were true-breeding and always produced progeny that had the same __________ as the parent. These strains were true breeding because they were __________ for the gene important for a specific trait. In other words, for these true-breeding strains, both chromosomes in the diploid pea plant carried the same __________ of the gene. Mendel started out examining the inheritance of a single trait at a time, and then moved on to examining two traits at once in a __________ cross. His studies examining the inheritance of two traits in one cross allowed him to discover what is now known as Mendel’s law of __________ assortment. Geneticists can study the inheritance of specific traits in humans by analyzing a __________, which shows the phenotypes of different family members over several generations for a particular trait.

ANS: Gregor Mendel studied pea plants and developed some very important ideas about how genes are inherited.

These studies used plant strains that were true-breeding and always produced progeny that had the same phenotype as the parent. These strains were true breeding because they were homozygous for the gene important for a specific trait. In other words, for these true-breeding strains, both chromosomes in the diploid pea plant carried the same al-

lele of the gene. Mendel started out examining the inheritance of a single trait at a time, and then moved on to examining two traits at once in a dihybrid cross. His studies examining the inheritance of two traits in one cross allowed him to discover what is now known as Mendel’s law of independent assortment. Geneticists can study the inheritance of specific traits in humans by analyzing a pedigree, which shows the phenotypes of different family members over several generations for a particular trait.

DIF: Easy REF: 19.3 OBJ: 19.3.b Define what it means for an organism to be true-breeding, and describe the appearance of the offspring Mendel observed when crossbreeding two true-breeding plants. | 19.3.c Distinguish genotype and phenotype. | 19.3.g Explain how pedigrees can be used to study the inheritance of recessive traits in human populations. MSC: Understanding 16. Cystic fibrosis results from mutations in a single gene that lies on Chromosome 7. Only homozygous mutant (ff) individuals are sick; homozygous wild-type (FF) and heterozygous (Ff) individuals are healthy. A healthy married couple has one child with cystic fibrosis and the wife is pregnant with a second child.

A. What is the genotype of the mother? The father?

B. What is the chance that the second child will have cystic fibrosis?

ANS: A. The genotypes of the mother and father are the same: Ff. The only way that a child can have the disease is if both parents are carriers of the mutant cystic fibrosis gene.

B. The chance that the second child will have cystic fibrosis is one-quarter. The chance that the mother will transmit her mutant f allele to the offspring is one-half, multiplied by an equal chance that the father will transmit his mutant f allele to the offspring: 1/2 × 1/2 = 1/4. (Having one child with cystic fibrosis does not change the probability of having another child with the disease.)

DIF: Easy REF: 19.3 OBJ: 19.3.c Distinguish genotype and phenotype. | 19.3.e Define heterozygosity and homozygosity and describe how they relate to phenotype. MSC: Applying 17. Sickle-cell anemia is caused by a mutant allele of a hemoglobin gene. Individuals with two mutant alleles have sicklecell anemia. Individuals homozygous and heterozygous for the mutant gene are more resistant to malaria than those with two wild-type alleles. Given this information, would you classify this mutant allele as behaving as a dominant, recessive, or codominant allele?

ANS: The classification of all mutations depends on the phenotypic feature under consideration. With regard to the sickle-cell anemia phenotype, the mutant allele is recessive because a heterozygous individual has the same healthy phenotype as a homozygous wild-type individual. With regard to the malaria phenotype, the mutant allele is dominant because a heterozygous individual has the same phenotype as a homozygous mutant individual; namely, resistance to malaria. With regard to the phenotype as a whole, the two alleles could be said to be co-dominant. DIF: Easy REF: 19.5 OBJ: 19.5.e Describe how genetic and environmental factors can interact to give rise to a “disease phenotype.” MSC: Applying 18. You are given two true-breeding strains of hamster. One strain has white fur color and the other has a dark brown fur color. When you cross the white fur strain to the dark brown–fur strain, you obtain F1 progeny that have a light brown fur color. When you cross the F1 progeny with each other, 25% of the F2 generation have white fur, 25% have dark brown fur, and 50% have light brown fur. How many genes crucial for fur coloration differ between the two starting strains? Explain your answer.

Extra credit: Propose a molecular mechanism for how fur color is determined in this species of hamster.

ANS: One gene involved in determining fur color differs between these two strains. The white allele is not fully recessive to the dark brown allele. Therefore, any animal that is heterozygous (and has one white allele and one dark brown allele) will have a light brown fur color. Thus, the white allele and the dark brown allele are co-dominant. When two heterozygous animals are crossed together, 25% of the offspring will be homozygous for one allele, 25% will be homozygous for the other allele, and 50% will be heterozygous, exactly what is observed in the cross you performed.

Extra credit: Any reasonable answer is fine. There are many possible answers, including the following.

1. The dark brown and white alleles are alleles of a gene involved in synthesizing pigment. Hamsters with the white allele do not produce any pigment; hamsters with the dark brown allele produce pigment. When hamsters are heterozygous, they have one chromosome that carries the white allele and does not produce any pigment. Therefore, they produce half the amount of pigment needed normally and thus are light brown instead of dark brown.

2. The dark brown and white alleles are alleles of a gene involved in a microtubule motor that is important for transporting the pigment to the correct place in the cell. Hamsters with the white allele produce a defective version of the microtubule motor and thus no pigment is properly transported, leading to white fur. The hamsters with the dark brown allele produce a properly functioning microtubule motor. When hamsters are heterozygous, they have one chromosome that carries the white allele and one chromosome that carries the dark brown allele; therefore, only half the amount of functioning microtubule motor is being produced in the heterozygote. Because the number of microtubule motors is limiting in the cell, half the number of properly functioning microtubule motors can only transport half the amount of pigment, leading to a light brown fur.

DIF: Hard REF: 19.3 OBJ: 19.3.e Define heterozygosity and homozygosity and describe how they relate to phenotype. | 19.3.f State Mendel’s law of segregation and explain how it gives rise to the phenotypic ratios he observed in the F2 generation. MSC: Applying 19. Conditional alleles are mutant gene versions encoding proteins that can function normally at the permissive condition but are defective at the restrictive condition. One commonly used condition is temperature. Conditional alleles are especially useful to geneticists because they permit the study of essential genes. At the permissive temperature, the organism lives normally. When the organism is shifted to the nonpermissive temperature, the effect of inactivating the gene can be studied. Which of the three types of mutation shown in Figure 19-19 is most likely to lead to a conditional allele? Explain your answer.

Figure 19-19

ANS: A single nucleotide substitution is most likely to cause a conditional allele. Nucleotide addition and nucleotide deletion will usually shift the reading frame of the protein, leading to a string of many amino acid substitutions (and an altogether different protein) and possibly even the formation of a stop codon, which would result in the premature truncation of the protein. In contrast, a single nucleotide substitution would be more likely to result in an amino acid substitution that would disrupt protein function in a conditional manner. For example, this amino acid substitution may af-

fect protein folding at high temperature as a result of a greater charge; at the permissive temperature, this protein may still fold properly and function normally.

DIF: Hard REF: 19.4 OBJ: 19.4.a Outline the steps involved in the classical genetic approach to studying gene activity, and explain how mutagens can be used to speed the process. MSC: Applying 20. You are studying a diploid yeast strain that normally uses glucose as an energy source but can use maltose when no glucose is present. You are interested in understanding how this yeast strain metabolizes maltose as an alternative energy source. You isolate the genes involved in maltose metabolism by screening for yeast that cannot grow when maltose is the sole energy source. You find six different mutants, all of which are recessive, and name these alleles mal1, mal2, mal3, mal4, mal5, and mal6. Next, you isolate gametes from the homozygous diploid mutant yeast strains and perform crosses between the different strains to do complementation analysis, because you wish to determine whether the mutations are likely to affect the same or different genes. Your results are shown in Table 19-20.

Table 19-20

In how many genes are you likely to have isolated mutations? Which alleles seem to affect the same genes? Explain your answer.

ANS: Three genes, mal1, mal2, and mal6, are mutations in one gene; mal3 and mal5 are mutations in a second gene; and mal4 is the only mutation you identified in a third gene. These results can be deduced from the complementation analysis. For example, the diploid yeast containing the mal1 and mal6 alleles cannot grow on maltose-containing medium, and therefore these are likely to be alleles of the same gene. However, the diploid yeast containing the mal1 and the mal4 alleles do grow on maltose-containing medium, and these are therefore likely to be alleles of different genes. DIF: Hard REF: 19.4 OBJ: 19.4.b Describe how genetic screens are used to identify mutants with a phenotype of interest. | 19.4.d Review how a complementation test can be used to determine whether two mutations affect the same gene. MSC: Applying 21. Gene A is located near gene B on Chromosome 13 in humans. A mutation in the germ line of an individual with the haplotype AB generates gametes with the genotype Ab. Many descendants of this founder individual carry the b mutation, which predisposes carriers to high blood pressure. Initially, all descendants who inherit the b mutation also inherit the neighboring A allele. Through the generations, fewer and fewer descendants with the b mutation carry the A allele, and instead they have the a allele. (Individuals with A and a are equally healthy and fit.) Explain how the b and A alleles are separated.

ANS: Eventually, the b mutation will be separated from the A allele by meiotic recombination. Meiotic recombination exchanges portions of homologous chromosomes, and thereby generates great diversity among the gametes of each

individual. The locations of the one to five exchanges per chromosome during meiosis in humans are more or less random. Thus, each passing generation increases the cumulative likelihood of a recombinational crossover between any two neighboring genes (A and b). Such a recombinational crossover in an individual heterozygous for both genes (Ab and aB) will separate two alleles that were originally linked (yielding AB and ab).

DIF: Hard REF: 19.5 OBJ: 19.5.d Review how an analysis of haplotype blocks can be used to estimate when a particular mutation or allele arose in the human population. MSC: Applying 22. Shown in Figure 19-22 is a genetic pedigree of a family with several members affected by a heritable disease.

Affected individuals are shown in black and healthy individuals are shown in white. Males are shown as boxes and females as circles. Can a single mutation explain the pattern of inheritance? Is the mutation responsible for the disease dominant or recessive? Is the mutation carried on the X chromosome, the Y chromosome, or an autosome?

Figure 19-22

ANS: Yes, a single mutation can explain the pattern. The mutation responsible for the disease is recessive, because two unaffected parents can produce affected offspring. The mutation is probably carried on the X chromosome, because males (having only one X chromosome, derived from one of the two X chromosomes of the mother) seem more likely to have the disease than their sisters (having two X chromosomes, one from the mother and one from the father). If the mutation were on an autosome, male and female offspring would have equal probability of having the disease. If the mutation were located on the Y chromosome, all males in the family would be affected.

DIF: Hard REF: 19.3 OBJ: 19.3.g Explain how pedigrees can be used to study the inheritance of recessive traits in human populations. MSC: Applying 23. Do you AGREE or DISAGREE with the following statement? Explain your answer.

A trait that is found at a low frequency in the population must be a recessive trait.

ANS: Disagree. The frequency of a trait in a population has nothing to do with its dominance or recessiveness. To test dominance or recessiveness, the segregation of this trait must be observed. For example, the defective version of the gene involved in Huntington’s disease occurs at a relatively low frequency in the population but behaves in a dominant fashion to cause disease.

DIF: Easy REF: 19.5 OBJ: 19.5.e Describe how genetic and environmental factors can interact to give rise to a “disease phenotype.” MSC: Understanding 24. You are trying to map a human gene thought to be involved in cat allergies. Because you know this gene is on Chromosome 20, you decide to examine the linkage of several SNPs located on Chromosome 20 with respect to the gene involved in cat allergies. You have obtained DNA from 10 individuals, and you know whether they are allergic to cats. Your SNP results are shown in Table 19-24.

Table 19-24 (+ indicates the presence of SNP)

A. Which SNP is most likely to be tightly linked to the gene involved in cat allergies? Explain your answer.

B. Of the SNPs tested above, which is likely to be the next closest to the gene responsible for the allergic state?

Why? ANS: A. SNP3 is most likely to be tightly linked to the gene involved in cat allergies because it is found in all people who are allergic to cats and in none of the people who are not allergic to cats. SNP3 might lie within the critical gene, or simply near it.

B. SNP4 is probably the next closest to the gene involved in cat allergies because it is found in all except one person who is allergic to cats and in none of the people who are not allergic to cats. SNP1, SNP2, and SNP5 seem to segregate randomly both in people who are allergic to cats and in those who are not, and therefore are probably distant from the gene involved in cat allergies.

DIF: Hard REF: 19.5 OBJ: 19.5.l Review how SNPs can be used to produce a genetic linkage map and to facilitate the search for alleles that predispose to disease. MSC: Applying 25. Any two human beings typically have an estimated 0.1% difference in their nucleotide sequences, which is equivalent to about 3 million nucleotide differences. These differences are the basis of the SNPs used to construct genetic linkage maps. Some of these SNPs actually lie in the region of the DNA that codes for the protein, yet they have no effect on the phenotype of individuals carrying the SNP on both homologous chromosomes. Explain how some SNPs can lie within the portion of the DNA that codes for the protein and yet have no discernible effect on the protein’s activity.

ANS: Because of the redundancy of the genetic code, in which more than one codon can code for the same amino acid, some single-nucleotide changes may not cause changes in the amino acid coded for by that codon. Furthermore, some amino acid substitutions are neutral (for example, the substitution of a small, uncharged amino acid for another amino acid with similar properties)—that is, they have no perceptible effect on the function of the protein.

DIF: Moderate REF: 19.5 OBJ: 19.5.l Review how SNPs can be used to produce a genetic linkage map and to facilitate the search for alleles that predispose to disease. MSC: Applying 26. You decide to carry out genetic association studies and identify a SNP variant that is found significantly more often in individuals who have schizophrenia than in those who are not affected. This SNP is found within an intron of the SZP gene.

A. Can you deduce that an abnormality of the SZP gene is a cause of increased risk of schizophrenia?

B. Can you say whether the SNP variant itself is a cause?

ANS: A. No, you cannot be sure of this. The SZP gene would be a prime suspect, but the abnormality causing the heightened risk of schizophrenia might well lie instead in some other nearby gene.

B. No, you cannot say for certain. Most point mutations in introns have no functional effect, but some can be functionally important. For example, the intronic SNP might alter an enhancer element that is lying within the intron and is involved in the regulation of SZP transcription; or it might affect the splicing of the SZP gene transcripts.

DIF: Moderate REF: 19.5 OBJ: 19.5.k Outline how genome-wide association studies can be used to search for genes that predispose individuals to common diseases. MSC: Applying 27. Figure 19-27 is a diagram of chromosomes during meiosis.

Figure 19-27

A. On the diagram, indicate which label lines correspond to the following items: (1) sister chromatids, (2) homologous chromosomes, (3) bivalent, (4) chiasma.

B. On the figure, draw as small circles (oooo) the cohesin “glue” that is released in meiotic division I, and draw as small crosses (xxxx) the cohesin glue that is released in meiotic division II. ANS: A. (1) a; (2) c; (3) b; (4) d B. See Figure 19-27A.

Figure 19-27A

DIF: Easy REF: 19.2 OBJ: 19.2.d Describe the chromosome composition of a bivalent. | 19.2.h Describe a chiasma and indicate how many are typically found in each bivalent. | 19.2.k Differentiate how cohesins are degraded during anaphase in meiosis I and II. MSC: Remembering 28. In some fungi, cell division during meiosis gives rise to an ordered spore sac containing a row of four haploid spores, as shown in Figure 19-28A. The position of each spore within the sac reflects its relation to its neighbors; in other words, spores that result from the same meiosis II division are positioned next to each other. You notice that a strain of the fungus produced by crossing a dark-colored strain with a light-colored strain gives rise mostly to spore sacs as shown in Figure 19-28B, with a few spore sacs like those in Figure 19-28C. Indicate whether the following statements are CORRECT or INCORRECT. Explain your reasoning.

Figure 19-28

A. Meiosis I and meiosis II in the fungus occur in the reverse order from that which occur in humans.

B. Recombination in the fungus can occur during prophase I. C. Recombination in the fungus cannot occur during prophase II. D. The spore sacs in Figure 19-28C result from recombination between the centromere and the gene responsible for spore color.

E. Recombination has occurred between the gene responsible for spore color and the end of the chromosome arm.

ANS: During normal meiosis, the chromosomes are distributed as shown in Figure A19-28A (the dark-colored chromosome carries the allele for dark pigment). The homologs are separated during meiotic division I and the sister chromatids are separated during meiotic division II. This pattern of chromosome segregation, in the absence of recombination, always gives the pattern of spores that was shown in the question (Figure 19-28B). To yield the unusual spore sacs, there must be recombination between the two homologous chromosomes. Such recombination always occurs when both homologs are in the same cell (that is, in prophase I of meiosis).

Figure 19-28A

A. Incorrect. If meiosis I and meiosis II were reversed, we would expect a completely random pattern of spores in spore sacs, because the two homologs could go to either cell in the second division, as shown in Figure 19-28C. A reversal of meiosis I and II cannot explain the observation that most of the spore sacs appear like those in Figure 19-28B. B. Correct. The unusual spore sacs must result from recombination between the homologs carrying different alleles of the color gene. Recombination between the homologs occurs during prophase I. C. Incorrect. The results shown cannot distinguish whether recombination has occurred during prophase II. During prophase II, there is only one homolog in each cell; if recombination took place between the two sister chromatids of this homolog we would not be able to detect it: if there had been no recombination in meiosis I, the sister chromatids would be identical and crossing-over between them would cause no genetic change; and if recombination

had occurred in meiosis I, further recombination in meiosis II would still lead only to the same three types of spore sac shown in Figure 19-28C. D. Correct. Segregation of chromosomes is mediated by microtubules that pull on the centromeres of chromosomes.

Thus, unusual segregation patterns arise from recombination (and thus the loss of a covalent connection) between the centromere and the gene of interest.

E. Incorrect. If recombination had taken place only between the gene for spore color and the end of that chromosome arm, it would look as though no recombination had taken place (Figure 19-28B) because the centromere would still be covalently connected to the gene of interest. (See also the answer to part D.) DIF: Hard REF: 19.2 OBJ: 19.2.a Outline how rounds of DNA replication and division produce a haploid nucleus from the nucleus of a diploid germline cell. MSC: Analyzing 29. In mammals, there are two sex chromosomes, X and Y, which behave like homologous chromosomes during meiosis.

Normal males have one X chromosome and one Y chromosome, and normal females have two X chromosomes.

Males with an extra Y chromosome (XYY) are found occasionally. Which of the following could give rise to such an XYY male? Explain your answer.

A. nondisjunction in the first meiotic division of spermatogenesis; normal meiosis in the mother B. nondisjunction in the second meiotic division of spermatogenesis; normal meiosis in the mother C. nondisjunction in the first meiotic division of oogenesis; normal meiosis in the father D. nondisjunction in the second meiotic division of oogenesis; normal meiosis in the father ANS: B Nondisjunction in one or the other meiotic division will give rise to gametes with two sex chromosomes instead of one.

Because the only source of Y chromosomes is the father, the gametes that produce an XYY male must be X (a normal egg) and YY (an abnormal sperm). Nondisjunction in the second meiotic division of spermatogenesis could give rise to a YY gamete that fertilizes a normal egg and thereby creates an individual with the karyotype XYY (see Figure 1929A). Nondisjunction in the first meiotic division of spermatogenesis would result in two XY sperm and two sperm with no sex chromosome (choice A). Nondisjunction during oogenesis cannot yield an individual with two Y chromosomes (choices C and D).

Figure 19-29A

DIF: Moderate REF: 19.2 OBJ: 19.2.n Explain how nondisjunction gives rise to aneuploid gametes, and recall the consequences of this type of genetic error. MSC: Applying 30. Your friend has obtained some pea seeds from the Abbey of St. Thomas in Brno, where Gregor Mendel worked. He is very excited because not only did he obtain some yellow and green pea seeds from true-breeding plants (like the

ones used in Mendel’s famous experiment), he was also able to obtain some purple pea seeds from a true-breeding plant. First, your friend takes the true-breeding yellow and green pea seeds, repeats the cross that Gregor Mendel did, and obtains the same results: he sees 100% yellow-seeded pea plants in the F1 generation, and 75% yellowseeded pea plants and 25% green-seeded pea plants in the F2 generation. His results are illustrated in Figure 19-30A. Your friend then decides to set up two more crosses. For cross #2, he crosses the true-breeding purple-seeded pea plants to the true-breeding yellow-seeded pea plants. The results from this cross are shown in Figure 19-30B. Next, for cross #3, he crosses the true-breeding purple-seeded pea plants to the true-breeding green-seeded pea plants.

These results are shown in Figure 19-30C.

Figure 19-30

Given these results, if you were to take the purple-seeded pea plants produced in the F1 generation in cross #2 and cross them to the purple-seeded pea plants produced in the F1 generation of cross #3, what do you expect that the phenotype of the progeny would look like? Explain your answer.

ANS: You would expect the progeny to be 75% purple-seeded plants and 25% yellow-seeded plants. The crosses are diagrammed in Figure 19-30A, where we designate the allele causing purple seeds as P, the allele causing yellow seeds as Y, and the allele causing green seeds as G. Note that P, Y, and G are all alleles of the same gene.

Figure 19-30A

DIF: Hard REF: 19.3 OBJ: 19.3.b Define what it means for an organism to be true-breeding, and describe the appearance of the offspring Mendel observed when crossbreeding two true-breeding plants. | 19.3.f State Mendel’s law of segregation and explain how it gives rise to the phenotypic ratios he observed in the F2 generation.

MSC: Applying

CHAPTER 20

Cell Communities: Tissues, Stem

Cells, and Cancer EXTRACELLULAR MATRIX AND CONNECTIVE TISSUES 20.1.a Review the relationship among cells, tissues, and organs.

20.1.b Contrast the extracellular matrix and cell wall, and recall what roles these materials have in the supportive tissues of animals and plants.

20.1.c Compare, in terms of structure and composition, the cell walls found in wood and those found in a leaf.

20.1.d Describe the driving force for plant cell growth.

20.1.e Review the composition of a plant cell wall and identify the role of cellulose microfibrils in this matrix.

20.1.f

Differentiate between the primary and secondary cell walls.

20.1.g Explain how the orientation of cellulose microfibrils in a plant cell wall influences the shape of a plant tissue.

20.1.h Outline how cellulose is produced by plant cells and how the orientation of the microfibrils is regulated.

20.1.i

Compare the extracellular matrix of connective tissues such as bone or tendon to that of muscle or epidermis.

20.1.j

Describe the distribution of cells in a connective tissue.

20.1.k Identify the component that provides tensile strength in the connective tissues of animals, and outline what gives different connective tissues their distinctive characteristics.

20.1.l

Compare the structures of collagen molecules, collagen fibrils, and collagen fibers.

20.1.m Distinguish the locations and functions of fibroblasts, osteoblasts, and osteoclasts.

20.1.n Outline how collagen molecules are synthesized, secreted, and assembled into fibrils.

20.1.o Explain how collagen fibrils become oriented in developing tissues.

20.1.p Review how integrins allow cells to attach to collagen in the extracellular matrix, and explain how these attachments are dynamically formed and broken.

20.1.q Describe the structure of glycosaminoglycans and summarize how they allow connective tissue to resist compression.

20.1.r

Describe the structure of proteoglycans and outline their functions in the extracellular matrix.

EPITHELIAL SHEETS AND CELL JUNCTIONS 20.2.a Define an epithelium and describe the four main types of epithelia.

20.2.b Outline the main functions common to epithelia and list some functions of specialized epithelia.

20.2.c Indicate the composition of the basal lamina.

20.2.d Review how epithelial sheets are polarized in structure and function.

20.2.e Summarize the structure and function of tight junctions.

20.2.f

Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function.

20.2.g Compare gap junctions and plasmodesmata in terms of structure, location, and function.

STEM CELLS AND TISSUE RENEWAL 20.3.a List the mixture of cell types that form tissues such as skin and describe the function of each cell type.

20.3.b Review the three main factors that contribute to tissue stability.

20.3.c Contrast cell turnover rates in different tissues, such as skin, bone, blood, and intestinal epithelium.

20.3.d Summarize how stem cells divide to produce a continuous supply of terminally differentiated cells.

20.3.e Describe the general appearance, distribution, and abundance of stem cells in an adult mammal.

20.3.f

Contrast how cell replacement occurs in the skin epidermis with replacement in the intestinal epithelium.

20.3.g Provide examples of stem cells that give rise to several types of differentiated cells.

20.3.h Outline how different signaling mechanisms contribute to maintaining the complex organization of a stem -cell system such as that of the intestine.

20.3.i

Define pluripotency and summarize how embryonic stem cells can be induced to differentiate into a variety of cell

types.

20.3.j

Present potential obstacles to the use of human embryonic stem cells for the replacement of damaged tissues in disorders such as Parkinson’s disease or diabetes.

20.3.k Review how induced pluripotent stem cells are produced and compare their properties to those of embryonic stem cells.

20.3.l

Explain how induced pluripotent stem cells are currently being used to study human disease, cell fate, and the

development of organs.

CANCER 20.4.a Recall the two heritable properties that define cancer cells.

20.4.b Contrast benign tumors, malignant tumors, and metastases.

20.4.c Present the epidemiological evidence that supports a strong role for environmental factors in causing human cancers.

20.4.d List several environmental factors involved in promoting the development of cancer.

20.4.e Outline the potential sources of mutations that could contribute to cancer.

20.4.f

Distinguish passenger mutations from cancer-critical or driver mutations.

20.4.g Explain why cancer is most often a disease of old age.

20.4.h Review how genetic instability can be produced and its role in cancer progression.

20.4.i

Outline how tumors evolve through repeated rounds of mutation, proliferation, and natural selection.

20.4.j

Summarize the characteristics that distinguish cancer cells from normal cells and how they provide cancer cells with a competitive advantage.

20.4.k Differentiate between oncogenes and tumor suppressor genes, and outline the types of genetic events that alter the activity of each.

20.4.l

Identify the key regulatory pathways that are altered in almost all human cancers.

20.4.m Explain how loss of the tumor suppressor gene APC can give rise to colorectal cancer, both in families predisposed to the condition and in patients with no family history of the disease.

20.4.n Describe how an accumulation of mutations can turn a polyp in the lining of the colon into an invasive or metastatic cancer.

20.4.o Outline how antibodies can be used to identify a protein’s binding partners, for example, proteins that interact with tumor suppressor APC. 20.4.p Review how APC regulates the activity of the Wnt signaling pathway and how inactivation of APC drives the formation of polyps.

20.4.q Recall how mutations in the gene-encoding beta-catenin can cause a similar effect as mutations in APC. 20.4.r

Explain how a lack of normal cell-cycle control mechanisms or DNA damage response may help make cancer

cells particularly vulnerable to therapeutic intervention.

20.4.s Summarize how the immune system can be used to help kill tumor cells.

20.4.t

Describe how the product of a specific oncogene can be targeted therapeutically.

MULTIPLE CHOICE 1. Both multicellular plants and animals have a. cells capable of locomotion. b. cells with cell walls. c. a cytoskeleton composed of actin filaments, microtubules, and intermediate filaments. d. tissues composed of multiple different cell types.

ANS: D Plant cells do not have intermediate filaments and are not capable of locomotion. Animal cells do not have cell walls.

DIF: Easy REF: 20.1 OBJ: 20.1.a Review the relationship between cells, tissues, and organs. MSC: Remembering 2. Which of the following statements about plant cell walls is TRUE? a. The microtubule cytoskeleton directs the orientation in which cellulose is deposited in the cell wall. b. The molecular components of the cell wall are the same in all plant tissues. c. Because plant cell walls are rigid, they are not deposited until the cell has stopped growing. d. The cellulose found in cell walls is produced as a precursor molecule in the cell and delivered to the extracellular space by exocytosis.

ANS: A Different plant cells can have specially adapted types of walls. Most newly formed plant cells have a primary cell wall that is thin and can expand to accommodate cell growth. Cellulose is synthesized on the outer surface of the cell.

DIF: Easy REF: 20.1 OBJ: 20.1.e Review the composition of a plant cell wall and identify the role of cellulose microfibrils in this matrix. MSC: Understanding 3. Which of the following molecules is NOT found in plants? a. cellulose b. lignin c. collagen d. pectin ANS: C DIF: Easy REF: 20.1 OBJ: 20.1.e Review the composition of a plant cell wall and identify the role of cellulose microfibrils in this matrix. MSC: Remembering 4. Which of the following statements about cellulose is FALSE? a. Cellulose synthase enzyme complexes are integral membrane proteins. b. An array of microtubules guides the cellulose synthase complex as it moves in the membrane. c. The sugar monomers necessary for the synthesis of a cellulose polymer are transported across the plasma membrane. d. Microtubules are directly attached to the outside surface of the plasma membrane to form tracks that help orient the cellulose polymers.

ANS: D In plants, the microtubules are directly attached to the plasma membrane inside the cell to form tracks that guide the movement of cellulose synthase and thus affect the orientation of the cellulose polymers outside the cell.

DIF: Easy REF: 20.1 OBJ: 20.1.h Outline how cellulose is produced by plant cells and how the orientation of the microfibrils is regulated. MSC: Understanding 5. Which of the following is NOT an example of a connective tissue? a. bone b. the layer of photoreceptors in the eye c. the jellylike interior of an eye d. cartilage ANS: B The layer of photoreceptors in the eye is considered an epithelial sheet.

DIF: Easy REF: 20.1 OBJ: 20.1.j Describe the distribution of cells in a connective tissue. MSC: Remembering 6. Which of the following statements about animal connective tissues is TRUE? a. Enzymes embedded in the plasma membrane synthesize the collagen in the extracellular matrix extracellularly. b. In connective tissue, the intermediate filaments within the cells are important for carrying the mechanical load. c. Cells can attach to a collagen matrix by using fibronectin, an integral membrane protein. d. Proteoglycans can resist compression in the extracellular matrix.

ANS: D Collagen is synthesized within the cell and reaches the extracellular environment through exocytosis. In connective

tissues, the extracellular matrix carries the mechanical load. Cells attach to the collagen matrix with the use of integrins, which span the plasma membrane; integrins then interact with fibronectin, which binds to collagen.

DIF: Easy REF: 20.1 OBJ: Describe the structure of proteoglycans and outline their functions in the extracellular matrix. MSC: Understanding 7. Which of the following statements about collagen is FALSE? a. Collagen synthase organizes the mature collagen molecules into ordered collagen fibrils. b. Collagen is synthesized as procollagen and secreted to the outside of the cell in a secretory vesicle. c. The terminal procollagen domains are cleaved by a protease in the extracellular space. d. Cells can break down a collagen matrix using matrix proteases.

ANS: A Mature collagen can self-assemble into ordered collagen fibrils, and does not require a synthase for this process.

DIF: Easy REF: 20.1 OBJ: 20.1.n Outline how collagen molecules are synthesized, secreted, and assembled into fibrils. | 20.1.o Explain how collagen fibrils become oriented in developing tissues. MSC: Understanding 8. Fibroblasts organize the collagen of the extracellular matrix by a. cutting and rejoining the fibrils. b. processing procollagen into collagen. c. twisting fibrils together to make ropelike fibers. d. pulling the collagen into sheets or cables after it has been secreted.

ANS: D DIF: Easy REF: 20.1 OBJ: 20.1.n Outline how collagen molecules are synthesized, secreted, and assembled into fibrils. MSC: Remembering 9. A cell can crawl through a tissue because of the transmembrane __________ proteins that can bind to fibronectin outside of the cell. a. integrin b. collagen c. gap junction d. claudin ANS: A DIF: Easy REF: 20.1 OBJ: 20.1.p Review how integrins allow cells to attach to collagen in the extracellular matrix, and explain how these attachments are dynamically formed and broken. MSC: Remembering 10. Which of the following statements about integrins is FALSE? a. Integrins use adaptor proteins to interact with the microtubule cytoskeleton. b. Integrins can switch to an activated state by binding to an extracellular matrix molecule. c. Integrins can switch to an activated state by binding to an intracellular protein. d. An activated integrin molecule takes on an extended conformation.

ANS: A DIF: Easy REF: 20.1 OBJ: 20.1.p Review how integrins allow cells to attach to collagen in the extracellular matrix, and explain how these attachments are dynamically formed and broken. MSC: Understanding 11. Proteoglycans in the extracellular matrix of animal tissues a. chiefly provide tensile strength. b. allow cartilage to resist compression.

c. are linked to microtubules through the plasma membrane. d. are polysaccharides composed of glucose subunits.

ANS: B DIF: Easy REF: 20.1 OBJ: 20.1.r Describe the structure of proteoglycans and outline their functions in the extracellular matrix. MSC: Remembering 12. Which of the following statements is FALSE? a. Proteoglycans can act as filters to regulate which molecules pass through the extracellular medium. b. The negative charge associated with proteoglycans attracts cations, which cause water to be sucked into the extracellular matrix. c. Proteoglycans are a major component of compact connective tissues but are relatively unimportant in watery tissues such as the jellylike substance in the interior of the eye. d. Glycosaminoglycans are components of proteoglycan.

ANS: C Proteoglycans are MORE important in the interior of the eye, where collagen is very scarce, than in compact connective tissues such as bone and tendon, whose structure is dominated by collagen.

DIF: Easy REF: 20.1 OBJ: 20.1.q Describe the structure of glycosaminoglycans and summarize how they allow connective tissue to resist compression. | 20.1.r Describe the structure of proteoglycans and outline their functions in the extracellular matrix. MSC: Understanding 13. A basal lamina a. is a thin layer of connective-tissue cells and matrix underlying an epithelium. b. is a thin layer of extracellular matrix underlying an epithelium. c. is attached to the apical surface of an epithelium. d. separates epithelial cells from each other.

ANS: B DIF: Easy REF: 20.2 OBJ: 20.2.c Indicate the composition of the basal lamina.

MSC: Remembering 14. Tight junctions a. allow small, water-soluble molecules to pass from cell to cell. b. interact with the intermediate filaments inside the cell. c. are formed from claudins and occludins. d. are found in cells in connective tissues.

ANS: C DIF: Easy REF: 20.2 OBJ: 20.2.e Summarize the structure and function of tight junctions.

MSC: Remembering 15. Adherens junctions a. can be used to bend epithelial sheets into tubes. b. are most often found at the basal surface of cells. c. are found only in adult tissues. d. involve fibronectin and integrin interactions.

ANS: A DIF: Easy REF: 20.2 OBJ: 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. MSC: Remembering

16. At desmosomes, cadherin molecules are connected to a. actin filaments. b. intermediate filaments. c. microtubules. d. gap junctions.

ANS: B DIF: Easy REF: 20.2 OBJ: 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. MSC: Remembering 17. Hemidesmosomes are important for a. tubulation of epithelial sheets. b. linkages to glycosaminoglycans. c. forming the basal lamina. d. attaching epithelial cells to the extracellular matrix.

ANS: D DIF: Easy REF: 20.2 OBJ: 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. MSC: Remembering 18. A major distinction between the connective tissues in an animal and other main tissue types such as epithelium, nervous tissue, or muscle is the a. ability of connective-tissue cells such as fibroblasts to change shape. b. amount of extracellular matrix in connective tissues. c. ability of connective tissues to withstand mechanical stresses. d. numerous connections that connective-tissue cells make with each other.

ANS: B Cells in nonconnective tissues can change shape (for example, muscle cells when they contract), and can withstand mechanical stress (for example, cells of the epidermis). Cells in connective tissue tend to be scattered throughout the extracellular matrix and make few or no contacts with each other, unlike nerve cells or epithelial cells.

DIF: Easy REF: 20.1 OBJ: 20.1.i Compare the extracellular matrix in connective tissues such as bone or tendon and that of muscle or epidermis. MSC: Analyzing 19. Which of the following statements about gap junctions is FALSE? a. Gap junctions are made of connexons. b. Molecules up to 1000 daltons in molecular mass can move across gap junctions. c. Because gap junctions only allow uncharged molecules to pass through, they are not used for electrically coupling cells. d. Gap junctions can close in response to extracellular signals.

ANS: C DIF: Easy REF: 20.2 OBJ: 20.2.g Compare gap junctions and plasmodesmata in terms of structure, location, and function. MSC: Remembering 20. Which type of junction involves a connection to the actin cytoskeleton? a. adherens junctions b. desmosomes c. tight junctions d. gap junctions

ANS: A DIF: Easy REF: 20.2 OBJ: 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. | 20.2.g Compare gap junctions and plasmodesmata in terms of structure, location, and function. MSC: Remembering 21. Which type of junction contributes the most to the polarization of epithelial cells? a. adherens junctions b. desmosomes c. tight junctions d. gap junctions ANS: C DIF: Easy REF: 20,2 OBJ: 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. MSC: Remembering 22. Cadherins a. are used to transfer proteins from one cell to another. b. mediate cell–cell attachments through homophilic interactions. c. are abundant in the plant cell wall. d. bind to collagen fibrils.

ANS: B DIF: Easy REF: 20.2 OBJ: 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. MSC: Remembering 23. Plasmodesmata a. permit small molecules to pass from one cell to another. b. are found only in animal cells. c. are closed by the neurotransmitter dopamine. d. provide tensile strength.

ANS: A DIF: Easy REF: 20.2 OBJ: 20.2.g Compare gap junctions and plasmodesmata in terms of structure, location, and function. MSC: Remembering 24. The plasmodesmata in plants are functionally most similar to which animal cell junction? a. tight junction b. adherens junction c. gap junction d. desmosome ANS: C DIF: Easy REF: 20,2 OBJ: 20.2.g Compare gap junctions and plasmodesmata in terms of structure, location, and function. MSC: Remembering 25. Cells that are terminally differentiated a. will undergo apoptosis within a few days. b. can no longer undergo cell division. c. are unable to move. d. no longer produce RNAs. ANS: B Terminally differentiated cells no longer divide. The life of a terminally differentiated cell depends on the cell type; alt-

hough terminally differentiated cells in the intestine typically die within a few days, neurons in the brain can last for a lifetime. Some terminally differentiated cells can still move and transcribe and translate genes.

DIF: Easy REF: 20.3 OBJ: 20.3.d Summarize how stem cells divide to produce a continuous supply of terminally differentiated cells. MSC: Remembering 26. When a terminally differentiated cell in an adult body dies, it can typically be replaced in the body by a stock of a. proliferating precursor cells. b. cells more apically located than the terminally differentiated cells. c. Wnt proteins. d. induced pluripotent cells.

ANS: A Proliferating precursor cells already within the tissue are typically used to replace terminally differentiated cells. Induced pluripotent cells are produced in the laboratory and are not typically used to replace adult terminally differentiated cells within the body.

DIF: Easy REF: 20.3 OBJ: 20.3.d Summarize how stem cells divide to produce a continuous supply of terminally differentiated cells. MSC: Remembering 27. An adult hemopoietic stem cell found in the bone marrow a. will occasionally produce epidermal cells when necessary. b. can produce only red blood cells. c. can undergo self-renewing divisions for the lifetime of a healthy animal. d. will express all the same transcription factors as those found in an unfertilized egg.

ANS: C Stem cells undergo self-renewing divisions. Adult stem cells are specialized, and thus an adult hemopoietic stem cell does not produce epidermal cells and also stably expresses the transcriptional regulators that ensure that its progeny will become blood cells. An adult hemopoietic stem cell has the ability to produce all the many types of cell in the blood.

DIF: Easy REF: 20.3 OBJ: 20.3.g Provide examples of stem cells that give rise to several types of differentiated cells.

MSC: Remembering 28. A pluripotent cell a. can only be produced in the laboratory. b. can give rise to all the tissues and cell types in the body. c. can only give rise to stem cells. d. is considered to be terminally differentiated.

ANS: B A pluripotent cell can give rise to all tissues and cell types in the body, including stem cells. A fertilized egg is a pluripotent cell and thus pluripotent cells can be found in nature. A terminally differentiated cell is the opposite of a pluripotent cell.

DIF: Easy REF: 20.3 OBJ: 20.3.i Define pluripotency and summarize how embryonic stem cells can be induced to differentiate into a variety of cell types. MSC: Remembering 29. Mouse embryonic stem (ES) cells

a. can only be grown in the laboratory. b. can give rise to all tissues and cell types in the body except germ cells. c. can be made in the lab from human iPS cells. d. come from the inner cell mass of early embryos.

ANS: D Mouse ES cells are cells that are dissociated from the inner cell mass of early embryos. These cells are produced naturally in embryos and can be isolated and grown in the lab. ES cells can give rise to all tissues and cell types of the body, including germ cells. An organism’s genes determine what an embryo grows up to become, and thus human iPS cells can only be used to generate human cells.

DIF: Easy REF: 20.3 OBJ: 20.3.e Describe the general appearance, distribution, and abundance of stem cells in an adult mammal. MSC: Remembering 30. Induced pluripotent stem (iPS) cells a. can be created by the expression of a set of key genes in most somatic cell types, including cells derived from adult tissues. b. require a supply of donor egg cells, as is the case for embryonic stem cells. c. can differentiate into a greater variety of adult tissues than embryonic stem cells. d. have been used to create human clones.

ANS: A iPS cells do not require donor egg cells and are made from adult (and not embryonic) cells. Embryonic stem cells can differentiate into all tissue and cell types. Human clones have not been created using iPS cells.

DIF: Easy REF: 20.3 OBJ: 20.3.k Review how induced pluripotent stem cells are produced and compare their properties to those of embryonic stem cells. MSC: Remembering 31. The artificial introduction of three key __________ into an adult cell can convert the adult cell into a cell with the properties of ES cells. a. chromosomes b. viruses c. hormones d. transcription factors ANS: D DIF: Easy REF: 20.3 OBJ: 20.3.k Review how induced pluripotent stem cells are produced and compare their properties to those of embryonic stem cells. MSC: Remembering 32. Which of the following statements about organoids is FALSE? a. Organoids can be produced from ES and iPS cells. b. Organoids can only be made for organs that are made up of a single type of differentiated cell. c. Organoids are self-assembling. d. Organoids can form multilayered structures.

ANS: B The cells in an organoid can differentiate into multiple organ-specific cell types.

DIF: Easy REF: 20.3 OBJ: 20.3.l Explain how induced pluripotent stem cells are currently being used to study human disease, cell fate, and the development of organs. MSC: Understanding

33. A malignant tumor is more dangerous than a benign tumor because a. its cells are proliferating faster. b. it causes neighboring cells to mutate. c. its cells attack and phagocytose neighboring normal tissue cells. d. its cells invade other tissues.

ANS: D DIF: Easy REF: 20.4 OBJ: 20.4.b Contrast benign tumors, malignant tumors, and metastases.

MSC: Understanding 34. A metastasis is a. a secondary tumor in a different part of the body that arises from a cell from the primary tumor. b. a cell that is dividing in defiance of normal constraints. c. a part of the primary tumor that has invaded the surrounding tissue. d. the portion of the cancerous tumor that displays genetic instability.

ANS: A All tumors, benign or malignant, divide in defiance of normal constraints. A malignant tumor will invade the surrounding tissue while a metastasis is found at a distant site from the original tumor. The genetic instability seen in cancer cells is not limited to metastases.

DIF: Easy REF: 20.4 OBJ: 20.4.i Outline how tumors evolve through repeated rounds of mutation, proliferation, and natural selection. MSC: Remembering 35. Which of the following statements about cancer is FALSE? a. Viruses cause some cancers. b. Tobacco use is responsible for more than 20% of all cancer deaths. c. A mutation in even a single cancer-critical gene is sufficient to convert a normal cell into a cancer cell. d. Chemical carcinogens cause cancer by changing the nucleotide sequence of DNA. ANS: C Multiple mutations are required to convert a normal cell into a cell that has all the properties needed to make it cancerous.

DIF: Easy REF: 20.4 OBJ: 20.4.e Outline the potential sources of mutations that could contribute to cancer.

MSC: Understanding 36. Which of the following genetic changes cannot convert a proto-oncogene into an oncogene? a. A mutation that introduces a stop codon immediately after the codon for the initiator methionine. b. A mutation within the coding sequence that makes the protein hyperactive. c. An amplification of the number of copies of the proto-oncogene, causing overproduction of the normal protein. d. A mutation in the promoter of the proto-oncogene, causing the normal protein to be transcribed and translated at an abnormally high level.

ANS: A A mutation that introduces a stop codon immediately after the codon for the initiator methionine will result in no protein being produced. Proto-oncogenes become oncogenes when they acquire mutations that cause them to act in a dominant hyperactive manner.

DIF: Easy REF: 20.4 OBJ: 20.4.e Outline the potential sources of mutations that could contribute to cancer. | 20.4.k

Differentiate between oncogenes and tumor suppressor genes, and outline the types of genetic events that alter the activity of each. MSC: Understanding 37. Which of the following statements about tumor suppressor genes is FALSE? a. Gene amplification of a tumor suppressor gene is less dangerous than gene amplification of a proto-oncogene. b. Cells with one functional copy of a tumor suppressor gene will usually proliferate faster than normal cells. c. Inactivation of tumor suppressor genes leads to enhanced cell survival and proliferation. d. Individuals with only one functional copy of a tumor suppressor gene are more prone to cancer than individuals with two functional copies of a tumor suppressor gene.

ANS: B Cells with one copy of a tumor suppressor gene should behave normally, because tumor suppressor genes generally work in a recessive manner.

DIF: Easy REF: 20.4 OBJ: 20.4.k Differentiate between oncogenes and tumor suppressor genes and outline the types of genetic events that alter the activity of each. MSC: Understanding 38. Ras is a GTP-binding protein that is often defective in cancer cells. A common mutation found in cancer cells causes Ras to behave as though it were bound to GTP all the time, which will cause cells to divide inappropriately. From this description, the normal Ras gene is a/an a. tumor suppressor. b. oncogene. c. proto-oncogene. d. gain-of-function mutation.

ANS: C The normal Ras gene is a proto-oncogene. Only the mutated form of Ras is an oncogene and a gain-of-function mutation.

DIF: Moderate REF: 20.4 OBJ: 20.4.e Outline the potential sources of mutations that could contribute to cancer. | 20.4.k Differentiate between oncogenes and tumor suppressor genes and outline the types of genetic events that alter the activity of each. MSC: Understanding 39. APC is a tumor suppressor and acts in the Wnt signaling pathway to prevent the TCF complex from turning on Wntresponsive genes. Mice that lack the gene encoding TCF4 do not have the ability to maintain the pool of proliferating gut stem cells needed to renew the gut lining. What do you predict will happen in mice that lack the APC gene? a. Like the mice lacking TCF4, they will not be able to renew the gut lining. b. They will have inappropriate proliferation of gut stem cells. c. Mice lacking the APC gene will have a hyperactive Wnt receptor even though there is no Wnt signal. d. Mice lacking the APC gene will be like normal healthy mice, since APC is a tumor suppressor and thus not needed unless there is a tumor present.

ANS: B Without the APC gene (and its protein product), gut stem cells will inappropriately proliferate. Mice lacking the APC gene have a phenotype that is the opposite to what is seen in mice lacking TCF4, since APC inhibits TCF. APC acts downstream of the Wnt signal and should not affect the Wnt receptor. Tumor suppressor genes are genes that can cause tumors when both copies are inactivated in a diploid cell; these tumor suppressor genes typically have other

normal functions in the cell.

DIF: Easy REF: 20.4 OBJ: 20.4.k Differentiate between oncogenes and tumor suppressor genes, and outline the types of genetic events that alter the activity of each. | 20.4.m Explain how loss of the tumor suppressor gene APC can give rise to colorectal cancer, both in families predisposed to the condition and in patients with no family history of the disease. | 20.4.p Review how APC regulates the activity of the Wnt signaling pathway and how inactivation of APC drives the formation of polyps. MSC: Understanding

MATCHING 1. Match each protein with the correct label in Figure 20-1.

Figure 20-1

1. integrin 2. actin 3. collagen 4. fibronectin 1. ANS: C DIF: Easy REF: 20.1 OBJ: 20.1.p Review how integrins allow cells to attach to collagen in the extracellular matrix, and explain how these attachments are dynamically formed and broken.

MSC: Remembering 2. ANS: D DIF: Easy REF: 20.1 OBJ: 20.1.p Review how integrins allow cells to attach to collagen in the extracellular matrix, and explain how these attachments are dynamically formed and broken.

MSC: Remembering 3. ANS: A DIF: Easy REF: 20.1 OBJ: 20.1.p Review how integrins allow cells to attach to collagen in the extracellular matrix, and explain how these attachments are dynamically formed and broken.

MSC: Remembering 4. ANS: B DIF: Easy REF: 20.1 OBJ: 20.1.p Review how integrins allow cells to attach to collagen in the extracellular matrix, and explain how these attachments are dynamically formed and broken.

MSC: Remembering

2. Match each name below with the best depiction of the type of epithelial sheet shown in Figure 20-2. Each name and each picture will only be used once.

Figure 20-2

1. stratified __________ 2. columnar __________ 3. cuboidal __________ 4. squamous __________ 1. ANS: C DIF: Easy REF: 20.2 OBJ: 20.2.a Define an epithelium and describe the four main types of epithelia.

MSC: Remembering 2. ANS: B DIF: Easy REF: 20.2 OBJ: 20.2.a Define an epithelium and describe the four main types of epithelia.

MSC: Remembering 3. ANS: A DIF: Easy REF: 20.2 OBJ: 20.2.a Define an epithelium and describe the four main types of epithelia.

MSC: Remembering 4. ANS: D DIF: Easy REF: 20.2 OBJ: 20.2.a Define an epithelium and describe the four main types of epithelia.

MSC: Remembering 3. Match the labeled parts of Figure 20-3 with the phrase that best matches the part. Each part will only be used once.

Figure 20-3

1. basal lamina __________

2. apical surface __________ 3. cell junction __________ 4. connective tissue __________ ANS: 1. B DIF: Easy REF: 20.2 OBJ: 20.2.a Define an epithelium and describe the four main types of epithelia.

MSC: Remembering 2. A DIF: Easy REF: 20.2 OBJ: 20.2.a Define an epithelium and describe the four main types of epithelia.

MSC: Remembering 3. C DIF: Easy REF: 20.2 OBJ: 20.2.a Define an epithelium and describe the four main types of epithelia.

MSC: Remembering 4. D DIF: Easy REF: 20.2 OBJ: 20.2.a Define an epithelium and describe the four main types of epithelia.

MSC: Remembering 4. Match the molecules (List 2) with the cell structures in which they are involved (List 1). A cell structure may be listed more than once or not at all.

1. ANS: A DIF: Easy REF: 20.2 OBJ: 20.2.c Indicate the composition of the basal lamina. | 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. | 20.2.g Compare gap junctions and plasmodesmata in terms of structure, location, and function. MSC: Remembering 2. ANS: B DIF: Easy REF: 20.2 OBJ: 20.2.c Indicate the composition of the basal lamina. | 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. | 20.2.g Compare gap junctions and plasmodesmata in terms of structure, location, and function. MSC: Remembering 3. ANS: E DIF: Easy REF: 20.2 OBJ: 20.2.c Indicate the composition of the basal lamina. | 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. | 20.2.g Compare gap junctions and plasmodesmata in terms of structure, location, and function. MSC: Remembering 4. ANS: C DIF: Easy REF: 20.2 OBJ: 20.2.c Indicate the composition of the basal lamina. | 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. | 20.2.g Compare gap junctions and plasmodesmata in terms of structure, location, and function. MSC: Remembering 5. ANS: G DIF: Easy REF: 20.2 OBJ: 20.2.c Indicate the composition of the basal lamina. | 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. | 20.2.g Compare gap junctions and plasmodesmata in terms of structure, location, and function. MSC: Remembering 6. ANS: G DIF: Easy REF: 20.2 OBJ: 20.2.c Indicate the composition of the basal lamina. | 20.2.f Distinguish ad-

herens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. | 20.2.g Compare gap junctions and plasmodesmata in terms of structure, location, and function. MSC: Remembering 7. ANS: A DIF: Easy REF: 20.2 OBJ: 20.2.c Indicate the composition of the basal lamina. | 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. | 20.2.g Compare gap junctions and plasmodesmata in terms of structure, location, and function. MSC: Remembering 5. Match the appropriate cell type found in the mammalian skin with the best description of its function.

Function: A. dispose of dying cells B. organize connective tissue C. combat infection D. provide electrical insulation for axons E. bring oxygen and remove waste Cell Type: 1. fibroblasts 2. Schwann cells 3. macrophages 4. blood vessels 5. lymphocytes 1. ANS: B DIF: Easy REF: 20.3 OBJ: 20.3.a List the mixture of cell types that form tissues such as skin and describe the function of each cell type. MSC: Remembering 2. ANS: D DIF: Easy REF: 20.3 OBJ: 20.3.a List the mixture of cell types that form tissues such as skin and describe the function of each cell type. MSC: Remembering 3. ANS: A DIF: Easy REF: 20.3 OBJ: 20.3.a List the mixture of cell types that form tissues such as skin and describe the function of each cell type. MSC: Remembering 4. ANS: E DIF: Easy REF: 20.3 OBJ: 20.3.a List the mixture of cell types that form tissues such as skin and describe the function of each cell type. MSC: Remembering 5. ANS: C DIF: Easy REF: 20.3 OBJ: 20.3.a List the mixture of cell types that form tissues such as skin and describe the function of each cell type. MSC: Remembering

SHORT ANSWER 1. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once. actin

lignin

pectin

cellulose

membranous

primary

collagen

microtubule

secondary

epidermis

nuclear

tertiary

lamin

osmosis

Plants are sedentary and thus their cells have different needs from those of cells found in motile animals. For example, in plant cells, __________ generates the turgor pressure that drives cell growth. Plants have cell walls, but cell growth is possible in the developing tissue because the __________ cell walls are expandable. The __________ cell walls are deposited once growth has stopped, and can be specially adapted to their function. Fibers made from __________ (the most abundant organic macromolecule on Earth) are found in plant cell walls, and provide tensile strength. In woody tissues, the __________ in the cell walls makes the tissue more rigid and waterproof. The deposition of the cell wall is directed by the __________ cytoskeleton.

ANS: Plants are sedentary and thus their cells have different needs from those of cells found in motile animals. For example, in plant cells, osmosis generates the turgor pressure that drives cell growth. Plants have cell walls, but cell growth is possible in the developing tissue because the primary cell walls are expandable. The secondary cell walls are deposited once growth has stopped, and can be specially adapted to their function. Fibers made from cellulose (the most abundant organic macromolecule on Earth) are found in plant cell walls, and provide tensile strength. In woody tissues, the lignin in the cell walls makes the tissue more rigid and waterproof. The deposition of the cell wall is directed by the microtubule cytoskeleton. DIF: Easy REF: 20.1 OBJ: 20.1.f Differentiate between the primary and secondary cell wall. | 20.1.c Compare the cell wall found in wood and in a leaf in terms of structure and composition. MSC: Understanding 2. Indicate the direction in which the plant cell shown in Figure 20-2 is most likely to grow. The black lines indicate the direction of the cellulose microfibrils around the cell. Explain your answer.

Figure 20-2

ANS: Vertically. Cellulose fibers are highly resistant to stretching, so a plant cell tends to grow, under the stimulus of turgor pressure, in a direction perpendicular to the orientation of the fibers in the cell wall. DIF: Moderate REF: 20.1 OBJ: 20.1.g Explain how the orientation of cellulose microfibrils in a plant cell wall influences the shape of a plant tissue. MSC: Applying 3. Indicate whether the following molecules are found in plants, animals, or both.

A. intermediate filaments B. cell walls C. microtubules D. cellulose E. collagen ANS: A. animals B. plants C. both D. plants E. animals DIF: Easy REF: 20.1 OBJ: 20.1.b Contrast the extracellular matrix and cell wall, and recall what roles these materials

have in the supportive tissue of animals and plants. MSC: Analyzing 4. Do you agree or disagree with the following statement? Explain your answer.

Like many other extracellular proteins, newly synthesized collagen molecules undergo post-translational processing inside the cell to convert them into their mature form; they are then secreted and self-assemble into fibrils in the extracellular space.

ANS: Disagree. The cell secretes newly synthesized collagen molecules in an immature form as procollagen, and the peptides at the ends of the procollagen molecules then have to be cleaved off in the extracellular space before fibril assembly can occur. This process ensures that collagen fibrils will not assemble prematurely inside the cell.

DIF: Easy REF: 20.1 OBJ: 20.1.n Outline how collagen molecules are synthesized, secreted, and assembled into fibrils. MSC: Evaluating 5. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; use each word or phrase only once.

Adherens junctions

Gap junctions

Highway junctions

Desmosomes

Hemidesmosomes

Tight junctions

__________ join the intermediate filaments in one cell to those in the neighboring cell. __________ anchor intermediate filaments in a cell to the extracellular matrix. __________ involve cadherin connections between neighboring cells and are anchorage sites for actin filaments. __________ permit the passage of small molecules from one cell to its adjacent cell. __________ prevent the leakage of molecules between adjacent cells.

ANS: Desmosomes join the intermediate filaments in one cell to those in the neighboring cell. Hemidesmosomes anchor intermediate filaments in a cell to the extracellular matrix. Adherens junctions involve cadherin connections between neighboring cells and are anchorage sites for actin filaments. Gap junctions permit the passage of small molecules from one cell to its adjacent cell. Tight junctions prevent the leakage of molecules between adjacent cells.

DIF: Easy REF: 20.2 OBJ: 20.2.e Summarize the structure and function of tight junctions. | 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function.

MSC: Understanding 6. Label the five different types of cell–cell junction shown in Figure 20-6 and identify the apical and basal surfaces of the epithelium.

Figure 20-6

ANS: 1, apical surface; 2, tight junction; 3, adherens junction; 4, desmosome junction; 5, gap junction; 6, hemidesmosome junction; 7, basal surface DIF: Easy REF: 20.2 OBJ: 20.2.e Summarize the structure and function of tight junctions. | 20.2.f Distinguish adherens junctions, desmosomes, and hemidesmosones in terms of structure, location, and function. MSC: Remembering 7. What are the main structures providing tensile strength in the following?

A. animal connective tissue B. animal epidermis C. plant cell walls ANS: A. collagen fibers B. intermediate filaments C. cellulose fibers DIF: Easy REF: 20.1 OBJ: 20.1.k Identify the component that provides tensile strength in the connective tissues of animals, and outline what gives different connective tissues their distinctive characteristics. | 20.1.e Review the composition of a plant cell wall and identify the role of cellulose microfibrils in this matrix. MSC: Remembering 8. Name the three key mechanisms important for maintaining the organization of cells into tissues.

ANS: 1. cell communication 2. selective cell–cell adhesion 3. cell memory DIF: Easy REF: 20.1 OBJ: 20.1.a Review the relationship between cells, tissues, and organs.

MSC: Remembering

9. Place the following in order of their replacement times, from shortest to longest.

A. epidermal cell B. nerve cell C. bone matrix D. red blood cell E. cell lining the gut ANS: E, cell lining the gut (few days) < A, epidermal cell (1 or 2 months) < D, red blood cell (4 months) < C, bone matrix (10 years) < B, nerve cell (lifetime) DIF: Moderate REF: 20.3 OBJ: 20.3.c Contrast cell turnover rates in different tissues, such as skin, bone, blood, and intestinal epithelium. MSC: Analyzing 10. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase can be used only once. apical

crypt

organic

basal

extracellular

single-layered

blood vessel

hyaluronan

stratified

bone marrow

neuronal

villus

The location of stem cells in various tissues varies. In the lining of the small intestine, the different types of intestinal cells are arranged in a continuous __________ epithelium. The epithelium is arranged such that the cells that are in the underlying connective tissue are in the region called the __________, and is where the stem cell population resides. In contrast, the epidermis is a/an __________ epithelium, with stem cells in the __________ layer. Blood contains man circulating cell types which originate from a stem cell population found in the __________.

ANS: The location of stem cells in various tissues varies. In the lining of the small intestine, the different types of intestinal cells are arranged in a continuous single-layered epithelium. The epithelium is arranged such that the cells that are in the underlying connective tissue are in the region called the crypt, and is where the stem cell population resides. In contrast, the epidermis is a stratified epithelium, with stem cells in the basal layer. Blood contains many circulating cell types which originate from a stem cell population found in the bone marrow. DIF: Easy REF: 20.3 OBJ: 20.3.f Contrast how cell replacement occurs in the skin epidermis and intestinal epithelium.

MSC: Understanding 11. A stem cell divides into two daughter cells. One of the daughter cells goes on to become a terminally differentiated cell. What is the typical fate of the other daughter cell?

ANS: The other daughter cell typically remains a stem cell. DIF: Easy REF: 20.3 OBJ: 20.3.d Summarize how stem cells divide to produce a continuous supply of terminally differentiated cells. MSC: Understanding 12. Your friend is a pioneer in ES cell research. In her research, she uses an ES cell line that originated from an inbred strain of laboratory mice called FG426. She has just figured out methods that allow her to grow an entire liver from an ES cell and has successfully grown 10 livers. She demonstrates that the newly grown livers are functional by successfully transplanting one of the new livers into a FG426 laboratory mouse.

You are particularly excited about this, because you have a sick pet mouse called Squeaky. You are very attached to Squeaky, as you found him when you were out camping in New Hampshire. Unfortunately, Squeaky has developed liver disease and will not live much longer without a liver transplant. After you see your friend on TV talking about her

new method for growing mouse livers, you immediately grab your cell phone to ask her whether Squeaky could have one of the newly grown livers. Just as you are about to dial your friend, you remember something you learned in cell biology and realize that instead, you should ask your friend about possibly creating mouse iPS cells for Squeaky’s benefit.

A. Why do you think that one of the newly grown livers may not work in Squeaky?

B. Explain how the use of iPS cells could solve this problem.

ANS: A. For organ transplantation to be successful, the donor and the recipient should be as close a genetic match as possible to minimize the risk of immunological rejection. Because you found Squeaky in the fields of New Hampshire, Squeaky is likely to have genetic differences from the FG426 inbred laboratory mice that would cause the livers to be rejected.

B. The use of iPS cells may help save Squeaky if the iPS cells were generated from Squeaky himself and therefore would be genetically identical to Squeaky. You hope that your friend will be successful with using iPS cells derived from Squeaky and can grow a functional liver for him.

DIF: Hard REF: 20.3 OBJ: 20.3.l Explain how induced pluripotent stem cells are currently being used to study human disease, cell fate, and the development of organs. MSC: Applying 13. Cancer is a disease of enhanced proliferation and cell survival. DNA repair mechanisms are normally important for cell survival. When a cell senses DNA damage, the cell cycle is inhibited until the damage is fixed. Given the importance of DNA repair mechanisms, how can their failure lead to the production of cancer cells with a competitive advantage over normal cells?

ANS: A cell with a defect in its DNA repair mechanisms will have an increased mutation rate, thus increasing its chances to acquire further mutations that give it (and its progeny) a competitive growth advantage. DIF: Moderate REF: 20.4 OBJ: 20.4.j Summarize the characteristics that distinguish cancer cells from normal cells and how they provide cancer cells with a competitive advantage. MSC: Understanding 14. A certain mutation in the receptor for epidermal growth factor (EGF) causes the mutated receptor protein to send a positive signal along the associated intracellular signaling pathway even when the EGF ligand is not bound to it. This signal leads to abnormal cell proliferation in the absence of growth factor. On the basis of this information, would you class the gene for the EGF receptor as a tumor suppressor gene or a potential oncogene? Explain your answer.

ANS: The mutation described leads toward cancerous cell behavior (excessive proliferation) by making the gene product hyperactive. The mutant gene is therefore, by definition, an oncogene. This effect is seen even if only one copy of the gene is affected; in other words, the mutation is dominant, as is typical for an oncogene. Mutations that delete an EGF receptor gene would be expected to have either no effect or an inhibiting effect on cell division. Thus, the normal EGF receptor is classed as a potential oncogene (a proto-oncogene). DIF: Moderate REF: 20.4 OBJ: 20.4.k Differentiate between oncogenes and tumor suppressor genes and outline the types of genetic events that alter the activity of each.

MSC: Understanding 15. Ras is a GTP-binding protein that is often defective in cancer cells. A signal from a growth factor through a receptor tyrosine kinase often stimulates normal cells to divide. When the receptor tyrosine kinase binds the growth factor, Ras is stimulated to bind GTP. Ras in turn activates proteins that promote cell proliferation. A common mutation in

cancerous cells causes Ras to behave as though it were bound to GTP all the time.

A. Why is this mutation advantageous to cancerous cells?

B. Your friend decides that the signaling pathway involving the Ras protein is a good target for drug design, because the Ras protein is often defective in cancer cells. Your friend designs a drug that will turn off the receptor tyrosine kinase by preventing it from dimerizing. Do you think that this drug will affect cells that have a defective Ras protein that acts as if it were always bound to GTP? Why or why not?

ANS: A. An Ras mutation that causes Ras to behave as though it were bound to GTP all the time is advantageous to cancer cells because Ras is then activated and turns on the activities of proteins required for cell proliferation. If the cell-proliferation proteins are always turned on, the cancer cell will be able to proliferate at an unregulated rate, outgrowing its normal neighbors. The ability to proliferate in a signal-independent fashion is one of the hallmarks of a cancer cell.

B. Unfortunately, a drug that blocks activation of the receptor that activates Ras will be unlikely to have a useful effect on a cell containing mutant Ras protein that behaves as though it were constantly activated. Because Ras acts downstream of the receptor, the activating mutation makes its effect felt regardless of the state of the receptor on which Ras activation would normally depend: mutant Ras that is always active is no longer dependent on the receptor for activation. Therefore, blocking the ability of the receptor to dimerize and activate Ras will probably not affect the cells containing the mutant Ras protein.

DIF: Hard REF: 20.4 OBJ: 20.4.j Summarize the characteristics that distinguish cancer cells from normal cells and how they provide cancer cells with a competitive advantage. MSC: Applying 16. People who inherit one copy of the Rb (retinoblastoma) gene that is normal and one copy that is mutated—that is, people who are heterozygous for Rb—have a greatly increased risk of cancer.

Given this information, do you agree or disagree with the following statement? Explain your answer.

The Rb mutation must have a dominant effect, which means that it must result in an increase in Rb function. Thus, Rb in its mutant form must be an oncogene.

ANS: Disagree. It is true that the Rb mutation is dominant, in the sense that a person who is heterozygous (inherits one normal copy and one mutant copy of the gene) is likely to show the mutant trait (that is, will be cancer-prone).

However, this does not mean that the mutation in Rb causes an increase in Rb gene function; in fact, the opposite is true—the propensity for cancer arises from a loss of Rb gene function. Therefore, Rb should be classified as a tumor suppressor gene and not as an oncogene. Most people have two functional Rb genes in each of their cells. Thus, for one of their cells to turn cancerous by losing Rb function, both copies of the gene in that cell must be inactivated or lost, which is a two-step process. However, in a person born lacking one copy of the Rb gene, each cell is only one step away from a complete loss of Rb function. Consequently, such a person has a high risk that at least one of the cells in the body will undergo a mutation that precipitates cancer. In this way, at the level of the whole person, the Rb loss-of-function mutation is dominant, even though at the level of the individual cell, it is recessive.

DIF: Hard REF: 20.4 OBJ: 20.4.j Summarize the characteristics that distinguish cancer cells from normal cells and how they provide cancer cells with a competitive advantage. | 20.4.k Differentiate between oncogenes and tumor suppressor genes, and outline the types of genetic events that alter the activity of each. MSC: Evaluating 17. Figure 20-17 shows a sequence of mutations that might underlie the development of colorectal cancer. Explain why

the loss of p53 is advantageous to cancerous cells.

Figure 20-17

ANS: When DNA is damaged, the protein p53 is activated and stabilized. When p53 is active, it stops the cell cycle to give the cell time to repair its damaged DNA, or, if that is not feasible, it causes the cell to commit suicide by apoptosis. Cells lacking p53 will continue to replicate their DNA, will avoid suicide, and will go through cell division even when the DNA has been damaged, producing mutant daughter cells. Repeat ed rounds of cell division under these circumstances perpetuate the genetic damage and allow still more mutations to occur. Some of these newly accumulated mutations will give the cell an increased ability to survive, proliferate, and metastasize, resultin g in invasive cancer.

DIF: Hard REF: 20.4 OBJ: 20.4.m Explain how loss of the tumor suppressor gene APC can give rise to colorectal cancer, both in families predisposed to the condition and in patients with no family history of the disease. | 20.4.r Explain how a lack of normal cell-cycle control mechanisms or DNA damage response may help make cancer cells particularly vulnerable to therapeutic intervention. MSC: Understanding 18. Drugs that block the function of oncogenic proteins hold great promise in the fight against cancer. Should cancer researchers also be attempting to design drugs that will interfere with the products of tumor suppressor genes?

Explain.

ANS: Oncogenic proteins lead toward cancer, because they have excessive or unregulated activity in comparison with the corresponding normal proteins. Blocking this activity with a drug molecule that simply clogs the active site of the oncogenic protein will remove the danger. For a tumor suppressor gene, the danger lies in a loss of function, and there is generally no simple way for a drug molecule to restore a protein function that has been lost. It is therefore hard to see how we could achieve any useful effect on cancer by means of drugs that interfere with tumor suppressor gene products. A drug that simply inhibited their function would be expected to promote, not cure, cancer.

DIF: Hard REF: 20.4 OBJ: 20.4.k Differentiate between oncogenes and tumor suppressor genes and outline the types of genetic events that alter the activity of each. MSC: Evaluating 19. Rb is a tumor suppressor gene; its normal function is to help restrain cell division. Loss of both copies of Rb is a causative factor in some kinds of cancer. You propose to treat these cancers by injecting the patients with a viral

vector that carries a copy of the Rb gene and has the ability to infect all the cells of the body, thereby artificially driving expression of Rb in all the cells, including the cancer cells. But your colleague replies, “No! You’ll kill the patient, because you will halt cell division throughout the body.”

A. Why would halting cell division throughout the body kill a full-grown adult?

B. Is your colleague right in thinking that forced expression of Rb in every cell will halt all cell division?

ANS: A. Many vital tissues in the adult body, including the lining of the gut, the epidermis, and the system of blood cells, require continual renewal, which is dependent on stem cells and cell division. Cessation of cell division will lead to a disappearance of these tissues, with fatal consequences.

B. No, your colleague is not right to assume this. As explained in Chapter 18 (Figure 18-14), cells that express the Rb gene can proliferate: whether or not they do so depends on whether the Rb protein is phosphorylated (allowing cell division) or unphosphorylated (blocking cell division).

DIF: Hard REF: 20.4 OBJ: 20.4.k Differentiate between oncogenes and tumor suppressor genes and outline the types of genetic events that alter the activity of each. MSC: Applying 20. In 1971, Dr. Judah Folkman published the “angiogenic hypothesis” suggesting that a tumor cannot grow beyond 1–2 millimeters without the development of new blood vessels (angiogenesis) to provide access to oxygen and nutrients.

During the 1990s, it was discovered that vascular endothelial growth factor (VEGF) stimulates the proliferation and migration of the cells that form blood vessels, thereby inducing the formation of new blood vessels. VEGF binds to specific receptor tyrosine kinases (RTKs) on the cell surface and causes the RTKs to dimerize and become active, initiating an intracellular signaling cascade that stimulates cell division and inhibits apoptosis. Many cancer cells secrete high levels of VEGF, and increased VEGF expression in a tumor is correlated with a poor medical outcome for the patient.

Some evidence suggests that blocking VEGF-dependent signaling may prevent the formation of new blood vessels and lead to the death of immature blood vessels without disturbing mature blood vessels. You work for a biotechnology company that seeks to create anticancer drugs that prevent the growth of tumors and/or cause tumors to shrink, while leaving normal cells relatively untouched. After learning about VEGF, you have a bright idea for a new mechanism of action for a potential anticancer drug. How might you design this drug to block angiogenesis?

ANS: The findings on VEGF signaling suggest several strategies for blocking angiogenesis in and around tumors, which might be a powerful weapon in the anticancer arsenal. Several mechanisms of drug action may prevent angiogenesis, including the four possibilities below.

1. The drug may prevent the production or secretion of VEGF by tumor cells.

2. The drug may bind free VEGF in the extracellular space, thus lowering the effective concentration of VEGF and preventing it from binding to the RTKs on epithelial cells in blood vessels (indeed, the antiangiogenesis drug Avastin® functions in this way).

3. The drug may bind to RTKs and prevent binding to VEGF or dimerization.

4. The drug may block the intracellular signaling cascade triggered when VEGF binds and activates RTKs. DIF: Hard REF: 20.4 OBJ: 20.4.j Summarize the characteristics that distinguish cancer cells from normal cells and how they provide cancer cells with a competitive advantage. MSC: Creating 21. Your studies in cell biology have revealed how APC, a protein encoded by a tumor suppressor gene that is frequently

inactivated in people with colorectal cancer, functions in the Wnt signaling pathway (see Figure 20-21). This has inspired you to study Wnt signaling. You would like to design a drug to treat people with colorectal cancer. Given the pathway shown in Figure 20-21 and the knowledge that most human colorectal tumors harbor mutations in the APC gene, name a protein from the pathway that would be a good target for an activity-blocking anticancer drug. Explain your answer.

Figure 20-21

ANS: There are two possible answers: β-catenin and TCF. Because most colorectal cancers consist of cells that lack APC, drugs that interfere directly with the Wnt signal protein, with its receptor in the cell membrane, or with the intracellular signaling protein that acts upstream of APC will not be very useful. If APC is missing, large quantities of active β-catenin will be produced even in the presence of such drugs. However, drugs that block β-catenin or TCF activity would be good candidates, provided they do not interfere with vital functions of these proteins elsewhere in the body.

DIF: Hard REF: 20.4 OBJ: 20.4.m Explain how loss of the tumor suppressor gene APC can give rise to colorectal cancer, both in families predisposed to the condition and in patients with no family history of the disease. | 20.4.q Recall how mutations in the gene encoding beta-catenin can cause a similar effect as mutations in APC. MSC: Applying