TEST BANK for Molecular Biology of the Cell 7th Edition by Alberts,Heald,Johnson,Morgan,Raff et Al

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Molecular Biology of the Cell 7th Edition Bruce Alberts 2023

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Molecular Biology of the Cell 7th Edition by Bruce Alberts, Rebecca Heald, Alexander Johnson, David Morgan, Martin Raff, Keith Roberts, Peter Walter, John Wilson & Tim Hunt.

ISBN-10 0393884821 ISBN-13 978-039388482

Test bank for Molecular Biology of the Cell 7th Edition by Bruce Alberts, Rebecca Heald, Alexander Johnson, David Morgan, Martin Raff, Keith Roberts, Peter Walter, John Wilson & Tim Hunt

ISBN-10 0393884821 ISBN-13 978-039388482

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TABLE OF CONTENTS Chapter 1: Cells, Genomes, and the Diversity of Life

Chapter 2: Cell Chemistry and Bioenergetics

Chapter 3: Proteins

Chapter 4: DNA, Chromosomes, and Genomes

Chapter 5: DNA Replication, Repair, and Recombination

Chapter 6: How Cells Read the Genome: From DNA to Protein

Chapter 7: Control of Gene Expression

Chapter 8: Analyzing Cells, Molecules, and Systems

Chapter 9: Visualizing Cells and Their Molecules

Chapter 10: Membrane Structure

Chapter 11: Small-Molecule Transport and Electrical Properties of Membranes

Chapter 12: Intracellular Organization and Protein Sorting

Chapter 13: Intracellular Membrane Traffic

Chapter 14: Energy Conversion and Metabolic Compartmentation: Mitochondria and Chloroplasts

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Chapter 15: Cell Signaling

Chapter 16: The Cytoskeleton

Chapter 17: The Cell Cycle

Chapter 18: Cell Death

Chapter 19: Cell Junctions and the Extracellular Matrix

Chapter 20: Cancer

Chapter 21: Development of Multicelluar Organisms

Chapter 22: Stem Cells in Tissue Homeostasis and Regeneration

Chapter 23: Pathogens and Infection

Chapter 24: The Innate and Adaptive Immune Systems

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Molecular Biology of the Cell 7e-Alberts

Chapter 1: Cells, Genomes, and the Diversity of Life 1. What is NOT true of all cells? a. They have membrane transport proteins. b. They synthesize proteins on the ribosome. c. They replicate their genome by DNA polymerization. d. They transcribe their genetic information by RNA polymerization. e. They use RNA as a template for genomic DNA polymerization.

2. Imagine a segment of DNA (within a gene) encoding a certain amount of information in its nucleotide sequence. When this segment is fully transcribed into mRNA and then translated into protein, in general, what happens? a. The protein sequence would carry more information compared to the DNA and mRNA sequences, because its alphabet has 20 letters. b. The protein sequence would carry less information compared to the DNA and mRNA sequences, because several codons can correspond to one amino acid. c. The amount of information in the mRNA sequence is lower, because the mRNA has been transcribed using only one of the DNA strands as the template. d. The amount of information in the mRNA sequence is higher, because several mRNA molecules can be transcribed from one DNA molecule.

3. Scientists discover more than 10,000 new species of living organisms every year.

What is shared between all of these organisms? a. They are made of cells, whose nuclei enclose their DNA. b. They obtain their energy from sunlight. c. They produce and use adenosine triphosphate (ATP). d. Their genome contains at least 1000 genes.

Molecular Biology of the Cell 7e-Alberts 4. What is NOT true of the plasma membrane? a. It is totally impermeable. b. It contains proteins that move materials across the plasma membrane. c. It contains enzymes that catalyze reactions within the cell. d. It is made of molecules that are both hydrophilic and hydrophobic. e. It allows waste materials to be removed from the cell.

5. What is NOT a source of movement in a cell? a. Random thermal motions b. Gliding in a specific direction without energy input c. Motor proteins moving large assemblies d. Ions moving across a membrane due to ion pumps e. Proteins moving the cell forward

6. Which of the following is NOT a source of movement in a cell? a. Random thermal motions b. Gliding in a specific direction without energy input c. Motor proteins moving large assemblies d. Ions moving across a membrane due to ion pumps e. Proteins moving the cell forward

7. Which of the following processes that happens inside of a cell does NOT normally require consumption of free energy by the cell? a. Replication of the genetic material b. Import of nutrients from the environment c. Diffusion of small molecules within the cell d. Regulation of gene expression

Molecular Biology of the Cell 7e-Alberts e. Synthesis of enzymes that catalyze cellular reactions

8. What is NOT true regarding the tree of life? a. Most bacteria and archaea have 1000 to 6000 genes in their genomes. b. Eukaryotes are more similar to archaea than to bacteria with respect to the proteins that act on their DNA c. Most bacteria and archaea have genome sizes between 1 and 10 million nucleotide pairs, whereas eukaryotic genomes can be millions of times larger. d. Archaeal species were thought to belong to the eukaryotic world before sequence analysis placed them in a separate domain of life. e. Photosynthetic bacteria are thought to be the ancestors of the eukaryotic chloroplasts.

9. Three bacteria samples are harvested, each from different boiling volcanic mud pots on three different continents. In order to assess their evolutionary relationships, the researcher should do what? a. Identify the structures they use for motility. b. Compare their nutrient sources. c. Analyze the differences in their rRNA genes. d. Compare the molecular makeup of their outer cell wall. e. Identify the other species living nearby in the same ecosystem.

10. Several newly discovered organisms have recently had their genomes sequenced for the first time. Which sample is most likely a eukaryote?

Sample Name

Number of genes in the genome

S. ciencus

2554

S. tudentus

4352

Molecular Biology of the Cell 7e-Alberts P. fessora

26056

R. searcherus

6042

G. aduationa

2043

a. S. ceincus b. S. tudentus c. P. fessora d. R. searcherus e. G. aduationa

11. What would you NOT expect to find in a bacterial cell? a. Swimming using flagella b. Having a cell wall around the plasma membrane c. ATP production in mitochondria d. Protein production on the ribosome e. Sexual exchange of DNA with other bacteria

12. Imagine two spherical cells, one of which is 5000 times larger in volume than the other. The smaller is a prokaryote, and the larger cell is a eukaryote with 20% of its volume confined in a spherical nucleus. If the diameter of the prokaryotic cell is 0.7 micrometers, what is the diameter of the nucleus in the eukaryotic cell in micrometers? a. 5000 b. 1000 c. 7 d. 0.14

Molecular Biology of the Cell 7e-Alberts 13. What is true about carbon and nitrogen from the environment? a. They are easily used by all organisms living in that environment. b. They can be ingested and absorbed by cells for direct use in biomolecules. c. They are not necessary for life because they are inorganic. d. They must be fixed into organic molecules to be utilized by cells. e. They can only be found near volcanic vents in the earth’s crust.

14. To trace family relationships between distantly related organisms such as humans, algae, bacteria, and archaea, in what regions should their genomes be compared? a. Those that evolve rapidly b. Those that have a higher mutation rate c. Those that code for proteins d. Those where mutations are rarely tolerated e. Those where most mutations are selectively neutral

15. What is the order of the following evolutionary landmarks, from the oldest to the most recent?

I. Divergence of human and bird lineages II. Divergence of human and chimpanzee lineages III. Divergence of A. thaliana lineage from the conifer’s lineage IV. Divergence of fish and insect lineages I. IV, III, II, I II. IV, I, III, II III. I, IV, III, II IV. I, IV, II, III

Molecular Biology of the Cell 7e-Alberts 16. Laboratory strains of the model organism Escherichia coli that are resistant to antibiotics are very often used in research laboratories as well as in the biotechnology industry. If cultures of such bacteria were allowed to contaminate the environment uncontrollably, it is possible that at some point, pathogenic bacteria such as Neisseria meningitidis (which causes meningitis and can lead to death, especially in children) could acquire the same antibiotic-resistant gene, causing a meningitis outbreak that is difficult to treat. In this scenario, which of the following mechanisms is a more likely source of the antibiotic-resistant gene in N. meningitidis? a. Random new gene generation b. Intragenic mutation c. Gene duplication d. DNA segment shuffling e. Horizontal gene transfer

17. Which statement about a virus is true? a. It is a type of cell. b. It has genetic material made of proteins. c. It can only infect a single host species. d. It can act as a vector for gene transfer. e. It cannot persist in its host for more than one cell generation.

18. What does NOT typically involve horizontal gene transfer? a. Sexual reproduction in humans b. Bacteriophage infection of bacteria c. The evolutionary history of the eukaryotic cell d. The accidental duplication of a small region of a bacterial chromosome followed by cell division

Molecular Biology of the Cell 7e-Alberts e. Introduction of plasmids into bacteria in a laboratory

19. Gene duplication can give rise to homologous genes that are part of gene families. For example, there are six actin genes in the genome of most mammalian species. In humans, the ACTB gene, which encodes a cytoskeletal actin, is expressed ubiquitously, while ACTC1 is expressed mainly in cardiac cells. Although bacteria lack the eukaryotic cytoskeletal organization, the bacterial MreB gene bears recognizable sequence similarity to mammalian actin genes and codes for a protein that is similar to actin in structure and function. Which statement about these genes is true? a. ACTB is homologous to ACTC1 but not to MreB. b. ACTB is orthologous to ACTC1 but not to MreB. c. ACTB is paralogous to ACTC1 but not to MreB. d. MreB is orthologous to ACTB but not to ACTC1. e. ACTB is paralogous to both ACTC1 and MreB.

20. Out of nearly 5000 protein-coding gene families, there is a set of nearly 300 conserved gene families that are found in species from all domains of life. What is one of the general functions assigned to these gene families? a. The majority of them function in cell-to-cell signaling. b. The majority of them are poorly characterized. c. More than one-third of them are involved in translation or amino acid transport and metabolism. d. More than one-half of the shared families are involved in DNA replication and transcription. e. Nearly all of them are involved in energy production and carbohydrate metabolism.

21. Which structure is exclusively found in eukaryotic cells? a. Plasma membrane

Molecular Biology of the Cell 7e-Alberts b. Cell wall c. Chromosome d. Ribosome e. Lysosome

22. Which group of living organisms has the highest variation in haploid genome size? a. Mammals b. Fish c. Fungi d. Protozoa e. Prokaryotes

23. Mitochondria and chloroplasts are thought to have evolved from free-living aerobic bacteria that were engulfed by an ancestral anaerobic cell and established a mutually beneficial (symbiotic) relationship with it. Which statement about these organelles is NOT true? a. They are similar in size to small bacteria. b. They have their own circular genomic DNA. c. They have their own ribosomes. d. They have their own transfer RNAs. e. They are found in all eukaryotes.

24. In terms of cellular lifestyle, different kingdoms of life can be likened to hunters, farmers, and scavengers. Which statement regarding this scheme is true? a. The ancestral eukaryotic cell was a farmer, but it turned into a hunter once it acquired mitochondria.

Molecular Biology of the Cell 7e-Alberts b. Plant cells are considered scavengers, because their cell wall does not allow them to move. c. Most protozoa are hunters, whereas animal cells are farmers. d. The ancestral eukaryotic cell was a hunter, but upon acquiring chloroplasts it made the transition into farming. e. Fungi are scavengers without mitochondria.

25. Comparing the genomes of present-day mitochondria or chloroplasts with the genomes of their corresponding bacteria reveals that these organelles do not have many of the genes that are essential for their function. For instance, they lack the many genes that are required for DNA replication. What has happened to these genes? a. They have been lost during evolution, since the organelles no longer rely on DNA replication. b. The required genes are kept in the nucleus, but many have evolved by gene transfer from the organelle. c. These genes have undergone mutations and have changed beyond recognition, but are still present in the organelle. d. The organelles do not replicate their DNA; they import new DNA from the nucleus. e. The required genes are on plasmids that are separate from the organelle’s genome.

26. Based on the variation of genome size and gene number in the organisms presented in the following graph, which organism has the highest number of genes per unit length of their genome? (Note the logarithmic scale.)

Molecular Biology of the Cell 7e-Alberts

a. H. sapiens b. M. musculus c. A. thaliana d. C. elegans e. E. coli

27. All cells in a multicellular organism have normally developed from a single cell and share the same genome, but can nevertheless be wildly different in their shape and function. Which of the following in the eukaryotic genome is NOT responsible for this cell-type diversity? a. The genes that encode transcription regulatory proteins

Molecular Biology of the Cell 7e-Alberts b. The regulatory sequences that control the expression of genes c. The genes that code for molecules involved in receiving cellular signals d. The genes that code for molecules involved in sending cellular signals to other cells e. The duplication and deletion of genes in each cell’s genome

28. Didinium nasutum is a single-celled eukaryote that can hunt and feed on other living cells. It has an elaborate anatomy with beating cilia, a “mouth opening,” an “anal aperture,” and a set of contractile bundles; it can also shoot “darts” to paralyze its prey. What group of living cells does D. nasutum represent? a. Protozoa b. Yeasts c. Algae d. Animals e. Archaebacteria

29. A researcher is interested in discovering the function of gene RS0324 from a new yeast strain. By comparing the sequence of the gene to the other genomic DNA sequences in a public database, the researcher has found that gene RS0324 is homologous to cpeb genes in other organisms which produce a protein that helps with RNA translation. What should the researcher do next? a. Nothing. They already know that the gene helps in RNA translation. b. Use mutations of the gene to analyze whether the RNA translation function is disrupted. c. Add the gene sequence to a new model organism and see what happens. d. Replicate the gene sequence many times in a test tube to understand what the DNA is doing. e. Grow mouse models with extra copies of the gene sequence to analyze whether more RNA translation is occurring.

Molecular Biology of the Cell 7e-Alberts

30. Which statement is NOT true about bacteriophages? a. They are a category of viruses. b. They infect bacteria. c. They have been used to study many molecular biological processes. d. They have no effect on the earth’s ecology. e. They are too small to see even using a light microscope.

31. Which statement regarding Escherichia coli is true? a. Most of our understanding about mitosis comes from studies on this model organism. b. It is a rod-shaped bacterium that can only grow in the gut of humans and other vertebrates. c. Two strains of E. coli can differ by up to 0.1% in their genomes. d. E. coli strain K-12 encodes about 4300 proteins. e. The E. coli (strain K-12) genome is about 430 million nucleotide pairs long.

32. A mutation in the cdc28 gene in the budding yeast Saccharomyces cerevisiae causes cell-cycle arrest, giving rise to unbudded cells that look like “dumbbells.”

Treatment of wild-type cells with nocodazole, a drug that destabilizes some cytoskeletal polymers, leads to a similar phenotype. Based only on these observations, which statement is true regarding cdc28? a. cdc28 codes for a master regulatory kinase that phosphorylates other proteins. b. Nocodazole binds to the protein coded by the cdc28 gene. c. The product of the cdc28 gene is responsible for resistance to nocodazole. d. The product of the cdc28 gene is involved in cell-cycle regulation.

Molecular Biology of the Cell 7e-Alberts e. The product of cdc28 destabilizes the same cytoskeletal polymers that nocodazole also destabilizes.

33. This model organism is used to study various cell processes such as regulation of the eukaryotic cell cycle. Mutants are available for every gene in its exceptionally small genome. It can live indefinitely in either a haploid or a diploid state. Which characteristic most likely describes this organism? a. It can reproduce only asexually. b. It is a fungus. c. It lacks a cell wall. d. Its cell cycle is typically much slower than that of human cells. e. It contains chloroplasts.

34. While knowing the sequence of a genome is a start to understanding the functions inside a cell, many researchers also are interested in monitoring the expression levels of all the genes in a genome. What would be an application of this technique? a. Analyzing if a gene is similar in sequence to any other genes b. Understanding how a cell responds to being placed in a tumor-promoting environment versus its normal environment c. Understanding the evolutionary relationships between two species d. Analyzing how large a certain genome is e. Discovering if a disorder is genetic or not

35. This model organism is used to study various eukaryotic cell and developmental processes such as cell division and cell death. Its hermaphrodite adult is composed of exactly 959 somatic (non-germ) cells, the lineage of each of which has been worked out with great precision. It is approximately 1 mm long. What is the best description of this organism? a. It is a vertebrate.

Molecular Biology of the Cell 7e-Alberts b. It is a plant pathogen that destroys many crops. c. Its genome codes for a few thousand genes. d. It can fly. e. It can be frozen indefinitely in a state of suspended animation.

36. Judged by the average number of nucleotide-pair differences per 1000 nucleotide pairs, which pair shows the highest difference? a. The genomes of S. cerevisiae and M. musculus b. The genomes of two different E. coli strains c. The ribosomal RNA genes from humans and E. coli d. The transfer RNA genes in a human and M. musculus e. The genomes of two humans

37. This model organism is used to study various cell and developmental processes such as the growth of developing body parts in the right place and with the correct shape. It develops from a fertilized egg to an adult in a little over a week, and has been a favorite of geneticists for almost a century. Some of its cells have giant chromosomes whose banding patterns have been extremely helpful in classical genetic studies. What is the best description of this organism? a. It is a vertebrate. b. Its genome is only 10 million nucleotide pairs long. c. There are many more duplicate genes in this organism compared to humans. d. Although useful for genetic studies, the molecular mechanisms governing its development are irrelevant to human development. e. It normally only reproduces sexually.

38. This model organism is particularly well suited for studying developmental

Molecular Biology of the Cell 7e-Alberts processes in higher animals. It develops from a fertilized egg to an adult in only two to three months, and its body is transparent for the first two weeks, making it easy to observe cell behavior during development. What is NOT a characteristic that describes this organism? a. It is a vertebrate. b. It is well suited for genetic analysis. c. Its early stages of development occur outside of the mother’s body. d. Its genome size is almost half that of humans. e. It can fly.

39. What is NOT a purpose of the 27 proteins produced by the SARS-CoV-2 genome? a. Packaging for the viral genome b. Replicating the viral genome inside the host cell c. Assuring that the viral genes are properly translated into protein d. Evading the host cell defenses e. Decreasing the host cell’s oxygen supply

40. Why do viruses support the “RNA world” hypothesis of how life originated? a. The smallest viruses contain only a single-stranded RNA molecule. b. Viruses use RNA molecules as catalysts. c. Viruses target and attack host RNA molecules. d. Viruses are made exclusively of RNA. e. Some viruses contain double-stranded RNA.

41. Which of the following descriptions match RNA, and which match DNA?

Molecular Biology of the Cell 7e-Alberts a. RNA b. DNA

1. It is mainly found as a long, double-stranded molecule.

2. It contains the sugar ribose.

3. It normally contains the bases thymine, cytosine, adenine, and guanine.

4. It can normally adopt distinctive folded shapes.

5. It can be used as the template for protein synthesis.

42. Which of the following descriptions match messenger RNAs, ribosomal RNAs, and transfer RNAs, respectively? a. Messenger RNAs b. Ribosomal RNAs c. Transfer RNAs

1. They contain codons.

2. They contain anticodons.

3. They are (covalently) attached to amino acids.

4. They are at the core of a complex that carries out protein synthesis.

43. In the following paragraph, fill in the blanks with either Archaea, Bacteria, or Eukaryotes.

Methanococcus jannaschii is an anaerobic thermophilic microbe that belongs to the domain _____(I)______ and is found in extreme environments such as hydrothermal vents. It lives at temperatures near the boiling point of water and at pressures over 200 times higher than at sea level. Its genome sequence, identified in 1996, revealed that most of its metabolic processes are similar to those in _____(II)______, while its genetic machinery is more similar to that of _____(III)______. a. Archaea b. Bacteria c. Eukaryotes

1. (I)

Molecular Biology of the Cell 7e-Alberts 2. (II) 3. (III)

44. Which of the following descriptions match lithotrophic, organotrophic, and phototrophic organisms, respectively? a. Lithotrophic b. Organotrophic c. Phototrophic

1. They feed on other living organisms or their organic products.

2. They are responsible for the current oxygen-rich atmosphere of the Earth.

3. They are all known to be prokaryotic.

4. They are the primary energy converters in hydrothermal vents in the ocean floor.

45. If animals are found in an isolated ecosystem, there should be photosynthetic organisms in that ecosystem as well. a. True b. False

46. While oceans cover approximately 70% of the earth’s surface, only 1% of the earth’s biomass is present there. Considering that the largest contributor to earth’s biomass is plants, why do you think most of the earth’s biomass is terrestrial instead of in the oceans?

47. A scientist has four different samples: a rod-shaped bacterial cell with flagella from a thermal vent deep in the ocean, a spherical bacterial cell from the glaciers high in the Himalayas, a yeast cell from bread, and a single-celled protist with cilia from a pond. Explain the best way to understand the evolutionary relationships between these four organisms. Why does this method work better than others?

48. All the cells in a multicellular organism contain the same DNA despite the massive differences in those cells’ appearance and function. Explain how this is

Molecular Biology of the Cell 7e-Alberts possible.

49. In this chapter, several model organisms are discussed. Based on what you have learned about why researchers use these organisms, explain what sorts of qualifications make for a good model organism.

50.

What are the benefits of studying “the human genome?” People from a certain ethnic group tend to have more commonalities in their genomes than they do with those outside of their ethnic group. If researchers intend for their genomic studies to benefit all groups of people, why might it be important to ensure that there is adequate representation from all ethnic groups in human genome databases?

Justify your response in 2-3 sentences.

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 01 1. Answer:

E

2. Answer:

B

3. Answer:

C

4. Answer:

A

5. Answer:

B

6. Answer:

B

7. Answer:

C

8. Answer:

D

9. Answer:

C

10. Answer:

C

11. Answer:

C

12. Answer:

C

13. Answer:

D

14. Answer:

D

15. Answer:

B

16. Answer:

E

17. Answer:

D

18. Answer:

D

19. Answer:

C

20. Answer:

C

Molecular Biology of the Cell 7e-Alberts 21. Answer:

E

22. Answer:

D

23. Answer:

E

24. Answer:

D

25. Answer:

B

26. Answer:

E

27. Answer:

E

28. Answer:

A

29. Answer:

B

30. Answer:

D

31. Answer:

D

32. Answer:

D

33. Answer:

B

34. Answer:

B

35. Answer:

E

36. Answer:

A

37. Answer:

E

38. Answer:

E

39. Answer:

E

40. Answer:

A

41. Answer:

1. B 2. A 3. B

Molecular Biology of the Cell 7e-Alberts 4. A 5. A 42. Answer:

1. A 2. C 3. C 4. B

43. Answer:

1. A 2. B 3. C

44. Answer:

1. B 2. C 3. A 4. A

45. Answer:

B

46. Answer:

Plants contain chloroplasts and are photosynthetic. This means that they require sunlight in order to produce their own organic molecules. Most of the ocean is not accessible to sunlight, and therefore plants cannot grow there. Thus, most plant biomass is found on land. Plants are also major producers of organic molecules due to their ability to photosynthesize, and a large amount of plant material is necessary to provide organic molecules to the many organotrophic organisms that share the same ecosystems.

47. Answer:

These four specimens, though all single cells, are very different. We have two bacteria, a fungus, and a protist.

While many would be tempted to characterize based on appearance, habitat or lifestyle, genomic analysis will be the best method to analyze the evolutionary relationships between these four samples. The number of differences in their DNA sequences (and specifically their genes for rRNA) will help determine how close each organism’s evolutionary relationship is. The more differences there are in their rRNA genes, the more distantly they are related, meaning the longer ago their last common ancestor lived.

Molecular Biology of the Cell 7e-Alberts 48. Answer:

The differences in different cell types in a multicellular organism result from different genes being expressed in different amounts or different times. This is based on the different signals received by those cells and then different responses to those signals. These are all controlled by regulatory elements acting on the DNA that determine when and how much of a gene will be transcribed.

49. Answer:

There are numerous ways to answer this question. Below are a few examples: Model organisms are usually plentiful in number and easy to care for. This makes them cheap for a researcher to use over and over again.

Model organisms usually reproduce quickly. This allows researchers to produce more of the organisms easily for more experiments and/or generational studies.

Model organisms have certain qualities that make them easy to study. For example, large eggs, large neurons, known chromosomal patterns, and transparency.

Model organisms have been the subjects of many other studies and much is already known about them. This allows a researcher to be able to ask better questions and design better experiments using the organism.

50. Answer:

Studying the human genome can help us understand how people with certain genetic makeups will respond to certain drugs, how they are more or less susceptible to disease, and provide better understanding of human history.

However, if certain groups are overrepresented in human genome databases, it is possible that the information we gain from these studies will primarily benefit those populations and exclude other groups that are underrepresented. This is because we will not have a number of data points needed to understand how those underrepresented people groups relate to the study being done.

Molecular Biology of the Cell 7e-Alberts

Chapter 2: Cell Chemistry and Bioenergetics 1. Which chemical do you NOT expect to be readily dissolved in water? a. Uric acid b. Hexane c. Glycerol d. Ethanol e. Potassium chloride

2. Weak noncovalent attractions in the cell can be very strong in a nonaqueous environment. Some of these attractions are as strong as covalent interactions in a vacuum (their bond energy is approximately 340 kJ/mole), but become more than 25 times weaker (their bond energy becomes approximately 13 kJ/mole) in water.

What type of attraction shows this phenomenon? a. Electrostatic attractions b. Hydrogen bonds c. van der Waals attractions d. Hydrophobic force

3. The bond energies associated with noncovalent attractions in the cell are too weak to resist disruption by thermal motion. However, cellular macromolecules can interact specifically AND strongly with each other (or fold by themselves) merely via such interactions. How is this possible? a. The bond energies increase radically when two interacting molecules approach each other. b. The interacting molecules also fortify their binding via covalent bonds to keep them from dissociation. c. Many weak bonds together in a complementary geometry can afford a strong binding. d. The cell lowers its internal temperature to reduce thermal motion of

Molecular Biology of the Cell 7e-Alberts molecules and enhance the weak attractions.

4. What is the pH of a 10−8 M solution of hydrochloric acid? Round the pH value to the nearest whole integer. a. 8 b. 7 c. 6 d. 5 e. 4

5. The cell can change the pH of its internal compartments using membrane transport proteins that pump protons into or out of a compartment. How many protons should be pumped into an endocytic vesicle that is 10−15 liters in volume and has a neutral pH in order to change the pH to 5? Avogadro’s number is 6 × 1023. Omit complications such as the membrane potential, buffers, and other cellular components. a. 6000 b. 60,000 c. 120,000 d. 600,000 e. 6,000,000

6. What trait is typical for a hydrophobic molecule? a. It is able to form hydrogen bonds with itself but not with water. b. It is able to form hydrogen bonds with water. c. It is charged. d. It is hard to dissolve in a solvent.

Molecular Biology of the Cell 7e-Alberts e. It is incapable of interacting favorably with water.

7. Which statement regarding a fatty acid molecule in water is true? a. It is positively charged at physiological pH, but can become neutral when the pH is high enough. b. It is positively charged at physiological pH, but can become neutral when the pH is low enough. c. It is negatively charged at physiological pH, but can become neutral when the pH is high enough. d. It is negatively charged at physiological pH, but can become neutral when the pH is low enough. e. It is not charged at physiological pH.

8. The amino acid serine has an amino group, a carboxyl group, and a hydroxyl group. Which of the following better represents the structure of this amino acid at neutral pH? a. It is positively charged at physiological pH, but can become neutral when the pH is high enough. b. It is positively charged at physiological pH, but can become neutral when the pH is low enough. c. It is negatively charged at physiological pH, but can become neutral when the pH is high enough. d. It is negatively charged at physiological pH, but can become neutral when the pH is low enough. e. It is not charged at physiological pH.

9. The amino acid serine has an amino group, a carboxyl group, and a hydroxyl group. Which of the following better represents the structure of this amino acid at neutral pH?

Molecular Biology of the Cell 7e-Alberts

a.

b.

c.

d.

e.

10. The concentrations of H3O+ and OH− are kept relatively equal inside of the cell due to what? a. The presence of buffers b. Enzymes that destroy excess H3O+ and OH− c. Diffusion of H3O+ and OH− through the cell membrane d. Digestion of H3O+ and OH− by lysosomes

Molecular Biology of the Cell 7e-Alberts e. Utilization of H3O+ and OH− in cellular respiration

11. Which of the following elements is not normally found in cells? a. Copper b. Iron c. Silver d. Cobalt e. Zinc

12. Which of the following is NOT true of carbon? a. It has four electrons in its outermost electron shell. b. It has four vacancies in its outermost electron shell. c. It can form C—C bonds which are highly unstable. d. It forms the basis of all organic molecules. e. It can form four covalent bonds.

13. The three families of cellular macromolecules are polymerized and depolymerized by a general mechanism involving water. Each of them has a set of monomers whose polymerization changes the total free energy of the system. Which of the following statements is true regarding these macromolecules? a. Each polymerization step requires free-energy input and proceeds by the consumption of one water molecule. b. Each depolymerization step requires free-energy input and proceeds by the consumption of one water molecule. c. Each polymerization step requires free-energy input and proceeds by the release of one water molecule. d. Each depolymerization step requires free-energy input and proceeds by the

Molecular Biology of the Cell 7e-Alberts release of one water molecule.

14. What is the correct order of a low to a high contribution to the total mass of an E. coli bacterium? a. Sugars, nucleic acids, proteins, water b. Nucleic acids, sugars, proteins, water c. Sugars, proteins, nucleic acids, water d. Nucleic acids, proteins, sugars, water

15. Macromolecules can adopt a vast number of conformations due to what action, but many macromolecules are held in specific shapes due to what mechanism? a. Thermal motion, covalent bonds b. Thermal motion, water interactions c. Proton movement, covalent bonds d. Proton movement, noncovalent interactions e. Thermal motion, noncovalent interactions

16. Two different proteins transiently associate with one another in the cell. Which of the following would NOT be a means by which these two proteins come together? a. Electrostatic interactions b. Hydrogen bonds c. Van der Waals interactions d. Covalent bonds e. Hydrophobic force

17. What is the correct summarization of the overall process of photosynthesis?

Molecular Biology of the Cell 7e-Alberts a. CO2 + O2 → H2O + sugars b. CH2O + CO2 + O2 → H2O + sugars c. CO2 + H2O → H2 + CO2 d. CO2 + H2O → O2 + sugars

18. Regarding reactions involving oxidation and reduction, which statement is true? a. The carbon atom is more oxidized in formaldehyde (CH2O) than in methanol (CH3OH). b. Oxidation of food in all organisms requires oxygen. c. A molecule is oxidized if it gains an electron (plus a proton) in a reaction. d. A dehydrogenation reaction is a reduction. e. In an organic molecule, the number of C—H bonds increases as a result of oxidation.

19. Enzymes are the cell’s catalyst crew. They make the life of the cell possible by carrying out various reactions with astounding performance. Which of the following is NOT true regarding cellular enzymes? a. Enzymes lower the activation energy of the reactions that they catalyze. b. Enzymes can specifically drive substrate along certain reaction pathways. c. Enzymes can push energetically unfavorable reactions forward by coupling them to energetically favorable reactions. d. Enzymes are proteins, but RNA catalysts, called ribozymes, also exist. e. Enzymes can change the equilibrium point for reactions that they catalyze.

20. Regarding cellular metabolism, which statement is true? a. A living organism decreases the entropy in its surroundings.

Molecular Biology of the Cell 7e-Alberts b. During catabolism, heat is generated, and the cell uses this heat to perform work during anabolism. c. The heat released by an animal cell as part of its metabolic processes comes from the bond energies in the foodstuffs that are consumed by the animal. d. Living organisms defy the second law of thermodynamics, but still obey the first law.

21. In the following diagram showing the reaction pathway for a simple singlesubstrate enzymatic reaction, which of the quantities corresponds to the activation energy of the forward reaction?

a. (a − b) b. (a + b) c. (a − c) d. (a + c) e. (b − c)

22. In the following diagram showing the distribution of thermal energy in a population of substrate molecules, the energy thresholds indicated by numbers represent what?

Molecular Biology of the Cell 7e-Alberts

a. The activation energy at high and low temperature b. The reaction rate at high and low pH c. The activation energy with and without an enzyme d. The reaction rate at high and low substrate concentrations e. The activation energy at high and low substrate concentrations

23. The molecules inside the cell constantly collide with other molecules and diffuse through the cytoplasm in a random walk. The average net distance traveled by such a molecule after a certain time period t is proportional to the square root of t, that is, (t)0.5, as well as to its diffusion coefficient. If, on average, it takes a molecule 100 milliseconds to travel a net distance of 0.5 µm from its starting point, how long would it normally take for the same molecule to travel a net distance of 5 µm from the same starting point? a. 0.2 seconds b. 0.3 seconds c. 1 second d. 10 seconds e. 0.32 seconds

24. A reaction can proceed spontaneously if what occurs?

Molecular Biology of the Cell 7e-Alberts a. It increases the disorder in the universe. b. It forms a chemical bond. c. It results in the loss of an H2O molecule. d. It requires an input of energy. e. It does useful work for the cell.

25. The enzyme phosphoglucose isomerase converts glucose 6-phosphate to its isomer fructose 6-phosphate in the second step of glycolysis. The equilibrium constant for the reaction is 0.36. Evaluating the ΔG° of the reaction (ΔG° = −5.9 × log Keq), decide which conclusion is true. a. The ΔG° is negative, therefore the reaction proceeds in the forward direction. b. The ΔG° is negative, but whether or not the reaction proceeds would depend on ΔG, not ΔG°. c. The ΔG° is positive, but in a cell that is active in glycolysis, the reaction can still proceed in the forward direction. d. The ΔG° is positive, therefore the reaction proceeds in the reverse direction.

26. The equilibrium constant for the reaction that breaks down each molecule of substrate A to one molecule of B and one molecule of C is equal to 0.5. Starting with a mixture containing only molecules A at 1 M concentration, what will be the concentration of molecule A after reaching equilibrium under these conditions? a. 0.5 M b. 0.25 M c. 0.125 M d. 0.333 M e. 0.667 M

Molecular Biology of the Cell 7e-Alberts

27. In the first reaction of the glycolytic pathway, the enzyme hexokinase uses ATP to catalyze the phosphorylation of glucose, yielding glucose 6-phosphate and ADP. The ΔG° value for this reaction is −17 kJ/mole. The enzyme glucose 6phosphatase catalyzes a “reverse” reaction, in which glucose 6-phosphate is converted back to glucose, and a phosphate is released. The ΔG° value for this reaction is −14 kJ/mole. What is the ΔG° value for the following reaction: ATP + H2O → ADP + Pi a. −3 kJ/mole b. +3 kJ/mole c. −31 kJ/mole d. +31 kJ/mole e. −15.5 kJ/mole

28. Imagine the reaction A → B with a negative ΔG value under experimental conditions. Which statement is true about this reaction? a. The reaction is energetically unfavorable. b. The reaction proceeds spontaneously and rapidly under these conditions. c. Increasing the concentration of B molecules would increase the ΔG value (toward more positive values). d. The reaction would result in a net decrease in the entropy (disorder) of the universe. e. The reaction cannot proceed unless it is coupled to another reaction with a positive value of ΔG.

29. A cellular enzyme catalyzes the catabolic reaction shown below. Its coenzyme is shown in the box. Which statement regarding this reaction is correct?

Molecular Biology of the Cell 7e-Alberts

a. The substrate is reduced in this reaction and the coenzyme is converted from state 1 to state 2. b. The substrate is oxidized in this reaction and the coenzyme is converted from state 1 to state 2. c. The substrate is reduced in this reaction and the coenzyme is converted from state 2 to state 1. d. The substrate is oxidized in this reaction and the coenzyme is converted from state 2 to state 1.

30. Which of the following represents an “activated” carrier molecule? a. AMP b. NADH c. NAD+ d. NADP+

Molecular Biology of the Cell 7e-Alberts e. CoA

31. Imagine that in an ATP molecule, the phosphate closest to the sugar ring is P1, the next phosphate away from the ring is P2, and the phosphate farthest from the ring is P3. In the hydrolysis of ATP to ADP, which bond is going to be broken to release energy? a. P1 with the C5 from the sugar b. P1 with P2 c. P2 with P3 d. P3 with P1 e. Any bond can be hydrolyzed in this process

32. In order to perform the synthesis reaction M—H + N—OH → Q, energy from ATP is used. What is a likely intermediate in this reaction? a. M—(PO4)3 b. N—O—PO3 c. Q—O—PO3 d. M—PO3—N e. N—(PO3) 3

33. Despite their overall similarity, NADH and NADPH are not used indiscriminately by the cell. What are the distinctive features of these two carrier molecules? a. NADPH has an extra phosphate near its nicotinamide ring, giving it distinct electron-transfer properties. b. In the cell, NADH is usually in excess over NAD+, but NADP+ is usually in excess over NADPH. c. NADH is normally involved in anabolic reactions, whereas NADPH is

Molecular Biology of the Cell 7e-Alberts normally involved in catabolism. d. Both NADPH and NADH are recognized by the same enzymes with similar affinities, since the extra phosphate group in NADPH is not involved in such recognition. e. In the cell, NADH is found mostly in the form that acts as an oxidizing agent, whereas NADPH is found mostly in the form that acts as a reducing agent.

34. In an enzymatic reaction involving NADH or NADPH, reduction of a substrate accompanies the oxidation of these carrier molecules to NAD + or NADP+, respectively. What else typically happens in such a reaction? a. A molecule of water is released to the solution upon completion of the reaction. b. A proton is released during the oxidation of the carriers. c. A proton is taken up by the substrate that is being reduced. d. A proton is taken up by the carrier molecule that is being oxidized. e. A phosphate group is transferred to the substrate.

35. What is the reaction performed on the molecule labeled as substrate in the following diagram? What is the name of the activated carrier?

Molecular Biology of the Cell 7e-Alberts

a. This is a methylation reaction and the activated carrier is ATP. b. This is a methylation reaction and the activated carrier is Sadenosylmethionine. c. This is a carboxylation reaction and the activated carrier is ATP. d. This is a carboxylation reaction and the activated carrier is carboxylated biotin. e. This is an acetylation reaction and the activated carrier is acetyl CoA.

36. ATP is the main energy currency in cells, and it can especially be used to drive condensation reactions that produce macromolecular polymers. How does ATP normally catalyze the condensation reaction, which by itself is energetically unfavorable? a. It transfers its terminal phosphate to an enzyme and is released as ADP. b. It transfers its two terminal phosphates to an enzyme, and is released as AMP. c. It covalently attaches to both of the substrates. d. It transfers either one or two terminal phosphate(s) to one of the substrates and is released as either ADP or AMP. e. It covalently attaches to the enzyme, forming an enzyme–AMP adduct.

Molecular Biology of the Cell 7e-Alberts

37. Macromolecules in the cell can be made from their monomers using one of two polymerization schemes. One is called head polymerization, in which the reactive bond required for polymerization is carried on the end of the growing polymer. In contrast, in tail polymerization, the reactive bond is carried by each monomer for its own incorporation. In the figure, indicate the polymerization scheme and the type of macromolecule.

a. Head polymerization of a protein b. Tail polymerization of a protein c. Head polymerization of a polysaccharide d. Head polymerization of a nucleic acid e. Tail polymerization of a nucleic acid

38. What is the end product of glycolysis in the cytoplasm of eukaryotic cells? How many carbon atoms does the molecule have? a. Acetyl CoA; it has two carbon atoms attached to coenzyme A b. Phosphoenolpyruvate; it has three carbon atoms

Molecular Biology of the Cell 7e-Alberts c. Glucose 6-phosphate; it has six carbon atoms d. Pyruvate; it has three carbon atoms e. Glyceraldehyde 3-phosphate; it has three carbon atoms

39. Which description matches glycolysis rather than the Krebs cycle? a. It oxidizes acetyl CoA to CO2. b. In eukaryotic cells, it is carried out in the cytosol. c. It produces FADH2. d. α-Ketoglutarate, one of its intermediates, is used to synthesize the amino acid glutamic acid.

40. Which molecule is an intermediate in glycolysis rather than in the tricarboxylic acid cycle? a. Fumarate b. Malate c. Phosphoenolpyruvate d. Succinate

41. The substrate for the glycolytic enzyme glyceraldehyde 3-phosphate dehydrogenase is glyceraldehyde 3-phosphate (with one phosphate group) while its product is 1,3-bisphosphoglycerate (with two phosphate groups). Where does the extra phosphate group come from? a. From combining two molecules of the substrate b. ATP c. Fructose 1,6-biphosphate d. Pi

Molecular Biology of the Cell 7e-Alberts e. NADH

42. Which of the following is NOT performed in steps 6 and 7 of glycolysis, catalyzed by the enzymes glyceraldehyde-3-phosphate dehydrogenase and phosphoglycerate kinase, respectively? a. The oxidation of an aldehyde to a carboxylic acid b. Production of both ATP and NADH c. Coupling the oxidation of a C—H bond to the activation of carrier molecules d. Catalyzing the only glycolytic reactions that create a high-energy phosphate linkage directly from inorganic phosphate e. Loss of an inorganic phosphate as ATP is hydrolyzed to ADP

43. Arsenate is a toxic ion that can interfere with both glycolysis and oxidative phosphorylation. Arsenate resembles Pi (inorganic phosphate) and can replace it in many enzymatic reactions. One such reaction is catalyzed by glyceraldehyde 3-phosphate dehydrogenase in step 6 of glycolysis. Upon completion of the reaction, instead of the normal product, 1,3-bisphosphoglycerate, the mixed anhydride 1-arsenato-3-phosphoglycerate is formed; this undergoes rapid spontaneous hydrolysis into arsenate plus 3-phosphoglycerate, the latter being a normal product of step 7 in glycolysis. What would be the effect of arsenate poisoning in glycolysis? a. It results in more ATP and NADH molecules generated for every glucose molecule. b. It results in fewer ATP molecules generated per glucose molecule, but NADH generation is not directly affected. c. It brings glycolysis to an abrupt stop. d. It results in fewer ATP and NADH molecules generated per glucose molecule. e. It does not affect the number of ATP or NADH molecules generated per glucose molecule.

Molecular Biology of the Cell 7e-Alberts 44. The following molecules must be sorted based on the amount of energy that is released when their phosphate bond is hydrolyzed as indicated. Choose the correct order. The molecule with the highest amount of hydrolysis energy will be on the left. a. 1,3-bisphosphoglycerate when hydrolyzed to 3-phosphoglycerate, phosphoenolpyruvate when hydrolyzed to pyruvate, ATP when hydrolyzed to ADP, glucose 6-phosphate when hydrolyzed to glucose b. Phosphoenolpyruvate when hydrolyzed to pyruvate, 1,3bisphosphoglycerate when hydrolyzed to 3-phosphoglycerate, ATP when hydrolyzed to ADP, glucose 6-phosphate when hydrolyzed to glucose c. 1,3-bisphosphoglycerate when hydrolyzed to 3-phosphoglycerate, phosphoenolpyruvate when hydrolyzed to pyruvate, glucose 6-phosphate when hydrolyzed to glucose, ATP when hydrolyzed to ADP d. Phosphoenolpyruvate when hydrolyzed to pyruvate, 1,3bisphosphoglycerate when hydrolyzed to 3-phosphoglycerate, glucose 6phosphate when hydrolyzed to glucose, ATP when hydrolyzed to ADP

45. Under anaerobic conditions, glycolysis provides most of the ATP that the cell needs. In animal cells, pyruvate, the end product of glycolysis, is converted to lactic acid by lactate dehydrogenase, as shown below: CH3(CO)COO− + X → CH3(CHOH)COO− + Y What is the correct carrier pair (in place of X and Y) in this reaction? a. X is (ADP + Pi), and Y is (ATP) b. X is (NADP+), and Y is (NADPH + H+) c. X is (NAD+), and Y is (NADH + H+) d. X is (NADH + H+), and Y is (NAD+) e. X is (NADP++ H+), and Y is (NADPH)

46. Regarding energy production and storage in plants and animals, which statement is true?

Molecular Biology of the Cell 7e-Alberts a. Plant and animal cells make starch for long-term energy storage. b. Most of the ATP in a plant cell has been generated in the chloroplast and transported to other parts of the cell. c. Oxidation of one gram of starch releases more energy than oxidation of fat, but since starch absorbs a lot of water, it is not as efficient as fat in energy storage. d. Animals, but not plants, can store fats in the form of triacylglycerol (triglyceride). e. Plant seeds often contain large amounts of fats and starch.

47. During a fasting state, the body breaks down amino acids to oxidize in the citric acid cycle instead of sugars. What other molecules can be readily oxidized in the citric acid cycle in many types of cells? a. ATP b. Ketone bodies c. Lactic acid d. Ethanol e. CO2

48. What are the molecules that normally supply carbon and oxygen atoms, respectively, for the citric acid cycle? a. Oxaloacetate, oxaloacetate b. Acetyl CoA, O2 c. Oxaloacetate, O2 d. Acetyl CoA, H2O e. Pyruvate, pyruvate

Molecular Biology of the Cell 7e-Alberts Below is a schematic drawing of the citric acid cycle. Answer the following questions about this cycle.

49. What is the molecule indicated by a question mark? a. O2 b. ATP c. H2O d. H+ e. Pyruvate

50. In the citric acid cycle, which steps produce CO2 as a by-product? a. Steps 1 and 3 b. Steps 2 and 3 c. Steps 3 and 4 d. Steps 3 and 5

Molecular Biology of the Cell 7e-Alberts e. Steps 4 and 5

51. In the citric acid cycle, which steps produce either NADH or FADH2? a. Steps, 1, 3, 4, and 8 b. Steps 1, 2, 4, and 8 c. Steps 3, 4, 6, and 8 d. Steps 2, 3, 4, and 8 e. Steps 3, 4, 6, and 8

52. Aconitase catalyzes an isomerization reaction in the citric acid cycle, in which H2O is first removed and then added back to the substrate. Which step is catalyzed by this enzyme? a. Step 1 b. Step 2 c. Step 3 d. Step 4 e. Step 5

53. The electron carriers NADH and FADH2 donate their electrons to the electrontransport chain in the inner mitochondrial membrane, leading to ATP synthesis powered by an H+ gradient across the membrane. If, on average, the oxidation of each NADH or FADH2 molecule in this pathway results in the production of 2.5 and 1.5 molecules of ATP, respectively, how many ATP (and GTP) molecules are produced on average as a result of the complete oxidation of one molecule of acetyl CoA in the mitochondrion? Consider only the citric acid cycle and oxidative phosphorylation. a. 10 b. 12

Molecular Biology of the Cell 7e-Alberts c. 13.5 d. 14.5 e. 15

54. Fatty acids are stored in fat droplets in the form of triacylglycerols. a. True b. False

55. In the course of glycolysis, one molecule of NADH is formed per molecule of glucose. a. True b. False

56. Oxaloacetate is only used in the citric acid cycle. a. True b. False

57. Nitrogen fixation occurs in the mitochondria in most animal cells to generate amino acids. a. True b. False

58. For each of the following pairs, indicate whether they interact via hydrogen bonds or ionic bonds, or if they do not favorably interact. a. Hydrogen bonds b. Ionic bonds c. Do not favorably interact

1. ATP and Mg2+

Molecular Biology of the Cell 7e-Alberts 2. Urea and water 3. Glucose and the enzyme hexokinase (which uses glucose as a substrate) 4. A phospholipid tail and inorganic phosphate

59. Indicate which description(s) match lipids, nucleic acids, polysaccharides, or proteins, respectively. a. Lipids b. Nucleic acids c. Polysaccharides d. Proteins

1. Their monomers contain phosphorus and nitrogen.

2. They constitute almost half of the cell’s dry mass.

3. They are the main constituent of all cellular membranes.

4. They are largely hydrophobic and can store energy.

60. In the following diagram, match each part of the diagram with its best cellular counterpart.

a. Falling rocks b. Paddle wheel c. Bucket of water

Molecular Biology of the Cell 7e-Alberts d. Hydraulic machines

1. Activated carrier molecule 2. Energetically unfavorable reactions 3. Energetically favorable reactions 4. Enzymes

61. Consider the diagram below. In 2-3 sentences, explain why heat energy must leave the cell.

62.

The following molecules (A to E) must be sorted based on the oxidation of the carbon atom, from higher to lower oxidation states. The letter corresponding to the highest oxidation level will be on the left. Your answer should be a five-letter string, for example, ABCDE.

Molecular Biology of the Cell 7e-Alberts

63. Sort the following molecules from a low to high rate of diffusion inside the cytosol.

Your answer would be a four-letter string composed of letters A to D only, for example, ADCB. A. Myoglobin (a protein) B. Glycine (an amino acid) C. Ribosome (a protein–RNA complex) D. CO2

64. The folding of proteins can be considered a simple conversion from the unfolded to the natively folded state. At about 27°C (or 300 K), the free-energy change of folding for a particular protein is measured to be −40 kJ/mole. If the enthalpy

Molecular Biology of the Cell 7e-Alberts change (ΔH) of folding is −640 kJ/mole, what is the entropy change (ΔS) of folding for this protein?

65. Why does ΔG° need to be determined under the same set of standard conditions for every reaction in order for different reactions to be comparable? Does a cell meet these standards?

66. The free-energy change (ΔG) for a simple reaction, A → B, is 0 kJ/mole at 37°C when the concentrations of A and B are 10 M and 0.1 M, respectively. What is the free-energy change for the reaction when the concentrations of A and B are instead 0.01 M and 1 M, respectively? Recall that ΔG° = –5.9 × log(Keq).

67. In the absence of oxygen, pyruvate undergoes fermentation, forming products that are removed from the cell. Explain why this process needs to occur even though glycolysis alone can produce ATP.

68.

Study the diagram below of how fatty acids are mobilized for energy use in animals. When the fatty acids leave the cell and move into the bloodstream, which is aqueous, they must bind to the protein serum albumin in order to be transported. Knowing the chemical composition of a fatty acid, why do you think this is the case?

69. Both fatty acids and glucose can be utilized as energy sources by being oxidized through the citric acid cycle. Why does one fatty acid molecule produce much

Molecular Biology of the Cell 7e-Alberts more ATP than one glucose molecule?

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 02 1. Answer:

B

2. Answer:

A

3. Answer:

C

4. Answer:

B

5. Answer:

A

6. Answer:

E

7. Answer:

D

8. Answer:

C

9. Answer:

C

10. Answer:

A

11. Answer:

C

12. Answer:

C

13. Answer:

C

14. Answer:

A

15. Answer:

E

16. Answer:

D

17. Answer:

D

18. Answer:

A

19. Answer:

E

20. Answer:

C

Molecular Biology of the Cell 7e-Alberts 21. Answer:

A

22. Answer:

C

23. Answer:

D

24. Answer:

A

25. Answer:

C

26. Answer:

A

27. Answer:

C

28. Answer:

C

29. Answer:

B

30. Answer:

B

31. Answer:

C

32. Answer:

B

33. Answer:

E

34. Answer:

C

35. Answer:

D

36. Answer:

D

37. Answer:

E

38. Answer:

D

39. Answer:

B

40. Answer:

C

41. Answer:

D

42. Answer:

E

Molecular Biology of the Cell 7e-Alberts 43. Answer:

B

44. Answer:

B

45. Answer:

D

46. Answer:

E

47. Answer:

B

48. Answer:

D

49. Answer:

C

50. Answer:

C

51. Answer:

C

52. Answer:

A

53. Answer:

A

54. Answer:

A

55. Answer:

B

56. Answer:

B

57. Answer:

B

58. Answer:

1. B 2. A 3. A 4. C

59. Answer:

1. B 2. D 3. A 4. A

Molecular Biology of the Cell 7e-Alberts 60. Answer:

1. C 2. D 3. A 4. B

61. Answer:

Heat energy leaves the cell because it is given off by the chemical reactions that happen in the cell. These reactions decrease entropy within the cell, but increase the entropy of the universe by releasing a high level of thermal energy out of the cell, which contributes to higher random particle motion outside the cell and thus, increases disorder.

62. Answer:

DAECB

63. Answer:

CABD

64. Answer:

−2 kJ/mole/K

65. Answer:

Several aspects of ΔG° rely on variable conditions. These include temperature and the concentration of the reactants and products. Since these can be different depending on the environment surrounding each reaction, ΔG is not able to be compared between reactions unless those variables are held constant. This allows for only the intrinsic properties of the molecules themselves to affect ΔG° instead of the environment. Cells do not operate under standard conditions.

66. Answer:

+23.6 kJ/mole

67. Answer:

Fermentation oxidizes NADH back into NAD+, and in the process, reduces pyruvate into one of the products secreted from the cell. This is because in order for glycolysis to continue and to produce enough ATP for use by the cell, it needs a steady supply of NAD+ to be reduced as glucose is oxidized. So the NADH formed during glycolysis must be regenerated back into NAD+ for glycolysis to continue.

68. Answer:

Fatty acids are hydrophobic, due to their many C—C and C—H bonds, and therefore do not readily interact with hydrophilic water molecules. Due to the hydrophobic affect, they have a tendency to stick together. When they bind to serum albumin in the blood, the protein can interact with both the hydrophobic fatty acids and the

Molecular Biology of the Cell 7e-Alberts water molecules. This allows the fatty acids to move throughout the body in the blood instead of forming large clumps which could clog the vessels.

69. Answer:

Glucose gets broken down into two pyruvates that each get oxidized through one turn of the citric acid cycle, meaning one glucose produces two cycles’ worth of NADH, FADH2 and GTP. However, fatty acids usually have many more carbons compared to glucose and can only be fully oxidized through the citric acid cycle. Thus, this requires many turns and produces many more molecules of NADH, FADH2 and GTP than glucose.

Molecular Biology of the Cell 7e-Alberts

Chapter 3: Proteins 1. At physiological pH, what is the approximate net charge of a hexapeptide with the following amino acid sequence: Asp-Val-Ile-Glu-Arg-Ser? a. −2 b. −1 c. 0 d. +1 e. +2

2. Which of the amino acids listed has the most limited combinations of phi (φ) and psi (ψ) angles in its Ramachandran plot?

a.

b.

Molecular Biology of the Cell 7e-Alberts

c.

d.

e.

3. Which of the following pairs of amino acid residues would you expect to form ionic bonds? a. Glutamic acid and glutamine b. Arginine and lysine c. Lysine and glutamic acid

Molecular Biology of the Cell 7e-Alberts d. Tryptophan and tyrosine e. Tyrosine and glutamine

4. Which of the following stretches of amino acid residues would you expect to find in the interior of protein molecules? a. Ala-Asp-Asp-Tyr-Arg b. Gly-Lys-Ser-Pro-Thr c. Phe-Glu-Gln-Glu-Asn d. Ala-Val-Leu-Ile-Trp e. Gly-Tyr-His-Arg-His

5. Which of the following is NOT the role of molecular chaperones in the folding of cellular proteins? a. They assist proteins in folding into their correct conformations. b. They help prevent formation of protein aggregates. c. They specify the final three-dimensional shape of proteins. d. They catalyze the folding of proteins in the crowded environment of the cell. e. They make the protein-folding process in the cell more reliable.

6. The observation that proteins often renature into their original conformations after they have been unfolded by denaturing solvents implies what? a. The information needed to specify the three-dimensional shape of a protein is encoded in its amino acid sequence. b. The cell does not need molecular chaperones for survival. c. The final folded structure of a protein is usually NOT the one with the lowest free energy. d. Each protein folds into several different conformations inside the cell.

7. Protein secondary structure elements such as α helices and β sheets constitute the major regular folding patterns in proteins. Regarding these elements, which statement is true? a. Hydrogen-bonding between the amino acid side chains defines the type of secondary structure.

Molecular Biology of the Cell 7e-Alberts b. A certain short amino acid sequence always adopts the same secondary structure. c. Only a few specific amino acid sequences can adopt these repetitive structures. d. The folding patterns result from hydrogen bonding between the N—H and C=O groups in the polypeptide backbone.

8. What does the figure below depict an example of?

a. Primary structure b. Secondary structure c. Tertiary structure d. Quaternary structure e. Hydrophobic structure

9. Imagine a cellular protein composed of 3000 amino acid residues in one continuous polypeptide chain. This protein almost certainly has what characteristic? a. It is extracellular. b. It is globular. c. It is multidomain. d. It is composed of mostly α-helical regions. e. It is intrinsically disordered.

Molecular Biology of the Cell 7e-Alberts

10. Which of the following is true of ALL protein structures? a. They are all intrinsically disordered. b. They all form due to disulfide bonds. c. They all require a chaperone to correctly fold. d. They are all dynamic due to thermal energy. e. They all contain an active site.

11. The number of ways in which protein domains fold in nature is limited. Which of the following is a better estimate of this number? a. 200 b. 2,000 c. 200,000 d. 2,000,000 e. 20 million

12. You have purified a multisubunit extracellular protein that has several interchain disulfide bonds. Which of the following chemicals would you add to your purified protein mixture if you wanted to eliminate the disulfide bonds? a. NaCl, a salt b. SDS, an ionic detergent and denaturing agent c. H2O2, an oxidizing reagent d. Tris, a buffering agent e. DTT, a reducing agent

13. Which of the following is NOT true regarding the members of a protein family in general? a. They have similar three-dimensional conformations. b. They share an ancestry, that is, they are homologs. c. They can functionally replace each other. d. Their gene sequence is less well conserved than their structure. e. Over evolutionary time scales, the family has expanded mainly through

Molecular Biology of the Cell 7e-Alberts gene duplication events.

14. Many viruses have large capsids in the form of a hollow sphere, made of hundreds of identical protein subunits. Which of the following is NOT an advantage of having coats made of several copies of only a few subunits? a. Assembly can be readily regulated. b. Disassembly can be readily regulated. c. It requires a smaller amount of genetic information. d. The effect of mistakes in protein synthesis on the overall assembly is minimized. e. It increases the likelihood of cell binding and infection.

15. Which of the following is NOT true of intrinsically disordered regions in proteins? a. They are relatively rare to find in proteins. b. They are 30 amino acids long or longer. c. They are continuously moving due to thermal energy. d. They are frequently found using bioinformatics methods. e. They have important functions in the cell.

16. The human estrogen receptor is a symmetrical dimeric nuclear protein that can regulate gene expression by binding to a DNA sequence called an estrogen response element (ERE) near the promoter of its target genes. Each subunit of the receptor binds to about six base pairs of DNA. Which of the following sequences is a likely candidate for the ERE? The sequences are written in the 5′to-3′ direction. The letter N represents any of the four DNA bases. a. AGGTCANNNTGACCT b. AGGTCANNNAGGTCA c. AGGTCANNNACTGGA d. AGGTCANNNATATAT e. AGGCCTNNNTCATGA

17. Stable β-sheet aggregates can form from many proteins, forming intertwined cross-beta strands that have the potential to kill cells or damage tissues. Which of

Molecular Biology of the Cell 7e-Alberts the following is NOT true regarding these aggregates? a. They form almost exclusively in the cells of the nervous system. b. Different types of such aggregates can form from the same protein. c. Their formation is associated with conditions such as Parkinson’s disease and Kuru. d. They can form spontaneously, but also can be triggered to form by an infection with the same aggregate. e. Some healthy cells form these aggregates to store their secretory proteins.

18. Evolutionary tracing has identified clusters of the most invariant amino acids in the SH2 domain family. Which of the following is true regarding these sites? a. Mutation in these sites generally keeps the protein functional. b. The invariant amino acids generally have negatively charged side chains. c. These sites correspond to the binding site for peptides containing phosphorylated tyrosine. d. The amino acids at these sites are generally hydrophobic. e. These sites have evolved fast because they have critical functions.

19. An amino acid residue that is not part of the active site of an enzyme and does not interact with the ligand is nevertheless critical for ligand binding and is highly conserved. Which of the following is NOT a possible explanation for why this occurs? a. This residue is critical for the correct folding and the placement of ligandbinding site residues. b. The residue helps with restricting the access of water to the ligand-binding site. c. This residue can affect the chemical properties of the residues in the ligand-binding site. d. The residue is large and physically blocks the ligand from the active site.

20. An enzyme acts on a tyrosine residue in a target protein to create a binding site for the SH2 domain. This enzyme is most specifically what? a. An isomerase

Molecular Biology of the Cell 7e-Alberts b. A nuclease c. A phosphatase d. A kinase e. A synthase

21. Protein A can bind to each of the proteins B or C. The association rate constants are the same for forming the AB and the AC complexes. However, the dissociation rate constant for AB is 100 times higher than that for AC. Given that every tenfold increase in the equilibrium constant (of the association reaction) corresponds to about −5.9 kJ/mole difference in the standard free-energy change for the reaction (ΔG°), what is the value of (ΔG°AB − ΔG°AC) in kJ/mole? a. −11.8 b. −5.9 c. 0 d. +5.9 e. +11.8

22. The turnover number for an enzyme is equivalent to the number of substrate molecules processed per second per enzyme molecule. To a test tube containing a 100 mM concentration of its substrate, you have added an enzyme at a final concentration of 10 µM, and have measured the rate of the reaction to be approximately 500 µM/sec. If the Km for the binding of the enzyme to this substrate is about 100 mM, what is the turnover number? a. 100 b. 500 c. 1,000 d. 5,000 e. 10,000

23. For a simple enzymatic reaction that involves only one substrate and follows Michaelis–Menten kinetics, the changes in the concentrations of substrate, product, free enzyme, and the enzyme–substrate complex over the course of the reaction are depicted by solid curves in the graph below. Which curve (1 to 4) corresponds to [ES]?

Molecular Biology of the Cell 7e-Alberts

a. Curve 1 b. Curve 2 c. Curve 3 d. Curve 4

24. Which of the regulatory interactions 1 to 5 depicted in the following diagram is NOT an example of a negative feedback regulation?

a. 1 b. 2 c. 3 d. 4 e. 5

Molecular Biology of the Cell 7e-Alberts 25. The phosphofructokinase (PFK) enzyme is one of the key players in the glycolytic pathway in which glucose eventually breaks down into pyruvate, some ATP is generated, and some NAD+ is reduced. PFK catalyzes the committed step in the pathway and is under extensive regulation. Which of the following compounds would you expect to activate PFK? a. ADP b. NADH c. Pyruvate d. ATP

26. Two ligands, A and B, bind to two different conformations of the enzyme X. The ligand A is the enzyme’s substrate, whereas ligand B binds to a remote allosteric site. Which of the following is a consequence of this arrangement? a. Binding of A to X does not affect the affinity of X for binding to B. b. Binding of B to X does not affect the rate of reaction catalyzed by X. c. Binding of A to X increases the affinity of X for B. d. Binding of B to X decreases the affinity of X for A. e. Binding of B to X has a large effect on the binding of A to X, but binding of A to X has a small effect on X–B binding.

27. What does phosphorylation of a protein by a protein kinase do? a. It adds two positive charges to the protein. b. It activates the protein. c. It deactivates the protein. d. It can create a binding site for other proteins. e. It requires the hydrolysis of two molecules of ATP per phosphorylated residue.

28. Enzymes can catalyze cellular reactions through various mechanisms. Which statement regarding enzymes is NOT true? a. They can provide the chemical groups necessary for simultaneous acid and base catalysis. b. They can form covalent bonds with the substrate during catalysis.

Molecular Biology of the Cell 7e-Alberts c. They can strain a substrate to force it toward a specific transition state. d. They accelerate a cellular reaction by destabilizing the transition state.

29. In interactions between proteins, each hydrogen bond contributes to the free energy of binding by about −4 kJ/mole. If two proteins bind to each other through nine hydrogen bonds, six of which are eliminated when one of these proteins is mutated, how much would you expect the equilibrium constant for their binding to change as a result of the mutation? (ΔG° = −5.9 × log Keq) a. Increase sixfold b. Decrease fourfold c. Decrease by six orders of magnitude d. Increase 1 million-fold e. Decrease by four orders of magnitude

30. The enzyme lysozyme catalyzes the cutting of a polysaccharide chain through hydrolysis. Which of the following is NOT true regarding the catalytic cycle for this enzyme? a. It involves acid catalysis. b. It involves base catalysis. c. It involves strain catalysis. d. It involves covalent catalysis. e. It involves metal ion catalysis.

31. In the following schematic diagram of a simple signaling pathway, protein Z regulates the activity of protein X, which is an upstream protein kinase, through a negative feedback loop. Which of the following better describes protein Z?

Molecular Biology of the Cell 7e-Alberts

a. It is a protein kinase that is activated by phosphorylation. b. It is a protein kinase that is inactivated by phosphorylation. c. It is a protein phosphatase that is activated by phosphorylation. d. It is a protein phosphatase that is inactivated by phosphorylation.

32. A viral version of the Src kinase called v-Src is found in some retroviruses. Unlike the cellular Src, the v-Src kinase is constitutively active, and can drive the cell into uncontrolled growth and tumor formation. Which of the following molecular differences between the two versions of Src is more likely to be responsible for this? a. v-Src lacks the active-site tyrosine residue. b. v-Src lacks a lobe of the kinase domain. c. v-Src has multiple inhibitory tyrosine phosphorylation sites in the kinase domain. d. v-Src lacks the C-terminal tail that can bind to Src’s SH2 domain. e. v-Src has a longer C-terminal tail.

33. A polyubiquitin chain has been attached to a protein. The ubiquitin molecules are linked together via isopeptide bonds between Lys48 of one molecule and the

Molecular Biology of the Cell 7e-Alberts carboxyl end of the next one. What is expected to happen to this protein? a. To be a part of chromatin b. To undergo proteasomal degradation c. To be involved in DNA repair d. To be targeted to endocytic vesicles

34. Which statement regarding the marking of a protein by polyubiquitylation to signify degradation is true? a. It requires the hydrolysis of one ATP molecule to ADP per polyubiquitin chain. b. It involves covalent attachment of the target protein to the E1 enzyme. c. It is carried out by the proteasome complex. d. It is typically done on an arginine residue in the target protein. e. It involves the recognition of the target protein by an E2–E3 ligase.

35. The SCF ubiquitin ligase can recognize and mark various target proteins at different stages of the cell cycle. What is true about this complex? a. Different F-box subunits recognize different target proteins. b. The F-box subunit and the E2 ubiquitin-conjugating enzyme are at the opposite ends of the C-shaped molecule. c. A scaffold protein arranges the other subunits such that the two ends of the complex are separated by a gap. d. The use of interchangeable parts such as the F-box subunits makes economical use of the genetic information and allows for rapid evolution of new functions. e. The largest subunit has GTPase activity.

36. The compound GDPNP is a GTP analog that can bind to GTPases in the same way as GTP. However, unlike GTP, it cannot undergo the hydrolysis reaction that normally releases Pi. What is the result of this deficiency? a. GDPNP and GDP bind to the same conformation of a GTPase. b. A GDPNP-bound Ras is constitutively active. c. A GDPNP-bound EF-Tu cannot bind to tRNA molecules.

Molecular Biology of the Cell 7e-Alberts d. Binding of GDPNP to EF-Tu allows multiple cycles of tRNA binding and release without the use of free energy.

37. Switch proteins that bind and hydrolyze GTP are ubiquitous cell regulators in a wide variety of molecular processes. Which of the following statements is NOT true regarding these proteins? a. EF-Tu is an example of such proteins. b. GTP hydrolysis by the GTPase generally renders it inactive. c. Activation of the GTPase involves addition of a phosphate group to its bound guanine nucleotide. d. Guanine nucleotide exchange factors generally activate the GTPases. e. Guanine nucleotide exchange factors accelerate the release of bound GDP from the GTPases.

38. Consider the proteins Ras, Src, and kinesin. Which of the following is NOT something they have in common? a. They are allosteric proteins. b. They can have two or more distinct conformations. c. They can bind to ATP (or GTP). d. They can hydrolyze ATP to ADP (or GTP to GDP). e. They contain many repeating subunits to form a large protein complex.

39. In terms of molecular function, what do Ras and myosin have in common? a. They normally bind to GTP. b. They couple the hydrolysis of a bound nucleoside triphosphate to protein movements. c. They switch between two distinct conformations controlled by protein phosphorylation and dephosphorylation. d. They are multisubunit proteins and contain a scaffold subunit.

40. Which of the following is NOT a function for intrinsically disordered protein sequences? a. High-specificity binding to other proteins

Molecular Biology of the Cell 7e-Alberts b. Cell signaling through covalent modification of the protein sequence c. Tethering to hold interacting proteins in close proximity d. Formation of a diffusion barrier from a dense network of such sequences e. Holding a substrate in a specific conformation

41. Many macromolecular complexes in the cell contain scaffold proteins. Which of the following is NOT something that these proteins do to benefit the cell? a. They can enhance the rate of critical cellular reactions. b. They can hold the many subunits of a large complex together. c. They can confine and concentrate a specific set of interacting proteins to a particular cellular location. d. They can provide a large macromolecular complex with either flexibility or rigidity. e. They can replace the phospholipid bilayer in the plasma membrane.

42. Which of the following is NOT true of biomolecular condensates? a. They are membrane-bound. b. They are caused by many short-lived, weak interactions. c. They are formed by large biomolecular polymers. d. They behave like a liquid. e. They are easy to disassemble.

43. What is the purpose of biomolecular condensates? a. To streamline cell signaling pathways b. To disassemble the DNA in the nucleus for mitosis c. To digest pathogens d. To compartmentalize a subset of the cell’s macromolecules e. To form water pockets in the midst of the cytosol

44. The formation of biomolecular condensates can be compared to what? a. The separation of oil and water b. The flowing of a liquid stream

Molecular Biology of the Cell 7e-Alberts c. The joining together of complex building blocks d. The powerhouse of the cell e. The mixing of a salad

45. What is true of phase separation, or demixing? a. It is associated with a favorable change in free energy. b. It has a larger entropic cost for large polymers than for small molecules. c. It involves an unfavorable free-energy change due to the entropic cost of gathering molecules together. d. It involves no change in free energy. e. It involves a change in temperature.

46. In order for a polymer to phase separate, what must happen? a. It must reach a threshold concentration. b. It must be at least a certain length. c. It must be unable to form weak bonds. d. It must be at a certain temperature. e. It must be placed in a specific cellular location.

47. A protein is drawn in the following simplified diagram undergoing a set of covalent modifications including the addition of a chain of ubiquitin protein monomers (U) to one of its lysine side chains, a phosphate moiety (P) to its tyrosine side chain, and a lipid farnesyl (F) to its cysteine side chain. What is the correct order of these modifications (1 to 3) in the diagram?

Molecular Biology of the Cell 7e-Alberts

a. Lipid farnesyl to its cysteine side chain, chain of ubiquitin protein monomers to one of its lysine side chains, phosphate moiety to its tyrosine side chain b. Lipid farnesyl to its cysteine side chain, phosphate moiety to its tyrosine side chain, chain of ubiquitin protein monomers to one of its lysine side chains c. Chain of ubiquitin protein monomers to one of its lysine side chains, lipid farnesyl to its cysteine side chain, phosphate moiety to its tyrosine side chain d. Chain of ubiquitin protein monomers to one of its lysine side chains, phosphate moiety to its tyrosine side chain, lipid farnesyl to its cysteine side chain

48. Imagine a protein that can be independently phosphorylated on any of its 10 tyrosine residues and acetylated on either of its 2 lysine residues. In principle, how many different combinations of these modifications are possible for this

Molecular Biology of the Cell 7e-Alberts protein? a. 210 b. 220 c. 1010 d. 1220 e. 212

49. Designing new protein structures involves what process? a. Using random generation of DNA sequences to form new proteins b. The use of computational methods to predict the lowest energy state of a sequence of amino acids c. Creating new amino acids to make new structures d. Many years of slow evolutionary change e. Guess-and-check methods where every amino acid is placed at every location in a growing polypeptide chain

50. You’re trying to design a new small protein, 40 amino acids in length. What is the theoretical number of possible conformations this small protein could adapt? a. Infinite b. 1600 c. 403 d. 340 e. 40

51. For each of the following visual representations from left to right, indicate whether the protein structure is composed of only α helices, only parallel β sheets, only antiparallel β sheets, α helices plus parallel β sheets, or α helices plus antiparallel β sheets as repetitive secondary structure elements.

Molecular Biology of the Cell 7e-Alberts

a.

b.

c.

Molecular Biology of the Cell 7e-Alberts

d.

1. Only α helices 2. Only antiparallel β sheets 3. α helices plus parallel β sheets 4. α helices plus antiparallel β sheets

52. Which of the following amino acid residues have acidic side chains: Alanine, cysteine, aspartic acid, alanine, glutamic acid?

The structural formulas for the 20 naturally occurring amino acid residues are shown in the panel below. Answer the following question(s) with the help of this panel.

Molecular Biology of the Cell 7e-Alberts

53. Which of the amino acids shown above has the most limited combinations of phi (φ) and psi (ψ) angles in its Ramachandran plot? Write down the one-letter abbreviation for it, for example, A.

54. Which amino acid residues shown above have acidic side chains? Write down the one-letter abbreviations for them, in alphabetical order, for example, ACGNV.

55. What is the sequence from the N- to the C-terminus? Use only the one-letter abbreviations, for example, AGCNT.

Molecular Biology of the Cell 7e-Alberts

56. What is the sequence from the N- to the C-terminus? Use only the one-letter abbreviations, for example, AGCNT.

Molecular Biology of the Cell 7e-Alberts

57.

What is the sequence from the N- to the C-terminus? Use only the one-letter abbreviations, for example, AGCNT.

Molecular Biology of the Cell 7e-Alberts

58. A folded, globular protein is mixed with a high concentration of urea. Urea molecules are capable of forming many H-bonds very easily. When the structure of the protein is analyzed, it is found that it has been denatured. Why has this occurred?

59. In an α helix, each amino acid residue is rotated by about +100° and raised by about 0.15 nm relative to the previous residue. This means that each full turn of the helix rises by how many nm parallel to the helix’s axis?

60. Imagine that 1 L of a solution containing each of the 20 naturally occurring amino acids at 50 mM concentration each (total concentration of 1 M) is allowed to polymerize in a perfectly stepwise fashion such that at each step, a random

Molecular Biology of the Cell 7e-Alberts amino acid can be incorporated into a growing polypeptide. The steps are repeated, until eventually the solution is only composed of 40-mers (and virtually all of the monomers have been used). What fraction of all 40-mers can possibly be present in this solution? Round your number to four decimal places.

(Avogadro’s number is 6 × 1023.)

61. You are a researcher who has recently identified the sequence of a new protein of unknown function. You also have access to the massive public database of over 100 million protein sequences. How might this be helpful in elucidating the function of your new protein and why does this work?

62. A certain protein is rendered nonfunctional by a mutation in an amino acid in the interior of the protein. This amino acid is not in any binding sites of the protein and is not involved in any chemistry performed by the protein. How could this mutation have such a large effect on the functionality of the protein?

63. A certain enzyme is synthesized in vitro and mixed with its substrate under physiological conditions. However, no catalysis occurs. Assuming there is nothing wrong with the sequence of the protein or its ability to fold, what might lead to the enzyme being nonfunctional? How could you test this?

64. You are a researcher trying to find ways to reduce the amount of microplastics in the ocean. You decide to try to design a new enzyme that could digest microplastics into natural organic compounds. You know of several different enzymes that are already capable of digesting long carbon-based polymers similar to plastics. How could you use this information to help you design your new enzyme?

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 03 1. Answer:

B

2. Answer:

D

3. Answer:

C

4. Answer:

D

5. Answer:

C

6. Answer:

A

7. Answer:

D

8. Answer:

B

9. Answer:

C

10. Answer:

D

11. Answer:

B

12. Answer:

E

13. Answer:

C

14. Answer:

E

15. Answer:

A

16. Answer:

A

17. Answer:

A

18. Answer:

C

19. Answer:

D

20. Answer:

D

Molecular Biology of the Cell 7e-Alberts 21. Answer:

E

22. Answer:

A

23. Answer:

C

24. Answer:

E

25. Answer:

A

26. Answer:

D

27. Answer:

D

28. Answer:

E

29. Answer:

E

30. Answer:

E

31. Answer:

D

32. Answer:

D

33. Answer:

D

34. Answer:

E

35. Answer:

E

36. Answer:

B

37. Answer:

C

38. Answer:

E

39. Answer:

B

40. Answer:

E

41. Answer:

E

42. Answer:

A

Molecular Biology of the Cell 7e-Alberts 43. Answer:

D

44. Answer:

A

45. Answer:

C

46. Answer:

A

47. Answer:

B

48. Answer:

A

49. Answer:

B

50. Answer:

D

51. Answer:

1. B 2. C 3. D 4. A

52. Answer:

Alanine and cysteine

53. Answer:

P

54. Answer:

DE

55. Answer:

CELL

56. Answer:

SCIENCE

57. Answer:

GRPCNQFYC

58. Answer:

Secondary, tertiary, and quaternary structures rely on noncovalent interactions between amino acid side chains and backbones. This includes a high number of hydrogen bonds. When urea is introduced into the solution, it is able to make H-bonds with the amino acids that replace the Hbonds amino acids should be forming with each other.

This causes the protein structure to denature.

59. Answer:

0.54

Molecular Biology of the Cell 7e-Alberts 60. Answer:

0.0000

61. Answer:

You should compare the new protein’s sequence with those of the database. Similarities in sequence would likely mean similarities in structure, and similarities in structure imply similarities in function. This process works because protein domains are repeated throughout the genome and usually maintain similar function in different proteins.

62. Answer:

This amino acid is most likely necessary for the structure of the protein and thus the orientation of the amino acids at critical sites of the protein such as binding sites or active sites. If these amino acids are moved even slightly, it could alter the binding and activity of the protein by disrupting the interactions of those amino acids.

63. Answer:

This enzyme may need a coenzyme or cofactor for its function. This is normally available in the environment that the protein functions in, but in the lab, they were not a part of the solution. This can be tested by adding in other small molecules or even whole cell lysate to see if this would allow the enzyme to perform its normal catalysis.

64. Answer:

You could study the domains of the enzymes that you already are aware of that digest substances similar to plastic. Once you understand how the structures of each of the domains of the proteins function, you can use those sequences to make your new protein. You would need to change the active sites of any old domains to be able to bind plastic and digest it. You would also need to make sure that bringing several domains together would still allow those domains to retain their structure and function.

Molecular Biology of the Cell 7e-Alberts

Chapter 4: DNA, Chromosomes, and Genomes 1. In the secondary structure of a DNA double helix, a purine (A or G) always base pairs with a pyrimidine (T or C). Why is this base-pair alignment critical to the structure of DNA? a. Pairing a purine with a pyrimidine ensures that the proper number of hydrogen bonds is generated. b. Pairing a purine with a pyrimidine ensures that the distance between the backbone strands is constant. c. Pairing a purine with a pyrimidine ensures that the proper complementary base is utilized during replication. d. Biochemically, it is impossible to generate a base pair between two purines or two pyrimidines. e. The only way to generate complementary strands is if a purine is paired with a pyrimidine.

2. A DNA nucleotide pair has an average mass of approximately 660 daltons.

Knowing the number of nucleotides in the human genome, how many picograms of DNA are there in a diploid human nucleus? Avogadro’s number is 6 × 10 23. a. 7 pg b. 15 pg c. 10 pg d. 17 pg e. 6 pg

3. Which feature of DNA underlies its simple replication procedure? a. The fact that it is composed of only four different types of bases b. The antiparallel arrangement of the double helix c. The complementary relationship in the double helix

Molecular Biology of the Cell 7e-Alberts d. The fact that there is a major groove and a minor groove in the double helix

4. Which of the following correlates the best with biological complexity in eukaryotes? a. Number of genes per chromosome b. Number of chromosomes c. Number of genes d. Genome size (number of nucleotide pairs)

5. What is a feature of the two chromosomes in each of the 22 homologous pairs in our cells? a. They have the exact same DNA sequence. b. They are derived from one of our parents. c. They show identical banding patterns after Giemsa staining. d. They usually bear different sets of genes.

6. How does the genome of yeast typically compare to the human genome? a. The yeast genome has more repetitive DNA. b. The yeast genome has longer genes. c. The yeast genome has more introns. d. The yeast genome has longer chromosomes. e. The yeast genome has a higher fraction of coding DNA.

7. Each chromosome in a human eukaryotic cell is composed of how many independent strands of DNA? a. 1

Molecular Biology of the Cell 7e-Alberts b. 2 c. 23 d. 46 e. Over 200

8. What property of genome diversity accounts for a majority of the variation in genome size, even among related species? a. The number of chromosomes present in the nucleus b. If the chromosome is linear or circular c. The number and pattern of banding d. The total number of genes present e. The amount of noncoding DNA

9. Chromosome 3 contains nearly 200 million nucleotide pairs of our genome. If this DNA molecule could be laid end-to-end, how long would it be? The distance between neighboring base pairs in DNA is typically around 0.34 nm. a. About 7 mm b. About 7 cm c. About 70 mm d. About 7 m e. About 70 cm

10. Centromeres, telomers, and origins of replication are all elements of chromosome structure, but what are these elements exactly? a. They are banding patterns that facilitate protein interactions with the chromosome. b. They are patterns of nucleosomes that compact to form these observable

Molecular Biology of the Cell 7e-Alberts structures. c. They are all repetitive nucleotide sequences that are contained within the chromosome itself. d. They are histone modifications that signal unique chromatin structures. e. They are descriptors of the heterochromatic state of the chromatin.

11. The eukaryotic chromosomes are organized inside the nucleus with a huge compaction ratio of several thousand-fold and is further accentuated during the mitotic cycle. What interaction is responsible for such a tight packaging? a. The various histone proteins that wrap and fold the DNA b. The nuclear envelope which encapsulates the chromosomes c. The nuclear matrix that provides a firm scaffold d. The nucleosome that is the core chromatin structure

12. In assembling a nucleosome, normally specific histone dimers first combine to form a tetramer, which then further combines with two of another type of histone dimers to form the octamer. Name these two histone dimers in order of how they combine. a. H1–H3; H2A–H2B b. H3–H4; H2A–H2B c. H2A–H2B; H1–H3 d. H2A–H2B; H3–H4 e. H1–H2; H3–H4

13. The chromatin remodeling complexes play an important role in chromatin regulation in the nucleus. The ability of these complexes to remodel the chromatin structure is dependent on which enzymatic function? a. They can slide nucleosomes on DNA.

Molecular Biology of the Cell 7e-Alberts b. They have ATPase activity. c. They can interact with histone chaperones. d. They can remove core histone subunits. e. They can substitute core histone subunits.

14. To study the chromatin remodeling complex SWR1, a researcher has prepared arrays of nucleosomes on long DNA strands that have been immobilized on magnetic beads. These nucleosomes are then incubated with an excess of the H2AZ–H2B dimer (which contains the histone variant H2AZ) in the presence or absence of SWR1 with or without ATP. She then separates the bead-bound nucleosomes (bound fraction) from the rest of the mix (unbound fraction) using a magnet, elutes the bound fraction from the beads, and performs SDS-PAGE on the samples. This is followed by a Western blot using an antibody specific to the H2AZ protein used in this experiment. The results are shown below, with the presence (+) or absence (−) of ATP, SWR1, or the H2AZ–H2B dimer in each incubation reaction indicated at the top of the corresponding lane.

a. SWR1 deposits H2AZ histones into the nucleosome arrays. b. SWR1 function is not ATP-dependent. c. The antibody used in this experiment binds to the SWR1 complex. d. The SWR1 complex has a preference for unbound H2AZ.

15. Nucleosomes that are positioned like beads on a string over a region of DNA can interact to form higher orders of chromatin structure. Which factor is not essential to the formation of the 30-nm chromatin fiber from these nucleosomes?

Molecular Biology of the Cell 7e-Alberts a. Interactions that involve the histone tails of neighboring nucleosomes b. Interaction of the linker histone H1 with each nucleosome c. Binding of proteins to DNA or the histones d. ATP-dependent function of chromatin remodeling complexes e. The G-C content of nucleotides within the minor groove of the DNA

16. Which statement regarding heterochromatin in a typical mammalian cell is true? a. About 1% of the nuclear genome is packaged in heterochromatin. b. The DNA in heterochromatin contains all of the inactive genes in a cell. c. Genes that are packaged in heterochromatin are permanently turned off. d. The different types of heterochromatin share an especially high degree of compaction. e. Heterochromatin is highly concentrated in the centromeres but not the telomeres.

17. The position effect variegation (PEV) phenotype described in this chapter can be used to identify new genes that regulate heterochromatin formation. For instance, strains of Drosophila melanogaster with the White variegation phenotype have been subjected to mutagenesis to screen for dominant mutations (in other genes) that either enhance or suppress PEV, meaning the mutations result in either lower or higher red pigment production, respectively. Which of the following mutations is expected to be an enhancer of variegation? a. A mutation that results in the loss of function of the fly’s HP1 (heterochromatin protein 1) gene b. A loss-of-function mutation in a gene encoding a histone deacetylase that deacetylates lysine 9 on histone H3 c. A gain-of-function mutation in a gene encoding a histone methyl transferase that trimethylates lysine 9 on histone H3, resulting in a hyperactive form of the enzyme d. A gain-of-function mutation in a gene encoding a histone acetyl transferase

Molecular Biology of the Cell 7e-Alberts that normally acetylates lysine 9 on histone H3, resulting in higher expression of the protein

18. Imagine a chromosome translocation event that brings a gene encoding a histone acetyl transferase enzyme from its original chromosomal location to a new one near heterochromatin. Which scenario is definitely NOT going to happen? a. The gene gets silenced due to heterochromatin expansion, leading to the misregulation of gene expression for a number of critical genes. b. The translocation event also brings along a chromatin barrier that can prevent heterochromatin expansion into the gene, and there is no phenotypic anomaly. c. Since the gene encodes a histone acetyl transferase, it resists heterochromatin expansion by acetylating its own histones. d. The level of the gene product decreases due to a position effect, leading to an imbalance in the chromatin state of the cell that results in the activation of programmed cell death.

19. What is a characteristic of the acetylation of lysines on the histone tails? a. It loosens the chromatin structure because it adds positive charges to the histone. b. It recruits the heterochromatin protein HP1, resulting in the establishment of heterochromatin. c. It can be performed on methylated lysines only after they are first demethylated. d. It is sufficient for the formation of an open chromatin structure. e. It is a covalent modification and is thus irreversible.

20. Following cellular differentiation, a region of the genome that had expressed the histone tag H3K9ac and was transcriptionally active is seen to become transcriptionally silent. What change to the histone code most likely had this effect?

Molecular Biology of the Cell 7e-Alberts a. The H3K9ac mark is replaced with a H3K9me3 mark. b. The H3K9ac mark is replaced with a H3K27ac mark. c. The H3K9ac mark is retained and an additional H3K9me3 mark is added. d. The H3K9ac mark is replaced with a H4K4me3 mark.

21. Most histone modifications are covalent modifications to already synthesized histones. Why do you think this characteristic is necessary for proper chromatin regulation? a. Histone modifications need to be permanent in order to ensure the cell can function properly. b. Histone modifications are irreversible, as the only way to regulate chromatin function is through histone variants. c. Histone modifications are applied by various different proteins and are therefore static once generated. d. Histone modifications need to be dynamic in order to effectively influence chromatin function. Covalent modifications can be easily added or replaced by enzymes. e. Histone modifications are irreversible once the chromatin has been condensed.

22. What is a characteristic of a chromatin “reader complex?” a. It is always coupled to a “writer complex” and spreads specific chromatin modifications. b. It can recognize any histone code. c. It binds tightly to the chromatin only when a specific set of histone marks is present. d. It can only bind to a single specific histone mark. e. It has at least five protein subunits.

Molecular Biology of the Cell 7e-Alberts 23. Why are barrier DNA and the proteins that are able to associate with barrier DNA regions of the genome essential for proper chromatin structure? a. Barrier proteins can block the progression of the reader-writer complex and prevent the spread of heterochromatin. b. Barrier proteins can block the progression of the reader-writer complex and prevent the spread of euchromatin. c. Barrier proteins can block the integration of histones into nucleosomes thus interfering with chromatin structure. d. Barrier proteins can recruit histone methyltransferases and block the progression of euchromatin. e. The barrier DNA is a region that is unable to be incorporated into nucleosomes and therefore blocks the formation of heterochromatin.

24. In human cells, what is true of the alpha satellite DNA repeats? a. They have a specific sequence indispensable for the seeding event that leads to chromatin formation. b. They can be seen to be packaged into alternating blocks of chromatin, one of which contains the histone H3 variant CENP-A. c. They are sufficient to direct centromere formation. d. They are necessary for centromere formation. e. They are unique to the centromeric region of the chromosome.

25. Variations in the structure of DNA, such as loops, or the chromatin organization such as euchromatin or heterochromatin, can influence gene expression and can be inherited by subsequent generations even though these changes do not result in alterations to the sequence of nucleotides. What is this type of inheritance called? a. Epigenetic inheritance b. Transmorphic inheritance c. Mendelian inheritance

Molecular Biology of the Cell 7e-Alberts d. Formative inheritance

26. What is true about lampbrush chromosomes? a. They are transcriptionally inactive. b. They are readily observed in the oocytes of humans and insects. c. They have thousands of duplicated DNA molecules arranged side by side. d. They are mitotic chromosomes with two sister chromatids attached together only at the centromere. e. They are thought to have a structure that is relevant to mammalian chromosomes in interphase.

27. What are the consequences for the cell of the “fractal globule” arrangement of chromosome segments? a. The chromosomes cannot be condensed maximally this way. b. The neighboring regions of DNA are furthest from each other in the threedimensional space. c. The ability of the chromatin fiber to fold and unfold efficiently is maintained. d. Dense packing is permitted, but the ability to easily fold and unfold the chromatin is prohibited.

28. Findings from a number of experiments on human chromatin have suggested that the DNA in our chromosomes is organized into loops of various lengths.

Approximately how long is a typical loop (in nucleotide pairs of DNA)? a. 50 b. 2000 c. 100,000 d. 10 million

Molecular Biology of the Cell 7e-Alberts e. 50 million

29. Why are polytene chromosomes useful for studying chromatin? a. They are smaller than regular chromosomes and easier to manipulate. b. They lack heterochromatin. c. They have distinct visible banding patterns. d. They can make polyploid cells.

30. Five major types of chromatin were identified in studies performed on Drosophila melanogaster cells, although much more remains to be learned about chromatin diversity and dynamics. Which of the following is correct regarding these findings in Drosophila? a. These results were obtained using the 3C technique, which determines the positions of loops in the chromatin. b. According to these results, there are four types of heterochromatin and only one type of euchromatin. c. The polycomb form of chromatin belongs to the euchromatin type. d. In addition to these five major types of chromatin, there seem to be additional minor types as well. e. The pattern of chromatin types in the chromosomes is constant across the different cell types in a multicellular organism.

31. What would you expect to observe when an SMC protein complex is loaded onto a prokaryotic chromosome at the origin of replication? a. The removal of repetitive DNA sequences from the origin b. The breaking of hydrogen bonds between complementary strands of DNA at the origin c. The hydrolysis of ATP to initiate DNA replication at the origin

Molecular Biology of the Cell 7e-Alberts d. The close association of sister chromatids e. The formation of a DNA loop and the subsequent segregation of the two daughter DNA molecules away from each other

32. What would most reliably suggest that a genomic sequence is functionally important? a. The presence of a long open reading frame in the sequence b. Multispecies conservation of the sequence c. Low copy number variation of the sequence d. The presence of active chromatin marks over the sequence

33. The genetic information carried by a cell is passed on, generation after generation, with astonishing fidelity. However, genomes are still altered over evolutionary time scales, and even their overall size can change significantly.

Which genome-altering event has increased the size of the mammalian genome the most? a. Transposition b. Point mutation c. Chromosomal deletion d. Chromosomal inversion e. Chromosomal translocation

34. Conservation of genomic sequences between humans and chickens is mainly due to a specific type of selection, whereas the conservation observed between humans and chimpanzees is mostly due to the short time available for mutations to accumulate. What type of selection is responsible and what position on the DNA sequences of synonymous codons are nearly identical between humans and chimpanzees? a. Purifying; third

Molecular Biology of the Cell 7e-Alberts b. Homologous; third c. Pacifying; first d. Purifying; first e. Directional; third

35. The globin gene family in mammals, birds, and reptiles is organized into α- and βglobin gene clusters that are located on two different chromosomes. In most fish and amphibians, however, the globin genes are close to each other on one chromosome. At which point (A to E) in the following simplified phylogenetic tree is a chromosomal translocation likely to have happened that placed the α- and βglobin genes on two separate chromosomes?

a. Point A b. Point B c. Point C d. Point D e. Point E

36. To discover genes that have undergone accelerated evolution in the human lineage, you compared the amino acid sequences of dozens of proteins from orthologous protein-coding genes in humans, chimpanzees, and mice. For each gene, you build an unrooted phylogenetic tree in which the branch lengths (a, b, or c) correspond to the number of amino acid substitutions in that branch, as

Molecular Biology of the Cell 7e-Alberts depicted below. Primates and rodents diverged ~90 million years ago, and humans and chimpanzees diverged ~5.5 million years ago. For each individual gene shared by the three species, you therefore define the “normalized substitution rate” parameters h and k as h = (a/5.5)/[c/([2 × 90] − 5.5)], and k = (b/5.5)/[c/([2 × 90] − 5.5)]. Based on these definitions, which genes are more likely to be responsible for “uniquely human” traits?

a. Genes with very high h and k values b. Genes with very low h and k values c. Genes with very high h values but not very high k values d. Genes with very high k values but not very high h values

37. The molecular clock for molecular evolution suggests that the rate of change for DNA sequences is nearly constant, yet different categories of DNA accumulate changes at different rates. Which category would you predict has the lowest rate of DNA alterations? a. Exons of essential genes b. Regulatory sequences that are not subject to purifying selection c. The third position of synonymous codons d. Pseudogenes e. Introns that lack splicing regulatory sequences

38. Most fish genomes are at least 1 billion nucleotide pairs long. However, the genome of the puffer fish Fugu rubripes is quite small at only about 0.4 billion

Molecular Biology of the Cell 7e-Alberts nucleotide pairs, even though the number of Fugu genes is estimated to be comparable to that of its relatives which have larger genomes. What do you think mainly accounts for the Fugu genome being this small? a. Evolutionary advantage of extremely small exon sizes in the Fugu lineage b. Unusual disappearance of all intronic sequences from the Fugu genome c. Increased abundance of transposable elements in the Fugu genome d. Increased occurrence of mitotic whole-chromosome loss in the Fugu lineage e. Low relative rate of DNA addition compared to DNA loss in the Fugu lineage

39. Imagine a human protein containing 33 repeats of a simple domain arranged in tandem. In contrast, a homolog found in bacteria contains only one domain. What is the minimum number of duplication events that can account for the evolution of this protein since our divergence from bacteria? a. .6 b. 6 c. 50 d. 5 e. 16

40. Through random events an essential gene along with most of its regulatory sequences is duplicated. What is one outcome of this duplication event that is not expected to be observed over time? a. The duplicated copy is lost due to accumulated deleterious mutations. b. Both genes are lost to the accumulation of deleterious mutations. c. Both genes are mutated and retained due to synergistic mutations ensuring that both genes are required for function. d. The nature of the original gene is retained and the duplicated copy diverges

Molecular Biology of the Cell 7e-Alberts to acquire a novel function. e. The two genes retain their original function, but the expression is diverged into different tissues.

41. Assume two isolated human communities with 500 individuals in each. If the same neutral mutation happens at the same time in two individuals, one from each community, what is the probability that it will be eventually fixed in both of the populations? How would the result change if the two communities fully interbreed? a. 10−6, 10−12 b. 10−3, 10−6 c. 10−6, 10−3 d. It is not possible for the mutation to become fixed with the given population size.

42. In a double-stranded DNA molecule, one of the chains has the sequence CCCATTCTA when read from the 5′ to the 3′ end. There are no C residues in the other chain. a. True b. False

43. Transposable elements occupy almost 10% of our genome. a. True b. False

44. The histone fold consists of three α helices. a. True b. False

Molecular Biology of the Cell 7e-Alberts 45. You have performed a chromosome conformation capture (3C) experiment to study chromatin looping at a mouse gene cluster that contains genes A, B, and C, as well as a regulatory region R. In this experiment, you performed in situ chemical cross-linking of chromatin, followed by cleavage of DNA in the nuclear extract with a restriction enzyme, intramolecular ligation, and cross-link removal.

Finally, a polymerase chain reaction (PCR) was carried out using a forward primer that hybridizes to a region in the active B gene, and one of several reverse primers, each of which hybridizes to a different location in the locus. The amounts of the PCR products were quantified and normalized to represent the relative cross-linking efficiency in each analyzed sample. You have plotted the results in the graph below. The same experiment has been done on two tissue samples: Fetal liver (represented by light gray lines) and fetal muscle (dark gray lines).

Interaction of the A, B, and C genes with the regulatory R region is known to enhance expression of these genes. Based on your results, the B gene would be predicted to have higher expression in the fetal liver compared to the fetal muscle tissue.

a. True b. False

46. Regarding a genome and its evolution, all of the “ultraconserved” elements found in the human genome have been shown to encode long noncoding RNAs. a. True b. False

Molecular Biology of the Cell 7e-Alberts 47. If the frequency of a point mutation in a population is only 0.1%, with no mutation at this site in the rest of the population, then the variation is NOT considered to constitute a single-nucleotide polymorphism. a. True b. False

48. Indicate whether each of the following descriptions better applies to a centromere, a telomere, or an origin of replication. a. Centromere b. Telomere c. Origin of replication

1. It contains repeated sequences at the ends of the chromosomes.

2. It is NOT generally longer in higher organisms compared to yeast.

3. Each eukaryotic chromosome has many such sequences.

4. There are normally two such sequences in each eukaryotic chromosomal DNA molecule.

5. There is normally one such sequence per eukaryotic chromosomal DNA molecule.

6. It is where DNA duplication starts in S phase.

7. It attaches the chromosome to the mitotic spindle via the kinetochore structure.

49. Indicate whether each of the following descriptions better matches the major histones or the histone variants. a. Major histones b. Histone variants

1. They are more highly conserved over long evolutionary time scales.

2. They are present in much smaller amounts in the cell.

3. They are synthesized primarily during the S phase of the cell cycle.

4. Their incorporation often requires histone exchange.

5. They are often inserted into already-formed chromatin.

6. They are assembled into nucleosomes just behind the replication fork.

Molecular Biology of the Cell 7e-Alberts 50. Indicate whether each of the following histone modifications is generally associated with active genes or silenced genes. a. Active genes b. Silenced genes

1. H3 lysine 9 acetylation 2. H3 serine 10 phosphorylation 3. H3 lysine 4 trimethylation 4. H3 lysine 9 trimethylation

51. Indicate whether each of the following histone modifications adds a negative charge to the histone, removes a positive charge from the histone, or does neither of these. a. Adds a negative charge to the histone b. Removes a positive charge from the histone c. Neither adds nor removes

1. H3 lysine 9 acetylation 2. H3 serine 10 phosphorylation 3. H3 lysine 4 trimethylation 4. H3 lysine 9 trimethylation

52. The centromeric regions in the fission yeast Schizosaccharomyces pombe are wrapped by nucleosomes containing the CENP-A histone H3 variant, and are flanked by clusters of tRNA genes that separate them from the surrounding pericentric heterochromatin. If the tRNA clusters are removed from this region, the HP1-bound heterochromatin spreads further to cover the centromeric regions.

The tRNA genes are transcribed by strong RNA polymerase III promoters, which can associate with transcription factors and recruit chromatin-modifying enzymes.

Based on these observations, indicate which blanks in the paragraph below correspond to each of the following phrases: HS4 element, chromatin boundaries, histone acetyl transferases, tRNA genes, CENP-A-containing histones.

The ______(I)______ is/are not sufficient to prevent heterochromatin expansion to the centromeric regions. Instead, the ______(II)______ is/are acting as ______(III)______ in S. pombe, similar to the role of the ______(IV)______ in the β-globin locus in chickens and humans. Likely candidates for the histone-

Molecular Biology of the Cell 7e-Alberts modifying enzymes recruited by the RNA polymerase III complexes is/are ______(V)______. a. HS4 element b. Chromatin boundaries c. Histone acetyl transferases d. tRNA genes e. CENP-A-containing histones

1. I 2. II 3. III 4. IV 5. V

53. For each of the following classifications, indicate whether you would expect to find an actively transcribed gene in the first category or the second. a. First category b. Second category

1. 1: Heterochromatin, or 2: euchromatin 2. 1: Chromosome puffs, or 2: condensed chromosome bands 3. 1: Nuclear periphery, or 2: the center of the nucleus 4. 1: Within the chromosome territory, or 2: extended out of the territory 5. 1: Apart from, or 2: close to actively transcribed genes within the nucleus 6. 1: 11-nm “beads-on-a-string” fibers, or 2: 30-nm fibers

54. In each of the following comparisons, indicate whether the molecular clock is expected to tick faster on average in the first or the second case. a. First case b. Second case

1. 1: The exons, or 2: the introns of a gene

Molecular Biology of the Cell 7e-Alberts 2. 1: The mitochondrial, or 2: the nuclear DNA of vertebrates 3. 1: The first, or 2: the third position in synonymous codons 4. 1: A gene, or 2: its pseudogene counterpart

55. The regions of synteny between the chromosomes of two species can be visualized in dot plots. In the example shown in the following graph, a chromosome of a hypothetical species A has been aligned with the related chromosome in species B. Each dot in the plot represents the observation of high sequence identity between the two aligned chromosomes in a window located at the two corresponding chromosome positions. A series of close dots can make a continuous line. Choosing a sufficiently large window size allows a “clean” dot plot with solid lines that show only the long stretches of identity. This allows ancient large-scale rearrangements to be identified. Several chromosomal events can be detected in such dot plots. Indicate which feature (a to g) in the dot plot is best explained by each of the following events. Each letter should be used only once.

a. A b. B c. C d. D e. E f. F

Molecular Biology of the Cell 7e-Alberts g. G

1. A duplication that exists in both species 2. A duplication in species A only 3. A triplication in species B only 4. An inversion without relocation 5. An inversion combined with relocation 6. A deletion in species A 7. A translocation in species A from a different chromosome

56.

Indicate which numbered feature (1 to 5) in the schematic drawing of the DNA double helix corresponds to each of the following: • Hydrogen-bonding • Covalent linkage • Phosphate group • Nitrogen-containing base • Deoxyribose sugar Your answer would be a five-digit number composed of digits 1 to 5 only, for example, 52431.

57. Complete the DNA sequence below such that the final sequence is identical to

Molecular Biology of the Cell 7e-Alberts that of the complementary strand.

58. For the Human Genome Project, cloning of large segments of our genome was first made possible by the development of yeast artificial chromosomes, which are capable of propagating in the yeast Saccharomyces cerevisiae just like any of the organism’s 16 natural chromosomes. In addition to the cloned human DNA, these artificial vectors were made to contain three elements that are necessary for them to function as a chromosome. What are these elements?

59.

Indicate which feature in the schematic drawing of a chromatin fiber corresponds to each of the following: • Nucleosome core particle • Linker DNA • Histone octamer • Non-histone protein Your answer would be a four-digit number composed of digits 1 to 4 only, for example, 2431.

60. What are the two mechanisms through which the ATP-dependent chromatin remodeling complexes are able to reorganize nucleosome structure?

Molecular Biology of the Cell 7e-Alberts

61. Each histone subunit has a core component and an amino terminal tail domain and while both parts of the histone are potential targets for covalent modification, the tail domains have been shown to be the most frequent recipients of modification. Why do you suppose the histone tail domains are such efficient targets for modifications that result in modified chromatin structure?

62. Hi-C has shown that heterochromatic DNA is often found localized to the periphery of the nucleus, but this localization is also dependent on condensation of the heterochromatin. How is it that heterochromatin condensation and localization to the nuclear periphery are coupled?

63. If species A in the distance matrix represents human, indicate which of the other species (B to D) represents chimpanzee, gorilla, and orangutan, respectively.

Your answer should be a three-letter string composed of letters B, C, and D only, for example, DCB.

64. The following tree can be constructed from these distances assuming a constant molecular clock, meaning that the length of each horizontal branch corresponds to evolutionary time as well as to the relative genetic distance from the common ancestor that gave rise to that branch.

Molecular Biology of the Cell 7e-Alberts

Indicate which one of the species in the matrix (B to D) corresponds to branches 1 to 3, respectively. Your answer should be a three-letter string composed of letters B, C, and D only, for example, DCB.

65. Genetic duplication has resulted in significant variation between closely related species, but novel genes can be created without the duplication of a complete, functional gene. What are two mechanisms by which duplication, but not of complete genes can create novel proteins?

66. The platypus is one of five extant species of egg-laying mammals, and the only member of its genus and species. How would you explain the relationship between the platypus and other groups of mammals? Before forming your response, consider the figure below, which shows the relative divergence time of vertebrate groups and patterns of gene regulatory changes.

Molecular Biology of the Cell 7e-Alberts

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 04 1. Answer:

B

2. Answer:

A

3. Answer:

C

4. Answer:

C

5. Answer:

C

6. Answer:

E

7. Answer:

A

8. Answer:

E

9. Answer:

B

10. Answer:

C

11. Answer:

A

12. Answer:

B

13. Answer:

B

14. Answer:

A

15. Answer:

E

16. Answer:

D

17. Answer:

C

18. Answer:

C

19. Answer:

C

20. Answer:

A

Molecular Biology of the Cell 7e-Alberts 21. Answer:

D

22. Answer:

C

23. Answer:

A

24. Answer:

B

25. Answer:

A

26. Answer:

E

27. Answer:

C

28. Answer:

C

29. Answer:

C

30. Answer:

D

31. Answer:

E

32. Answer:

B

33. Answer:

A

34. Answer:

A

35. Answer:

C

36. Answer:

C

37. Answer:

A

38. Answer:

E

39. Answer:

B

40. Answer:

B

41. Answer:

C

42. Answer:

A

Molecular Biology of the Cell 7e-Alberts 43. Answer:

B

44. Answer:

A

45. Answer:

A

46. Answer:

B

47. Answer:

A

48. Answer:

1. B 2. B 3. C 4. B 5. A 6. C 7. A

49. Answer:

1. A 2. B 3. A 4. B 5. B 6. A

50. Answer:

1. A 2. A 3. A 4. B

51. Answer:

1. B 2. A 3. C

Molecular Biology of the Cell 7e-Alberts 4. C 52. Answer:

1. D 2. C 3. E 4. B 5. A

53. Answer:

1. B 2. A 3. B 4. B 5. B 6. A

54. Answer:

1. B 2. A 3. B 4. B

55. Answer:

1. B 2. C 3. F 4. E 5. D 6. A 7. G

56. Answer:

43215

57. Answer:

TCTAAAG

Molecular Biology of the Cell 7e-Alberts 58. Answer:

centromere, origin of replication, telomere

59. Answer:

2431

60. Answer:

The chromatin remodeling complexes can either slide the DNA along existing histone octamers, or the complex can exchange the histone octamer for other histone variants.

61. Answer:

The tail domains extend out of the nucleosome structure and are more accessible to enzymatic modification, even when the chromatin is compacted. Some of the tails, such as H3, closely interact with DNA as it enters and leaves the structure of the nucleosome thereby influencing compaction.

62. Answer:

Histone condensation is aided by the association of HP1 with the heterochromatin marked nucleosomes.

HP1 then deforms the nucleosomes and results in further compaction of the heterochromatin. The heterochromatic nucleosomes also interact with the nuclear lamina which support the nuclear envelope.

63. Answer:

BDC

64. Answer:

BDC

65. Answer:

Novel proteins could be generated by duplication and rearrangement of exons, such as is seen with immunoglobulin and other protein domain. Novel proteins could also be generated by the duplication of regulatory elements that alter the post-translational modification of the duplicated gene product.

66. Answer:

The last common ancestor between all mammalian species and other egg laying groups probably shared similar regulation of genes involved in the reception of extracellular proteins. The ancestors of the non-platypus mammals then evolved changes in the regulation of post-translational modification proteins separate from the platypus.

Molecular Biology of the Cell 7e-Alberts

Chapter 5: DNA Replication, Repair, and Recombination 1. The mutation rate in bacteria is about 3 nucleotide changes per 10 billion nucleotides per cell generation. Under laboratory conditions, bacteria such as Escherichia coli can divide and double in number about every 40 minutes. If a single Escherichia coli cell is allowed to exponentially divide for 10 hours in this manner, how many mutations would you expect to observe on average in the genome (4.5 million nucleotide pairs) of each of the resulting bacteria compared to the original cell? Assume all mutations are neutral; that is, they do not affect the cell-division time. a. Fewer than 0.001 b. About 0.02 c. One or two d. About 10 e. About 100

2. Which statement regarding the mutation rate of genomic DNA in different organisms is true? a. Human cells have a much higher mutation rate compared to bacteria when the rate is normalized to a single round of replication over the same length of DNA. b. Mutation rates limit the number of essential genes in an organism’s genome. c. Mutations in the somatic cells cannot be lethal. d. Even if the mutation rate was 10 times higher than its current value, germcell stability in humans would not have been affected.

3. A replication fork is shown schematically below. The strand labeled A is called the __________________ strand.

Molecular Biology of the Cell 7e-Alberts

a. Template b. Primer c. Leading d. Lagging

4. On average, errors occur in DNA synthesis only once in every 10 billion nucleotides incorporated. Which of the following does NOT contribute to this high fidelity of DNA synthesis? a. Complementary base-pairing between the nucleotides b. “Tightening” of the DNA polymerase enzyme around its active site to ensure correct pairing before monomer incorporation c. Exonucleolytic proofreading by the 5′-to-3′ exonuclease activity of the enzyme to correct mispairing even after monomer incorporation d. A strand-directed mismatch repair system that detects and resolves mismatches soon after DNA replication

5. The nuclear DNA polymerases in human cells have which trait? a. They polymerize about 1000 nucleotides per second during DNA replication in vivo. b. They are incapable of 3′-to-5′ exonuclease activity. c. They are capable of 3′-to-5′ DNA polymerase activity.

Molecular Biology of the Cell 7e-Alberts d. They have a single active site that is used for both polymerization and editing. e. They are unable to initiate polymerization de novo (i.e., in the absence of a primer).

6. During DNA replication in the cell, DNA primase makes short primers that are then extended by the replicative DNA polymerases. Which statement about these primers is true? a. They are made from deoxyribonucleotide triphosphates. b. They provide a 3′-hydroxyl group for the DNA polymerases to extend. c. They are made more frequently in the leading strand than the lagging strand. d. They are joined to the neighboring DNA by DNA ligase.

7. DNA ligases are used in both DNA replication and repair to seal breaks in the DNA. But DNA damage can result in single- or double-strand breaks that are not normal ligase substrates. These need to be processed first before a ligase can act on them. One of the enzymes that is recruited to some of such breaks is called PNK. It has two separate activities on the DNA, both of which can help provide a canonical ligase substrate. Which activity would you expect PNK to have in this context? a. 5′ kinase (phosphorylation of a free 5′-OH group) and 3′ kinase b. 5′ phosphatase (dephosphorylation to create a free 5′-OH group) and 3′ phosphatase c. 3′ kinase and 3′ phosphatase d. 5′ phosphatase and 3′ kinase e. 5′ kinase and 3′ phosphatase

8. What is the main source of the free energy for the mechanical work performed by DNA helicases during DNA replication in our cells?

Molecular Biology of the Cell 7e-Alberts a. The hydrogen-bonding energy in the DNA double helix b. Thermal energy in the nucleus c. ATP hydrolysis by the helicase d. The energy of SSB proteins binding to single-stranded DNA e. ATP hydrolysis by DNA topoisomerases

9. During DNA replication, what is the role of the single-strand DNA-binding (SSB) proteins? a. They are generally found more on the leading strand than the lagging strand where they can impede synthesis. b. They bind cooperatively to single-stranded DNA and cover the bases to prevent base-pairing. c. They prevent the folding of the single-stranded DNA. d. They bind cooperatively to short hairpin helices that readily form in the single-stranded DNA.

10. This protein is present at every replication fork and prevents DNA polymerase from dissociating, but does not impede the rapid movement of the enzyme. Which statement regarding this protein is true regarding? a. It self-assembles onto DNA at the replication fork. b. It is assembled on DNA as soon as DNA polymerase runs into a doublestrand region of DNA. c. Its assembly normally follows the synthesis of a new primer by the DNA primase. d. It disassembles from DNA as soon as DNA polymerase runs into a doublestrand region.

11. At bacterial replication forks, the template for the lagging strand is thought to loop around. This looping would allow the lagging-strand polymerases to move along

Molecular Biology of the Cell 7e-Alberts with the rest of the replication fork instead of in the opposite direction. The singlestrand parts of the loops are bound by the single-strand DNA-binding (SSB) proteins. (1) As the Okazaki fragment closest to the replication fork is synthesized, how does the size of the loop between its polymerase and the helicase change? (2) What happens to the size of the SSB-bound part of that loop?

a. (1) Increases; (2) increases b. (1) Increases; (2) decreases c. (1) Decreases; (2) increases d. (1) Decreases; (2) decreases e. (1) Decreases; (2) does not change

12. In DNA replication, a clamp loader is required to place the sliding clamp on the double strand for its binding to DNA polymerase. Where in this process is ATP hydrolysis required? a. For the clamp loader to bind to the clamp

Molecular Biology of the Cell 7e-Alberts b. For the clamp loader to open the clamp c. To move the clamp to the DNA d. To load the clamp around DNA and release the clamp loader e. For DNA polymerase to bind to the clamp

13. The Dam methylase is responsible for methylating the adenine base in GATC sequences in Escherichia coli. Imagine two E. coli strains, one without any active Dam methylase, and the other with a hyperactive version of the enzyme that operates faster than the wild-type enzyme. Which of these strains would you expect to show a “mutator” phenotype? a. Both of the strains b. Neither of them c. Only the first strain d. Only the second strain

14. Type I topoisomerases enzymatically act to nick the DNA backbone to relieve accumulated strain. What is the role of tyrosine in this mechanism at the enzyme’s active site? a. Covalently binding to a DNA hydroxide group b. Covalently binding to a DNA phosphate group c. Interacting noncovalently with a DNA hydroxide group d. Interacting noncovalently with a DNA phosphate group

15. Which feature is common between the replication origins in Escherichia coli and Saccharomyces cerevisiae? a. They both normally exist in one copy per genome. b. Both are specified by DNA sequences of tens of thousands of nucleotide pairs.

Molecular Biology of the Cell 7e-Alberts c. Both contain sequences that attract initiator proteins, as well as stretches of DNA rich in A-T base pairs. d. Both contain GATC repeats that are methylated to prevent the inappropriate “firing” of the origin.

16. DNA polymerases prefer deoxyribonucleotides approximately one million times more than ribonucleotides, but ribonucleotides are incorporated into the growing DNA strand every few thousand nucleotides. What causes this? a. Mismatch repair proteins incorporate ribonucleotides into the DNA they fix. b. The DNA polymerase synthesizing the lagging strand does not have the same fidelity as the polymerase synthesizing the leading strand. c. There is no mechanism by which to remove ribonucleotides incorporated by DNA primase. d. There is a lower concentration of deoxyribonucleotides in the cell compared to their ribonucleotide counterparts.

17. What do the enzymes topoisomerase I and topoisomerase II have in common? a. They both have nuclease activity. b. They both create double-strand DNA breaks. c. They both require ATP hydrolysis for their function. d. They both can create winding (tension) in an initially relaxed DNA molecule.

18. In the following schematic drawing, two DNA molecules are shown before and after the action of a protein that is also involved in the process of DNA replication.

What is this protein called?

Molecular Biology of the Cell 7e-Alberts

a. DNA ligase b. DNA helicase c. DNA polymerase I d. DNA topoimerase I e. DNA topoisomerase II

19. You have found a strain of Escherichia coli that has an unusually short doubling time of only 15 minutes, despite the fact that its complete DNA replication should take almost 35 minutes. You also find that there is only one replication origin on its chromosome from which two forks originate, just like the normal process described for E. coli. However, you discover that the origin of replication in this strain has a significantly shorter “refractory period,” resulting in the reactivation of the origin before the previous round of replication is over. Based on this model, if you examine the chromosomes of this strain (under conditions of fast growth), how many replication forks would you expect to observe per chromosome on average? a. Two, just like the wild-type strain b. Four c. Six d. Eight e. Fourteen

20. Which statement regarding eukaryotic replication origins is true? a. They are composed of specific, short DNA sequences, wrapped by

Molecular Biology of the Cell 7e-Alberts histones. b. Firing of a single origin or replication is mediated by dephosphorylation of components bound to it. c. Helicases are loaded onto the origin facing toward each other on opposite strands and pass each other as they begin to move away from the origin. d. They simultaneously fire once and only once during the cell cycle. e. They contain a stretch of DNA that is rich in Gs and Cs.

21. In eukaryotes, a protein is ubiquitinylated when two replication forks run into each other. This event causes the disassembly of the replication fork. Which protein is ubiquitinylated? a. DNA polymerase b. Helicase c. DNA ligase d. The sliding clamp e. The sliding clamp loader

22. What is a trait of the telomerase enzyme in human cells? a. It has an RNA component. b. It extends the telomeres by its RNA polymerase activity. c. It polymerizes the telomeric DNA sequences without using any template. d. It removes telomeric DNA from the ends of the chromosomes. e. It creates the “end-replication” problem.

23. If this protein complex does not function normally, the ends of the eukaryotic chromosomes would activate the cell’s DNA damage response, causing chromosomal fusions and other genomic anomalies. What is this protein complex

Molecular Biology of the Cell 7e-Alberts called? a. Telomerase b. T-loop c. ORC d. Shelterin e. RecA

24. Which of the following schematic drawings better depicts the end of mammalian chromosomal DNA?

a.

b.

c.

d.

e.

25. Which spontaneous lesion in DNA occurs most frequently in a mammalian cell? a. Depurination

Molecular Biology of the Cell 7e-Alberts b. Cytosine deamination c. Guanine oxidation d. Guanine alkylation e. Depyrimidination

26. DNA glycosylases constitute an enzyme family found in all three domains of life.

What is a function of this battery of enzymes? a. Add sugar moieties to DNA b. Remove sugar moieties from DNA c. Add a purine or pyrimidine base to DNA d. Remove a purine or pyrimidine base from DNA e. Remove a nucleotide from DNA

27. What are the products of deamination of cytosine and 5-methyl cytosine, respectively? a. Thymine and uracil b. Thymine in both cases c. Uracil and thymine d. Uracil in both cases e. Xanthine and hypoxanthine

28. Upon heavy damage to the cell’s DNA, the normal replicative DNA polymerases may stall when encountering damaged DNA, triggering the use of backup translesion polymerases. What can these backup polymerases do? a. Employ 3′-to-5′ exonucleolytic proofreading activity b. Continue to synthesize the new DNA strand well past the area of damaged DNA

Molecular Biology of the Cell 7e-Alberts c. Create mutations even on undamaged DNA d. Fail to recognize specific DNA damage such as a thymine dimer in order to add the appropriate nucleotide, restoring the original sequence

29. This protein folds into a doughnut shape that can encircle DNA. It can load on the DNA only when the DNA is broken in both strands, so that the DNA can thread through the hole in the protein. Which protein matches this description? a. PCNA, the sliding clamp for DNA polymerases at the replication forks b. Ku, the protein that recognizes DNA ends and can initiate nonhomologous end joining c. MCM, the helicase critical for the initiation and elongation of replication d. Topoisomerase II, which can create or relax superhelical tension in DNA e. RecA/Rad51, which carries out strand invasion in homologous recombination

30. Which repair pathway can accurately repair a double-strand break? a. Base excision repair b. Nucleotide excision repair c. Direct chemical reversal d. Homologous recombination e. Nonhomologous end joining

31. In which phases of the eukaryotic cell cycle does homologous recombination repair often occur to repair DNA damage? a. G1 and S phases b. S and G2 phases c. G2 and M phases

Molecular Biology of the Cell 7e-Alberts d. M and G1 phases e. G1 and G2 phases

32. Before homologous recombination can occur to repair a double strand break, a nuclease must resect, or “chew back” one of the two strands, leaving an overhang. (1) In which direction does the resection occur? (2) Which strand is left protruding? a. (1) 3′ → 5′; (2) 3′ overhang b. (1) 3′ → 5′; (2) 5′ overhang c. (1) 5′ → 3′; (2) 3′ overhang d. (1) 5′ → 3′; (2) 5′ overhang

33. If the template for the leading strand contains a single-strand break, this can turn into a double-strand break upon replication fork collapse. After resection of the broken strand, which strand can be used as a template for homologous recombination repair of the break? a. Original strand of the leading strand b. Original strand of the lagging strand c. New strand of the leading strand d. New strand of the lagging strand

34. Which statement regarding homologous recombination and its regulation is NOT true? a. Loss of heterozygosity can occur if a broken chromosome is repaired using a sister chromatid instead of its homologous chromosome. b. Repair of double-strand breaks by homologous recombination is favored shortly before DNA replication. c. Homologous recombination can rescue broken or stalled replication forks in

Molecular Biology of the Cell 7e-Alberts S phase. d. Excessive use of homologous recombination by human cells can lead to cancer.

35. In the following schematic drawing of a Holliday junction that undergoes branch migration, cutting at which combination of the sites a to d would generate a crossover?

a. a and b b. a and c c. a and d d. b and c e. b and d

36. During meiosis, homologous recombination can result in gene conversion. Which of the following could NOT contribute to a possible gene conversion event? a. Branch migration of Holliday junctions b. Capture of second strand by the overhang c. DNA synthesis using a homologous chromosome as a template for doublestrand break repair

Molecular Biology of the Cell 7e-Alberts d. Mismatch repair near sites of crossover e. Release of invading strand before DNA synthesis

37. In meiosis, a crossover in one position is thought to inhibit crossing over in the neighboring regions. This regulatory mechanism does what? a. Results in a very uneven distribution of crossover points along each chromosome b. Ensures that even small chromosomes undergo at least one crossover c. Controls how the Holliday junctions are resolved d. Behaves randomly, resulting in 50% of double-strand breaks produced during meiosis being resolved as crossovers

38. What group of mobile genetic elements is largely responsible for the resistance of the modern strains of pathogenic bacteria to common antibiotics? a. DNA-only transposons b. Retroviral-like retrotransposons c. Nonretroviral retrotransposons d. Site-specific recombinases

39. Which characteristic specifically describes the type of DNA-only transposon known as a helitron, setting it apart from other DNA-only transposons? a. Can be recognized by the presence of short inverted repeats at each end b. The encoded transposase functions as both a sequence-specific nuclease and as a helicase c. Leaves double-strand breaks in the donor chromosome d. Can move by a cut-and-paste mechanism

Molecular Biology of the Cell 7e-Alberts 40. SARS-CoV-2 is a [+]-strand RNA virus. Knowing this, what can we infer? a. RNA can immediately be translated to protein by the cell machinery. b. RNA needs to be used as a template to first make complementary RNA before translation can occur. c. A genome must be integrated into the host cell’s genome by reverse transcriptase before proteins can be expressed. d. Enzymes use a cut-and-paste mechanism to integrate its genome into the host’s chromosome.

41. Coronaviruses, such as SARS-CoV-2, have a proofreading ability in their replicase complex. Which effect does this proofreading have? a. Coronaviruses have a higher mutation rate than other RNA viruses. b. Coronaviruses must maintain a smaller genome than other RNA viruses. c. Coronaviruses must maintain a smaller number of essential genes compared to other RNA viruses. d. Nucleoside triphosphate analogs do not work as well as antiviral drugs against coronaviruses.

42. Which statement regarding retroviral-like retrotransposons is true? a. They encode both a reverse transcriptase and an RNA polymerase. b. They have directly repeated long terminal repeats at their two ends when integrated into chromosomal DNA. c. Their genomic RNA can be translated to produce viral coat proteins. d. They leave double-strand breaks in the original donor DNA. e. The Alu element in our genome is an example of retroviral-like retrotransposons.

43. Which statement regarding long and short interspersed nuclear elements is NOT

Molecular Biology of the Cell 7e-Alberts true? a. Each of them encodes a reverse transcriptase. b. They rely on the cellular transcription machinery to produce their RNA transcript. c. They use one of the strands in the target DNA as a primer to synthesize their DNA. d. Together, they make up about 40% of our genome. e. They can move into new regions of genome that do not have any homology with their DNA.

44. Below are different mobile genetic elements that are found in our genome. Which type appears to still be active and move in our genome, accounting for a detectable fraction of human mutations? a. DNA-only transposons b. Retroviral-like retrotransposons c. Nonretroviral retrotransposons d. Helitrons

45. “Phase variation” helps protect the bacterium Salmonella enterica against the immune system of its host by switching the orientation of a certain promoter.

What is true about this process? a. It is carried out through a DNA transposition mechanism. b. It is irreversible. c. It can often result in the excision of the promoter from the chromosome altogether. d. It is mediated by a conservative site-specific recombinase.

46. The sliding clamp and the DNA helicase that function at the replication fork both

Molecular Biology of the Cell 7e-Alberts have three-dimensional structures resembling a ring with a central hole through which DNA is threaded. The clamp has a wider hole in its structure than the DNA helicase. a. True b. False

47. Both replication forks in a replication bubble are normally active in replication. a. True b. False

48. In strand-directed mismatch repair in eukaryotes, discrimination of which strand is to be repaired is determined by the orientation of the clamp loader. If a MutS/MutL complex binds to a recently completed Okazaki fragment and scans the DNA in the direction of the polymerase that just completed the synthesis of that particular fragment, the MutS/MutL complex will encounter the polymeraseface of the sliding clamp.

a. True b. False

49. Examples of two general types of DNA damage are shown in the following drawing. Type (1) is more common in our cells.

Molecular Biology of the Cell 7e-Alberts

a. True b. False

50. The nucleotide excision repair pathway involves recognition of distortions in the DNA double helix rather than specific base changes. a. True b. False

51. In contrast to vertebrates, there is very little DNA methylation in the genomes of invertebrates such as Drosophila melanogaster and Caenorhabditis elegans. On average, approximately one dinucleotide out of every 256 in the D. melanogaster genome is a CG dinucleotide. a. True b. False

52. In human cells, the predominant pathway to repair double-strand breaks is homologous recombination, in which the broken ends are simply rejoined with the concomitant loss of a few nucleotides. This leaves scars at the breakage sites. a. True b. False

Molecular Biology of the Cell 7e-Alberts

53. The RecA/Rad51 protein carries out strand exchange during homologous recombination. RecA hydrolyzes ATP upon binding to the invading DNA strand. a. True b. False

54. Holliday junctions can form during meiotic recombination, sometimes in pairs. a. True b. False

55. DNA nucleotides can be damaged by undergoing deamination events, yielding several nonstandard bases that can be detected by DNA repair pathways. For each of the following naturally occurring bases, what type of base would you expect to see created after its spontaneous deamination?

a.

b.

c.

Molecular Biology of the Cell 7e-Alberts

d.

e.

1. A non-standard base not normally included nucleic acids 2. A base not normally included in DNA but standard to RNA 3. A base standard to DNA but not normally included in RNA 4. A base standard to both DNA and in RNA 5. No deamination is possible

56. In Escherichia coli, replication of DNA occurs concurrently with transcription. This means collisions between replication forks and RNA polymerases are inevitable.

Depending on the orientation of the genes, collisions can be rear-end (when both machines are traveling in the same direction) or head-on (when they are traveling in opposite directions). In the following paragraph, identify which of the options (A through E) should appear in the numbered descriptions listed below. Do not use an option more than once.

In a rear-end collision, the ______ (I)______ of RNA polymerase typically collides with the ______ (II) ______ in the replication fork. In contrast, in a head-on collision, the ______ (III) ______ of RNA polymerase hits the ______ (IV) ______ in the fork. a. Front edge (of RNA polymerase) b. Rear edge (of RNA polymerase) c. DNA helicase d. Leading-strand DNA polymerase

1. I 2. II 3. III

Molecular Biology of the Cell 7e-Alberts 4. IV

57. Indicate whether each of the following DNA lesions is typically repaired via the base excision or nucleotide excision repair pathway. a. Base excision b. Nucleotide excision

1. Deaminated cytosines 2. Depurinated residues 3. Thymine dimers 4. Bulky guanine adducts

58. Sort the following steps in the order in which they normally occur during the process of repairing double-strand breaks by homologous recombination.

a.

b.

c.

d.

e.

Molecular Biology of the Cell 7e-Alberts

f.

1. DNA synthesis using original DNA as the template 2. DNA synthesis using undamaged DNA as the template 3. Ligation 4. Nuclease digestion (resection) 5. Release of the invading strand 6. Strand invasion

59. Indicate whether each of the following mobile elements in Drosophila be best described as a DNA-only transposon, a retroviral-like retrotransposon, or a nonretroviral retrotransposon. a. DNA-only transposon b. Retroviral-like retrotransposon c. Nonretroviral retrotransposon

1. P elements: these have inverted terminal repeats and move with the help of a transposase enzyme.

2. Copia elements: these have directly repeated long terminal repeats and move with the help of a reverse transcriptase and an integrase.

3. F elements: these are long interspersed nuclear elements (LINEs).

4. Mariner elements: these move via a DNA intermediate.

60. As shown in the following drawing, a researcher has engineered three pairs of LoxP sites (for conservative site-specific recombination) in a region that contains three reporter genes coding for red, yellow, or cyan fluorescent proteins, respectively. Each type of LoxP sequence (shown as a black, gray, or white arrowhead) is specific, meaning it does not recombine with the other types of LoxP sequences.

Upon Cre recombinase activation, depending on which recombination event occurs first (which we assume is random), a number of possible combinations of reporters can remain in the final DNA. For each of the following combinations,

Molecular Biology of the Cell 7e-Alberts indicate whether it (A) can or (B) cannot result from this recombination scheme.

Do not consider the reintegration of excised DNA, which happens very rarely.

a. Can result b. Cannot result

1. Red and yellow 2. Red only 3. Yellow only 4. Cyan only 5. Yellow and cyan 6. Red and cyan

61. Compare and contrast the proteins involved in the bacterial and the eukaryotic replication forks. How would you describe the differences, or even similarities, between the two?

62. Mismatch repair is coupled to DNA replication in both prokaryotes and eukaryotes. For this coupling, where would the difference between taxonomic domains lie? What part of the coupling is the same? Explain your answer in 2-3 sentences.

63. What combination of the following events normally prevents the origins of replication in the yeast Saccharomyces cerevisiae from “firing” more than once during the cell cycle? Your answer is a two-letter string composed of letters A to E only, for example, AB. Order the letters in your answer alphabetically.

A. The helicase-loading proteins Cdc6 and Cdt1 are only active in S phase.

B. The helicase Mcm1 can be delivered to the origin recognition complex (ORC) only in S phase.

C. The ORC can only become active (by dephosphorylation) in G 1 phase.

D. The helicase can only become active (by phosphorylation) in S phase.

E. The ORC can bind to the origin only in S phase.

Molecular Biology of the Cell 7e-Alberts 64. Telomerase, rather than DNA polymerase, replicates the ends of linear chromosomes. Why is it necessary that there be an alternative mechanism for this replication, and what would occur if there were no other mechanism?

65.

In cells that are exposed to sunlight, ultraviolet (UV) light can result in covalent linkage between two adjacent DNA bases. If not repaired in time, such dimers can lead to inheritable mutations. Consider the sequence 5′GGTATGATCATTATAA-3′ in the chromosome of a cell that is exposed to intense sunlight. How many possible dimers can form in the region of DNA double helix corresponding to this sequence?

66. How would you describe the process by which “fork reversal” can faithfully synthesize daughter DNA strands without first having DNA damage repaired?

Assume the damage is on the template for the leading strand.

67. The following genomic region contains gene X, facing to the left, gene Y, facing to the right, and a single promoter. Colorimetric assays exist to detect the expression of these genes—if gene X is expressed, a colorless compound will be converted to a blue compound; if gene Y is expressed, a red compound will be converted to a different colorless compound. Site-specific recombination can occur between the arrows indicated (showing region and direction of homology).

If both assays are used to measure gene expression and the result is a clear solution, explain the expected results of the assays if they are conducted after a site-specific recombination event here.

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 05 1. Answer:

B

2. Answer:

B

3. Answer:

D

4. Answer:

C

5. Answer:

E

6. Answer:

B

7. Answer:

E

8. Answer:

C

9. Answer:

C

10. Answer:

C

11. Answer:

A

12. Answer:

D

13. Answer:

A

14. Answer:

B

15. Answer:

C

16. Answer:

D

17. Answer:

A

18. Answer:

E

19. Answer:

C

20. Answer:

C

Molecular Biology of the Cell 7e-Alberts 21. Answer:

B

22. Answer:

A

23. Answer:

D

24. Answer:

D

25. Answer:

A

26. Answer:

D

27. Answer:

C

28. Answer:

C

29. Answer:

B

30. Answer:

D

31. Answer:

B

32. Answer:

C

33. Answer:

D

34. Answer:

B

35. Answer:

E

36. Answer:

E

37. Answer:

B

38. Answer:

A

39. Answer:

B

40. Answer:

A

41. Answer:

D

42. Answer:

B

Molecular Biology of the Cell 7e-Alberts 43. Answer:

A

44. Answer:

C

45. Answer:

D

46. Answer:

A

47. Answer:

A

48. Answer:

B

49. Answer:

A

50. Answer:

A

51. Answer:

B

52. Answer:

A

53. Answer:

B

54. Answer:

A

55. Answer:

1. A 2. B 3. C 4. A 5. E

56. Answer:

1. C 2. A 3. B 4. D

57. Answer:

1. A 2. A 3. B

Molecular Biology of the Cell 7e-Alberts 4. B 58. Answer:

1. E 2. C 3. F 4. A 5. D 6. B

59. Answer:

1. A 2. B 3. C 4. A

60. Answer:

1. E 2. B 3. B 4. A 5. B 6. A

61. Answer:

While there is a single DNA polymerase used at the bacterial replication fork, there are three (alpha, delta, and epsilon) used at the eukaryotic fork. Alpha DNA polymerase contains the RNA primase. The helicases are present on different strands—the lagging strand template in prokaryotes and the leading strand template in eukaryotes. Finally, the clamp loader in prokaryotes coordinates synthesis, providing connection to each DNA polymerase active at the replication fork. It is currently unknown how these are coordinated in eukaryotes.

62. Answer:

In prokaryotes, mismatch repair uses the methylation of GATC sites to determine which strand is the original and which contains the error, while in eukaryotes, it uses the face of the sliding clamp to determine the same. In both domains, DNA polymerase and DNA ligase are needed at the end of the processes, and these are readily present

Molecular Biology of the Cell 7e-Alberts during DNA replication.

63. Answer:

CD

64. Answer:

DNA polymerase cannot replicate the ends of linear chromosomes because it requires the 3′-hydroxyl group of a duplexed nucleotide chain in order to begin synthesis.

There would need to be a priming site upstream of the free, 3′ end of the DNA strand. Without a separate mechanism for replicating these ends, the ends of the chromosome would be lost.

65. Answer:

4

66. Answer:

When a replication fork stalls because of unrepaired DNA damage, a specialized helicase peels the new DNA strands from the parent strands, and the daughter strand generated from the lagging strand is temporarily used as a template for leading strand synthesis. The normal fork then reforms, and the leading strand DNA, synthesized from the new lagging strand, is past the point of damage on the original template.

67. Answer:

The assays will result in both a red and a blue color, or purple. The “before” results being clear indicates that the promoter is driving expression of Y and not X. Because the arrows are facing in opposite directions, the homology is such that an inversion will occur after site-specific recombination. This inversion will result in the expression of X and not Y.

Molecular Biology of the Cell 7e-Alberts

Chapter 6: How Cells Read the Genome: From DNA to Protein 1. DNA and RNA polymerase differ in all of the following except which of the following? a. The nucleotide substrates they incorporate b. Their requirement for a prime c. Their error rate d. The type of chemical reaction they catalyze e. Their processivity

2. What enzyme is depicted in the following schematic drawing?

a. DNA polymerase b. RNA polymerase c. Ribosome d. Reverse transcriptase e. Topoisomerase

3. Due to their high transcription rate, active ribosomal RNA (rRNA) genes can be easily distinguished in electron micrographs of chromatin spreads. They have a characteristic “Christmas tree” appearance, where the DNA template is the “trunk” of the tree and the nascent RNA transcripts form closely packed “branches.” At

Molecular Biology of the Cell 7e-Alberts the base of each branch is an RNA polymerase extending that branch, while RNA processing complexes at the tip of the branch form terminal “ornaments.” Would you expect the top of the tree to represent the beginning or end of the rRNA gene? Would the “ornaments” be at the 3′, 5′, or both ends of the nascent rRNA molecules? a. End; 3′ b. End; 5′ c. Beginning; either 3′ or 5′ d. Beginning; 3′ e. Beginning; 5′

4. Which of the following types of noncoding RNA chiefly functions in the processing and chemical modification of ribosomal RNAs (rRNAs)? a. Small nuclear RNAs (snRNAs) b. Small nucleolar RNAs (snoRNAs) c. Small interfering RNAs (siRNAs) d. Transfer RNAs (tRNAs) e. MicroRNAs (miRNAs)

5. For the bacterial transcription machinery, which of the following mRNA sequences would you expect to constitute a potent transcriptional termination signal? Note that the two underlined regions in each sequence are complementary to each other. a. 5′... UGGCCCAGUCGGAAGACUGGGCCUUUUGUUUU...3′ b. 5′... UGGCCCAGUCGGAAGACUGGGCCCGCGGAGCU...3′ c. 5′... UUUUGUUUUAGGCCCAGUCGGAAGACUGGGCCA...3′ d. 5′... CGCGGAGCUAGGCCCAGUCGGAAGACUGGGCCA...3′

Molecular Biology of the Cell 7e-Alberts 6. The sequence of a region of DNA around the 5′ end of a gene in Escherichia coli is shown below. The −10 hexamer and the transcription start site are highlighted.

What would be the sequence of the first 10 nucleotides of the mRNA transcribed from this gene?

5′…GCGCTTGGTATAATCGCTGGGGGTCAAAGAT…3′ a. AUAUUAGCGA b. CCAGUUUCUA c. UAUAAUCGCT d. GGGGUCGCUA e. GGUCAAAGAU

7. You have identified an RNA-binding protein that can bind directly (with varying affinities) to mRNA molecules that bear the following sequences near the 3′ splice sites of their exons. Based on these results, what is the five-nucleotide consensus sequence for the binding site of this protein?

a. AAGCG b. CCAGA c. AACGC d. GGCAC e. CCGAG

8. The transcript for which of the following noncoding RNA in our cells is expected to undergo 5′ cap addition after transcription?

Molecular Biology of the Cell 7e-Alberts a. 5S rRNA b. miR-21 (a microRNA) c. tRNAPhe d. 5.8S rRNA e. 18S rRNA

9. This large and complex general transcription factor has a DNA helicase activity that exposes the template for RNA polymerase II transcription. It also has a kinase activity that phosphorylates the C-terminal domain of the polymerase on Ser5 leading to promoter clearance. What is it? a. TFIIB b. TFIID c. TFIIE d. TFIIF e. TFIIH

10. This general transcription factor recognizes the TATA box in RNA polymerase II promoters. Which statement applies to it? a. It is the only single-subunit general transcription factor. b. It is able to introduce a rather sharp kink in the double helix upon binding to DNA. c. It is responsible for the phosphorylation of the RNA polymerase CTD during transcription initiation. d. It is also responsible for the recognition of the BRE element in the promoter.

11. The general transcription factor, TFIID, contains a TATA-binding protein subunit to align the transcription start point. What other feature of transcription does TFIID

Molecular Biology of the Cell 7e-Alberts carry out? a. Deformation of the DNA to prepare the region for transcription b. Positioning of RNA polymerase c. Unwinding DNA d. Releasing RNA polymerase from the promoter e. Phosphorylating the CTD on RNA Polymerase II

12. How does a eukaryotic cell deal with the superhelical tension in its genomic DNA resulting from the activity of RNA polymerases? a. DNA gyrase introduces negative supercoils, keeping the DNA under constant tension. b. The RNA polymerases are allowed to rotate freely around their templates during transcription, leading to the relaxation of the tension. c. DNA topoisomerases rapidly remove the superhelical tension caused by transcription. d. The nucleosomes adjust the tension by binding to positively supercoiled regions behind a moving RNA polymerase.

13. Of the following proteins or protein complexes, which one does NOT typically interact with an elongating RNA polymerase II? a. Histone-modifying enzymes b. Capping enzymes c. Chromatin remodeling complexes d. Mediator complexes e. Histone chaperones

14. Which region of RNA Polymerase II is phosphorylated repeatedly, allowing it to

Molecular Biology of the Cell 7e-Alberts recruit RNA processing machinery to the emerging transcript? a. N-terminal domain b. Catalytic core c. Mediator d. C-terminal domain e. Ribonucleotide triphosphate uptake channel

15. Comparing mRNA molecules from human and Escherichia coli cells, which of the following is typically NOT true? a. A human mRNA has a special 5′ cap, while a bacterial mRNA does not. b. A human mRNA has a poly-A tail, while a bacterial mRNA does not. c. A human mRNA undergoes alternative splicing, while a bacterial mRNA does not. d. A human mRNA contains noncoding sequences, while a bacterial mRNA does not. e. A typical human mRNA encodes one protein, while many bacterial mRNAs encode several different proteins.

16. Eukaryotic pre-mRNAs undergo a number of modifications such as capping at the 5′ end. What is a characteristic of a 5′ cap? a. It consists of a modified terminal adenine nucleotide. b. It has a 3′-to-5′ linkage between the terminal nucleotide and the 5′ end of the pre-mRNA. c. It contains a triphosphate bridge between the terminal base and the 5′ end of the pre-mRNA. d. It carries a negative charge in the terminal base due to methylation. e. It is identical for all mRNAs that are transcribed by RNA polymerase II.

Molecular Biology of the Cell 7e-Alberts

17. After the first and before the second chemical step of RNA splicing, the intron of the pre-mRNA is still covalently connected where? a. To the 3′ exon and has an internal branch in the shape of a lariat b. To the 3′ exon and is linear c. To the 5′ exon and has an internal branch in the shape of a lariat d. To the 5′ exon and is linear e. To both of its flanking exons and is linear

18. What is true about the catalytic site of the spliceosome complex for RNA splicing? a. It remains in active conformation as it processes reactions across multiple pre-mRNAs processively. b. It is formed by the U2 and U6 snRNAs. c. It only processes the first transesterification reaction, not the second. d. It requires more ATP in order to catalyze the two splice reactions than are consumed by the rest of the RNA rearrangements in the active spliceosome. e. It can “run backward” and reinsert exons that have been previously removed.

19. In the following qualitative histogram, which two curves better correspond to human exon and intron length distributions, respectively?

Molecular Biology of the Cell 7e-Alberts

a. Curves 1 and 2 b. Curves 2 and 1 c. Curves 2 and 3 d. Curves 3 and 2 e. Curves 3 and 1

20. To ensure the fidelity of splicing, the spliceosome uses all of these mechanisms EXCEPT what? a. Hydrolyzing ATP to undergo complex rearrangements b. Examining the splicing signals on the pre-mRNA several times c. Assembling on the pre-mRNA co-transcriptionally d. Taking advantage of “exon definition”

Molecular Biology of the Cell 7e-Alberts e. Comparing the transcript back to the DNA template strand

21. A primary mRNA transcript with three exons is depicted below. Which mature mRNA products of this transcript is a result of exon skipping?

a.

b.

c.

d.

Molecular Biology of the Cell 7e-Alberts

e.

22. The enzyme poly-A polymerase is responsible for adding 3′ poly-A tails to eukaryotic mRNAs. What is a trait of this enzyme? a. It cuts the mRNA after recognition of the cleavage/polyadenylation signal

by CstF and CPSF proteins. b. It polymerizes the tail using an RNA template that is part of the enzyme. c. It is extremely processive. d. It normally adds about 1000 A nucleotides to the mRNA. e. It uses ATP as a substrate.

23. Which of the following better describes a typical, actively translated mRNA in its journey from the nucleus to the cytosol? a. Initially in a circular conformation, the mRNA linearizes, enters the cytosol

Molecular Biology of the Cell 7e-Alberts (5′ end first), and remains linear. b. Initially in a circular conformation, the mRNA linearizes, enters the cytosol (3′ end first), and remains linear. c. Initially linear, the mRNA enters the cytosol (5′ end first), and adopts a circular conformation. d. Initially linear, the mRNA enters the cytosol (3′ end first), and adopts a circular conformation. e. Initially linear, the mRNA enters the cytosol (3′ end first), and remains linear.

24. As an mRNA molecule is processed in the nucleus, it loses some proteins and binds to new ones, some of which are used in mRNA surveillance pathways. The presence of which of the following molecules on an mRNA is a signal that the mRNA is still NOT ready for nuclear export? a. Cap-binding complex b. Exon junction complex c. snRNPs used in splicing d. Poly-A-binding proteins e. SR proteins

25. The nucleolus is a dynamic subcompartment within the nucleus and its size varies depending on the circumstances. In which of the following cells would you NOT expect to see the nucleoli? a. A yeast cell undergoing DNA replication b. A human neuron in a quiescent (G0) state c. A human macrophage active in phagocytosis d. A fruit fly embryonic nucleus in the G2 phase of the cell cycle e. A mouse embryonic cell in the metaphase stage of mitosis

Molecular Biology of the Cell 7e-Alberts

26. Which of the following is true about the nucleolus? a. It is an intranuclear membrane-bound organelle. b. It is of uniform size across cell types. c. It excludes DNA from its structure. d. It remains intact during cell division. e. It is a macromolecular condensate.

27. Which statement is true for Cajal bodies in the eukaryotic cell nucleus? a. They are stockpiles of fully mature snRNPs and other RNA processing components. b. They can only be observed by electron microscopy. c. They are absolutely required in all cell types. d. They speed up the maturation and assembly of snRNPs and snoRNPs. e. They are the main sites of pre-mRNA splicing.

28. Several mechanisms contribute to the diversity of the mRNAs and proteins encoded by a single gene in our genome. Which of the following is normally NOT one of them? a. Alternative choice of polyadenylation sites b. Alternative choice of translation initiation sites c. Alternative choice of transcription initiation sites d. Alternative choice of the reading frames e. Alternative choice of splice sites

29. Which genes are transcribed in the nucleoli within the eukaryotic nucleus?

Molecular Biology of the Cell 7e-Alberts a. tRNA b. rRNA c. mRNA d. snRNA e. telomerase RNA

30. No matter where translation begins, only one polypeptide sequence pattern can be obtained from the translation of an mRNA chain with the sequence (AC) n; that is, an mRNA with the sequence 5′...ACACACACAC...3′. In contrast, an mRNA chain with the sequence (ACG)n can be translated to polypeptide sequences of three distinct patterns, depending on the choice of the reading frame. Using similar judgment, how many different polypeptide sequence patterns can be obtained from an mRNA chain with the sequence (ACGU) n? a. 1 b. 2 c. 3 d. 4 e. 5

31. The anticodon sequence of a phenylalanine-specific tRNA is 5′-GAA-3′. Given the wobble base-pairing rules presented in the following table, how many different codons can be “read” by this tRNA? Wobble base in the codon

Possible anticodon bases

U

A, G, or I

C

G or I

A

U

Molecular Biology of the Cell 7e-Alberts G

C

a. 1 b. 2 c. 3 d. 4 e. 5

32. Which statement regarding the molecule depicted is true?

a. Its gene is transcribed by RNA polymerase I. b. There are about 50 genes in our genome encoding this type of molecule. c. This molecule normally undergoes various covalent modifications. d. It is normally composed of about 20 monomers. e. Its gene transcript normally undergoes alternative splicing.

33. tRNA splicing differs from mRNA splicing in eukaryotic cells. What do that they have in common?

Molecular Biology of the Cell 7e-Alberts a. Both can remove introns. b. Both are carried out by proteins only. c. Both create a lariat intermediate. d. Both involve RNA endonuclease and RNA ligase activities. e. Both proceed via transesterification reactions.

34. The modified nucleotide shown below is NEVER found in mature what?

a. tRNAs b. Ribosomes c. Spliceosomes d. snoRNAs e. DNA

35. Each aminoacyl-tRNA synthetase activates a certain amino acid by attaching it to its corresponding tRNA(s) through a high-energy linkage between which two groups? a. The amino group of the amino acid and the 3′ hydroxyl group of the tRNA b. The amino group of the amino acid and the 5′ hydroxyl group of the tRNA c. The carboxyl group of the amino acid and the 3′ or 2′ hydroxyl group of the tRNA

Molecular Biology of the Cell 7e-Alberts d. The carboxyl group of the amino acid and the 5′ hydroxyl group of the tRNA e. The amino group of the amino acid and the 3′ or 2′ hydroxyl group of the tRNA

36. Activation of an amino acid by an aminoacyl-tRNA synthetase involves what covalent linkage? a. AMP to the amino acid b. ADP to the amino acid c. AMP to the tRNA d. ADP to the tRNA e. ADP to the enzyme

37. In order to prevent errors in the protein sequence due to a tRNA carrying the wrong amino acid, a tRNA synthetase will preferentially bind to a particular amino acid target. What happens next? a. The specificity of this mechanism is sufficient to achieve near-perfect accuracy. b. It checks the amino acid by moving it into a second editing pocket that excludes the correct amino acid while allowing access to closely related amino acids. c. Misfolds are targeted for degradation in the case of mismatch. d. It releases and binds to a different tRNA if the amino acid is incorrect. e. It alters the tRNA anticodon to an appropriate sequence to match the amino acid.

38. An elongating ribosome is bound to appropriate tRNAs in both the A and the P sites and is ready for peptidyl transfer. What happens next? a. The carboxyl end of the polypeptide chain is released from the P-site tRNA and joined to the free amino group of the amino acid linked to the A-site

Molecular Biology of the Cell 7e-Alberts tRNA. b. The amino end of the polypeptide chain is released from the P-site tRNA and joined to the free carboxyl group of the amino acid linked to the A-site tRNA. c. The carboxyl end of the amino acid is released from the A-site tRNA and joined to the free amino group of the polypeptide chain linked to the P-site tRNA. d. The amino end of the amino acid is released from the A-site tRNA and joined to the free carboxyl group of the polypeptide chain linked to the Psite tRNA.

39. Which of the following nucleotides is hydrolyzed in both transcription and in translation elongation? a. ATP b. GTP c. TTP d. UTP e. CTP

40. How fast does a bacterial ribosome move on an mRNA? a. At about 2 nucleotides per second, significantly lower than the speed of the RNA polymerase. b. At about 5 nucleotides per second, comparable to the speed of the RNA polymerase. c. At about 10 nucleotides per second, significantly lower than the speed of the RNA polymerase. d. At about 20 nucleotides per second, comparable to the speed of the RNA polymerase. e. At about 60 nucleotides per second, comparable to the speed of the RNA polymerase.

Molecular Biology of the Cell 7e-Alberts

41. What feature is common between the bacterial and eukaryotic ribosomes in translation initiation? a. They both use an initiator tRNA that carries formylmethionine. b. They both bind to the 5′ end of the mRNA and move forward to find the start codon. c. They both recognize the start codon by interacting with the Shine–Dalgarno sequence. d. They both interact with various translation initiation factors.

42. The specific transfer RNA used for the incorporation of selenocysteine in proteins recognizes the UGA codon, which is normally a translation stop codon. What prevents this tRNA from reading through all the other “legitimate” UGA stop codons in the cells and causing a massive, disastrous recoding? a. The cells that have selenocysteine in their proteins do not normally use the UGA stop codon and always use the other two stop codons (UAA and UAG) instead. b. This tRNA is made in such a small amount that its side effects are negligible. c. This tRNA only interacts with the UGA codon in the P site of the ribosome, and therefore does not interfere with the normal function of the codon in translation termination, which takes place in the A site. d. The mRNAs encoding the selenocysteine-containing proteins also contain additional sequences required for the recoding event.

43. In a gene that normally contains three exons, which of the following changes probably will NOT activate the nonsense-mediated mRNA decay pathway? The sizes of exons 1 to 3 are 100, 150, and 200 nucleotide pairs, respectively. a. A nonsense mutation in exon 1 b. A nonsense mutation in exon 2

Molecular Biology of the Cell 7e-Alberts c. A mutation in the first intron resulting in the inclusion of a large intronic fragment in the mature mRNA d. A frameshift mutation in exon 1 e. A mutation in exon 2 leading to its loss through exon skipping

44. On the ribosome, the mRNA is read from which end to which end, and the polypeptide chain is synthesized from which terminus to which terminus? a. 5′ to 3′; C- to N-terminus b. 5′ to 3′; N- to C-terminus c. 3′ to 5′; C- to N-terminus d. 3′ to 5′; N- to C-terminus

45. This complex uses ATP and forms an “isolation chamber” for its misfolded protein substrates, allowing them to fold to their native conformation. What is it called? a. A proteasome b. A ribosome c. A mediator d. A chaperonin e. An exosome

46. What is NOT a function of the 19S cap in the proteasome complex? a. Recognizing the proteins that will be degraded b. Unfolding the proteins that will be degraded and threading them into the central 20S cylinder for digestion c. Removing the polyubiquitin chain from the proteins that will be degraded d. Proteolytic cleavage of the inserted protein, resulting in free amino acids

Molecular Biology of the Cell 7e-Alberts e. Hydrolyzing ATP to make the digestion process highly efficient

47. In which of the following regions would you expect to find a chaperone bound to the protein that it helps fold into the correct confirmation? a. A hydrophobic region b. A hydrophilic region c. C-terminal end d. N-terminal end

48. Regarding how proteins are tagged for and undergo degradation, which statement is TRUE? a. Tagged misfolded proteins are destroyed by a protease called ubiquitin. b. Proteins that cannot be unfolded are imported into the proteosome in their folded state. c. The presence of ubiquitin signals proper protein conformation. d. If the protein cannot be linearized and imported into the proteosome in the first attempt, it is de-ubiquitylated and released. e. Degradation of a tagged misfolded protein is accomplished by the proteasome complex, which occurs in the cytosol.

49. In humans, nearly 80% of proteins are acetylated on their N-terminal residue, a modification known to be recognized by a specific E3 enzyme, which directs the ubiquitylation of the protein for rapid degradation. Does this mean that all of these proteins are actively degraded at any given time? a. Yes; this high turnover rate ensures that their activity is under tight control. b. Yes; but they are not fully degraded in this way and can still function as protein fragments. c. No; the destruction signal can be buried in the interior of the protein or bound to other proteins.

Molecular Biology of the Cell 7e-Alberts d. No; the E3 enzyme recognizing this mark is inactive most of the time.

50. Which of the following occurs in the nucleus? a. Initiation of protein synthesis b. Proteosome degradation c. mRNA polyadenylation d. Protein folding e. Post-translational mRNA degradation

51. In the following representation of a single-stranded RNA molecule, the nucleotide residues are drawn as gray circles, while base-pairing interactions between them are indicated by dashed lines. Describe the structure of this RNA molecule.

a. One hairpin loop and one pseudoknot b. One pseudoknot only c. Three hairpin loops and a four-stem junction d. Two hairpin loops and a three-nucleotide bulge e. Three hairpin loops

52. Along with the established functional diversity of RNA, the presence of biologically active peptides that do not require ribosomal production supports which of the following hypotheses? a. Macromolecules first emerged and became biologically functional in the order: RNA, protein, DNA.

Molecular Biology of the Cell 7e-Alberts b. The first DNA probably encoded small peptides—not RNA. c. Catalysis was a property of protein-based enzymes before it appeared in ribozymes. d. ATP is the oldest energy substrate molecule to exist. e. In the course of evolution, protein synthesis could only occur after DNA became the main repository of genetic information.

53. Which of the following is an advantage to living things using DNA, rather than RNA, to carry hereditary information? a. It is easier to synthesize deoxyribose than ribose. b. DNA’s deoxyribose backbone is more chemically stable. c. Uracil is harder for the cell to produce than thymine. d. DNA can have catalytic functionality. e. RNA uses more nucleotides.

54. The exon junction complexes mark the sites on the mRNA where splicing has not been successful. a. True b. False

55. According to the RNA world hypothesis, RNA in primitive cells was responsible for storing genetic information, while proteins were responsible for the catalysis of chemical reactions. a. True b. False

56. How many ATP molecules must be hydrolyzed by the proteasome for the digestion of a protein that has been tagged for degradation with a polyubiquitin chain?

Molecular Biology of the Cell 7e-Alberts a. None; the digestion does not require ATP hydrolysis and is exergonic. b. One c. Two d. Ten e. Many; the number depends on the specific protein.

57. The ribosome can detect correct Watson–Crick base-pairing between the codon and anticodon in the A site, and consequently trigger GTP hydrolysis by EF-Tu. a. True b. False

58. A protein may undergo several cycles of hsp70 binding and release while still being synthesized on the ribosome. a. True b. False

59. What is the correct order of events in which transcription initiation in Escherichia coli takes place? a. Step 1 b. Step 2 c. Step 3 d. Step 4

1. Abortive initiation trials 2. σ Factor release from the RNA polymerase holoenzyme 3. Binding of the holoenzyme to the promoter in the “closed” complex 4. Formation of the transcription bubble

60. Would you expect the following to be located (or take place) in the large or the small ribosomal subunit?

Molecular Biology of the Cell 7e-Alberts a. Large ribosomal subunit b. Small ribosomal subunit

1. Peptide bond formation 2. Codon–anticodon interaction 3. The path of the mRNA 4. The polypeptide exit tunnel

61. The below diagram shows the overall flow of genetic information in all living cells.

What is the name of the process indicated with a question mark?

62. In the following schematic drawing, a transcribing RNA polymerase is moving toward the right on the DNA template as indicated. If the polymerase is allowed to rotate freely while the DNA is kept from doing so, would the enzyme rotate in a right-handed or left-handed fashion around the template?

63. Describe the likely outcome of the following scenario: A mutation in the CTD of RNA Polymerase II caused a decrease in the affinity for cleavage stimulation factor (CstF) and cleavage and polyadenylation specificity factor (CPSF).

64. In an experimental cell line, a scientist uses a highly targeted, weak pulse of energy to physically disrupt the contents of the nucleus without damaging the DNA. They observe that ribosome, snoRNP, and snRNA production decrease and then resume. In 2–3 sentences, explain how this is likely to occur.

Molecular Biology of the Cell 7e-Alberts

65. A researcher conducts an experiment in which cultured neurons from a mouse brain are exposed to the antibiotic puromycin. In order to induce a strengthening of the cell’s response, the researcher stimulates the cell with chemical neurotransmitters paired with repeated electrical pulses. In the end, the researcher finds that the unexposed cells develop stronger responses to neurotransmitter, while exposed cells fail to do so. The unexposed cells have more transmitter receptor channel proteins on their surface. What common feature of antibiotics may have led to this result?

66. How does the ribosome-associated quality control mechanism benefit the cell?

67.

Your friend is studying the sequence conservation around the branch-point site on a subset of pre-mRNAs in Macaca mulatto (rhesus macaque) and has since sent you the following sequence logo of the region of interest. However, they have forgotten to indicate the position of the branch-point nucleotide itself within this region. On a scale from 1–8, on which position do you think the branch-point nucleotide lies? Write down your answer as a number between 1 and 8.

Molecular Biology of the Cell 7e-Alberts

68. Compare and contrast the mechanisms by which bacteria versus eukaryotes typically initiate mRNA translation.

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 06 1. Answer:

D

2. Answer:

B

3. Answer:

E

4. Answer:

B

5. Answer:

A

6. Answer:

E

7. Answer:

C

8. Answer:

B

9. Answer:

E

10. Answer:

B

11. Answer:

A

12. Answer:

C

13. Answer:

D

14. Answer:

D

15. Answer:

D

16. Answer:

C

17. Answer:

A

18. Answer:

B

19. Answer:

A

20. Answer:

E

Molecular Biology of the Cell 7e-Alberts 21. Answer:

B

22. Answer:

E

23. Answer:

C

24. Answer:

C

25. Answer:

E

26. Answer:

E

27. Answer:

D

28. Answer:

D

29. Answer:

B

30. Answer:

A

31. Answer:

A

32. Answer:

C

33. Answer:

A

34. Answer:

E

35. Answer:

C

36. Answer:

A

37. Answer:

B

38. Answer:

A

39. Answer:

B

40. Answer:

E

41. Answer:

D

42. Answer:

D

Molecular Biology of the Cell 7e-Alberts 43. Answer:

E

44. Answer:

B

45. Answer:

D

46. Answer:

D

47. Answer:

A

48. Answer:

E

49. Answer:

C

50. Answer:

C

51. Answer:

D

52. Answer:

A

53. Answer:

B

54. Answer:

B

55. Answer:

B

56. Answer:

E

57. Answer:

A

58. Answer:

A

59. Answer:

1. C 2. D 3. A 4. B

60. Answer:

1. A 2. B 3. B

Molecular Biology of the Cell 7e-Alberts 4. A 61. Answer:

Translation

62. Answer:

Right-handed

63. Answer:

RNAs transcribed by RNA polymerase II would be more likely to proceed past the termination sequence, resulting in an increase in aberrant mRNAs (snoRNAs, miRNAs, siRNAs, lncRNAs, and snRNAs would also be affected).

While improper mRNA transcripts would likely be targeted for nonsense-mediated decay—and recycled rather than processed to be released to the nucleus—consequently there could be an increase in aberrant protein production.

This has the potential to stress the cell or even become fatal if it were to occur frequently.

64. Answer:

The nucleolus, the site of ribosome synthesis, and Cajal bodies, the site of snRNA and sno RNP maturation, are both subnuclear condensates of the necessary cellular machinery to carry out these functions. Consequently, the components readily and spontaneously re-condense out of the nuclear materials after a physical disruption (similarly to how they do so after cell division.)

65. Answer:

Some antibiotics, such as puromycin, act by preventing protein synthesis. If neurons require de novo synthesis of receptors in order to increase their sensitivity, and are thus prevented from synthesizing additional receptors in response to stimulation, the receptors at the surface will not increase and the response will not change.

66. Answer:

Quality control mechanisms eliminate damaged mRNAs from the pool to be translated, reduce the production of aberrant proteins, and release functional ribosome components to assemble on and translate a more appropriate target.

67. Answer:

7

68. Answer:

In eukaryotes, mRNA with a 5′ cap and poly-A tail is introduced to the ribosome small ribosomal subunit, which is loaded with the initiator tRNA-methionine. The small subunit scans along the mRNA until it encounters the first consensus initiation (Kozak) sequence. Bacteria do not use mRNA with a 5′ cap or poly-A tail, use formylmethionine, and initiate translation at a Shine-Dalgarno sequence

Molecular Biology of the Cell 7e-Alberts anywhere within the mRNA.

Molecular Biology of the Cell 7e-Alberts

Chapter 7: Control of Gene Expression 1. In the following schematic graph of a hypothetical set of RNA-seq data, the number of reads is plotted for a region of chromosome containing two genes, from samples obtained from two different tissues (liver and brain). Which gene (X or Y) do you think is more likely a “housekeeping” gene? Which region (1 or 2) within gene Y most likely corresponds to an exon?

a. Gene X; region 1 b. Gene X; region 2 c. Both genes; region 1 d. Gene Y; region 1 e. Gene Y; region 2

2. Studying the expression of a transcription regulatory protein in two cell types, you have performed experiments showing that the mRNA encoding the protein is present at comparable levels in the cytosol of both cell types. However, based on the expression of its target genes, you suspect that the protein activity might be significantly different in the two cell types. Which step in expression of the gene encoding this protein is more likely to be differentially controlled in these cell types? a. Transcription b. Translation c. mRNA transport d. mRNA degradation

Molecular Biology of the Cell 7e-Alberts

3. In analysis using two-dimensional gel electrophoresis of the proteins expressed in different cell types, the number of spots that are different in different cells usually exceeds the number of common spots, and even the common spots can still have different intensities. The spots representing which of the following proteins would you expect to be among the common spots when compared across several cell types? a. RPL10 (a ribosomal protein) b. HBA1 (a hemoglobin subunit) c. Insulin (a hormone) d. Tyrosine aminotransferase (a metabolic enzyme)

4. Which of the following does NOT have an effect on the time and place that a certain gene is transcribed in the cell? a. The type of cis-regulatory sequences associated with it b. The relative position of cis-regulatory sequences associated with it c. The arrangement of various cis-regulatory sequences associated with it d. The specific combination of transcription regulators present in the nucleus e. The specific combination of transcription regulators present in the cytoplasm

5. The majority of transcription regulators make sequence-specific contacts with DNA in the major groove. In the two diagrams below, where are the contact surfaces that are exposed in the major groove?

Molecular Biology of the Cell 7e-Alberts

a. 1, 3 b. 1, 4 c. 2, 3 d. 2, 4

6. Considering the diagrams below that show hydrogen-bond donors and acceptors as well as hydrophobic groups in four DNA base pairs, which of the following do you think is the most difficult to accomplish by DNA-binding proteins?

a. Distinguishing between A-T and T-A in the major groove b. Distinguishing between A-T and C-G in the major groove

Molecular Biology of the Cell 7e-Alberts c. Distinguishing between C-G and G-C in the major groove d. Distinguishing between C-G and G-C in the minor groove e. Distinguishing between C-G and T-A in the minor groove

7. Regarding protein subunits that interact specifically with DNA sequences, which of the following statements is correct? a. They typically recognize sequences of two to three nucleotide pairs in length. b. They do so via hydrogen bonds, ionic bonds, and hydrophobic interactions. c. They typically form about five weak interactions at the protein–DNA interface. d. They often bind loosely to DNA.

8. Which DNA-binding motif uses a secondary structure that is different from the others listed in order to recognize DNA bases? a. The helix–turn–helix motif b. The leucine zipper c. The zinc finger motif d. The helix–loop–helix motif e. The β-sheet DNA recognition motif

9. A DNA-binding protein recognizes a specific eight-nucleotide sequence in DNA. Assuming that its binding is perfectly specific, how many binding sites are expected to exist for it in human genomic DNA, which is composed of about 6 × 109 nucleotide pairs? What about if the protein forms a homodimer that binds to an inverted repeat? Note that the target sequence can be oriented either way in the double-stranded DNA. a. About 200,000; about 300

Molecular Biology of the Cell 7e-Alberts b. About 2000; about 30 c. About 2000; about 15 d. About 200,000; about 3 e. About 200,000; about 1.5

10. For a DNA-binding protein that binds cooperatively to DNA by fairly weak homodimerization, the binding of this protein to DNA can be described in which way? a. It shows a rectangular hyperbola binding curve. b. It occurs such that only a portion of the DNA sequence may be occupied, as opposed to an all-or-none manner. c. It occurs because the protein molecules are predominantly dimers in solution. d. It is generally inhibited by nucleosomes bound to the DNA.

11. The Trp operon in Escherichia coli encodes the components necessary for tryptophan biosynthesis. In the presence of the amino acid in a bacterium, what happens? a. The tryptophan repressor is bound to the tryptophan operator. b. The tryptophan repressor is bound to bacterial RNA polymerase. c. The expression of the tryptophan repressor is shut off. d. The operon genes are expressed.

12. Under which condition is the Lac operon in Escherichia coli fully turned on? a. Low glucose and low lactose levels b. Low glucose but high lactose levels c. High glucose but low lactose levels

Molecular Biology of the Cell 7e-Alberts d. High glucose and high lactose levels e. Low cAMP and high glucose levels

13. Consider a cis-regulatory sequence in the Escherichia coli chromosome that is located thousands of nucleotide pairs upstream of the gene that it regulates. If the regulatory sequence is mutated to become nonfunctional, the introduction of the wild-type regulatory sequence on a plasmid fails to regulate the gene. What does this imply? a. The regulatory sequence encodes a regulatory protein that binds near the promoter of the target gene and controls RNA polymerase binding. b. The regulation of the target gene involves looping out of the intervening DNA, and the promoter of the cis-regulatory sequence must be on the same chromosome. c. The regulatory sequence can bind directly to the RNA polymerase. d. The regulatory sequence cannot bind to a protein.

14. Transcription regulation has similarities and differences in bacteria and in eukaryotes. Based on this knowledge, which statement is correct? a. Most bacterial genes are regulated individually, whereas most eukaryotic genes are regulated in clusters. b. The rate of transcription for a eukaryotic gene can vary in a much wider range than for a bacterial gene (which is, at most, only about 1000-fold). c. DNA looping for gene regulation is the rule in bacteria but the exception in eukaryotes. d. Transcription regulators in both bacteria and eukaryotes usually bind directly to RNA polymerase. e. The default state of both bacterial and eukaryotic genomes is transcriptionally active.

15. Which of the following is NOT directed by transcription activators in eukaryotic

Molecular Biology of the Cell 7e-Alberts cells in order to provide a more accessible DNA for the transcription machinery? a. Nucleosome remodeling b. Histone removal c. Histone replacement d. Histone modifications e. Histone addition

16. To prevent spurious transcription from a gene, acetylation of histones—which is carried out by histone acetyl transferases ahead of a moving RNA polymerase II—is quickly reversed by histone deacetylases and histone methyl transferases in the wake of the polymerase, leaving a trail of specific methylated histones.

Which of the following curves do you think better represents the distribution of this specific histone methylation mark with respect to a gene?

a. Curve A b. Curve B c. Curve C d. Curve D

Molecular Biology of the Cell 7e-Alberts 17. In order to rapidly respond to an extracellular signal, human RNA polymerases are frequently paused where? a. At the end of the first exon b. At intron/exon boundaries c. Just after the promoter d. In the 3′ untranslated region

18. In single-molecule tracking experiments, scientists see that individual transcription regulators have high mobility in the nucleoplasm and very low mobility when they are bound to DNA. Sometimes, this low mobility persists for minutes, while other times, it only lasts a few seconds. Persistent low mobility means that the regulator has bound to a. its matching trans-regulatory site. b. its matching cis-regulatory site. c. its weakly matching trans-regulatory site. d. its weakly matching cis-regulatory site.

19. Two transcription activators cooperate to recruit a coactivator to a cis-regulatory sequence and activate transcription of a nearby gene. If each of the activators increases the affinity of the coactivator for the reaction site (and therefore the rate of transcription) by 100-fold, how much would you expect the affinity to increase when both activators are bound to DNA compared to when none is bound? a. tenfold b. 100-fold c. 200-fold d. 10,000-fold

20. Transcriptional regulators, coactivators, and co-repressors are held in loose

Molecular Biology of the Cell 7e-Alberts proximity with weak interactions as small, biomolecular condensates. What effect does this have? a. Allowing these proteins to quickly dissociate immediately after transcription initiation b. Blocking further regulation of the gene under their control c. Maintaining a stable transcription bubble d. Making transcription initiation an efficient and dynamic process e. Silencing transcription of neighboring genes

21. Which mechanism of eukaryotic gene repression is the most difficult to derepress? a. Recruitment of histone methyl transferase b. Recruitment of histone deacetylases c. Direct interaction with the general transcription factors d. DNA methylation e. Masking the activation surface

22. Persistent displacement of nucleosomes is associated with which of the following? a. Tissue-specific genes b. Housekeeping genes c. Repressed genes d. Mutated genes e. Methylated genes

23. In the following schematic diagram of a region of a mammalian genome, genes A

Molecular Biology of the Cell 7e-Alberts to D (triangles) are located in between a number of insulator elements (white squares) and barrier sequences (black squares). If the cis-regulatory sequence (oval) is bound by an abundant repressor protein, which gene would you expect to be expressed at a high rate in this cell?

a. Gene A b. Gene B c. Gene C d. Gene D

Expression of the Eve gene in fruit fly embryos is under the control a regulatory module containing binding sites for the activators Bicoid and Hunchback and the inhibitors Giant and Krüppel. This results in seven possible areas in which stripes of Eve can occur (five labeled A through E, one anterior to A, and one posterior to E), where concentrations of the two activators are high and the two inhibitors low.

A reporter gene has been placed under the control of this module. Answer the following question based on these findings.

Molecular Biology of the Cell 7e-Alberts

24. In the following schematic diagrams of an early Drosophila embryo, in which region would you expect to find the reporter protein put under the control of the regulatory module mentioned above? a. Region A b. Region B c. Region C d. Region D e. Region E

25. What would you expect to happen to the pattern of reporter expression in flies that lack the gene encoding Giant? a. It would be expressed in all seven stripes.

Molecular Biology of the Cell 7e-Alberts b. It would be expressed in stripe 5 (region D) only. c. It would expand to cover a broad anterior region of the embryo. d. It would fail to express efficiently in the stripe 2 (region A) region. e. It would be expressed throughout the whole embryo.

26. What would you expect to happen to the pattern of reporter expression in flies that lack the gene encoding Bicoid? a. It would be expressed in all seven stripes. b. It would be expressed in stripe 5 (region D) only. c. It would expand to cover a broad anterior region of the embryo. d. It would fail to express efficiently in the stripe 2 (region A) region. e. It would be expressed throughout the whole embryo.

27. In the Eve regulatory modules, the binding sites for Giant and Krüppel are usually close to, or even overlapping with, those of Bicoid and Hunchback. This implies that Eve expression is regulated by _____________ between the activators and regulators. a. Feedback regulation b. Cooperation c. Competition d. Feed-forward regulation

28. As signal is passed from outside of the cell through to the nucleus, eukaryotic cells employ a variety of mechanisms. What is most likely? a. Masking of an activation domain. b. Insertion of a transcriptional regulator into the plasma membrane.

Molecular Biology of the Cell 7e-Alberts c. Covalent modification involving dephosphorylation. d. A ligand-binding event. e. Disruption of a complex between a DNA-binding subunit and an activation subunit.

29. A single transcriptional regulator can coordinate the expression of many different genes, e.g., the glucocorticoid receptor. After binding to the glucocorticoid hormone, the receptor binds to which of the following? a. A cis-regulatory element controlling genes to be transcribed but not to other proteins b. A cis-regulatory element controlling genes to be transcribed and then recruits other regulating proteins, allowing them to bind to their cisregulatory elements c. Both a cis-regulatory element controlling genes to be transcribed and another transcriptional regulator already bound to the DNA d. Another transcriptional regulator already bound to the DNA but not to a cisregulatory element

30. Which network motif is the most prevalent way that transcriptional regulators ensure that daughter cells have cell memory so they become terminally differentiated? a. Positive feedback loop b. Negative feedback loop c. Indirect positive feedback loop d. Feed-forward loop e. Flip-flop device

31. Considering coherent and incoherent feed-forward loops, eight different designs are possible using three transcriptional regulator components and involving

Molecular Biology of the Cell 7e-Alberts activation (positive) and inhibition (negative) regulation only. In all such loops, A would regulate both B and C, and B would regulate only C. How many of these designs constitute a coherent loop?

a. 4 b. 6 c. 2 d. 3 e. 7

32. Regarding genomic imprinting, what is true? a. It is an epigenetic phenomenon. b. It occurs in most vertebrates. c. It always involves inactivation of genes through direct DNA methylation. d. It can “unmask” recessive alleles but cannot “mask” dominant ones.

33. In mammals, X inactivation occurs when more than one X chromosome is present in a somatic cell. Why does this occur? a. To cause phenotypic differences between the sexes

Molecular Biology of the Cell 7e-Alberts b. To avoid nondisjunction of sex chromosomes c. To allow a higher rate of expression from the active X chromosome d. To achieve dosage compensation

34. You have engineered the X chromosomes in female mice such that one X chromosome expresses green fluorescent protein when active, while the other expresses red fluorescent protein. You have used these mice to study cancer in females. You know that each tumor is a clonal cellular proliferation, meaning all of its proliferating cells are descendants of a single original cancer-causing cell.

Unless X-chromosome inactivation is perturbed in tumors, what is the expected outcome? a. All tumor cells in one mouse should express the same fluorescent protein (either red or green), but tumor cells from different mice can show either red or green fluorescence. b. The cells in any tumor should all express the same fluorescent protein (either red or green), but independently derived tumors in the same mouse can show either green or red fluorescence. c. Different cells within each tumor can express different fluorescent proteins, and the tumors would therefore show yellow fluorescence, but each cell shows either red or green fluorescence. d. Each cell can express both fluorescent proteins and would therefore emit yellow fluorescence, and the tumors would glow in yellow as well. e. Different tumors would show red, yellow, or green fluorescence.

35. Two copies of the gene A exist in a diploid mammal. However, only one copy is expressed in every somatic cell. Different somatic cells in the body of each organism have inactivated one or the other allele in a seemingly random fashion, but when they divide, the daughter cells inherit the choice of the inactive allele faithfully. What is this an example of? a. X-inactivation b. Genomic imprinting

Molecular Biology of the Cell 7e-Alberts c. Loss of heterozygosity d. Monoallelic gene expression e. Alternative gene splicing

36. In exon skipping, an exon is spliced out of the RNA transcript along with its flanking introns. In the following schematic diagram, exons are colored dark blue, and introns are colored yellow. Light blue indicates possible exons. In the diagram of five pairs of alternative RNA splicing patterns, which pair depicts exon skipping only?

a.

b.

c.

d.

e.

37. Imagine a gene encoding a pre-mRNA with 12 exons, 10 of which can be alternatively spliced in vivo. Assuming that alternative splicing for this RNA only occurs through exon skipping, how many different proteins can possibly be made from this pre-mRNA as a result of alternative splicing? a. About 10 b. About 20 c. About 100 d. About 200 e. About 1000

Molecular Biology of the Cell 7e-Alberts

38. What normally results when a 3′ splice site is joined to a downstream 5′ splice site? a. An intron is spliced out. b. A circular RNA molecule is created. c. An exon is removed with two flanking introns. d. An intron is retained in the mature mRNA. e. Alternative mRNA splicing occurs.

39. Stimulated B lymphocytes switch from the synthesis of membrane-bound to secreted antibody molecules by increasing the concentration of a subunit of the trimeric CstF complex that cleaves and polyadenylates mRNAs. How does this up-regulation of CstF bring about the production of soluble antibodies? a. It favors a strong polyadenylation site in the immunoglobulin primary transcript, creating a longer antibody molecule that is secreted. b. It activates a weak polyadenylation site in the immunoglobulin primary transcript and prevents splicing, creating a shorter molecule that is secreted. c. It favors a strong polyadenylation site in the immunoglobulin primary transcript, creating a shorter antibody molecule that is secreted. d. It activates a weak polyadenylation site in the immunoglobulin primary transcript, creating a longer antibody molecule that is secreted. e. It favors a strong polyadenylation site in the immunoglobulin primary transcript, aborting translation and creating a shorter antibody molecule that is secreted.

40. A researcher studying the human immunodeficiency virus (HIV) has infected human T cells with wild-type virus or Rev-deficient virus. She extracts cytoplasmic RNA from the cells and separates each isolated RNA mixture by agarose-gel electrophoresis. She then uses northern blot analysis using HIV-specific probes

Molecular Biology of the Cell 7e-Alberts to identify those bands that contain viral RNA. The results are presented in the following schematic drawing. Which lane (1 or 2) corresponds to cells infected with the Rev-deficientvirus? What is the size of the full-length HIV genome?

a. Lane 1; about 9000 nucleotides b. Lane 1; about 4000 nucleotides c. Lane 1; about 2000 nucleotides d. Lane 2; about 9000 nucleotides e. Lane 2; about 2000 nucleotides

41. Some viruses encode a protease that cleaves the translation initiation factor eIF4G, rendering it unable to bind to eIF4E. What is the consequence of this cleavage? a. It shuts down most of the cellular translation machinery, which causes the release of virus particles. b. It favors viral protein synthesis because IRES-dependent translation initiation is inhibited. c. It shuts down most host-cell protein synthesis and diverts the translation machinery to IRES-dependent initiation, thus favoring viral protein synthesis. d. It favors viral protein synthesis by shutting down translation from uORFs. e. It shuts down IRES-dependent translation, forcing the virus into latency.

Molecular Biology of the Cell 7e-Alberts

42. Gcn4 is a protein kinase that can phosphorylate and inactivate the initiation factor eIF2. The Gcn4 mRNA contains several short upstream open reading frames (uORFs) that negatively regulate its translation initiation. The phosphorylation of eIF2 by Gcn4, therefore, can control Gcn4 expression in which of the following? a. A negative feedback loop b. A positive feedback loop c. A feed-forward loop d. A flip-flop device

43. Under iron-starvation conditions in the cell, what happens? a. Cytosolic aconitase binds to the 3′ UTR of the mRNA encoding the transferrin receptor. b. Cytosolic aconitase binds to the 3′ UTR of the mRNA encoding ferritin. c. Cytosolic aconitase blocks translation of the mRNA encoding the transferrin receptor. d. The mRNA encoding the transferrin receptor is cleaved by an endonuclease.

44. Which of the following classes of noncoding RNAs is NOT directly involved in RNA interference? a. miRNA b. snoRNA c. piRNA d. siRNA

45. What is the function of RNA-dependent RNA polymerases in RNAi?

Molecular Biology of the Cell 7e-Alberts a. They prevent the spread of the RNAi pathway by replicating the target RNAs. b. They help amplify the RNAi response by replicating the target RNAs. c. They produce additional copies of the siRNAs to ensure that the RNAi response is sustained and spread. d. They are viral proteins that prevent the spread of RNAi by preferentially replicating siRNA sponges.

46. In many animals, siRNAs can only function within the cell in which the siRNA is introduced. In contrast, the worm Caenorhabditis elegans can be fed with bacteria that synthesize double-stranded RNAs, and the RNAi spreads throughout the animal. A genetic screen to identify genes involved in systemic RNAi led to the discovery of Sid-1, a transmembrane protein expressed in most tissues in the adult worm. Loss-of-function mutations in the gene encoding Sid-1 restrict the RNAi activity to the cells surrounding the digestive tract after siRNA feeding.

Ectopic expression of Sid-1 in Drosophila melanogaster cells that normally lack systemic RNAi and are unable to take up siRNAs from the medium, enables these cells to passively take up siRNA. What do these findings suggest? a. Sid-1 is necessary and sufficient for systemic RNAi and siRNA uptake, respectively. b. Sid-1 is necessary but not sufficient for systemic RNAi and siRNA uptake, respectively. c. Sid-1 is not necessary but is sufficient for systemic RNAi and siRNA uptake, respectively. d. Sid-1 is not necessary and not sufficient for systemic RNAi and siRNA uptake, respectively.

47. Which of the following is NOT a covalent base modification commonly found in RNA?

Molecular Biology of the Cell 7e-Alberts

a.

b.

c.

d.

e.

48. Many differentiated plant cells can be fully de-differentiated and give rise to an entire plant. a. True b. False

Molecular Biology of the Cell 7e-Alberts

49. Dimerization of transcription regulators increases the chance of binding to nonspecific sequences. a. True b. False

50. DNA methylation makes repression of gene expression “leakier.” a. True b. False

51. Methylated cytosine can be accidentally deaminated to produce thymine, leading to a C-to-T transition. a. True b. False

52. Riboswitches are often located near the 3′ end of mRNAs, and therefore fold after the rest of the mRNA. a. True b. False

53. RNA editing in animal cells mainly takes place through adenine or cytosine deamination. a. True b. False

54. Any factor that affects the translation efficiency of an mRNA tends to have the opposite effect on its degradation. a. True b. False

Molecular Biology of the Cell 7e-Alberts 55. P-bodies are membrane-enclosed organelles scattered throughout the cytoplasm. a. True b. False

56. The following schematic drawing represents the activation of transcription for a eukaryotic gene (gene X). Indicate what component in the drawing corresponds to each of the following.

a. Component A b. Component B c. Component C d. Component D

1. Mediator 2. cis-Regulatory sequence 3. A general transcription factor 4. “Spacer” DNA that may encode lncRNAs

Molecular Biology of the Cell 7e-Alberts 57. In the following schematic drawings showing four different network motifs in gene transcription circuits, indicate whether each of the motifs 1 to 4 corresponds to a feed-forward loop, flip-flop device, negative feedback loop, or positive feedback loop.

a. Feed-forward loop b. Flip-flop device c. Negative feedback loop d. Positive feedback loop

1. Motif 1 2. Motif 2 3. Motif 3 4. Motif 4

58. When unfolded proteins accumulate in the lumen of the endoplasmic reticulum due to inefficient folding conditions, a membrane-bound protein kinase is activated and phosphorylates a subunit of the translation initiation factor eIF2.

Indicate which of the following events would or would not occur as a result. a. Would occur b. Would not occur

1. eIF2 binds tightly to the ribosome.

2. Translation initiation is stimulated.

3. The GDP-bound form of eIF2 accumulates.

Molecular Biology of the Cell 7e-Alberts 4. eIF2B is activated.

59. Which of the following descriptions better matches miRNAs, and which siRNAs, in animal cells? a. miRNAs b. siRNAs

1. They usually cleave their target RNAs using the slicing activity of the Argonautes.

2. They typically show perfect complementarity with their target RNA. 3. They can bind to the RITS as well as the RISC complex.

4. They seem to be the more ancient form of small noncoding RNAs.

60. Which of the following enzymatic activities corresponds to A, B, or C in the following schematic drawing of DNA methylation events in the human genome?

a. Event A b. Event B c. Event C

1. DNA demethylation 2. Maintenance methyl transferase enzyme 3. De novo methyl transferase enzyme

Molecular Biology of the Cell 7e-Alberts 61. Comparing the bacterial CRISPR system and the eukaryotic RNAi mediated by siRNAs, indicate whether each of the following CRISPR components is most analogous to Argonaute, siRNA, or target mRNA in the RNAi pathway. a. Argonaute b. siRNA c. Target mRNA

1. Cas 2. Viral DNA 3. crRNA

62. Consider the descriptions below for three different lncRNAs. Would you expect them to function in cis or in trans? a. In cis b. In trans

1. The tsiX RNA is transcribed from the X-inactivation center in the X chromosome that will not be inactivated. It is antisense with respect to Xist, repressing its transcription and function.

2. The ANRIL lncRNA is an antisense transcript in a gene cluster expressing inhibitors of cell-cycle proteins. It silences the entire cluster by recruiting heterochromatin proteins of the Polycomb group.

3. The PTENP1 lncRNA is similar to an mRNA that encodes the protein PTEN and contains many of the miRNA response elements present in PTEN mRNA; it therefore protects the PTEN mRNA from miRNA-mediated silencing by acting as a decoy.

63. The following sequence logo represents the preferred cis-regulatory sequences of an imaginary transcription regulator that functions as a dimer. Would you expect this sequence to be recognized by a homodimer or a heterodimer? Explain your answer in 2–3 sentences.

Molecular Biology of the Cell 7e-Alberts

64. At one time, it was thought that cell differentiation was irreversible, but now scientists have discovered how to create iPS cells (induced pluripotent stem cells). What is needed to induce these iPS cells to differentiate into some particular cell type? Describe your answer in 2–3 sentences.

65. The two feed-forward loops below contain three transcription regulators (A, B, and C), where A receives the input signal and C generates the output.

Considering the two feed-forward designs, would you expect the coherent or the incoherent loop to have a stable output and respond best to input signals that are above a certain threshold? Explain your answer in 2–3 sentences.

66. In a certain strain of E. coli, a scientist found a mutation in the operator of lac operon such that the repressor could not bind as strongly as in the wild-type. How

Molecular Biology of the Cell 7e-Alberts does this strain differ in lac operon expression compared to wild-type when cells are grown in neither glucose nor lactose? Explain your answer in 2–3 sentences.

67.

For the three transcriptional regulators A, B, and C, would you expect a coherent or an incoherent loop to generate the following response pattern? In this example, A is stimulated by the input, and the transcription of C is measured as the output.

68. Beckwith-Wiedemann syndrome in humans is characterized by “overgrowth” in childhood, sometimes leading to unusually large parts of the body. An imprinted gene cluster on chromosome 11 is associated with this disease. The cluster contains several genes including Igf2 and H19. Igf2 encodes an insulin-like growth factor that is maternally imprinted, that is, the maternal copy is not expressed. However, the DNA methylation pattern of this locus is not different between the two homologous chromosomes. On the other hand, H19 is also imprinted, and its methylation pattern does differ between the two parental

Molecular Biology of the Cell 7e-Alberts chromosomes. H19 is transcribed into a noncoding RNA that appears to silence the transcription of the Igf2 gene in cis. Would you expect the H19 locus to be hypermethylated in the maternally inherited chromosome or paternally inherited chromosome? Explain your answer in 2–3 sentences.

69. The alternative splicing of a certain transcript can result in the production of two mRNA isoforms, one predominant in muscle cells and the other in neurons. The gene contains an exon that is skipped in muscle cells but retained in neurons.

You create a mutant version of the gene in which the 3′ splice site near this exon is deleted. However, when you introduce this into a culture of neural cells, an even longer pre-mRNA is produced, consistent with the activation of a secondary splice site located near the deleted one. Is the secondary splice site within the exon or its neighboring intron?

70. In the following pedigree, females and males are indicated by circles and squares, respectively, and the presence of a rare disease caused by a loss-offunction mutation in an imprinted autosomal gene is indicated in black. Is the gene maternally imprinted, that is, not expressed from the maternally inherited chromosome, or is it paternally imprinted?

71. Immediately after fertilization of the Drosophila egg, the transcriptional regulator Bicoid is translated such that a gradient of the protein forms with the highest concentration at the anterior of the cell. It has been found that the protein gradient stems from the localization of the mRNA for the protein at the anterior tip. What is bringing about the localization of this mRNA? Explain your answer in 2–3 sentences.

Molecular Biology of the Cell 7e-Alberts

72.

You have transfected HeLa cells with a gene encoding the human immunodeficiency virus (HIV) protein Rev and have induced the expression of the protein. You incubate the cell culture in the presence or absence of leptomycin B and later measure the localization of Rev inside the cells by immunofluorescence microscopy. Leptomycin B specifically inhibits the cellular protein Crm1, a nuclear transport receptor that is essential for the normal function of Rev in the HIV life cycle. Your results are tabulated below. Which condition (1 or 2) do you think corresponds to the presence of leptomycin B? Explain your answer in 2–3 sentences.

Condition

Cytoplasmic Rev (%)

Nuclear Rev (%)

1

30

70

2

3

97

73. A scientist has been able to isolate their eukaryotic mRNA of interest to conduct ribosome binding experiments in a cell-free in vitro system. As part of this, they use an oligonucleotide that perfectly hybridizes to the first AUG, including 10 nucleotides both up- and downstream of this sequence. Surprisingly, they find that even under high concentrations of this oligonucleotide, the ribosome translates a peptide, although it is not the peptide expected from the open reading frame following the first AUG. What process is likely occurring? Explain your answer in 2–3 sentences.

74. Analysis of miRNA sequences has revealed an asymmetry in the stability of the double-stranded RNA precursor. The strand showing lower thermodynamic stability near its 5′ end (nucleotides 2 to 6 in the mature guide strand) is normally selected as the guide strand, and the other strand is usually degraded. In the following miRNA precursor, which strand (1 or 2) do you think will be incorporated

Molecular Biology of the Cell 7e-Alberts into the active RNA-induced silencing complex (RISC)? Explain your answer in 2– 3 sentences.

75. A certain region of a mammalian genome is transcribed at low but sustained levels using two flanking promoters. The RNA products are annealed together, processed, and used to recruit the RITS complex to this region, as well as chromatin modifying enzymes, RNA-dependent RNA polymerase, and a Dicer enzyme. Indicate whether this pathway is more closely related to that seen in miRNA, piRNA, or siRNA interference pathways. Explain your answer in 2–3 sentences.

76. Piwi-interacting RNAs (piRNAs) are particularly important in inactivating transposons in the germ line. Why is the genome particularly sensitive to transposon movement in these cells?

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 07 1. Answer:

D

2. Answer:

B

3. Answer:

A

4. Answer:

E

5. Answer:

B

6. Answer:

D

7. Answer:

B

8. Answer:

E

9. Answer:

E

10. Answer:

D

11. Answer:

A

12. Answer:

B

13. Answer:

B

14. Answer:

B

15. Answer:

E

16. Answer:

B

17. Answer:

C

18. Answer:

B

19. Answer:

D

20. Answer:

D

Molecular Biology of the Cell 7e-Alberts 21. Answer:

D

22. Answer:

B

23. Answer:

C

24. Answer:

B

25. Answer:

C

26. Answer:

D

27. Answer:

C

28. Answer:

D

29. Answer:

C

30. Answer:

A

31. Answer:

A

32. Answer:

A

33. Answer:

D

34. Answer:

B

35. Answer:

D

36. Answer:

A

37. Answer:

E

38. Answer:

B

39. Answer:

B

40. Answer:

D

41. Answer:

C

42. Answer:

B

Molecular Biology of the Cell 7e-Alberts 43. Answer:

A

44. Answer:

B

45. Answer:

C

46. Answer:

A

47. Answer:

E

48. Answer:

A

49. Answer:

B

50. Answer:

B

51. Answer:

A

52. Answer:

B

53. Answer:

A

54. Answer:

A

55. Answer:

B

56. Answer:

1. D 2. B 3. A 4. C

57. Answer:

1. D 2. A 3. C 4. B

58. Answer:

1. B 2. B

Molecular Biology of the Cell 7e-Alberts 3. A 4. B 59. Answer:

1. B 2. B 3. B 4. B

60. Answer:

1. C 2. A 3. B

61. Answer:

1. A 2. C 3. B

62. Answer:

1. A 2. A 3. B

63. Answer:

Heterodimer, because homodimeric transcription regulators typically have a symmetrical arrangement which is reflected in their target DNA sequence in the form of an inverted repeat. The sequence logo shown is not in agreement with an inverted repeat, or even a direct repeat; the DNA is therefore expected to be recognized by a heterodimer.

64. Answer:

In short, they would need a particular set of master transcription regulators that set up the transcriptional profile characteristic of a specific cell type. These profiles can be produced by signaling molecules such as growth factors.

65. Answer:

Coherent, because C is activated in two ways by the same input signal; the positive effect caused by one activation is stabilized by the positive effect of the other.

With the incoherent loop, C is receiving opposite regulation effects from the same signal, which, depending on the parameters of the system, will be less likely to

Molecular Biology of the Cell 7e-Alberts produce a stable effect.

66. Answer:

There will be virtually no expression in wild-type but some expression in the mutant. In the absence of glucose, both strains will have enough cAMP bound to CAP for CAP to bind to its cis-regulatory sequence and activate transcription. Without lactose, there will not be enough allolactose to remove the repressor from the operator in the wild-type strain, but in the mutant strain, the repressor will not bind well, allowing some transcription of the operon.

67. Answer:

Incoherent, because as the positive effect of the signal is passed from A to B, then from B to C, the transcription of C increased over time. Because A also has a negative effect on C directly, the transcription of C is later reduced even though the signal persists. With a coherent loop, we would expect the transcription of C to increase and remain increased as long as the signal persisted.

68. Answer:

Paternally inherited chromosome, because on the maternally inherited chromosome, H19 is transcribed and silences Igf2. This does not occur on the paternal chromosome. As a result, Igf2 is maternally imprinted with no maternal-specific methylation.

69. Answer:

The secondary splice site is within the intron, because the cryptic splice site resides within the upstream intron. This results in the production of a longer mRNA when this splice site is chosen.

70. Answer:

Maternally imprinted, because if a disease-causing gene is maternally imprinted, carrier females silently pass it to the next generation. Their sons may have affected children, but their daughters would not.

71. Answer:

The mRNA is being trapped by an anchor protein at the anterior tip of the egg. Because the mRNA is located exclusively there, there is also generalized degradation of those copies that do not make it to the anterior tip. The cytoskeleton assists in transporting the mRNA to this location.

72. Answer:

Condition 2, because when Crm1 is inhibited by leptomycin B, it will no longer be able to carry out its nuclear export function. This will then inhibit the export of Rev, resulting in its nuclear accumulation. We expect to

Molecular Biology of the Cell 7e-Alberts see an accumulation of Rev where it would get stuck if Crm1 were inhibited, and more Rev is localized to one place in condition 2.

73. Answer:

The ribosome is translating from an internal ribosome entry site (IRES). Although the first AUG is inhibited by competition from the oligonucleotide, there is another start site somewhere in the mRNA, allowing translation of a different peptide.

74. Answer:

Strand 1 will be incorporated, because of the presence of mismatches and even a bulge along the duplex. Strand 1 contains mostly A-U base pairs, as well as a mismatch, near its 5′ end (left). Strand 2 contains mostly C-G pairs near its 5′ end (right).

75. Answer:

More closely related to siRNA interference pathways, because RNA interference can direct heterochromatin formation, altering the packaging of nuclear chromatin and expression of genes. Although Dicer is employed in miRNA interference pathways, neither the RITS complex nor RNA-dependent RNA polymerase are. The processing enzymes found in piRNA pathways are entirely different from Dicer.

76. Answer:

During gametogenesis, most histone modifications and DNA methylation events that normally keep transposons under wraps are lost. With the transposons more free to move, the cell employs piRNAs and associated proteins to base-pair with the transposons and block translocation.

Molecular Biology of the Cell 7e-Alberts

Chapter 8: Analyzing Cells, Molecules, and Systems 1. To create cellular factories for monoclonal antibody production, cells derived from an immortalized B lymphocyte cell line (M) are cultured together with antibodysecreting B lymphocytes (B) on a so-called HAT medium. The medium contains an inhibitor of DHFR, an enzyme essential for de novo nucleotide biosynthesis.

Additionally, it contains substrates for the only alternative nucleotide biosynthetic pathway called the salvage pathway. The enzyme HGPRT is essential for the salvage pathway. One of the cells mentioned above lacks the gene encoding one of the enzymes mentioned. Considering the role of the HAT medium in selecting for hybrid cells that are both immortalized and secrete antibodies, which cell (B or M) do you think is missing an enzyme? Which enzyme is missing in this cell? a. M; DHFR b. M; HGPRT c. B; DHFR d. B; HGPRT

2. How are antibody-producing hybridoma cell lines immortalized to provide an unlimited source of monoclonal antibodies? a. By ectopic expression of telomerase b. By transformation with a retrovirus c. By cell fusion d. By overproduction of introduced oncogenes e. By irradiation

3. What is NOT an advantage of monoclonal antibodies over antisera? a. They can be produced in higher quantities. b. They can be produced with higher purity. c. They have a more uniform specificity.

Molecular Biology of the Cell 7e-Alberts d. They can be made against molecules that constitute only a minor component of a complex mixture. e. They can bind to multiple antigenic sites on a macromolecule.

4. Choose the correct order of the following cellular components. The order should illustrate the centrifugal force required to sediment the components by preparative ultracentrifugation, from low to high force required. a. Lysosomes, nuclei, ribosomes (free cytosolic), microsomes b. Lysosomes, nuclei, microsomes, ribosomes (free cytosolic) c. Nuclei, lysosomes, ribosomes (free cytosolic), microsomes d. Nuclei, lysosomes, microsomes, ribosomes (free cytosolic)

5. Due to friction with molecules of the surrounding solution, sedimenting particles in a mixture reach a terminal velocity when subjected to ultracentrifugation. The sedimentation coefficient (S) for a particle is defined as the ratio of terminal velocity (vt) to the centrifugal acceleration applied to the particle (ac).

S = vt / ac If each Svedberg unit (S value) is defined as 10−13 seconds, approximately how long would it take for a free 50S ribosomal subunit to travel the length of a 9 cm ultracentrifuge tube subjected to a centrifugal force of 50,000 times gravity (g)?

Recall that g is approximately 10 m/s2. a. 1 min b. 10 min c. 50 min d. 5 h e. 10 h

6. Which of the following descriptions applies to velocity sedimentation rather than equilibrium sedimentation?

Molecular Biology of the Cell 7e-Alberts a. It separates subcellular components based on their buoyant density. b. It can conveniently separate 15N-containing and 14N-containing DNA. c. It uses a shallow sucrose gradient to stabilize the sedimentation bands against convection. d. In this method, subcellular components may either move up or down to form discrete bands.

7. For the two following graphs, the same purified protein has been subjected to velocity sedimentation and gel filtration, as indicated. Two absorption peaks are visible in each graph, corresponding to the monomeric and dimeric forms of the protein. Which peaks correspond to the protein dimer?

a. 1 and 3 b. 1 and 4 c. 1 and 2 d. 2 and 3 e. 2 and 4

Molecular Biology of the Cell 7e-Alberts 8. In purifying proteins by column chromatography, elution can be performed in different ways, depending on the type of matrix. For example, bound proteins can be eluted by gradually increasing or decreasing salt concentration in the solution applied to the column. For which of the following matrices do you think increasing and decreasing salt concentration, respectively, are used for elution? a. Ion-exchange; hydrophobic b. Affinity; ion-exchange c. Hydrophobic; affinity d. Hydrophobic; ion-exchange e. Ion-exchange; gel-filtration

9. Which of the following chromatography methods resembles immunoprecipitation in the purification of proteins from mixtures? a. Ion-exchange chromatography b. Gel-filtration chromatography c. Hydrophobic interaction chromatography d. Affinity chromatography

10. How is tandem affinity purification distinct from other purification schemes such as simple affinity or ion-exchange chromatography? a. In its use of antibodies b. In its use of proteases c. In its use of glutathione S-transferase d. In its use of viral peptides e. In its use of limited washing steps

11. Which of the following is NOT true of SDS-PAGE of proteins?

Molecular Biology of the Cell 7e-Alberts a. The proteins are unfolded while being separated. b. The proteins are separated by size, mostly independent of their native charge. c. Large proteins move more slowly. d. An ionic detergent is used. e. It requires the use of antibodies for visualization.

12. Migration of proteins during separation by SDS polyacrylamide-gel electrophoresis (SDS-PAGE) is generally related to the logarithm of their molecular weight, as depicted in the qualitative graph below. If a protein mixture containing proteins of evenly distributed molecular weights, for example, 10, 20, 30, or 40, is separated by SDS-PAGE, which of the following staining patterns would you expect to observe?

a. Pattern A b. Pattern B c. Pattern C d. Pattern D

13. Where do individual peptides get fragmented in a tandem mass spectrometry

Molecular Biology of the Cell 7e-Alberts (MS/MS) set-up? a. Electrospray ion source b. Mass filter c. Inert gas chamber d. Mass analyzer e. Detector

14. Which mass spectrometry instrument is most suitable for determining all the proteins present in an organelle? a. MALDI-TOF b. MS/MS c. LC-MS/MS d. GC-MS

15. The binding of protein A to two other proteins, B and C, has been studied with fluorescence anisotropy, and the results of the measurements have been plotted in the following schematic diagram. Given equal kon rate constants, which complex do you think has a longer half-life? What is the value of Ka (the association constant) for this complex?

Molecular Biology of the Cell 7e-Alberts a. A-B; 105 L/mol b. A-B; 10−5 L/mol c. A-C; 1.5 × 105 L/mol d. A-C; 6.7 × 104 L/mol e. A-C; 6.7 × 10−4 L/mol

16. Hydrodynamic measurements can be used to calculate the approximate size and shape of a large protein complex using what method? a. Visualizations from an electron microscope b. Binding of antibodies with fluorescent probes c. An electric current which pulls the protein through a gel matrix d. Ionization and electrospray e. The velocity of the complex as it moves under centrifugal force and the migration of the complex as it moves through a gel-filtration column

17. You have discovered a mutant protein that forms promiscuous disulfide bonds with many other cellular proteins in the endoplasmic reticulum. Using an antibody against the protein, you immunoprecipitate proteins that interact with it. You then separate these proteins using a special type of two-dimensional polyacrylamidegel electrophoresis (PAGE): in this method, separation in the first dimension is carried out by SDS-PAGE in the absence of β-mercaptoethanol, while separation in the second dimension is carried out by SDS-PAGE in the presence of βmercaptoethanol. The result is shown in the following schematic drawing.

Apparent molecular weights are indicated on the left. What do you think is the molecular weight of the mutant protein? What is the molecular weight represented by the question mark?

Molecular Biology of the Cell 7e-Alberts

a. 15K; 45K b. 30K; 45K c. 15K; 30K d. 30K; 30K e. 30K; 60K

18. What is a benefit of using NMR to determine protein structure rather than X-ray crystallography? a. NMR can solve protein structures in aqueous solutions, while X-ray crystallography requires the protein to form a solid crystal. b. NMR can solve the structure of large protein complexes while X-ray crystallography can only be done on small protein fragments. c. NMR requires small, dilute concentrations of protein while X-ray crystallography needs large amounts of highly concentrated protein. d. NMR can be performed without the use of magnets, while X-ray crystallography requires a large magnetic field. e. NMR allows for calculation of a three-dimensional protein structure while Xray crystallography cannot.

Molecular Biology of the Cell 7e-Alberts

19. You have carried out a genetic screen in a species of fish and have identified a new gene in this organism. You have cloned and sequenced the gene, but do not know anything about the biochemical function of the gene product. Which of the following is the most sensible next step in order to get clues about the function? a. Purifying the protein product and determining its structure by x-ray diffraction b. Screening for small molecules that perturb gene function c. Using the BLAST algorithm to scan sequence databases for similar sequences d. Using MS/MS to sequence the gene product without the need for purification e. Mutating the gene and performing reverse genetics

20. The EcoRI restriction enzyme recognizes and cuts at the sequence GAATTC. Such an enzyme is expected to cut a typical double-stranded DNA at 1 in every how many base pairs? a. About 400 b. About 4000 c. About 40,000 d. About 400,000 e. About 4 million

21. The AgeI restriction enzyme recognizes the sequence ACCGGT and cuts each strand between the first (A) and the second (C) nucleotide. This can be represented as A|CCGGT, with the vertical bar indicating the cleavage position. If a DNA fragment is cut with AgeI at both ends, cutting a plasmid with which of the following enzymes would then allow insertion of the fragment into the plasmid by simple ligation of sticky ends? a. FseI, with the recognition sequence GGCCGG|CC

Molecular Biology of the Cell 7e-Alberts b. AatI, with the recognition sequence AGG|CCT c. BstZI, with the recognition sequence C|GGCCG d. MroI, with the recognition sequence T|CCGGA

22. Your friend has cut a 3000-nucleotide-pair circular plasmid with either or both restriction enzymes A and B, and has separated the digestion products by electrophoresis. The results are shown in the schematic drawing below, with the approximate size of the fragments indicated above each band. Based on these results, what is the closest distance (in nucleotide pairs) between an A restriction site and a B restriction site on this plasmid?

a. 400 b. 600 c. 1000 d. 1200 e. 1600

23. Which of these describes bacterial artificial chromosomes (BACs)? a. They are derived from the E. coli chromosome. b. They are present in high copy numbers per cell. c. They can maintain DNA sequences of hundreds of thousands of nucleotide

Molecular Biology of the Cell 7e-Alberts pairs. d. They are generally not stable.

24. In a forensic lab, DNA samples from a mother (M) and her child (C) have been analyzed at three short tandem repeat (STR) loci using PCR and gel electrophoresis. The same analysis has been performed for five men (1 to 5).

According to the results below, which men can be eliminated as possible candidates for the biological father of the child?

a. 123 b. 1235 c. 235 d. 2345 e. 1345

25. You would like to clone a gene by PCR. The sequence of the two ends of the coding strand for the gene is provided below. Which following pair of primers would direct the amplification of this gene? All sequences are written from 5′ to 3′.

ATGAAATCTACGTTTCAC……CCCCCAGTACCCCCCTTA a. ATGAAATCTACGTTTCAC ; CCCCCAGTACCCCCCTTA b. ATGAAATCTACGTTTCAC ; TAAGGGGGGTACTGGGGG c. TACTTTAGATGCAAAGTG; ATTCCCCCCATGACCCCC

Molecular Biology of the Cell 7e-Alberts d. TACTTTAGATGCAAAGTG ; CCCCCAGTACCCCCCTTA e. GTGAAACGTAGATTTCAT; ATTCCCCCCATGACCCCC

26. A protein made from an expression vector can be so plentiful as to account for about 10% of total cell protein. You would like to produce a protein of interest that can be expressed at such a level in Escherichia coli. You plan to induce the expression and collect the cells when cell density in the culture medium reaches about 109 cells/mL. Proteins make up about 15% of total cell weight, and each E. coli cell typically has a mass of about 1 picogram, that is, 10−12 g. Assuming that almost 50% of the protein is lost during purification, how many liters of culture should you prepare in order to obtain 15 milligrams of pure protein? a. 2.00 b. 1.00 c. 0.25 d. 4.00

27. Which of the following descriptions applies to genomic DNA libraries rather than cDNA libraries? a. Their production requires use of reverse transcriptase. b. They are enriched in protein-coding genes. c. Different sequences are represented in them based on their transcription level. d. They contain intronic sequences.

28. Which of the following allows many smaller strands of synthetic DNA with complementary ends to be joined together into one long strand of DNA? a. Gibson assembly b. DNA ligation

Molecular Biology of the Cell 7e-Alberts c. Polymerase chain reaction d. Dideoxy sequencing e. Illumina sequencing

29. The specificity of nucleic acid hybridization is remarkable. Even a single mismatch between a probe and a target sequence can be detected depending on the hybridization conditions. This sensitivity can be employed in applications such as SNP haplotyping. Consider a probe that is complementary to a 20-nucleotidelong genomic region including a single polymorphic nucleotide. The probe is perfectly complementary to the more common allele, but has one mismatch in the case of the minor allele. The temperature-dependent hybridization of the probe to the single-strand genomic fragment is schematically presented in the following graph. To detect hybridization, a fluorescent dye is added to the reaction. The dye fluoresces strongly only when intercalated between the bases of double-stranded DNA, and is therefore used to quantify double-strand formation or dissociation.

Which “melting curve” (1 or 2) in the graph do you think corresponds to the more common SNP allele? Which curve shows a higher melting temperature?

a. 1; 1 b. 1; 2 c. 2; 1 d. 2; 2

30. In Sanger sequencing, does each labeled DNA molecule have a fluorescently labeled dideoxyribonucleotide at its 5′ end or its 3′ end? Do the labeled

Molecular Biology of the Cell 7e-Alberts nucleotides lack the 2′ or 3′ hydroxyl group? a. 3′ end; 2′ hydroxyl group only b. 3′ end; 3′ hydroxyl group only c. 5′ end; 2′ hydroxyl group only d. 5′ end; 3′ hydroxyl group only e. 3′ end; both hydroxyl groups

31. Which of the following is NOT characteristic to SMRT sequencing? a. It uses a single DNA polymerase at the bottom of a well. b. It uses deoxynucleotide triphosphates with a fluorescent tag. c. It can read DNA strands up to tens of kilobases long. d. It only works on circular DNA. e. It causes a brief pulse of colored fluorescence based on the nucleotide that binds.

32. Which of the following is NOT true about nanopore sequencing? a. It requires deoxyribonucleotides that have a fluorescent tag. b. It can be performed in a handheld device. c. It utilizes an electric current to pull DNA though a protein pore. d. It can detect modified nucleotides. e. It can be used for extremely long strands of DNA.

33. There are six possible reading frames for any double-stranded DNA. In the following schematic drawing of a genomic region in Escherichia coli, stop codons in each of these frames are indicated by black bars. Choose the reading frame in which the open reading frame (ORF) is translated in this region.

Molecular Biology of the Cell 7e-Alberts

a. 3 b. 2 c. 1 d. −1 e. −2 f. −3

34. A portion of RNA-seq data obtained from two tissue samples is plotted in the following schematic diagram. Which region corresponds to an alternative exon?

a. Region A b. Region B c. Region C d. Region D e. Region E

Molecular Biology of the Cell 7e-Alberts

35. The p53 gene is an important tumor suppressor gene in our genome. It codes for a transcription regulatory protein that can assemble into tetramers. A number of mutations in p53 render the protein inactive. These mutant protein molecules form mixed tetramers with wild-type p53 proteins and, in this case, the tetramer is inactive. The mutations can therefore be categorized as a. Null b. Gain-of-function c. Recessive lethal d. Dominant-negative e. Suppressor

36. The wild-type nematode Caenorhabditis elegans has a bent body with the shape of a wave. In the presence of increasing concentrations of ethanol, however, the body bend is progressively diminished. Loss-of-function mutations in SLO-1, a gene that codes for an ion channel, confer ethanol resistance and allow the worm to maintain its body bend up to higher ethanol concentrations. The channel function is defined based on the ability of the channel to conduct ions across the membrane. According to these observations, would you expect the channel to open or close in the presence of ethanol in wild-type worms? Would a gain-offunction mutation in SLO-1 tend to increase or decrease body bend? a. Open; increase b. Open; decrease c. Close; increase d. Close; decrease

37. You have studied mutations in a transcription regulatory protein from Bacillus subtilis by an in vivo reporter gene assay carried out in Escherichia coli. The cisregulatory sequence that the protein normally recognizes was inserted upstream of a gene encoding an enzyme that can produce a visible blue product. E. coli cells were then transformed with the reporter construct along with the gene

Molecular Biology of the Cell 7e-Alberts encoding one of the various mutant forms of the transcription regulatory protein.

Since the transcription regulatory protein is an inhibitor, the colonies turn blue only when the mutant protein is nonfunctional and unable to inhibit gene expression. Interestingly, you have identified a number of “cold-sensitive” loss-offunction mutations that show the mutant phenotype at 22°C but not at 37°C. Which of the following correctly describes the phenotype of these cold-sensitive mutants? a. Their colonies turn blue at permissive (low) temperatures. b. Their colonies turn blue at permissive (high) temperatures. c. Their colonies turn blue at nonpermissive (low) temperatures. d. Their colonies turn blue at nonpermissive (high) temperatures.

38. For a complementation test to work, the mutations under study must be of what type? a. Recessive b. Dominant c. Gain-of-function d. Null e. Conditional

39. You have isolated five mutations (1 to 5) in the yeast Saccharomyces cerevisiae that make the haploid cells unable to grow in the absence of histidine. Each haploid mutant can be mated with any of the other ones, forming diploid cells that either can (+) or cannot (−) grow in the absence of histidine, as indicated in the following complementation table. How many complementation groups do these mutations represent? Each complementation group typically corresponds to a separate gene.

Molecular Biology of the Cell 7e-Alberts

a. 1 b. 2 c. 3 d. 4 e. 5

40. Loss-of-function mutations in either imp or yfgL genes in Escherichia coli result in an increased permeability of the outer membrane to small molecules and therefore confer antibiotic sensitivity, as the antibiotics can enter the cell more readily. Interestingly, double mutants, in which both genes are defective, are resistant to antibiotics. Similar to the wild-type strain, but unlike the single mutants, they can grow on media containing antibiotics. This is the opposite of what would happen in what other condition? a. Synthetic lethality b. Complementation c. Synergy d. A recessive mutation e. A gain-of-function mutation

41. In a bacterium that is normally cylindrical in shape, loss of either of two genes A and B results in round cells. Additionally, gain-of-function mutations in gene A can rescue the phenotype of gene B loss-of-function mutants. What phenomenon

Molecular Biology of the Cell 7e-Alberts best describes this observation? The product of which gene is more likely to function upstream of the other one in determining cell shape in this bacterium? a. Suppression; gene A b. Suppression; gene B c. Complementation; gene A d. Complementation; gene B

42. Consider a biochemical pathway in petal cells of a plant, in which a white precursor is turned into a final red pigment through a yellow and an orange intermediate by the activity of three enzymes encoded by genes A, B, and C. The petal color in plants harboring functional (capital letter) or nonfunctional (lowercase letter) alleles of these genes is presented in the following table. What do you think is the order in which the enzymes act in the pathway?

Genotype Petal color

a. ABC b. ACB c. BAC d. BCA

ABC

Red

ABc

Orange

AbC

White

Abc

White

aBC

Yellow

aBc

Yellow

abC

White

Abc

White

Molecular Biology of the Cell 7e-Alberts

43. Susceptibility to drug and alcohol addiction is partially genetic. For example, the human dopamine receptor DRD2 is associated with cocaine and alcohol abuse; individuals carrying an A nucleotide, instead of the more common C nucleotide, at a certain position within the sixth intron of the DRD2 gene are at a significantly increased risk of abuse. This variation in sequence is an example of what? a. SNP b. CNV c. An indel d. Epistasis

44. Which statement applies to single nucleotide polymorphisms? a. They are blocks of DNA that are passed from parent to child. b. They are rare mutations that always have a deleterious effect on a protein gene product. c. They are only found in individuals with disease. d. They are common sequence variants that are passed down from one generation to the next. e. They are generated in individuals by environmental exposure to mutagenic substances.

45. What is one difference between polymorphisms and rare DNA variants? a. Polymorphisms are changes in protein-coding genes, while rare DNA variants are changes in non-coding regions of DNA. b. Polymorphisms arose early in human history, while rare DNA variants are more recent. c. Polymorphisms affect one gene, while rare DNA variants affect many genes.

Molecular Biology of the Cell 7e-Alberts d. Polymorphisms are always benign, while rare DNA variants are always harmful. e. It takes only one polymorphism to cause a disease, but it takes several rare DNA variants to cause a disease.

46. Which of the following descriptions better applies to forward rather than reverse genetics? a. It involves new gene discovery by genetic screens. b. It may also be called genome editing. c. In this approach, mutations are introduced deterministically in a known gene. d. It is involved in creating knockout mice lacking a particular gene.

47. You have created transgenic mice that carry the gene encoding green fluorescent protein (GFP) under the control of a ubiquitous promoter. The GFP-encoding gene, however, is inactive unless recombination by Cre removes a blocking sequence between the promoter and the gene. These mice have been mated with those harboring the Cre recombinase under the control of the insulin promoter. The Cre recombinase is fused to the estrogen receptor, which keeps the fusion in the cytosol unless it binds to tamoxifen, in which case it allows nuclear entry. After mating, you select adult mice harboring both of these transgenes and inject them with either tamoxifen or water. Which of the following would you expect to observe later as a result when you examine brain and pancreas tissue samples under a fluorescence microscope? a. GFP signal is observed in the brain in the presence of tamoxifen only. b. GFP signal is observed in the brain in the absence of tamoxifen only. c. GFP signal is observed in the pancreas in the presence of tamoxifen only. d. GFP signal is observed in the pancreas in the absence of tamoxifen only. e. GFP signal is observed in the brain in either the presence or the absence of tamoxifen.

Molecular Biology of the Cell 7e-Alberts

48. You mated male and female heterozygous Oct4 knockout mice to study the function of this important gene. Among the progeny, one-third were homozygous wild type and the rest were heterozygous. You then repeated the cross, but this time you genotyped several embryos at different stages of early development and grouped them as either homozygous wild type (+/+), heterozygous (+/−), or homozygous null (−/−). You have summarized your results in the following table, in which the number of embryos in different groups and at different developmental stages are presented. According to these results, complete loss of Oct4 is what type of lethal?

a. NOT lethal b. Synthetic lethal c. Zygotic lethal d. Embryonic lethal

49. Which of the following is NOT correct regarding both Cas9 and EcoRI? a. They are both endonucleases. b. They both create double-strand breaks in DNA. c. They both recognize their target sequences with the help of guide RNAs. d. They are both part of bacterial defense mechanisms against foreign DNA. e. They are both greatly useful in manipulating DNA and studying gene expression and function.

50. A pool of five barcoded yeast mutants, each deleted for a different gene (genes A

Molecular Biology of the Cell 7e-Alberts to E), has been grown under three different conditions (1 to 3). After the growth phase was completed, genomic DNA was isolated from the entire pool and the relative amount of each mutant was determined by quantifying the amount of DNA containing each barcode. The results are presented in the following graph.

Which gene is essential under condition 1 but not under conditions 2 or 3?

a. Gene A b. Gene B c. Gene C d. Gene D e. Gene E

51. In which one of the following organisms is systemic RNAi possible through feeding the animal with the RNA? a. Caenorhabditis elegans b. Drosophila melanogaster c. Mus musculus d. Homo sapiens

52. Which of the following is NOT a limit on the use of RNA interference in studying

Molecular Biology of the Cell 7e-Alberts gene expression and function? a. RNAi does not efficiently inactivate all genes. b. Certain tissues are resistant to the action of RNAi. c. Off-target effects are common in RNAi treatments. d. Foreign RNA is being reverse transcribed into DNA.

53. What information can be gained from in situ hybridization but not a reporter gene experiment? a. The location of non-coding RNAs in a cell b. The sequence of the DNA being transcribed c. Whether or not a specific sequence is being transcribed d. If a protein is properly folded or not e. The life-span of a specific gene product

54. The following schematic graph shows the result of two quantitative RT-PCR experiments. Total mRNA from two tissue samples (1 and 2) was isolated and subjected to RT-PCR using primers designed to amplify a tissue-specific gene.

Added to the reaction was a fluorescent dye that fluoresces only when bound to double-stranded DNA. According to the graph, which tissue has a higher level of this mRNA? By how much?

a. About 2-fold higher in tissue 1

Molecular Biology of the Cell 7e-Alberts b. About 1000-fold higher in tissue 1 c. About 2-fold higher in tissue 2 d. About 1000-fold higher in tissue 2

55. Which of the following methods provides the most sensitive and accurate way of measuring and comparing mRNA levels of a set of genes in two different tissues?

The genes are expressed at relatively low levels in these tissues, and do not necessarily code for protein. a. RNA-seq b. RT-PCR c. DNA microarray analysis d. ChIP-seq e. Ribosome profiling

56. Which technique generates a complete list of all major mRNAs expressed in every cell type in an organism? a. Illumina sequencing b. Nanopore sequencing c. Single-cell RNA sequencing d. Ribosome profiling e. RT-PCR

57. The results of two sets of chromatin immunoprecipitation experiments on two cell types (1 and 2) are shown in the following simplified graphs for the same genomic region containing the first two exons of a gene. Antibodies against a transcription regulatory protein (X) or against a subunit of RNA polymerase II were used in the ChIP-seq experiments, as indicated. Do you think the protein is a transcriptional activator or repressor? The results from a parallel RNA-seq experiment are

Molecular Biology of the Cell 7e-Alberts provided in the third graph. Which profile (a or b) in the graph would you expect to correspond to cell type 1?

a. Activator; profile a b. Repressor; profile a c. Activator; profile b d. Repressor; profile b

58. Consider a protein composed of only 21 amino acids encoded by a short ORF that is located between two known genes in a mammalian genome. Due to its length, the ORF had not been annotated as a gene. Which of the following methods can demonstrate that it is indeed a gene that is actively translated to protein? a. RNA-seq b. RT-PCR c. DNA microarray analysis d. ChIP-seq

Molecular Biology of the Cell 7e-Alberts e. Ribosome profiling

59. RNA-seq and ribosome profiling experiments have been carried out on the same cells. The following simplified graphs show the results for the same genomic region containing a gene with three exons. Which graph do you think corresponds to the RNA-seq results? What feature is represented by the arrow?

a. Graph 1; stop codon b. Graph 1; polyadenylation site c. Graph 2; stop codon d. Graph 2; polyadenylation site

60. Below are the steps in the common procedure to create transgenic plants. Which order reflects the correct sequence of events? a. Removal of leaf tissue from a plant, callus growth, Agrobacterium infection, shoot/root induction b. Removal of leaf tissue from a plant, Agrobacterium infection, callus growth, shoot/root induction c. Removal of leaf tissue from a plant, Agrobacterium infection, shoot/root induction, callus growth d. Removal of leaf tissue from a plant, callus growth, shoot/root induction, Agrobacterium infection

61. Consider the binding of a transcription regulatory protein (A) to a promoter DNA

Molecular Biology of the Cell 7e-Alberts (p) according to the following scheme.

A + p → A:p At steady state, which of the following formulae correctly yields the bound promoter fraction, that is, [A:p]/([A:p] + [p])? The association constant is designated as K. a. [A]/(K + [A]) b. K[A]/(1 + K[A]) c. K[A]/(K + [A]) d. [A]/(1 + K[A])

62. In the following schematic graph, the fraction of promoters (p x) bound to the regulatory protein A is plotted at various concentrations of A. Which of the following quantities represents X in the graph?

a. [Total px] b. [px] c. kon / koff d. koff / kon e. kon[px] / koff

63. Consider a transcription regulatory protein (A) that can bind to the promoter (p) of

Molecular Biology of the Cell 7e-Alberts a target gene. After reaching an initial steady state, the concentration of A is suddenly increased tenfold. The following graph shows the formation of A:p over time after this change. What is the slope of the curve at time zero?

a. 2 koff b. 2 kon c. 5 koff d. 5 kon e. 9 koff

64. Everything else being equal, if the mean lifetime of a protein is doubled, its concentration at steady state would do what? a. Increase approximately twofold b. Increase less than twofold c. Increase over twofold d. Decrease approximately twofold e. Decrease less than twofold

65. Consider two proteins A and B with equal transcription and translation rates. The mean lifetime of A, however, is twice as long as that of B. Starting with no protein, what is the time it takes for a protein to reach 99% of its steady-state concentration?

Molecular Biology of the Cell 7e-Alberts a. The same for proteins A and B b. Two times longer for A compared to B c. Two times longer for B compared to A d. Less than twofold longer for A compared to B e. Less than twofold longer for B compared to A

66. You have set up a system of differential equations to describe the concentration of a protein as a function of time based on transcription, translation, and degradation rates. You simulate two conditions (1 and 2) that are identical except that the transcription rate for the protein is set tenfold higher in one of them (1) compared to the other (2). Starting from a concentration of zero, what is the time required to reach 50% of the final steady-state concentration? a. The same for conditions 1 and 2 b. Tenfold longer for 1 compared to 2 c. Tenfold longer for 2 compared to 1 d. Less than tenfold longer for 1 compared to 2 e. Less than tenfold longer for 2 compared to 1

67. Which equation better describes the steady-state concentration of a protein (X) that is regulated by the simple binding of a transcription repressor (R) to its cisregulatory DNA sequences? β is the transcription rate, m is the translation rate, K is the association constant in the binding of R to the regulatory sequences, and τX is the mean lifetime of the protein X. a. βmτX / (1 + K[R]) b. βmτX / (1 + [R]) c. βmτX K / (1 + K[R]) d. βmτX [R] / (1 + K[R])

Molecular Biology of the Cell 7e-Alberts e. βmτX [R] / (K + [R])

68. Consider a genetic network consisting of gene A whose product activates the transcription of gene R. Imagine that the concentration of protein A is suddenly changed by a significant amount. The following graphs show how the concentration of protein R would then change over time to reach steady state again. The two curves (1 and 2) in each graph correspond to systems that have different transcription rates for A. For one of the two graphs (A or B) a negative feedback loop is present, in which protein R represses the transcription of gene A. Which conclusion is NOT correct according to these results?

a. The negative feedback loop makes the system slower in reaching the steady state. b. With negative feedback, the steady-state concentration is less sensitive to fluctuations in kinetic rate constants. c. Higher transcription rates for gene A result in lower steady-state levels of protein R both with and without negative feedback. d. With negative feedback, the response to activating input is decreased.

69. Consider a promoter sequence with n binding sites for an activator protein. Which Hill coefficient (h) describes noncooperative binding, that is, independent, noninteracting binding sites? a. h = n b. h = 1 c. h = 0

Molecular Biology of the Cell 7e-Alberts d. h = n −1 e. h < 0

70. Bistability in a biological system can be accompanied by hysteresis. Consider switchlike cellular responses to a signal molecule, as shown in the following graphs. In each graph, the solid curve corresponds to the change in cellular response when the signal concentration is being increased over time, whereas the dashed curve corresponds to the change when the signal concentration is being decreased. Which graph do you think corresponds to a system with a greater degree of hysteresis? What mechanism is responsible for this behavior?

a. Graph A; negative feedback b. Graph A; positive feedback c. Graph B; negative feedback d. Graph B; positive feedback

71. Some feed-forward motifs are capable of generating a brief pulse of gene activation in response to a sustained input signal: in such motifs, the signal activates an activator (A) and a repressor (R), both of which control the expression of the same gene. The quantitative output of the gene depends in part on the relative affinities of A and R for binding to the cis-regulatory sequences of the target gene. Under which of the following conditions is a brief pulse of gene activation observed in response to a sustained input signal? KA and KR are the association constants for A and R, respectively. a. KA = KR

Molecular Biology of the Cell 7e-Alberts b. KA << KR c. KA >> KR d. KAKR > 1

72. Which of the following is NOT something statistical analysis will tell us? a. The magnitude of the noise in a dataset b. The mean value of a measurement that was taken many times c. The likely error in the mean value of our measurements d. The reason an error occurred in a set of measurements e. The likelihood an individual measurement will read a certain value

73. Indicate whether each of the following descriptions applies to transformed cell lines, nontransformed cell lines, or both. a. Transformed cell lines b. Nontransformed cell lines c. Both

1. They can be thawed and recultured after long-term storage in liquid nitrogen.

2. They often grow only when attached to a surface, such as that of a culture dish.

3. They proliferate to very high densities.

4. They usually cause tumors when injected into immunocompromised animals.

74. Three subunits of a protein complex have been resolved in a two-dimensional polyacrylamide gel, as shown in the schematic drawing below. Indicate whether each of the following better describes the a, b, or c subunit.

Molecular Biology of the Cell 7e-Alberts

a. a subunit b. b subunit c. c subunit

1. It has the smallest apparent molecular weight.

2. It has the highest negative charge.

3. It has the lowest isoelectric point.

75. Indicate whether each of the following descriptions better applies to the determination of atomic structure of proteins by x-ray diffraction or by nuclear magnetic resonance (NMR) spectroscopy. a. X-ray diffraction b. Nuclear magnetic resonance (NMR) spectroscopy

1. It is more suitable for complexes larger than about 100 kD. 2. It requires samples to crystallize.

3. It can better reveal dynamics of molecular motions.

4. It involves the calculation of electron-density maps.

5. It is used to determine the solution structure of proteins.

76. Indicate whether each of the following manipulations or procedures used in current recombinant DNA technology commonly rely on an enzyme or are done nonenzymatically. a. Enzymatically

Molecular Biology of the Cell 7e-Alberts b. Nonenzymatically

1. Cleavage of DNA at specific sites 2. DNA ligation 3. DNA cloning 4. Nucleic acid hybridization 5. Synthesis of DNA of any desired specific sequence 6. DNA sequencing

77. Indicate whether ordinary agarose-gel electrophoresis, polyacrylamide-gel electrophoresis, or pulsed-field gel electrophoresis is used to better separate each of the following DNA molecules. a. Agarose-gel electrophoresis b. Polyacrylamide-gel electrophoresis c. Pulsed-field gel electrophoresis

1. Small DNA oligonucleotides such as hybridization probes and PCR primers 2. Plasmid DNA 3. Bacterial chromosomes

78. The following schematic graph shows temperature change over time in a cycle of a PCR assay. Which steps (A to C) in the cycle correspond to hybridization (annealing), strand separation (denaturation), and DNA synthesis (extension), respectively?

a. Step A

Molecular Biology of the Cell 7e-Alberts b. Step B c. Step C

1. Hybridization (annealing) 2. Strand separation (denaturation) 3. DNA synthesis (extension)

79. You have devised genetic screens to identify gain- and loss-of-function mutations in the Escherichia coli DNA ligase gene. One of the screens (1) seeks mutant cells that are infected more readily with a ligase-deficient strain of T4 phage. The other screen (2) seeks mutant cells that have increased radiation sensitivity at high temperatures. Indicate whether each of the following descriptions better applies to screen 1 or screen 2. a. Screen 1 b. Screen 2

1. It would identify gain-of-function DNA ligase mutants.

2. It would identify loss-of-function DNA ligase mutants.

3. It would identify conditional mutants.

80. You have used a reporter gene system to investigate the contribution to gene expression of three cis-regulatory DNA sequences (A to C) from a key developmental gene. Reporter gene expression from constructs containing all three or only a subset of the regulatory sequences is summarized in the diagram below. According to these findings, indicate whether each of the following conclusions is or is not acceptable.

Molecular Biology of the Cell 7e-Alberts

a. Acceptable b. Not acceptable

1. Region B is likely to normally bind an activator and turn the gene on.

2. Region C is sufficient for activation of gene expression.

3. Region A is necessary for activation of gene expression.

4. Region C is necessary for activation of gene expression.

81. The schematic graph below represents the nullclines (solid curves) and a few selected trajectories (dotted curves) for a system composed of two proteins (X and Y) with a complex regulatory behavior. According to the graph, indicate whether each of the following statements is or is not acceptable.

Molecular Biology of the Cell 7e-Alberts

a. Acceptable b. Not acceptable

1. The system is bistable.

2. There are three steady states in the system.

3. The system contains positive feedback.

4. The system contains delayed negative feedback.

82. Indicate which of the following network motifs represent negative feedback, positive feedback, incoherent feed-forward, and coherent feed-forward, respectively.

a.

Molecular Biology of the Cell 7e-Alberts

b.

c.

d.

1. Negative feedback 2. Positive feedback 3. Incoherent feed-forward 4. Coherent feed-forward

83. Indicate whether each of the following better applies to a coherent or incoherent feed-forward motif. a. Coherent feed-forward motif

Molecular Biology of the Cell 7e-Alberts b. Incoherent feed-forward motif

1. It can provide cellular responses with robustness despite perturbations in the input signal.

2. A cellular pathway with such a motif generally requires a sustained (long) input signal to generate a full response.

3. A cellular pathway with such a motif can generate a brief pulse of response in the presence of a sustained input signal.

84. You have performed an SDS-PAGE followed by a Western blot in order to detect the presence of a specific protein in two new cell lines. You have decided to use a brand-new antibody that has been generated against this protein. As a positive control, you also run on the gel an extract from a cell line that you know contains the protein of interest. However, when you look at the resulting blot, there are no visible protein bands. Assuming the SDS-PAGE gel was run correctly, as well as the imaging process, what might have gone wrong with the blotting procedure to give this result? Explain your reasoning.

85.

In an unfolded (random coil) protein, amino acid residues are exposed to the solvent and share more or less the same environment, whereas each residue in a folded protein has its own unique neighborhood. This fact can be exploited in using NMR to study protein folding. The two schematic diagrams below represent two-dimensional NMR spectra for the same protein in either its folded (native) or unfolded state. The chemical shifts, which depend on the local neighborhood of each atom, are plotted in these diagrams. Which diagram (A or B) do you think corresponds to the folded protein? Explain your answer in 2–3 sentences.

Molecular Biology of the Cell 7e-Alberts

86.

Each cycle of PCR doubles the amount of DNA synthesized in the previous cycle.

A pair of primers is used to direct the specific synthesis of a desired DNA segment. The following diagram shows the products of PCR amplification after three cycles. The DNA strands of the desired segment are indicated in yellow boxes in the diagram. As shown, at the end of the third cycle, 8 out of 16 strands have the desired length and are boxed. After four more cycles, that is, at the end of the seventh cycle, how many strands of DNA are expected to be synthesized?

How many of them have the desired length?

Molecular Biology of the Cell 7e-Alberts

87. Traditionally, inserting a PCR product into a plasmid required the use of restriction endonucleases. However, recently it has become the case that utilizing a Gibson assembly allows one to avoid this requirement. In 2-3 sentences, explain how the Gibson assembly functions to avoid the need for restriction endonucleases.

88. After treatment with mutation-inducing chemicals, a mutant Drosophila. melanogaster fruit fly has been identified that displays neuromuscular abnormalities that are similar to those seen in Parkinson’s disease. How can a researcher go about discovering which gene or genes have been affected to produce these effects?

89. A genome-wide association study has been conducted in order to search for SNPs associated with heart disease. One SNP was discovered that had much higher correlation with heart disease. However, this SNP was not located in a protein-coding gene. Explain how this SNP could be associated with a disease without directly affecting the sequence of a protein product.

Molecular Biology of the Cell 7e-Alberts 90. The body uses insulin, a small protein, as a hormone to control blood sugar levels. Originally, insulin was harvested from pig pancreas, and a small amount of insulin was produced during the process, creating great expense. However, for pharmaceutical use we now have a cheaper insulin that is produced by bacteria.

Name two modern techniques that contributed to this important medical advancement, and explain how these techniques enable bacterial expression of a gene as well as the purification of the gene product. (Ignore any need for posttranslational modifications for the activation of insulin.)

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 08 1. Answer:

B

2. Answer:

C

3. Answer:

E

4. Answer:

D

5. Answer:

E

6. Answer:

C

7. Answer:

D

8. Answer:

A

9. Answer:

D

10. Answer:

B

11. Answer:

E

12. Answer:

C

13. Answer:

C

14. Answer:

C

15. Answer:

A

16. Answer:

E

17. Answer:

C

18. Answer:

A

19. Answer:

C

20. Answer:

B

Molecular Biology of the Cell 7e-Alberts 21. Answer:

D

22. Answer:

C

23. Answer:

C

24. Answer:

D

25. Answer:

B

26. Answer:

A

27. Answer:

D

28. Answer:

A

29. Answer:

D

30. Answer:

E

31. Answer:

D

32. Answer:

A

33. Answer:

D

34. Answer:

D

35. Answer:

D

36. Answer:

B

37. Answer:

C

38. Answer:

A

39. Answer:

C

40. Answer:

A

41. Answer:

B

42. Answer:

C

Molecular Biology of the Cell 7e-Alberts 43. Answer:

A

44. Answer:

D

45. Answer:

B

46. Answer:

A

47. Answer:

C

48. Answer:

D

49. Answer:

C

50. Answer:

D

51. Answer:

A

52. Answer:

D

53. Answer:

A

54. Answer:

B

55. Answer:

B

56. Answer:

C

57. Answer:

D

58. Answer:

E

59. Answer:

D

60. Answer:

B

61. Answer:

B

62. Answer:

D

63. Answer:

E

64. Answer:

A

Molecular Biology of the Cell 7e-Alberts 65. Answer:

B

66. Answer:

A

67. Answer:

A

68. Answer:

B

69. Answer:

B

70. Answer:

D

71. Answer:

B

72. Answer:

D

73. Answer:

1. C 2. B 3. A 4. A

74. Answer:

1. B 2. A 3. A

75. Answer:

1. A 2. A 3. B 4. A 5. B

76. Answer:

1. A 2. A 3. A 4. B

Molecular Biology of the Cell 7e-Alberts 5. B 6. A 77. Answer:

1. B 2. A 3. C

78. Answer:

1. B 2. A 3. C

79. Answer:

1. A 2. B 3. B

80. Answer:

1. B 2. B 3. A 4. B

81. Answer:

1. A 2. A 3. A 4. A

82. Answer:

1. D 2. B 3. C 4. A

83. Answer:

1. A 2. A

Molecular Biology of the Cell 7e-Alberts 3. B 84. Answer:

The most obvious answer is that the new antibody did not bind the protein of interest. Since the control must have the protein of interest present, if the antibody bound, one should have seen a band in the control sample. Another potential issue is that the electrophoresis to move the proteins to the blotting membrane did not pull them to the membrane. This would result in no protein on the membrane and no protein available to bind to the antibody.

85. Answer:

Diagram B corresponds to the folded protein, because in the unfolded, random-coil state, variability in the environment surrounding the nuclei, for example, hydrogens, is decreased significantly. This is reflected in the distribution of peaks in the nuclear magnetic resonance (NMR) spectra, which is broader for the folded state (B).

86. Answer:

256,240

87. Answer:

When using a Gibson assembly, a sequence of DNA (15–40 bp) is added to each end of the desired PCR product that is complementary to the sequence that surrounds the cut sites within the plasmid. This is done by adding the complementary sequence to the sequence of the PCR primers. The PCR product and the plasmid are then treated with a 5′ exonuclease which leaves complementary sticky ends available for pairing between the plasmid and the PCR product.

88. Answer:

A researcher could sequence the genome of the mutant fruit fly and compare it to a genome from a wild-type fruit fly. There may be many differences in the genomes due to the effects of the chemical, but a possible mutant gene can be tested by intentionally mutating that gene in other fruit flies and monitoring those flies for similar phenotypes to the original mutant.

89. Answer:

One possibility is that this SNP could be located in a block of frequently inherited DNA, and could signal that a protein-changing DNA variant is nearby. Another is that the SNP could be located in a regulatory sequence of DNA that affects the expression of a functional gene product related to the disease.

Molecular Biology of the Cell 7e-Alberts 90. Answer:

Students may make a case for various answers. Below are two sample responses: • DNA cloning using a bacterial plasmid can be used to express a gene in bacteria. First, the plasmid is linearized with restriction endonucleases and a gene is ligated into the plasmid, which is re-circularized. When these plasmids are transformed into bacteria, they are replicated as the bacteria replicate. Thus, many copies of the DNA gene, in this case, the gene for insulin, is replicated as well.

• Once the insulin proteins are produced in bacteria, they need to be separated and purified from the bacterial cells. This can be done using affinity column chromatography. The column can be filled with antibodies for insulin, which as the cell lysate is added to the column, will bind strongly and specifically to insulin. The other components of the cell lysate can be washed away, and the insulin can be eluted.

Molecular Biology of the Cell 7e-Alberts

Chapter 9: Visualizing Cells and Their Molecules 1. If an average globular protein was of the size of a tennis ball, a typical animal cell would be as large as which of the following? a. Cubicle b. Room c. Tennis court d. Stadium e. City

2. In which of the following microscopy techniques are oblique rays of light focused on the specimen? a. Bright-field microscopy b. Dark-field microscopy c. Phase-contrast microscopy d. Differential-interference-contrast microscopy

3. The light used to excite a fluorescent molecule carries ___________ energy and has a ___________ wavelength compared to the light that is then emitted from the molecule. a. Greater; longer b. Greater; shorter c. The same amount of; shorter d. Less; longer e. Less; shorter

4. A biology student is attempting to observe the spindle microtubules in a dividing

Molecular Biology of the Cell 7e-Alberts cell with a fluorescence microscope. During this process, the cellular structure is absorbing light at a particular wavelength, and emitting a photon of comparatively ___________ energy and ___________ wavelength. a. Increased; longer b. Decreased; longer c. Decreased; shorter d. Identical; shorter e. Increased; shorter

5. Scientists have been researching patients that have a difficult time digesting lactose, as there is a mutation in the gene for the enzyme β-galactosidase. They hypothesize that affected individuals have an absence of β-galactosidase in their digestive tracts. What would be the best experiment to test this hypothesis? a. Use immunohistochemistry to tag β-galactosidase with GFP and view samples with a fluorescence microscope. b. Use bright-field microscopy to look at the colon cell structures. c. Have patients eat dairy to determine if they get sick or are tolerant. d. Apply phase-contrast to enhance the contrast of light microscopy images of colon cells from biopsies.

6. Your lab is studying a new protein that is thought to be involved in setting up the spindle apparatus in mitosis. You hypothesize the protein will recruit microtubules to the centrioles, and want to design an experiment to visualize and track the movement of your protein during division in real time. What technique will be most appropriate for this experiment? a. Create a transgenic cell line that expresses a GFP-tagged novel protein and RFP-tagged microtubules, and record video from a confocal microscope. b. Image cells in multiple stages of mitosis using TEM with immunoconjugated gold particles to locate your protein of interest.

Molecular Biology of the Cell 7e-Alberts c. Fix and treat cells with dyes to counterstain the mitotic spindle and use indirect immunohistochemistry to locate your protein of interest. d. Use phase-contrast microscopy to show the spindle structure in high contrast and an immunoconjugated dye to locate your protein of interest. e. Image cells at each stage of division with SEM to visualize the contours of cellular cleavage.

7. Two segments (S1 and S2) in a viral protein are suspected to be responsible for the nuclear localization of the protein in infected human cells. You have engineered a plasmid to encode the green fluorescent protein (GFP) fused to either or both of these peptide segments, and have introduced the plasmid into the cells. After the expression of the fusion protein is induced, you visualize the cells using a fluorescence microscope equipped with filters appropriate for detection of GFP. Your results are presented in the following schematic drawings in which the GFP signal is represented in shades of gray. Which of the following is more consistent with these observations?

Molecular Biology of the Cell 7e-Alberts a. Both S1 and S2 are required for nuclear localization. b. S1 is required and sufficient for nuclear localization. c. S2 is required and sufficient for nuclear localization. d. S1 is required but not sufficient for nuclear localization. e. S2 is required but not sufficient for nuclear localization.

8. Given the absorption and emission spectra of three fluorescent dyes in the simplified diagrams below, which pair of dyes is better suited for a fluorescence resonance energy transfer (FRET) study?

a. AB b. BC c. AC d. BA e. CB

9. Protein A is marked with a fluorescent marker. When excited with green light, it emits blue light. Protein B is also marked with a fluorescent marker. When it is excited by blue light, it emits violet light. When these proteins come into close

Molecular Biology of the Cell 7e-Alberts proximity to one another while excited by green light, violet light will be emitted.

What is this an example of? a. Fluorescence resonance energy transfer b. Fluorescence recovery after photobleaching c. Fluorescence protein excitation d. Fluorescence-tagged resolution

10. Consider an engineered chimeric protein made from fusion of three proteins: a blue fluorescent protein (BFP), a calmodulin-binding peptide, and a green fluorescent protein (GFP). Calmodulin is an abundant calcium-binding protein in eukaryotes. Once bound to calcium ions, it can recognize the calmodulin-binding peptide in the fusion protein, change conformation, wrap around the peptide, and bring the BFP and GFP components in close proximity. This results in fluorescence resonance energy transfer (FRET) between BFP and GFP. Accordingly, the fusion protein is which of the following? a. A luminescent ion-sensitive indicator that red-shifts its emission wavelength in the presence of calcium b. A luminescent ion-sensitive indicator that increases its emission in the presence of calcium c. A genetically encoded calcium indicator that red-shifts its emission wavelength in the presence of calcium d. A genetically encoded calcium indicator that increases its emission in the presence of calcium

11. Which of the following statements best describes "deconvolution?" a. The process by which micrographs are spliced together and brightened in order to increase visibility b. The process by which the resolution of a micrograph is digitally enhanced by a sharpening algorithm c. The process by which several micrographs are combined in order to

Molecular Biology of the Cell 7e-Alberts produce a three-dimensional image d. The process by which out-of-focus light is computationally subtracted from a series a micrographs in order to produce a sharp, three-dimensional image

12. Which microscopy set-up uses a longer wavelength of light than usually excites a particular fluorophore? Which one allows researchers to peek deeper into biological samples? a. Single-photon; single-photon b. Single-photon; two-photon c. Two photon; single photon d. Two-photon; two-photon

13. What type of microscope allows micrographs to be taken in different focal planes then reconstructed into a three-dimensional image? a. Fluorescence microscope b. Confocal microscope c. Scanning electron microscope d. Transmission electron microscope e. Phase-contrast microscope

14. Stimulated Emission Depletion (STED) microscopy can improve the resolution of fluorescence microscopy. How is this accomplished? a. STED increases fluorescence, allowing detection of objects previously too small to be seen. b. STED modulates the color of fluorescence, allowing differentiation between small molecules. c. STED decreases the size of fluorescence stimulation, eliminating point

Molecular Biology of the Cell 7e-Alberts spread overlap. d. STED increases the contrast of fluorescence, emphasizing the differences between small molecules.

15. Single-molecule detection by fluorescence microscopy is limited by the presence of an excess of out-of-focus fluorescent molecules. How does a TIRF microscope uniquely overcome this limitation? a. By removal of the out-of-focus molecules by selective destruction b. By using a deconvolution algorithm that reverses the convolution of signals due to the out-of-focus molecules c. By a confocal set-up that eliminates out-of-focus signals d. By exciting only in-focus molecules via an evanescent field e. By using an objective lens with extremely high numerical aperture

16. The presence of which element provides the sample with the lowest electron density? a. Osmium b. Lead c. Uranium d. Carbon e. Gold

17. You have generated strains of Drosophila melanogaster that are expected to show interesting developmental phenotypes such as misplaced organs in the adult fly. However, some of these phenotypes are not readily seen with light microscopy. You therefore fix each mutant fly, dry it, coat it with a thin layer of gold, and place the entire fly on a stationary electron microscope platform for viewing. What type of imaging technique are you using?

Molecular Biology of the Cell 7e-Alberts a. SEM b. TEM c. CryoEM d. Tomographic EM

18. Electron microscopy samples are often chemically fixed before dehydration, resin-embedding, and sectioning. But they can also be “fixed” by freezing in a way that precludes ice-crystal formation, to ensure minimal damage to the original cell structures. Which is NOT a useful method to achieve noncrystalline freezing? a. High-pressure cooling b. Plunging into liquid nitrogen c. Spraying with a jet of liquid propane d. Contact with a copper block cooled by liquid helium e. Timed freezers that drop the temperature slowly by 1°C per second

19. Which microscopy technique can solve the problem of superposition of structures in transmission electron microscopy (TEM)? a. SEM b. STED c. EM tomography d. Superresolution fluorescence microscopy e. Light microscopy

20. Which of the following is NOT correct regarding cryo-electron microscopy? a. It does not require shadowing or negative staining. b. It involves obtaining images from many—sometimes tens of thousands—

Molecular Biology of the Cell 7e-Alberts individual molecules that may or may not all be in the same orientation. c. Subnanometer resolutions, for example, 0.5 nm or better, can be achieved by this technique. d. It can only be used to see the exterior surface of molecules and complexes. e. Atomic models obtained from x-ray crystallography can be fitted into the density envelope obtained from this technique.

21. CryoEM uses single particle reconstruction to determine very accurate atomic structures of protein complexes. Which of the following best describes the process of single particle reconstruction? a. Images are obtained from multiple bombardments of the specimen with single electrons, which are individually recorded with high precision. b. The specimen is bombarded with extremely high numbers of particles to gather high-resolution data. c. The specimen is rotated and captured from a few different angles to gather three-dimensional data. d. Images of the individual atoms are obtained from all directions and reassembled in a panorama. e. Images from many identical molecules are computationally superimposed and averaged to reveal fine detail in common.

22. What is an advantage of tandem electron/light microscopy techniques such as CLEM? a. Localizing a specific molecule in its full, detailed cellular context b. High speed and ease of sample preparation c. Requires inexpensive, easily maintained equipment d. Generation of live images that show movement of molecules within highresolution structures e. Requires relatively little computing power and time to process images

Molecular Biology of the Cell 7e-Alberts

23. The limit of resolution for conventional light microscopy is approximately 0.4 µm, corresponding to the wavelength of violet light. a. True b. False

24. PALM and STORM techniques reduce the width (or “spread”) of the point spread function. a. True b. False

25. Depending on acceleration voltage, the resolution limit of an electron microscope can be as small as 0.05 nm. a. True b. False

26. In the diagram below, a logarithmic scale of sizes is shown. Indicate which of the sizes indicated (A to H) better corresponds to the dimensions of each of the following.

Molecular Biology of the Cell 7e-Alberts

a. 20 μm b. 2 μm c. 2 nm

1. A bacterium 2. An animal cell 3. A small molecule

27. Would you use a fluorescent organic molecule, in situ hybridization, or a coupled fluorescent protein to visualize the cells and their molecules in each of the following cases? a. Fluorescent biosensor

Molecular Biology of the Cell 7e-Alberts b. in situ hybridization c. Fluorescent protein

1. You would like to see where in the early Drosophila embryo the mRNA encoding a certain transcription regulator is located.

2. You would like to see the nuclei and count them in an early mouse embryo.

3. You would like to visualize chromosome 3 in a human cell culture derived from a patient’s tissue, based on specific sequences present on this chromosome.

4. You would like to observe the oscillations in Ca 2+ ions inside a fertilized frog egg.

5. You would like to compare the localization of two transcription regulatory proteins in cultured human T cells.

28. Two approaches have been devised to deal with the problem of blurring in light microscopy with thicker samples. Which of the following descriptions better applies to confocal design, and which to image deconvolution? a. Confocal design b. Image deconvolution

1. It is normally faster.

2. It requires a higher degree of sample illumination.

3. It can be used to obtain images from relatively deeper parts of the specimen.

29. Which of the following descriptions better applies to SIM, STED, and STORM/PALM superresolution techniques, respectively? a. SIM b. STED c. STORM/PALM

1. It switches on and off individual fluorophores at random over time to accurately determine their position.

2. It creates a moiré pattern from the interference of the illuminating pattern

Molecular Biology of the Cell 7e-Alberts and the sample features.

3. It doubles the resolution of conventional fluorescence microscopy.

4. It limits excitation to the fluorophores that are located at the center of the focal point by using a doughnut-shaped beam in addition to the excitation beam.

30. Which of the following descriptions better applies to the scanning, and which to the transmission electron microscopy techniques? a. Scanning b. Transmission

1. It generally has a greater depth of field.

2. It is usually smaller, cheaper, and simpler.

3. It detects electrons that are scattered or emitted from the specimen.

4. It is used to create electron-microscope tomograms.

31. When the gene encoding a certain cytoskeleton protein is deleted, the resulting mutant cells round up and do not form their normal appendages. These mutants can be rescued when a gene encoding an N-terminal green fluorescent protein (GFP) fusion of the protein is expressed, but not when a gene encoding a Cterminal GFP fusion is expressed. Which fusion protein (N or C) is appropriate to use in studying cellular localization and activity? Explain your answer in 2 to 3 sentences.

32. Imagine a transcription regulatory protein (X) that is known to shuttle back and forth between nucleus and cytosol in an oscillatory pattern. Protein Y is a nuclear protein that can bind to X to create a dimer that binds to DNA. You have fused protein X to green fluorescent protein (GFP) and protein Y to blue fluorescent protein (BFP), and have measured fluorescence resonance energy transfer (FRET) and non-FRET signals in the nucleus at different time points, as indicated in the following simplified plot. At which time period (1 or 2) do you think protein X is in the nucleus? BFP can be excited at 440 nm, and emits maximally at 470 nm.

GFP is excited at 470 nm and emits maximally at 500 nm. Explain your reasoning in 2 to 3 sentences.

Molecular Biology of the Cell 7e-Alberts

33. How can a genetically encoded biosensor signal the presence of a molecule using Förster/fluorescence resonance energy transfer (FRET)? Furthermore, how might the concentration of a ligand of interest be measured using this technique?

In your response, describe the necessary components of the biosensor, and how it functions as a detector.

34.

A certain GTP-binding protein can exist in two main states. When bound to GDP, it is mostly cytosolic. In its GTP-bound form, however, it associates with the cytosolic face of the endoplasmic reticulum (ER) membrane, where it hydrolyzes the bound GTP after a short delay and is released again into the cytosol. You have created and expressed green fluorescent protein (GFP) fusions of the wildtype protein, as well as that of a mutant protein that does not bind GTP as readily as the wild type. You then perform a fluorescence recovery after photobleaching (FRAP) experiment by photobleaching a small area of the ER membrane and measuring GFP fluorescence recovery over time. According to the results below, which curve (1 or 2) do you think corresponds to the wild-type fusion protein?

Molecular Biology of the Cell 7e-Alberts

35. What sample preparation technique uses fixation of molecules within an expanding hydrogel to increase the space between the individual components?

Identify this technique, and indicate why it is beneficial in comparison to traditional sample preparation.

36. What are some limitations of traditional fluorescent microscopy when studying live cells and tissues in situ, for example, cells in early embryos? How can light-sheet microscopy be used to address these challenges instead, despite its own challenges?

37. Total internal reflection fluorescence (TIRF) uses the energy field of reflected light to activate only a small layer of fluorophores at the margin of a sample. What are the benefits of such a method compared to traditional fluorescence illumination?

38. In your laboratory, you are studying a receptor protein found in the plasma membrane of neurons that has been genetically linked to familial epilepsy. In affected families, the gene for the receptor contains distinctive mutations that are predicted to alter the three-dimensional structure of the receptor in the region where it binds its neurotransmitter ligand. What is a microscopy technique that could be used to investigate this mutation? Defend your choice to utilize this technique.

Molecular Biology of the Cell 7e-Alberts

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 09 1. Answer:

D

2. Answer:

B

3. Answer:

B

4. Answer:

B

5. Answer:

A

6. Answer:

A

7. Answer:

C

8. Answer:

A

9. Answer:

A

10. Answer:

C

11. Answer:

D

12. Answer:

D

13. Answer:

B

14. Answer:

C

15. Answer:

D

16. Answer:

D

17. Answer:

A

18. Answer:

E

19. Answer:

C

20. Answer:

D

Molecular Biology of the Cell 7e-Alberts 21. Answer:

E

22. Answer:

A

23. Answer:

A

24. Answer:

B

25. Answer:

A

26. Answer:

1. B 2. A 3. C

27. Answer:

1. B 2. A 3. B 4. A 5. C

28. Answer:

1. A 2. A 3. A

29. Answer:

1. C 2. A 3. A 4. B

30. Answer:

1. A 2. A 3. A 4. B

Molecular Biology of the Cell 7e-Alberts 31. Answer:

The N-terminal fusion protein will be more useful in experiments. This is because it is functional, replacing the knock-out gene and restoring normal cell growth/structure. The C-terminal fusion protein does not restore the cell’s normal phenotype, implying that presence of the GFP tag at this location renders it nonfunctional in some way. Its localization and activity in the cells may be similarly aberrant.

32. Answer:

Protein X is in the nucleus during time period 2.

FRET will occur when the BFP’s emission photon, with a wavelength of 470nm, excites the GFP and leads to green emission at 500nm. This is possible when protein X and Y are dimerized in the nucleus. When the blue wavelength is detected more strongly (time period 1), there is no FRET occurring and protein X is most likely in the cytosol.

33. Answer:

A sensor requires two fluorophores that will undergo FRET when located in close proximity, often connected by a hinge domain that exhibits a large conformational change in response to the binding of a ligand to indicate its presence. This moves the fluorophores into position for FRET to occur (or out of position to end the FRET energy transfer). Thus, the fluorescence color will indicate whether the ligand is present or absent.

Concentration can be correlated to intensity of the color that indicates the ligand is present.

34. Answer:

Curve 1 corresponds to the wild-type fusion protein, because since the mutant protein cannot efficiently bind GTP, it takes longer to recover the fluorescence in the ER membrane after photobleaching.

35. Answer:

The technique described is expansion microscopy. It allows a conventional fluorescence microscope to localize smaller structures with greater specificity than the limits of resolution would otherwise allow. The components maintain their relative position within the expanded structure.

36. Answer:

Tissue damage due to prolonged exposure to light and out-of-focus fluorescence in samples

Molecular Biology of the Cell 7e-Alberts with depth are limiting factors in traditional fluorescent microscopy. Light-sheet microscopy can reduce photobleaching by targeting the light beam through a thin section of the sample, exciting only the labeled molecules at that depth.

Imaging through many planes can digitally recreate three-dimensional structures. Sample preparation does not require fixation, so live cells and tissues can be examined and monitored over time. Light-sheet microscopy can also be combined with superresolution or sample expansion techniques, although otherwise it is limited to the minimal resolution capability of fluorescence microscopy. It also requires a specialized laser light source so that the “sheet” of light can be targeted and moved across the depth of the sample, and computational systems to build 3D representations from the captured images.

37. Answer:

By activating fewer fluorophores than direct light, there is less propensity for photobleaching, preserving the samples for further imaging. More targeted activation of the molecules that are directly in the plane of focus for visualization reduces the interference from excess fluorescence, improving resolution.

38. Answer:

CryoEM for single particle reconstruction would be an effective technique to use. As an EM technique, it has the appropriate level of resolution. It is particularly useful for integral membrane proteins (such as the candidate transmembrane receptor) that are difficult to investigate with X-ray crystallographic techniques.

Aligned high-resolution imaging of many individual receptors, and comparing results for the wild-type to the mutated receptor, can elucidate the structural differences between the two.

Ligand-bound and -unbound states can also be compared in this manner.

Molecular Biology of the Cell 7e-Alberts

Chapter 10: Membrane Structure 1. What is the typical thickness of a lipid bilayer such as the plasma membrane of our cells? a. 0.5 nm b. 5 nm c. 50 nm d. 100 nm e. 500 nm

2. In the following schematic drawing of an abundant plasma membrane phosphoglyceride, which part is positively charged?

a. Part A b. Part B c. Part C d. Part D e. Part E

Molecular Biology of the Cell 7e-Alberts 3. Which of the following is NOT correct regarding the molecule whose structural formula is shown below?

a. It is an amphiphilic molecule. b. It is a sterol. c. It makes the membrane less permeable to small hydrophilic molecules. d. It is found in membranes of virtually all living cells. e. It affects the fluidity of the lipid bilayer.

4. Which of the following is NOT naturally found in a eukaryotic membrane? a. Cholesterol b. Phosphatidylinositol c. Sphingomyelin d. Ganglioside GM1 e. Octylglucoside

5. The compound isobutane, formula HC(CH₃)₃, is added to water. How will it interact with the water molecules? a. The partial negative charge on water will interact with the partial positive on

Molecular Biology of the Cell 7e-Alberts isobutane. b. The partial positive charge on water will interact with the partial negative on isobutane. c. The nonpolar patches on water will interact with the nonpolar patches on isobutane. d. There will be no interaction, because water will form a cage around isobutane.

6. The motion of lipid molecules in a synthetic bilayer can be studied by various techniques. Which of the following has been observed in these systems? a. Phospholipids diffuse rapidly within and between the two leaflets of a bilayer. b. An average lipid molecule can diffuse the length of about 2 micrometers in a fraction of a millisecond. c. The flip-flops are very rare for phospholipids but cholesterol molecules flipflop more often. d. Within a bilayer, lipid molecules rarely rotate about their long axis, but diffuse laterally at very high rates.

7. Why do liposomes not fuse with one another spontaneously when suspended in an aqueous environment? a. Because fusion requires a large number of flip-flops, which are very rare. b. Because the hydration shell of the polar head groups of the lipids needs to be removed. c. Because fusion requires micelle formation. d. Because of the rapid lateral diffusion and rotation of the lipid molecules.

8. Which of the following lipids do you expect to be a canonical scramblase substrate in the plasma membrane?

Molecular Biology of the Cell 7e-Alberts a. Ganglioside GM1 b. Cholesterol c. Glycosylphosphatidylinositol d. Phosphatidylethanolamine e. Galactocerebroside

9. Analysis of lipid composition of a eukaryotic membrane revealed that lipid composition is complex, because membranes are comprised of how many unique lipid varieties? a. 5–10 b. 10–50 c. 100–500 d. 500–2000 e. 30,000–100,000

10. Which of the following changes would you expect to increase the phase transition temperature of a synthetic bilayer composed of phosphatidylserine? a. Incorporation of phospholipids with longer fatty acid chains b. Introduction of double bonds in the fatty acids c. Addition of cholesterol d. Removal of serine from the head group

11. Which of the following is correct regarding the composition of various biological membranes? a. Bacterial plasma membranes are often composed of one main type of phospholipid and lack cholesterol.

Molecular Biology of the Cell 7e-Alberts b. Cholesterol in the eukaryotic plasma membrane induces phase transition to the gel state. c. Inositol phospholipids are the most abundant lipids in the endoplasmic reticulum membrane. d. The mitochondrial and bacterial membranes are rich in glycolipids. e. Yeast cells synthesize more fatty acids with cis-double bonds when the temperature in the environment rises.

12. Many cells store lipids in droplets of varying sizes. Which statement best describes these droplets? a. They are enclosed by a phospholipid monolayer (instead of a bilayer). b. They mostly store cholesterol and phospholipids. c. They are produced by and released from the Golgi apparatus. d. They have mostly protein-free bilayer membranes. e. They are composed primarily of charged amphiphilic lipids.

13. Lipid droplets have proteins embedded in their membrane. How do these proteins arrive there? a. They are translocated into the membrane after budding from the ER. b. They are synthesized by a ribosome docked on the droplet. c. They are found in the membrane of a vesicle which will fuse with the droplet. d. They are aggregated in the cytosolic leaflet of the ER before budding.

14. Which statement best describes the two monolayers of the plasma membrane in a human red blood cell? a. They have different overall electrical charges, with negatively charged phospholipids enriched in the inner monolayer.

Molecular Biology of the Cell 7e-Alberts b. They have the same abundance of phosphatidylinositol. c. They exchange phospholipids only through spontaneous flip-flops. d. They both contain glycolipids. e. They both contain gangliosides.

15. Which of the following is NOT a function of a glycolipid? a. Protect the cell against degradative enzymes b. Participate in the cell recognition process c. Protect the cell against low pH d. Alter the electrical field across the plasma membrane e. Assist in solubility of proteins near the plasma membrane in the cytosol

16. Which of the following is preferentially found in the cytosolic leaflet of the plasma membrane in mammalian cells? a. Cholesterol b. Phosphatidylserine c. Phosphatidylcholine d. Sphingomyelin e. Glycolipid

17. Glycolipids, such as gangliosides, are best described in which way? a. They contain oligosaccharide chains with positively charged residues. b. They are found to constitute about 20% of the total lipid mass in the plasma membrane of neurons. c. They are found in the cytosolic leaflet in cellular membranes.

Molecular Biology of the Cell 7e-Alberts d. They affect the electrical environment of the membrane.

18. Which of the following proteins is the most likely to be directly involved in transducing a signal from the extracellular side of the plasma membrane to the intracellular side?

a. Protein A b. Protein B c. Protein C d. Protein D e. Protein E

19. Transmembrane proteins are best characterized in which way? a. They are typically exposed only to one side of the membrane. b. They can be released from the membrane by a gentle extraction procedure such as salt treatment. c. They are often further attached to the membrane via a GPI anchor. d. They are sometimes covalently attached to a fatty acid chain that inserts

Molecular Biology of the Cell 7e-Alberts into the membrane. e. They cannot contain β sheets in the part of their structure that interacts with the membrane interior.

20. Regarding lipid-anchored membrane proteins, which type of bilayer anchor features both a saturated fatty acid and a thioester linkage between the anchor and the peptide? a. Myristoyl anchor b. Palmitoyl anchor c. Farnesyl anchor d. Geranylgeranyl anchor e. Glycosylphosphatidylinositol anchor

21. Which protein is NOT in the same superfamily as the other four? The other four share an overall architecture composed of seven closely packed transmembrane helices. a. Bacteriorhodopsin b. Channelrhodopsin c. G-protein-coupled receptor d. Aquaporin e. Rhodopsin

22. The nicotinic acetylcholine receptor is a cation channel in the plasma membrane of some neurons. It is composed of five subunits, each of which is a transmembrane protein. The hydropathy plot for each mature subunit is qualitatively represented in the following diagram. How many membranespanning α helices do you expect to exist in each subunit?

Molecular Biology of the Cell 7e-Alberts

a. 1 b. 2 c. 3 d. 4 e. 5

23. A scientist wants to study only the cytosolic domains of a membrane-bound protein. To do this, they want to engineer a cell to only express the cytosolic portion of the protein. Which type of protein will they NOT be able to study in this manner? a. Single-pass membrane proteins

Molecular Biology of the Cell 7e-Alberts b. Multipass membrane proteins c. Proteins tethered to the membrane by a fatty acid d. Proteins embedded in the cytosolic leaflet of the membrane

24. Integrins are single-pass integral membrane proteins in the plasma membrane of animal cells and are involved in the interaction of the cell with the surrounding extracellular matrix. Which of the following descriptions do you think matches the transmembrane part of an integrin molecule? a. It forms a β barrel. b. It is an α helix that is bent in the middle. c. It is about 10 amino acids long, with every other amino acid side chain being hydrophobic. d. It folds in a conformation with maximal intrachain hydrogen-bonding. e. It is about 100 amino acids long.

25. What do all β-barrel transmembrane proteins have in common? a. The number of β strands. b. The diameter of the barrel. c. The number of negative peaks in their hydropathy plots. d. The general function, that is, membrane transport. e. The structural rigidity compared to α-helical transmembrane proteins.

26. A given transmembrane protein functions as an ion channel and has the secondary structure of a beta barrel, forming a cylinder in the membrane. Which total number of beta strands is the most unlikely to support ion transport? a. 8 beta strands b. 12 beta strands

Molecular Biology of the Cell 7e-Alberts c. 16 beta strands d. 20 beta strands e. 22 beta strands

27. Under certain conditions, about 20 molecules, on average, are predicted to exist in each β-octylglucoside detergent micelle. If the critical micelle concentration for this detergent is about 20 mM under these conditions, at what total βoctylglucoside concentration are about half of the detergent molecules found in micelles? a. 1 mM b. 10 mM c. 20 mM d. 21 mM e. 40 mM

28. While examining the crystal structure of a membrane protein, you find several phospholipid molecules bound to the protein. You know that these lipids can be described in which of the following ways? a. They are thought to destabilize many membrane proteins. b. They may enhance the crystallization of the bound membrane proteins. c. They interact nonspecifically with the protein. d. They will most likely have head groups of a single size and the same charge.

29. On the plasma membrane of the archaeon Halobacterium salinarum, bacteriorhodopsin is found in high density in patches of purple membrane. Which statement best describes this protein? a. It is a light-gated anion channel.

Molecular Biology of the Cell 7e-Alberts b. It is a single-pass transmembrane protein. c. It uses the energy stored in the proton gradient to transport small molecules. d. It changes conformation in response to light.

30. What is the most energetically unfavorable movement that a membrane protein can experience in the lipid bilayer? a. Flip-flop b. Rotational diffusion c. Lateral diffusion d. Flexion

31. The lateral diffusion of a plasma membrane protein has been studied using the fluorescence recovery after photobleaching (FRAP) technique in a cell under normal conditions, and the following plot has been obtained. Under a different condition, (e.g., when a signaling pathway is turned on), the protein binds to a few other proteins and protein complexes in the plasma membrane. As a result, its lateral diffusion coefficient is reduced and a significant fraction of the molecules of this protein are immobilized in large aggregates. How would you expect the FRAP curve to change under these new conditions? The dotted line in each graph is the original curve shown for comparison.

Molecular Biology of the Cell 7e-Alberts

a.

b.

c.

Molecular Biology of the Cell 7e-Alberts

d.

e.

32. Which of the following changes is more likely to further confine a membrane protein within a corral established by the cortical cytoskeletal network? a. Increased temperature b. Protease cleavage of the extracellular domain of the protein c. Binding to an extracellular ligand d. Protease cleavage of the cytosolic domain of the protein e. Binding of the protein to other transmembrane proteins

33. In the following schematic drawing of the cytosolic side of human red blood cell plasma membrane, which of the labeled proteins is spectrin?

Molecular Biology of the Cell 7e-Alberts a. Protein A b. Protein B c. Protein C d. Protein D e. Protein E

34. In intestinal epithelial cells, the different plasma membrane domains are separated from each other by special barriers. These barriers are a result of proteins of the apical surface forming a large aggregate together that excludes proteins of the other domains. a. True b. False

35. For each membrane protein in the following schematic drawings, would the cytoplasmic side of the membrane more likely be on the left or on the right? a. Left b. Right

1.

Molecular Biology of the Cell 7e-Alberts

2.

3.

4.

36. In the schematic drawing below, a protein (circle) is present on one side of a membrane in a high-enough quantity to induce curvature in this region of the membrane. In each of the following scenarios, indicate whether the protein would induce positive or negative curvature in the membrane.

Molecular Biology of the Cell 7e-Alberts

a. Positive b. Negative

1. The protein inserts a hydrophobic domain into the membrane leaflet that is facing it.

2. The protein is a phospholipase that cleaves the head group in specific phospholipids.

3. The protein binds cooperatively to multiple phospholipids with very large head groups.

37. In your own words, explain why some amphiphilic molecules form micelles in aqueous environments while others form bilayers.

38. A scientist studying the nature of phospholipids in bilayers adds fluorescently labeled sphingomyelin in a 2:2:1 ratio with unlabeled phosphatidylcholine and cholesterol to form artificial liposomes. They notice fluorescent patches rather than a uniformly fluorescent membrane as expected. What phenomenon are they likely observing, and why does it occur?

39.

Bacteriorhodopsin has seven transmembrane α helices, with nonstructured loops connecting the α helices. These loops find themselves either in the cytosol or in the extracellular space. What is the chemical nature of the amino acid residues found in the α helices and in the loops? Support your answer by explaining why your described chemical nature is true.

Molecular Biology of the Cell 7e-Alberts

40. Fluorescence recovery after photobleaching (FRAP) is a technique by which the fluorescent labels on particular protein molecules in an area of a cell are permanently “bleached” with a laser beam. During the subsequent “recovery” phase, nonbleached fluorescent proteins move into the photo-bleached area.

What are two reasons for why there might be no recovery? Consider how a scientist would differentiate between these two reasons.

41. There is a high concentration of the same cone-shaped protein clustered within one leaflet of a membrane. What do you expect to happen to the membrane, and why?

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 10 1. Answer:

B

2. Answer:

A

3. Answer:

D

4. Answer:

E

5. Answer:

D

6. Answer:

C

7. Answer:

B

8. Answer:

D

9. Answer:

D

10. Answer:

A

11. Answer:

A

12. Answer:

A

13. Answer:

D

14. Answer:

A

15. Answer:

E

16. Answer:

B

17. Answer:

D

18. Answer:

A

19. Answer:

D

20. Answer:

B

Molecular Biology of the Cell 7e-Alberts 21. Answer:

D

22. Answer:

D

23. Answer:

B

24. Answer:

D

25. Answer:

E

26. Answer:

A

27. Answer:

E

28. Answer:

B

29. Answer:

D

30. Answer:

A

31. Answer:

C

32. Answer:

E

33. Answer:

A

34. Answer:

B

35. Answer:

1. A 2. A 3. B 4. A

36. Answer:

1. A 2. B 3. A

37. Answer:

The differing property is the overall shape of the molecule. Cone-shaped amphiphilic molecules will form micelles while cylindrical molecules, such as phospholipids, will form bilayers.

Molecular Biology of the Cell 7e-Alberts 38. Answer:

Transient lipid rafts are forming. The sphingomyelin is associated with other molecules of sphingomyelin rather than being evenly distributed within the bilayer. This is likely due to the longer hydrocarbon tails in the sphingolipid molecules and how they interact with cholesterol.

39. Answer:

The residues found in the helices will be hydrophobic because they must interact with the hydrophobic portion of the phospholipids in the membrane. (Although it is hydrophobic, proline is unlikely to be present in an α helix.) The residues found in the loops will more likely be hydrophilic because they will be interacting with the polar head groups of the phospholipids as well as with the aqueous environments of the cytosol and extracellular space.

40. Answer:

The fluorescently labeled molecule may be immobilized or may be restricted to a very small area of the cell. A scientist may employ a single-particle tracking technique which labels only one molecule with antibodies coupled to fluorescent dyes or tiny gold particles and, tracking their movement by video microscopy, then determine if it truly is immobile or simply has limited mobility.

41. Answer:

The membrane will curve or bend. The shape of the proteins causes the area of one side of the membrane to be larger than the other leaflet. This induces curvature and deforms the overall shape of the membrane around them as the phospholipids interact with the proteins.

Molecular Biology of the Cell 7e-Alberts

Chapter 11: Small-Molecule Transport and Electrical Properties of Membranes 1. Why do cells not have membrane transport proteins for O 2? a. Because they need to keep the oxygen concentration low inside the reducing environment of the cell b. Because oxygen can dissolve in water and leak in via water channels c. Because oxygen can dissolve in the lipid bilayer and diffuse in and out rapidly without the need for a transporter d. Because oxygen is transported in and out of the cell in special oxygencarrying proteins such as hemoglobin e. Because oxygen transport across a membrane is energetically unfavorable

2. For which of the following ions is the intracellular concentration typically higher than the extracellular concentration? a. Sodium b. Calcium c. Magnesium d. Potassium e. Chloride

3. Imagine a small synthetic vesicle made from pure phospholipids enclosing an interior lumen containing 1 mM glucose and 1 mM sodium chloride. If the vesicle is placed in pure water, which of the following happens faster? a. Na+ diffuses out. b. Cl– diffuses out. c. H2O diffuses in.

Molecular Biology of the Cell 7e-Alberts d. Glucose diffuses out. e. Sodium chloride diffuses out.

4. After release into the synaptic cleft, the neurotransmitter dopamine is actively taken up by the cells via specific dopamine transporters. The drug cocaine interferes with this process and is therefore called a reuptake inhibitor. The inhibition of the transporter at a certain cocaine concentration is qualitatively represented in the following graph. In transport kinetics, Vmax describes the maximal rate of transport. Based on this graph, which of the following describes the effect of cocaine on the kinetics of dopamine reuptake by the transporter?

a. In the presence of cocaine, Vmax is unaffected, but the apparent affinity of the transporter for dopamine is reduced. b. In the presence of cocaine, Vmax is reduced, but the apparent affinity of the transporter for dopamine is unaffected. c. In the presence of cocaine, Vmax is enhanced, but the apparent affinity of the transporter for dopamine is unaffected. d. In the presence of cocaine, both Vmax and the apparent affinity of the transporter for dopamine are enhanced. e. In the presence of cocaine, both Vmax and the apparent affinity of the transporter for dopamine are unaffected, but the transporter is nevertheless inhibited.

5. If the concentration of both solutes (shown as small circles and squares) is higher

Molecular Biology of the Cell 7e-Alberts outside of the cell, which of the following transporters mediates primary active transport when actively transporting both solutes in the directions indicated?

a. Transporter A b. Transporter B c. Transporter C d. Transporter D e. Transporter E

6. In contrast to transporters, the channel proteins in cellular membranes do what? a. Interact strongly with the solute(s) that they transport b. Undergo a conformational change every time they transport a solute c. Can only mediate passive transport d. Form pores that are always open

7. The lactose permease in Escherichia coli is a H+-lactose symporter that mediates the inward active transport of lactose if this sugar is present in the environment.

Which of the following is true about this transporter? a. It has a twofold pseudosymmetrical structure. b. Lactose and H+ bind to two different conformations of the transporter. c. The transporter goes through an intermediate state in which the bound lactose is open to both sides of the membrane. d. If lactose and H+ concentrations are changed sufficiently, the transporter

Molecular Biology of the Cell 7e-Alberts can act as an H+-lactose antiporter.

8. What is an accurate description of an ion channel? a. An ion channel always mediates active transport. b. An ion channel transports a variety of ions. c. An ion channel is typically several orders of magnitude faster than a transporter. d. An ion channel is infrequently gated. e. An ion channel usually requires ATP hydrolysis.

9. A hypothetical, secondary active transporter protein is found in the plasma membrane of a cell. With the gradients indicated, what will be the direction of movement of the two molecules?

a. Circles and hexagons will both move into the cytoplasm. b. Circles and hexagons will both move out of the cytoplasm. c. Circles will move into the cytoplasm and hexagons will move out. d. Circles will move out of the cytoplasm and hexagons will move in.

10. The following possibilities represent potential transmembrane transport proteins.

Which of the following matches the conformation commonly found in active transporters?

Molecular Biology of the Cell 7e-Alberts

a.

b.

c.

d.

e.

11. Which of the following plasma membrane active transporters would NOT directly contribute to the regulation of cytosolic pH? a. A Na+-driven Cl-–HCO3- exchanger b. A Na+–H+ exchanger c. A P-type H+-ATPase d. A Na+-independent Cl-–HCO3- exchanger e. A Na+-K+ pump

Molecular Biology of the Cell 7e-Alberts

12. Many amino acids in our diet are absorbed via the transcellular transport pathway by the intestinal epithelial cells. This process requires ATP hydrolysis by which of the following? a. The Na+–amino acid symporters in the apical domain of the plasma membrane b. The Na+–amino acid antiporters in the apical domain of the plasma membrane c. The Na+-K+ pumps in the basal and lateral domains of the plasma membrane d. The amino acid carriers in the basal domain of the plasma membrane e. F-type ATPases in the apical domain of the plasma membrane

13. A particular family of ATPases is structurally related to the turbine-like pumps that acidify lysosomes and vesicles; however, they usually function in reverse, generating ATP from ADP and Pi using proton gradients across membranes.

What is this family called? a. P-type pumps b. ABC transporters c. V-type pumps d. F-type pumps e. Permeases

14. Which of the following normally functions to lower the pH of the cytosol? a. A Na+–H+ exchanger in the plasma membrane. b. A Na+-driven Cl––HCO3– exchanger in the plasma membrane. c. A Na+-independent Cl––HCO3– exchanger in the plasma membrane.

Molecular Biology of the Cell 7e-Alberts d. A V-type ATPase in the lysosomal membrane. e. Both answers A and B above.

15. The concentration of the dipeptide leucyl-leucine (Leu-Leu) is higher intracellularly compared to the cell’s environment. The protein Ptr2p facilitates the uptake of this peptide. What type of transport does Ptr2p engage in? a. Channel-mediated passive transport b. Transporter-mediated passive transport c. Active transport, affected by the electrochemical gradient of the dipeptide d. Active transport, with no dipeptide electrochemical gradient effect e. Simple diffusion

16. Which of the following transporters is NOT electrogenic? a. Na+–glucose symporter b. Ca2+-pump c. Na+-K+ pump d. Na+-independent Cl––HCO3– exchanger e. Bacteriorhodopsin

17. If the phosphorylation domain and the actuator domain were covalently crosslinked in the closed position, in which part of the pumping cycle would the sarcoplasmic Ca2+ pump become stuck? a. Open, facing the cytosol b. Open, facing the SR lumen c. Occluded, occupied by 2 Ca2+

Molecular Biology of the Cell 7e-Alberts d. Occluded, occupied by 2 H+ e. Occluded, but empty

18. There is an auxiliary transport system associated with ABC transporters found in the plasma membrane of bacteria with double membranes. This auxiliary system consists of a porin embedded in the cell’s outer membrane, and a substratebinding protein in the periplasmic space. As a solute moves from the cell exterior to the cytosol, how must its concentration change? Answers below are in the following order: cell exterior, periplasmic space, cytosol. a. Low, high, low b. High, low, high c. High, middle-range, low d. Low, middle-range, high e. Middle-range, low, high

19. Which of the following is a pump that hydrolyzes two ATP molecules per transport cycle? a. The cystic fibrosis transmembrane conductance regulator protein b. The multidrug resistance protein c. The Na+-K+ pump d. The Ca2+-pump e. The V-type ATPase

20. What is the reason for why an aquaporin does not pass ions (such as Na +) through? a. The Na+ ions are too small to pass through the pore rapidly enough. b. The Na+ ions are too large to pass through the narrow pore.

Molecular Biology of the Cell 7e-Alberts c. The Na+ ions do not favorably interact with the wall of the pore and therefore cannot be dehydrated. d. Two Asn residues in the channel prevent the passage of Na+ ions. e. A Na+ ion must become fully hydrated to pass through.

21. Aquaporin has a pair of key asparagine residues located on the wall almost halfway through its pore. These residues simultaneously bind to the oxygen atom of a passing water molecule, rendering it unavailable for hydrogen bonding. As a result, what happens? a. The channel can conduct negatively charged (but not positively charged) ions. b. The channel can conduct glycerol as well as water. c. The channel cannot conduct glycerol. d. The channel cannot conduct protons. e. The channel cannot conduct chloride ions.

22. The Nernst equilibrium potential for an ion that is 10 times more concentrated in the cytosol compared to the extracellular fluid is about −60 mV. How much would the potential be if the extracellular concentration decreases 100-fold with no change in the intracellular concentration? a. −6000 mV b. −6 mV c. −600 mV d. −20 mV e. −180 mV

23. Sanshool is a natural compound found in Sichuan pepper, and the cause for the numbing sensation that one experiences in the mouth. It is also known to inhibit a

Molecular Biology of the Cell 7e-Alberts subset of potassium leak channels involved in maintaining the resting membrane potential in sensory neurons. How do you think sanshool affects these sensory neurons? a. It elevates the resting potential (to less negative values) and makes it easier to excite the neuron. b. It elevates the resting potential and makes it harder to excite the neuron. c. It lowers the resting potential and makes it easier to excite the neuron. d. It lowers the resting potential and makes it harder to excite the neuron. e. It causes hyperpolarization and decreases the spontaneous firing of the neuron.

24. Which of the following is correct regarding the electrical properties of cellular membranes? a. Setting up (or changing) the membrane potential requires changing the concentrations of the ions in the microenvironment adjacent to both sides of the membrane. b. The resting potential in most animal cells is between 20 mV and 120 mV (positive inside). c. If the membrane is impermeable to an ion, the membrane potential approaches the equilibrium potential for that ion. d. The resting potential decays immediately following the inhibition of the Na +K+ pump by a drug.

25. A membrane potential is plotted over time in the following graph. The Nernst equilibrium potentials for four ions (A to D) that affect the membrane potential in this system are indicated in the graph. What happened in the time period indicated by a question mark?

Molecular Biology of the Cell 7e-Alberts

a. Channels conducting ion A opened b. Channels conducting ion B opened c. Channels conducting ion C opened d. Channels conducting ion D opened e. Channels conducting ions A and C opened

26. Why does a potassium channel conduct K+ ions several orders of magnitude better than Na+ ions? a. The Na+ ion is too large to pass through the channel pore. b. The hydrated Na+ ion occupies a larger volume compared to the hydrated K+ ion, and is too large to pass through the channel pore. c. The Na+ ion is too small to interact with the channel in a way that facilitates the loss of water from the ion. d. The Na+ ion cannot bind to the high-affinity K+-binding sites in the channel pore.

27. Which of the following is true of the bacterial K+ channel? a. The pore helices have a partial positive charge angled away from the pore.

Molecular Biology of the Cell 7e-Alberts b. The vestibule is on the extracellular side of the pore. c. The extracellular-facing portions of the channel are positively charged. d. The cytosolic-facing portions of the channel are positively charged. e. The selectivity filter is lined with partial positive charges.

28. MscS and MscL are mechanosensitive channels that can be characterized in which of the following ways? a. They constitute the majority of channel proteins in the plasma membrane of auditory hair cells in the human cochlea. b. They open in response to membrane tension. c. They open at the same osmotic pressure, but differ in their ion selectivity. d. They are highly ion selective.

29. A neuron’s repetitive firing rate is limited by an absolute refractory period, during which a new action potential cannot be generated. Which event is chiefly responsible for this limit? a. Opening of K+ leak channels b. Opening of voltage-gated Na+ channels c. Opening of transmitter-gated cation channels d. Inactivation of voltage-gated K+ channels e. Inactivation of voltage-gated Na+ channels

30. If the extracellular concentration of sodium ions is artificially decreased for a neuron, then what happens? a. Both the resting potential and the peak of the action potential rise significantly.

Molecular Biology of the Cell 7e-Alberts b. The resting potential increases but the action potential peak does not. c. The resting potential hardly changes but the action potential peak is raised. d. The resting potential decreases but the action potential peak is raised. e. The resting potential hardly changes but the action potential peak is lowered.

31. Which of the following situations that take place within a neuron results in action potentials that have extended depolarization phases? a. Inhibiting the voltage-gated Na+ channels b. Stimulating the voltage-gated K+ channels c. Compromising the inactivation mechanism in voltage-gated Na+ channels d. Compromising the inactivation mechanism in voltage-gated K+ channels

32. In a voltage-clamp experiment on a squid giant axon, the membrane potential is clamped at +20 mV, as shown below. Within the same time frame, which of the curves, A to D, best represents qualitatively how the total membrane current changes? A positive value of the current represents an outward flow of cations.

Molecular Biology of the Cell 7e-Alberts

a. Curve A b. Curve B c. Curve C d. Curve D

33. There is a wide variety of structural and functional diversity among voltage-gated cation channels, although they are found throughout eukaryotes. Which statement is true about their evolutionary relatedness? a. All voltage-gated Na+ and K+ channels are related, but they are independent from Ca2+ channels. b. All voltage-gated K+ and Ca2+ channels are related, but they are independent from Na+ channels. c. All voltage-gated Na+ and Ca2+ channels are related, but they are independent from K+ channels. d. All voltage-gated Na+, K+, and Ca2+ channels are related within a kingdom, but they are not related between kingdoms.

Molecular Biology of the Cell 7e-Alberts e. All voltage-gated Na+, K+, and Ca2+ channels are related.

34. A certain neuron in a mouse brain is firing about 300 times per second, whereas almost all of its neighboring neurons are firing at much lower rates of only a few times per second. What does this imply? a. This neuron expresses more depolarizing (Na+ and Ca2+) than hyperpolarizing (K+) channels compared to most of its neighbors. b. The mouse is experiencing an epileptic seizure. c. The refractory period of its voltage-gated Na+ channels is at least 300 ms. d. It releases excitatory neurotransmitters at its axon terminals. e. It is probably of a different type compared to most of its neighbors.

35. In neurons, the synaptotagmin-1 protein is localized to presynaptic axon terminals. When activated, it facilitates the fusion of neurotransmitter-containing synaptic vesicles to the plasma membrane, leading to neurotransmitter release into the synaptic cleft. Which of the following would you expect to directly activate synaptotagmin-1 for this function? a. Binding to Mg2+ b. Binding to Ca2+ c. Binding to phosphatidylcholine d. Binding to K+ e. Binding to the K+ leak channels

36. At synapsis between neurons, the signal is passed from the presynaptic cell to the postsynaptic cell. As the signal moves from one cell to the next, what type of signal exists at each of the following: Nerve terminal, synaptic cleft, target cell? a. Chemical, chemical, electrical

Molecular Biology of the Cell 7e-Alberts b. Chemical, electrical, chemical c. Electrical, electrical, chemical d. Electrical, chemical, electrical e. Electrical, chemical, chemical

37. Which of the following better represents the process of adaptation mediated by the voltage-gated Ca2+ channels and Ca2+-activated K+ channels? The vertical axis in each graph is the membrane potential in the postsynaptic neuron (in millivolts), while the horizontal axis is the time of constant stimulation (in milliseconds).

a.

b.

c.

d.

Molecular Biology of the Cell 7e-Alberts

e.

38. Once expressed in a neuron in the mouse brain and subjected to a flash of light, how does channelrhodopsin affect the membrane potential? a. It causes neurotransmitter release, leading to an action potential in the postsynaptic neuron. b. It increases the membrane permeability to cations and makes the membrane potential more negative. c. It depolarizes the membrane, leading to the opening of voltage-gated Na+ channels. d. It hyperpolarizes the membrane, leading to the inactivation of voltage-gated K+ channels. e. It changes the membrane potential such that voltage-gated K+ channels but not voltage-gated Na+ channels are activated.

39. The membrane potential in a particular hyperpolarized cell is measured to be −70 mV. For each of the following channels in this cell, the calculated Nernst equilibrium potential for the corresponding ion is presented. Indicate whether each channel is driven to pass the ion into or out of the cell. a. Into the cell b. Out of the cell

1. K+ channel; the K+ equilibrium potential is −60 mV. 2. Na+ channel; the Na+ equilibrium potential is +70 mV. 3. Ca2+ channel; the Ca2+ equilibrium potential is +130 mV. 4. Cl− channel; the Cl– equilibrium potential is −90 mV.

Molecular Biology of the Cell 7e-Alberts 40. Indicate whether the transmitter-gated channels for calcium, chloride, potassium, and sodium at a chemical synapse usually open in response to excitatory or inhibitory neurotransmitters. a. Excitatory neurotransmitters b. Inhibitory neurotransmitters

1. Calcium 2. Chloride 3. Potassium 4. Sodium

41. The nicotinic acetylcholine receptor is an ionotropic receptor that can be activated by nicotine, a stimulant drug found in cigarettes. Additionally, α-bungarotoxin, a neurotoxin found in the venom of some snakes, binds preferentially and tightly to the closed conformation of the receptors. Another drug, donepezil, is a potent acetylcholinesterase inhibitor used to treat patients with Alzheimer’s disease. If applied at a neuromuscular junction, indicate whether each of the compounds acetylcholine, nicotine, α-bungarotoxin, and donepezil would enhance or suppress the Ca2+ influx in the muscle cells. a. Enhance b. Suppress

1. Acetylcholine 2. Nicotine 3. α-bungarotoxin 4. Donepezil

42. Which graph better represents the change in the relative conductance of Na + and K+ channels during an action potential in a neuronal membrane?

Molecular Biology of the Cell 7e-Alberts

a.

b.

c.

d.

e.

43. The inactivation rate of voltage-gated Na+ and K+ channels can regulate the

Molecular Biology of the Cell 7e-Alberts firing frequency of neurons. Which of the following combinations results in the highest firing frequency? a. Rapid inactivation of both Na+ and K+ channels b. Rapid inactivation of Na+ channels, slow inactivation of K+ channels c. Slow inactivation of Na+ channels, rapid inactivation of K+ channels d. Slow inactivation of both Na+ and K+ channels

44. The aggregate current crossing a patch of plasma membrane represents the degree to which each individual channel is open at that time. a. True b. False

45. Which of the following descriptions matches an ABC transporter, a P-type pump, and a V-type pump? a. ABC transporter b. P-type pump c. V-type pump

1. The pumps in this family are phosphorylated at a key Asp residue in each transport cycle.

2. This family is the largest among membrane transport proteins and includes some channels as well as pumps.

3. The pumps in this family are responsible for the acidification of synaptic vesicles.

4. The sodium-potassium pump is a member of this family.

5. The multidrug resistance protein is a member of this family.

46. Astroglial cells in the brain can actively import the amino acid glutamate from the synaptic cleft, convert it to the amino acid glutamine, and release the glutamine into the cleft to be subsequently taken up by the presynaptic neurons. The neurons then convert it back to glutamate. Why is this function of glial cells important?

Molecular Biology of the Cell 7e-Alberts a. Because it competes with and replaces the direct uptake of glutamate by the neurons and thus regulates neuronal function b. Because neurons cannot produce the amino acids glutamine and glutamate and rely on the glial cells for their supply c. Because glutamate cannot be taken up directly by the neurons d. Because glutamate is an excitatory neurotransmitter and should be cleared from the synaptic cleft rapidly after its release e. Because glutamate import through glutamate–Na+ symporters in the glial cells is required for the maintenance of Na+ concentration in the synaptic cleft

47. Myelination of axons in the peripheral nervous system does what? a. It is carried out by oligodendrocytes. b. It insulates the axons to assist in current leakage. c. It increases the speed of action potential propagation through heat conduction. d. It conserves energy because the active excitations are restricted to the nodes of Ranvier.

48. What is NOT common between long-term potentiation and long-term depression? a. They are both important for synaptic plasticity in normal individuals. b. They both require the activation of the glutamate-gated NMDA receptors. c. They both require a rise in the intracellular concentration of Ca2+. d. They both result in an enhanced postsynaptic response. e. They can both last for a long time.

49. Dopamine is a neurotransmitter involved in the reward pathways in the brain, and

Molecular Biology of the Cell 7e-Alberts its decreased activity has been associated with diseases such as Parkinson’s disease and attention deficit/hyperactivity disorder (ADHD). Which of the following drugs is NOT a likely candidate to treat these diseases? a. Methylphenidate, a dopamine reuptake inhibitor b. Amphetamine, a dopamine-releasing agent that triggers the release of dopamine into the synaptic cleft c. Carbidopa, which enhances the availability of the dopamine synthesis precursors in the brain d. Forskolin, a sensitizer of dopamine receptors e. Chlorpromazine, a dopamine antagonist that binds to and inhibits dopamine receptors

50. What would cause prolonged signaling in a muscle cell at the neuromuscular junction? a. An acetylcholine reuptake inhibitor b. An acetylcholinesterase inhibitor c. An acetylcholine receptor antagonist d. Botulinum toxin, which prevents synaptic vesicles from fusing e. The downregulation of the acetylcholine receptor

51. The permeability of a protein-free lipid bilayer to various molecules depends on their properties. Sort the following in order of low to high permeability from left to right. Your answer would be a six-digit number composed of digits 1 to 6, for example, 123456.

1. O2 2. ATP 3. RNA oligonucleotide 4. Na+

Molecular Biology of the Cell 7e-Alberts 5. Glucose 6. Urea

52. In animal cells, the Na+-K+ pump transports both Na+ and K+ across the plasma membrane to establish electrochemical gradients. Using information about ion gradients, explain how this transporter is electrogenic. Explain how the antiport of positively charged ions can establish electrical potential across the membrane.

53. Most eukaryotic ABC transporters are involved in exporting small molecules from the cytosol. In this subset of the ABC transporters, does the small molecule bind better to the ATP-bound or the ATP-free transporter? Hint: The ATP-bound transporter “faces outward.”

54. Neither aqueous pores nor active transporters are gated. Why are—and, furthermore, why must—ion channels be gated?

55. In animal cells, the electrochemical gradient of K+ is nearly at equilibrium. Explain how this balance is achieved between the two components of the electrochemical gradient and what class of proteins is responsible for its maintenance.

56. How does the structure of a stretch-activated Piezo channel support its function?

57. Compare and contrast the similarities and differences between the signaling mechanisms of ionotropic and metabotropic neurotransmitter receptors.

58. Neuromuscular transmission involves the sequential activation of several ion channels. How would you order the sequence of actions A through E to correctly lead to muscle contraction? Your response should be a five-letter string, for example, ABCDE. A. Opening of voltage-gated Na+ channels in the muscle cell plasma membrane

Molecular Biology of the Cell 7e-Alberts B. Opening of Ca2+ release channels in the sarcoplasmic reticulum membrane C. Opening of ionotropic acetylcholine receptors on the muscle cell plasma membrane D. Opening of voltage-gated Ca2+ channels in the T tubules E. Opening of voltage-gated Ca2+ channels in the motor neuron

59. How would you order the sequence of actions A through E to correctly lead to long-term potentiation (LTP)? Your response should be a five-letter string, for example, ABCDE. A. Activation of AMPA receptors B. Glutamate release into the synaptic clefts C. Activation of calcium-dependent protein kinases in the postsynaptic cell D. Postsynaptic membrane depolarization E. Activation of NMDA receptors

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 11 1. Answer:

C

2. Answer:

D

3. Answer:

C

4. Answer:

A

5. Answer:

E

6. Answer:

C

7. Answer:

A

8. Answer:

C

9. Answer:

C

10. Answer:

B

11. Answer:

E

12. Answer:

C

13. Answer:

D

14. Answer:

C

15. Answer:

D

16. Answer:

D

17. Answer:

C

18. Answer:

B

19. Answer:

B

20. Answer:

C

Molecular Biology of the Cell 7e-Alberts 21. Answer:

D

22. Answer:

E

23. Answer:

A

24. Answer:

A

25. Answer:

D

26. Answer:

C

27. Answer:

A

28. Answer:

B

29. Answer:

E

30. Answer:

E

31. Answer:

C

32. Answer:

C

33. Answer:

E

34. Answer:

E

35. Answer:

B

36. Answer:

D

37. Answer:

D

38. Answer:

C

39. Answer:

1. A 2. A 3. A 4. A

40. Answer:

1. A

Molecular Biology of the Cell 7e-Alberts 2. B 3. B 4. A 41. Answer:

1. A 2. A 3. B 4. A

42. Answer:

B

43. Answer:

A

44. Answer:

B

45. Answer:

1. B 2. A 3. C 4. B 5. A

46. Answer:

D

47. Answer:

D

48. Answer:

D

49. Answer:

E

50. Answer:

B

51. Answer:

324561

52. Answer:

In animal cells, the electrochemical gradients of Na+ and K+ are established by the Na+-K+ pump with a higher concentration of Na+ outside the cell and a higher concentration of K+ inside. Each ion only carries a single positive charge, but because the pump exchanges 3 Na+

Molecular Biology of the Cell 7e-Alberts for 2 K+ each cycle, an electrical potential more positive outside can be established.

53. Answer:

ATP-free transporter

54. Answer:

Because of their more rigid structure, the central pore of ion channels does not change conformation significantly.

Instead, channels must have gates that can be conformationally changed to be opened or closed. If these channels were always open, the ion gradients could not be maintained.

55. Answer:

K+ leak channels allow K+ in or out of the cell, down its electrochemical gradient. The concentration of K+ is higher inside the cell, providing momentum for K+ to move outward, is counterbalanced by a higher positive charge outside the cell, providing momentum for K+ to move inward. These two forces are balanced as K+ moves through the K+ leak channels.

56. Answer:

The mono-trimer channel has three extended arms (or wings), joining in a central pore. The arms bend the membrane into an inward-facing dome, which flattens as the membrane is stretched. This flattening causes a conformational change, opening the pore and allowing passage of the particular ion.

57. Answer:

Ionotropic receptors are transmitter-gated ion channels, and as such, a very quick, simple, and brief signal is relayed (the ion signal can be turned on and off very quickly). In contrast, metabotropic receptors are GPCRs that are slower to generate a cellular response and therefore also slower to turn the signaling pathway off.

Both types of receptors bind to neurotransmitters such as acetylcholine, glycine, glutamate, or GABA as their ligand.

58. Answer:

ECADB

59. Answer:

BADEC

Molecular Biology of the Cell 7e-Alberts

Chapter 12: Intracellular Organization and Protein Sorting 1. Consider a human cell such as a hepatocyte. Which of the following compartments occupies a larger volume in the cell? a. Cytosol b. Nucleus c. Endoplasmic reticulum d. Mitochondria e. Peroxisomes

2. Consider a human liver hepatocyte. Among the following membranes, which one has the largest total area? a. Plasma membrane b. Nuclear inner membrane c. Mitochondrial outer membrane d. Rough ER membrane e. Smooth ER membrane

3. A significant fraction of total membrane area in the eukaryotic cell encloses the lumen of which cytosolic organelle? a. Ribosomes b. Nucleus c. Golgi apparatus d. Endoplasmic reticulum e. Mitochondria

Molecular Biology of the Cell 7e-Alberts 4. Liquid–liquid phase separation is important in the assembly of membrane-less compartments called biomolecular condensates. The formation of this intricate network is reliant on what kind of molecular forces? a. Gravitational b. Covalent bonds c. Monovalent interactions d. Glycosylation e. Hydrogen bonds

5. Cyclin B1, a key cell-cycle regulatory protein in vertebrates, is mostly cytosolic before mitosis. Early in mitosis, however, the protein is phosphorylated by certain protein kinases and consequently accumulates in the nucleus. How can phosphorylation bring about nuclear accumulation of this protein? a. Phosphorylation within the nuclear localization signal inhibits the function of the signal. b. Phosphorylation within the nuclear export signal enhances the function of the signal. c. Phosphorylation within the nuclear export signal interferes with the function of the signal. d. Phosphorylation elsewhere on the protein enhances binding to cytosolic proteins.

6. Imagine a protein that has been engineered to contain a nuclear localization signal, a nuclear export signal, a C-terminal peroxisomal targeting sequence, and a canonical endoplasmic reticulum (ER) signal sequence. With all of these signals, where would you expect to find the protein after its synthesis? a. Cytosol b. Nucleus c. Shuttling between the cytosol and the nucleus d. Peroxisomes

Molecular Biology of the Cell 7e-Alberts e. Endoplasmic reticulum

7. A geneticist has devised a strategy to study protein translocation into the endoplasmic reticulum (ER) in yeast cells. She is interested in two different signal sequences that are thought to operate via slightly different translocation mechanisms. Using genetic engineering, she has fused the first signal sequence to a protein whose cytosolic expression is absolutely necessary for cell survival in the selective medium but is inactive when in the ER. In the same cell, she has also fused the second signal sequence to a toxic protein whose cytosolic expression leads to cell lysis but is harmless when in the ER. Whereas wild-type cells undergo lysis upon the expression of these fusion proteins, she has been able to identify viable mutants, each of which has a loss-of-function mutation in a gene encoding a protein involved in membrane translocation. The products of these genes are probably involved in what? a. The general transport of proteins into the ER, regardless of the type of signal sequence b. The transport of proteins with the first signal sequence but not the second one c. The transport of proteins with the second signal sequence but not the first one d. The transport of proteins with a novel signal sequence, i.e., neither the first signal sequence nor the second one

8. The production of new organelles requires the DNA that encodes the organelle specific proteins and what else? a. Digestive enzymes b. Preexisting protein translocators c. Organelle structure d. Modified lipids e. Secretory proteins

9. Proteins of the protein disulfide isomerase (PDI) family share common domains

Molecular Biology of the Cell 7e-Alberts that harbor an active-site CXXC motif (C = cysteine; X = one of several residues).

During the course of a redox reaction, one of the active-site cysteines in its reduced form can attack a disulfide bond in a substrate protein (that could itself be another PDI family member) to form a mixed disulfide between the enzyme and its substrate. This is then attacked by the other active-site cysteine, releasing the substrate in reduced form. The reverse of these reactions can occur instead in order to make disulfide bonds in target proteins. CXXA mutant PDI family proteins—in which one of the active-site cysteines is mutated to an alanine—can be trapped in the intermediate mixed disulfide state for an elongated time, facilitating the identification of their substrate proteins.

Using these mutations in each of three interacting PDI family proteins—A, B, and C (one mutation at a time)—you have discovered that the mutant B interacts with A and C, while mutant A interacts with C but not with B. Finally, mutant C interacts with neither of the other two. The following diagram shows the thermodynamically favorable flow of electrons (plus protons) in the cascade involving these three proteins. Based on your results above, what proteins correspond to 1, 2, and 3 in the diagram, respectively?

a. ABC b. BAC c. ACB d. BCA

10. Which statement is true about the signal-recognition particle (SRP)? a. It is a heterodimeric protein. b. It transiently inhibits translation and polypeptide elongation by binding to and inhibiting the elongation factors. c. It accompanies the nascent polypeptide all the way into the ER lumen.

Molecular Biology of the Cell 7e-Alberts d. It binds GTP. e. It is permanently attached to the cytosolic face of the ER membrane, thus bringing the ribosomes into close proximity of the translocon.

11. You set up an in vitro translation system containing the entire translation machinery but devoid of any component of the endoplasmic reticulum (ER) targeting machinery. To this system, you can add mRNA encoding either a 20 kD secretory protein or a 20 kD cytosolic protein. You perform in vitro translation in the presence of radioactively labeled methionine, with or without the addition of saturating amounts of SRP or microsomes, as indicated below. After separating the protein products by SDS-PAGE, and visualizing the radioactivity by autoradiography, you obtain the following results. The presence or absence of each component in the reaction is indicated at the top of the corresponding lane(s) by + and –, respectively. The numbers on the left indicate the apparent molecular mass (×1000) of spots on the gel. Which protein (X or Y) is the secretory protein? Which of the reactions (1 or 2) contained SRP?

a. X; 1 b. X; 2

Molecular Biology of the Cell 7e-Alberts c. Y; 1 d. Y; 2

12. The Sec61 translocator facilitates the translocation of proteins that contain a signal sequence. What organelle are these proteins destined for? a. Golgi b. Nucleus c. ER d. Chloroplast e. Lipid

13. Which of the following is NOT a likely ER signal sequence recognized by the signal-recognition particle? All sequences are written with their N-terminus on the left. a. Met-Lys-Leu-Ser-Leu-Val-Ala-Ala-Met-Leu-Leu-Leu-Leu-Ser-Ala-Ala-ArgAla b. Met-Glu-Met-Phe-Gln-Gly-Leu-Leu-Leu-Leu-Leu-Leu-Leu-Leu-Ser-Met-GlyGly-Thr-Trp-Ala c. Met-Lys-Ala-Lys-Leu-Leu-Val-Leu-Leu-Tyr-Ala-Phe-Val-Ala-Gly-Asn d. Met-Met-Ala-Ala-Gly-Pro-Arg-Thr-Ser-Leu-Leu-Leu-Ala-Phe-Ala-Leu-LeuCys-Leu-Pro-Trp-Thr-Gln-Val-Val e. Met-Leu-Ser-Leu-Arg-Gln-Ser-Ile-Arg-Phe-Phe-Lys-Pro-Ala-Thr-Arg-ThrLeu-Ser-Ser-Arg-Tyr-Leu

14. When the hydrophobic transmembrane α helix at the C-terminus of an ER tailanchored protein emerges from the ribosome, the tail-anchored binding domain (TABD) of the Get3 ATPase binds to it. The job of TABD in Get3 is similar to that of the flexible hydrophobic domain in the SRP54 protein of the signal-recognition particle, in that it also has to recognize a degenerate set of hydrophobic αhelices. Considering the corresponding SRP4 domain, the Get3 binding site is

Molecular Biology of the Cell 7e-Alberts composed mostly of what type of residues? a. Tyrosine and tryptophan b. Serine and methionine c. Glycine and cysteine d. Methionine and glycine e. Aspartic acid and glutamate

15. The following diagram depicts the topology of a multipass transmembrane protein in the endoplasmic reticulum (ER) membrane. Which set of helices act as stoptransfer signals in this protein?

a. 1, 2, 3 b. 4, 5, 6 c. 1, 3, 5 d. 2, 4, 6 e. 1, 6

16. Consider the following hydropathy plot for a multipass transmembrane protein in the plasma membrane. Are the protein termini expected to be located inside (I) or outside (O) of the cell? Which of the regions indicated as 2 and 3 in the plot is expected to have more positively charged residues? The position of positively charged residues near the first transmembrane helix is indicated by an asterisk on the plot.

Molecular Biology of the Cell 7e-Alberts

a. I; 2 b. I; 3 c. O; 2 d. O; 3

17. Mitochondrial hsp70 relates to matrix protein import as what relates to posttranslational ER protein import? a. Ribosome b. Sec61 c. Sec63

Molecular Biology of the Cell 7e-Alberts d. BiP e. PDI

18. A protein is covalently attached to glycosylphosphatidylinositol. Which of the following is typically NOT true regarding this protein? a. The linkage of the anchor to the C-terminus of the protein occurs in the ER. b. The attachment of the anchor coincides with cleavage of a C-terminal transmembrane segment of the protein precursor. c. The protein is likely to be an ER resident, helping with the folding of nascent imported proteins. d. A phospholipase can cleave the protein from the membrane. e. The anchor affects the localization of the protein in the membrane.

19. Consider a transmembrane protein with the following topology that has an internal signal sequence (helix 1). If you fuse a canonical ER signal sequence at the Nterminus of this protein, how would you expect the topology to change? The ER lumen is at the bottom in all drawings. For simplicity, assume that the effect of charged residues flanking the transmembrane helices is negligible in this case.

Molecular Biology of the Cell 7e-Alberts

a.

b.

c.

d.

e.

Molecular Biology of the Cell 7e-Alberts

20. Calnexin, a calcium-dependent chaperone localized to the ER membrane, binds to folded proteins that contain what type of sugar modification? a. O-linked glycosidic linkage b. N-linked oligosaccharide c. O-linked oligosaccharide d. N-linked glycosidic linkage e. N-linked glucosidase

21. Rough microsomes can be subjected to a “salt extraction” procedure in which a high salt concentration is used to remove membrane-associated ribosomes and peripheral proteins. Such salt-extracted microsomes are known to be translocation-incompetent, meaning that when present co-translationally in vitro, they fail to protect translated proteins from protease digestion. However, adding back an 11S particle (S is the sedimentation coefficient) purified from the saltwash fraction is sufficient to restore the protein-translocation activity of the saltextracted microsomes. Which of the following do you think is true regarding the 11S particle? a. It is composed of 21 proteins. b. It is a digestion product of ER-associated ribosomes. c. It is normally assembled in the nucleus and exported to the cytoplasm by exportins. d. It is an ER integral membrane protein that can interact with the translocon. e. It requires high salt concentration for its function in vivo.

22. Misfolded proteins in the ER may actively undergo any of the following EXCEPT what? a. Binding to chaperones such as BiP to allow unfolding and refolding b. Binding to chaperones such as calnexin to prevent aggregation

Molecular Biology of the Cell 7e-Alberts c. Being transported back to the cytosol for degradation d. Being expelled to the Golgi apparatus for disposal e. Being glycosylated and binding to lectins

23. Organelle-mediated compartmentalization of the eukaryotic cell allows for the segregation and regulation of various reactions. Mechanisms have evolved to facilitate the exchange of materials, including the formation of membrane contact sites (MCS). Which of the following describes the main function of MCSs? a. Cytoskeleton networking b. Lipid biosynthesis c. Tethering one organelle to another organelle d. Neuron synapsis e. Apoptosis

24. Betaoxidation is the process of catabolizing fatty acids into acetyl CoA. Identify the cell organelle(s) responsible for this process in mammalian cells. a. Phospholipids and plasmalogens b. Mitochondria and peroxisomes c. Glyoxymes and chloroplasts d. Chloroplasts and mitochondria e. Peroxisomes and ER

25. Which reaction is normally catalyzed by the enzyme catalase in the peroxisome? a. 2O2– + 2H+ → H2O2 + O2 b. FMNH2 + O2 → FMN + H2O2 c. C2H5OH (ethanol) + NAD+ → C2H4O (acetaldehyde) + NADH + H+

Molecular Biology of the Cell 7e-Alberts d. 2H2O2 → 2H2O + O2 e. Both C and D

26. Below is an illustration of a chloroplast. What is the name of the area identified by a question mark?

a. Intermembrane space b. Inner membrane c. Thylakoid d. Stroma e. Thylakoid space

27. Certain cysteine-containing proteins of the mitochondrial intermembrane space are imported from the cytosol with the help of the Mia40 protein via a disulfide relay system. What drives the unidirectional import of these proteins? Are these proteins reduced or oxidized at their cysteine residues upon import? a. ATP hydrolysis; oxidized b. The electron-transport chain; oxidized c. ATP hydrolysis; neither oxidized nor reduced d. ATP hydrolysis; reduced e. The electron-transport chain; reduced

Molecular Biology of the Cell 7e-Alberts

28. Thylakoids are membrane-enclosed compartments found in the chloroplasts of photosynthetic eukaryotic cells. Translocation of proteins into the thylakoid lumen or membrane can occur by which pathway? a. Nuclear import pathway b. Endocytic pathway c. IRE1 pathway d. Sec pathway e. Unfolded protein response pathway

29. Which of the following proteins or protein complexes is directly required for the targeting of mitochondrial inner membrane multipass proteins, such as metabolite transporters, whose signal sequence is normally not cleaved after import? a. TIM22 b. TIM23 c. OXA d. Mia40 e. SAM

30. The import of nuclear-encoded mitochondrial proteins requires the use of protein translocators. Proteins spanning the mitochondrial outer membrane utilize the TOM complex for proper localization. Identify the translocator below responsible for the import of inner mitochondrial membrane proteins. a. TOM b. Hsp70 c. TIM23 d. OXA1

Molecular Biology of the Cell 7e-Alberts e. TIM22

31. Regarding protein import in a photosynthetic eukaryotic cell, targeting to which location requires two different signal peptides? a. Chloroplast b. Lysosome c. Nucleus d. Rough ER e. Golgi Apparatus

32. The mitochondria evolved through an endosymbiotic relationship between a Gram-negative ancestor and an ancestral host cell. The outer membrane of mitochondria and Gram-negative bacteria both contain β-barrel proteins, evidence of this evolutionary relationship. After synthesis in the cytosol, β-barrel proteins are transported to which mitochondrial compartment first? a. Mitochondrial inner membrane b. Mitochondrial outer membrane c. Chloroplast thylakoid d. Mitochondrial matrix e. Mitochondrial intermembrane space

33. Consider a transcription regulatory protein that has both a nuclear localization signal (NLS) and a nuclear export signal (NES) and is normally found both in the nucleus and in the cytosol at comparable concentrations. This protein has a highaffinity binding partner in the nucleus. Upon activation of a certain signaling pathway, the binding protein is ubiquitylated and degraded. What is the result? a. The transcription regulatory protein accumulates in the nucleus. b. The transcription regulatory protein accumulates in the cytosol.

Molecular Biology of the Cell 7e-Alberts c. The distribution of the transcription regulatory protein does not change, but the expression of its target genes may be altered. d. The distribution of the transcription regulatory protein does not change, and the expression of its target genes is not necessarily changed as a result of the degradation event.

34. According to the model for nuclear transport described in this chapter, what do you think would happen if you could artificially limit all Ran-GAP activity to the nucleus and all Ran-GEF activity to the cytosol? a. Proteins containing an NLS would be actively exported from the nucleus, while NES-containing proteins would be actively imported. b. Both import and export of nuclear proteins would be stalled, as they lose their directionality. c. Protein import into the nucleus would be reversed, but export would be unaffected. d. Protein import into the nucleus would be stalled, but export would be unaffected. e. Nothing would change; this is the normal Ran-GAP and Ran-GEF distribution.

35. The formation of a stable ternary complex involving Ran GTPase, a nuclear transport receptor, and a cargo protein occurs during which process(es)? a. Both nuclear import and export b. Nuclear import, but not export c. Nuclear export, but not import d. Neither nuclear import nor export

36. Almost all eukaryotic cells have plastids, but only plant cells have chloroplasts capable of photosynthesis. a. True

Molecular Biology of the Cell 7e-Alberts b. False

37. The signal-recognition particle is not the only factor that evaluates the authenticity of endoplasmic reticulum (ER) signal sequences in proteins. The Sec61 complex is also able to recognize the signal sequences and “opens” after binding to them.

Even single-point mutations within the signal sequence of a protein can render the protein unable to enter the ER efficiently, as the Sec61 complex does not readily open in response to the mutant sequence. However, “suppressor” mutations in genes encoding components of the translocation pathway, including the Sec61 subunits, can partially restore the wild-type localization of proteins with mutant signal sequences. Many such suppressor mutations (also called prl mutations) map to or near the “plug” domain in the Sec61 translocon. These mutations, including the deletion of the entire plug, generally result in destabilization of the closed conformation of the translocon and favor its open conformation.

Your friend has mutated a certain residue in the plug domain of the yeast Sec61.

She has just finished measuring the translocation efficiency of an ER protein with either a wild-type or a mutant (defective) signal sequence, either in wild-type or in her Sec61-mutant cells, and has obtained the following results. Based on these early results, the Sec61 mutation shows a prl phenotype.

Molecular Biology of the Cell 7e-Alberts

a. True b. False

38. Most proteins synthesized in the rough ER are N-glycosylated, and some of them require this modification for their correct folding. a. True b. False

39. All peroxisomal proteins are encoded in the nucleus. a. True b. False

Molecular Biology of the Cell 7e-Alberts 40. Signal sequences that target precursor proteins to the mitochondrial matrix form an α helix in which positively charged residues cluster near its N-terminus, while uncharged or hydrophobic residues cluster near the other side. a. True b. False

41. Tom40 is a nuclear-encoded essential subunit of the TOM complex in the outer mitochondrial membrane. It is a β-barrel protein that forms the pore through which precursor proteins enter the intermembrane space from the cytosol. Tom40 is translocated through the inner membrane as a precursor. a. True b. False

42. Most nuclear import receptors contain unstructured domains with FG-repeats. a. True b. False

43. Ribosomal proteins pass through the nuclear pore complexes twice: They are imported into the nucleus after synthesis and exported from the nucleus after assembly with ribosomal RNA. a. True b. False

44. Indicate whether each of the following descriptions better matches the rough or smooth endoplasmic reticulum (ER). a. Rough endoplasmic reticulum b. Smooth endoplasmic reticulum

1. It mostly has a tubular appearance.

2. It contains the transitional ER. 3. It is coated by ribosomes.

4. It can be specialized for functions such as detoxification and lipid metabolism.

Molecular Biology of the Cell 7e-Alberts 5. Sarcoplasmic reticulum in muscle cells is one of its specialized forms.

45. Indicate whether the C-terminus or the N-terminus of each of the following proteins is expected to be located in the cytosol upon membrane integration of the protein. a. C-terminus b. N-terminus

1. A single-pass transmembrane protein that has one N-terminal signal sequence and one internal stop-transfer signal 2. A single-pass transmembrane protein that has one internal signal sequence that is preceded by a patch of positively charged residues 3. An ER tail-anchored protein

46. Indicate whether the location of each of the following is topologically equivalent to the cytosol or the extracellular space. a. Cytosol b. Extracellular space

1. Ribosomes 2. Chromatin 3. Lysosomal hydrolases 4. Calcium ions in the ER 5. Peroxisomal catalase

47. Indicate whether each of the following transport processes occurs via the mechanisms described as gated transport, transmembrane transport, or vesicular transport. a. Gated transport b. Transmembrane transport c. Vesicular transport

1. Import into nucleus

Molecular Biology of the Cell 7e-Alberts 2. Export from nucleus 3. Import into mitochondria 4. Return from Golgi to ER 5. Return from ER to cytosol

48. Sorting signals are essential in the import of molecules in the cell. Which of the following scenarios would lead to the correct sorting of molecules to the nucleus? a. Phosphorylation causes a conformational change exposing the nuclear receptor tail. b. In a single pore, a nuclear localization signal (NLS)-containing protein is imported, while at the same time a nuclear export signal (NES)-containing protein is exported. c. A nuclear import receptor is modified in the cytosol. d. A protein containing an NLS is imported into the mitochondria.

49. In the following schematic diagram of a nuclear pore complex, on which side is the cytosol located? What is the approximate diameter of the pore, as indicated by the vertical bar?

a. Right; 1 µm b. Right; 0.1 µm c. Left; 1 µm d. Left; 0.1 µm

50. Consider two cells, A and B. Both are approximately spherical, but cell A is a

Molecular Biology of the Cell 7e-Alberts bacterium with a diameter of only about 1 µm, while the diameter of the eukaryotic cell B is about 10 µm. If the plasma membrane in the eukaryotic cell comprises only about 2% of the total cell membrane, then cell B has a higher ratio of total cell membrane to volume. a. True b. False

51. Indicate whether each of the following descriptions better matches the ATF6, the IRE1, or the PERK branch of the ER unfolded protein response. a. ATF6 b. IRE1 c. PERK

1. It involves a noncanonical cytoplasmic splicing process.

2. Its sensor is a latent transcription regulator.

3. It involves regulated proteolysis of the sensor protein in the Golgi apparatus.

4. Its sensor bears both kinase and endoribonuclease activities.

52. Indicate whether each of the following occurs on the cytosolic side of the ER membrane, on the luminal side of the ER membrane, or in the Golgi apparatus. a. Cytosolic side b. Luminal side c. Golgi apparatus

1. Assembly of GlcNAc oligosaccharides on dolichol phosphate 2. Synthesis of phosphatidylcholine 3. Synthesis of sphingomyelin 4. Ubiquitylation of misfolded ER protein 5. Synthesis of phosphatidic acid

53. Indicate whether each of the following descriptions refers to protein import into mitochondria, chloroplasts, or both.

Molecular Biology of the Cell 7e-Alberts a. Mitochondria b. Chloroplasts c. Both

1. ATP and GTP hydrolysis drive translocation into the organelle.

2. The organelle has an extra compartment that requires extra signal sequences for protein targeting.

3. Transport through the double membrane is driven in part by an H+ gradient across the inner membrane.

4. Imported precursor proteins have amphiphilic N-terminal signal sequences that are usually removed after use.

5. Hsp70 family chaperones inside the organelle assist in protein translocation during import.

54. In the following graph, the magnitude of concentration difference across the nuclear pore complexes (NPCs) is plotted for four molecules (A to D) as a function of time, starting from an arbitrarily chosen initial concentration difference.

Indicate which curve corresponds to each of the following molecules.

a. Curve A b. Curve B

Molecular Biology of the Cell 7e-Alberts c. Curve C d. Curve D

1. A large protein that is being actively transported across the NPC 2. A small water-soluble molecule 3. A small protein composed of a few dozen residues 4. A large protein that is NOT actively transported into or out of the nucleus

55. Scientists believe that the predecessors of the first eukaryotic cell were simple unicellular prokaryotes. These ancestral bacteria or achaeal cells had a plasma membrane but lacked internal membrane structures. Describe the events that ultimately lead to the internal membrane organization in eukaryotic cells.

56. The pyrenoid contains a condensate known to possess potent levels of the enzyme ribulose 1,5-bisphosphate carboxylase/oxygenase (Rubisco), which in turn plays a central role in carbon fixation during photosynthesis. The membrane tubules interwoven throughout this condensate hold carbonic anhydrase, a second enzyme that converts bicarbonate (HCO3 –) into CO2. In two or three sentences, explain why this process is often seen in algae living in aquatic environments rather than land plants.

57. Mitochondrial protein homeostasis (proteostasis) is crucial for the maintenance of cell health. Thus, the import of nuclear-encoded mitochondrial proteins from the cytosol are essential, serving as the result of the formation of a multi-layered regulatory system. Decreases in the protein-folding capacity, increased abundance of complex folded proteins, or formation of aggregates of nondegradable proteins results in the disruption of proteostasis. How does the cell regulate this response?

58. Post-translational modifications such as phosphorylation is essential to the regulation of many protein-mediated cellular processes, including the regulation of protein function and signal transduction. In two or three sentences, discuss the role that phosphorylation plays in regulating the formation and dissolution of biomolecular condensates.

Molecular Biology of the Cell 7e-Alberts 59. Sort the following events as they occur during the peroxisomal protein import cycle, starting with the release of cargo from Pex5. Your answer should be a string of letters, for example, ABCDE. A. Pex5 deubiquitylation B. Pex5 ubiquitylation C. Docking and translocation of the cargo protein along with Pex5 D. Pex5 export from the peroxisome with the help of ATPases Pex1 and Pex6 E. Pex5 binding to a cargo protein containing a C-terminal peroxisomal targeting sequence

60.

Which subset of the following is directly involved in driving protein import into the mitochondrial matrix space? Choose all correct sources. Your answer would be a string composed of letters A to G only, in alphabetical order, for example, AE. A. ATP hydrolysis inside mitochondria B. ATP hydrolysis outside mitochondria C. GTP hydrolysis inside mitochondria D. GTP hydrolysis outside mitochondria E. Light F. Membrane potential across the inner membrane G. Membrane potential across the outer membrane

61. The nucleus of the cell is surrounded by a nuclear envelope perforated by 3000– 4000 nuclear pore complexes (NPCs). NPCs transport approximately 1000 macromolecules per second and can support simultaneous bidirectional transport. Small molecules diffuse through the nuclear pore at a faster rate than larger molecules, such as nucleic acids and large proteins. With this information in mind, answer the following: 1.) How does the nuclear pore complex facilitate the transport of large molecules? 2.) Discuss the main method used by the cell to transport macromolecules into the nucleus.

Molecular Biology of the Cell 7e-Alberts

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 12 1. Answer:

A

2. Answer:

D

3. Answer:

D

4. Answer:

E

5. Answer:

C

6. Answer:

E

7. Answer:

B

8. Answer:

B

9. Answer:

B

10. Answer:

D

11. Answer:

A

12. Answer:

C

13. Answer:

E

14. Answer:

D

15. Answer:

D

16. Answer:

A

17. Answer:

D

18. Answer:

C

19. Answer:

D

20. Answer:

B

Molecular Biology of the Cell 7e-Alberts 21. Answer:

C

22. Answer:

D

23. Answer:

C

24. Answer:

B

25. Answer:

D

26. Answer:

D

27. Answer:

B

28. Answer:

D

29. Answer:

A

30. Answer:

C

31. Answer:

A

32. Answer:

E

33. Answer:

B

34. Answer:

A

35. Answer:

C

36. Answer:

B

37. Answer:

B

38. Answer:

A

39. Answer:

A

40. Answer:

B

41. Answer:

B

42. Answer:

B

Molecular Biology of the Cell 7e-Alberts 43. Answer:

A

44. Answer:

1. B 2. B 3. A 4. B 5. B

45. Answer:

1. B 2. B 3. A

46. Answer:

1. A 2. A 3. B 4. B 5. B

47. Answer:

1. A 2. A 3. B 4. C 5. B

48. Answer:

B

49. Answer:

E

50. Answer:

A

51. Answer:

1. B 2. A 3. A

Molecular Biology of the Cell 7e-Alberts 4. B 52. Answer:

1. A 2. A 3. C 4. A 5. A

53. Answer:

1. B 2. B 3. A 4. C 5. C

54. Answer:

1. B 2. D 3. C 4. A

55. Answer:

Plasma membrane protrusions increased the capacity of the ancestral cells to exchange resources, resulting in the evolution of larger cells. This also led to the fusion of the membrane protrusions and created internal organization and membrane enclosures.

56. Answer:

The key to why this process is often seen in algae living in aquatic environments rather than land plants is the amount of CO2 available in the environment. On land, CO2 is plentiful, and thus there is no need for a pyrenoid to fix carbon.

57. Answer:

Similar to the unfolded protein response found in the ER, the mitochondria also have an unfolded protein response. When the cell senses an imbalance in calcium levels, what results is a protective response. This response includes activation of signaling pathways, induction of key transcriptional changes of nuclear genes, and the upregulation in the expression of mitochondrial chaperones and proteases.

Molecular Biology of the Cell 7e-Alberts 58. Answer:

When a ligand binds to its membrane receptor, the cytosolic tail of the receptor is phosphorylated. This phosphorylation can lead to a conformational change in the receptor or other proteins associated with the receptor. This modification, along with surrounding amino acids, forms a specific binding site for various cytosolic and membrane proteins, many of which are multivalent. The multivalent proteins interact with each other to drive the formation of a condensate that has distinctive signaling properties. When the key sites become dephosphorylated, the condensate disassembles and signaling stops.

59. Answer:

BDAEC

60. Answer:

ABF

61. Answer:

1. Receptor-mediated mechanisms are used to transport macromolecules to move from the cytosol to the nucleus.

2. Both the nuclear compartment and the cytosol maintain different protein compositions. This generates a gradient that results in the movement of proteins from the cytosol to the nucleus and from the nucleus to the cytosol.

Molecular Biology of the Cell 7e-Alberts

Chapter 13: Intracellular Membrane Traffic 1. Consider a perfectly assembled clathrin cage that is composed of 12 pentagons and 20 hexagons. It resembles a soccer ball. How many clathrin heavy chains are there in this cage? a. 185 b. 160 c. 180 d. 165

2. Adaptor proteins select cargo proteins that will be incorporated into clathrincoated vesicles. Which of the following is not a characteristic of an adaptor protein, such as AP2? a. Can induce membrane curvature only after clathrin molecules bind b. Acts as a coincidence detector, assembling only when a number of specific requirements are met c. Binds to phosphoinositides in the cytosolic leaflet of the membrane of the transport vesicle d. Alternates between a locked cytosolic form and an unlocked membranebound form

3. Phosphoinositides mark different cellular membranes and play key roles in protein trafficking inside the cell. Among them, PI(4,5)P 2 is involved in receptormediated endocytosis as well as phagocytosis. Which statement about this particular phosphoinositide is NOT true? a. It is bound by the adaptor protein AP2. b. It is found in specific regions on the cytosolic face of the plasma membrane. c. It is depleted from clathrin-coated vesicles to promote their uncoating.

Molecular Biology of the Cell 7e-Alberts d. It possesses a total of two phosphate groups.

4. If this protein is unable to hydrolyze its bound GTP, invaginated clathrin-coated pits accumulate but fail to pinch off from the plasma membrane. In neurons, long vesicle necks collared by the protein are observed and presynaptic endocytosis is blocked. Which of the following characteristics is inconsistent with this protein’s identity? a. Contains a phosphoinositide-binding domain and a GTPase domain b. Induces membrane curvature via BAR domains c. May change the membrane lipid composition by recruiting lipid-modifying enzymes d. May directly distort the membrane using the energy from GTP hydrolysis

5. All of the following components bind the Rab proteins that guide and dock transport vesicles at their correct target membranes. If its Rab-binding domain was mutated, which component would be most likely to prevent transport vesicles from associating with actin filaments or microtubules? a. Coat proteins b. Tethering proteins c. Motor proteins d. GTP e. v-SNAREs

6. In the highly simplified diagrams below, the energy landscape of secretory vesicle fusion to the plasma membrane is modeled to represent various hypothetical conditions. Which diagram BEST represents a membrane fusion event mediated

by SNARE complexes?

Molecular Biology of the Cell 7e-Alberts

a.

b.

c.

Molecular Biology of the Cell 7e-Alberts

d.

7. The transmembrane protein Tango1 is a packaging protein that helps some secretory proteins leave the endoplasmic reticulum (ER) after synthesis.

Knocking down Tango1 by RNA interference impairs the incorporation of procollagen VII, but not procollagen I, into transport vesicles destined for the Golgi apparatus. The protein contains a lumenal N-terminal SH3 domain and a cytosolic C-terminal proline-rich domain. Which of the following proteins would you expect to interact with the N-terminal and C-terminal domains of Tango1, respectively? a. Procollagen VII; COPI components b. Procollagen VII; COPII components c. Procollagen I; COPI components d. Procollagen I; COPII components e. COPI components; procollagen I

8. You know that a particular sequence at the C-terminus of the lectin ERGIC53 enhances its exit from the endoplasmic reticulum (ER). You create two mutant versions of the protein, one without the suspected sequence (1) and one in which the sequence is replaced with an ER retention signal (2). You transfect a fibroblast cell line with a plasmid that encodes either wild-type ERGIC53 (0) or one of the engineered versions of it (1 or 2). After inducing the expression of the proteins, you lyse the cells and either treat the lysate with endoglycosidase H (Endo H) or leave the lysate untreated, as indicated below. You then separate the proteins by SDS-PAGE and perform a Western blot to detect the bands corresponding to ERGIC53. Your results are represented in the following drawing.

Molecular Biology of the Cell 7e-Alberts Which lanes (A to C) do you expect to correspond to each of the proteins 0, 1, and 2, respectively?

a. CAB b. ACB c. CBA d. ABC e. BAC

9. Which of the following descriptions better applies to v-SNAREs rather than tSNAREs? a. They are usually located on the target membrane.

Molecular Biology of the Cell 7e-Alberts b. They are usually composed of three proteins. c. They are composed of a single polypeptide chain. d. They bind a cognate SNARE on the vesicle membrane.

10. The association of a v-SNARE with its cognate t-SNARE to form a trans-SNARE complex is an energetically favorable interaction that drives membrane fusion.

Curiously, ATP is not directly required for trans-SNARE complex formation. How then is ATP involved in the membrane fusion process? a. ATP is hydrolyzed by NSF in order to pry apart trans-SNARE complexes and thus “reset” the v-SNARE and t-SNARE components for future membrane fusion events. b. A separate ATP-driven pump removes the water from between the membranes. c. Upon binding to each other, the SNARE proteins hydrolyze ATP to change their conformations and draw the two membranes closer together. d. Rab proteins hydrolyze ATP to release their tethers and allow the transSNARE complex to take over. e. ATP hydrolysis is used to form active t-SNARE trimers on the target membrane.

11. Which one of the following scenarios would allow a protein to leave the ER via its typical trafficking pathway? a. The protein’s ER exit signal is obscured by BiP. b. The protein has not yet completed folding. c. The protein is misfolded and marked for transport to the cytosol. d. The protein is a fully assembled complex with all of its polypeptide subunits. e. The protein is bound to the chaperone calnexin.

12. The cytoplasmic C-terminal KKXX sequence of transmembrane proteins interacts

Molecular Biology of the Cell 7e-Alberts with which one of the following? a. t-SNAREs b. v-SNAREs c. AP2 d. COPI coatomers e. COPII coatomers

13. In the following schematic diagram depicting the formation of vesicular tubular clusters between the endoplasmic reticulum (ER) and the cis Golgi network, what major coat proteins are indicated by 1 and 2, respectively?

a. COPI; COPII b. COPI; clathrin c. COPII; COPI d. COPII; clathrin e. Clathrin; COPII

14. Which of the following mechanisms prevents resident proteins in the ER and Golgi from entering transport vesicles and leaving their parent organelle?

Molecular Biology of the Cell 7e-Alberts a. Low concentrations (sub-millimolar) b. Protein aggregation c. Enzyme inactivation d. KDEL retrieval signals e. pH-sensitive protein interaction

15. The Golgi apparatus is made up of an ordered series of compartments. To process N-linked oligosaccharides, different Golgi compartments contain different enzymes that sequentially alter the sugar chains. Which Golgi cisternae are normally responsible for the addition of galactose and sialic acid, respectively, to complex oligosaccharides? a. cis cisterna; medial cisterna b. cis cisterna; trans cisterna c. Medial cisterna; trans cisterna d. Medial cisterna; medial cisterna e. trans cisterna; trans cisterna

16. Consider two N-linked oligosaccharide chains on the same protein. The first chain contains three mannose residues and three negatively charged sialic acid residues, as well as other residues. The second chain contains only two Nacetylglucosamine and eight mannose residues. Which chain would NOT be fully accessible to the processing enzymes in the Golgi apparatus? Which chain is Endo H-sensitive? a. Chain one; chain one b. Chain one; chain two c. Chain two; chain one d. Chain two; chain two

Molecular Biology of the Cell 7e-Alberts 17. Which of the following is an accurate description of one of the many purposes of glycosylation? a. Glycosylation discourages protein folding. b. The glycosylation state of a protein can determine its fate along the secretory pathway. c. Glycosylation makes a protein more sensitive to digestion by proteases. d. Glycosylated proteins are generally more flexible.

18. All but one of the following statements are similarities shared by M6P and KDEL receptors. Which statement represents a difference between the two? a. They both shuttle back and forth between different membrane-enclosed compartments. b. They are both transmembrane proteins. c. They both release their soluble targets at lower pH. d. They both prevent the escape of proteins to the cell exterior by the “default” pathway.

19. All but one of the following factors are required for sorting proteins like lysosomal hydrolases to lysosomes (via endosomes). Which factor is NOT involved in this process? a. M6P receptor b. GlcNAc phosphotransferase c. ATP-driven proton pump d. Clathrin coat e. COPI coat

20. What is the effect of defective or missing N-acetylglucosamine phosphotransferase on lysosomal protein sorting?

Molecular Biology of the Cell 7e-Alberts a. Lysosomal proteins are secreted from the cell. b. Lysosomal proteins are retained in the Golgi network. c. Lysosomal proteins are retained in the ER. d. Lysosomal proteins remain tightly bound to M6P receptors. e. Nonfunctional lysosomal proteins accumulate in the lysosome.

21. Insulin is a secretory protein made by the β cells in the pancreas. All but one of the following statements about insulin are true. Which statement is false? a. Insulin is released by the β cells in response to increased intracellular Ca 2+ concentration. b. Insulin is synthesized first as preproinsulin. c. Insulin is stored as aggregates inside secretory vesicles in the β cells. d. Insulin undergoes secretion before proteolytic cleavage.

22. Which of the following scenarios is consistent with a defect in the pathway for secretory vesicle transport? a. The vesicles fuse with the plasma membrane in response to a transient increase in the concentration of Ca2+ in the cytosol. b. The vesicles secrete their cargo when the cell receives a stimulatory signal like an action potential or hormone. c. The vesicles fuse with the plasma membrane and release their cargo immediately upon reaching their destination. d. The vesicles remain poised at, but not fused to, the plasma membrane when they first arrive there from the TGN. e. The vesicles follow microtubule tracks from the TGN to their site of export at the plasma membrane.

23. The membranes of synaptic vesicles contain specialized proteins such as

Molecular Biology of the Cell 7e-Alberts neurotransmitter transporters and the v-SNARE synaptobrevin. What happens to these membrane proteins when synaptic vesicles fuse with the plasma membrane? a. They are released into the synaptic cleft along with the neurotransmitters. b. They are retrieved from the plasma membrane and returned to the TGN for further processing. c. They momentarily become part of the plasma membrane and then are recycled almost immediately by endocytosis to form new synaptic vesicles. d. They remain within and contribute new functionality to the plasma membrane. e. They are removed from the plasma membrane by membrane-bound proteases.

24. How does the affinity of M6P receptor proteins for the mannose 6-phosphate marker change between the TGN and early endosome? Which coat protein is mainly responsible for their transport from the TGN to the endosome? a. The affinity is higher in the TGN; clathrin b. The affinity is higher in the TGN; retromer c. The affinity is higher in endosomes; clathrin d. The affinity is higher in endosomes; retromer

25. Which of the following descriptions better applies to the cisternal maturation model rather than the vesicle transport model for the organization of the Golgi apparatus? a. Golgi cisternae are static compartments. b. Golgi cisternae exchange material by both forward- and backward-moving transport vesicles. c. A cis Golgi cisterna becomes a medial cisterna which becomes a trans cisterna. d. Any protein that passes through the Golgi apparatus should be

Molecular Biology of the Cell 7e-Alberts incorporated into transport vesicles several times.

26. An animal cell has been wounded and has a small rupture in its plasma membrane. Which of the following is more likely to happen next? a. The cell rapidly cleaves by cytokinesis. b. The rate of receptor-mediated endocytosis is increased. c. The rate of exocytosis is increased. d. The rate of pinocytosis is increased.

27. Your research team recently acquired a new antibody from a commercial supplier. The antibody recognizes clathrin heavy chains. Regarding studying endocytosis, these anti-clathrin antibodies would find useful a list of materials from those outlined below. Which list of materials would that be? a. Solutes, macromolecules, and other nutrients b. Lipid rafts associated with cavin proteins c. Invaginations that provide a reservoir of additional plasma membrane d. Extracellular fluid taken up nonselectively e. Large particles ingested selectively

28. How can the formation of tubular structures, as opposed to spherical vesicles, in the early endosomes facilitate recycling of membrane proteins such as cellsurface receptors? a. By increasing surface area-to-volume ratio b. By decreasing surface area-to-volume ratio c. By enhancing the formation of intralumenal vesicles d. By preventing the formation of intralumenal vesicles

Molecular Biology of the Cell 7e-Alberts 29. What happens during maturation of early endosomes to late endosomes? a. The vacuolar domain of the endosome is shed, whereas the tubular domain is retained. b. The endosome migrates along actin filaments away from the cell interior. c. The lumen of the endosome becomes more acidic. d. Intralumenal vesicles disappear.

30. Which of the following processes is topologically similar to the formation of intralumenal vesicles in multivesicular bodies? a. Fission of peroxisomes b. Formation of clathrin-coated vesicles c. Virus entry into a host cell d. Cytokinesis

31. Lysosomes are the principal site of cellular digestion. What is one of their characteristics? a. They normally maintain a pH of about 2.0 to 2.5. b. They contain F-type ATPases that pump protons out of the organelles. c. They contain heavily glycosylated membrane proteins. d. They are homogeneous in size and shape. e. They receive all of their components from endocytic processes.

32. How many protons [or, more accurately, hydronium ions (H 3O+)] are there in a spherical lysosome that is about 0.6 µm in diameter and maintains an interior pH of about 5.0? Recall that the volume of a sphere of radius r is calculated as V = 4/3 × π × r3, and that pH = −log [H+]. Avogadro’s constant is approximately 6 × 1023 molecules/mole.

Molecular Biology of the Cell 7e-Alberts a. About 700 ions b. About 7000 ions c. About 70,000 ions d. About 700,000 ions e. About 7,000,000 ions

33. Why are lysosomes so heterogeneous in terms of both function and morphology? a. They are formed from relatively random processes that do not follow consistent pathways. b. They vary greatly in the amount and nature of the material they digest. c. Some contain acid hydrolases for digestion, others do not. d. They vary widely in pH, from mildly acidic to mildly alkaline. e. The material they digest is acquired randomly and often includes both intraand extracellular components.

34. All but one of the following are functions and/or characteristics of plant vacuoles.

Which of the following is NOT consistent with these organelles? a. They store nutrients and waste products. b. They help increase cell size and maintain turgor pressure. c. They are used for degradation of cytoplasmic components. d. They help maintain cytosolic pH. e. They comprise a relatively small proportion of the cell’s volume.

35. You are studying the cellular basis of petal coloration in the flowering plant Ipomoea tricolor. These plants have colorful petals due to the presence of pHsensitive vacuolar anthocyanins that change color from red/purple in acidic pH to blue at neutral pH. You treat petals with either vanadate (a specific inhibitor of P-

Molecular Biology of the Cell 7e-Alberts ATPases) or bafilomycin (a specific inhibitor of V-ATPases) or both and compare the color with that of control petals, obtaining the results shown in the table below.

Which of the following conclusions is consistent with these observations?

Treatment Color Control

Red

Vanadate

Blue

Bafilomycin

Blue

Vanadate + bafilomycin Blue

a. Both P- and V-ATPases are required to sufficiently acidify the vacuoles in petal cells. b. P-ATPases are sufficient for acidification of the vacuoles in petal cells. c. V-ATPases are sufficient for acidification of the vacuoles in petal cells. d. Neither P- nor V-ATPases are necessary for vacuole acidification in petal cells.

36. Which of the following pathways does NOT directly deliver materials to lysosomes? a. Endocytosis b. Exocytosis c. Phagocytosis d. Autophagy e. Macropinocytosis

37. Which term best describes the endocytic process depicted in the schematic diagram below?

Molecular Biology of the Cell 7e-Alberts

a. Receptor-mediated endocytosis b. Pinocytosis c. Macropinocytosis d. Phagocytosis e. Endocytosis

38. A mitochondrion has just been engulfed by a cup-shaped isolation membrane as it undergoes mitophagy. Before lysosomal fusion, how many membranes separate the cytosol from the matrix of the engulfed mitochondrion? a. 2 b. 3 c. 4 d. 5

39. Which of the following explains in part how the mTOR complex 1 regulates nonselective autophagy? a. It is activated by decreases in carbohydrate levels in the cell. b. When active, it stimulates the ATG1 kinase complex to initiate

Molecular Biology of the Cell 7e-Alberts autophagosome formation. c. It recruits ATG8 to autophagy-related vesicles. d. It is inhibited when the levels of certain amino acids become limiting. e. When active, it facilitates the homotypic fusion of autophagocytic vesicles.

40. In Drosophila melanogaster, loss-of-function mutations in either Pink1 or Parkin show similar phenotypes including impaired ability to fly, male sterility, and degeneration of dopaminergic neurons. Transgenic overexpression of Parkin in mutants lacking Pink1 significantly ameliorates the loss-of-function phenotype, but overexpression of Pink1 cannot rescue the Parkin loss-of-function phenotype.

According to these findings, which protein is more likely to act upstream of the other one? Assuming that the Parkin loss-of-function phenotype is merely due to defects in autophagy, would you expect Parkin overexpression to also rescue an ATG9 loss-of-function phenotype? a. Parkin acts upstream of Pink1; yes b. Parkin acts upstream of Pink1; no c. Pink1 acts upstream of Parkin; yes d. Pink1 acts upstream of Parkin; no

41. Reconstitution of vesicle transport in cell-free systems was historically carried out in the presence of isolated “donor” and “acceptor” Golgi stacks. The donor stacks are isolated from cells expressing a viral protein and lacking a processing enzyme. The viral protein can be transported to the acceptor stacks (that do contain the processing enzyme) and be processed only in the presence of an energy source (such as ATP) and a cytosolic fraction. The transport vesicles that mediate this process are visualized by microscopy. Various compounds can be added to this system to help understand the mechanism of transport. Two such compounds were used in early experiments in the 1980s, both of which blocked transport of the viral protein and resulted in the accumulation of transport vesicles in the cytoplasm. However, following treatment with one of these compounds (A), the accumulated vesicles appeared to be coated, while those resulting from treatment with the other compound (B) did not.

Based on this information, if one of these compounds is a nonhydrolyzable GTP analog and the other one is N-ethylmaleimide, then compound A better

Molecular Biology of the Cell 7e-Alberts represents the GTP analog than does compound B. a. True b. False

42. Large molecules such as collagen fibers of about 300 µm in length are highly packed and coiled to fit into normal COPII vesicles of about 80 nm in diameter. a. True b. False

43. Rab5 and Rab7 constitute a Rab cascade in the process of endosome maturation. One of the Rab5 effectors is a Rab7-GEF, while one of the Rab7 effectors is a Rab5-GAP. According to this explanation, early endosomes contain more Rab5 than Rab7 and late endosomes contain more Rab7 than Rab5. a. True b. False

44. The glycosylation state of a protein can determine its fate along the secretory pathway. a. True b. False

45. You have engineered a fusion protein composed of a cis Golgi resident protein and the green fluorescent protein. Similarly, you have made a fusion protein composed of a trans Golgi protein and the red fluorescent protein. You express these proteins in cells and follow individual Golgi cisternae using fluorescence microscopy. You observe that each individual cisterna emits a green, faint yellow, or red fluorescent signal at different times: it initially shows green fluorescence, but the green fluorescent signal fades away with time, and instead red fluorescence appears in the same cisterna. Additionally, you do not observe any red fluorescent cisterna that becomes green over time.

These observations agree better with the vesicle transport model rather than the cisternal maturation model for the organization of Golgi stacks. a. True b. False

Molecular Biology of the Cell 7e-Alberts

46. Aspartic acid residues resemble phosphorylated serine residues in proteins and also carry a similar negative charge. As a result, a protein in which serine residues are mutated to aspartic acid residues may mimic the function of the corresponding wild-type protein when those serines are phosphorylated.

Given that phosphorylation on serine residues in a number of Golgi matrix proteins occurs during mitosis, you would expect that mutation of these serines to aspartic acids would prevent Golgi stack formation in interphase. a. True b. False

47. Proteins destined for secretory vesicles for regulated secretion often aggregate in the lumen of the TGN. a. True b. False

48. A protein that is not normally expressed in polarized epithelial cells has been induced in these cells and has been engineered such that a GPI anchor is attached to it as it is processed in the ER. You would expect to find this protein in the apical domain of the plasma membrane in these cells. a. True b. False

49. A mutation that impairs the migration of LDL receptors into clathrin-coated pits would cause depletion of blood LDL levels. a. True b. False

50. In mammals, colostrum (or “first milk”) is produced by the mother in late pregnancy and shortly after giving birth to feed the newborn. In addition to nutrients, it is particularly rich in antibodies that are absorbed through the intestinal epithelium and support the weak immune system of the infant.

Occasionally, some infants suffer from either respiratory acidosis or alkalosis. In acidosis, the blood pH becomes acidic due to lack of sufficient ventilation, while

Molecular Biology of the Cell 7e-Alberts the opposite happens in alkalosis due to hyperventilation. Considering the molecular mechanism of transcytosis in intestinal epithelia, you would expect acidosis to interfere more with antibody absorption by transcytosis in these infants. a. True b. False

51. Indicate whether each of the following descriptions better applies to COPI-, COPII-, clathrin-, or retromer-coated vesicles. a. COPI-coated vesicles b. COPII-coated vesicles c. Clathrin-coated vesicles d. Retromer-coated vesicles

1. They mediate transport from the ER to the cis Golgi network.

2. They retrieve proteins from endosomes to the Golgi apparatus.

3. They are involved in retrograde transport in the Golgi apparatus.

4. Their coat protein forms a three-legged structure called a triskelion.

52. Match each of the indicated types of coat proteins in the diagram below with its corresponding COPI, COPII, clathrin, and retromer label.

Molecular Biology of the Cell 7e-Alberts a. A b. B c. C d. D

1. COPI 2. COPII 3. Clathrin 4. Retromer

53. Indicate whether each of the following descriptions better applies to COPI-, COPII-, or clathrin-coated vesicles. a. COPI-coated vesicles b. COPII-coated vesicles c. Clathrin-coated vesicles

1. They are uncoated by an hsp70 family protein, which is stimulated by the binding of auxilin.

2. They keep their coat proteins for a relatively long time, until they dock onto their target membrane.

3. Their uncoating depends on activation of an ARF-GAP. 4. Their uncoating depends in part on activation of a Sar1-GAP

54. A schematic drawing of the secretory and endocytic pathways is presented below. Indicate which component in the drawing (A to J) corresponds to each of the following.

Molecular Biology of the Cell 7e-Alberts

a. A b. B c. C d. D e. E f. F g. G h. H i. I j. J

1. Early endosome 2. Late endosome 3. ER 4. Lysosome

Molecular Biology of the Cell 7e-Alberts 5. cis Golgi cisterna 6. Medial Golgi cisterna 7. trans Golgi cisterna 8. cis Golgi network (CGN) 9. trans Golgi network (TGN) 10.Secretory vesicle

55. In the simplified diagram below depicting exocytosis of synaptic vesicles, which components (A to D) correspond to each of the following? Assume that an action potential has just arrived at the nerve terminal.

a. Component A b. Component B c. Component C d. Component D

1. Ca2+ 2. Complexin 3. Synaptotagmin 4. SNARE complex

56. Indicate whether each of the following descriptions better applies to receptormediated endocytosis, phagocytosis, pinocytosis, or macropinocytosis. a. Receptor-mediated endocytosis b. Phagocytosis

Molecular Biology of the Cell 7e-Alberts c. Pinocytosis d. Macropinocytosis 1. This “cell-drinking” pathway is mediated by clathrin-coated pits, or in some instances by caveolin and cavin proteins.

2. This process selectively captures and concentrates specific macromolecules within clathrin-coated pits and vesicles.

3. Some specialized cells use this “cell eating” pathway to engulf large particles that ultimately are degraded within lysosomes.

4. This nonselective process of fluid uptake utilizes the actin cytoskeleton to restructure the plasma membrane.

57. Indicate whether each of the following descriptions better applies to a BAR domain, the outer shell of COPI coatomer, or a dynamin oligomer. a. BAR domain b. COPI coatomer c. Dynamin oligomer

1. It resembles a crescent.

2. It resembles a collar.

3. It resembles a cage.

58. Indicate whether each of the following membrane proteins releases its protein ligand, remains bound to its protein ligand, or is degraded in the lysosome along with its protein ligand following receptor-mediated endocytosis. a. LDL receptor b. Transferrin receptor c. EGF receptor

1. Releases its protein ligand 2. Remains bound to its protein ligand 3. Degraded in the lysosome along with its protein ligand

59. Sort the following events to reflect the order in which they occur during the

Molecular Biology of the Cell 7e-Alberts formation of vesicles from the ER destined for the Golgi apparatus. Your answer will be a four-letter string composed of letters A to D only, for example, DACB. A. Sar1 GTP hydrolysis B. Sar1 GTP binding C. Sar1–Sec23 binding D. Sar1 membrane association

60.

Sort the following events to reflect the order in which they occur during vesicle docking onto a target membrane, starting with an inactive Rab in the cytosol.

Your answer will be a four-letter string composed of letters A to D only, for example, DACB. A. Rab is bound to its effector (tethering protein) on the target membrane.

B. Rab is bound to its Rab-GDI. C. Rab is bound to the membrane in its GTP-bound form.

D. Rab dissociates from the membrane.

61. Discuss the similarities and differences between the fusion of a vesicle with its target membrane and the fusion of an enveloped virus with its host cell membrane.

62. Explain how signaling receptors like opioid and growth factor receptors are downregulated. Begin your answer with an activated receptor in the plasma membrane and conclude with its degradation in the lysosome.

63. Compare and contrast the three pathways of protein sorting in the trans Golgi network: The constitutive pathway, the regulated pathway, and the endosomal pathway.

64. You identify and isolate a defective proteoglycan from the mucosal lining of the human esophagus. The defect appears to consist of a serine-to-alanine amino

Molecular Biology of the Cell 7e-Alberts acid substitution. Describe how this substitution might affect the proteoglycan’s processing in the Golgi apparatus prior to its secretion.

65. Summarize the various purposes and mechanisms used that cells carry out phagocytosis to ingest large particles. Be sure to include both protozoans and professional phagocytes like macrophages in your answer.

66. Sort the following events to reflect the order in which they occur during engulfment of an antibody-coated microorganism by a neutrophil in the blood.

Your answer will be a four-letter string composed of letters A to D only, for example, DACB. A. Rho-GEF activation B. Activation of Fc receptors C. Local PI(4,5)P2 production D. Local PI(3,4,5)P3 production

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 13 1. Answer:

C

2. Answer:

A

3. Answer:

D

4. Answer:

B

5. Answer:

C

6. Answer:

A

7. Answer:

B

8. Answer:

A

9. Answer:

C

10. Answer:

A

11. Answer:

D

12. Answer:

D

13. Answer:

C

14. Answer:

B

15. Answer:

E

16. Answer:

D

17. Answer:

B

18. Answer:

C

19. Answer:

E

20. Answer:

A

Molecular Biology of the Cell 7e-Alberts 21. Answer:

D

22. Answer:

C

23. Answer:

C

24. Answer:

A

25. Answer:

C

26. Answer:

C

27. Answer:

A

28. Answer:

A

29. Answer:

C

30. Answer:

D

31. Answer:

C

32. Answer:

A

33. Answer:

B

34. Answer:

E

35. Answer:

A

36. Answer:

B

37. Answer:

C

38. Answer:

C

39. Answer:

D

40. Answer:

D

41. Answer:

A

42. Answer:

B

Molecular Biology of the Cell 7e-Alberts 43. Answer:

A

44. Answer:

A

45. Answer:

B

46. Answer:

A

47. Answer:

A

48. Answer:

A

49. Answer:

B

50. Answer:

A

51. Answer:

1. B 2. D 3. A 4. C

52. Answer:

1. D 2. C 3. A 4. B

53. Answer:

1. C 2. B 3. A 4. B

54. Answer:

1. H 2. I 3. A 4. J 5. C

Molecular Biology of the Cell 7e-Alberts 6. D 7. E 8. B 9. F 10. G 55. Answer:

1. A 2. C 3. B 4. D

56. Answer:

1. C 2. A 3. B 4. D

57. Answer:

1. A 2. C 3. B

58. Answer:

1. A 2. B 3. C

59. Answer:

BDCA

60. Answer:

BCAD

61. Answer:

The fusion of a vesicle with its target membrane is mediated

by SNARE proteins, whereas the fusion of an enveloped virus with its host cell membrane requires viral fusion proteins. SNARE-mediated fusion involves proteins in both membranes, while viral fusion requires factors provided by the virus. The two fusion mechanisms are similar in that both include a compaction process that brings the membranes close together to drive out water and coalesce the two

Molecular Biology of the Cell 7e-Alberts bilayers.

62. Answer:

The activated receptors are tagged on their cytolic domains with one or a few copies of ubiquitin (mono- or multiubiquitylation, respectively). The ubiquitin tags then are bound by ubiquitin-binding proteins that collect the modified receptors into clathrin-coated pits for endocytosis. and delivery to the early endosome. Here, with the help of ESCRT complexes, the ubiquitin tags prevent the modified receptors from recycling to the plasma membrane. Instead, the receptors are taken up into intralumenal vesicles within the endosome (now a multivesicular body). This process deactivates the signaling pathway as the endosome matures into a lysosome for receptor degradation.

63. Answer:

Proteins that follow the constitutive pathway lack specific signals for post-Golgi trafficking and thus by default are routed directly to the cell surface. This pathway operates continually in all cells. In specialized cells, secretory proteins that follow the regulated pathway have specific signals that cause them to aggregate within the TGN for packaging and storing within secretory vesicles. The vesicles undergo exocytosis when the cell is stimulated to secrete their contents. Proteins like lysosomal hydrolases that follow the endosomal pathway receive a special mannose 6-phosphate sugar on their N-linked oligosaccharide groups. This signal routes the protein to endosomes that are destined to become lysosomes.

64. Answer:

Proteoglycans receive O-linked glycosaminoglycan chains in the Golgi prior to their secretion. The glycan chains are added to the hydroxyl groups of serine or threonine residues in the core protein. Substitution of a serine with an alanine will prevent the protein from receiving all of its O-linked sugars because alanine lacks a hydroxyl group in its side chain.

65. Answer:

In single-celled protozoa like amoeba, the process of phagocytosis is a feeding mechanism. The cells ingest nutritious debris and other microorganisms into large endocytic vesicles called phagosomes. The phagosomes then fuse with lysosomes and the resulting digestion products are transported to the cytosol as food. In multicellular organisms like humans, professional phagocytes like macrophages and neutrophils use phagocytosis to defend against infection and to remove senescent and dead cells. The general processes of phagosome formation, lysosome fusion, and digestion are

Molecular Biology of the Cell 7e-Alberts similar to what occurs within protozoans.

66. Answer:

BACD

Molecular Biology of the Cell 7e-Alberts

Chapter 14: Energy Conversion and Metabolic Compartmentation: Mitochondria and Chloroplasts 1. You have isolated mitochondria from a liver tissue sample, suspended them in a hypotonic buffer that causes them to swell and burst their outer membrane, and then added sucrose to a final concentration of about 25%. You then layer the suspension on a density gradient, ultracentrifuge, collect the different fractions, and analyze their protein content. Which proteins would you expect to be highly enriched in the lower density and higher density fractions, respectively? a. Cytochrome oxidase; porin b. Cytochrome oxidase; ATP synthase c. ATP synthase; cytochrome oxidase d. Porin; ATP synthase

2. Regarding the fusion and fission of mitochondria, which statement is NOT correct? a. They both require dynamin-related GTPases. b. Fission normally proceeds in a single step in which both outer and inner membranes are severed. c. They both proceed in a single step in which both outer and inner membranes are fused or severed. d. They both require GTP hydrolysis for force generation.

3. A respiring mitochondrion will passively export to the cytoplasm which of the following? a. O2 b. H+ c. ADP d. CO2

Molecular Biology of the Cell 7e-Alberts

4. In addition to their respiratory function, mitochondria have other important roles in cellular metabolism. Which process is NOT carried out mainly by mitochondria? a. Biosynthesis of cardiolipin b. Biosynthesis of fatty acids c. Biosynthesis of heme d. Biosynthesis of iron–sulfur clusters

5. Which phospholipid species, synthesized in the mitochondrial inner membrane, contributes structurally to the curvature of the crista membrane? a. Phosphatidylcholine b. Cardiolipin c. Phosphatidylinositol d. Phosphatidylserine e. Phosphatidylethanolamine

6. Compared to the highly energetic single-step combustion of gaseous hydrogen and oxygen combining into water, oxidative phosphorylation allows the cell to store approximately how much of the energy released by the single step reaction? a. Equal b. Double c. Half d. One-third e. One-tenth

7. The membrane potential across the inner mitochondrial membrane is −180mV,

Molecular Biology of the Cell 7e-Alberts consisting of a difference of −150mV electrical charge relative to the cytosol/intermembrane space, and −30mV attributed to the pH difference. You experimentally insert an electrode into the mitochondrial matrix and inject a current to alter the voltage. What injected current would eliminate the proton-motive force? a. −180mV b. −30mV c. 30mV d. 150mV e. 180mV

8. Which electron transport cofactor type has the lowest redox potential? a. Iron-sulfur clusters b. Quinones c. Heme cytochromes d. Copper atoms e. Flavin mononucleotides

9. In what way is Complex II similar to Complex I in the electron transport chain? a. It pumps protons across the inner mitochondrial membrane. b. It is composed of a membrane arm and a matrix arm. c. It accepts electrons from FADH2. d. It donates electrons, reducing ubiquinone to ubiquinol. e. It also carries out a reaction in the citric acid cycle.

10. Which of the following is a hypothesized benefit for the association of electron transport chain proteins into supercomplexes in the inner mitochondrial membrane?

Molecular Biology of the Cell 7e-Alberts a. Prevention of redox side reactions to avoid oxidative damage to the cell. b. Eliminating the need for electron transfer via mobile carriers to speed up electron transfer. c. Spreading the redox reactions over a larger surface of inner membrane to increase proton pumping efficiency. d. Creating a strongly positively charged local microenvironment to attract more electrons. e. Forming proton wires to transmit H+ more efficiently across the inner membrane.

11. The free-energy change of ATP hydrolysis to ADP inside the cell is about −50 kJ/mole. The standard free-energy change for the same reaction is about −30 kJ/mole. This necessitates that [ATP] inside the cell be which of the following? a. Higher than that under standard conditions b. Greater than [ADP] c. Greater than [ADP] × [Pi] d. Greater than [ADP] × [Pi] × Keq e. Greater than [ADP] × [Pi] / Keq

12. The free-energy change for ATP hydrolysis to ADP inside the cell is about −50 kJ/mole. The complete cellular oxidation of glucose to carbon dioxide and water has a free-energy change of about −3000 kJ/mole. What is the efficiency of the cellular metabolism in harvesting this amount of free energy in the form of ATP synthesis? a. About 10% b. About 20% c. About 30% d. About 50%

Molecular Biology of the Cell 7e-Alberts e. About 70%

Consider the structure of the F1Fo ATP synthase shown in the below schematic drawing and answer the following questions.

13. With about 7000 c-ring revolutions per minute, on average how many moles of these pumps should be working at each moment in an entire human body to make about 50 kilograms of ATP per day? The molecular weight of ATP is about 500. a. About 300 millimoles b. About 3 millimoles c. About 30 micromoles d. About 0.3 micromoles

14. As seen from the F1 head in the matrix, the rotation of the c ring is counterclockwise in the ATP-synthesis mode of the motor and clockwise in the ATP-hydrolysis mode of the motor. As shown in the above drawing, the “a subunit” (labeled 2) is part of the stator embedded in the membrane, and it forms the entry and exit sites for the protons that are bound by the rotating c ring. How do you think the two half-channels of the a subunit are arranged as seen from the c ring? In 2–3 sentences, explain whether your choice is A or B, as illustrated in the below drawing.

Molecular Biology of the Cell 7e-Alberts

15. In Escherichia coli, the c ring of ATP synthase is composed of 12 c subunits. For one fully productive revolution of this ring, how many protons are transported across the membrane for each molecule of ATP produced? Your answer should be an answer that is rounded to the nearest integer, for example, 17.

16. ATP synthase molecules in mitochondria form dimers that are localized mostly to sharp cristae ridges. What is most likely to happen if subunits of the synthase that are required for dimerization are mutated in yeast? a. Oxygen consumption would increase. b. Cristae ridges would become sharper. c. Cell growth would increase. d. Monomeric ATP synthases would distribute randomly over the inner membrane. e. ATP synthesis would speed up.

17. In the following schematic drawing of the inner and outer mitochondrial membranes, in what region (A to D) would you expect to find more ATP synthase dimers?

a. Region A b. Region B

Molecular Biology of the Cell 7e-Alberts c. Region C d. Region D

18. In actively respiring mitochondria, where in the following schematic drawing of the inner and outer mitochondrial membranes would you expect to find the lowest local pH?

a. Region A b. Region B c. Region C d. Region D e. Region E

19. The synthetic toxin 2,4-dinitrophenol can uncouple ATP synthesis from mitochondrial respiration by increasing the permeability of the inner membrane to protons. What would be the effect of dinitrophenol treatment on the amount of ATP produced by mitochondria and on the rate of ATP transport across the inner membrane, respectively? a. Positive; positive b. Positive; negative c. Negative; positive

Molecular Biology of the Cell 7e-Alberts d. Negative; negative

20. Which of the following molecules cannot serve as the terminal electron acceptor in bacterial electron-transport chains? a. Oxygen b. Sulfate c. Fumarate d. Nitrite e. NAD+

21. Which of the following descriptions applies to dark reactions rather than light reactions in plant chloroplasts? a. They produce the sugar glyceraldehyde 3-phosphate. b. They involve the electron-transfer chain embedded in the thylakoid membrane. c. They involve O2 production. d. They generate ATP.

22. In the following simplified diagram of the Calvin–Benson–Bassham cycle (Calvin cycle), which step (A to E) is catalyzed by the abundant enzyme Rubisco?

Molecular Biology of the Cell 7e-Alberts

a. Step A b. Step B c. Step C d. Step D e. Step E

23. What is true about photorespiration in plants? a. It occurs under conditions of high CO2. b. It results from ribulose 1,5 bisphosphate carboxylase using HCO 3- instead of CO2 as a reactant. c. It can be greatly reduced in C4 species by using ATP to enrich CO2 in specialized photosynthetic cells. d. It tends to occur in cool, damp conditions. e. It is another term for the “light reactions” of photosynthesis.

Molecular Biology of the Cell 7e-Alberts 24. Which pair of molecules carries out equivalent functions in chloroplasts and mitochondria, respectively? a. Cytochrome b6-f complex and cytochrome c b. Plastoquinone and chloroplast ATP synthase c. Cytochrome b6-f complex and cytochrome c reductase (Complex III) d. Antenna complex and NADH dehydrogenase (Complex I) e. Photosystem II and succinate dehydrogenase (Complex II)

25. When the special pair in a photosystem is excited by a quantum of light, charge separation can occur. Where does this take place? What is the charge of the ionized chlorophyll? a. In the antenna complex; positive b. In the antenna complex; negative c. In the reaction center; positive d. In the reaction center; negative

26. Regarding light-harvesting complexes in plant chloroplasts, which of the following is false? a. They contain chlorophyll and other pigments. b. They are found in both photosystem I and photosystem II. c. They can protect the cell by preventing the generation of reactive oxygen species. d. They can carry out charge separation. e. They do not contain special pair chlorophyll.

27. Photosynthesis provides the strongest known electron donor as well as the strongest

Molecular Biology of the Cell 7e-Alberts known electron acceptor in all biology. What do you think the donor and the acceptor are, respectively? a. Excited P700; excited P680 b. Excited P700; ionized P680 c. Ionized P700; excited P680 d. Ionized P700; ionized P680

28. The electrons used in carbon fixation by chloroplasts ultimately come from what? a. Atmospheric oxygen b. Atmospheric carbon dioxide c. Water d. Glyceraldehyde 3-phosphate

29. The water-splitting step in photosynthesis does which of the following? a. It occurs on the stromal side of the thylakoid membrane. b. It is catalyzed by an iron–sulfur cluster. c. It consumes H+ and therefore contributes to the proton gradient across the thylakoid membrane. d. It generates all the O2 in the Earth’s atmosphere.

30. The cytochrome b6-f complex that transfers electrons between the two photosystems in photosynthesis is structurally and functionally similar to which of the complexes in the mitochondrial electron-transport chain? a. NADH dehydrogenase b. Succinate dehydrogenase

Molecular Biology of the Cell 7e-Alberts c. Cytochrome c reductase d. Cytochrome c oxidase e. ATP synthase

31. Which component of the chloroplast electron transport chain donates electrons to reduce ferredoxin? a. Photosystem II b. Antenna chlorophyll c. P680 chlorophyll d. Plastocyanin e. Photosystem I

32. Which of the following does NOT occur in the normal process of light-driven production of ATP and NADPH in plant chloroplasts? a. The cytochrome b6-f complex pumps protons from the stroma into the thylakoid lumen. b. The oxygen-evolving enzyme generates protons in the thylakoid lumen. c. The ferredoxin-NADP+ reductase generates protons in the thylakoid lumen. d. The ATP synthase transports protons from the thylakoid lumen to the stroma.

33. Some photosynthetic bacteria have only one type of photosystem in their plasma membrane. Which of the following is likely to be true regarding such bacteria? a. They absorb purple light only. b. Their photosystem is unrelated to those in cyanobacteria and plants. c. They do not use water as the electron donor.

Molecular Biology of the Cell 7e-Alberts d. They lack ATP synthase.

34. What is the main reason for why atmospheric oxygen levels did not rise immediately after the mass production of the gas by photosynthetic bacteria? a. Because oxygen was rapidly consumed by aerobic microorganisms. b. Because oxygen was dissolved in water. c. Because large amounts of ferrous ions reduced the oxygen molecules. d. Because oxygen was rapidly recycled by early photosynthetic bacteria.

35. Which of the following is NOT universally encoded by the mitochondrial DNA? a. Small ribosomal RNA b. Large ribosomal RNA c. A cytochrome oxidase subunit d. Transfer RNA

36. Two haploid budding yeast cells are allowed to mate. One of them carries a mutation in its mitochondrial DNA that makes the yeast cell resistant to an antifungal drug. If the resulting diploid zygote is allowed to propagate (in the absence of the drug), how do you predict that the fraction of drug-resistant cells will change over time in the population? Choose the best curve (A to F) from the following graph. Note that the “population” is only composed of one cell at the beginning of the experiment. The curves are smoothened to cancel stochastic fluctuations that happen at such low population sizes.

Molecular Biology of the Cell 7e-Alberts

a. Curve A b. Curve B c. Curve C d. Curve D e. Curve E

37. In mammals, mitochondrial gene inheritance can be described by all of the following, except which statement? a. They are inherited in a non-Mendelian fashion. b. They are maternally inherited.

Molecular Biology of the Cell 7e-Alberts c. They are cytoplasmically inherited. d. They are biparentally inherited. e. They are enriched in oocytes.

38. Which of the following probably occurred first on Earth? a. H2O photosynthesis b. H2S photosynthesis c. O2 respiration d. Eukaryotic photosynthesis

39. Who has a different mitochondrial DNA in a family? a. The son b. The father c. The mother d. The daughter e. The maternal grandmother

40. Which description applies to organelle genomes? a. They are seen in mitochondria, chloroplasts, and peroxisomes. b. They encode all the proteins necessary for independent function of the organelle. c. They include the proteinaceous components of organelle ribosomes. d. They may have lost some genes to the nuclear genome and retained others, in what may be an ongoing evolutionary process. e. They preferentially retain all hydrophobic proteins, which are too energetically

Molecular Biology of the Cell 7e-Alberts costly to import from the cytosol.

41. The outer mitochondrial membrane is freely permeable to ions and small molecules. a. True b. False

42. Plant cells can generate starch granules in their chloroplasts using glyceraldehyde 3phosphate. a. True b. False

43. The first photosynthetic cyanobacteria evolved before the first aerobic proteobacteria. a. True b. False

44. Regarding the genetic systems of mitochondria and chloroplasts, the genetic systems of these organelles are most similar to extremophilic archaea. a. True b. False

45. Which description better matches the outer mitochondrial membrane, the intermembrane space, the crista membrane portion of the inner mitochondrial membrane, and the matrix, respectively? a. Outer mitochondrial membrane b. Intermembrane space c. Crista membrane portion of the inner mitochondrial membrane d. Matrix

Molecular Biology of the Cell 7e-Alberts 1. It is where NADH is produced.

2. It is composed of 75% protein by weight.

3. It is where the respiratory chain is located.

4. It has the same electrochemical potential for H+ as the cytoplasm.

46. Which of the following descriptions better applies to cytochrome c oxidase complex, cytochrome c reductase complex, or NADH dehydrogenase complex? a. Cytochrome c oxidase complex b. Cytochrome c reductase complex c. NADH dehydrogenase complex

1. It is the largest of the three respiratory complexes.

2. It accounts for the majority of O2 uptake in most cells.

3. It employs the Q cycle to increase proton pumping.

4. It contains separate modules for electron transport and proton pumping.

47. Which descriptions better apply to photosystem I, and which better apply to photosystem II? a. Photosystem I b. Photosystem II

1. It reduces plastoquinone.

2. It uses a manganese cluster to oxidize water.

3. It is confined to unstacked thylakoids.

4. It contains the special pair P680.

48. Which of the following statements better describes ATP production in chloroplasts, and which in mitochondria? a. Chloroplasts b. Mitochondria

1. The electrochemical gradient that drives ATP production is dominated by the

Molecular Biology of the Cell 7e-Alberts electrical component.

2. It involves a significant pH difference across the membrane.

3. The ATP synthase molecules form dimers.

49. Which of the following descriptions better applies to the genome of mitochondria, and which to the nuclei in vertebrates? a. Mitochondria b. Nuclei

1. It has a higher percentage of noncoding DNA. 2. It has a stricter codon usage.

3. Its genes are present at higher copy numbers per cell.

4. It has infrequent regulatory DNA sequences.

50. Consider the schematic drawings below of a mitochondrion and a chloroplast. Which letters (A to D) correspond to which features?

a. Region A b. Region B c. Region C d. Region D

Molecular Biology of the Cell 7e-Alberts

1. DNA 2. Stroma 3. Matrix 4. Thylakoid

51. Consider the remarkable structure of the F1Fo ATP synthase shown in the following schematic drawing. Answer the following question(s) based on the structure.

a. Rotary b. Stationary

1. Component 1 2. Component 2 3. Component 3 4. Component 4 5. Component 5

52. Which of the following descriptions better applies to chloroplasts, and which to mitochondria? a. Chloroplasts b. Mitochondria

1. Extensive editing occurs on their RNAs.

Molecular Biology of the Cell 7e-Alberts 2. They utilize dynamin-like GTPases to divide from the outside.

3. The organization of gene clusters in their genome is strikingly similar to cyanobacteria.

4. They typically have larger genomes.

5. They have a more ancient endosymbiotic relationship with their host.

53. Suppose there is a mutation that reverses the insertion orientation of all innermembrane-bound mitochondrial proteins EXCEPT ATP synthase. Assume mobile carrier ETC components are also present in the matrix space. How would you predict the synthase to respond?

54.

Organisms that express a gene for alternative oxidase (AOX) have mitochondria that are able to resist the toxin cyanide. Cyanide typically blocks respiration by preventing complex IV (cytochrome c oxidase) from reducing O2. If AOX is reduced by ubiquinol, describe the likely impact of cyanide exposure on ATP production in cyanide-resistant mitochondria. (Note: In the figure below, dashed arrows represent processes that are not occurring under conditions where both AOX and cyanide are present in the mitochondrion.)

Molecular Biology of the Cell 7e-Alberts 55. Sort the following molecules to reflect the order in which they carry electrons in the respiratory chain. Your answer would be a seven-letter string composed of letters A to G only, for example, GFADCBE. A. NADH dehydrogenase complex B. Cytochrome c oxidase complex C. Cytochrome c reductase complex D. Cytochrome c E. Ubiquinone F. NADH G. H2O

56. Cyanide is a toxin that blocks respiration at the end of the mitochondrial electron transport chain by preventing electron transfer from Complex IV (cytochrome c oxidase) to reduce O2. Would the presence of cyanide be likely to lead to an increase in ubisemiquinone radicals? Why or why not?

57. Rotenone is a pesticide that disrupts mitochondrial function by inhibiting the transfer of electrons from Complex I (NADH dehydrogenase) to ubiquinone in the electron transport chain. Would the presence of rotenone be likely to lead to an increase in ubisemiquinone radicals? Why or why not?

58. If a spontaneous mutation in a mitochondrion resulted in a 50% reduction of the ATP/ADP antiporter's activity, what would be the consequence for that mitochondrion's ADP and ATP levels in the matrix?

59. A facultative anaerobic bacterium can survive both aerobic and anaerobic conditions.

Under both conditions, the cell maintains a proton-motive force across the plasma membrane that drives many cellular processes including nutrient import. Under aerobic conditions, the proton-motive force is produced by a respiratory chain. What is the mode of action of the plasma membrane ATP synthase in this bacterium under aerobic conditions: ATP-hydrolysis mode or ATP-synthesis mode? Explain your answer in 2–3 sentences.

60. Sort the following molecules to reflect the order in which they transfer electrons in

Molecular Biology of the Cell 7e-Alberts noncyclic photophosphorylation in plants. Your answer would be an eight-letter string composed of letters A to H only, for example, FCDEHGBA. A. Ferredoxin-NADP+ reductase B. Plastoquinone C. Ferredoxin D. Cytochrome b6-f complex E. H2O F. Plastocyanin G. Photosystem I H. Photosystem II

61. Order the following metabolic innovations to reflect the most likely order in which they evolved during biological evolution. Your answer would be a four-letter string composed of letters A to D only, for example, BDCA. A. Using light energy to generate reducing power B. Splitting of water molecules (an electron source) C. Fermentation of organic material D. Reduction of molecular oxygen (a terminal electron acceptor)

62. Mitotic segregation of organelles can lead to so-called homoplasmy, when all organelles in each cell carry the same genome. In contrast, heteroplasmy describes the presence of organelles within the same cell that are different with respect to their genomic sequences. Would you expect a defect in the mitochondrial fusion machinery to favor homoplasmy or heteroplasmy? Explain your answer in 2–3 sentences.

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 14 1. Answer:

D

2. Answer:

C

3. Answer:

D

4. Answer:

B

5. Answer:

B

6. Answer:

C

7. Answer:

E

8. Answer:

A

9. Answer:

D

10. Answer:

A

11. Answer:

E

12. Answer:

D

13. Answer:

C

14. Answer:

Illustration B is the answer, because in the ATP-synthesis mode, protons move from the cristae space into the matrix, which rotates the c ring counterclockwise. Consequently, as seen in part B, protons enter the right half-channel, bind to a c subunit, rotate along with the c ring, and only exit from the left half-channel after a complete turn of the ring. If the channels were arranged as in part A, the protons would pass through the membrane only after a small rotation of the ring. Many more protons would need to pass through the ATP synthase for every ATP molecule that is synthesized by the F1 head.

15. Answer:

4

Molecular Biology of the Cell 7e-Alberts 16. Answer:

D

17. Answer:

D

18. Answer:

E

19. Answer:

D

20. Answer:

E

21. Answer:

A

22. Answer:

E

23. Answer:

C

24. Answer:

C

25. Answer:

C

26. Answer:

D

27. Answer:

B

28. Answer:

C

29. Answer:

D

30. Answer:

C

31. Answer:

E

32. Answer:

C

33. Answer:

C

34. Answer:

C

35. Answer:

D

36. Answer:

B

37. Answer:

D

Molecular Biology of the Cell 7e-Alberts 38. Answer:

B

39. Answer:

B

40. Answer:

D

41. Answer:

A

42. Answer:

A

43. Answer:

A

44. Answer:

B

45. Answer:

1. D 2. C 3. C 4. B

46. Answer:

1. C 2. A 3. B 4. C

47. Answer:

1. B 2. B 3. A 4. B

48. Answer:

1. B 2. A 3. B

49. Answer:

1. B 2. B

Molecular Biology of the Cell 7e-Alberts 3. A 4. A 50. Answer:

1. A 2. C 3. B 4. D

51. Answer:

1. A 2. B 3. B 4. B 5. A

52. Answer:

1. A 2. B 3. A 4. A 5. B

53. Answer:

ATP synthase will now act as a V-type ATPase. If the electron transport chain is flipped, the matrix will be acidified and the proton gradient would not be able to power ATP synthesis. The ATP/ADP antiport is also flipped, so ATP will likewise be trafficked inside the matrix, allowing the fueling of the F-type synthase in the opposite direction, as a V-type proton pump.

54. Answer:

ATP production will be reduced, but not eliminated, as the proton motive force will be maintained solely by the participation of complex I (NADH dehydrogenase) and the proton-pumping action of complexes III and IV (cytochrome c reductase and cytochrome c oxidase) will not participate in creating the proton concentration gradient. While the protonpumping capacity of the ETC would be reduced by approximately 60% (based on the numbers of protons each complex is capable of pumping across the membrane), the driving force for protons would still be maintained by the

Molecular Biology of the Cell 7e-Alberts voltage difference across the inner membrane, so the actual reduction in ATP production would likely not reach a 60% decrease.

55. Answer:

FAECDBG

56. Answer:

No, because preventing electrons from exiting the ETC will likely result in all elements of the chain remaining in a fully reduced state. It is more likely that the mobile carrier molecules would persist in a fully reduced state (as ubiquinol) since they would be unable to donate electrons due to a lack of oxidized acceptors.

57. Answer:

No, because ubisemiquinone would still be able to accept electrons from the Q cycle in complex III (cytochrome c reductase) and be fully reduced. Additionally, electrons would still be able to enter the ETC because of complex II (succinate dehydrogenase) acting as an alternate donor.

58. Answer:

ADP levels would decrease, and ATP levels would increase.

The ATP/ADP antiporter works to rapidly send ATP out of the mitochondrial matrix, and replace it with ADP as a substrate for the F-type ATP synthase. If its kinetics were slowed, ATP would remain in the matrix longer, building up in concentration. ADP, meanwhile, would be imported at a reduced rate and might not keep up with consumption by the synthase, resulting in lower ADP concentration. Both of these events would impact the energetics of the synthase, and of the entire cell, as the ATP/ADP+Pi product/reactant balance shifted away from maximum efficiency.

59. Answer:

ATP-synthesis mode, because under aerobic conditions, the proton gradient established by the respiratory chain can drive ATP synthesis. On the other hand, under anaerobic conditions, the ATP synthase pumps protons out (by ATP hydrolysis) to maintain the gradient.

60. Answer:

EHBDFGCA

61. Answer:

CABD

62. Answer:

It is expected that a defect in the mitochondrial fusion machinery would favor homoplasmy, because mitochondrial fusion works to ensure an even distribution of mitochondrial DNA throughout the cell. Fusion can reduce the effect of mitotic segregation and favor heteroplasmy. Thus, a defect

Molecular Biology of the Cell 7e-Alberts in fusion should favor homoplasmy.

Molecular Biology of the Cell 7e-Alberts

Chapter 15: Cell Signaling 1. A cell expresses a transmembrane protein that is cleaved at the plasma membrane to release an extracellular fragment. The fragment binds to receptor proteins on nearby cells and activates signaling pathways resulting in altered gene expression patterns in the cells. What form of intercellular signaling does this represent? a. Contact-dependent signaling b. Paracrine signaling c. Synaptic signaling d. Endocrine signaling e. Autocrine signaling

2. Consider the signaling pathway depicted below, which involves two different receptors. Assume that a protein is active only when it is activated but not inhibited by its upstream signaling molecules. Activation and inhibition are indicated by normal and blunt arrows, respectively. Under which of the following conditions is target gene expression induced?

Molecular Biology of the Cell 7e-Alberts

a. Only in the presence of both signals (1 and 2) b. In the presence of either or both signals (1 or 2) c. In the presence of either signal but not both d. Only in the absence of both signals e. Only in the presence of signal 2 but not signal 1

3. Some signaling pathways in the cell are chiefly based on inhibition rather than activation. For example, proteins that initiate programmed cell death by apoptosis are inhibited by anti-apoptotic proteins, which can in turn be inhibited by a group of pro-apoptotic proteins, and so on. Consider the following networks of inhibitory interactions, taking note that each protein is inactive if at least one of its inhibitors is active. In which pathway is the activation of either upstream protein (1 or 2)

Molecular Biology of the Cell 7e-Alberts sufficient to inhibit apoptosis?

a.

b.

c.

d.

4. Ion-channel-coupled receptors found in nerve and muscle cells are regulated by what kind of small molecule? a. Hormones b. Neurotransmitters c. Insulin d. Cyclic AMP and calcium e. Serine/Threonine

Molecular Biology of the Cell 7e-Alberts

5. Which of the following events normally activates a GTP-binding protein? a. GTP hydrolysis by protein b. Activation of an upstream GTPase-activating protein c. Activation of an upstream guanine nucleotide exchange factor d. Phosphorylation of a bound GDP molecule by an upstream phosphorylase e. Pi release after GTP hydrolysis

6. In which of the following schematic drawings of signaling pathways does the activation of the receptor lead to gene expression? Activating and inhibitory steps are indicated with (+) and (−), respectively.

a.

b.

c.

d.

e.

Molecular Biology of the Cell 7e-Alberts

7. The Src kinase can be regulated by at least two mechanisms. Kinases of the CSK family inactivate Src by phosphorylating a C-terminal tyrosine residue. On the other hand, binding of activating ligands (including some activated receptor tyrosine kinases, or RTKs) results in autophosphorylation at a tyrosine residue near the active site, stimulating Src activity. Full activation of Src therefore requires both dephosphorylation at the inhibitory tyrosine and binding to an RTK. Which of the following logic gates (A to E) correctly models the activation of Src kinase (output) as a function of the activity of the upstream regulators (inputs)?

Inputs

Output of logic gates

CSK activity

RTK activity

OFF

OFF

→

OFF

ON

→ OFF OFF

ON

OFF

→ OFF OFF OFF ON

ON

ON

→ OFF

A

B

C

ON

ON

OFF OFF OFF

ON

ON

b. Logic gate B c. Logic gate C d. Logic gate D e. Logic gate E

8. Which of the following is NOT a common second messenger in cell signaling?

b. Cyclic adenosine monophosphate c. Diacylglycerol

E

OFF OFF OFF

OFF OFF ON

a. Logic gate A

a. Ca2+

D

Molecular Biology of the Cell 7e-Alberts d. Tyrosine e. Inositol trisphosphate

The following graphs are plotted over time for five cell types, which in turn are treated with three different concentrations of a signal molecule. The cellular response is measured in real time by the concentration of a certain active effector protein, and during the time period indicated by the horizontal gray bar, the signal molecule is present in the culture medium at a concentration of 1 nM (dotted curve), 5 nM (gray curve), or 25 nM (solid curve).

9. Which of the cell types shows the lowest sensitivity to the signal?

Molecular Biology of the Cell 7e-Alberts

a.

b.

c.

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d.

e.

10. Which of the cell types shows a response with the highest persistence?

a.

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b.

c.

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d.

e.

11. Which of the cell types shows the fastest signal adaptation?

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a.

b.

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c.

d.

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e.

12. Which of the cell types shows the widest dynamic range of signal concentration?

a.

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b.

c.

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d.

e.

13. In the following simplified diagram, the activation of a hypothetical allosteric protein is plotted as a function of effector molecule concentration. Which curve corresponds to a condition in which activation requires the highest number of effector molecules bound to each protein?

Molecular Biology of the Cell 7e-Alberts

a. Curve A b. Curve B c. Curve C d. Curve D e. Curve E

14. Positive and negative feedback loops are common regulatory elements in cell signaling. What is the result of negative feedback loops in the cell? a. Local pulses, waves, and cell polarization b. Regulation of circadian rhythms, limitations in output signaling, modifications in signaling response, and switchlike responses c. Increased input levels, amplifying a signal, alterations in signaling response, and synthesis of bistable switches d. Cell polarization, local pulses, synthesis of bistable switches, and increased input levels e. Adaptive behavior, local pulses, waves, and amplifying a signal

15. The cholera toxin ADP ribosylation of the α subunit of stimulatory G protein (G s)

Molecular Biology of the Cell 7e-Alberts blocks GTP hydrolysis, but the pertussis toxin ADP ribosylation of the α subunit of inhibitory G protein (Gi) prevents interaction with the receptor. What is the effect of these toxins on the concentration of intracellular cAMP? a. Cholera toxin tends to increase cAMP concentration, whereas pertussis toxin tends to decrease cAMP concentration. b. Cholera toxin tends to decrease cAMP concentration, whereas pertussis toxin tends to increase cAMP concentration. c. They both tend to increase cAMP concentration. d. They both tend to decrease cAMP concentration.

16. Which of the following is a major consequence of activation of phospholipase C-β (PLCβ) by the Gq trimeric GTPase? a. Elevation of intracellular cAMP levels, leading to the activation of protein kinase A b. Elevation of PIP3 levels in the plasma membrane, leading to the activation of protein kinase B c. Elevation of intracellular Ca2+ levels, leading to the activation of protein kinase C d. Elevation of IP3 in the plasma membrane, leading to the activation of protein kinase D e. Elevation of intracellular cGMP levels, leading to the activation of protein kinase G

17. Which of the following molecular mechanisms is NOT used in the cell in order to produce abrupt responses to a signal? a. Activation of a protein only when bound to multiple molecules of an activating ligand b. Activation of a protein only when it is phosphorylated at multiple sites c. Activation of a protein by simultaneously activating its activator(s) and

Molecular Biology of the Cell 7e-Alberts inactivating its inhibitors d. A positive feedback loop in which an activated protein positively regulates its activator e. A negative feedback loop in which an activated protein inhibits its activator without delay

18. A certain effector protein can be activated by phosphorylation at a key tyrosine residue. An upstream kinase rapidly phosphorylates this tyrosine in the presence of a signal. However, the kinase also phosphorylates and activates a slow-acting phosphatase that can dephosphorylate the phosphotyrosine. Which curve in the following graph would you expect to represent the activity of the effector molecule over time? The input signal is present during the period indicated in gray. The dashed line represents the response in the absence of the phosphatase.

a. Curve A b. Curve B c. Curve C d. Curve D e. Curve E

19. What is true for the α subunit of a heterotrimeric G protein? a. A G-protein-coupled receptor (GPCR) acts as a guanine nucleotide

Molecular Biology of the Cell 7e-Alberts exchange factor (GEF), whereas a regulator of G protein signaling (RGS) can act as a GTPase-activating protein (GAP). b. A GPCR acts as a GAP, whereas an RGS can act as a GEF. c. Both a GPCR and an RGS can act as a GEF. d. Both a GPCR and an RGS can act as a GAP.

20. Eukaryotic signal transduction was essential to the development of complex multicellular organisms. Which two signaling pathways are responsible for the eukaryotic response to changes in the environment? a. Calcium signaling and GPCR b. IP3 and Akt signaling c. JAK/STAT signaling d. Inflammatory response e. G-protein and phospholipid signaling

21. What is NOT part of a negative feedback mechanism in adaptation to light in retinal rod cells? a. Phosphorylation of G-protein-coupled receptor b. Binding of arrestin to G protein c. Stimulation of guanylyl cyclase by decreased Ca2+ levels d. Binding of regulator of G protein signaling to transducin

22. Consider visual transduction in rod photoreceptors in the vertebrate retina. Which of the following steps does NOT normally amplify the signal in this pathway? a. Activation of transducin by active rhodopsin b. Blockage of Na+ influx by cation-channel closure

Molecular Biology of the Cell 7e-Alberts c. Cation-channel closure due to cGMP depletion d. Depletion of cGMP by active cGMP phosphodiesterase

23. A genetic form of “night blindness” (poor vision in dim light) is caused by mutations in genes encoding rhodopsin kinase (RK) and arrestin. The mutations reduce the dynamic range of light perception, and the affected individuals have a prolonged light response and adapt very slowly to low light (when entering a dark theater room, for example). Which of the following mutations do you expect to be the cause? a. Loss-of-function RK mutations and loss-of-function arrestin mutations b. Loss-of-function RK mutations and gain-of-function arrestin mutations c. Gain-of-function RK mutations and loss-of-function arrestin mutations d. Gain-of-function RK mutations and gain-of-function arrestin mutations

24. How is the activation of epidermal growth factor receptor (EGFR) different from that of insulin receptor (IR)? a. EGFR is a dimer in its inactive and active forms, whereas IR dimerizes only when active. b. EGFR activation requires dimerization, whereas IR can become activated as a monomer. c. EGFR is activated by transautophosphorylation, whereas IR activation is phosphorylation-independent and is a simple consequence of receptor dimerization. d. Unlike IR activation, EGFR activation involves an asymmetric arrangement of the intracellular kinase domains.

25. You grow a tumor-derived cell line in a suspension culture, in which the cells are seen either as single cells or in cell clusters. The cells have been engineered to express green fluorescent protein (GFP) under the control of the mitogenactivated protein kinase (MAP kinase) pathway. To your culture media, you add

Molecular Biology of the Cell 7e-Alberts different concentrations of fibroblast growth factor (FGF)—which is known to activate the MAP kinase pathway in these cells—and briefly incubate the cells before harvesting them. Under a microscope, you also collect hundreds of individual cell clusters from each suspension. You then disaggregate the cells in all of your samples and use a FACS (fluorescence-activated cell sorting) machine to measure the GFP signal intensity in individual cells. Finally, you plot the results as histograms, as shown below:

a.

b.

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c.

d.

e.

26. What two cell-surface receptors are represented in the two simplified diagrams below (from left to right)?

Molecular Biology of the Cell 7e-Alberts

a. TGFβ receptor and TNF receptor b. Cytokine receptor and TNF receptor c. TNF receptor and TGFβ receptor d. TGFβ receptor and cytokine receptor e. Cytokine receptor and TGFβ receptor

27. Consider a signaling protein that is only made up of one SH2 domain and two SH3 domains. This protein is most likely which of the following? a. A monomeric G protein b. A guanine nucleotide exchange factor c. A kinase associated with receptor tyrosine kinase signaling d. An adaptor protein e. A negative regulator of receptor tyrosine kinase signaling

28. The SH2 domain of STAT proteins plays a key role in cytokine receptor signaling.

Where can this domain bind to phosphotyrosine?

Molecular Biology of the Cell 7e-Alberts a. On an activated receptor molecule, or (intramolecularly) on the same STAT molecule b. On a Janus kinase molecule, or (intramolecularly) on the same STAT molecule c. On an activated receptor molecule, or on another STAT molecule in a dimer d. On a Janus kinase molecule, or on another STAT molecule in a dimer e. On the same STAT molecule or on another STAT molecule in a dimer

29. What is the major way by which the monomeric G protein Ras is activated in receptor tyrosine kinase signaling? a. Activation of Ras-GAP b. Activation of Ras-GEF c. Inactivation of Ras-GAP d. Inactivation of Ras-GEF e. Inactivation of Ras-GDI

30. Which of the following proteins lacks a PH domain? a. Sos (a Ras-GEF) b. Grb2 (an RTK signaling adaptor) c. Pleckstrin (a protein kinase C substrate) d. Akt (protein kinase B) e. PDK1 (phosphoinositide-dependent kinase)

31. What is the effect of using scaffold proteins on precision and amplification capacity in cell signaling?

Molecular Biology of the Cell 7e-Alberts a. Both precision and amplification have improved. b. Precision is improved, but amplification is limited. c. Precision is compromised, but amplification is improved. d. Both precision and amplification are limited.

32. Which of the following phosphoinositides is the preferred binding target for a PH domain-containing protein?

a.

b.

c.

d.

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e.

33. What is a characteristic of PI 3-kinase? a. It is a membrane-associated tyrosine kinase. b. It activates PDK1 by phosphorylating a serine residue on the protein. c. It is counteracted by PTEN phosphatase. d. It is only known to be activated by receptor tyrosine kinases. e. It inhibits Akt.

34. Proteins of the Src family (e.g., Src, Fyn, Lck, and Hck) have important roles in several signaling pathways in mammals. What is a characteristic of a typical Src family protein? a. It is a cytosolic tyrosine kinase. b. It does not contain SH2 and SH3 domains in addition to a kinase domain. c. It has ionically attached lipid chains. d. It is located on the extracellular side of the plasma membrane. e. It contains both serine and lysine residues.

35. Activated STATs can induce the expression of SOCS proteins, which in turn inhibit the JAK-dependent phosphorylation of the STATs. Under sustained serum stimulation of mouse fibroblast cells, a number of oscillations can be observed as a consequence of this delayed negative feedback loop, as shown in the following simplified graph. According to this graph, approximately how long is the delay in the feedback response, and what process is mainly responsible for it?

Molecular Biology of the Cell 7e-Alberts

a. 1 hour; transcription b. 2 hours; transcription c. 1 hour; mRNA export d. 1 hour; mRNA translation e. 2 hours; mRNA translation

36. Signal transducers and activators of transcription (STATS) are latent transcription regulators, meaning they require activation in order to function. What is the name of the domain that is responsible for STAT function in the nucleus? a. Ras b. MAP c. JAK d. SH2 e. RTK

37. Which of the following mechanisms is NOT employed by I-Smads to negatively regulate Smad signaling?

Molecular Biology of the Cell 7e-Alberts a. Competing with R-Smads for binding sites on the receptor, decreasing RSmad phosphorylation b. Recruiting the ubiquitin ligase Smurf, leading to receptor degradation c. Binding to caveolin, inhibiting the formation of caveolae d. Binding to the co-Smad, Smad4, inhibiting it e. Recruiting protein phosphatases, leading to receptor inactivation

38. How does the expression of Delta on the surface of a cell activate the expression of certain genes in the nucleus of its neighboring cell? a. Delta binding activates Notch, which activates a transcriptional activator through the JAK–STAT pathway. b. Delta binding leads to the stabilization of a cytoskeleton-associated transcriptional activator. c. Delta binding releases the intracellular tail of Notch, which enters the nucleus and converts a transcriptional repressor into a transcriptional activator. d. Delta binding leads to Notch-mediated recruitment of protein complexes to the plasma membrane, resulting in the degradation of a transcriptional repressor. e. Delta binding leads to the proteolytic cleavage of Notch and inhibition of its activity as a transcriptional repressor, leading to the activation of target genes.

39. Which of the following represents the active form of β-catenin in cells stimulated with Wnt? a. Phosphorylated by GSK3 and CK1 b. Bound to APC c. Bound to Groucho d. Bound to LRP

Molecular Biology of the Cell 7e-Alberts e. Bound to the ER

40. Regulation of the Ci (Cubitus interruptus) protein in the Hedgehog signaling pathway is reminiscent of that of β-catenin in the Wnt pathway. Which of the following features is shared among these two proteins? a. Both proteins are ubiquitylated when the signal (Hedgehog or Wnt) is present. b. Both proteins are able to enter the nucleus only in the absence of the signal. c. Both proteins can be phosphorylated by GSK3 and CK1 in the absence of the signal. d. Both proteins are fully degraded by the proteasome in the absence of the signal.

41. Which of the following occurs inside the cell in the presence but NOT the absence of stimulation with a Hedgehog signal? a. Sequestration of Smoothened in intracellular vesicles b. Accumulation of unprocessed Ci in the nucleus c. Phosphorylation of Smoothened d. Recruitment of Fused and Costal2 to the plasma membrane

42. The Hedgehog protein functions in the proliferation of cells and histogenesis associated in early vertebrate development. From the below lists, which includes the three genes that encode the Hedgehog proteins found in animals? a. Wnt, LEF1/TCF, and Myc b. Sonic, Desert, and Indian Hedgehog c. Desert, GPCR, and Gli d. NFκB, Dorsal, and Sonic

Molecular Biology of the Cell 7e-Alberts e. Sonic, Indian Hedgehog, and NFκB

43. In the following simplified diagram of the NFκB-dependent signaling pathway, which boxes (A to C) correspond to the NFκB, IκB, and IKK proteins, respectively?

a. BCA b. BAC c. CBA d. CAB e. ABC

44. You have studied the circadian clock in a certain mouse strain by monitoring the physical activity, for example, wheel-running, of a few mice over an extended period of time. For the first week, you kept the mice under normal 12 hr light /12 hr dark cycles; they were then transferred to constant darkness. You have plotted the average activity of the mice in the following graph. The light and dark conditions are represented with white and gray backgrounds, respectively. The activity in each day is plotted in black in one row of the graph, and consecutive days are arranged from the top to the bottom. The horizontal axis represents the time of day. Based on these results, what is the most accurate estimate for the free-running period of the circadian clock in this mouse strain?

Molecular Biology of the Cell 7e-Alberts

a. About 25 hr b. About 24.5 hr c. About 24 hr d. About 23.5 hr e. About 23 hr

45. What are the consequences of the accumulation and degradation of Tim and Per, two transcription regulatory proteins? a. Increase in Notch-responsive genes, as Notch migrates to the nucleus b. Regulation of NFκB, as a result of the degradation of Inositol κB c. Decrease in Wnt and Hedgehog signaling, resulting in a decrease in Tim d. Repression in the transcription of Tim and Per genes e. Increases in enzyme coupled cell surface receptors, resulting in increased levels of Per

Molecular Biology of the Cell 7e-Alberts

46. In the following schematic diagrams of ethylene signaling in plants, 1 denotes ethylene, 2 denotes ethylene receptor, 3 denotes CTR1, 4 denotes EIN3, and 5 represents the expression of ethylene-responsive genes. Which diagram better depicts the signaling pathway? Activation and inhibition are indicated as normal and blunt arrows, respectively. a. b. c. d. e.

47. What is true of ethylene receptors in plant cells? a. They are located primarily at the plasma membrane. b. They are single-pass transmembrane proteins. c. They interact with CTR1, which is closely related to the monomeric G protein Ras. d. They have a copper-containing ethylene-binding domain.

48. What is a characteristic of all nuclear receptors? a. They are cytosolic proteins that enter the nucleus upon ligand binding. b. They have ligand-binding and DNA-binding domains and can directly bind to DNA. c. They are transcriptional activators when bound to their ligand. d. They bind to steroid hormones.

Molecular Biology of the Cell 7e-Alberts e. They are transcriptional repressors in the absence of their ligand.

49. Regarding the evolution of complex multicellularity, which of the following statements is TRUE? a. Unicellular organisms evolved from complex multicellular ancestors. b. The plants’ lineage acquired mitochondria before the divergence from animals. c. Animal cells migrate during cell signaling, increasing cell complexity. d. Different genes are activated and repressed in different cells. e. Complex multicellularity evolved independently in bacteria and plants.

50. Arabidopsis thaliana seedlings show positive phototropism: They bend and grow toward the source of (blue) light. However, phototropism is not observed if the seedlings are treated with an auxin efflux inhibitor called NPA. Knowing that auxin stimulates elongation of the cells in the growing shoot, these observations are consistent with a model in which, in the presence of a lateral light, what happens? a. Auxin efflux transporter proteins become localized to the side of the cell that is facing the light source. b. Auxin efflux transporter proteins become localized to the side of the cell that is away from the light source. c. Auxin influx transporter proteins become localized to the side of the cell that is facing the light source. d. Auxin influx transporter proteins become localized to the side of the cell that is away from the light source. e. Auxin efflux transporter proteins become uniformly distributed around the cell.

51. In a plant’s life cycle, hormones play a critical role in the regulation of growth and behavioral processes. Which of the following hormones function in phototropism and gravitropism?

Molecular Biology of the Cell 7e-Alberts a. GTPase-activating hormone b. Auxin c. Estrogen d. Raf MAP kinase e. Receptor tyrosine kinase

52. In plants, there is a cytoplasmic serine/threonine kinase that is activated by red light and inactivated by far-red light. What is its name? a. Phototropin b. Photoprotein c. Cryptochrome d. Phytochrome

53. Once activated, a GPCR molecule can activate multiple molecules of G protein to amplify the incoming signal. a. True b. False

54. NO normally decreases cGMP concentration by activating cGMP phosphodiesterase. a. True b. False

55. The immediate early genes turn on a few hours after activation of the RTK. a. True b. False

Molecular Biology of the Cell 7e-Alberts 56. mTOR in complex 1 contains the protein raptor, is sensitive to rapamycin, and stimulates cell growth. a. True b. False

57. The ubiquitin ligase c-Cbl ubiquitylates activated RTKs. a. True b. False

58. Receptor tyrosine kinases are the largest class of cell-surface receptors in plant cells. a. True b. False

59. Indicate whether each of the following descriptions applies to visual signal transduction, olfactory signal transduction, or both. a. Visual signal transduction b. Olfactory signal transduction c. Both

1. It involves cGMP as a second messenger.

2. A rise in cyclic mononucleotide concentration follows the arrival of the signal.

3. The signal is received by a G-protein-coupled receptor.

4. An incoming signal leads to membrane hyperpolarization.

60. Genetic screens for enhancers and suppressors of temperature-sensitive loss-offunction mutations in Sevenless (Sev) in Drosophila melanogaster led to the discovery of some other genes involved in the Ras–MAP-kinase pathway. Would you expect loss-of-function mutations in genes encoding each of the following proteins to be an enhancer or suppressor of Sev? Note that loss-of-function mutations can be just partial defects and not necessarily null mutations.

Molecular Biology of the Cell 7e-Alberts a. Enhancer b. Suppressor

1. Sos 2. Ras 3. Ras-GAP 4. Raf

61. In the following schematic diagram showing the initial signaling steps in growth cone collapse in motor neurons, indicate which protein corresponds to each lettered area.

a. Area A b. Area B c. Area C d. Area D e. Area E

1. Ephrin

Molecular Biology of the Cell 7e-Alberts 2. Ephexin 3. RhoA GTPase 4. Cytoplasmic tyrosine kinase 5. Activated receptor

62. In the following diagram showing five parallel signaling pathways downstream of a receptor tyrosine kinase (RTK) and a G-protein-coupled receptor (GPCR), indicate which boxes correspond to the following proteins.

a. Box A b. Box B c. Box C d. Box D e. Box E

1. Cyclic AMP

Molecular Biology of the Cell 7e-Alberts 2. Protein kinase C 3. Sos 4. PI(3,4,5)P3 5. Ca2+

63. Cancer-related genes can be classified into two major groups. Proto-oncogenes encode proteins that normally act in promoting cell growth and proliferation. Gainof-function mutations in proto-oncogenes can therefore transform the cells into a cancerous state. Tumor suppressor genes, on the other hand, encode proteins that normally function to keep cell proliferation in check. Thus, loss-of-function mutations in tumor suppressors can lead to cancer by eliminating the inhibitory effect of these genes. Mutations in genes encoding the components of the canonical Wnt signaling pathway are commonly found in colorectal tumors.

Indicate whether the gene for each of the following is a proto-oncogene or tumor suppressor. a. Proto-oncogene b. Tumor suppressor

1. Wnt 2. APC 3. Axin 4. β-Catenin

64. Indicate whether each of the following descriptions applies to phytochromes, cryptochromes, or phototropins. a. Phytochromes b. Cryptochromes c. Phototropins

1. They respond to red light.

2. They are found in animals as well as in plants.

3. They are flavoproteins.

4. They are dimeric, cytosolic serine/threonine protein kinases that are normally activated by autophosphorylation.

Molecular Biology of the Cell 7e-Alberts 65. Consider a signaling protein that can be phosphorylated by a dual-specificity protein kinase at two independent sites (one tyrosine and one serine residue), such that each site is phosphorylated in about 1% of the protein molecules under normal conditions. Phosphorylation at the first site has a short half-life, since a fast tyrosine phosphatase removes the phosphate soon after the residue is phosphorylated, whereas dephosphorylation at the second site—carried out by a serine/threonine phosphatase—is relatively slow.

Upon activation of a signaling pathway, the concentration of the dual-specificity kinase increases several-fold rapidly. The phosphorylation state of the tyrosine and serine residues following stimulation is compared in the following graph.

Which curve (1 or 2) do you think corresponds to the serine residue?

66. Sort the following events to reflect the normal order in which they occur in Gprotein-coupled receptor signaling leading to transcription of genes with cAMP response elements. Your answer should be a four-letter string composed of letters A to E only, for example, ACBD. A. Binding of CREB to PKA B. Binding of cAMP to PKA C. Dissociation of PKA into catalytic and regulatory subunits D. Activation of adenylyl cyclase

67. Sort the following events into the order that they take place in phototransduction in response to a flash of light. Your answer would be a four-letter string composed of letters A to E, for example, BCDEA. A. The cGMP level drops.

Molecular Biology of the Cell 7e-Alberts B. The cGMP level rises.

C. The Ca2+ level drops.

D. Retinal isomerizes to an all-trans configuration.

E. Gt exchanges its bound nucleotide.

68. Intracellular Ca2+ oscillations, initiated by the activation of IP3 receptors and orchestrated via intricate positive and negative feedback loops, can be subject to frequency modulation, an example of which is the increase in oscillation frequency with increasing stimulus strength. Caffeine is known to bind to and sensitize the ryanodine receptors, resulting in their opening in response to lower calcium ion concentrations. Do you think caffeine would tend to increase or decrease the frequency of calcium ion oscillations in stimulated cells? Explain your answer in 2–3 sentences.

69. Nuclear receptors are regulated through the binding of intracellular proteins. On the other hand, consider steroid hormones. What role do intracellular receptors play in relation to steroid hormones, and more specifically, the regulation of gene expression? Describe your response in 2–3 sentences.

70. Though cyclopamine is known to cause birth defects as a cancer therapeutic, recently it has been nonetheless used to treat basal cell carcinoma of the skin.

For what cancers are cyclopamine used as a treatment and why is it preferred?

71. Consider specialized cell types, which are essential for a host response against infection. For example, Natural Killer (NK) cells are activated when inhibitory signals are attenuated, a process that in turn results in the activation of cellular receptors. With regard to the proliferation and differentiation of NK cells, how does the inositol phospholipid pathway play an important role?

72. The cell-mediated immune response in multicellular organisms is dependent on the activation and interaction of proteins, known as cytokines, and their cytokine receptors. These receptors are phosphorylated and influence downstream transcription regulation in the cell. How does cytokine concentration regulate the

Molecular Biology of the Cell 7e-Alberts gene expression of cytokine-induced genes?

73. The phosphorylation state of a human protein follows the oscillatory pattern shown in the first graph below. The protein is phosphorylated by two kinases (1 and 2), both of which are activated in an oscillatory pattern. Kinase 1 oscillates in congruence with the circadian rhythm, while kinase 2 oscillates at a higher frequency and is independent of the circadian clock. The second graph shows the phosphorylation state of the same protein in mutant cells lacking one of the kinases mentioned above. Which kinase (1 or 2) do you think is missing in these mutant cells?

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 15 1. Answer:

B

2. Answer:

C

3. Answer:

B

4. Answer:

B

5. Answer:

C

6. Answer:

A

7. Answer:

C

8. Answer:

D

9. Answer:

E

10. Answer:

A

11. Answer:

C

12. Answer:

C

13. Answer:

C

14. Answer:

B

15. Answer:

C

16. Answer:

C

17. Answer:

E

18. Answer:

B

19. Answer:

A

20. Answer:

E

Molecular Biology of the Cell 7e-Alberts 21. Answer:

B

22. Answer:

C

23. Answer:

A

24. Answer:

D

25. Answer:

C

26. Answer:

E

27. Answer:

D

28. Answer:

C

29. Answer:

B

30. Answer:

B

31. Answer:

B

32. Answer:

E

33. Answer:

C

34. Answer:

A

35. Answer:

C

36. Answer:

D

37. Answer:

C

38. Answer:

C

39. Answer:

E

40. Answer:

C

41. Answer:

A

42. Answer:

B

Molecular Biology of the Cell 7e-Alberts 43. Answer:

C

44. Answer:

D

45. Answer:

D

46. Answer:

E

47. Answer:

D

48. Answer:

B

49. Answer:

D

50. Answer:

B

51. Answer:

B

52. Answer:

D

53. Answer:

A

54. Answer:

B

55. Answer:

B

56. Answer:

A

57. Answer:

A

58. Answer:

B

59. Answer:

1. A 2. B 3. C 4. A

60. Answer:

1. A 2. A 3. B

Molecular Biology of the Cell 7e-Alberts 4. A 61. Answer:

1. A 2. C 3. E 4. D 5. B

62. Answer:

1. A 2. C 3. D 4. E 5. B

63. Answer:

1. A 2. B 3. B 4. A

64. Answer:

1. A 2. B 3. B 4. A

65. Answer:

Curve 2

66. Answer:

DBCA

67. Answer:

DEACB

68. Answer:

Caffeine would tend to increase the frequency, because the effect of channel sensitization to Ca2+ is comparable to that of increased Ca2+ concentration, that is, higher-frequency oscillations. Furthermore, the frequency and amplitude of the Ca2+ spikes can be modulated by other mechanisms.

Molecular Biology of the Cell 7e-Alberts 69. Answer:

Nuclear receptors bind to specific DNA sequences upstream of the gene that is regulated by the specific ligand. Meanwhile, steroid hormones, such as cortisol, Vitamin D, ecdysone, and thyroid hormones bind to intracellular nuclear receptors. The binding of the ligand triggers a conformational change in the receptor, allowing the receptor to interact with regulatory proteins. These interactions then result in the expression of genes or the inhibition of gene expression.

70. Answer:

Cyclopamine, often isolated from the meadow lily, is a steroidal alkaloid and a naturally occurring chemical that is often used to treat cancers that are associated with increased Hedgehog signaling. This is the case, because the chemical tightly binds to and inhibits Smoothened, which results in the down regulation of Hedgehog signaling.

71. Answer:

The inositol signaling pathway is activated when a signal molecule binds to the GPCR of a trimeric G protein. This ligand-receptor interaction results in the activation of phospholipase C, which in turn leads to the increase of cellular Ca²⁺. Modification of the inositol phospholipids regulates the attenuation or promotion of signaling pathways that determine NK development, function, survival, and/or trafficking in the host.

72. Answer:

Cytokines bind to cytokine receptors and are associated with either one or multiple JAKs (JAK1, JAK2, JAK3, and Tyk2). The binding of a cytokine to its receptor leads to a conformational change in the JAKs, leading to the phosphorylation of the tyrosine kinase domain in enzyme. The phosphorylated JAKs then phosphorylate tyrosines on the cytoplasmic tails of cytokine receptors. If the concentration of extracellular cytokine is low, the expression of genes associated with the cytokine-induced response will be downregulated.

73. Answer:

Kinase 2

Molecular Biology of the Cell 7e-Alberts

Chapter 16: The Cytoskeleton 1. Which cytoskeletal filament is abundant in an animal cell nucleus? a. Actin filaments b. Microtubules c. Septins d. Intermediate filaments e. Spectrin filaments

2. Which statement is consistent with the roles of cytoskeletal filaments in establishing and maintaining large-scale cellular polarity in polarized epithelial cells? a. Polarity information is conveyed primarily by only one of the three types of cytoskeletal filaments. b. Microtubules are attached to the adhesive structures that connect adjacent cells to each other. c. All three types of cytoskeletal filaments work together to maintain critical differences between the apical and basolateral surfaces of the cell. d. Intermediate filaments are responsible for transporting newly synthesized materials between the apical and basolateral surfaces of the cell. e. Actin filaments form a dense, gel-like matrix near the volumetric center of the cell.

3. All but one of the following are generalized functions of accessory proteins that bind to cytoskeletal filaments. Which one is NOT a function of these accessory proteins? a. They change the kinetics of filament assembly and disassembly. b. They bind to ribosomes to regulate the synthesis of cytoskeletal subunits.

Molecular Biology of the Cell 7e-Alberts c. They respond to extracellular and intracellular signals. d. They harness energy to generate force for the movement of materials. e. They determine the sites of assembly of new filaments.

4. Shruthi is studying the behaviors of motor proteins moving along cytoskeletal filaments. She uses an in vitro translation system to synthesize a motor protein that has a single molecule of green fluorescent protein (GFP) attached to its cargo-binding tail domain. She adds the GFP-tagged motor proteins to glass microscope slides containing short segments of intact microtubules and then bathes the whole preparation in an ATP-containing solution that has the same viscosity as the cytoplasm of a typical animal cell. When she observes the experiment with a fluorescence microscope, she sees the fluorescent motor proteins moving along the microtubules. What will she see if she adds an ATPquenching enzyme, that is, an enzyme that removes all ATP, while continuing to observe the sample? a. The motor proteins will stop and then rebound slightly as if hitting a wall. b. The motor proteins will glide to a gentle halt. c. The motor proteins will travel away from the microtubules along random paths. d. The motor proteins will stutter (stop and start repeatedly) before coming to a halt. e. The motor proteins will stop dead in their tracks.

5. The Reynolds number is the ratio of inertial to viscous forces acting on an object.

Which of the following has the smallest Reynolds number? a. Ribosome b. Single-celled protozoan c. Blue whale d. Bacterium e. Small fish

Molecular Biology of the Cell 7e-Alberts

6. Persistence length for a cytoskeletal filament is the minimum filament length at which random thermal fluctuations are likely to cause it to bend. Which of the following comparisons are true, considering the persistence lengths of (a) an actin filament, (b) a bundle of cross-linked actin filaments, and (c) a microtubule? a. (a > b) and (a > c) b. (a > b) and (a < c) c. (a < b) and (a > c) d. (a < b) and (a < c) e. (a > b) and (b > c)

7. How many protofilaments are found in a typical actin filament? a. Four b. Thirty-two c. Two d. Nine e. Thirteen

8. In the polymerization in vitro of actin filaments and microtubules from their subunits, what does the “lag phase” correspond to? a. Nucleation b. Reaching steady state c. Nucleotide exchange d. ATP or GTP hydrolysis e. Treadmilling

Molecular Biology of the Cell 7e-Alberts 9. The time courses of seeded actin polymerization under two different conditions are compared in the following graph. If the overall kon for polymerization is known to be the same under both of these conditions, which curve—(1) or (2)— corresponds to the condition with a higher koff rate constant? Which one corresponds to a higher Cc for polymerization?

a. Curve (1); curve (1) b. Curve (1); curve (2) c. Curve (2); curve (1) d. Curve (2); curve (2) e. Curve (2); both curves have the same Cc

10. You have prepared actin filament seeds in a microfluidic chamber through which you then pass either unlabeled or fluorescently labeled actin subunits in alternation several times, keeping the total monomer concentration constant. You then observe the resulting filaments under a fluorescence microscope. If the filaments appear as drawn schematically in the example below (in which fluorescence is indicated by black color), which end do you think is the plus end— (a) or (b)? Was the concentration of free actin subunits below or above the critical concentration (Cc) for polymerization at the barbed end?

Molecular Biology of the Cell 7e-Alberts

a. (a); below b. (b); below c. (a); above d. (b); above

11. If the concentration of free subunits is C, under which condition does the growth of a cytoskeletal filament proceed spontaneously? a. C > 1/kon b. C < 1/koff c. C > kon/koff d. C > koff/kon e. C > Cc×kon/koff

12. Consider the ATP-bound and ADP-bound forms of actin and the polarized nature of the actin filaments. In the following diagram that shows the various actin polymerization rate constants (konand koff values), which monomer corresponds to an ADP-bound actin incorporated at the plus end?

Molecular Biology of the Cell 7e-Alberts

a. Monomer A b. Monomer B c. Monomer C d. Monomer D

13. In the following graph of actin elongation rates under different subunit concentrations, which of the following corresponds to the slope of the line?

a. kon b. koff c. kon/koff d. koff/kon

Molecular Biology of the Cell 7e-Alberts e. kon – koff

14. According to the following graph, which shows the elongation rate at the plus and minus ends of actin filaments as a function of actin subunit concentration, at what concentration (A to E) does the total length of the filament remain more or less constant with time, that is, steady-state treadmilling occurs?

a. Concentration A b. Concentration B c. Concentration C d. Concentration D e. Concentration E

15. Which of the following drugs binds along assembled actin filaments and stabilizes them? a. Cytochalasin B b. Phalloidin c. Nocodazole d. Colchicine

Molecular Biology of the Cell 7e-Alberts e. Latrunculin

16. Which of the following is an actin homolog? a. NPF b. Arp3 c. Thymosin d. Profilin e. Formin

17. The actin-nucleating protein formin has flexible “whiskers” containing binding sites that help recruit actin subunits in order to enhance polymerization by this protein.

What protein would you expect to bind to these sites? a. Thymosin b. Profilin c. Cofilin d. Gelsolin e. Tropomodulin

18. Which of the following pairs of actin-binding proteins cannot bind to the same actin filament simultaneously? a. Gelsolin and tropomodulin b. Tropomyosin and tropomodulin c. Profilin and tropomodulin d. Cofilin and CapZ e. Formin and CapZ

Molecular Biology of the Cell 7e-Alberts

19. In the following graph, the elongation rate of pure actin filaments as a function of actin subunit concentration is shown as a dashed line. Which one of the other five lines (A to E) shows what happens when a plus-end capping protein such as CapZ is present?

a. Line A b. Line B c. Line C d. Line D e. Line E

20. Actin filaments that are held together by the cross-linking protein fimbrin are not contractile. What is the probable cause for why this is true? a. The very weak cross-linking by this protein cannot convert myosin II activity into a coherent contraction. b. Fimbrin arranges the actin filaments in parallel bundles in which all the plus ends point to the same direction. c. The very tight packing of actin filaments by this small protein excludes myosin II filaments and other large proteins. d. Fimbrin arranges the actin filaments into gel-like networks in which myosin

Molecular Biology of the Cell 7e-Alberts II activity does not produce contraction. e. Fimbrin is a large protein that binds to several actin filaments and resists contraction.

21. Each of the following actin-binding proteins is capable of linking two filaments together because it possesses two or more separate actin-binding sites. Which of the proteins has its actin-binding sites separated by the greatest distance in three-dimensional space? a. Spectrin b. Filamin c. α-Actinin d. Fimbrin e. Myosin II

22. All but one of the following describe functions of the ERM family of proteins, for example, ezrin, radixin, and moesin. Which one is NOT a function of these actinbinding proteins? a. They bind to and organize the cortical actin cytoskeleton. b. They interact with transmembrane proteins. c. They affect cortical stiffness and cell shape. d. They affect the localization and activity of cell signaling molecules. e. They attach vesicles and organelles to the cell cortex.

23. Cofilin binds preferentially to ADP-containing actin filaments rather than to ATPcontaining filaments. Consequently, this protein does which of the following? a. Competes with profilin for actin binding b. Binds to older actin filaments

Molecular Biology of the Cell 7e-Alberts c. Binds to the plus end of actin filaments d. Stabilizes actin filaments e. Binds to newer actin filaments

24. Which of the following is responsible for nucleating the highly branched networks of actin filaments that are found within lamellipodia? a. ERM proteins b. WAVE proteins c. Formin d. Arp 2/3 complex e. γ-TuRC

25. Which of the following is NOT an example of a crawling cell? a. Macrophage b. Osteoclast c. Primary tumor cell d. Neural crest cell e. Sperm cell

26. How is membrane protrusion by blebbing different from that by lamellipodia? a. Blebbing usually occurs on a rigid substratum such as glass, whereas lamellipodia form on pliable substrata. b. Blebbing is mostly observed in vitro, whereas lamellipodia are observed both in vitro and in vivo. c. Blebbing does not involve myosin II activity, but formation of lamellipodia does.

Molecular Biology of the Cell 7e-Alberts d. Blebbing requires loss of membrane interaction with actin filaments, whereas lamellipodia formation requires a close interaction between the two.

27. Consider an actin subunit that has just been incorporated into an actin filament at the leading edge of a lamellipodium. Before its ATP is hydrolyzed, how does its distance from the leading front edge of the plasma membrane change over time?

How does its distance from the F-actin minus end change over time? a. Decreases; decreases b. Decreases; remains constant c. Decreases; increases d. Increases; decreases e. Increases; remains constant

28. The following figure shows a side view of a fibroblast exhibiting mesenchymal cell migration. In which direction (1 or 2) is the fibroblast migrating? What is the approximate length of the cell indicated by the scale bar?

a. Direction 1; about 20 µm b. Direction 1; about 2 µm c. Direction 2; about 20 µm d. Direction 2; about 2 µm e. Either direction; about 2 µm

29. In lamellipodia, actin polymerization is nucleated by _____(1) ______, while

Molecular Biology of the Cell 7e-Alberts depolymerization is catalyzed by _____(2) ______. a. (1) formin; (2) gelsolin b. (1) Arp 2/3 complex; (2) gelsolin c. (1) formin; (2) cofilin d. (1) Arp 2/3 complex; (2) cofilin e. (1) formin; (2) capping protein

30. If myosin II heads are attached to a glass slide and actin filaments are allowed to bind to them, the filaments will glide on the surface to where? a. Toward their plus end in the presence of ATP b. Toward their plus end in the presence of GTP c. Toward their minus end in the presence of ATP d. Toward their minus end in the presence of GTP e. Toward their plus end in the presence of ADP

31. In the presence of an ATP analog that can bind myosin normally but cannot be hydrolyzed, what takes place? a. A myosin cannot bind tightly to an actin filament and cannot move on it. b. A myosin cannot be released from an actin filament and cannot move on it. c. A myosin performs the power stroke but cannot be released from an actin filament. d. A myosin performs the power stroke but cannot bind tightly to an actin filament. e. A myosin is able to perform normally.

32. Consider the structure of a sarcomere as it appears in a low-magnification

Molecular Biology of the Cell 7e-Alberts electron micrograph. Which of its features does NOT shorten during skeletal muscle contraction? a. The dark band b. The light band c. The distance from the M line to the Z disc d. The distance between two consecutive Z discs e. The region containing the titin molecules

33. After an animal dies, its muscles start to stiffen before the decomposition of tissues relaxes the muscles again. Which statement best explains this muscle stiffening, that is, rigor mortis? a. The myosin II heads in muscle fibers remain attached to actin filaments due to the absence of Ca2+ in these cells. b. The ATPase activity of muscle myosin II is inhibited by the elevated Ca 2+. c. The myosin II heads in muscle cells remain attached to actin filaments due to the elevated Ca2+ in the muscle fibers. d. Titin molecules unfold, preventing muscle relaxation. e. The myosin II heads in muscle cells remain attached to actin filaments due to covalent cross-linking.

34. Skeletal muscle cells consume enormous amounts of ATP. Which pair of proteins is responsible for hydrolyzing most of this ATP? a. Myosin heavy chain and Ca2+-release channel b. Myosin light chain and Ca2+-release channel c. Myosin light-chain kinase and sarcoplasmic reticulum Ca2+-pump d. Myosin heavy chain and sarcoplasmic reticulum Ca2+-pump

Molecular Biology of the Cell 7e-Alberts e. Myosin light chain and sarcoplasmic reticulum Ca2+-pump

35. In which structure are actin–myosin II bundles regulated by the troponin complex? a. Stress fiber b. Circumferential belt c. Contractile ring d. Adherens junction e. Skeletal muscle myofibril

36. Different myosin motors have different “step lengths” on actin depending on their function. Which of the following step lengths would you expect to belong to a processive myosin motor that carries large cellular cargoes such as endocytic vesicles? a. About 2.8 nm b. About 8 nm c. About 13 nm d. About 27 nm e. About 36 nm

37. A small unique insert found near the end of the motor domain in myosin VI has been linked to the singular ability of this myosin to move toward the pointed, that is, the minus end of actin filaments. If the unique insert is removed by genetic engineering, the resulting mutant myosin VI is a plus-end directed motor. This observation indicates what? a. The unique insert is necessary and sufficient for minus-end directed movement. b. The unique insert is necessary for motor activity.

Molecular Biology of the Cell 7e-Alberts c. The unique insert is necessary for plus-end directed movement. d. The unique insert is not sufficient for minus-end directed movement. e. The unique insert is necessary for minus-end directed movement.

38. The duty ratio for a motor protein is defined as the fraction of time in each cycle of activity in which the motor head is bound with high affinity to its cytoskeletal track.

Which of the following is reasonable given this definition? a. Myosin V has a higher duty ratio than myosin II, because Pi release is the rate-limiting step in its cycle. b. Myosin II has a higher duty ratio than myosin V, because Pi release is the rate-limiting step in its cycle. c. Myosin V has a higher duty ratio than myosin II, because ADP release is the rate-limiting step in its cycle. d. Myosin II has a higher duty ratio than myosin V, because ADP release is the rate-limiting step in its cycle.

39. A microtubule is a hollow tube composed of 13 parallel protofilaments. Each protofilament is constructed of αβ-tubulin heterodimers that are stacked end-toend such that the α- and β-tubulin subunits alternate along the length of the protofilament. Each α- and β-tubulin subunit binds a molecule of GTP. What types of interactions occur between laterally-associated protofilaments in an intact microtubule (you may disregard the irregular “seam” that runs along the length of the microtubule)? a. Covalent interactions between monomers of different types b. Noncovalent interactions between monomers of the same type c. Covalent interactions between monomers of the same type d. Noncovalent interactions between bound GTP molecules e. Noncovalent interactions between monomers of different types

Molecular Biology of the Cell 7e-Alberts 40. In the following graph that shows changes in the lengths of two microtubules over time, which time point corresponds to a catastrophe for both microtubules? Which trace corresponds to a microtubule with greater dynamic instability?

a. t1; trace (a) b. t1; trace (b) c. t2; trace (a) d. t2; trace (b)

41. In contrast to growing microtubules, shrinking microtubules have what? a. A GTP cap at their plus end b. Strong lateral interactions at their plus ends c. Curved protofilaments at their plus ends d. An inability to be rescued unless microtubule-stabilizing proteins bind and inhibit depolymerization

42. Vinblastine is a tubulin-binding alkaloid found in the Madagascar periwinkle plant.

Its ability to inhibit microtubule polymerization makes it a potent anti-cancer chemotherapeutic drug. For which of the following drugs could vinblastine be

Molecular Biology of the Cell 7e-Alberts used as an alternative if your goal is to provide a different drug with the same mode of action? a. Cytochalasin B b. Taxol c. Latrunculin d. Colchicine e. Phalloidin

43. The γ-tubulin ring complex is to microtubules what the _______________ is to actin filaments. Which of the following fills in the blank? a. Arp 2/3 complex b. Dynactin complex c. Troponin complex d. Formin e. Contractile ring

44. What is the major microtubule-organizing center in animal cells? a. The γ-tubulin ring complex b. The centrosome c. The cell cortex d. The primary cilium e. The spindle pole body

45. Which of the following matches a type of cell or cellular structure with the location of the plus ends of its microtubules?

Molecular Biology of the Cell 7e-Alberts a. Yeast – nuclear envelope b. Epithelial cell – apical end c. Axon of neuronal cell – cell body d. Myotube – nuclear envelope e. Fibroblast – cell periphery

46. Which of the following proteins do you expect to be enriched near the plus end of microtubules? a. Dynein b. XMAP215 c. γ-Tubulin d. Katanin e. Plectin

47. One molecule of the tubulin-sequestering protein stathmin binds to two tubulin heterodimers, and phosphorylation of stathmin causes it to release its bound tubulin. How does this function impact microtubule dynamics? a. Phosphorylated stathmin increases the likelihood that microtubules will exhibit dynamic instability. b. Unphosphorylated stathmin increases the pool of tubulin heterodimers that are available for polymerization. c. Unphosphorylated stathmin decreases the effective concentration of tubulin heterodimers that are available for polymerization. d. Phosphorylated stathmin decreases the rate of microtubule elongation. e. Unphosphorylated stathmin makes it harder for growing microtubules to switch to a shrinking state.

Molecular Biology of the Cell 7e-Alberts 48. Which of the following descriptions applies to myosin but not to kinesin and dynein motor proteins? a. They have larger structures than the other two. b. They are generally faster than the other two. c. They are structurally unrelated to the other two. d. They walk on a different cytoskeletal filament than the other two. e. They are all minus-end directed.

49. Unlike a myosin head, a kinesin head is tightly bound to its cytoskeletal track when bound to what? a. ATP b. ADP c. No nucleotide d. GTP e. GDP

50. Dynamitin is a subunit of the dynactin complex. Its overexpression leads to the disassembly of the complex, and it is therefore considered a dynactin inhibitor.

Which of the following processes would happen if dynamitin is overexpressed in a cell? a. The motile cilia (if any) would stop beating because the axonemal dyneins could not attach to their neighboring microtubules. b. The Golgi apparatus would fragment and become dispersed because the cytoplasmic dyneins could not associate with the Golgi membranes. c. The endoplasmic reticulum would collapse because it could not recruit enough kinesins. d. The nuclear envelope would disintegrate because the nuclear lamins could not interact with the cytoplasmic cytoskeleton.

Molecular Biology of the Cell 7e-Alberts e. The cortex would lose its integrity because the Arp 2/3 complex would also be inhibited.

51. Using time-lapse fluorescence microscopy, you have recorded the transport of fluorescently labeled mitochondria in a region of an axon in a fruit-fly larva; you have plotted the results in the following graph, in which the vertical axis is time (from the top to the bottom), the horizontal axis is the position along the axon, and each trace represents one mitochondrion. Is the retrograde transport or the anterograde transport of mitochondria faster in these axons? Does the trace indicated by an asterisk correspond to a mitochondrion that was transported by a dynein or a kinesin?

a. Anterograde; kinesin b. Anterograde; dynein c. Retrograde; kinesin d. Retrograde; dynein

52. “Headless” kinesin mutants only contain the stalk (middle) and tail domains and can therefore dimerize with their wild-type kinesin partners. However, since they lack the motor (head) domain, the resulting dimers are unable to carry out processive transport of their cargoes and the mutation thus behaves as “dominant negative,” meaning that the mutant not only is nonfunctional, but can also interfere with the function of its wild-type counterparts. If a headless mutant of a kinesin heavy chain involved in melanosome movement is overexpressed in fish

Molecular Biology of the Cell 7e-Alberts melanocytes, what would you predict happens in these cells? Recall that melanosome movement in fish melanocytes is the result of the coordinated action of both kinesin and dynein motors. a. Pigment dispersion would be inhibited and there would be more tug-of-war between the kinesin and dynein motors. b. Pigment aggregation would be inhibited and there would be more tug-ofwar between the kinesin and dynein motors. c. Pigment dispersion would be inhibited and there would be less tug-of-war between the kinesin and dynein motors. d. Pigment aggregation would be inhibited and there would be less tug-of-war between the kinesin and dynein motors.

53. In the presence of ATP in a flagellum, an axonemal dynein that is interacting through its tail with the A microtubule of a peripheral doublet can push this doublet toward the ____(1)____ of the flagellum, but due to the presence of linking proteins such as ____ (2) ____, this force is converted into a bending motion. a. (1) tip; (2) nexin b. (1) base; (2) nexin c. (1) tip; (2) nebulin d. (1) base; (2) nebulin e. (1) tip; (2) tau

54. Which statement best describes keratins? a. They are localized inside the nucleus. b. They impart mechanical strength to epithelial tissues. c. They are bundled in the epidermis to make skin more flexible. d. They are expressed exclusively in skeletal, cardiac, and smooth muscle cells.

Molecular Biology of the Cell 7e-Alberts e. They provide strength and stability to neuronal axons.

55. What are the typical consequences of mutations in keratins and their associated proteins? a. Neurodegeneration as a result of interference with normal axonal transport b. Muscle development defects as a result of sarcomere disorganization c. Skeletal and cardiac abnormalities as a result of a weakened nuclear envelope d. Cornea disorders as a result of cell rupture from mechanical trauma e. Developmental defects as a result of abnormal signaling related to the primary cilia

56. In a cross section of a vertebrate axon, longitudinally organized cytoskeletal proteins can be seen as dots, as shown in the schematic drawing below. What type of filaments do these dots represent?

a. Microtubules and microfilaments b. Microtubules and vimentins c. Microtubules and neurofilaments d. Microtubules and keratins e. Microtubules and septins

Molecular Biology of the Cell 7e-Alberts

57. SUN and KASH proteins embedded in the nuclear envelope provide connections between the organization of the nucleus and the cytoplasm. Which of the following pairs of proteins do NOT bind directly to each other in these connections? a. Nuclear lamina and SUN proteins b. Plectins and KASH proteins c. KASH proteins and actin filaments d. KASH proteins and SUN proteins e. SUN proteins and cytoplasmic intermediate filaments

58. In the budding yeast Saccharomyces cerevisiae, septin proteins form filaments in the neck region between a mother cell and its growing bud to help coordinate protein distribution during cell division. All but one of the following describe additional features of septin proteins. Which option is NOT a characteristic of septins? a. They can polymerize to form paired filaments and sheets. b. They bind GTP to fold properly. c. They recruit the machinery that forms the contractile ring during cytokinesis in animal cells. d. They contribute to subcellular organization and vesicle dynamics at the plasma membrane. e. They form long polarized strands that are similar in structure to actin filaments.

59. Bacteria contain homologs of cytoskeletal filament subunits with what characteristic? a. They do not have those of the intermediate filaments. b. These homologs are incapable of nucleotide binding and hydrolysis.

Molecular Biology of the Cell 7e-Alberts c. They are less diverse in their function relative to their eukaryotic counterparts. d. They can have different functions to those of their eukaryotic counterparts. e. They are dispensable for cell growth and proliferation.

60. The following schematic shows a polarized neutrophil that is migrating toward a collection of bacterial peptides. In what region (a or b) is Rac activity highest?

What other member of the Rho family is most active at the other region?

a. Region (a); Rho b. Region (a); Cdc42 c. Region (b); Rho d. Region (b); Cdc42

61. Plant cells lack microtubules. a. True b. False

62. The pathogen Listeria monocytogenes can hijack the actin cytoskeleton within human cells and then use it to move around the viscous cytoplasm.

Understanding this, true or false: This movement can be reconstituted and studied in vitro by placing the bacteria in a mixture of actin, formin, gelsolin, and capping protein.

Molecular Biology of the Cell 7e-Alberts a. True b. False

63. All dynein motors move toward the minus end of microtubules. a. True b. False

64. All microtubules in an axon are normally oriented in the same direction. a. True b. False

65. Indicate which structures are based on actin filaments, microtubules, and intermediate filaments, respectively. a. Actin filaments b. Microtubules c. Intermediate filaments

1. The cell cortex 2. The mitotic spindle 3. The nuclear lamina 4. Cilia 5. Filopodia

66. Which of the following changes occurring during mitosis in a fibroblast is the result of the reorganization of actin filaments, microtubules, and intermediate filaments? a. Actin filaments b. Microtubules c. Intermediate filaments

1. The cell rounds up, that is, it assumes a more spherical shape.

Molecular Biology of the Cell 7e-Alberts 2. The endoplasmic reticulum collapses.

3. The Golgi apparatus fragments.

4. The nuclear envelope disperses.

5. The primary cilium is resorbed (disappears).

6. The contractile ring forms and constricts.

67. Which of the actin-binding proteins in the following schematic drawing represents tropomyosin, α-actinin, Arp 2/3 complex, CapZ, and myosin?

a. Actin-binding protein A b. Actin-binding protein B c. Actin-binding protein C d. Actin-binding protein D e. Actin-binding protein E

1. Tropomyosin 2. α-actinin 3. Arp 2/3 complex 4. CapZ 5. Myosin

68. Primary cilia are motile. a. True b. False

Molecular Biology of the Cell 7e-Alberts

69. Which of the following descriptions match microtubules, and which match intermediate filaments? a. Microtubules b. Intermediate filaments

1. They form hollow structures with multiple lateral interactions.

2. They form strong structures that are more resilient than the other two cytoskeletal filaments.

3. Their subunits bind GTP and hydrolyze it.

4. They form coiled-coil interactions between the subunits.

70. Which of the following structures is based on actin filaments, microtubules, and intermediate filaments, respectively? a. One-dimensional arrangement b. Two-dimensional arrangement c. Three-dimensional arrangement

1. Stress fibers 2. Invadopodia 3. Lamellipodia 4. Filopodia 5. Pseudopodia

71. Which microtubule-associated proteins (MAPs) correspond to the following descriptions? a. EB1 b. Kinesin-1 c. Kinesin-13 d. Katanin

Molecular Biology of the Cell 7e-Alberts 1. This is a microtubule-severing protein that can release microtubules from the microtubule-organizing centers.

2. It increases the frequency of catastrophe by deforming microtubule protofilaments.

3. This is a conventional motor protein that moves toward the plus end of a microtubule.

4. It recognizes the structure of a growing microtubule end and binds to it, helping other proteins to also bind to the plus end.

72. Which of the descriptions matches actin filaments, microtubules, and intermediate filaments? a. Actin filaments b. Microtubules c. Intermediate filaments

1. They have the smallest persistence length.

2. They lack polarity.

3. They have the highest tensile strength.

73. Which of the following descriptions matches Cdc42, Rac, or Rho from the Rho family of monomeric GTPases, respectively? a. Cdc42 b. Rac c. Rho

1. It activates formin and stimulates the extension of filipodia.

2. It inhibits myosin II activity and induces the formation of branched actin networks in lamellipodia.

3. When constitutively active, it induces the formation of many prominent stress fibers.

74. The following diagram shows two signaling pathways initiated by activated GTPases of the Rho family. Which letters (A to E) match to which structures/outcomes in the diagram?

Molecular Biology of the Cell 7e-Alberts

a. Letter A b. Letter B c. Letter C d. Letter D e. Letter E

1. Rac-GTP 2. Rho-GTP 3. Lamellipodia formation 4. Stress fiber formation 5. Stress fiber inhibition

75. Like many biological polymers such as DNA, RNA, and protein, all three types of cytoskeletal filaments are polymeric assemblies of subunits. Whereas subunits of many biological polymers are joined with covalent bonds, the cytoskeletal subunits are held together by weak noncovalent interactions. Explain the advantage of using noncovalent interactions as compared to stronger covalent bonds for forming the three types of cytoskeletal filaments.

76. You have used “optical tweezers” to study the mechanics of myosin movement in

Molecular Biology of the Cell 7e-Alberts a newly discovered member of the myosin II subfamily. Your results indicate that each myosin exerts a force of about 3 pN along the length of the actin filament, and displaces the filament by about 10 nm in each cycle of ATP hydrolysis.

Assuming that the free-energy change for ATP hydrolysis is −50 kJ/mol under your experimental conditions, what is the efficiency of the myosin motor in converting the free energy to mechanical work? Remember that under a constant force F (in newtons), work (W; in joules) is calculated as W = F × d, where d is displacement (in meters) in the direction of the force. Avogadro’s number is approximately 6 × 1023 molecules/mole. Your answer should be written as a percentage with no decimals, for example, 99%.

77.

What is the correct sequence of events that takes place during smooth muscle cell contraction? You may assume that the smooth muscle cell has just received a stimulatory signal. Your response should be a five-letter string composed of letters that correspond to individual events, e.g., ABCDE. (A) Activation of the myosin light-chain kinase (B) Interaction of myosin head with actin (C) Phosphorylation of myosin (D) Calcium release into the cytoplasm (E) Activation of calmodulin

78. A dimeric kinesin-1 molecule has 8-nm steps and can move at rates of about 1 µm/sec. Olympic 100-meter sprinters typically run at about 180 steps per minute and can reach speeds of about 42 km/h. With the same step size, if the Olympic runner had a step frequency of a kinesin-1 molecule, how fast could she run?

Your answer should be written in km/h, for example, 52 km/h.

79. Describe the arrangement of microtubules and associated proteins in a flagellum or cilium. Explain how the proteins work together to produce motion, and then predict what would happen if the structure was treated with the protein-digesting enzyme trypsin. You may assume the enzyme leaves the microtubules intact.

Molecular Biology of the Cell 7e-Alberts 80. A 2013 study showed that the overexpression of filaggrin in epidermal cells correlates with the severity of symptoms associated with chronic idiopathic urticaria (hives). State the function of filaggrin and propose a link between its function and the persistent itchy red welts that characterize this condition.

81. Describe the role of the small GTP-binding protein Cdc42 in the formation of a bud in a replicating yeast cell and then provide a hypothesis for how bud formation will be affected by a null mutation in the Cdc42 guanine nucleotide exchange factor (GEF).

82. Crumbs proteins are responsible for establishing the apical domain in polarized epithelial cells. This domain also contains the minus ends of microtubules that help polarize the secretory pathway. You hypothesize that it may be the Crumbs proteins themselves that are causing the microtubules to align in this manner.

Considering what you know about microtubule nucleation, propose a target for the Crumbs proteins and describe how it would allow the Crumbs proteins to influence microtubule alignment.

83. List in the correct order the series of cytoskeletal events that occurs in a newly fertilized egg of the nematode worm C. elegans. Assume that the paternal sperm nucleus has just entered a symmetrical egg cell. Your response should be a fiveletter string composed of letters that correspond to individual events, e.g., ABCD. (A) Asymmetric localization of PAR proteins (B) Sperm centrosome nucleates a microtubule aster (C) Uneven first cell division along anterior-posterior axis (D) Loss of Rho activity decreases myosin II contractility

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 16 1. Answer:

D

2. Answer:

C

3. Answer:

B

4. Answer:

E

5. Answer:

A

6. Answer:

D

7. Answer:

C

8. Answer:

A

9. Answer:

D

10. Answer:

D

11. Answer:

D

12. Answer:

D

13. Answer:

A

14. Answer:

B

15. Answer:

B

16. Answer:

B

17. Answer:

B

18. Answer:

E

19. Answer:

D

20. Answer:

C

Molecular Biology of the Cell 7e-Alberts 21. Answer:

A

22. Answer:

E

23. Answer:

B

24. Answer:

D

25. Answer:

E

26. Answer:

D

27. Answer:

D

28. Answer:

C

29. Answer:

D

30. Answer:

C

31. Answer:

A

32. Answer:

A

33. Answer:

C

34. Answer:

D

35. Answer:

E

36. Answer:

E

37. Answer:

E

38. Answer:

C

39. Answer:

B

40. Answer:

D

41. Answer:

C

42. Answer:

D

Molecular Biology of the Cell 7e-Alberts 43. Answer:

A

44. Answer:

B

45. Answer:

E

46. Answer:

B

47. Answer:

C

48. Answer:

D

49. Answer:

A

50. Answer:

B

51. Answer:

A

52. Answer:

C

53. Answer:

B

54. Answer:

B

55. Answer:

D

56. Answer:

C

57. Answer:

E

58. Answer:

E

59. Answer:

D

60. Answer:

C

61. Answer:

B

62. Answer:

B

63. Answer:

A

64. Answer:

A

Molecular Biology of the Cell 7e-Alberts 65. Answer:

1. A 2. B 3. C 4. B 5. A

66. Answer:

1. A 2. B 3. B 4. C 5. B 6. A

67. Answer:

1. D 2. B 3. C 4. E 5. A

68. Answer: 69. Answer:

B 1. A 2. B 3. A 4. B

70. Answer:

1. A 2. C 3. B 4. A

Molecular Biology of the Cell 7e-Alberts 5. C 71. Answer:

1. D 2. C 3. B 4. A

72. Answer:

1. C 2. C 3. C

73. Answer:

1. A 2. B 3. C

74. Answer:

1. B 2. A 3. D 4. C 5. E

75. Answer:

Though strong, it takes considerably more time and energy to break and form covalent bonds than it does to do the same for noncovalent interactions. Thus, noncovalent interactions allow for the rapid subunit assembly and disassembly exhibited by all three types of cytoskeletal filaments.

76. Answer:

36%

77. Answer:

DEACD

78. Answer:

1750 km/h

79. Answer:

Microtubules and associated proteins are arranged into an axoneme that includes nine microtubule doublets organized in a ring around a pair of single microtubules. Nexin and spoke proteins form cross-links that connect the microtubule

Molecular Biology of the Cell 7e-Alberts doublets to each other and to the central pair, respectively.

Dynein motor proteins form bridges between the adjacent doublets. When activated, the dynein motors slide one doublet relative to another, but the sliding force is converted into a bending motion due to the nexin cross-links.

Alternating waves of dynein activation that travel down both sides of the axoneme produce the back-and-forth motion characteristic of flagella and cilia. If the axoneme was treated with trypsin, then the enzyme would break the nexin crosslinks and prevent the microtubule doublets from bending relative to one another.

80. Answer:

Filaggrin is a keratin-bundling accessory protein of intermediate filaments that gives the epidermis its special toughness. Overexpression of filaggrin may promote excessive bundling of keratin filaments and an overall decrease in the pliability of the outer skin layers. This abnormally “stiff” epidermis may trigger inflammatory responses that lead to itchy red welts on the skin.

81. Answer:

In budding yeast cells, Cdc42-GDP is distributed uniformly on the plasma membrane. Occasionally, one of these molecules will exchange its GDP for GTP. Cdc42-GTP recruits the protein kinase PAK, and PAK recruits a scaffolding protein that brings Cdc42’s GEF to the membrane. The GEF causes nucleotide exchange at additional Cdc42 molecules to produce an active Cdc42-GTP cluster at the cell membrane. This cluster recruits and activates the formin proteins that stimulate rapid production of actin filaments. These filaments polarize the cell and promote bud formation by allowing myosin V motors to deliver materials to the bud site. A null mutation in the Cdc42 GEF would halt bud site identification by preventing the accumulation of a Cdc42-GTP cluster.

82. Answer:

A logical target for the Crumbs proteins is the γ-tubulin ring complex (γ-TuRC). This complex is responsible for nucleating microtubules from their minus ends. Thus, if Crumbs proteins recruited γ-TuRC’s to their location within the apical domain of the cell, the ring complexes could polymerize microtubules toward the basolateral membrane to align the filaments appropriately. (MTOC proteins, centrosome proteins, and other minus end-directed proteins like augmin or CAMSAPs also may be suggested as logical answers, provided the student explains how the suggested target allows Crumbs proteins to orient microtubules with their minus ends at the apical domain.)

Molecular Biology of the Cell 7e-Alberts 83. Answer:

BDAC

Molecular Biology of the Cell 7e-Alberts

Chapter 17: The Cell Cycle 1. Consider two mammalian cells, one in G1 and the other in G0 (stationary) phase.

If they are stimulated to pass the restriction point by the addition of an extracellular proliferation signal, but the signal is then immediately removed, what would you expect to happen? a. Both cells will replicate their DNA. b. Only the G1 cell will replicate its DNA. c. Only the G0 cell will replicate its DNA. d. Only the G1 cell will start to replicate its DNA, but will stop halfway through the replication and will not reach G2. e. Neither of the cells will replicate their DNA.

2. During G1 phase, the cell checks whether the external conditions are favorable for cell division. If the conditions are unfavorable, what happens? a. The cell delays progress through G1. b. The cell continues to progress through G1. c. The cell enters S phase. d. The cell dies. e. The cell irreversibly enters G0 phase.

3. In the following schematic diagram of a typical eukaryotic cell cycle, choose two major time points (among A to E) at which the cell-cycle control system normally arrests the cycle if DNA damage is detected.

Molecular Biology of the Cell 7e-Alberts

a. B and D b. A and B c. C and D d. B and C e. A and C

4. During a flow cytometry analysis of the cell cycle of a proliferating cell population, the distribution of cells obtained shows which of the following? a. A fewer number of cells in G1 than in G2+M b. A greater number of cells in G1 than in G2+M c. G2+M is longer than G1 d. The number of cells is similar in G1 and G2+M phase e. The number of cells is similar in G1 and S phase

Molecular Biology of the Cell 7e-Alberts

5. Resveratrol is a natural compound found in red grapes (and red wine) and is thought to have beneficial effects in mammals, such as preventing tumor growth and delaying age-related diseases. In vitro, resveratrol and its derivatives have been shown to cause cell-cycle arrest in S phase. You have analyzed the DNA content of cultured cells in the presence and absence of these drugs using fluorescence-activated cell sorting. Compared with control cells (dashed line), which of the following curves do you think better represents the DNA content of cells treated with these compounds?

a.

b.

Molecular Biology of the Cell 7e-Alberts

c.

d.

e.

You have been studying the effect of loss-of-function mutations in the Cdk inhibitor protein (CKI) p21. You add the drug fucoxanthin to cell cultures harboring either wild-type or mutant versions of the p21 gene. Fucoxanthin is known to induce cell-

Molecular Biology of the Cell 7e-Alberts cycle arrest in G1. After a day, you add the thymidine analog BrdU to the culture media, collect the cells after an hour, treat them with anti-BrdU antibody and the fluorescent DNA stain DAPI, and finally subject them to fluorescence-activated cell sorting (FACS). The FACS data can be viewed as a two-dimensional dot plot composed of thousands of dots, in which each cell is represented by one dot at coordinates that correspond to the intensities of the DAPI fluorescence signal (X axis) and BrdU fluorescence signal (Y axis) for that cell. With reference to the simplified dot plot below, generated from your experiment, answer the following question(s).

6. Which sequence of boxed regions (1, 2, or 3) in the FACS plots corresponds better to the phases of the cell cycle? a. 231 b. 213 c. 321 d. 312 e. 123

7. Which one of the FACS plots (A or B) would you expect to correspond to the lossof-function p21 mutants? a. Plot A b. Plot B

Molecular Biology of the Cell 7e-Alberts

8. Which of the time points (A to E) in the following schematic drawing of the mammalian cell cycle represents the restriction point?

a. Time point A b. Time point B c. Time point C d. Time point D e. Time point E

9. Which of the following simplified diagrams better shows the changes in the concentrations of three major cyclin–Cdk complexes (G1/S-Cdk, S-Cdk and MCdk) in the cell in different stages of the cell cycle?

a.

Molecular Biology of the Cell 7e-Alberts

b.

c.

d.

e.

10. M-cyclins activate Cdks that do what? a. Cdks that stimulate chromosome duplication b. Cdks that trigger progression through Start c. Cdks that stimulate entry into mitosis at the G 2/M transition d. Cdks that promote DNA replication e. Cdks that remain at elevated levels until mitosis

11. The major cyclins and Cdks of vertebrates associated with G 1/S Cdk are a. Cyclin D; Cdk4, Cdk6 b. Cyclin E; Cdk2

Molecular Biology of the Cell 7e-Alberts c. Cyclin A; Cdk2, Cdk1 d. Cyclin B; Cdk1 e. Cyclin D; Cdk2

12. Which of the following descriptions does not apply to the Wee1 protein? a. Its loss in fission yeast results in small cell size. b. It activates M-Cdk complexes. c. This protein kinase inhibits Cdk activity. d. The phosphatase Cdc25 dephosphorylates sites phosphorylated by Wee1. e. M-Cdk activates Cdc25 and inhibits Wee1.

13. Anaphase-promoting complex, or cyclosome (APC/C), triggers the destruction of which of the following? a. Securin b. Cohesin c. Cdc20 d. Separase e. G1/cyclins

14. How is Cdc20–APC/C similar to Cdh1–APC/C? a. They are both active throughout interphase. b. They are both inhibited by M-Cdk. c. They both inhibit M-Cdk activity. d. They are both activated suddenly at the onset of mitosis.

Molecular Biology of the Cell 7e-Alberts e. They are both inactivated soon after anaphase.

15. In most mammalian cells, low M-cyclin protein levels are maintained during G1 phase. What is mainly responsible for maintaining these low levels? a. Cdc20–APC/C b. Cdh1–APC/C c. Skp2–SCF d. β-trCP–SCF e. p27

16. Which of the following is a substrate of the ubiquitin ligase SCF? a. Securin b. p27 c. Cdt1 d. S-cyclins e. APC/C

17. Which of the following events occurs in mitotic metaphase? a. Nuclear envelope breakdown b. Nuclear envelope reassembly c. Chromosome attachment to spindle microtubules for the first time d. Chromosome alignment at the spindle equator e. Mitotic spindle assembly

Molecular Biology of the Cell 7e-Alberts 18. Which of the following occurs mainly in S phase of the cell cycle? a. DNA helicase activation b. DNA helicase deposition on DNA at the replication origins c. Assembly of new DNA helicases at replication origins d. Licensing of replication origins e. Chromosome segregation

19. The antitumor drug paclitaxel promotes the assembly of microtubules and inhibits their depolymerization. What is the consequence of this effect on the cell cycle? a. The induction of cell-cycle progression b. The arrest of cell-cycle progression at M phase c. The promotion of chromosome segregation in mitosis d. The blockage of S-phase progression e. The suppression of cell death

20. Consider two kinesin motor proteins at the mitotic spindle midzone: kinesin-5 is a tetrameric motor that walks toward the plus end of both microtubules to which it is attached via its motor domains; kinesin-14, on the other hand, walks toward the minus end of one microtubule while it is attached to another microtubule via its tail domain. How do these motors affect the length of the spindle? a. They both work to shorten the spindle. b. Kinesin-5 works to shorten the spindle whereas kinesin-14 works to lengthen it. c. Kinesin-5 works to lengthen the spindle whereas kinesin-14 works to shorten it. d. They both work to lengthen the spindle.

Molecular Biology of the Cell 7e-Alberts 21. How is centrosome duplication similar to DNA replication? a. Only centrosome duplication uses a semiconservative mechanism. b. They are initiated at around the same time in the cell cycle, near the G2/M transition. c. They are both controlled in such a way that they replicate twice and only twice per cell cycle. d. They both involve separation of sister copies in mitosis. e. Chromosomes must replicate once to ensure that the cell enters mitosis with precisely one copy.

22. Which of the following is more directly driven by active M-Cdk? a. Centrosome maturation b. Centrosome duplication c. Nuclear envelope reassembly d. Inactivation of APC/C e. Cell cleavage

23. The cell cycle can be arrested in mitosis when a single sister-chromatid pair is mono-oriented on the mitotic spindle. If at this point a glass microneedle is used to pull the mono-oriented chromosome toward the pole to which it is not attached, the cell proceeds to anaphase. What does this observation confirm? a. Mono-oriented chromosomes are stable. b. Mechanical tension drives the formation of bi-oriented chromosomes. c. Bi-oriented chromosomes activate the spindle assembly checkpoint. d. Lack of mechanical tension at the kinetochore in at least one chromosome prevents entry into anaphase. e. Mechanical tension at the kinetochore in at least one chromosome is

Molecular Biology of the Cell 7e-Alberts required for entry into anaphase.

24. What is true about the kinetochore, a large structure formed at the centromeric region of each eukaryotic chromosome? a. It impedes the attachment of the chromosomes to the spindle poles. b. It captures spindle microtubules in mitosis. c. It can bind one microtubule in an animal cell. d. It is attached to a microtubule by interactions with one copy of the Ndc80 complex. e. It can bind more than10 microtubules in budding yeast.

25. Which one of the following chromosomes has formed stable attachment(s) to the spindle microtubules?

a.

b.

c.

Molecular Biology of the Cell 7e-Alberts

d.

e.

26. Imagine a prometaphase chromosome pair that is attached to one spindle pole.

Which of the following would happen if both arms of the chromosome are severed with a strong laser beam? a. All chromosome fragments (the centromere-containing fragment and the arm fragments) would be pushed away from the pole. b. Only the two arm fragments would be pulled toward the pole. The kinetochore-containing fragment would remain stationary. c. The kinetochore-containing fragment would be pulled toward the pole. The two arm fragments would move away from the pole. d. The kinetochore-containing fragment would move away from the pole, but the two arm fragments would be pulled toward the pole.

27. Below is a schematic drawing of a vertebrate prometaphase chromosome as seen under a microscope. Is the chromosome more likely to be closer to pole A or pole B?

Molecular Biology of the Cell 7e-Alberts

a. Pole A b. Pole B

28. Mutant Saccharomyces cerevisiae cells that lack the gene encoding securin can divide normally by mitosis, without significant chromosome segregation defects.

Cells harboring a nondegradable version of securin, on the other hand, arrest in metaphase as expected, since they cannot activate separase to enter anaphase.

Similarly, cells lacking Cdc20 arrest in metaphase, since they cannot activate APC/C. Finally, cells lacking both securin and Cdc20 arrest in anaphase: they manage to separate sister chromatids, but do not progress much further. What do these results suggest about activity in wild-type cells? a. Degradation of securin is necessary to trigger sister-chromatid separation. b. Degradation of securin is sufficient to trigger sister-chromatid separation. c. Cdc20–APC/C is NOT necessary for sister-chromatid separation. d. Cdc20–APC/C is NOT necessary for later events in anaphase.

29. If cells that have started mitosis are treated with nocodazole, a drug that depolymerizes microtubules, what would you expect to happen? Where would you expect Mad2 protein to be localized? a. The cells would arrest in prometaphase; Mad2 would be localized to the spindle poles. b. The cells would arrest in telophase; Mad2 would be localized to the spindle poles.

Molecular Biology of the Cell 7e-Alberts c. The cells would immediately enter anaphase, finish mitosis, and enter G 1; Mad2 would be localized to the spindle poles. d. The cells would arrest in prometaphase; Mad2 would be localized to almost all kinetochores. e. The cells would arrest in telophase; Mad2 would be localized to almost all kinetochores.

30. Treatment of dividing cells with a low dose of the antifungal drug benomyl, which destabilizes microtubules, slows down correct spindle assembly. But at such doses, the spindle is eventually formed, and the cells survive. However, mutations in some genes confer benomyl sensitivity: the mutant cells die because they fail to arrest the cell cycle in the presence of unattached kinetochores and progress through anaphase, with disastrous consequences. Which of the following would you expect to be a benomyl-sensitive mutant? a. A mutation causing the overexpression of Mad2 b. A loss-of-function mutation in the gene encoding Cdc20 c. A loss-of-function mutation in the gene encoding a kinase that inhibits Cdc20–APC/C d. A mutation causing the overexpression of a tubulin subunit

31. To study chromosome movement during anaphase in mammals, you have injected fluorescently labeled tubulin subunits into cells such that microtubules can be seen under a fluorescence microscope. You then use a relatively strong laser beam to bleach the fluorescent dyes in a limited area of a metaphase cell, as indicated in the schematic diagram below, and follow the progression of mitosis under the microscope by time-lapse imaging. You then use the images to measure the change in distance between various components and plot the results in the graph below. Answer the following questions based on these results.

Molecular Biology of the Cell 7e-Alberts

Which is dominant in this cell: Anaphase A or anaphase B? Which force is dominant in chromosome movement in this cell: Polar ejection force, microtubule flux, or kinetochore microtubule plus-end depolymerization? a. Anaphase A; polar ejection force b. Anaphase B; polar ejection force c. Anaphase A; microtubule flux d. Anaphase B; kinetochore microtubule plus-end depolymerization

32. Which of the following motor proteins is more directly involved in anaphase B? a. Kinesin-5 on interpolar microtubules and dynein on kinetochore microtubules b. Kinesin-13 on kinetochore microtubules and dynein on astral microtubules c. Kinesins 4 and 10 on interpolar and astral microtubules and dynein on kinetochore microtubules

Molecular Biology of the Cell 7e-Alberts d. Kinesin-5 on interpolar microtubules and kinesins 4 and 10 on interpolar and astral microtubules e. Kinesin-5 on interpolar microtubules and dynein on astral microtubules

33. The microtubule-binding protein Patronin binds to the minus ends of microtubules at the spindle pole and protects them from the effect of catastrophe factors. If Patronin activity is inhibited by injecting an anti-Patronin antibody into Drosophila melanogaster embryos prior to cellularization, anaphase B is suppressed, and the spindles are significantly shorter. Which of the forces (1 or 2) in the following schematic diagram do you think is mostly responsible for anaphase B in this organism at this stage? Does the effect of Patronin inhibition resemble that of kinesin-5 inhibition or overactivation?

a. 1; kinesin-5 inhibition b. 1; overactivation c. 2; kinesin-5 inhibition d. 2; overactivation

34. In the following schematic diagram of an early Drosophila embryo, which step (A to C) corresponds to cellularization?

Molecular Biology of the Cell 7e-Alberts

a. Step A b. Step B c. Step C

35. Nuclear division without cytokinesis occurs in which of these mammalian cell types? a. Megakaryocytes b. Osteoblasts c. Macrophages d. Splenocytes e. Fibroblasts

36. Which stage of meiosis is usually the longest? a. Prophase I b. Prometaphase I c. Telophase I d. Prophase II e. Prometaphase II

Molecular Biology of the Cell 7e-Alberts

37. What is the principal function of mitogens? a. To stimulate cell growth b. To promote cell survival c. To stimulate cell division d. To induce cell death e. To suppress apoptosis

38. Formin nucleates the growth of parallel actin bundles in the cell. Additionally, myosin motor activity is positively regulated by phosphorylation. The monomeric G protein RhoA is important in cytokinesis because it directly or indirectly does what? a. Activates formins and inactivates myosin light-chain phosphatase b. Inactivates formins and activates myosin light-chain kinases c. Activates formins as well as myosin light-chain phosphatases d. Inactivates formins as well as myosin light-chain kinases

39. Which division in meiosis is more similar to mitosis? In which division do sister chromatids separate from each other? a. Meiosis I; meiosis I b. Meiosis I; meiosis II c. Meiosis II; meiosis I d. Meiosis II; meiosis II

40. In many multicellular organisms, loss of the gene encoding shugoshin leads to sterility, suggesting defects in meiosis. What would you expect to occur in meiotic cells that lack shugoshin?

Molecular Biology of the Cell 7e-Alberts a. The homologs fail to separate in anaphase I. b. The sister chromatids fail to separate in anaphase II. c. All chromatids separate prematurely in anaphase I. d. Removal of cohesion between homolog arms fails in prophase I. e. Removal of cohesion between sister chromatids fails in prophase II.

41. Which one better supports cell proliferation when added to fibroblast cultures: serum or plasma? This activity is due to the presence of mitogens in this fluid.

What is responsible for making these mitogens? a. Serum; red blood cells b. Serum; platelets c. Plasma; red blood cells d. Plasma; white blood cells e. Plasma; platelets

42. Which of the following is an inhibitory extracellular signal for cell proliferation? a. EGF b. TGFβ c. Erythropoietin d. IGF e. PDGF

43. Which of the following cell populations in our body has the highest mitotic index? a. Neurons b. Hepatocytes

Molecular Biology of the Cell 7e-Alberts c. Red blood cells d. Fibroblasts e. Skeletal myocytes

44. Which protein is the product of an immediate early gene expressed following mitogenic stimulation of cell-cycle entry? a. E2F b. Rb c. Myc d. G1-cyclins

45. Which of the following events occurs at the onset of S phase in a mitogenstimulated vertebrate cell? a. Activation of Cdh1–APC/C b. Activation of Rb c. Activation of geminin d. Activation of Cdc20

46. Which of the following events contributes to driving the mammalian cell past the restriction point of the cell cycle? a. Phosphorylation of Cdh1–APC/C by M-Cdk b. Induction of CKIs that target S-Cdks c. Repression of E2F gene expression by active E2F protein d. Dephosphorylation of Rb by G1-Cdk, G1/S-Cdk, and S-Cdk

Molecular Biology of the Cell 7e-Alberts 47. In animal cells, G1-Cdk complexes play a key function in what? a. The activation of E2F proteins b. The binding of retinoblastoma protein family members to E2F c. The inhibition of the transcriptional activity of E2F d. The dephosphorylation of retinoblastoma protein e. M-cyclin levels

48. p53 binds to DNA in a regulatory region of which gene? a. p27 gene b. p19 gene c. p16 gene d. p21 gene e. p14 gene

49. DNA damage induces the recruitment of various protein kinases to the site of damage and activates the signaling pathway involved in cell-cycle arrest. In this process, a CKI binds and inactivates G1/S-Cdk and S-Cdk complexes, arresting the cell in G1. For which mechanism is the CKI responsible? a. p16 b. p19 c. p21 d. p27 e. p14

50. A cell has been subjected to ultraviolet irradiation, causing a significant number of

Molecular Biology of the Cell 7e-Alberts mutations in the genome. Which of the following would you NOT expect to occur as a result? a. Activation of the protein kinase ATR b. Activation of the protein kinase Chk1 c. Inactivation of the protein phosphatase Cdc25 d. Binding of p53 to Mdm2 e. Stabilization of p53

51. You have synchronized a culture of HeLa cells so that the cells are all at the same stage in the cell cycle. You then treat the cells with serum containing nutrients and growth factors in the presence or absence of the drug rapamycin.

The distribution of cell sizes in the two cell populations is shown below. Based on these results, do you think rapamycin is an activator or inhibitor of the TOR complex? Is the S6 kinase expected to be up-regulated or down-regulated in the presence of rapamycin?

a. Activator; up-regulated b. Activator; down-regulated c. Inhibitor; up-regulated

Molecular Biology of the Cell 7e-Alberts d. Inhibitor; down-regulated

52. Chromosomes stabilize mitotic microtubules through activation of Ran monomeric G protein. a. True b. False

53. Regarding cytokinesis in animal cells, the force for cytokinesis is generated by kinesin motors on microtubule bundles that form the contractile ring. a. True b. False

54. Regarding eukaryotic cells, all recombination events lead to crossovers. a. True b. False

55. Replicative cell senescence usually arises due to the accumulation of mutations in genes encoding S- and M-cyclins. a. True b. False

56. Indicate whether each of the following phosphorylation events typically activates or inactivates the protein that is being phosphorylated. a. Activates b. Inactivates

1. Phosphorylation of Cdc25 by M-Cdk 2. Phosphorylation of PP2A-B55 by M-Cdk 3. Phosphorylation of Wee1 by M-Cdk 4. Phosphorylation of Greatwall by M-Cdk

Molecular Biology of the Cell 7e-Alberts 57. The following schematic diagram shows the activation of M-Cdk in mitosis.

Indicate which proteins below correspond to those indicated as A to D in the diagram.

a. A b. B c. C d. D

1. Wee1 2. CAK 3. Cdc25 4. M-cyclin

58. Indicate whether each of the following, when active, tends to activate or inactivate M-Cdks. a. Activate b. Inactivate

1. M-Cdk 2. Cdc20–APC/C 3. Cdh1–APC/C 4. Sic1

Molecular Biology of the Cell 7e-Alberts

59. The preprophase band, composed of microtubules and actin filaments, marks the site of cytokinesis in plant cells. a. True b. False

60. Use the following schematic diagram of a meiotic bivalent in diplotene in order to identify whether each of the following chromatid pairs have undergone DNA exchange. (Note: The final form of chromatid 3 is shown in gray for clarity only.)

a. Has undergone DNA exchange b. Has not undergone DNA exchange

1. Chromatids 1 and 3 2. Chromatids 1 and 4 3. Chromatids 2 and 3 4. Chromatids 2 and 4

61. Indicate whether each of the following descriptions better applies to mitogens, growth factors, or survival factors. a. Mitogens b. Growth factors c. Survival factors

Molecular Biology of the Cell 7e-Alberts

1. They unblock cell-cycle progression.

2. They suppress apoptosis.

3. They trigger a wave of G1/S-Cdk activity.

4. They inhibit the degradation of cellular macromolecules.

62. In a multicellular organism such as a mammal, loss-of-function mutations in many genes can contribute to the development of cancer. These genes are therefore called tumor suppressors. In their absence, the cell fails to stop progression through the cell cycle under conditions in which normal cells would arrest, paving the way for tumorigenesis. Indicate whether each of the following proteins is or is not expected to be the product of a tumor suppressor gene based on its function in the cell cycle. a. Is expected b. Is not expected

1. Rb 2. Myc 3. p53 4. p21

63. Sort the following schematic diagrams (A through F) to reflect the order of events in a typical eukaryotic M phase. An interphase cell in G 2 phase is drawn on the left for comparison. Your answer would be a six-letter string composed of letters A to F only, for example, BCEADF.

64. How would you describe the eukaryotic cell cycle phases and its transitions?

65. How do CKIs inhibit cyclin-Cdk complexes? Explain your reasoning in 2-3

Molecular Biology of the Cell 7e-Alberts sentences.

66. The drug nocodazole disrupts microtubule assembly/disassembly dynamics. If we perform a cell-cycle analysis by flow cytometry of treated cells following propidium iodine staining, what will the DNA profile show? As part of your analysis, draw a graph that reflects these results.

67. Using a protein that emits fluorescence when Ran-GTP is active, where do we see Ran activity during mitosis? Why do we have this expectation?

68. Cells harboring mutated mitogen receptors result in a constitutively active signaling pathway. How will this stimulate cell-cycle entry?

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 17 1. Answer:

A

2. Answer:

A

3. Answer:

D

4. Answer:

B

5. Answer:

B

6. Answer:

B

7. Answer:

A

8. Answer:

B

9. Answer:

A

10. Answer:

C

11. Answer:

B

12. Answer:

B

13. Answer:

A

14. Answer:

C

15. Answer:

B

16. Answer:

B

17. Answer:

D

18. Answer:

A

19. Answer:

B

20. Answer:

C

Molecular Biology of the Cell 7e-Alberts 21. Answer:

D

22. Answer:

A

23. Answer:

D

24. Answer:

B

25. Answer:

D

26. Answer:

C

27. Answer:

A

28. Answer:

A

29. Answer:

D

30. Answer:

C

31. Answer:

C

32. Answer:

E

33. Answer:

A

34. Answer:

C

35. Answer:

A

36. Answer:

A

37. Answer:

C

38. Answer:

A

39. Answer:

D

40. Answer:

C

41. Answer:

B

42. Answer:

B

Molecular Biology of the Cell 7e-Alberts 43. Answer:

D

44. Answer:

C

45. Answer:

C

46. Answer:

C

47. Answer:

A

48. Answer:

D

49. Answer:

C

50. Answer:

D

51. Answer:

D

52. Answer:

A

53. Answer:

B

54. Answer:

B

55. Answer:

B

56. Answer:

1. A 2. A 3. B 4. A

57. Answer:

1. C 2. B 3. D 4. A

58. Answer:

1. A 2. B

Molecular Biology of the Cell 7e-Alberts 3. B 4. B 59. Answer: 60. Answer:

A 1. B 2. B 3. A 4. B

61. Answer:

1. A 2. C 3. A 4. B

62. Answer:

1. A 2. B 3. A 4. A

63. Answer:

CFEDAB

64. Answer:

The eukaryotic cell cycle is divided into four sequential phases: G1, S, G2 , and M. During this process, a controller evaluates the information about the completion of cell-cycle events and different signals coming from the environment, deciding if the cell cycle continues or stops.

There are three transitions: Start (or the restriction point) in late G1, the G2/M transition, and the metaphase-toanaphase transition.

65. Answer:

CKIs bind to Cdks and cyclin complexes by modifying the disposition of the active site of the Cdks. As a consequence, the inactivation of these enzymes takes place.

66. Answer:

Cells treated with nocodazole will arrest in G2/M, and therefore result in the DNA profile showing a higher

Molecular Biology of the Cell 7e-Alberts proportion of cells in the G2/M phase of the cell cycle.

67. Answer:

In the metaphase, Ran activity is highest around the chromosomes, between the poles of the mitotic spindle. Its role is to stimulate the local nucleation and stabilization of microtubules around chromosomes.

68. Answer:

A constitutively active mitogenic signaling results in an increase of active G1-Cdk activity, which phosphorylates the retinoblastoma family of proteins. The inactivation of these proteins allows E2F family of transcription factors to control the expression of genes encoding G1/S-cyclins, Scyclins and E2F1. The expression of these genes will further enhance Rb protein phosphorylation. The constitutive activation of these pathways may be associated with carcinogenesis.

Molecular Biology of the Cell 7e-Alberts

Chapter 18: Cell Death 1. Which morphological change is NOT typically seen in a cell that is undergoing apoptosis? a. The cell rounds up. b. The nuclear envelope disassembles. c. The cell swells. d. Large cells break up into membrane-enclosed fragments. e. The nuclear chromatin breaks into fragments.

2. Regarding the functions of cell death in animal development, which of the following does NOT belong? a. To help sculpt hand and feet b. To eliminate abnormal cells c. To return the cell number in an organ to normal d. To eliminate misplaced cells

3. Which of the following is an executioner caspase? a. Caspase-8 b. Caspase-3 c. Caspase-5 d. Caspase-2 e. Caspase-9

4. Initiator and executioner caspases share all of the following features EXCEPT which of the following?

Molecular Biology of the Cell 7e-Alberts a. They are cysteine proteases (they have a cysteine residue at their active site). b. Their inactive form is a monomer. c. They undergo cleavage during activation. d. Their active form is a dimer. e. They are inhibited by IAPs.

5. v-FLIPs are viral proteins that were first identified as modulators of apoptosis; they contain two death effector domains, which are also found in some initiator caspases such as procaspase-8. These v-FLIP proteins can be recruited to the DISC through the binding of the death effector domain to similar domains in the adaptor proteins, but are otherwise catalytically inactive. What do you think is the effect of v-FLIP expression in the host cell? a. It promotes apoptosis mainly via the extrinsic pathway. b. It inhibits the extrinsic pathway of apoptosis. c. It activates only the mitochondrial pathway of apoptosis. d. It inhibits the intrinsic pathway of apoptosis. e. It enhances the caspase cascades in both the intrinsic and extrinsic pathways.

6. Fas receptors cluster and expose their death effector domain (DED) on the receptor tails, enabling them to bind to what? a. FADD b. Caspase-9 c. FLIP d. FLICE e. Apaf1

Molecular Biology of the Cell 7e-Alberts

7. In the intrinsic pathway of apoptosis, the mitochondria releases a protein into the cytosol, which binds to the adaptor protein Apaf1. Apaf1 then oligomerizes into a wheel-like assembly called an apoptosome, which recruits initiator caspase-9 proteins. What is the name of this protein? a. Cytochrome c b. FADD c. Fas d. Bad e. Bim

8. The intrinsic pathway of apoptosis dictates that proteins be released from which of the following? a. Endoplasmic reticulum b. Cell membrane c. Mitochondria d. Golgi e. Lysosome

9. Which of the following proteins activates the mitochondrial pathway of apoptosis? a. The tumor suppressor protein Bcl2, when activated in response to extensive DNA damage. b. The BH3-only protein Bid, when cleaved by the initiator caspase-9 (from the extrinsic pathway). c. The anti-IAP protein XIAP, when activated by dephosphorylation. d. The BH3-only protein Puma, when activated by phosphorylation. e. The BH3-only protein Bad, when activated by dephosphorylation.

Molecular Biology of the Cell 7e-Alberts

10. In cells irradiated with ultraviolet light, where is cytochrome-c located? a. Mitochondria b. Golgi c. Endoplasmic reticulum d. Nucleus e. Cytosol

11. Where is the caspase recruitment domain (CARD) present? a. Inactive caspase-8 monomer b. Active Apaf1 c. Cytochrome c d. Death effector domain (DED) e. FADD

12. Anti-apoptotic Bcl2 family proteins share which of the following Bcl2 homology (BH) domains? a. BH1 and BH2 b. BH2, BH3 and BH4 c. BH4, BH3, BH1 and BH2 d. BH3, BH1 and BH2 e. BH4 and BH3

13. Which of the following is among the pro-apoptotic Bcl2 family effectors? a. Bax

Molecular Biology of the Cell 7e-Alberts b. BclxL c. Bad d. Bcl2 e. Mcl1

14. Soon after the discovery of nerve growth factor (NGF), researchers injected newborn mice with rabbit antiserum (i.e., serum that contains antibodies) against NGF. They observed massive nerve cell death compared to appropriate control injections. Up to 99% of the neurons in some parts of the developing peripheral nervous system died after about a week of daily injections. What do these results suggest? a. NGF signaling sensitizes developing nerve cells to apoptotic signals. b. Developing neurons undergo apoptosis in the presence of rabbit proteins. c. Developing neurons require NGF for apoptosis. d. Developing neurons undergo cell death in the absence of NGF. e. NGF signaling is sufficient for survival of the developing neurons.

15. The pro-apoptotic BH3-only protein Bad is phosphorylated and inactivated by which protein? a. Akt b. PI3K c. PTEN d. PKC e. PDK1

16. Which of the following is characteristic of the phospholipid phosphatidylserine (PS)?

Molecular Biology of the Cell 7e-Alberts a. It accumulates in the nucleus of healthy cells. b. It accumulates on the cell surface of healthy cells. c. It is confined to the inner leaflet by an activated ATP-dependent aminophospholipid flippase (ATP11C). d. It is confined to the nucleus by an activated ATP-dependent aminophospholipid flippase (ATP11C). e. It is sufficiently removed from the cell surface by scramblase activation.

17. Apoptotic cells are efficiently phagocytosed by neighboring cells or macrophages.

What does NOT normally happen in this process? a. The apoptotic cell releases some of its cytoplasmic content to induce a local inflammatory response. b. The apoptotic cell exposes phosphatidylserine at its surface, which interacts with receptor proteins on the surface of phagocytes via “bridging” proteins. c. The apoptotic cell loses or inactivates “don’t eat me” signals. d. The apoptotic cell rounds up and detaches from its neighbors, which facilitates phagocytosis.

18. Apoptosis is the final fate of almost all cells in an adult animal. a. True b. False

19. In about half of human cancers, the gene that encodes the tumor suppressor protein p53 is mutated. a. True b. False

20. For each of the following proteins that are proteolytically cleaved in apoptosis,

Molecular Biology of the Cell 7e-Alberts indicate whether the cleavage is first carried out by an initiator or executioner caspase. a. Initiator b. Executioner

1. The initiator caspase-2 2. The executioner caspase-3 3. The BH3-only protein Bid 4. The endonuclease inhibitor iCAD 5. The nuclear protein Lamin A

21. The Bcl2 family is comprised of anti-apoptotic, BH3-only, and effector proteins. In the following diagram representing the regulation of the intrinsic pathway of apoptosis, what class of activated Bcl2 family proteins corresponds to boxes 1 to 3, respectively?

a. Anti-apoptotic proteins b. BH3-only proteins c. Effector proteins

1. Box 1 2. Box 2 3. Box 3

22. Indicate whether each of the following mutations would likely promote or inhibit apoptosis in cells harboring the mutation(s). a. Promote

Molecular Biology of the Cell 7e-Alberts b. Inhibit

1. Mutations in the pro-apoptotic effector Bcl2 family proteins Bax and Bak that prevent their association with the outer mitochondrial membrane.

2. A mutation in the BIR domain of the IAP protein DIAP1 that prevents binding to either caspases or anti-IAP proteins.

3. A mutation in the anti-IAP protein Reaper that prevents its binding to the IAP proteins.

4. A mutation in the CARD domain of caspase-9 that prevents its binding to Apaf1.

23. Depicted below are the results of inducing apoptosis in mouse thymus lymphocytes with dexamethasone, with DNA having been extracted at the timestamps listed across the top. Note the ladder-like pattern. Why is this particular pattern notable, and what is its importance with regard to analyzing the results of this experiment?

24. How would you describe the process of induction of mitochondrial outer membrane permeabilization (MOMP) by Bad? Explain your reasoning in 2-3 sentences.

Molecular Biology of the Cell 7e-Alberts 25. The phosphatase PTEN dephosphorylates the products of PI3K-catalyzed reactions, inhibiting activation of AKT. In 2–3 sentences, describe the expected consequence of an inactivating mutation of PTEN on AKT-dependent survival pathway.

26. Apoptotic cells expose aminophospholipids (phosphatidylserine and phosphatidylethanolamine) on the outer plasma membrane. What would be an apoptosis detection method based on this feature, and why?

27. What could be a possible strategy for designing a drug that acts as an apoptotic inducer? Defend your answer in 2-3 sentences.

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 18 1. Answer:

C

2. Answer:

D

3. Answer:

B

4. Answer:

B

5. Answer:

B

6. Answer:

A

7. Answer:

A

8. Answer:

C

9. Answer:

E

10. Answer:

E

11. Answer:

B

12. Answer:

C

13. Answer:

A

14. Answer:

D

15. Answer:

A

16. Answer:

C

17. Answer:

A

18. Answer:

B

19. Answer:

A

20. Answer:

1. A

Molecular Biology of the Cell 7e-Alberts 2. A 3. B 4. B 5. B 21. Answer:

1. B 2. A 3. C

22. Answer:

1. B 2. A 3. B 4. B

23. Answer:

One can see the ladder-like pattern characteristic to gel electrophoresis, a laboratory method that is helpful when one wishes to distinguish certain molecules (e.g., DNA) and analyze each fragment on an individual basis, including any unknown samples. The shorter the molecule, the faster it will migrate, and so one views the smaller DNA molecules quicker. With respect to this experiment in particular, it is important to analyze how activation of an executioner caspase leads to cleavage of iCAD.

24. Answer:

The BH3-only protein Bad indirectly induces MOMP by its binding to the anti-apoptotic Bcl2 family protein BclxL, liberating activated Bak. The oligomerization of activated Bak induces MOMP.

25. Answer:

The serine/threonine kinase AKT phosphorylates and inactivates the pro-apoptotic BH3-only protein Bad. The inactivating mutation of the phosphatase results in a constitutively active survival pathway.

26. Answer:

During apoptosis, activated executioner caspases cleave the phospholipid flippase. This impedes phosphatidylserine and phosphatidylethanolamine transfer from the outer to the inner leaflet. Additionally, these caspases also cleave and activate phospholipid scramblase, an enzyme responsible for transferring

Molecular Biology of the Cell 7e-Alberts plasma membrane phospholipids nonspecifically between inner and outer lipid leaflets. Both mechanisms contribute to the exposure of phosphatidylserine on the surface of apoptotic cells. Using fluorescent phosphatidylserinebinding proteins, such as annexin V, provides a detection method for the externalization of phosphatidylserine as an indicator of apoptosis.

27. Answer:

BH3-only proteins bind to the long BH3-binding groove in anti-apoptotic Bcl2 family proteins, inhibiting their mechanism of blocking apoptosis.

Therefore, drugs that display binding to pro-survival Bcl-2 proteins will promote cellular apoptosis.

Molecular Biology of the Cell 7e-Alberts

Chapter 19: Cell Junctions and the Extracellular Matrix 1. Which of the following cell junctions uses cadherin cell adhesion molecules to anchor the actin cytoskeleton? a. Tight junction b. Adherens junction c. Desmosome d. Hemidesmosome e. Gap junction

2. Which of the following is NOT an anchoring junction? a. Actin-linked cell–matrix junction b. Adherens junction c. Desmosome d. Hemidesmosome e. Gap junction

3. What is true of overproduction of cadherins such as E-cadherin? a. It is often found in cancers originating from epithelia. b. It is induced by the transcription regulatory protein Twist. c. It induces epithelial–mesenchymal transition. d. It leads to stronger cell–cell adhesion.

4. The formation of a large adhesion surface between two cells requires changes in the actin cytoskeleton that lead to a local reduction in cortical tension. How is this reduction in cortical tension achieved?

Molecular Biology of the Cell 7e-Alberts a. When the cells first interact, small clusters of cadherins activate intracellular proteases that break down the cells’ actin networks. b. The small GTPase Rac is activated to inhibit the formation of actin-myosin stress fibers and promote branched actin networks. c. Before they interact, the two cells secrete extracellular signaling molecules that trigger their actin cytoskeletons to restructure. d. The small GTPase Rho is activated to promote the formation of actinmyosin stress fibers. e. When the cells first interact, small clusters of cadherins activate calcium ion channels.

5. Your research lab utilizes in vitro binding assays to understand the interactions between cadherins and actin filaments in adherens junctions. One of your top students is close to identifying a specific amino acid motif on the cytoplasmic tail of cadherin that is responsible for binding β-catenin, a key adaptor protein. Her approach is to mix cadherin tail fragments containing various amino acid substitutions with purified β-catenin protein conjugated to magnetic beads. If a tail fragment variant binds to the β-catenin-coated beads, it will co-precipitate with the β-catenin pellet that forms when the beads are placed into a magnetic field; if the tail fragment doesn’t bind to β-catenin, it will stay in the supernatant.

The following table summarizes the data from a recent in vitro binding assay.

Which tail fragment variant(s) bind to β-catenin?

Tail Fragment Variant Supernatant Pellet 12345

a. Variant 2 only b. Variant 4 only c. Variants 1 and 4 only d. Variants 2 and 3 only e. Variants 2, 3, and 5 only

YNNYY

NYYNY

Molecular Biology of the Cell 7e-Alberts 6. Consider the role of the actin cytoskeleton in two cells attached to each other via adherens junctions. Which of the following statements is true? a. The forces generated in the two cells do not depend on myosin activity. b. The forces generated in one cell are normally balanced by similar forces in the neighboring cell. c. Contractile forces can strengthen the adherens junction by proteolytically cleaving associated catenin proteins. d. Myosin activity in one cell leads to a reduction in vinculin recruitment in the other cell.

7. Particularly abundant in cells of heart muscle and the epidermis, ________ are structurally similar to adherens junctions and contain a number of homologous components. However, they are linked to intermediate filaments instead of the actin cytoskeleton. a. Gap junctions b. Cadherins c. Desmosomes d. Catenins e. Tight junctions

8. In insects such as Drosophila melanogaster, electrical synapses are abundant in the ventral nerve cord. Unlike a chemical synapse, the cytosols of pre- and postsynaptic cells in an electrical synapse are connected, allowing action potentials to spread rapidly and without delay. What junctional proteins in the flies are chiefly responsible for this unique feature of the electrical synapses? a. Cadherins b. Claudins c. Connexins d. Innexins

Molecular Biology of the Cell 7e-Alberts e. Integrins

9. In the following schematic drawing of connexons as seen from the cytosol of a cell, what is the approximate diameter of each of the pores? Which of the indicated connexons (1 or 2) is more likely to have been incorporated more recently?

a. About 20 nm; connexon 1 b. About 4 nm; connexon 1 c. About 1.4 nm; connexon 1 d. About 4 nm; connexon 2 e. About 1.4 nm; connexon 2

10. Plasmodesmata in plant cells are functionally similar to what in animal cells? a. Adherens junctions b. Desmosomes c. Gap junctions d. Hemidesmosomes e. Tight junctions

Molecular Biology of the Cell 7e-Alberts

11. Which of the following is an accurate description of selectins? a. They mainly mediate cell–matrix attachments. b. They are members of the Ig superfamily of cell adhesion molecules. c. They do NOT require Ca2+ for their adhesive function, unlike integrins. d. They are carbohydrate-binding proteins. e. They mediate homophilic interactions with other selectins.

12. Malignant cancer cells that have entered the bloodstream can imitate lymphocytes in exiting the bloodstream and entering underlying tissues in the process of metastasis. How can this “lymphocyte mimicry” be summarized? a. An initial cadherin-dependent rolling followed by an integrin-dependent adhesion and emigration b. An initial cadherin-dependent rolling followed by an I-CAM-dependent adhesion and emigration c. An initial cadherin-dependent rolling followed by a selectin-dependent adhesion and emigration d. An initial selectin-dependent rolling followed by a cadherin-dependent adhesion and emigration e. An initial selectin-dependent rolling followed by an I-CAM-dependent adhesion and emigration

13. Which of the following cells is responsible for synthesizing and secreting the components of the extracellular matrix in most connective tissues in animals? a. Endothelial cells b. Lymphocytes c. Epithelial cells d. Fibroblasts

Molecular Biology of the Cell 7e-Alberts e. Platelets

14. How is hyaluronic acid different from other glycosaminoglycans of the extracellular matrix? a. It contains sulfate groups. b. It is covalently linked to proteins. c. It is not assembled in the Golgi apparatus. d. It is relatively small in mass. e. It contains a large variety of disaccharide units.

15. In the following schematic drawings of aggrecan and decorin, which one (1 or 2) corresponds to aggrecan? Which one (1 or 2) is synthesized primarily by chondrocytes in cartilage?

a. 1; 1 b. 1; 2 c. 2; 1 d. 2; 2

16. A giant aggrecan aggregate can have a molecular mass of 108 daltons or more, and can be as large as a bacterium. Its structure resembles a fuzzy pipe cleaner.

What is responsible for attaching the fuzzy “bristles” of the pipe cleaner to its wire core?

Molecular Biology of the Cell 7e-Alberts

a. Covalent bonds between aggrecan core protein and keratan or chondroitin sulfate b. Noncovalent bonds between aggrecan core protein and keratan or chondroitin sulfate c. Covalent bonds between aggrecan core protein and hyaluronan d. Noncovalent bonds between aggrecan core protein and hyaluronan

17. Which of the following extracellular matrix components contains a transmembrane protein? a. Type I collagen b. Perlecan c. Aggrecan

Molecular Biology of the Cell 7e-Alberts d. Syndecan e. Decorin

18. Collagens are extremely rich in which amino acids? a. Alanine and valine b. Glycine and proline c. Lysine and arginine d. Serine and threonine e. Tyrosine and phenylalanine

19. Type IV collagen molecules can interact at their C-termini to form dimers and at their N-termini to form tetramers. These interactions form a major part of the sheetlike basal lamina found in epithelial tissues. To which category does Type IV collagen belong? a. Fibrillar b. Fibril-associated c. Network-forming d. Proteoglycan core e. Anchoring fibril

20. Which collagens are the most common in our body? a. Fibrillar b. Fibril-associated c. Network-forming d. Proteoglycan core e. Anchorage fibril

Molecular Biology of the Cell 7e-Alberts

21. The nucleotide length and distribution of exons coding for collagen helical domains reveals a unique pattern. This pattern provides clues to the evolutionary history of the genes encoding the α chains of fibrillar collagens. What is this pattern? a. The length of most exons is a multiple of 100 nucleotides. b. Most exons code for 54 or multiples of 54 Gly-X-Y repeats. c. The length of most exons is a multiple of 30 nucleotides. d. Most exons code for 6 or multiples of 6 Gly-X-Y repeats.

22. Which of the following is responsible for stabilizing the triple-stranded helix structure of a collagen molecule? a. Covalent intermolecular cross-links between modified lysine side chains b. Noncovalent interchain hydrogen bonds between the hydroxyl groups of hydroxyproline and hydroxylysine c. Noncovalent interchain van der Waals attractions between aligned glycine residues d. Noncovalent interchain charge-charge interactions between lysine and aspartate side chains e. Covalent intramolecular cross-links between hydroxyproline side chains

23. Individual collagen polypeptide chains are synthesized on membrane-bound ribosomes and sorted to the extracellular space via the secretory pathway. The polypeptides undergo a series of post-translational modifications in transit to their final destination. Place the following collagen modifications in the correct order from earliest to latest.

1. Propeptide removal 2. Procollagen assembly 3. Fibril formation 4. Proline and lysine hydroxylation 1. 4321

Molecular Biology of the Cell 7e-Alberts 2. 2134 3. 4213 4. 1243 5. 4312

24. Which components of the extracellular matrix are responsible for its ability to resist compressive and tensile forces, respectively? a. Collagens and glycoproteins b. Glycoproteins and collagens c. Glycosaminoglycans and glycoproteins d. Glycosaminoglycans and collagens e. Collagens and glycosaminoglycans

25. Which of the following is true regarding fibril-associated collagens such as type IX collagen? a. They aggregate with one another to form long fibrils in the extracellular space. b. They lack the triple-stranded helical structure found in other collagens. c. They are more flexible than collagens of types I to IV. d. They are the most common among collagens.

26. This large glycoprotein of the extracellular matrix is associated with elastin fibers and is essential for their integrity. Its deposition in the developing connective tissues often precedes the appearance of elastin. It is part of the microfibril sheath that covers the elastic fibers. What is this protein? a. Collagen b. Fibronectin

Molecular Biology of the Cell 7e-Alberts c. Fibrillin d. Filamin e. Laminin

27. Multidomain glycoproteins in the extracellular matrix are large scaffold proteins that organize the matrix and help cells attach to it. They contain binding sites for many other macromolecules. All but one of the following are functions imparted by multidomain glycoproteins. Which function is NOT provided by them? a. They can cleave collagen fibrils to increase elasticity. b. They can bind to growth factors and influence cell signaling. c. They can bind to cell-surface receptors. d. They can serve as repellents to forbid cell migration. e. They can serve as tracks along which cells can migrate.

28. All but one of the following locations are consistent with where basal laminae reside in animals. Which location does NOT include a basal lamina? a. Surrounding individual muscle cells b. Surrounding the chondrocytes in cartilage c. Beneath epithelial cells d. Surrounding the Schwann cells that wrap around nerve axons e. Between endo- and epithelial cell layers in the kidney glomerulus

29. Neuromuscular junctions of vertebrates are special types of chemical synapses formed from close association between motor neuron axon terminals and muscle fibers. A thin layer of basal lamina separates the two cell membranes at the neuromuscular junction and is important for its organization and maintenance.

Which of the following would you NOT expect to find in the basal lamina in these structures?

Molecular Biology of the Cell 7e-Alberts a. Nidogen b. Laminin c. Type I collagen d. Type IV collagen e. Perlecan

30. Which of the following functions of the basal lamina explains why immune cells like macrophages and lymphocytes need to utilize specialized protease enzymes to break through it? a. It accelerates repair of damaged neuromuscular junctions. b. It serves as a filter for certain molecules. c. It separates the muscle and nerve cell membranes at a synapse. d. It acts as a selective barrier to the movement of cells. e. It provides a scaffold for the migration of regenerating cells.

31. A researcher has coated the wells in a multiwell plate with human plasma fibronectin. She introduces integrin-expressing cells into these wells in the presence of different concentrations of a synthetic peptide. She then washes the wells to remove unbound cells and counts the number of cells that remain attached to each well. The results of her experiment are presented in the following schematic graph. The synthetic peptide sequences used in these experiments were GRGDSPC (1) or GGDRSLN (2). Which one of these peptides (1 or 2) would you expect to correspond to Curve a in the graph? Which curve (a or b) would you expect if a synthetic peptide with the sequence GDGRSPC was used? C is cysteine, D is aspartic acid, G is glycine, L is leucine, N is asparagine, P is proline, R is arginine, and S is serine.

Molecular Biology of the Cell 7e-Alberts

a. 1; a b. 1; b c. 2; a d. 2; b

32. The following figure shows the subunit structure of an active integrin molecule and its interacting partners on both side of the cell membrane. Let’s assume this is the α7β1 integrin found in muscle cells. Mutant mice that cannot make the β 1 subunit die early in embryonic development. Curiously, mice that cannot make the α7 subunit survive but develop muscular dystrophy. Which of the following provides a logical hypothesis for why β1 mutants die but α7 mutants survive?

Molecular Biology of the Cell 7e-Alberts

a. The α7 subunit is specific to muscle cells. b. The β1 subunit binds multiple types of α subunits. c. There are more β1 subunits in cell membranes than there are α7 subunits. d. The α7 subunit binds to extracellular matrix proteins. e. The β1 subunit plays the primary role in bridging both extracellular and intracellular components.

33. The following schematic drawing shows two integrin molecules. Which molecule (a or b) is depicted in its active configuration? Which side of the plasma membrane (1 or 2) represents the cytosol?

Molecular Biology of the Cell 7e-Alberts

a. a; 1 b. a; 2 c. b; 1 d. b; 2

34. What happens in the active state of the integrin dimer? a. Both intracellular and extracellular binding sites are exposed. b. The intracellular binding sites are inaccessible, while the extracellular binding sites are exposed. c. The intracellular binding sites are exposed, while the extracellular binding

Molecular Biology of the Cell 7e-Alberts sites are inaccessible. d. Both intracellular and extracellular binding sites are inaccessible.

35. How does talin activate the inside-out signaling through integrin? a. It binds to both integrin subunits to bring them together in an interlocked conformation. b. It is a protein kinase that phosphorylates both integrin subunits and induces activation by conformational change. c. It competes with the α subunit for binding to the β subunit, therefore blocking an intimate α–β linkage. d. It binds to the α subunit and dissociates the dimer into individual subunits that are functionally active.

36. The chemical compound PMA is a potent stimulator of the epithelial-tomesenchymal transition (EMT) in epithelial cells. In EMT, epithelial cells lose their polarity and cell–cell adhesion capabilities and gain the ability to migrate through the extracellular matrix.

Consider the activities of integrin, connexin, matrix metalloprotease, and cadherin in PMA-treated epithelial cells. The activities of two of these proteins typically are up-regulated while those of the other two primarily are down-regulated. Which of the following pairs of proteins is up-regulated in PMA-treated epithelial cells? a. Cadherin, connexin b. Integrin, connexin c. Integrin, matrix metalloprotease d. Cadherin, matrix metalloprotease

37. Which of the following statements is characteristic to the focal adhesion kinase (FAK)? a. It is a cytosolic serine/threonine kinase. b. It binds directly to integrin dimers.

Molecular Biology of the Cell 7e-Alberts c. It can be phosphorylated by Src kinases. d. It prevents the relay of outside-in integrin signaling.

38. What is the most common additional polymer present in the plant secondary, but not primary, cell wall? a. Cellulose b. Pectin c. Lignin d. Starch e. Collagen

39. Which of the following structures is responsible for maintaining the solute imbalance that drives the formation of turgor pressure in living plant cells? a. Middle lamella b. Cellulose microfibril c. Primary cell wall d. Plasma membrane e. Secondary cell wall

40. Fillipa grows tomatoes in her home garden. When her plants become infested with insect pests, she uses a natural insecticide that contains EDTA, a divalent cation chelator, as a preservative. The product does a great job driving away the insects, but it has the undesirable side effect of causing her immature tomatoes to fall off their vines before they are ripe.

Consider your knowledge of plant cell wall structure. Which of the following provides a logical explanation for why Fillipa’s tomatoes fall before they ripen? a. The insects secrete digestive enzymes that weaken the cell walls in the tomato stems.

Molecular Biology of the Cell 7e-Alberts b. The EDTA sequesters the Ca2+ that is responsible for cross-linking the cell wall pectins that hold cells together. c. The EDTA attracts and binds water molecules that cause the pectins in the middle lamellae to swell and weaken. d. The weight of the insecticide places added stress on the cell walls that hold the tomatoes on the vines. e. The EDTA causes the cell wall’s cellulose microfibrils to dissociate into their component polysaccharides.

41. Which of the following descriptions better applies to cellulose microfibrils rather than pectins in plant cell walls? a. They are made into a network by cross-linking glycans. b. They are particularly abundant in the middle lamella. c. Divalent ions such as calcium can cross-link them to form gels. d. They contain many negatively charged galacturonic acid units.

42. Place the following plant cell wall components in order from closest to farthest from the plasma membrane. You may assume all four are present in a mature plant cell.

1. Newest layers of primary wall 2. Middle lamella 3. Secondary wall 4. Oldest layers of primary wall 1. 3142 2. 3412 3. 4132 4. 1432 5. 1423

Molecular Biology of the Cell 7e-Alberts 43. In the following schematic diagram of a cubic plant cell that has just left the meristem zone, the cellulose microfibrils are shown as black lines. In which direction (1 or 2) do you think the cell will elongate when the cell grows? If the cell wall is momentarily dissolved, in what direction (3 or 4) would water tend to flow after the turgor pressure has dissipated?

a. 1; 3 b. 1; 4 c. 2; 3 d. 2; 4

44. Cellulose is deposited onto the plant cell wall in highly ordered crystalline aggregates by the cellulose synthase complex embedded in the plasma membrane. Would you expect cell wall deposition to continue following treatment of plant cells with a drug that depolymerizes microtubules? Would you expect these cells to be able to switch the orientation of the microfibril pattern between successive lamellae? a. Yes; yes b. Yes; no c. No; yes d. No; no

Molecular Biology of the Cell 7e-Alberts 45. Cadherins are present in all multicellular organisms. a. True b. False

46. You have isolated epithelial cells from mouse mammary glands and have grown them in a three-dimensional gelatinous matrix in the presence or absence of a specific inhibitor that targets a transmembrane metalloprotease. The protease is capable of cleaving cadherin near the base of its extracellular domain to release an N-terminal fragment containing the extracellular cadherin domains. The distribution of the cells in the matrix, as seen under a light microscope, one day after the transplantation of the isolated tissue is presented in the following schematic drawings.

TRUE or FALSE: Condition 1 corresponds to the presence of the protease inhibitor.

a. True b. False

47. A researcher has grown monolayers of cells from either of two epithelial cell lines on a permeable supporting membrane that separates two electrolyte chambers.

One of the cell lines (A) has been derived from the mammalian kidney and forms epithelial sheets resembling those lining the renal collecting duct and the urinary bladder. The other (B) is derived from the mammalian small intestine.

The researcher has measured the voltage (V) and electrical current (I) across the monolayer and has plotted the results in the graph below. TRUE or FALSE: Line 2 in the graph corresponds to cell line A. Recall that current (I) is related to the applied voltage (V) by the equation I = V/R, in which R is the electrical resistance.

Molecular Biology of the Cell 7e-Alberts

a. True b. False

48. In the following schematic diagram, the estimated attractive or repulsive force between two adhesive cells is plotted as a function of distance between the cells.

The dashed curve represents a condition under which the cells express cadherin but not N-CAM. The other two curves represent conditions under which both cadherin and heavily sialylated N-CAM are expressed. In one of the conditions, however, the ionic strength of the medium has been artificially increased by addition of salt.

TRUE or FALSE: Curve 2 represents the high-salt condition.

Molecular Biology of the Cell 7e-Alberts

a. True b. False

49. The protein α-catenin is an intracellular tension-sensing molecule that links actin filaments to cadherins at cell–cell adherens junctions. Tension on these junctions mechanically extends (unfolds) α-catenin to expose an otherwise hidden binding site for vinculin; vinculin binding recruits additional actin filaments to stabilize the junction. Which of the following proteins plays an analogous tension-sensing role to α-catenin by linking actin filaments to β-integrins at cell–matrix junctions? a. Kindlin b. Talin c. Fibronectin d. Plectin e. Laminin

Molecular Biology of the Cell 7e-Alberts

50. Consider a sheet of epithelial tissue that invaginates to form a tube along the anterior–posterior axis of a developing vertebrate embryo. The tube is then closed concomitant with a “convergence and extension” process similar to that observed in germ-band extension in Drosophila melanogaster. During this process, the tube elongates and narrows before pinching off from the sheet.

TRUE or FALSE: Axis 2 corresponds to the anterior–posterior axis.

a. True b. False

51. Glycosaminoglycan chains in the extracellular matrix all contain Nacetylglucosamine or N-acetylgalactosamine. a. True

Molecular Biology of the Cell 7e-Alberts b. False

52. The elasticity of the components of the extracellular matrix can be visualized in a stress–strain curve, in which the tensile stress (corresponding to load) applied to a tissue sample is plotted against the strain (corresponding to deformation) measured in the tissue. In the following stress–strain graph, the elasticity of an intact connective tissue is represented by a dashed curve. The other two curves correspond to the same tissue that has been treated with either formic acid (to digest collagen) or trypsin (to digest elastin).

TRUE or FALSE: Curve 1 corresponds to trypsin treatment.

a. True b. False

53. Tension can increase the binding affinity of an integrin for its intracellular and extracellular ligands. a. True b. False

Molecular Biology of the Cell 7e-Alberts 54. EDTA is a metal-ion “chelator” that is commonly used to lower the concentration of free divalent cations in solution. In the lab, your friend has cultured a human cell line on elastic membranes that have been coated with fibronectin and attached to a device that can stretch the membrane. She has quantified the binding of talin to integrin under different conditions and has presented the results in the following schematic graph, which shows the binding in the presence or absence of EDTA in the cell-culture medium.

TRUE or FALSE: Experiment A was performed in the presence of EDTA.

a. True b. False

55. Integrins usually bind their ligands with a higher affinity compared to other cellsurface receptors. a. True b. False

Molecular Biology of the Cell 7e-Alberts 56. Platelets that express a constitutively active Rap1 are able to activate integrins even in the absence of thrombin stimulation. In contrast, cells expressing a binding-deficient version of talin are unable to perform integrin activation. Now consider platelets expressing both of these mutant proteins.

TRUE or FALSE: Rather than the Rap1 gain-of-function phenotype, you would expect to observe the talin loss-of-function phenotype. a. True b. False

57. Fibronectin molecules can assemble into fibrils only in the vicinity of cells. a. True b. False

58. Indicate whether each of the following descriptions better applies to connective or epithelial tissues. a. Connective tissues b. Epithelial tissues

1. Cells are distributed sparsely in the extracellular matrix.

2. Gap junctions are found rarely.

3. Direct cell–cell attachments are common.

4. Cells are associated tightly into sheets.

59. In the following schematic diagram of a simple columnar epithelium lining the digestive tract, indicate which position (A to F) along the apical–basal axis best corresponds to each of the following features.

Molecular Biology of the Cell 7e-Alberts

a. Position A b. Position B c. Position C d. Position D e. Position E f. Position F

1. Basal lamina 2. Cell apex 3. Adherens junctions 4. Gap junctions 5. Hemidesmosomes 6. Tight junctions

60. Cells expressing either N-cadherin, high levels of E-cadherin, or low levels of Ecadherin have sorted themselves out on a substratum in a cadherin-dependent manner as shown in the schematic drawing below. Which cells would you expect to correspond to white, gray, and black in the drawing, respectively?

Molecular Biology of the Cell 7e-Alberts

a. N-cadherin b. High levels of E-cadherin c. Low levels of E-cadherin

1. White 2. Gray 3. Black

61. Indicate whether each of the following descriptions better applies to adherens junctions, desmosomes, gap junctions, or tight junctions. a. Adherens junctions b. Desmosomes c. Gap junctions d. Tight junctions

1. They play a direct role in structural polarization of epithelial cells.

2. They couple cells both electrically and metabolically.

3. They are formed from hexameric hemichannels that can assemble into heterotypic complexes.

4. They create semipermeable paracellular pores.

62. Indicate whether each of the following descriptions better applies to the glycosaminoglycans, fibrous proteins, or glycoproteins of the extracellular matrix in animal cells. a. Glycosaminoglycans b. Fibrous proteins c. Glycoproteins

Molecular Biology of the Cell 7e-Alberts 1. They include laminin and fibronectin.

2. They primarily include collagens.

3. They include hyaluronan.

4. They are found in proteoglycans.

5. They occupy large volumes and form hydrated gels.

6. They constitute the major proteins of the matrix.

63. Indicate whether each of the following descriptions better applies to the intracellular or extracellular domain of integrins. a. Intracellular domain b. Extracellular domain

1. It binds to talin.

2. It binds to the RGD tripeptide motif.

3. It is bulkier.

4. It contains divalent cation-binding domains.

64. A cell’s response is not the same when placed on a rigid matrix compared to a soft matrix. Indicate whether each of the following is expected to occur in a cell that is placed on a rigid matrix or a soft matrix. a. Rigid matrix b. Soft matrix

1. More tension is generated in cell–matrix contacts.

2. More vinculin proteins would bind to the C-terminal tail of talin.

3. More talin proteins are found in their compact, folded conformation.

4. More actin filaments are recruited to the site of cell–matrix adhesion.

65. Indicate whether each of the following descriptions better applies to the plant cell wall or the extracellular matrix of animal cells. a. Plant cell wall b. Extracellular matrix of animal cells

Molecular Biology of the Cell 7e-Alberts 1. It is generally thicker and stronger.

2. It is more rigid.

3. It has higher nitrogen content.

66.

Cadherin molecules at the cell surface are often clustered side-to-side to create a molecular Velcro that attaches the cell to another cell or to the extracellular matrix. In the following simplified drawings, two cells are initially attached via such a Velcro. When the top cell moves in one of three different ways (A, B, or C) relative to the bottom cell, it is faced with a mechanical resistance from the cadherin interactions between the cells. The resistance against which type of movement do you think is the weakest? Your answer should be the single letter A, B, or C.

Molecular Biology of the Cell 7e-Alberts

67. How would you describe the general structure and function of zonula occludens (ZO) proteins?

68. When fibroblasts are mixed with a meshwork of randomly oriented collagen fibrils, the fibroblasts interact mechanically with the collagen fibrils and cause the meshwork to contract to a fraction of its initial volume. Now consider a modification of the experiment in which two clusters of fibroblasts are placed relatively far apart on the same collagen meshwork. What would you expect to see for both the fibroblasts and the collagen fibrils? How does the experiment lead you to this conclusion?

69. How do cells degrade (break down) the extracellular matrix? Provide two reasons for why this process is important.

70. How do peptides that contain an RGD sequence function as anti-clotting agents?

(R is arginine, G is glycine, and D is aspartate.)

71. How is turgor pressure generated in plant cells? List two of its functions for plants.

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 19 1. Answer:

B

2. Answer:

E

3. Answer:

D

4. Answer:

B

5. Answer:

E

6. Answer:

B

7. Answer:

C

8. Answer:

D

9. Answer:

C

10. Answer:

C

11. Answer:

D

12. Answer:

E

13. Answer:

D

14. Answer:

C

15. Answer:

D

16. Answer:

D

17. Answer:

D

18. Answer:

B

19. Answer:

C

20. Answer:

A

Molecular Biology of the Cell 7e-Alberts 21. Answer:

D

22. Answer:

B

23. Answer:

C

24. Answer:

D

25. Answer:

C

26. Answer:

C

27. Answer:

A

28. Answer:

B

29. Answer:

C

30. Answer:

D

31. Answer:

C

32. Answer:

E

33. Answer:

A

34. Answer:

A

35. Answer:

C

36. Answer:

C

37. Answer:

C

38. Answer:

C

39. Answer:

D

40. Answer:

B

41. Answer:

A

42. Answer:

A

Molecular Biology of the Cell 7e-Alberts 43. Answer:

C

44. Answer:

B

45. Answer:

B

46. Answer:

A

47. Answer:

B

48. Answer:

A

49. Answer:

B

50. Answer:

A

51. Answer:

A

52. Answer:

A

53. Answer:

A

54. Answer:

B

55. Answer:

B

56. Answer:

A

57. Answer:

A

58. Answer:

1. A 2. A 3. B 4. B

59. Answer:

1. F 2. A 3. C 4. D

Molecular Biology of the Cell 7e-Alberts 5. E 6. B 60. Answer:

1. C 2. A 3. B

61. Answer:

1. D 2. C 3. C 4. D

62. Answer:

1. C 2. B 3. A 4. A 5. A 6. B

63. Answer:

1. A 2. B 3. B 4. B

64. Answer:

1. A 2. A 3. B 4. A

65. Answer:

1. A 2. A 3. B

Molecular Biology of the Cell 7e-Alberts 66. Answer:

C

67. Answer:

The zonula occludens (ZO) proteins are large scaffold proteins that provide a structural support on which tight junctions are built. They are found on the intracellular side of the plasma membrane. Each ZO protein contains multiple protein-binding domains linked together in a flexible chain.

The PDZ domains bind the main tight junction protein claudin as well as other proteins, including other ZO proteins. An SH3 domain binds signaling proteins, a GK domain binds the tight junction protein occludin, and a proline-rich domain binds actin filaments. Together, these domains serve to integrate and organize the functions of proteins within networks of tight-junctional sealing strands.

68. Answer:

As the fibroblasts interact mechanically with the collagen fibrils, the latter will become organized into a compact band of aligned fibers that connect the two cell clusters. After the band of fibers is formed, the fibroblasts will migrate away from their clusters along the aligned collagen fibers. The experiment shows how fibroblasts influence fiber alignment in the extracellular matrix, and how the fibers in turn affect the distribution of fibroblasts within the matrix.

69. Answer:

In general, components of the extracellular matrix are broken down by proteases that act close to the cells that secrete them. This process is important because it allows cells embedded within the matrix to travel through it, and when necessary, elongate and divide. Degrading the matrix also allows cells to remove damaged components of the matrix and remodel it to adapt its properties to the changing needs of the tissue.

70. Answer:

The RGD sequence is an integrin-binding domain present in several extracellular matrix proteins like fibronectin, as well as in certain clotting factors. The latter bind to integrins on the cell membranes of platelets and in doing so cause them to aggregate and form clots in the blood. Peptides containing an RGD sequence interrupt these crucial interactions by competing for the integrin binding site and thus prevent the blood from coagulating.

71. Answer:

Turgor pressure is generated by the large osmotic imbalance that exists across the plasma membrane of plant cells. The extracellular environment and cell wall fluid are hypotonic with respect to the cell interior, so water moves across the cell membrane, creating hydrostatic pressure that pushes outward on it and the cell wall until an osmotic equilibrium is reached. The resulting turgor pressure

Molecular Biology of the Cell 7e-Alberts functions as the main force driving cell expansion during plant growth. It also provides much of the mechanical rigidity that allows plants to grow upward against the force of gravity.

Molecular Biology of the Cell 7e-Alberts

Chapter 20: Cancer 1. What is a benign neoplasia of cartilage called? a. Chondrocarcinoma b. Condroma c. Chondrosarcoma d. Adenochondroma e. Chondromelanoma

2. Among the following cancers, one is currently leading in the number of deaths in the United States and in the rest of the world. In the United States, it contributes to cancer mortality more than the next three killing cancers combined. Worldwide, it claims more than 1.5 million lives every year. Which cancer is this? a. Lung cancer b. Breast cancer c. Colon cancer d. Pancreatic cancer e. Stomach cancer

3. What is the most common cancer in humans? a. Sarcoma b. Leukemia c. Myeloma d. Lymphoma e. Carcinoma

Molecular Biology of the Cell 7e-Alberts 4. Which of the following is NOT a carcinogenic agent? a. X-rays b. UV light c. Tobacco d. Viruses e. Inherited genetic defect

5. Consider a healthy adult animal in which 1015 cell divisions have taken place since birth. Spontaneous mutations can occur at a rate of approximately one nucleotide out of about 10 billion nucleotides every time DNA is replicated. The animal has a diploid genome size of about 2 billion nucleotide pairs. Assuming that only about 5% of mutations occur within genes or gene regulatory sequences, and further assuming that about 0.1% of those may cause cancer, how many potential cancer-causing mutations has the animal been able to successfully suppress (i.e., been able to survive) during its lifetime? a. 10,000 b. 100,000 c. 10,000,000 d. 1,000,000,000 e. 10,000,000,000

6. Which cancer has the highest correlation between its lifetime risk and the division rate of its cell of origin? a. Basal cell b. Esophageal c. Glioblastoma d. Osteosarcoma e. Testicular

Molecular Biology of the Cell 7e-Alberts

7. Which characteristic is among the hallmarks of cancer cells? a. Maintenance of the limits to cell proliferation b. Unaltered cell metabolism c. Altered homeostasis that results in cells dividing at a faster rate, inducing apoptosis d. Manipulation of the tissue environment, inducing an immune response e. Evasion of cell-death signals

8. Compared to cells of a normal tissue, which of the following occurs less frequently in cells within a solid tumor? a. Apoptosis b. Necrosis c. Cell division d. Miotic recombination e. Stress

9. Which of the following is a heritable property of cancer cells? a. Cancer cells grow and divide in a controlled manner. b. Cancer cells invade and colonize territories normally reserved for other cells. c. Cancer cells stop growing when they come into contact with other cells. d. Cancer cells are identified and removed by the immune system. e. Cancer cells have a lifespan.

10. The requirement for accumulation of multiple mutations in cancer progression is

Molecular Biology of the Cell 7e-Alberts manifested in the normalized percentage of new cases of cancer diagnosed in different age groups. Which of the curves in the following graph better represents the incidence of human cancers as a function of age?

a. Curve A b. Curve B c. Curve C d. Curve D e. Curve E

11. Regarding cancer cells and their use of glucose, which statement is true? a. Cancer cells use glucose mainly in the oxidative phosphorylation pathway. b. Cancer cells use glucose to induce aerobic respiration. c. Cancer cells use glucose so that they can oxidize it into CO 2. d. Cancer cells use glucose to support the production of raw materials that enable cellular proliferation. e. Cancer cells use glucose mainly in the mitochondria.

Molecular Biology of the Cell 7e-Alberts

12. In medical oncology, PET (positron emission tomography) is used to selectively image tumors in the body and to monitor cancer progression and response to treatment. Before performing a PET scan, the patient should fast for at least several hours for blood glucose to be sufficiently low. At the time of the scan, the positron-emitting glucose analog fluorodeoxyglucose (FDG) is injected into the bloodstream and the patient is asked to wait for up to an hour while avoiding physical activity. Finally, the scanner moves slowly over the body to reveal the location of possible tumors. Why do you think the patient should avoid physical activity before the scan? a. To prevent the Warburg effect b. To accelerate glucose uptake by the tumor cells c. To prevent the absorption of the radioactive tracer by healthy tissues d. To promote fermentation in healthy tissues

13. Which statement regarding circulating tumor cells (CTCs) is correct? a. CTCs penetrate the lymphatic vessels. b. CTCs give rise to lymph-node metastases. c. CTC clusters form metastases at a significantly higher rate than single CTCs. d. CTCs are undetectable. e. CTCs form adhesions that in turn induce death signals.

14. Which protein is NOT encoded by a proto-oncogene? a. Src b. Ras c. EGF receptor d. Myc

Molecular Biology of the Cell 7e-Alberts e. E-cadherin

15. The immortalized nonmalignant mouse cell line NIH-3T3 was derived from normal mouse fibroblasts in the early 1960s. These cells are able to readily take up exogenous DNA and are prone to transformation by cancer-causing agents, including some retroviruses. DNA extracted from a human bladder carcinoma line is able to transform these cells, as judged by a significant increase in the number of foci (cell clumps) in the cell-culture plates when the DNA is added. The malignant cells contain human DNA, and the DNA can be shown by sequence analysis to contain a single mutant gene that is present in the original bladder carcinoma cell line. The gene codes for a monomeric G protein and was one of the first cancer-critical genes to be identified in this way. The protein encoded by this gene is which of the following? a. Src b. Ras c. Myc d. p53 e. Apc

16. You have analyzed a large set of human cancer-critical genes for a selected group of carcinomas, classifying each of the genes based on whether they are known to undergo somatic or germ-line mutations, as well as based on whether they are dominant or recessive. You then group them and plot the statistics in the following histograms. Which group (a or b) do you think represents somatic, as compared to germ-line, mutations? Which group (1 or 2) do you think represents dominant, as compared to recessive, mutations? (Note that the sum of percentages of somatic and germ-line mutations is more than 100%, since some genes are mutated in both somatic and germ cells).

Molecular Biology of the Cell 7e-Alberts

a. a; 1 b. a; 2 c. b; 1 d. b; 2

17. What is the most common mechanism behind the conversion of Ras into Ras oncogenes? a. Gene amplification b. Regulatory mutation c. Chromosome rearrangement d. Deletion in coding sequence e. Point mutation in coding sequence

18. Retinoblastoma is an early-onset cancer of the retina with a rapid progression, and is mostly diagnosed in children. In its hereditary form, multiple eye tumors

Molecular Biology of the Cell 7e-Alberts usually arise in both eyes, while the nonhereditary form usually causes fewer tumors in only one eye. Treatment may involve a combination of chemotherapy, radiotherapy, and other therapies and the majority of patients can be cured if given the right treatment. However, survivors of one form of retinoblastoma (and not the other form) have a markedly increased frequency of subsequent neoplasms that can lead to other cancers later in life, especially soft-tissue sarcomas. These patients should therefore be closely monitored throughout their lives. Which gene is affected by the primary driver mutation in this cancer as well as the later sarcomas? Which form of retinoblastoma do you think is associated with a higher risk of subsequent neoplasms? a. p53; hereditary b. Rb; hereditary c. p53; nonhereditary d. Rb; nonhereditary

19. Myelomas are cancers of blood plasma cells—white blood cells that are normally responsible for producing large quantities of antibodies. In the following Circos plot for myelomas in a hypothetical mammalian genome, the interchromosomal rearrangements are indicated by gray lines and variations in copy numbers are indicated in black. The positions of named genes are indicated with arrows. On which chromosome do you expect to find the antibody genes?

a. 1

Molecular Biology of the Cell 7e-Alberts b. 2 c. 3 d. 4 e. 5

20. Mutations in two important cancer-critical genes, encoding p53 and Rb, respectively, are commonly found in cancers. What type of mutations are these expected to be? a. Loss-of-function mutations in both genes b. Loss-of-function mutation in p53 and gain-of-function mutation in Rb c. Gain-of-function mutation in p53 and loss-of-function mutation in Rb d. Gain-of-function mutations in both genes

21. The effect of the deletion of one copy of the gene encoding p53 is different from the effects caused by other p53 mutations. For example, some loss-of-function mutations in the DNA-binding domain of p53 cripple its function as a transcription regulator. Such a mutation in only one copy of the p53 gene can be enough to confer a p53 loss-of-function phenotype, even when the other copy of the gene on the homologous chromosome is wild type. Why is this the case? a. p53 is a proto-oncogene. b. These mutations are recessive. c. p53 is a tumor suppressor. d. p53 forms a tetramer. e. p53 can induce apoptosis.

22. Which of the following induces the p53 tumor suppressor? a. Cell-cycle arrest

Molecular Biology of the Cell 7e-Alberts b. Senescence c. Apoptosis d. Oncogene activation e. Telomere lengthening

23. Genetically knocking out both copies of the p53 gene in rats renders what result? a. Embryonic lethality b. A lower malignancy rate, but the rats are otherwise seemingly normal c. A higher rate of cancer onset, but the rats are otherwise seemingly normal d. Increase in cell death by apoptosis, leading to developmental defects e. No effect unless the rats live outside of the laboratory and are exposed to various types of stress

24. Which of the following can lead to p53 stabilization and activation? a. Normoxia b. p53 mutation c. DNA damage d. E6 and E7 viral proteins e. Chromosomal stability

25. What is true about tumors, according to the cancer stem-cell model for tumor growth and propagation? a. Transit amplifying cells are incapable of cell division, which is reserved for the cancer stem cells. b. Transit amplifying cells constitute the great majority of the cells in the tumor.

Molecular Biology of the Cell 7e-Alberts c. If each stem cell divides to create one stem cell and one transit amplifying cell, the abundance of stem cells in the tumor will increase exponentially over time. d. Tumors are genetically heterogeneous, even though they show high phenotypic homogeneity. e. The transit amplifying cells, even though each acts transiently, carry the whole responsibility for maintenance of the tumor in the long term.

26. How is the Rb gene in retinoblastomas similar to the Apc gene in polyposis colon carcinomas? a. Both are proto-oncogenes. b. Both are mutated in one copy in all cells of cancer patients. c. Both are in a locus that shows loss of heterozygosity in the nonhereditary form of the cancer. d. Both should be inactivated in both copies to cause the nonhereditary form of the cancer. e. Both are mutated in more than 80% of colorectal cancers.

27. Mutation in which gene is most prevalent in human colorectal cancer cells? a. K-Ras b. β-Catenin c. Apc d. p53 e. MLH

28. The genotypes of 400 colorectal cancer tumors are tabulated below, where the number of tumors with or without mutations in each of the two cancer-critical genes, β-catenin and Apc, are indicated. Which row (a or b) corresponds to those with a mutant Apc gene? Which column (1 or 2) corresponds to those with a

Molecular Biology of the Cell 7e-Alberts mutant β-catenin gene?

a. a; 1 b. a; 2 c. b; 1 d. b; 2

29. You have karyotyped cells from two colorectal tumor samples, one from a hereditary nonpolyposis colorectal cancer (HNPCC) patient, and the other from a familial adenomatous polyposis coli (FAP) patient. One group of the karyotypes shows gross chromosomal abnormalities with extra or deleted chromosomes and several translocations and deletions. The other group, however, is almost normal, and comparable to noncancerous samples. Which group would you expect to have loss-of-function mutations in the DNA mismatch repair system genes MSH2 and MLH1 as their primary driver mutations? a. HNPCC, which has an almost normal karyotype b. HNPCC, which has a grossly abnormal karyotype c. FAP, which has an almost normal karyotype d. FAP, which has a grossly abnormal karyotype

30. In general, which of the following sequential barriers to metastasis is the easiest to overcome for cancer cells? a. Vessel entry through acquisition of local invasiveness

Molecular Biology of the Cell 7e-Alberts b. Exit from the blood into a remote tissue or organ c. Survival of cells in the foreign tissue d. Initial growth of cells in the foreign tissue e. Persistence of growth in the remote site

31. How do carcinoma cells that have acquired malignancy and started local invasiveness to begin metastasis behave? a. They decrease the expression of E-cadherin and undergo mesenchymal– epithelial transition. b. They increase the expression of E-cadherin and undergo mesenchymal– epithelial transition. c. They decrease the expression of E-cadherin and undergo epithelial– mesenchymal transition. d. They increase the expression of E-cadherin and undergo epithelial– mesenchymal transition.

32. What is a nonpreventable cause of cancer? a. Cigarette smoke b. UV radiation c. Viruses d. Diet e. Hereditary predisposition

33. Which of the following is estimated to be the leading cause of death from cancer in the United States? a. UV light b. Obesity

Molecular Biology of the Cell 7e-Alberts c. Smoking d. Alcohol consumption e. Sedentary lifestyle

34. Most DNA tumor viruses inhibit the products of a. Apc and Rb. b. Brca1 and Rb. c. p53 and Rb. d. Brca1 and Apc. e. Brca1 and p53.

35. What is true of carcinoma of the uterine cervix in humans? a. It can be largely prevented by vaccination. b. It is caused by a retrovirus. c. It is caused by activation of a viral Src kinase. d. It rapidly progresses (from uterine warts) to malignancy.

36. PARP inhibitors can efficiently kill many breast cancer cells that lack functional Brca1 or Brca2 genes. How do these drugs accomplish this? a. By increasing the ability of p53 in cancer cells to limit cell proliferation b. By inhibiting proteins that are normally inhibited by the Brca1 or Brca2 gene products c. By increasing the occurrences of homologous recombination d. By inhibiting a DNA repair pathway e. By increasing the ability of the cancer cells to repair the mutations in their cancer-critical genes

Molecular Biology of the Cell 7e-Alberts

37. Once the molecular aberrations in a cancer are understood, drugs can be designed with a rational approach to treat the cancer. Which of the following is NOT true regarding such drugs? a. The drug imatinib (Gleevec®) can inhibit the chimeric tyrosine kinase BcrAbl in chronic myelogenous leukemia (CML). b. Even if a protein is not the product of a cancer-critical gene, drugs that specifically block its activity can be effective in curing cancer. c. Imatinib is most effective in treating chronic myelogenous leukemia (CML) in its acute blast-crisis phase. d. Protein kinases have turned out to be relatively easy to inhibit with small molecules. e. The success of imatinib relies on the phenomenon of oncogene dependence in cancer cells.

38. Cancer cells grow and proliferate in defiance of normal restraints. a. True b. False

39. Most cancers originate from a single aberrant cell. a. True b. False

40. Cancer cells maintain their telomeres by inhibiting the enzyme telomerase. a. True b. False

41. The karyotype is often severely disordered in cancer cells. a. True b. False

Molecular Biology of the Cell 7e-Alberts

42. Knocking out Ras or Myc genes individually leads to a higher incidence of cancers in mice, and knocking out both genes simultaneously has an even stronger phenotype. a. True b. False

43. Invasive colorectal cancer cells usually metastasize to lymph nodes via lymphatic vessels and then into the bloodstream. a. True b. False

44. Molecule A is a more potent mutagen.

a. True b. False

45. Effective cures have already been found for some previously considered deadly cancers. a. True b. False

46. Indicate whether each of the following cancers can be best classified as a carcinoma, sarcoma, or neither of these.

Molecular Biology of the Cell 7e-Alberts a. Carcinoma b. Sarcoma c. Neither

1. Breast cancer 2. Lung cancer 3. Colorectal cancer 4. Myeloma

47. Indicate whether each of the following descriptions better applies to a cancer cell or a normal adult cell. a. Cancer cell b. Normal adult cell

1. Higher lactate production 2. Higher oxidative phosphorylation 3. Contact inhibition 4. Anchorage independence

48. Indicate whether each of the following descriptions better applies to protooncogenes or tumor suppressor genes. a. Proto-oncogenes b. Tumor suppressor genes

1. Cancer mutations in these genes are usually recessive.

2. Cancer mutations in these genes include gene duplications.

3. Cancer mutations in these genes are responsible for most hereditary cancers.

4. Cancer mutations in these genes are commonly in the form of nonsense (truncating) mutations that abort protein synthesis.

49. For each of the following genes involved in regulation of cell growth through the mTOR pathway, indicate whether the gene is activated or inactivated in cancer cells compared to normal healthy cells.

Molecular Biology of the Cell 7e-Alberts a. Activated b. Inactivated

1. mTOR 2. Akt 3. PTEN 4. PI3K

50. The following simplified diagram shows the typical sequence of genetic changes in a developing colorectal carcinoma. Which event corresponds to the following changes?

a. Event A b. Event B c. Event C

1. Activation of K-Ras 2. Loss of p53 3. Loss of Apc

51. Indicate whether each of the following viruses is mostly associated with cervical cancer, Kaposi’s sarcoma, liver cancer, or stomach cancer. a. Cervical cancer b. Kaposi’s sarcoma c. Liver cancer

Molecular Biology of the Cell 7e-Alberts d. Stomach cancer

1. Hepatitis-B virus (HBV) 2. Human immunodeficiency virus (HIV) 3. Human papillomavirus (HPV) 4. Helicobacter pylori

52. Indicate whether each of the following descriptions better applies to trastuzumab which targets Her2, imatinib which targets Bcr-Abl, or ipilimumab which targets the CTLA4 protein. a. Trastuzumab b. Imatinib c. Ipilimumab

1. It is NOT an antibody.

2. It counters the immunosuppressive microenvironment of tumors.

3. It does not bind to a cancer cell component.

53.

In the following simplified diagram of cell divisions in a multicellular species, the germ-cell and somatic-cell lineages are depicted. Which of the indicated cells represents the germ line? Your answer should be Cell 1 or Cell 2.

Molecular Biology of the Cell 7e-Alberts

54.

X-chromosome inactivation in female mammals occurs mostly randomly early in development, resulting in a heterogeneous cell population, with each cell having inactivated one or the other of its X chromosomes and passing on the same Xinactivation choice to its offspring. The inactivated X chromosome is generally hypermethylated and transcriptionally inactive.

You are studying a newly discovered type of colon tumor in women that has a morphology distinct from that of other colon adenomas. You extract chromosomal DNA from the tumor cells. You then either keep the DNA untreated or digest the DNA with a methylation-sensitive restriction enzyme that only cleaves its recognition DNA sequence if the sequence is not methylated. Finally, you amplify

Molecular Biology of the Cell 7e-Alberts by polymerase chain reaction (PCR) a locus on the X chromosome known to be polymorphic in length, that is, it is expected to be of different sizes in different X chromosomes. The locus has a restriction site for the mentioned enzyme, such that cleavage would prevent PCR amplification. You quantify the amount of PCR products corresponding to shorter and longer versions of the locus, and obtain the results shown in the following table. Do these data appear to be in better agreement with a monoclonal or a polyclonal origin of cancer? Explain your reasoning in 2–3 sentences.

55. Review the below figure and then compare the features of an invasive epithelial carcinoma against those of a normal epithelium. How do these changes occur over the course of tumor progression?

Molecular Biology of the Cell 7e-Alberts

56. How would you describe what happens in the interior of a large tumor? What makes it unique in comparison to other tumors?

57. The homologous chromosome pairs in our cells do not carry identical sequences in all loci. This heterozygosity (difference between the two copies) can be altered in cancer: in fact, loss of heterozygosity at many loci is observed in cancer cells, through an increase in either homozygosity (two identical copies) or hemizygosity, that is, loss of one copy. Researchers can take advantage of this loss of heterozygosity in cancer cells to identify genomic loci that contain cancer-critical genes. What type of gene would you expect to find in chromosomal regions with a loss of heterozygosity: Proto-oncogenes or tumor suppressor genes?

58. Three fundamental controls seem to have been subverted in essentially every type of cancer. Choose these three among the following regulatory axes. Your answer would be a three-letter string composed of letters A to F only, in alphabetical order, for example, BDF. A. Wnt pathway B. Rb pathway C. RTK/Ras/PI3K pathway D. p53 pathway E. Hippo pathway F. GPCR/PKA pathway

Molecular Biology of the Cell 7e-Alberts 59.

From an immortalized human HeLa cell line with wild-type p53 genes, you have derived a line that lacks both copies of the gene. You then treat the original and derived cells with the anticancer drug doxorubicin, which can activate the p53 pathway in the cell by stalling DNA replication forks and inducing double-strand breaks in DNA. You measure cell proliferation in the presence of different doses of the drug in each of the two cell lines, and plot the results as shown in the graph below. Which cell line (A or B) do you expect to be the original HeLa line? Your answer should be either single-letter A or B.

60. How would you describe the phenotype obtained after crossing a transgenic mouse strain carrying a Ras oncogne with a stain carrying a Myc oncogene? The transgenes in each strain are under the control of the mouse mammary tumor virus (MMTV) promoter.

61. Suppose you are studying tumor heterogeneity in a certain type of melanoma.

You have used fluorescence-activated cell sorting (FACS) to specifically isolate those melanoma tumor cells that either do (first category) or do not (second

Molecular Biology of the Cell 7e-Alberts category) express a specific marker present in normal stem cells in the tissue of origin, that is, the melanocyte stem cells. You implant the same number of cells from each of these categories into severely immunodeficient mice and compare the tumor-formation efficiencies after several weeks, which turn out to be significantly higher for the first category. You then analyze the new tumors using FACS, and find out that the majority of the cells in the tumors that originated from the first category of cells harbor the stem-cell marker, whereas the majority of the cells in the tumors that originated from the second cell category lack the marker, just like their respective founder cells. Yes or no: Do these observations support the existence of cancer stem cells? Support your reasoning in 2–3 sentences.

62. One of the most common hereditary conditions associated with colorectal cancer is hereditary nonpolyposis colorectal cancer (HNPCC), or Lynch syndrome. How would you describe the pathophysiology of (HNPCC)?

63. Why or why not is cancer a contagious disease?

64. The Ames test is used to test the mutagenicity of a compound suspected to be a carcinogen. In a simple form of the test, the carcinogen is first mixed with a rat liver extract. A disc of filter paper is soaked with this mixture and placed on a culture of a strain of Salmonella typhimurium that is defective in a gene involved in the synthesis of histidine, an amino acid that is essential for cell growth and proliferation. The strain is thus normally unable to grow into visible colonies when the histidine in the culture medium is depleted. In the presence of a mutagen, however, mutations (often “reverse mutations” in the same gene) can enable the bacteria to produce histidine on their own, and therefore grow into colonies. The results of the Ames test for three compounds A, B, and C—each used at the same concentration—are shown in the schematic diagram below. Colonies are indicated with black dots, and the disc is indicated with a white circle at the center of each plate. Which compound (A to C) appears to be a stronger mutagen in this assay? Your answer should be single letter A, B, or C.

Molecular Biology of the Cell 7e-Alberts

65.

You are studying the rising incidence of a certain subtype of cervical cancer in Oceania and are curious to know whether environmental factors are the dominant cause of the disease. You collect the incidence statistics from indigenous populations as well as from two different immigrant populations in three different countries, as shown in the following table. Do these data appear to be consistent with a dominant role of environmental risk factors or a genetic background for this type of cancer?

Molecular Biology of the Cell 7e-Alberts

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 20 1. Answer:

A

2. Answer:

A

3. Answer:

E

4. Answer:

E

5. Answer:

E

6. Answer:

A

7. Answer:

E

8. Answer:

A

9. Answer:

B

10. Answer:

E

11. Answer:

D

12. Answer:

C

13. Answer:

C

14. Answer:

E

15. Answer:

B

16. Answer:

A

17. Answer:

E

18. Answer:

B

19. Answer:

E

20. Answer:

A

Molecular Biology of the Cell 7e-Alberts 21. Answer:

D

22. Answer:

D

23. Answer:

C

24. Answer:

C

25. Answer:

B

26. Answer:

D

27. Answer:

C

28. Answer:

B

29. Answer:

A

30. Answer:

B

31. Answer:

C

32. Answer:

E

33. Answer:

C

34. Answer:

C

35. Answer:

A

36. Answer:

D

37. Answer:

C

38. Answer:

A

39. Answer:

A

40. Answer:

B

41. Answer:

A

42. Answer:

B

Molecular Biology of the Cell 7e-Alberts 43. Answer:

A

44. Answer:

B

45. Answer:

A

46. Answer:

1. A 2. A 3. A 4. C

47. Answer:

1. A 2. B 3. B 4. A

48. Answer:

1. B 2. B 3. A 4. B

49. Answer:

1. A 2. A 3. B 4. A

50. Answer:

1. B 2. C 3. A

51. Answer:

1. C 2. B 3. A

Molecular Biology of the Cell 7e-Alberts 4. D 52. Answer:

1. B 2. C 3. C

53. Answer:

Cell 1

54. Answer:

The data are in better agreement with a monoclonal origin of cancer, because the individual is heterozygous in the locus, represented in a ratio close to 1:1 for the PCR products of the longer to shorter alleles when an undigested DNA is used. When fully digested, however, only the methylated DNA is expected to be represented in the PCR products. Whereas the polyclonal healthy tissue sample still has a mostly unbiased 1:1 ratio, the monoclonal tumor cells (all with the same inactive X chromosome) have a skewed ratio.

55. Answer:

In the normal epithelium, dividing cells are strictly seen in the basal layer; however, epithelial cells also show high rates of proliferation and altered differentiation. Over the course of tumor progression, cancer cells move through or destroy the basal lamina, the layer on which the epithelium lies, in order to invade the underlying connective tissue.

These changes occur because in each stage of tumor progression, the cell acquires another mutation, which gives the tumor a greater chance to succeed in its environment and grow until malignancy takes place.

56. Answer:

Although cancer cells typically evade programmed cell death, if tumor cells grow rapidly in the interior of a large solid tumor, cells must compete for nutrients and oxygen.

This leads to death by necrosis. The access to these key elements is proportional to the vascular supply of the tissue.

57. Answer:

Tumor suppressor genes

58. Answer:

BCD

59. Answer:

B

60. Answer:

Usually, normal cells are not transformed into cancerous cells by the introduction of only one oncogene. The expression of oncogenes Ras and Myc in the offspring of

Molecular Biology of the Cell 7e-Alberts the cross will induce hyperproliferation and inhibit apoptosis, contributing to cell transformation and tumor growth in mouse mammary glands.

61. Answer:

No, these observations do not support the existence of cancer stem cells, because cancer stem cells are expected to constitute a minority of the tumor cell population. Even though the transplantation is more efficient with the subpopulation of cells that expresses the stem-cell marker, this may simply be a reflection of the heterogeneity in the original tumor, with these cells harboring a higher ability to found new tumors.

62. Answer:

HNPCC is associated with the existence of mutations in one copy of several genes related with the DNA mismatch repair system. Afflicted individuals may suffer a loss or inactivation of the remaining gene copy during their lives, resulting in an elevated spontaneous mutation rate. Cells that are genetically unstable are more likely to become malignant. The risk level is high without any increase in the number of colorectal polyps. HNPCC cells are difficult to detect due to their normal or almost normal appearance.

63. Answer:

Although several infectious agents are associated with cancer, cancer is not contagious. A significant minority of cancers are related to infections with viruses, bacteria, and parasites. These agents induce cellular transformation through different mechanisms, such as interfering with components of the cell cycle and programmed cell death control systems.

64. Answer:

B

65. Answer:

According to the data above, it is suggested that environmental risk factors play an important role in the development of a certain subtype of cervical cancer. For example, immigrants from Melanesia, a region with high incidence of this tumor, but living in Australia and Japan, present a lower incidence than the Melanesian population.

Moreover, Australia and Japanese immigrants in Melanesia have experienced a higher incidence of this specific tumor than in their country of origin.

Molecular Biology of the Cell 7e-Alberts

Chapter 21: Development of Multicellular Organisms 1. The figure below depicts the lineage from blastomere to a pancreatic β cell. What process, then, does the figure illustrate?

a. Proliferation b. Cell division c. Cell signaling d. Terminal differentiation e. Morphogenesis

2. Plants and animals use different developmental strategies and have very different

Molecular Biology of the Cell 7e-Alberts ways of life. In which of the following fundamental cellular processes during development are plants most different to animals? a. Cell proliferation b. Cell–cell interactions c. Cell specialization d. Cell movement

3. Even though the genome is the same in an animal's skin and muscle cell, what can account for the different structure and biochemistry observed in these two different cell types? a. Transcriptional regulators direct the expression of completely different genes to direct the development of the two cell types. b. Transcriptional regulators direct the expression of the same subset of genes, the proteins of some are then degraded. c. Transcriptional regulators drive the expression of similar but different gene sets across developmental time to direct the development of the two cell types. d. Post-translational modifications to the same set of gene products allows for the development of the two different cell types. e. Differential expression of Hox genes ensures that the two cell types develop differently.

4. Cortical rotation following fertilization in X. laevis places which pole (1) at the point of sperm entry, while Wnt11 mRNA is transported to which pole (2)? a. 1: anterior; 2: posterior b. 1: posterior; 2: anterior c. 1: animal; 2: vegetal d. 1: dorsal; 2: ventral e. 1: ventral; 2: dorsal

Molecular Biology of the Cell 7e-Alberts

5. Which of the following is NOT true regarding maternal-effect genes? a. Bicoid and Nanos are maternal-effect genes. b. A female homozygous for a loss-of-function maternal-effect gene mutation can be fully normal, but her offspring will show the phenotype. c. The offspring of a female homozygous for a loss-of-function maternal-effect gene mutation will show the phenotype regardless of the paternal genotype. d. The second-generation offspring of a male homozygous for a loss-offunction maternal-effect gene mutation can show the phenotype. e. They are expressed in a spatially restricted manner in the maternal organism and give rise to discrete segmented patterns in the embryo.

6. Which of the following groups of genes confers positional value to a group of cells and is capable of transforming the identity of one segment into another, such as transforming a limb segment into an antenna segment? a. Egg-polarity genes b. Pair-rule genes c. Hox genes d. Gap genes

7. Which of the following schematic drawings better depicts the regulatory network that maintains Engrailed (en), Hedgehog (hh), and Wingless (wg) expression following cellularization in a developing Drosophila embryo?

Molecular Biology of the Cell 7e-Alberts

a.

b.

c.

d.

e.

Molecular Biology of the Cell 7e-Alberts

8. Which of the following is the correct order of hierarchical regulation, from earliest acting to latest acting, of A-P pattering genes in Drosophila embryo? a. Egg-polarity genes, gap genes, pair-rule genes, segment polarity genes b. Egg polarity genes, segment polarity genes, gap genes, pair-rule genes c. Egg polarity genes, pair-rule genes, gap genes, segment polarity genes d. Gap genes, egg polarity genes, segment polarity genes, pair-rule genes e. Egg polarity genes, segment polarity genes, pair-rule genes, gap genes

9. In Drosophila melanogaster, the expression of genes Ultrabithorax and Antennapedia can normally be observed in the third thoracic segment which bears a pair of legs as well as a pair of halteres. Homeotic mutations associated with these two genes can give rise to remarkable disturbances in the organization of the adult fly: Two pairs of wings in the case of Ultrabithorax, and legs in the place of antennae in the case of Antennapedia. Would you expect these to be gain-of-function or loss-of-function mutations? a. Loss-of-function for both genes b. Loss-of-function for Ultrabithorax and gain-of-function for Antennapedia c. Gain-of-function for Ultrabithorax and loss-of-function for Antennapedia d. Gain-of-function for both genes

10. The gene clusters known as the Bithorax complex and the Antennapedia complex contain which of the following? a. A subset of segment-polarity genes b. Segmentation genes c. Homeobox-containing genes d. Genes encoding chromatin repressors e. Genes encoding chromatin remodelers

Molecular Biology of the Cell 7e-Alberts

11. Cactus is a maternal-effect gene coding for an inhibitory protein in the Drosophila Toll signaling pathway that binds to Dorsal in the cytosol and keeps it from nuclear entry. Would you expect maternal loss-of-function mutations in Cactus to give rise to dorsalized or ventralized embryos? What about loss-of-function mutations in Decapentaplegic (Dpp)? a. Dorsalized; dorsalized b. Dorsalized; ventralized c. Ventralized; dorsalized d. Ventralized; ventralized

12. Assuming that the only function of Short gastrulation (Sog) in fruit flies is to regulate Decapentaplegic (Dpp), which of the following genetic interactions would you NOT expect to observe? a. Mutant embryos lacking both Sog and Dpp phenotypically resemble those lacking only Dpp. b. Partial loss of Sog suppresses the phenotype of partial loss of Dpp. c. Gain of Dpp suppresses the phenotype of partial loss of Sog. d. Gain of Sog suppresses the phenotype of gain of Dpp.

13. The following graph qualitatively represents the gradients of Nodal and Lefty gene products in an early frog embryo. The position along which primary axis is defined by these gradients? Which curves correspond to these two proteins?

Molecular Biology of the Cell 7e-Alberts

a. D-V axis; curves 1 and 2 b. D-V axis; curves 1 and 3 c. A-V axis; curves 1 and 2 d. A-V axis; curves 1 and 3

14. The role of Chordin and Noggin in patterning in developing vertebrates is equivalent to that of what (1) in Drosophila? They are secreted from which (2) pole and antagonize the activity of bone morphogenetic factors that are secreted throughout the embryo? a. 1: Decapentaplegic; 2: dorsal b. 1: Decapentaplegic; 2: ventral c. 1: Short gastrulation; 2: dorsal d. 1: Short gastrulation; 2: ventral

15. Which of the following proteins is most concentrated near the dorsal rather than

Molecular Biology of the Cell 7e-Alberts the ventral side of the Drosophila embryo? a. Decapentaplegic b. Dorsal (nuclear fraction) c. Toll (active form) d. Twist

16. Through which mechanism do Polycomb and Trithorax ensure a cell remembers its positional information along the A-P axis? a. Trithorax maintains the expression of already expressed Hox genes while Polycomb represses Hox genes that are not yet expressed. b. Polycomb maintains the expression of already expressed Hox genes while Trithorax represses Hox genes that are not yet expressed. c. Both Polycomb and Trithorax imprint the chromatin to repress the expression of non-expressed Hox genes. d. Both Polycomb and Trithorax imprint the chromatin to ensure the expression of already expressed Hox genes. e. Both Polycomb and Trithorax imprint the chromatin to ensure only the positionally appropriate Hox genes are expressed.

17. If you know the relative positional value of a vertebrate Hox gene along the A-P axis of an animal, what other information can you assume about that same gene? a. That an identical Hox gene exists in the Drosophila b. That the vertebrate gene is regulated by the Trithorax complex c. The relative order of the vertebrate Hox gene along the animal's chromosome d. The exact time during development that the Hox gene will be expressed e. You cannot infer any additional information based on the local positional value of the Hox gene expression.

Molecular Biology of the Cell 7e-Alberts

18. Which of the following is NOT true regarding cell-type specification in developing animals? a. Members of the MyoD/myogenin family of transcription regulators drive cells to differentiate into muscle cells. b. Members of the Achaete/Scute family of transcription regulators drive the cells to form endothelial layers. c. Many proteins that induce the differentiation of particular cell types belong to the bHLH (basic helix–loop–helix) class of transcription regulators. d. Eyeless is both necessary and sufficient to generate eye structures composed of various specialized cell types in Drosophila. e. In vertebrates, loss of the Eyeless homolog Pax6 leads to an animal without eyes.

19. The result of Notch-mediated competitive lateral inhibition in a patch of wild-type cells is depicted on the left in the following schematic drawing, in which the black cells have become specialized. Consider a genetic mutation that interferes with intracellular Notch signaling in such a way that Notch can no longer regulate Delta effectively. Which drawing (1 or 2) better represents the outcome with cells bearing this mutation? Do the specialized cells (black hexagons) have active or inactive Notch at their surface?

Molecular Biology of the Cell 7e-Alberts

a. Drawing 1; active Notch b. Drawing 1; inactive Notch c. Drawing 2; active Notch d. Drawing 2; inactive Notch

20. Which of the following evolutionary changes better explains morphological differences between different animals despite many common molecular mechanisms governing their development? a. Changes in the types of genes responsible for key developmental processes b. Changes in regulatory DNA controlling expression of key developmental

Molecular Biology of the Cell 7e-Alberts genes c. Changes in DNA-binding domains of key transcription regulators involved in development d. Changes in coding sequences of key developmental genes e. Changes in copy number of key developmental genes

21. Which of the following is NOT true regarding vertebrate segmentation? a. The presomitic mesoderm retreats tailward as new somites are generated. b. The period of gene-expression oscillations in the presomitic mesoderm determines the size of each somite. c. If feedback delays in the gene-expression oscillations are increased, somite size will consequently increase as well. d. Notch-mediated lateral inhibition ensures that the segmentation clock is different in neighboring cells in the presomitic mesoderm.

22. The gene encoding Hes7 in mouse contains three introns. Any or all of these introns can be deleted to alter the delay associated with transcription and splicing of Hes7 mRNA. As a result of such deletions, either the oscillation frequency of this gene’s expression can change (result 1) or the oscillatory behavior can disappear altogether (result 2). One of these two results is obtained when only one intron is deleted from the gene, while the other result is obtained when all three introns are removed. Would you expect the oscillation frequency to increase or decrease if result 1 is obtained? Which result (1 or 2) would you expect to observe when all three introns are deleted? a. Increase; result 1 b. Increase; result 2 c. Decrease; result 1 d. Decrease; result 2

23. Consider the embryonic cerebral cortex of a mammal. Near which side of the

Molecular Biology of the Cell 7e-Alberts cortex are the cell bodies of radial glial cells located? To which side of the cortex do the first-born neurons eventually migrate? a. Inner surface; inner surface b. Inner surface; outer surface c. Outer surface; inner surface d. Outer surface; outer surface

24. What can you conclude about the developmental history of neurons that ultimately reside at the outer layer of the mammalian cerebral cortex? a. That these outer cells are the first population of cells that became specified. b. That these outer cells are the most differentiated neurons in the cortex. c. That these outer cells are the youngest and have migrated the furthest since they were formed. d. That these outer cells are the youngest and have migrated the shortest distance of any cortical neuronal cell. e. That these outer cells were created in the outermost layer of the cortex and have not migrated at all.

25. Lin4 and Let7 are two genes found in C.elegans that encode for what type of regulatory element? a. Constitutive transcription factors b. Cell-type specific transcription factors c. Transcriptional effectors d. Spliceosome RNAs e. miRNAs

26. In the embryo, as the rate of cell division slows down, the ratio of cytoplasm to

Molecular Biology of the Cell 7e-Alberts nuclei increases and the embryo begins to express genes from its own genome.

Which developmental phenomena is occurring? a. Gastrulation b. Maternal-zygotic transition (MZT) c. Metamorphosis d. Convergent extension e. Branching morphogenesis

27. TSH is a pituitary hormone that stimulates the production of thyroid hormone by the thyroid gland. This production can be blocked by sodium perchlorate (SP), which inhibits the cellular import of iodine necessary for thyroid hormone synthesis. What would be the effect of TSH or SP exposure, respectively, on the timing of metamorphosis in frog larvae? a. Premature metamorphosis; premature metamorphosis b. Premature metamorphosis; delayed metamorphosis c. Delayed metamorphosis; premature metamorphosis d. Delayed metamorphosis; delayed metamorphosis

28. Stromal cell-derived factor 1 (SDF1) is a secreted protein that plays a major role in guiding the migration of various cells during development. It binds to a Gprotein-coupled receptor on the surface of the migrating cell, which in turn activates a trimeric G protein containing a G i subunit. Pertussis toxin (PTX) modifies and inactivates the Gi subunit. What would you expect to observe if migrating cells are treated with PTX? a. They migrate toward the source of SDF1. b. They migrate away from the source of SDF1. c. They migrate in a certain direction chosen randomly. d. They move in small steps in random directions.

Molecular Biology of the Cell 7e-Alberts e. They lose their motility.

29. The Steel factor/Kit signaling pathway plays an important role in the migration of many types of cells during development as well as in the adult animal. Not surprisingly, loss-of-function mutations in Steel factor result in cell migration defects. Selective inactivation in these cells of which of the following proteins might be expected to rescue (i.e., partially restore to normal) the defective phenotype of Steel loss-of-function mutants? a. Kit, the receptor for Steel b. Bax, an essential apoptotic protein c. Endothelin-3, a survival factor for neural crest cells d. Fibronectin, a multi-adhesive protein of the extracellular matrix e. FGF4, a growth factor

30. In classical experiments done half a century ago, the cells of early frog embryos were disaggregated and later reaggregated in desired combinations. The cells managed to rearrange and sort themselves out into an overall arrangement similar to that of a normal embryo. This effect is mainly due to what mechanism? a. Chemotactic cell motility b. Asymmetric cell division c. Cell adhesion d. Convergent extension e. Lateral inhibition

31. In the hindbrain of the developing mouse embryo EphrinB2 and EphA4 are expressed in adjacent cells at the R4/R5 boundary. Suppose the expression of EphA4 was blocked in the developing R5. What effect would you predict to observe? a. There would be no effect on the development of the rhombomeres as the fate of the rhombomeres are already specified.

Molecular Biology of the Cell 7e-Alberts b. The boundary between R4 and R5 would not develop as ephrin-Eph signaling is required to sort out the boundary. c. No rhombomeres would develop in the affected animal. d. R4 would be duplicated into an additional R5 with a well-defined boundary between the two R5s. e. The affected animal would fail to develop at all since an animal cannot live without a brain.

32. During the process of convergent extension, populations of cells migrate in a coordinated manner. One of the most dramatic examples of convergent extension is observed coordinating the development of the vertebrate notochord. As the notochord develops, via convergent extension, which axis(axes) is(are) extended? a. Only the anterior-posterior axis b. Only the dorsal-ventral axis c. Only the proximal-distal axis d. Both the anterior-posterior and dorsal-ventral axis e. Both the dorsal-ventral and proximal-distal axis

33. The following schematic drawings show an epithelial sheet. Which example shows a mutant with planar-cell-polarity defects but no other defects?

a.

Molecular Biology of the Cell 7e-Alberts

b.

c.

d.

34. What happens during branching morphogenesis in lung development? a. FGF10 is secreted by epithelial cells at the tip of the growing epithelial tubes. b. FGF10 inhibits Shh production by the epithelial cells at the tip of the growing epithelial tubes. c. Shh inhibits FGF10 production by the epithelial cells at the tip of the growing epithelial tubes. d. Shh is produced by epithelial cells at the tip of the growing epithelial tubes.

35. Which cell behavior depicted below is involved in the tube formation that underlies the development of lungs and trachea?

Molecular Biology of the Cell 7e-Alberts

a.

b.

c.

d.

e.

Molecular Biology of the Cell 7e-Alberts

36. You have obtained leaf samples from three strawberry varieties A, B, C, and D. You isolate the cells, extract their nuclei, and stain them with propidium iodide, a fluorescent dye that binds quantitatively to DNA. You then use a fluorescenceactivated cell sorter (FACS) machine to sort the nuclei based on their fluorescence. Guessing from the results, presented in the following histograms, which variety would you expect to yield larger strawberries?

a. Variety A b. Variety B c. Variety C d. Variety D

37. After reaching sexual maturity, the nematode Caenorhabditis elegans normally doubles in size within about two weeks. This doubling is mainly due to what?

Molecular Biology of the Cell 7e-Alberts a. Cell division b. DNA replication without cell division c. Reduced cell death d. Deposition of extracellular matrix

38. Mutations in certain components of the cell-cycle machinery in Drosophila melanogaster can be used to slow down the rate of progression through the cell cycle in the wing imaginal discs of the fly larvae, without a major effect on the rate of cell growth. As a result, how would the wing in the mutant flies compare to wildtype flies? a. It would have a larger size (area) with a higher number of relatively normal cells. b. It would have the same size, but with a higher number of relatively smaller cells. c. It would have the same size, but with a smaller number of relatively larger cells. d. It would have a smaller size with the same number of relatively smaller cells. e. It would have a smaller size with a smaller number of relatively normal cells.

39. What are epithelia pouches of the Drosophila that grow through cellular proliferation during the larval period and then differentiate into the adult organs and extremities during the pupal period? a. Imaginary discs b. Imagery discs c. Integral discs d. Interpolar discs e. Imaginal discs

Molecular Biology of the Cell 7e-Alberts

40. Regarding animal development, a cell’s response to a signal depends on its exposure to other signals at that present time as well as in the past. a. True b. False

41. Generally, the more anterior of the Hox genes dominate (or suppress) posterior Hox genes. a. True b. False

42. Timing of developmental transitions can be coordinated by cell–cell interactions as well as globally by hormones. a. True b. False

43. Correlate the appropriate cellular process to the statement it best fits. a. Cellular proliferation b. Cellular differentiation c. Cellular morphogenesis

1. The division of one cell into many identical progeny.

2. The interaction between many cells which results in the adoption of new structures.

3. The change in identity of one or few cells in a uniform field which results in the cells adopting a unique specialization.

4. Is often caused by changes in gene expression as regulated by external cues.

44. The simplified drawing below depicts early stages of animal development.

Indicate which letter (A to E) in the drawing corresponds to each of the following terms.

Molecular Biology of the Cell 7e-Alberts

a. A b. B c. C d. D e. E

1. Blastula 2. Ectoderm 3. Endoderm 4. Mesoderm 5. Gastrula

45. Indicate whether each of the following organs or tissues arises from ectoderm, mesoderm, or endoderm. a. Ectoderm b. Mesoderm c. Endoderm

1. Blood 2. Liver and pancreas 3. Brain 4. Bone and cartilage

46. Imagine a morphogen gradient established from left to right in a field of cells in a

Molecular Biology of the Cell 7e-Alberts developing tissue, as shown in the following schematic diagram. Below a first threshold of morphogen concentration, cells do not respond to the morphogen and express gene “light gray” by default. Cells exposed to morphogen concentrations above this threshold respond by expressing gene “white” instead, while those exposed to even higher concentrations, above a second threshold, express gene “dark gray.” With no other change, if the diffusion rate of the morphogen is increased (by a modification that decreases its affinity for heparan sulfate proteoglycans, for example), the gradient profile changes from the gray curve to the black curve, as indicated. Under this new condition, indicate whether each of the following would be expected to increase, decrease, or remain unchanged in its range.

a. Increase b. Decrease c. Unchanged

1. Dark gray expression domain 2. White expression domain 3. Light gray expression domain

47. Indicate whether each of the following descriptions better applies to patterning by lateral inhibition, reaction-diffusion systems, or sequential induction.

Molecular Biology of the Cell 7e-Alberts a. Lateral inhibition b. Reaction-diffusion systems c. Sequential induction

1. It does NOT generate asymmetrical patterns from an initial noisy field.

2. It is based on short-range activation and long-range inhibition.

3. It can readily generate complex patterns resembling the spots of a leopard or stripes of a zebra.

4. It is commonly mediated by Notch signaling.

48. Indicate whether each of the following descriptions better applies to these poles established in an early embryo: anterior, posterior, dorsal, ventral, animal, or vegetal. a. Anterior b. Posterior c. Dorsal d. Ventral e. Animal f. Vegetal

1. It defines the part to become internalized in gastrulation.

2. It defines the parts to remain external.

3. It defines the location of the future head.

4. It defines the location of the future tail.

5. It defines the location of the future belly.

6. It defines the location of the future back.

49. Indicate whether each of the following descriptions better applies to egg-polarity genes, gap genes, pair-rule genes, or segment-polarity genes. a. Egg-polarity genes b. Gap genes c. Pair-rule genes

Molecular Biology of the Cell 7e-Alberts d. Segment-polarity genes

1. Mutations in these genes show a maternal effect.

2. Mutations in these genes leave the embryo with only half the number of normal segments.

3. Mutations in these genes produce a normal number of segments but with part of each segment replaced by a mirror-image duplicate of other parts of the segment.

4. Mutations in these genes can eliminate one or two groups of adjacent segments altogether.

50. In a developing Drosophila melanogaster embryo, a hierarchy of gene regulatory interactions subdivides the embryo to regulate progressively finer details of patterning. For each of the following proteins, indicate which expression pattern (1 to 4) in the schematic drawing below is more appropriate.

Molecular Biology of the Cell 7e-Alberts

a.

b.

c.

d.

1. Hunchback (product of a gap gene) 2. Engrailed (product of a segment-polarity gene) 3. Bicoid (product of an egg-polarity gene) 4. Even-skipped (product of a pair-rule gene)

51. In the following fate map for the Xenopus blastula, indicate which zone in the map (A to D) better corresponds to each of the following fates.

Molecular Biology of the Cell 7e-Alberts

a. Zone A b. Zone B c. Zone C d. Zone D

1. Lung 2. Blood 3. Brain 4. Skin

52. The qualitative graph below shows the molecular changes accompanying vernalization in the flowering plant Arabidopsis thaliana. Indicate which curve (A to D) in the graph better represents the temporal changes in each of the following variables.

Molecular Biology of the Cell 7e-Alberts

a. Curve A b. Curve B c. Curve C d. Curve D

1. Level of Coolair noncoding RNA 2. Level of Flowering locus C (FlC) gene product (transcriptional repressor) 3. Level of Flowering locus T (Ft) gene product (transcriptional activator) 4. Level of repressive chromatin marks at the FlC locus

53. Indicate whether each of the following conditions favors a larger or smaller tissue or body size. a. Larger b. Smaller

Molecular Biology of the Cell 7e-Alberts 1. Hippo overexpression 2. Insulin-like growth factor overexpression 3. Myostatin deletion 4. Growth hormone deficiency

54. Morphogens are an essential class of signaling molecules that direct a large array of developmental processes. Moreover, a single molecule is capable of inducing the development of a diverse array of tissue structures. How can a single morphogen exert a variety of effects across a uniform field of cells?

55. Sort the following organisms from the least to the most degree of asymmetry in the unfertilized egg (i.e., pre-defined axes of polarization). Your answer would be a three-letter string composed of letters A to C only, for example, BCA. A. M. musculus B. D. melanogaster C. X. laevis

56. Sort the following primary axes in the order that they are established during the development of Xenopus laevis. Your answer would be a three-letter string composed of letters A to C only, for example, ACB. A. A-P B. D-V C. A-V

57. During the development of the Drosophila embryo, the mRNAs and proteins of Decapentaplegic (Dpp) and Short gastrulation (Sog) contribute to the development of the dorsal-ventral axis. How do the expression patterns of Dpp and Sog mRNA mediate the pattern formation of the D-V axis, with respect to dorsal-medial-ventral structure?

58. Notch-driven lateral inhibition in a field of ectodermal cells results in some cells becoming specialized, while other cells in the field remain unspecialized. What predictions can you make with regard to the specialized cells in the field?

Molecular Biology of the Cell 7e-Alberts 59. Developmental time is perceived by the Drosophila neuroblasts due to changes in intracellular transcriptional regulators and then by asymmetric cell division. While investigating a novel Drosophila mutant, you discover an increase in the number of U4 neurons and associated glial cells; however, the ratio of neurons to glia is unaffected. What predictions can you make about the molecular nature of your mutant?

60. How can changes in CXCL12 expression alter the migratory route that primordial germ cells and other migratory cells (such as lymphocytes and neurons) travel from the tissue, where the cell is specified to the ultimate target tissue, which in some cases can be across the embryo?

61.

CFSE (carboxyfluorescein succinimidyl ester) is a cell-tracing molecule that, once inside a cell, is modified to yield a highly fluorescent molecule that is retained within the cell. Due to its stability, it can be used to trace cells and follow their divisions: each division dilutes the fluorescent dye twofold. You label T cells from a healthy mouse with CFSE and inject them into either a wild-type strain of mouse or a strain with T cell deficiency, both of which are immunologically compatible with the donor mouse. You later collect and analyze peripheral T cells from the recipient mice and draw the following histograms according to the fluorescent intensity per cell. Based on these results, do you think the total number of T cells in the mouse body is (1) controlled by intracellular programs (as in the thymus, for example) or is (2) regulated as a whole (as in the spleen)? Your response should be the single-digit 1 or 2.

Molecular Biology of the Cell 7e-Alberts

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 21 1. Answer:

D

2. Answer:

D

3. Answer:

C

4. Answer:

E

5. Answer:

E

6. Answer:

C

7. Answer:

A

8. Answer:

A

9. Answer:

B

10. Answer:

C

11. Answer:

D

12. Answer:

C

13. Answer:

C

14. Answer:

C

15. Answer:

A

16. Answer:

A

17. Answer:

C

18. Answer:

B

19. Answer:

B

20. Answer:

B

Molecular Biology of the Cell 7e-Alberts 21. Answer:

D

22. Answer:

B

23. Answer:

A

24. Answer:

C

25. Answer:

E

26. Answer:

B

27. Answer:

B

28. Answer:

D

29. Answer:

B

30. Answer:

C

31. Answer:

B

32. Answer:

A

33. Answer:

B

34. Answer:

D

35. Answer:

B

36. Answer:

B

37. Answer:

B

38. Answer:

C

39. Answer:

E

40. Answer:

A

41. Answer:

B

42. Answer:

A

Molecular Biology of the Cell 7e-Alberts 43. Answer:

1. A 2. C 3. B 4. B

44. Answer:

1. A 2. C 3. E 4. D 5. B

45. Answer:

1. B 2. C 3. A 4. B

46. Answer:

1. B 2. A 3. B

47. Answer:

1. C 2. B 3. B 4. A

48. Answer:

1. F 2. E 3. A 4. B 5. D 6. C

Molecular Biology of the Cell 7e-Alberts 49. Answer:

1. A 2. C 3. D 4. B

50. Answer:

1. B 2. D 3. A 4. C

51. Answer:

1. D 2. C 3. B 4. A

52. Answer:

1. B 2. A 3. D 4. C

53. Answer:

1. B 2. A 3. A 4. B

54. Answer:

The concentration of the morphogen at a target cell is sufficient to induce the development of a unique structure. A high level of the morphogen can generate one structure, while a lower level of the same morphogen will result in the development of a different structure.

55. Answer:

ACB

56. Answer:

CBA

Molecular Biology of the Cell 7e-Alberts 57. Answer:

Dpp mRNA is expressed along the dorsal surface of the embryo, while Sog mRNA is expressed in lateral regions.

Since a majority of Dpp protein will be dorsal and little sog will be present, dorsal structures will be specified due to high Dpp activity. In lateral regions where both Dpp and Sog are expressed, Sog will inhibit Dpp, and intermediate structures can then form due to lower Dpp activity. In ventral regions, no Dpp will be active, and thus ventral structures can form.

58. Answer:

You can expect that the specialized cells in the field will have expressed higher levels of the Notch ligand Delta, and thus will have lower levels of Notch activity than their neighbors. The specialized cells will also show positive regulation of Delta to continue to inhibit the specialization of neighboring cells.

59. Answer:

You can assume that the expression or activity of Pdm is prolonged or retained in progeny of the neuroblast. Once the neuroblast divides into the ganglion mother cell the asymmetric division of that cell is unaffected by your mutation.

60. Answer:

The migrating cell will follow the track of CXCL12 expression. As the localization of CXCL12 changes, the path that the migratory cell takes will also change.

61. Answer:

2

Molecular Biology of the Cell 7e-Alberts

Chapter 22: Stem Cells in Tissue Homeostasis and Regeneration 1. How does the process of self-renewal in stem cells differ from mitosis seen in typical somatic cells? a. The process of self-renewal always produces two cells that retain the properties of the original stem cell and do not differentiate. b. The process of self-renewal produces at least one cell that retains the properties of the original stem cell and does not differentiate. c. The process of somatic cell mitosis produces two cells that self-renew and become stem cells. d. The process of self-renewal produces two cells that retain the properties of the stem cell and begin to differentiate. e. The process of self-renewal is an asymmetrical division that produces two cells that differentiate into two different pathways.

2. When a multipotent stem cell divides, it is possible that the product of that division is no longer a stem cell. And yet—neither is it terminally differentiated. These cells are required to divide and produce the progenitors that will lead to terminally differentiated cells. What are these daughters of stem cells called? a. Predifferentiated cells b. Transit-amplifying cells c. Unprogrammed progenitor cells d. Unipotent cells e. Products of asymmetric stem cell division

3. Which statement is characteristic of stem cells? a. They always divide asymmetrically to give rise to two different daughter cells. b. They are terminally differentiated.

Molecular Biology of the Cell 7e-Alberts c. They can divide for the entire lifetime of the organism. d. They divide at a relatively fast rate.

4. Which of the following cells in a healthy adult human show the highest turnover? a. Neurons of the hippocampus b. Epithelial cells of the urinary tract c. Hepatocytes of the liver d. Epithelial cells of the small intestine e. Neutrophils of the blood

5. Each gut stem cell can be described in which way? a. Is pluripotent b. Divides approximately once a week. c. Gives rise to all four major epithelial cell types in the right proportions. d. Expresses Lgr5, just like the epithelial cells that derive from it

6. The maintenance of the epidermis is dependent upon stem cell activity. Which of the following is not characteristic of the epidermis, but is characteristic of the lining of the epithelium? a. The epidermis is stratified and squamous, that is, it forms squames. b. The stem-cell niche in epidermis is provided by the basal lamina and the underlying connective tissue. c. The transit amplifying cells in the epidermis move mostly perpendicular to the plane of the epithelium. d. The cells of the epidermis die long before they are shed. e. Differentiating cells rapidly divide (12-hour cell cycle) prior to differentiation.

Molecular Biology of the Cell 7e-Alberts

7. What is the best explanation for why satellite cells are difficult to observe through traditional lineage tracing experiments? a. Once activated, satellite cells divide rapidly leaving little time for them to be detected by traditional lineage tracing experiments. b. Satellite cells have mechanisms of exporting potentially toxic compounds, such as many lineage tracing molecules. c. Satellite cells are located deep below the basal lamina where it is difficult for lineage tracing compound to reach. d. Satellite cells are incapable of expressing GFP even when driven by a tissue-specific promotor. e. Satellite cells are quiescent until activated and therefore are not detectable by most lineage tracing experiments.

8. Which of the following is characteristic of all blood cells in our body? a. They have exceptionally short life spans—at most a month or so. b. They are produced only before adulthood. c. They are ultimately generated from multipotent hematopoietic stem cells. d. They are made from progenitors that themselves circulate in the blood.

9. A physician had prescribed a differential white blood cell count to help diagnose a sick child with sustained fever and weight loss. The results show a significant increase in the number of eosinophils beyond the normal range for healthy individuals. This is most likely due to what? a. Bacterial infection b. Parasitic infection c. Viral infection d. Kidney failure

Molecular Biology of the Cell 7e-Alberts e. Internal bleeding

10. How do erythrocytes differ from blood platelets? a. Erythrocytes lack nuclei while blood platelets are nucleated. b. Blood platelets are NOT derived from hematopoietic stem cells. c. Erythrocytes are derived from committed erythrocyte progenitor cells. d. There is no difference between the erythrocytes and blood platelets.

11. Using Cre recombination, the gene encoding stem cell factor (SCF) can be deleted in specific cell types in adult mice. Deletion of this gene in which of the following cells would you expect to lead to the most significant depletion of hematopoietic stem cells in the bone marrow? a. Osteoblasts b. Osteoclasts c. Stromal cells d. Hematopoietic stem cells e. Common myeloid precursor cells

12. What is a main function of natural killer (NK) cells? a. Kill invading bacteria b. Regulate activities of other leukocytes c. Kill tumor cells d. Make and secrete antibodies

13. Which of the following cell types in our body is typically renewed independently of stem cells?

Molecular Biology of the Cell 7e-Alberts a. Red blood cells b. Hepatocytes of the liver c. Epithelial cells of the epidermis d. Neutrophils of the blood

14. If some cells in an adult mammal are lost, they are lost forever and cannot be replaced. Which of the following cell types have such a limitation? a. Photoreceptive epithelium of the retina b. β Cells of pancreatic islets c. Hepatocytes of the liver d. Olfactory epithelium of the nose

15. In some adenomas of the colon, intestinal crypt cells appear to have proliferated abnormally to form small tumors known as polyps. Studies on the familial type of such a disease led to the identification of a major signaling pathway (which was aberrant in the polyp cells) as being involved in the maintenance of the gut stemcell compartment. What signaling pathway is this? Is it up-regulated or downregulated in colon cancers? a. MAPK pathway; up-regulated b. MAPK pathway; down-regulated c. Wnt pathway; up-regulated d. Wnt pathway; down-regulated e. Hedgehog pathway; down-regulated

16. This environment maintains a stem cell, and is also defined collectively by the secreted factors, the extracellular matrix, and the support cells. What environment is it? a. The stem cell tissue

Molecular Biology of the Cell 7e-Alberts b. The stem cell community c. The stem cell environment d. The stem cell niche e. The stem cell maintenance cues

17. If a stem cell were to divide asymmetrically with respect to cell-fate determinants, what can one predict about the products of that asymmetric division? a. One cell will retain the cell-fate determinant and self-renew, while the other cell will differentiate. b. Both cells will retain the cell-fate determinant and self-renew. c. Both cells will retain the cell-fate determinant and differentiate. d. One cell will retain the cell-fate determinant and differentiate, while the other will self-renew. e. Neither cell will retain the cell-fate determinant, yet both will differentiate.

18. If a planarian is irradiated to block all cell division and then amputated, the resulting fragments are incapable of regeneration. As an investigator, what would you need to add back to the planarian, prior to amputation, that would restore the regenerative ability of the animal? a. One unirradiated neoblast b. Hundreds of unirradiated neoblasts c. Any cell from an animal that can still regenerate d. Germ cells from an unirradiated donor e. Nothing will be able to restore the regeneration potential of the irradiated animal.

19. A researcher is studying limb regeneration in the newt Notophthalmus viridescens. She transplants skeletal muscle cells from a transgenic newt that

Molecular Biology of the Cell 7e-Alberts expresses green fluorescent protein (GFP) in all of its cells into the limb of a wildtype recipient newt. She then amputates the recipient limb right at the site of the graft, such that the stump contains GFP-expressing cells. She repeats this experiment several times. Where would you expect to detect GFP in the resulting regenerated limbs? a. Only in muscle cells b. Only in skin cells c. Only in nerve cells d. Only in muscle and skin cells

20. Stem cells can be obtained from self-renewing regions of an adult vertebrate brain, cultured in suspension or in monolayer, and later grafted back into a live animal. When implanted, stem cells isolated from mouse hippocampus do which of the following? a. Give rise to hippocampus neurons, even in a different location in the brain b. Give rise to neurons in another part of the brain, for example, in the olfactory bulb c. Are unable to proliferate unless implanted in the hippocampus, where they contribute to its neurons d. Give rise to any cell, for example, muscle cells when implanted into muscle tissue

21. An increasing number of lymphomas are being treated with “stem-cell transplantation” therapy that is composed of the following overall steps: Injection of cells into the blood, harvesting and isolating selected cells from the patient’s bone marrow, and heavy x-ray irradiation of the patient. Which of the following reflects the correct sequence of steps? a. Harvesting and isolating selected cells from the patient’s bone marrow, injection of cells into the blood, heavy x-ray irradiation of the patient b. Harvesting and isolating selected cells from the patient’s bone marrow, heavy x-ray irradiation of the patient, injection of cells into the blood

Molecular Biology of the Cell 7e-Alberts c. Heavy x-ray irradiation of the patient, harvesting and isolating selected cells from the patient’s bone marrow, injection of cells into the blood d. Heavy x-ray irradiation of the patient, injection of cells into the blood, harvesting and isolating selected cells from the patient’s bone marrow

22. Although quite inefficient, cloning by nuclear transplantation can be successful, as exemplified by cloning of the famous Dolly. Which of the following is not an implication of the early work with nuclear transplantation? a. Nuclei can be reprogrammed by cytoplasmic factors in a foreign cytoplasm. b. Epigenetic changes in somatic cells are not functionally irreversible. c. Even a differentiated nucleus contains a complete genome, capable of supporting the development of an entire organism. d. Somatic cells are capable of dedifferentiation.

23. The differentiation of multipotent hematopoietic stem cells into an erythrocyte requires the progression through which type of precursor cells? a. Lymphoid progenitors b. Monocytes c. Mast cell progenitors d. Megakaryocyte progenitors e. Myeloid progenitors

24. When hematopoietic stem cells from an old mouse are transplanted into a younger mouse with irradiated bone marrow, the transplanted older stem cells are unable to fully repopulate the host’s system. What is the best explanation for this observation? a. The younger stem cell niche is different from the older stem cell niche, and thus cannot support the older stem cells. b. The stem cells from the older donor are unable to survive in the younger

Molecular Biology of the Cell 7e-Alberts host’s stem cell niche. c. The stem cells from the older host are out-competed by the stem cells from the host for the bone marrow niche, and thus cannot divide. d. The stem cells from the older host have undergone cellular senescence, and thus cannot effectively divide. e. The stem cells from the older host are destroyed by the younger host’s immune system.

25. Which of the following statements is true regarding skeletal muscle development and regeneration in humans? a. Muscle is among those tissues that cannot be renewed once damaged. b. Myoblasts are multipotent stem cells. c. Muscle regeneration occurs as a result of stimulation of quiescent stem cells (satellite cells) in the periphery of muscle fibers. d. A muscle that has been repaired once cannot be repaired again.

26. Cells of the inner cell mass in an early mammalian embryo can be isolated and grown in culture. Regarding these cells, which of the following is NOT true? a. They are totipotent. b. They can be introduced into another developing embryo where their progeny become incorporated into the resulting adult animal, even into its germ line. c. They are stem cells. d. They can be manipulated in culture to give rise to almost any type of differentiated cell.

27. Which of the following transcription regulators constitute the master set of transcription regulatory proteins whose expression is most required for generating induced pluripotent stem cells and for maintaining pluripotency?

Molecular Biology of the Cell 7e-Alberts a. Myc b. Oct4 c. MyoD d. Nanog

28. Oct4 is a gene that encodes a transcriptional regulator that is expressed exclusively in ES cells. What happens to a cell when the expression of Oct4 is silenced? a. The cell immediately undergoes apoptosis. b. The cell immediately differentiates into a germ cell. c. The cell loses its ES cell characteristics. d. The cell up-regulates other genes to retain its ES cell characteristics. e. The cell fuses with neighboring cells to generate the ES cell niche.

29. Even though OSKM factors can induce the de-differentiation of fibroblasts, the resultant cell is not considered a stably reprogrammed iPS cell until this characteristic is observed. a. Expression of the endogenous OSKM factors b. Integration of the transgenic OSKM factors into the host’s genome c. The formation of embryoid bodies d. The ability to survive in the presence of G418 e. The ability to repopulate various tissues of the donor organism

30. Only cells that express a puromycin resistance (PR) gene can grow in the presence of the drug puromycin. Ganciclovir is a nontoxic drug that is converted to a lethal toxin in a cell that expresses thymidine kinase (TK). Thy-1 is a fibroblast-specific marker protein. Which of the following selection strategies is the most reasonable one to use in experiments that attempt to generate induced

Molecular Biology of the Cell 7e-Alberts pluripotent stem cells (iPS cells) from fibroblasts? a. Engineering fibroblasts to express PR under the control of the Nanog promoter, and selecting iPS cells in puromycin-containing media b. Engineering fibroblasts to express PR under the control of the Thy-1 promoter, and selecting iPS cells in puromycin-containing media c. Engineering fibroblasts to express TK under the control of the Nanog promoter, and selecting iPS cells in ganciclovir-containing media d. Engineering fibroblasts to express TK under the control of the Thy-1 promoter, and selecting iPS cells in ganciclovir-containing media e. Engineering fibroblasts to express TK under the control of the Thy-1 promoter, and selecting iPS cells in puromycin-containing media

31. What is the correct order in which the following events typically take place during the reprogramming of fibroblasts to induced pluripotent stem cells? a. Loss of fibroblast-specific marker protein, OSKM overexpression, gain of embryonic marker protein, sustained endogenous expression of Oct4 b. Loss of fibroblast-specific marker protein, OSKM overexpression, sustained endogenous expression of Oct4, gain of embryonic marker protein c. OSKM overexpression, loss of fibroblast-specific marker protein, gain of embryonic marker protein, sustained endogenous expression of Oct4 d. OSKM overexpression, gain of embryonic marker protein, loss of fibroblastspecific marker protein, sustained endogenous expression of Oct4

32. Which group of factors is thought to impede iPS cell generation from a differentiated cell, inasmuch as the reprogramming efficiency is increased when the activity of the factors is decreased? a. Histone acetyl transferases b. Histone deacetylases c. Chromatin remodelers d. DNA demethylases

Molecular Biology of the Cell 7e-Alberts

33. The reprograming of fibroblasts into iPS cells requires massive alteration to chromatin structure, without directly altering the cell’s DNA. However, it has been shown that one epigenetic modification of a cell’s DNA can increase reprogramming efficiency. What is this modification? a. The addition of methyl groups to specific miRNAs b. The addition of acetyl groups to DNA c. The removal of methyl groups from DNA d. The addition of methyl groups to DNA e. The removal of methyl groups from OSKM mRNAs

34. Which of the following tasks is possible with current stem-cell and reprogramming technology? a. Guiding ES cells to generate any desired adult cell type b. Generating functional whole organs from iPS cells for transplantation c. Forcing transdifferentiation of any terminally differentiated cell into another cell d. In vitro screening for drug discovery using patient-specific iPS-derived cells as disease models

35. Regarding the pattern of cell proliferation in the epithelium that forms the lining of the small intestine, there is a net movement of cells from the bottom of the crypts up toward the tip of the villi. a. True b. False

36. Monocytes typically give rise to macrophages and polymorphonuclear leukocytes. a. True b. False

Molecular Biology of the Cell 7e-Alberts

37. Most blood cells are derived from the pluripotent stem cells by a few cell divisions. a. True b. False

38. Regarding the regenerative capacity of the planarian Schmidtea mediterranea, when the animal is starved, some of its cells are cannibalized by other cells. a. True b. False

39. The schematic drawing below shows the cross section of a crypt in the gut epithelium of a mammal. Indicate whether each of the following descriptions better applies to the areas labeled A, B, or C. (Answers will be used more than once.)

a. Area A b. Area B c. Area C

1. The stem cells reside in this area.

Molecular Biology of the Cell 7e-Alberts 2. This area contains the most rapidly dividing cells.

3. Cells in this area would show the least radioactivity following a pulse of tritiated thymidine.

4. This area contains cells of the innate immune system.

40. In each of the following examples of contact signaling between adjacent cells in the mammalian gut epithelia, indicate whether the signaling is mainly mediated by Delta or ephrin signals. a. Delta signals b. Ephrin signals

1. Separation of crypt cell types from villus cell types 2. Regulation of stem-cell differentiation by Paneth cells 3. Differentiation of transit-amplifying cells into either secretory cells or absorptive cells

41. Indicate whether each of the following descriptions better applies to an asymmetric or a symmetric stem-cell division strategy. a. Asymmetric b. Symmetric

1. It predicts the number of stem cells in a population to remain strictly constant after several rounds of cell division.

2. Cytoplasmic factors that control stemness are localized to one side of the cell before cytokinesis.

3. It is a more flexible strategy and can better accommodate the effect of environmental factors.

4. It is more consistent with observations in gut stem cells and other stem cells.

42. Describe a cell found in the intestinal lining of a mammal that expresses Lgr5 and then answer the following question: When this cell divides, what will you then learn about the cells that result from this division?

43. A multipotent stem cell divides once in order to produce a transient amplifying cell

Molecular Biology of the Cell 7e-Alberts that can in turn produce thousands of differentiated progenies. What is the benefit of reduced stem cell divisions?

44. Erythrocytes are heterogeneous. a. True b. False

45. Sort the following cell types from high to low abundance in the blood (number of cells per unit volume) in a healthy individual. Your answer would be a four-letter string composed of letters A to D only, for example, BADC. A. Erythrocytes B. Neutrophils C. Basophils D. T lymphocytes

46. When stem cells of the Drosophila germ line divide, the cell that remains attached to the central hub remains a stem cell while the other daughter differentiates.

What process dictates the pattern of division that specifies the stem cell?

47. Neural plasticity is the ability of the brain to re-form various regions in response to either damage or changes in brain function. Recent studies have shown that endogenous neuronal stem cells contribute to the plasticity of two regions of the mammalian brain. Where in the brain are these stem cells located, and what two regions rely on these stem cells for plasticity?

48. When an individual is afflicted by a particular genetic disorder, we can now utilize their fibroblasts for analysis and treatment development. In what two ways is this possible?

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 22 1. Answer: B 2. Answer: B 3. Answer: C 4. Answer: D 5. Answer: C 6. Answer: E 7. Answer: E 8. Answer: C 9. Answer: B 10. Answer: C 11. Answer: C 12. Answer: C 13. Answer: B 14. Answer: A 15. Answer: C 16. Answer: D 17. Answer: A 18. Answer: A 19. Answer: A 20. Answer: B

Molecular Biology of the Cell 7e-Alberts 21. Answer: B 22. Answer: D 23. Answer: E 24. Answer: D 25. Answer: C 26. Answer: A 27. Answer: B 28. Answer: C 29. Answer: A 30. Answer: A 31. Answer: C 32. Answer: B 33. Answer: C 34. Answer: D 35. Answer: A 36. Answer: B 37. Answer: B 38. Answer: A 39. Answer:

1. C 2. B 3. A 4. C

40. Answer:

1. B

Molecular Biology of the Cell 7e-Alberts 2. A 3. A 41. Answer:

1. A 2. A 3. B 4. B

42. Answer: The Lgr5expressing cell is a multipotent stem cell of the intestinal crypt.

The products of this cell will differentiate into all of the epithelial cells of the gut while some will remain as stem cells.

43. Answer: Like all cells, a stem cell can only divide a limited number of times until it reaches replicative senescence. Once this point is reached, the stem cell will no longer efficiently divide.

Furthermore, by maintaining division rates of stem cells low, the risk of replicative stress (mutations) is reduced. If a stem cell acquires a mutation, that damage will be passed on to all the progeny and could affect many

Molecular Biology of the Cell 7e-Alberts levels of the cell lineage.

44. Answer: B 45. Answer: ABCD 46. Answer: Division of the stem cell is governed by the orientation of the mitotic spindle.

Following division, one daughter will remain in contact with the stem cell niche and remain a stem cell, while the other daughter is forced away from the niche and begins to differentiate.

47. Answer: Populations of neural stem cells can be found along the ventricle regions of the cerebral hemispheres.

These neuronal stem cells have been shown to be able to contribute to the olfactory bulb and the hippocampus.

48. Answer: The individual’s fibroblasts can be reprogrammed into iPS cells using OSKM factors. These iPS cells can then be used in two different ways.

The iPS cells can

Molecular Biology of the Cell 7e-Alberts be redifferentiated in vitro and used to identify diseasespecific drugs that can then be administered to the patient.

Another method would be to genetically repair the iPS cells, redifferentiate them into healthy cells, and transplant them back into the patient.

Molecular Biology of the Cell 7e-Alberts

Chapter 23: Pathogens and Infection 1. In order to initiate infection, Salmonella enterica secretes an effector protein by using a type III secretion system. With regard to the host cell, on which of the following does the secretion of this protein family have an impact? a. The production of extracellular growth factors b. The rearrangement of actin cross-linking proteins c. The ruffling of the host cell membrane through actin polymerization d. The cell-matrix adhesion e. The shifting of extracellular growth factors

2. Which of the following is a primary pathogen? a. Staphylococcus aureus b. Streptococcus pneumoniae c. Yersinia pestis d. Papillomavirus e. MRSA (methicillin-resistant S. aureus)

3. How has pathogenicity in pandemic strains of Vibrio cholerae been acquired? a. By transformation b. By transduction c. By conjugation d. By vertical gene transfer

4. You infect human epithelial cells in culture with either nonpathogenic bacteria or Vibrio cholerae, each without any further treatment, in the presence of MDC (an inhibitor of clathrin-dependent endocytosis) or in the presence of filipin (an

Molecular Biology of the Cell 7e-Alberts inhibitor of a clathrin-independent endocytic pathway). You then measure the intracellular concentration of cyclic AMP (in picomoles per milligram of total cell protein) and summarize the results in the following table. Which row (1 or 2) do you think corresponds to infection with V. cholerae? From these results, does cholera toxin enter the cell in clathrin-coated vesicles?

a. Row 1; yes b. Row 1; no c. Row 2; yes d. Row 2; no

5. Which of the following molecular characteristics of bacteria allow the host's innate immune system to recognize them as foreign? a. Phospholipids and peptides b. Lipopolysaccharide

Molecular Biology of the Cell 7e-Alberts c. Pathogen-associated peptides d. Pathogen-associated molecular patterns e. Lipopolysaccharide-associated peptides

6. Vibrio cholerae and Bacillus anthracis do which of the following? a. They secrete a lethal factor and an edema factor. b. They secrete toxins that enter the cell in endosomes and eventually enter the cytosol after reaching the endoplasmic reticulum. c. They raise cyclic AMP levels in the cytosol of their target cells. d. They secrete toxins that have enzymatic adenylyl cyclase activity.

7. The type III secretion system of pathogenic bacteria can be described in which way? a. It is a general secretion system found only in Gram-positive bacteria. b. It is used to inject effector proteins directly into the cytoplasm of host cells. c. It is similar to the eukaryotic secretion system. d. It is similar to the bacterial conjugation apparatus.

8. Which of the following descriptions better applies to protozoan rather than viral pathogens? a. They cannot be all grouped neatly into a single phylogenetic tree. b. They have the most complex life cycles. c. They are essentially fragments of nucleic acid wrapped in a protective shell of proteins and (in some cases) an outer membrane. d. They are responsible for diseases such as polio and smallpox.

Molecular Biology of the Cell 7e-Alberts 9. Viruses are obligate parasites, which rely on a host for survival. Which of the following represents the correct order of events in the process of SARS CoV2 virus replication? a. Release, replication, disassembly, transcription, and assembly b. Assembly, replication, entry, transcription, and release c. Attachment, assembly, transcription, release, and entry d. Entry, disassembly, replication, transcription, assembly, and release e. Entry, transcription, assembly, disassembly, and release

10. Which of the following illnesses is NOT associated with infection with the pathogen indicated? a. Cervical cancer associated with human papillomavirus infection b. Cancer in chickens associated with Rous sarcoma virus infection c. Colon cancer associated with Vibrio cholerae infection d. Stomach cancer associated with Helicobacter pylori infection e. Atherosclerosis associated with Chlamydia pneumoniae

11. How does Helicobacter pylori persist in the hostile environment of the stomach? a. By expressing adhesin proteins that bind to molecules on the surface of gastric epithelial cells b. By producing the enzyme urease, which locally neutralizes the gastric acid c. By chemotaxis, using flagella, toward gastric epithelial cells d. By producing toxins that target both epithelial cells and immune cells e. By attaching to gastric epithelial cells

12. Which of the following best describes the function of adhesion proteins during infection with pathogenic E. coli?

Molecular Biology of the Cell 7e-Alberts a. The recognition of epithelial cell surface glycolipids b. The recognition of antigen c. Binding of proteins and lipid grafts d. Attachment of E. coli to adhesion molecules e. Attachment and binding of E. coli to D-mannose modified proteins

13. You have recently identified a new bacterial toxin. The toxin is composed of protein and easily degraded. You also discover that this novel toxin is produced by enteropathogenic E. coli (EPEC) to target the intestinal epithelial layer. By what method does the toxin enter the host’s intestinal epithelium? a. The use of effector proteins b. The use of a type III secretion system c. The colonization of respiratory epithelium d. The use of a type IV secretion system e. The creation of pedestals

14. Which of the following conditions is NOT expected to provide protection against HIV-1 infection of helper T cells? a. A loss-of-function mutation in the CCR5 gene b. Treatment with enfuvirtide, a membrane-fusion inhibitor c. Treatment with concanamycin A, an inhibitor of endosomal proton pumps d. Treatment with AMD-070, a CXCR4 antagonist e. Treatment with PRO-542, a soluble decoy protein-containing CD4 sequences

15. Cytochalasin D (CyD) is a drug that binds to the plus end of actin filaments and prevents actin polymerization. Having identified mutations in actin that confer

Molecular Biology of the Cell 7e-Alberts resistance to CyD, a researcher sets out to study the role of the actin cytoskeleton in the invasion of mammalian host cells by two intracellular parasites: the protozoan Toxoplasma gondii and the bacterium Salmonella enterica. She infects wild-type or CyD-resistant host cells with either wild-type T. gondii, CyD-resistant T. gondii, or wild-type S. enterica, each in the presence (+) or absence (−) of CyD, and measures parasite internalization and intracellular proliferation as a result. The findings are summarized in the table below, in which High or Low levels of proliferation are indicated. According to these results, identify the row that represents the wild-type host. Also, indicate which columns (A to C) corresponds to infection with both the wild-type and CyD-resistant T. gondii?

a. ABC b. DCB c. DBC d. ABD e. ADC

16. Toxoplasma gondii is an intracellular eukaryotic parasite that can cause serious human infections. Which of the following is NOT true regarding this parasite?

Molecular Biology of the Cell 7e-Alberts a. It lyses its host cell to leave. b. The energy for its invasion into the host cell comes from actin polymerization in the host cell. c. Once inside the cell, it is surrounded by a host-cell membrane that is deprived of many host transmembrane proteins. d. Once inside the cell, it absorbs nutrients and small metabolites from the host cytosol. e. Its invasion mechanism is similar to that of malaria parasites.

17. Trypanosoma cruzi uses two alternative strategies to invade its host cells. To examine the contribution of each of these strategies to the pathogenicity, you culture mammalian cells that can be infected with this pathogen. The cells are engineered to express green fluorescent protein (GFP) fused to a plasma membrane protein, as well as red fluorescent protein (RFP) fused to a lysosomespecific transmembrane protein. You then either add only tiny latex beads (as a negative control) or add T. cruzi protozoa (to infect the cells). After 10 minutes, you fix the cells and examine them under a fluorescence microscope. You observe that a significant fraction (~30%) of phagosomes/vacuoles in sample 1 are RFP-positive (that is, they show RFP fluorescence at their membrane), whereas all phagosomes in sample 2 are RFP-negative. Which sample (1 or 2) is the one infected with T. cruzi? In this infected sample, would you expect the RFPnegative trypanosome-containing vacuoles to be GFP-positive or GFP-negative? a. Sample 1; GFP-positive b. Sample 1; GFP-negative c. Sample 2; GFP-positive d. Sample 2; GFP-negative

18. After gaining entry into the host cell by the zipper mechanism, the bacterium Listeria monocytogenes escapes phagosomes by secreting listeriolysin O. This protein is active at which level of pH, with what result? a. Neutral pH and cannot be degraded by the proteasome b. Neutral pH and is rapidly degraded by the proteasome

Molecular Biology of the Cell 7e-Alberts c. Acidic pH and cannot be degraded by the proteasome d. Acidic pH and is rapidly degraded by the proteasome

19. Which two processes are responsible for the virulence of the human immunodeficiency virus (HIV) virus? a. Isolation and drug resistance b. Drug resistance and replication c. Replication and selection d. Mutation and selection e. Competition and isolation

20. The human gut microbiota is best described as what? a. A population of organisms that live in the oral cavity b. The sum total of all microorganisms within the body c. The community of microbes that live in the gastrointestinal tract d. A population of organisms that live in the gastrointestinal tract e. The sum total of undigested carbohydrates that remain in the intestines

21. How are polio virus particles able to infect host cells? a. Non-mediated pinocytosis b. Receptor-mediated endocytosis c. Receptor-mediated phagocytosis d. Membrane fusion e. Production of endosomes in the ER

Molecular Biology of the Cell 7e-Alberts 22. Consider the movement in the host cell of two intracellular pathogens, the herpes simplex virus and the bacterium Listeria monocytogenes. Which description better applies to the virus rather than the bacterium? a. It moves by a “pushing” mechanism (as opposed to a “pulling” mechanism). b. It can collide with the plasma membrane and create long, thin protrusions from the cell that can then be engulfed by a neighboring cell. c. Its movement is powered by ATP hydrolysis by kinesin and dynein motors. d. Its movement relies on dynamic cytoskeletal filaments.

23. There are a variety of strategies that pathogens use to avoid antimicrobial autophagy, but which one does Francisella tularensis, otherwise known as rabbit fever,use? a. Secrete degradative enzymes b. Employ a protective shield to prevent detection c. Phagocytize into the host cell d. Rearrange bacterial genetic material e. Use microtubules to move around the host cell

24. In which of the following groups of viruses is an RNA-dependent RNA polymerase packaged as a structural protein in each newly made viral particle? a. DNA viruses b. Viruses with a positive [+] strand RNA genome c. Viruses with a negative [−] strand RNA genome d. Retroviruses e. Viruses with a part single- and part double-stranded DNA genome

25. Of the factors listed below, which is a result of the microbial genomic diversity

Molecular Biology of the Cell 7e-Alberts found in humans? a. An accumulation of simple organic molecules in the gut b. An abundance of metabolite production, expanding the physiological potential of humans c. Viral entry and replication Increased mucus production in the gastrointestinal tract d. Increased mucus production in the gastrointestinal tract e. Survival and replication of pathogenic bacteria

26. Most important pathogenic fungi show dimorphism, growing as either yeast or mold. a. True b. False

27. Mycobacterium tuberculosis can cause tuberculosis, a life-threatening lung infection, but can also infect an individual asymptomatically for years. It is not considered to be part of the normal flora and can infect healthy individuals upon exposure. It can only replicate in the host and thus mainly spreads by direct human contact. Mycobacterium tuberculosis is therefore what? a. A facultative primary pathogen b. A facultative opportunistic pathogen c. An obligate primary pathogen d. An obligate opportunistic pathogen

28. A point mutation in the gene encoding the β subunit of hemoglobin can lead to sickle-cell anemia. Due to the formation of mutant hemoglobin aggregates, red blood cells in affected individuals can adopt a sickle-like shape that interferes with their normal function. Interestingly, individuals carrying this mutation are less vulnerable to malaria infection by Plasmodium falciparum, which replicates inside red blood cells.

TRUE or FALSE: You would expect the frequency of the mutated hemoglobin

Molecular Biology of the Cell 7e-Alberts allele to be higher in West Africa rather than Europe. a. True b. False

29. Individuals with a defective CCR5 gene are more susceptible to HIV infection. a. True b. False

30. Whereas bacterial and eukaryotic pathogens evade the immune response mainly by genetic recombination, antigenic variation in most viruses occurs by errorprone replication mechanisms. a. True b. False

31. The number of human cells in our body is greater than the number of bacterial, fungal, and protozoan cells of our normal flora. a. True b. False

32. In Gram staining, crystal violet (a violet dye) is used to specifically stain Grampositive bacteria. After performing Gram staining on a bacterial sample taken from an infected animal tissue, you observe the results with a microscope. You find two major types of bacteria in the sample, as shown in the schematic drawing below. According to these results, indicate whether each of the following statements is correct or incorrect. a. Correct b. Incorrect

1. The Gram-positive bacteria in this sample are bacilli.

2. Crystal violet stains lipopolysaccharide (LPS).

3. The bacteria-stained violet in this sample have a thicker layer of peptidoglycan compared to the other bacteria.

Molecular Biology of the Cell 7e-Alberts 33. In the following simplified diagram, three mechanisms for the horizontal transfer of virulence genes to an avirulent bacterium are depicted. Indicate which mechanism (A to C) corresponds to conjugation, transduction, and transformation.

a. Mechanism A b. Mechanism B c. Mechanism C

1. Conjugation 2. Transduction 3. Transformation

34. Which of the following descriptions better applies to enveloped viruses, and which to nonenveloped viruses? a. Enveloped viruses b. Nonenveloped viruses

1. They include retroviruses.

2. They include adenoviruses.

3. To leave the cell, they normally lyse it.

4. They are more sensitive to treatment with detergents, heat, or drying.

Molecular Biology of the Cell 7e-Alberts 35. In each of the following organs, indicate whether the epithelium is kept almost sterile in a healthy human. a. Is kept almost sterile b. Is not kept almost sterile

1. Bladder 2. Cervix 3. Lower lung 4. Colon

36. To enter the host cell, intracellular bacterial pathogens can induce phagocytosis in cells that are normally nonphagocytic. This is done by two major mechanisms depicted in the following schematic diagrams (A and B). Which of the following descriptions better applies to mechanism A, and which to mechanism B?

a. Mechanism A b. Mechanism B

1. It is called the zipper mechanism.

2. It depends on invasin proteins on the surface of the bacterium that bind to their receptors on the surface of the host cell.

3. It depends on the injection of effector proteins into the host cell by a bacterial secretion system.

4. It resembles the process of macropinocytosis.

37. An intracellular pathogen uses one of three major strategies to survive and replicate once inside the host cell, as shown in the schematic drawing below.

Which strategy (1, 2, or 3) is employed by which pathogens for this purpose?

Molecular Biology of the Cell 7e-Alberts

a. Strategy 1 b. Strategy 2 c. Strategy 3

1. Listeria monocytogenes 2. Legionella pneumophila 3. Shigella flexneri

38. Modifications of membrane traffic in host cells by three bacterial pathogens are shown in the following schematic drawing. Indicate whether each of the following bacteria is represented by A, B, or C in the drawing.

Molecular Biology of the Cell 7e-Alberts

a. A b. B c. C

1. It is a common cause of food poisoning in humans.

2. It causes the life-threatening lung infection tuberculosis.

3. It is the cause of a type of pneumonia known as Legionnaire’s disease.

39. Indicate whether each of the following examples better applies to commensalism, mutualism, or parasitism in the interaction between a host and a microbe. a. Commensalism b. Mutualism c. Parasitism

1. Due to poor public sanitation, a child catches cholera.

2. A mouse is infected with a virus but shows no noticeable health defect whatsoever.

Molecular Biology of the Cell 7e-Alberts 3. Biotin and other vitamins produced by intestinal microbiota are regularly absorbed in the human intestine.

4. Bacteria on the skin of cattle produce antifungal compounds.

40. Many pathogens have evolved ways to infect their host and cause disease, though in the end they follow a similar path to spreading infection. Note the steps that a human pathogen takes in order to successfully cause disease pathogens, and then compare and contrast the various methods that bacteria use to cause disease in humans.

41. Sort the following events to reflect the order in which they typically occur in viral replication after a virus enters a cell. Your answer would be a four-letter string composed of letters A to D only, for example, BACD. A. Replication of the viral genome and transcription of viral genes B. Virus particle assembly C. Progeny virion release D. Virus particle disassembly

42. Trypanosoma brucei and Neisseria meningitidis are two pathogens that cause disease in humans. What are the advantages that give these pathogens the ability to adapt at an increased rate during infection?

43. The dimorphic nature of Histoplasma capsulatum, the causative agent of histoplasmosis, has made drug development difficult in the past. How does the role of morphology act as a virulence factor in the targeting and infection of host lung tissue?

44. In the bacterium Yersinia pestis, which causes bubonic plague, loss of an operon that is normally responsible for iron storage diminishes the ability of the cells to form biofilms and colonize inside the iron-rich gut of infected fleas. Would you expect the efficiency of transmission of plague to increase, decrease, or remain unchanged as a result of this mutation? Explain your answer in 2 to 3 sentences.

45. Sort the following events to reflect the order in which they take place during the

Molecular Biology of the Cell 7e-Alberts infection of intestinal epithelium by enteropathogenic E. coli. Your answer would be a five-letter string composed of letters A to E only, for example, BCDAE. A. Actin polymerization and pedestal formation B. Binding of intimin to Tir C. Folding of Tir in the plasma membrane D. Insertion of Tir into the epithelial cell E. Assembly of type III secretion system

46. Methicillin-resistant Staphylococcus aureus (MRSA) is one of the leading causes of nosocomial or hospital infections. What are the consequences of drug resistance in the prevention of MRSA infection?

47. Disturbances in the environment or environmental change is often caused by human influences and natural ecological processes. Consequences of these disturbances can be observed across biological scales, including individual organisms. How might the environment influence the human microbiota?

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 23 1. Answer:

C

2. Answer:

C

3. Answer:

B

4. Answer:

D

5. Answer:

D

6. Answer:

C

7. Answer:

B

8. Answer:

B

9. Answer:

D

10. Answer:

C

11. Answer:

B

12. Answer:

A

13. Answer:

B

14. Answer:

C

15. Answer:

C

16. Answer:

B

17. Answer:

A

18. Answer:

D

19. Answer:

D

20. Answer:

C

Molecular Biology of the Cell 7e-Alberts 21. Answer:

B

22. Answer:

C

23. Answer:

B

24. Answer:

C

25. Answer:

B

26. Answer:

A

27. Answer:

C

28. Answer:

A

29. Answer:

B

30. Answer:

A

31. Answer:

B

32. Answer:

1. B 2. B 3. A

33. Answer:

1. C 2. B 3. A

34. Answer:

1. A 2. B 3. B 4. A

35. Answer:

1. A 2. B 3. A

Molecular Biology of the Cell 7e-Alberts 4. B 36. Answer:

1. B 2. B 3. A 4. A

37. Answer:

1. A 2. B 3. A

38. Answer:

1. B 2. A 3. C

39. Answer:

1. C 2. A 3. B 4. B

40. Answer:

In order to successfully cause disease pathogens, human pathogens must do the following: 1.) Enter the host, 2.) find a nutritionally compatible niche in the host’s body, 3.) avoid or overcome the host’s innate and adaptive immune response, 4.) replicate, and 5.) exit one host and spread to another.

Bacteria cells use numerous methods to infect target host cells. Some of the methods include the zipper mechanism and the trigger mechanism. Bacteria induce phagocytosis by host cells by expressing and invasion protein or injecting effector proteins into the host cell. An example is Salmonella.

Salmonella contains SPI1 a type III secretion system that induces ruffling in the host cell.

41. Answer:

DABC

42. Answer:

These pathogens have evolved the ability to modify surface antigens during infection as a way of evading the host’s immune system. T. brucei expresses a variant-specific glycoprotein (VSG) on its surface, which elicits an antibody response from the host. During infection, T. brucei uses

Molecular Biology of the Cell 7e-Alberts inactive pseudogene copies of VSG to perform gene rearrangements that generate new variants of the VSG protein, thereby presenting novel variants to the immune system. In a similar way, N. meningitidis uses genetic recombination to vary the pilin protein used to attach to host cells. This, combined with its unusual ability to uptake DNA from the environment and incorporate it into its genome, results in an incredible variety of surface proteins, making it difficult to develop effective vaccinations against Neisseria infections.

43. Answer:

The ability of Histoplasma to switch morphology during its life cycle makes it difficult to treat. This pathogen grows as a mold at low temperatures in the soil and as a yeast at higher temperatures in the lung. Drugs that are effective at killing one form of the pathogen are often ineffective against the other, allowing part of the population to survive an antimicrobial treatment.

44. Answer:

The expectation is that the efficiency of transmission of plague would decrease, because the wild-type bacterium blocks the flea gut, preventing the flea from digesting its blood meal. This leads to repetitive attempts at biting and dissemination of the infection; thus, the mutant bacteria cannot block the flea gut and are therefore transmitted less efficiently.

45. Answer:

EDCBA

46. Answer:

Bactericidal drugs often work by disrupting the synthesis of molecules and macromolecules that are specific to bacteria and not found in their eukaryotic hosts. In this way, they inhibit bacterial biochemical processes without interfering with the physiology of the host. Drug resistance arises when genetic mutations that block the drug’s interaction with bacterial processes arise and are selected for. The block may be at the level of enzyme activity or the bacterium may acquire novel ways of pumping the drug out of the cell or of inactivating it. Drug-resistant mutations frequently become widespread through horizontal gene transfer.

47. Answer:

The human microbiota can be influenced by many factors in the environment. When environmental changes like natural disasters or climate change occur, humans may have less access to healthy, nutritious foods that will aid in the support of the gut microbiota through the provision of nutrients.

Stress also has a negative effect on the microbiota, as stressors caused by environmental changes have the ability to decrease the healthy microbes. Consequently, what takes

Molecular Biology of the Cell 7e-Alberts place is an increase to the susceptibility of that individual to disease. Pollution is another result of human interference that negatively affects the human microbiota if the human body is exposed to it.

Molecular Biology of the Cell 7e-Alberts

Chapter 24: The Innate and Adaptive Immune Systems 1. Which of the following groups of proteins are pattern recognition receptors? a. Toll-like receptors b. Interferon receptors c. Interleukin receptors d. R-type lectin receptors e. Retinoic acid receptors

2. Most cell-surface and intracellular PRRs stimulate which of the following transcriptional regulators? a. NFϰB b. CREB c. SMAD d. Gli e. ER

3. Which of the following are recognized by pattern recognition receptors as pathogen-associated molecular patterns? a. Single-stranded bacterial RNAs b. Methionine-containing proteins c. Methylated CpG motifs d. Viral lipids e. Bacterial flagella components

Molecular Biology of the Cell 7e-Alberts 4. The activation of PRRs induces the expression and releases a variety of extracellular signal molecules, such as pro-inflammatory cytokines. What is an example of an extracellular signal molecule? a. IL4 b. IL10 c. IL14 d. TGF-β e. TNF-α

5. Which cell type is NOT phagocytic? a. Macrophage b. Monocyte c. Neutrophil d. Lymphocyte e. Dendritic cell

6. A cell has been infected with an RNA virus. As a consequence, what will happen? a. It will reduce the expression of its surface class II MHC molecules. b. It will secrete type II interferons such as interferon-γ. c. It will shut down most protein translation and destroy most of its RNA. d. It will be induced to die by necrosis. e. It will detect viral single-strand RNA.

7. Consider two receptors, one inhibitory and one activating, on the surface of natural killer (NK) cells: KIR receptors that interact with certain class I MHC proteins, and special Fc receptors that recognize the tail region of IgG antibodies.

When NK cells are incubated with an immortalized B cell line derived by infection

Molecular Biology of the Cell 7e-Alberts of human B cells with Epstein–Barr Virus, the B cells are killed by NK cells.

Normal B cells, in contrast, are not efficiently killed by NK cells. In considering the following scenarios, choose which of the described changes are expected to suppress rather than enhance the cytotoxicity (killing activity) of the NK cells in this system. a. The cell line is transformed with human class I MHC genes, and the genes are expressed. b. An anti-CD23 IgG antibody is added to the cells. (CD23 is a known marker on the surface of the immortalized B cells.) c. The cells are treated with stibogluconate, an inhibitor of KIR receptor signaling. d. Type I interferons are added to the cells.

8. Which of the following is a key cell type involved in the antiviral response of type I interferons? a. Basophil b. Natural killer cell c. B cell d. Macrophage e. Neutrophil

9. How are natural killer (NK) cells different from cytotoxic T (T C) cells? a. NK cells induce apoptosis in their target cells, whereas T C cells are professional phagocytes. b. NK cells kill virus-infected cells, whereas TC cells kill cancerous cells. c. NK cells kill cells with a high level of class I MHC protein expression, whereas killing by TC cells requires low expression levels. d. NK cells respond quickly to a virus infection, whereas the activation of T C cells to become cytotoxic is a slow process.

Molecular Biology of the Cell 7e-Alberts e. NK cells are normally abundant in the tissue even before infection, whereas TC cells are mostly in the bloodstream.

10. Which of the following is a functional cellular link between innate and adaptative immunity? a. T cell b. Monocyte c. Macrophage d. Dendritic cell e. Natural killer cell

11. Where do T cells develop? a. Bone marrow b. Thymus c. Lymph nodes d. Spleen e. Tonsil

12. In contrast to a primary immune response, what is a characteristic of a secondary immune response? a. It has a longer lag period. b. It is stronger. c. It involves induction of effector cells to become memory cells. d. It takes longer to develop.

13. If mice are injected with antibodies against selectin homing receptors or against

Molecular Biology of the Cell 7e-Alberts the integrin ligand I-CAM1 (which is expressed on the surface of endothelial cells), they show immune defects, such as higher susceptibility to bacterial infections. Which of the following properties of lymphocytes at endothelial cells would you expect to observe after injection with both antibodies? a. Normal lymphocyte rolling and normal lymphocyte adhesion b. Impaired lymphocyte rolling but normal lymphocyte adhesion c. Normal lymphocyte rolling but impaired lymphocyte adhesion d. Impaired lymphocyte rolling and impaired lymphocyte adhesion

14. If heart surgery is necessary in a newborn, the thymus is sometimes removed to gain better access to the heart. If a lymph node of such a patient is later studied in a microscope, one particular area of the node is found to have an unusually low number of cells. Looking at the drawing below, what is the name of this area?

a. Lymphoid follicle b. Marginal sinus c. Paracortex d. Medullary sinus

Molecular Biology of the Cell 7e-Alberts

15. Which description better applies to the clonal deletion rather than receptor editing mechanism of immunological self-tolerance? a. It involves apoptosis of self-reactive cells. b. It is mostly limited to B cells. c. It involves DNA recombination. d. It is limited to central tolerance processes.

16. Why do we produce antibodies against other blood-group antigens even without having been exposed to those foreign blood cells before? It has been suggested that these antibodies are generated in response to minor infections (in early life) with microbes of the normal flora of our bodies, and that, in addition to binding to the microbial antigens, these antibodies can cross-react with the similar A- and Btype carbohydrate antigens of the ABO blood-group system. If this is indeed the case, blood plasma from a “germ-free” individual would react with what blood types? a. Blood of any ABO type b. Blood of no other ABO type c. Blood from O-type individuals only d. Blood from non-O-type individuals only e. Blood from AB-type individuals only

17. This is a major class of antibody that crosses the placenta. Which of the following fits this description? a. IgM b. IgE c. IgG d. IgD

Molecular Biology of the Cell 7e-Alberts e. IgA

18. This is a major class of antibody that is secreted into the blood in the early stages of a primary response to an antigen. Which of the following fits this description? a. IgD b. IgG3 c. IgG2 d. IgG1 e. IgM

19. You have prepared glass beads coated with rabbit anti-Ig antibodies that were isolated from rabbits immunized with mouse immunoglobulin (Ig). If you incubate a sample of mononucleated cells from mouse spleen with these beads, which of the following cell types would stick to the beads? a. Only macrophages b. Only B cells c. Only cytotoxic T cells d. B cells and cytotoxic T cells e. B cells and macrophages

20. Antigens and antibodies can form large, insoluble, cross-linked aggregates that precipitate from solution. The maximal amount of precipitant is observed at a certain antigen/antibody ratio; adding either excess antigen or excess antibody can prevent (or even reverse) the precipitation. Consider an antigen with multiple identical antigenic determinants that can be recognized by either bivalent or decavalent (i.e., pentameric) antibodies with binding sites that recognize the determinants. The amount of precipitation in the presence of a constant antigen concentration is plotted as a function of antibody concentration in the following qualitative graph. Which curve (1 or 2) in the graph do you think corresponds to the bivalent antibody? Would you expect to see such a bell-shaped curve if the

Molecular Biology of the Cell 7e-Alberts antibodies were monovalent?

a. Curve 1; yes b. Curve 2; yes c. Curve 1; no d. Curve 2; no e. Curve 2; only if the antigen was also monovalent

21. After having cereal and nut mix for breakfast, your friend realizes that the mix contains a nut to which she is allergic. Soon afterward, symptoms of an allergic reaction start to appear, and she is going to take an antihistamine pill to alleviate them. What happens during the allergic reaction in her body? a. Mast cells synthesize and secrete IgE antibodies into the blood. b. The allergen molecules bind to IgE antibodies that are bound to Fc receptors on the surface of basophils and mast cells. c. Histamine and other amines are recognized by IgE antibodies. d. Histamine binds to mast cells and induces IgE production.

22. How many hypervariable loops are there in a bivalent IgG antibody?

Molecular Biology of the Cell 7e-Alberts a. 6 b. 9 c. 12 d. 15 e. 18

23. Where are the hypervariable regions located in a typical antibody molecule? a. Only in the N-terminal region of the light chain b. Only in the C-terminal region of the light chain c. Only in the N-terminal region of the heavy chain d. Only in the C-terminal region of the heavy chain e. In the N-terminal region of both the light and the heavy chain

24. Consider two solution chambers of equal volume divided by a dialysis membrane that is impermeable to antibody molecules but permeable to small antigens.

Starting with an antigen concentration of 4.8 nM in both chambers, you add its specific IgG antibody to one of the chambers at 1 nM concentration. You then measure the concentration of the antigen in the other chamber after equilibrium is reached, which turns out to be about 4 nM. What is the association constant (Ka) for the binding of the antigen to each antigen-binding site of IgG? Assume independent antigen-binding sites in each antibody molecule. a. 1.0 × 109 liters/mole b. 0.8 × 109 liters/mole c. 4.8 × 109 liters/mole d. 4.0 × 109 liters/mole

25. Class switch recombination occurs when a B cell switches from making what to

Molecular Biology of the Cell 7e-Alberts what? a. IgM to one of the secondary classes of Ig b. IgD to one of the major classes of Ig c. IgD to IgM d. IgM and IgD to one of the secondary classes of Ig e. IgM to IgD

26. Normally, naïve cytotoxic and helper T cells interact for the first time with foreign antigens presented by dendritic cells located where? a. The thymus b. The bloodstream c. Peripheral lymphoid organs d. The bone marrow e. Inflamed tissues

27. Which of the following better describes cross-presentation of protein antigens by professional antigen-presenting cells to naïve TC cells? a. Presentation of intracellular antigens by class I MHC proteins b. Presentation of intracellular antigens by class II MHC proteins c. Presentation of extracellular antigens by class I MHC proteins d. Presentation of extracellular antigens by class II MHC proteins

28. In antigen presentation to helper T cells (T H cells), both the TCR and co-receptor proteins perform how? a. They are expressed on the dendritic cell. b. They bind to the class II MHC protein.

Molecular Biology of the Cell 7e-Alberts c. They bind to the presented peptide antigen. d. They bind to the class I MHC protein.

29. Which characteristic is unique to CD4? a. Signals through the Lck kinase b. Finds itself expressed on cytotoxic T cells c. Serves as the receptor for HIV infection d. Is a single-pass transmembrane protein with extracellular Ig-like domains e. Binds to invariant parts of the MHC protein

30. T cell receptors that do not interact at all with self-peptide–MHC complexes during T cell development in the thymus normally undergo what process? a. Positive selection b. Negative selection c. Receptor editing d. Clonal suppression e. “Death by neglect”

31. The transcription regulator AIRE (autoimmune regulator) plays a key role in developing immunological self-tolerance by promoting the production of various cellular proteins that are mainly expressed outside of the thymus. Which of the following cells have the highest level of AIRE? a. Double-negative thymocytes b. Double-positive thymocytes c. Lymphoid follicle cells in lymph nodes d. Thymus epithelial cells

Molecular Biology of the Cell 7e-Alberts e. Peyer’s patch cells

32. In the following schematic diagram, which curve (A to D) do you think better shows the qualitative relationship between the affinity of a developing T cell’s TCR for self-peptide–MHC complexes in the thymus and the survival chance of the cell?

a. Curve A b. Curve B c. Curve C d. Curve D

33. The early complement components form the membrane attack complex, whereas the late complement components cleave and activate C3. a. True b. False

34. Upon encountering its specific antigen in a peripheral lymphoid organ, a lymphocyte remains there to proliferate and differentiate into effector and memory cells. a. True

Molecular Biology of the Cell 7e-Alberts b. False

35. All the cell-surface immunoglobulin molecules that are produced by a single, mature, naïve B cell have the same antigen-binding site. a. True b. False

36. Indicate whether each of the following descriptions is shared or not shared between macrophages and neutrophils in destroying invading pathogens in vertebrates. a. Shared b. Not shared

1. They use a respiratory burst and create toxic reactive oxygen species to kill engulfed pathogens.

2. They are normally found in most tissues even before pathogen invasion.

3. They are long-lived and normally survive long after activation.

4. They secrete pro-inflammatory cytokines upon encountering invading pathogens.

37. Indicate whether each of the following descriptions better applies to an adaptive or innate immune response. a. Adaptive b. Innate

1. It is found in invertebrate and vertebrate animals as well as in plants.

2. It has a long-lasting memory.

3. It is highly specific to the particular invading pathogen.

4. It employs natural killer (NK) cells to induce apoptosis in infected host cells.

38. Indicate whether each of the following normally occurs before or after antigen stimulation of B cells. a. Before

Molecular Biology of the Cell 7e-Alberts b. After

1. V(D)J recombination 2. Class switch recombination 3. Somatic hypermutation 4. Junctional diversification

39. The schematic drawing below summarizes the differentiation of naïve helper T cells (TH cells) into various effector T cells, which occurs in a peripheral lymphoid organ. Indicate which effector cell (A to E) in the drawing is better described by each of the following descriptions.

a. Effector cell A b. Effector cell B c. Effector cell C d. Effector cell D

Molecular Biology of the Cell 7e-Alberts e. Effector cell E

1. It can also develop from thymocytes in the thymus, as well as from naïve TH cells.

2. It suppresses the development, activation, or function of most other immune cells.

3. It produces IFNγ and activates macrophages.

4. They reside in lymphoid follicles close to developing B cells.

40. Indicate whether each of the following cell-surface proteins is expressed mainly

by B cells, dendritic cells, or T cells. a. B cells b. Dendritic cells c. T cells

1. Co-stimulatory protein B7, which is recognized by CD28 2. Invariant CD3 complex 3. Inhibitory CTLA4 protein

41. Both B cell receptors and T cell receptors have domains with an immunoglobulin fold, which consists of a β-sandwich and an intrachain disulfide bond. a. True b. False

42. Which statement regarding herd immunity is NOT true? a. To be achieved, the vaccine must be viewed favorably by the public. b. Individuals are less susceptible to the disease. c. A majority of the population has become immune to the pathogen. d. The spread of infection has significantly slowed. e. We must prioritize herd immunity over vaccination.

Molecular Biology of the Cell 7e-Alberts 43. In the schematic drawing of an antibody molecule below, indicate which letter corresponds to each of the following features.

a. A b. B c. C d. D e. E

1. N-terminus 2. C-terminus 3. Light chain 4. Antigen-binding site 5. Disulfide bond

44. Mammals produce five major classes of immunoglobulins. Indicate whether each of the following descriptions better applies to the immunoglobulin A, D, E, G, or M class. a. IgA b. IgG

Molecular Biology of the Cell 7e-Alberts c. IgM d. IgD e. IgE

1. It binds to cells that secrete histamine.

2. It is the major antibody in secretions such as saliva.

3. It is the major antibody circulating in the blood.

4. It is found on the surface of naïve B cells along with IgM. 5. It forms wheel-like pentamers involved in the primary immune response.

45. Which statement best describes a negative selection mechanism of thymocytes in the thymus? a. No TCRs expressed b. TCRs with relatively weak recognition of self class I or class II MHC + self peptide c. TCRs with intermediate-strength recognition of self class II MHC + self peptide + IL2 d. TCRs with low binding affinity to self MHC e. TCRs with recognition of self class I or class II MHC + self peptide

46. Indicate whether each of the following descriptions better applies to class I or class II MHC proteins of vertebrates. a. Class I b. Class II

1. They mainly present intracellular peptide fragments.

2. They are found on most nucleated cells.

3. They present antigens mainly to helper or regulatory T cells.

4. They are recognized by CD4 co-receptors on helper and regulatory T cells.

47. Consider the following three diseases associated with activation-induced

Molecular Biology of the Cell 7e-Alberts deaminase (AID) and indicate whether you think each of them is accompanied by a higher or lower AID activity in the affected cells compared to normal cells. a. Higher b. Lower 1. Burkitt’s lymphoma is a cancer of B lymphocytes originating from the germinal center. In this cancer, chromosomal translocations are observed between the immunoglobulin heavy-chain genes and proto-oncogenes (genes whose overexpression can promote cancer), resulting in aberrant expression of the latter.

2. Diffuse large B cell lymphoma is a cancer of B lymphocytes in which, in addition to chromosomal translocations, high mutation rates are observed in many genes including proto-oncogenes.

3. Hyper IgM syndrome is a genetic immunodeficiency syndrome in which IgM antibodies are produced by B cells at high levels and for extended periods, while levels of IgA, IgE, and IgG antibodies are abnormally reduced.

48. Indicate whether each of the following descriptions better applies to B cell receptors or T cell receptors. a. B cell receptors b. T cell receptors

1. They can also be produced as secreted antibodies with the same antigen specificity.

2. They diversify by class switching and somatic hypermutation.

3. They often exist as heterodimers.

4. They have a relatively low antigen-binding affinity.

49. Gout patients have high uric acid levels in their blood and suffer from arthritis in their joints as a result of formation of intracellular and extracellular uric acid crystals. Would you expect glucocorticoids or tumor necrosis factor to be used to treat such patients? While similarly characterized as pro-inflammatory, why would you use one treatment over the other?

50. How do the late complement components assemble to form membrane attack complexes?

Molecular Biology of the Cell 7e-Alberts

51. How do clonal expansion and production of memory cells occur?

52. Mice whose immune systems had previously rejected a tissue transplant from a certain strain of donor mouse received another transplant shortly afterward, either from the same donor strain or a new one. All of the recipient mice eventually rejected the transplant. However, one group did so more quickly than the other group. Do you think this group received the graft from the original mouse strain or the new strain? Explain your answer in 2–3 sentences.

53.

A patient with persisting larger-than-normal lymph nodes is suspected to have a T cell lymphoma, a cancer of T lymphocytes. Abnormally enlarged lymph nodes are also found in local infections, when lymphocytes are activated to proliferate there.

In order to distinguish between cancer and infection, you take tissue samples from the patient, as well as from a healthy individual, and perform a so-called clonality test.

You extract DNA from the tissues and amplify a small chromosomal region that includes the T-cell-receptor β chain D-J junctions. Amplification is done by PCR, using primers that hybridize to specific DNA sequences flanking the junctions.

You then analyze the size distributions of the PCR products by capillary electrophoresis, which makes it possible to distinguish between DNA molecules with small differences in size. Based on the results, presented in the following schematic graphs, you conclude that, unfortunately, it is likely that the patient has developed a T cell lymphoma. Which graph (1 or 2) represents the patient’s test results? Justify your response in 2-3 sentences.

Molecular Biology of the Cell 7e-Alberts

54. Sort the following events to reflect the order in which they occur in the presentation of viral peptides to cytotoxic T cells by a virus-infected cell. Your answer would be a four-letter string composed of letters A to D only, for example, BDCA. A. Peptide transport into the ER lumen B. Partial proteasomal degradation of the viral protein into peptide fragments C. Peptide binding to class I MHC protein in the ER lumen D. The appearance of viral protein in the cytosol

55. From an evolutionary perspective, which mating pattern confers a higher fitness to a mammalian population: Preferred mating between pairs whose MHC genes are most similar or between those whose MHC genes are most dissimilar?

Explain your answer in 2–3 sentences.

56. When COVID-19 (SARS-CoV-2 coronavirus) was first detected and a global pandemic ensued, one of the main topics of conversation became the development of an effective vaccine. Today, we have not one but two: The DNA virus vector vaccine and the mRNA lipid nanoparticle vaccine. Compare and contrast these nucleic acid vaccines by explaining how they work to combat the COVID-19 disease, despite any differences in their process.

Molecular Biology of the Cell 7e-Alberts

Molecular Biology of the Cell 7e-Alberts

Answer Key Chapter 24 1. Answer:

A

2. Answer:

A

3. Answer:

E

4. Answer:

E

5. Answer:

D

6. Answer:

C

7. Answer:

A

8. Answer:

B

9. Answer:

D

10. Answer:

D

11. Answer:

B

12. Answer:

B

13. Answer:

D

14. Answer:

C

15. Answer:

A

16. Answer:

B

17. Answer:

C

18. Answer:

E

19. Answer:

E

20. Answer:

D

Molecular Biology of the Cell 7e-Alberts 21. Answer:

B

22. Answer:

C

23. Answer:

E

24. Answer:

A

25. Answer:

D

26. Answer:

C

27. Answer:

C

28. Answer:

B

29. Answer:

C

30. Answer:

E

31. Answer:

D

32. Answer:

C

33. Answer:

B

34. Answer:

A

35. Answer:

A

36. Answer:

1. A 2. B 3. B 4. A

37. Answer:

1. B 2. A 3. A 4. B

Molecular Biology of the Cell 7e-Alberts 38. Answer:

1. A 2. B 3. B 4. A

39. Answer:

1. E 2. E 3. A 4. C

40. Answer:

1. B 2. C 3. C

41. Answer:

A

42. Answer:

E

43. Answer:

1. D 2. E 3. B 4. A 5. C

44. Answer:

1. E 2. A 3. B 4. D 5. C

45. Answer: 46. Answer:

A 1. A

Molecular Biology of the Cell 7e-Alberts 2. A 3. B 4. B 47. Answer:

1. A 2. A 3. B

48. Answer:

1. A 2. A 3. B 4. B

49. Answer:

Glucocorticoids would be expected, because prostaglandin is a proinflammatory signal molecule, and tumor necrosis factor is a pro-inflammatory cytokine. In gout, NOD-like receptor (NLR)-dependent inflammasome assembly leads to an inflammatory response, which can be suppressed by drugs such as glucocorticoids.

50. Answer:

Once the large fragment of C3, called C3b, binds covalently to the surface of the pathogen, the cleavage of the early components results in the formation of C3b-containing proteolytic complexes on the pathogen membrane, catalyzing the subsequent steps in the complement cascade. Complement components C5a and C5b are produced, and C5b assembles with C6 and C7. The complex binds through C7 to the pathogen membrane, and also to C8 and C9. After the interaction with C9, a ring of C9 molecules constitutes a transmembrane aqueous channel in the pathogen membrane.

51. Answer:

A naïve B or T cell, stimulated by a specific antigen and co-stimulatory signals, proliferates and differentiates,

Molecular Biology of the Cell 7e-Alberts producing both effector and memory cells.

After a second exposure to the same antigen, in the presence of co-stimulatory signals, memory cells respond faster and more efficiently than naïve cells.

52. Answer:

The group received the graft from the original mouse strain. The rejection of a second graft from the original donor strain occurs more quickly, because memory B and/or T cells were produced in response to the first transplant. This results in a secondary adaptive immune response to the second transplant from the same mouse strain that is faster and stronger than the primary immune response elicited when the second transplant is from a different donor mouse strain.

53. Answer:

Graph 2 represents the patient’s results, because V(D)J recombination in T lymphocytes creates deletions (or insertions) of various numbers of nucleotide pairs at the junctions of the V, D, and J segments. In response to an infection, a number of T lymphocytes (that recognize various antigenic determinants associated with the pathogen) proliferate to create effector and memory cells, leading to lymph node enlargement. The different activated T cell clones normally differ in the exact number of deleted or inserted nucleotide pairs, which one can see in Graph 1. In contrast, cancer is a clonal disease; all malignant cells in a T cell lymphoma derive from a single original cell with its unique V(D)J junctions, which one sees in Graph 2.

54. Answer:

DBAC

55. Answer:

Those whose MHC genes are most dissimilar, because mating between animals whose MHC genes differ provides the population with greater advantages in terms of survivability and reproduction.

More specifically, due to variation, the population will experience an increased MHC polymorphism and a broader

Molecular Biology of the Cell 7e-Alberts capability for adaptive immune responses to pathogens, traits that are especially important when adapting to a new or changing environment.

56. Answer:

As the descriptor “nucleic acid vaccine” suggests, the DNA virus vector vaccine and the mRNA lipid nanoparticle vaccine have nucleic acids encode forms of the spike (S) protein, a mimicry of the act likewise performed by the SARS-CoV-2 coronavirus, which uses its own spike-like protrusions to bind to and enter human cells. However, the vaccines have marked differences. On the one hand, to produce the mRNA lipid nanoparticle vaccine, one must synthesize in vitro with nucleoside modifications, while the DNA virus vector are based on adenovirus vectors, which transfer DNA into the human cell’s nucleus and produce mRNAs. As for similarities, both vaccines are injected into an arm muscle so that lymphatic vessels can carry the vaccine to lymph nodes, where antigen-presenting cells absorb it, triggering its activation in the body. The effectiveness of both vaccines has also been proven to increase through second and even third immunizations, because increased immunity ensures that memory cells respond when vaccinated individuals encounter the virus.