Test Bank for Lehninger Biochemistry Core Concepts and Applications 1st Edition By Michael Cox, Aaro

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Chapter 01: Biochemistry Concepts and Themes 1. Which is the major feature that distinguishes eukaryotes from Bacteria and Archaea? a. DNA b. photosynthetic capability c. plasma membranes d. ribosomes e. the nucleus ANSWER: e 2. In a bacterial cell, the DNA is in the: a. cell envelope. b. cell membrane. c. nucleoid. d. nucleus. e. ribosomes.

ANSWER: c 3. In eukaryotes, the nucleus is enclosed by a double membrane called the: a. cell membrane. b. nuclear membrane. c. nucleolus. d. nucleoplasm. e. nucleosome.

ANSWER: b 4. Which is the range of typical diameters of animal and plant cells? a. 0.1 μm to 10 μm b. 0.3 μm to 30 μm c. 5 μm to 100 μm d. 10 nm to 300 nm e. 1 μm to 300 μm ANSWER: c 5. Which group of single-celled microorganisms has many members found growing in extreme environments? a. bacteria b. archaea c. eukaryotes d. heterotrophs e. None of the answers is correct.

ANSWER: b 6. Which is a list of organelles?

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Chapter 01: Biochemistry Concepts and Themes a. mitochondria, chromatin, endoplasmic reticulum b. peroxisomes, lysosomes, plasma membrane c. proteasomes, chromatin, lysosomes d. mitochondria, endoplasmic reticulum, peroxisomes e. All of the answers are correct.

ANSWER: d 7. Which list has the cellular components arranged in order of increasing size? a. nucleotide < DNA < nucleus < chromatin b. nucleotide < DNA < chromatin < nucleus c. nucleotide < chromatin < DNA < nucleus d. DNA < nucleotide < nucleus < chromatin e. DNA < chromatin < nucleus < nucleotide ANSWER: b 8. Which element is NOT among the four most abundant in living organisms? a. carbon b. hydrogen c. nitrogen d. oxygen e. phosphorus ANSWER: e 9. The four covalent bonds in methane (CH4) are arranged around carbon to give which geometry? a. linear b. tetrahedral c. trigonal bipyramidal d. trigonal planar e. trigonal pyramidal ANSWER: b 10. What functional groups are present on this molecule?

a. ether and aldehyde b. hydroxyl and aldehyde c. hydroxyl and carboxylic acid d. hydroxyl and ester Copyright Macmillan Learning. Powered by Cognero.

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Chapter 01: Biochemistry Concepts and Themes e. hydroxyl and ketone ANSWER: b 11. Humans maintain a nearly constant level of hemoglobin by continually synthesizing and degrading it. This is an example of a(n): a. dynamic steady state. b. equilibrium state. c. exergonic change. d. free-energy change. e. waste of energy.

ANSWER: a 12. If the free-energy change ΔG for a reaction is –46.11 kJ/mol, the reaction is: a. at equilibrium. b. endergonic. c. endothermic. d. exergonic. e. exothermic.

ANSWER: d 13. The breakage of a phosphoanhydride bond in ____________ is a major source of chemical energy to drive cellular reactions. a. acetyl triphosphate b. adenosine monophosphate c. adenosine triphosphate d. cytosine tetraphosphate e. uridine diphosphate ANSWER: c 14. Enzymes are biological catalysts that enhance the rate of a reaction by: a. decreasing the activation energy. b. decreasing the amount of free energy released. c. increasing the activation energy. d. increasing the amount of free energy released. e. increasing the energy of the transition state.

ANSWER: a 15. Hereditary information (with the exception of some viruses) is preserved in: a. deoxyribonucleic acid. b. membrane structures. c. nuclei.

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Chapter 01: Biochemistry Concepts and Themes d. polysaccharides. e. ribonucleic acid.

ANSWER: a 16. The three-dimensional structure of a protein is determined primarily by: a. electrostatic guidance from nucleic acid structure. b. noncovalent association with other proteins. c. hydrophobic interaction with lipids that provide a folding framework. d. modification during interactions with ribosomes. e. the sequence of amino acids in the protein.

ANSWER: e 17. Which statement is NOT a distinguishing feature of living organisms? a. There exists a high degree of organizational complexity. b. The structure of components influences their function. c. Organisms can reproduce themselves. d. Organisms can exist without interacting with their environment. e. Organisms change over time.

ANSWER: d 18. Which statement is NOT true regarding the plasma membrane? a. It is a physical barrier separating the inside of the cell from its surroundings. b. It is flexible, with a hydrophobic internal structure. c. The individual lipids and proteins of the plasma membrane are covalently linked. d. The plasma membrane incorporates newly made lipid and protein components as a cell grows. e. Cell division occurs without loss of the membrane integrity.

ANSWER: c 19. Which does NOT contain a double membrane? a. mitochondrion b. ribosome c. chloroplast d. endoplasmic reticulum e. Golgi body ANSWER: b 20. Which is NOT found in both animal and plant cells? a. ribosome b. Golgi complex c. endoplasmic reticulum d. vacuole Copyright Macmillan Learning. Powered by Cognero.

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Chapter 01: Biochemistry Concepts and Themes e. mitochondrion ANSWER: d 21. Which are the four MOST abundant elements in living organisms? a. carbon, hydrogen, oxygen, iron b. carbon, hydrogen, nitrogen, oxygen c. carbon, hydrogen, phosphorous, oxygen d. carbon, nitrogen, phosphorous, oxygen e. carbon, hydrogen, sulfur, oxygen ANSWER: b 22. Which ranking correctly describes the rigidity of the red bond (the central bond) shown in the figure?

a. 2 = most rigid, 3 = least rigid b. 1 = most rigid, 5 = least rigid c. 4 = most rigid, 3 = least rigid d. 2 = most rigid, 1 = least rigid e. 4 = most rigid, 1 = least rigid ANSWER: c 23. Which list of descriptive terms for biological molecules is placed in correct order from smallest to largest? a. monomer, oligomer, polymer b. monomer, multimer, macromer c. oligomer, monomer, polymer d. polymer, oligomer, monomer e. metamer, oligomer, polymer ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 01: Biochemistry Concepts and Themes 24. Which types of molecules can serve as informational macromolecules in cells? a. proteins b. nucleic acids c. oligosaccharides d. both proteins and nucleic acids e. proteins, nucleic acids, and oligosaccharides ANSWER: e 25. Which statement about living systems is NOT true? a. Living organisms can be described as an open system. b. Living organisms maintain a more or less constant composition at maturity. c. Living systems are in equilibrium with their surroundings. d. Living systems exist in a dynamic steady state. e. Living systems have efficient mechanisms to convert chemical energy from one form into another.

ANSWER: c 26. In an oxidation-reduction reaction, the reactant that is oxidized _____, and the reactant that is reduced _____. a. is energized; is de-energized b. is de-energized; is energized c. loses electrons; gains electrons d. gains electrons; loses electrons e. maintains the same number of electrons; loses electrons ANSWER: c 27. The joining of two amino acids via a peptide bond (the process of protein synthesis) has a positive ΔG value. What does this imply? a. Forming a peptide bond is endergonic and must be coupled to another reaction. b. Forming a peptide bond is exergonic and must be coupled to another reaction. c. Forming a peptide bond is spontaneous and does not need to be coupled to another reaction. d. Forming a peptide bond is spontaneous and can sometimes be coupled to another reaction. e. Forming a peptide bond increases the entropy of a system.

ANSWER: a 28. Reaction 1 has a ΔG° of –12.3 kJ/mol, and reaction 2 has a ΔG° of –23.4 kJ/mol. Which statement is true of these two reactions? a. Reaction 1 occurs faster. b. Reaction 2 occurs faster. c. Both reactions occur at the same rate. d. Reaction 2 will not occur. e. It is impossible to know which reaction occurs faster with this information.

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Chapter 01: Biochemistry Concepts and Themes ANSWER: e 29. Which statement is true when Keq >> 1? a. ΔG° is large and negative b. ΔG° is large and positive c. ΔG° is small and negative d. ΔG° is small and positive e. the value of ΔG° is independent of Keq ANSWER: a 30. The typical three-dimensional structure of a protein in a cell, or _____ conformation, is critical to a protein's function. a. native b. molecular c. chaperone d. macromolecular e. high-affinity ANSWER: a 31. The similarities of gene sequences and metabolic pathways across the three domains of life are evidence that: a. all modern organisms are derived from a common evolutionary progenitor. b. multiple evolutionary progenitors converged to a single evolutionary model. c. cross-species genetic transfer happens with ease. d. evolution ceases when an organism is successful in its niche. e. All of the above.

ANSWER: a 32. All cells are surrounded by a plasma membrane composed of lipid and protein molecules. What is the function of the plasma membrane?

ANSWER: The plasma membrane acts as a barrier to the free passage of inorganic ions and most other charged or polar compounds into or out of the cell. It contains proteins that can transport specific ions or molecules. Other membrane proteins act as receptors that transmit signals from the outside to the inside of the cell.

33. Most cells of higher plants have a cell wall outside the plasma membrane. What is the function of the cell wall?

ANSWER: The cell wall provides a rigid, protective shell for the cell. It is porous, allowing water and small molecules to pass readily. It is also rigid enough to resist the swelling of the cell caused by the accumulation of water (see Fig. 1-4).

34. What is the difference, if any, between cytosol and cytoplasm?

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Chapter 01: Biochemistry Concepts and Themes ANSWER: The cytoplasm is the internal volume enclosed by the plasma membrane; the cytosol is the aqueous portion of the cytoplasm.

35. Draw the structures of the functional groups in their nonionized forms for (a) hydroxyl, (b) carboxyl, (c) amino, (d) phosphoryl in their nonionized forms.

ANSWER:

36. What is the underlying, organizing biochemical principle that results in the chemical similarity of virtually all living things? Given this biochemical similarity, how is the structural and functional diversity of living things possible?

ANSWER: Living things are composed primarily of macromolecules, polymers of simple compounds of just a few different types. The properties of these polymers are determined by their sequence of monomers and these can be combined in many different ways. Diversity is thus achieved through the nearly limitless variety of sequences that can exist when amino acids are linked to form proteins, nucleotides are linked to form nucleic acids, and monosaccharides are linked to form polysaccharides. Branching in the latter can contribute additional heterogeneity. Each type of organism constructs a unique set of macromolecules from these monomeric units, resulting in the structural and functional diversity among species.

37. Name two functions of (a) proteins, (b) nucleic acids, (c) polysaccharides, (d) lipids.

ANSWER: Many answers are possible including: (a) proteins function as enzymes, structural elements, signal carriers, transporters; (b) nucleic acids store and transmit genetic information and act as both structural and catalytic elements; (c) polysaccharides serve as energy-yielding fuel stores and cellular and extracellular structural and recognition elements; (d) lipids function as membrane components, fuel stores, and cellular signals.

38. Proteins are constantly being synthesized in a living cell. Why doesn't the number of protein molecules become too great for the cell to contain, leading to cell destruction?

ANSWER: The proteins in a cell are continuously being synthesized and degraded. The cell maintains a dynamic steady state in which the amount of each protein remains fairly constant at the level required under given conditions.

39. Describe the relationship between a living organism and its surroundings in terms of both matter and energy.

ANSWER: Living organisms are open systems and exchange both matter and energy with their surroundings.

They are not at equilibrium with their surroundings; that is, the exchange of matter and energy with the surroundings is not constant and equal in both directions. To maintain this situation, the organism must acquire energy from its surroundings, either in the form of chemical energy or directly from sunlight.

40. The free-energy change for the formation of a protein from the individual amino acids is positive and is thus an endergonic reaction. How, then, do cells accomplish this process?

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Chapter 01: Biochemistry Concepts and Themes ANSWER: The endergonic (thermodynamically unfavorable) reaction is coupled to an exergonic (thermodynamically favorable) reaction through a shared intermediate, so that the overall freeenergy change of the coupled reactions is negative (the overall reaction is exergonic).

41. Instant cold packs get cold when the contents (usually solid urea and liquid water) are mixed, producing an aqueous solution of urea. Although this process is clearly spontaneous, the products are colder than the reactants. Explain how this is possible in terms of the difference between ΔG and ΔH. ANSWER: Since the dissolution reaction is spontaneous, the ΔG must be negative. Since the reaction absorbs heat, the ΔH must be positive. Given ΔG = ΔH – TΔS, this is possible if the ΔS is very large and positive, as one would expect for a solid dissolving.

42. (a) On the reaction coordinate diagram, label the transition state and the overall free-energy change (ΔG) for the uncatalyzed reaction A→B. (b) Is this an exergonic or endergonic reaction? (c) Draw a second curve showing the energetics of the reaction if it were enzyme-catalyzed.

ANSWER: (a) and (c) see Fig. 1.15 (b) exergonic reaction 43. Briefly, compare and contrast the three domains of life.

ANSWER: Bacteria and Archaea consist of unicellular organisms, while Eukarya contains both single- and multicellular organisms. Bacteria and Archaea have been found in all kinds of environments.

Archaea historically were characterized from extreme environments. Neither bacteria nor archaea have chromosomes separated from the cytoplasm by a nuclear membrane, but eukaryotes have chromosomes inside a membrane-bounded nucleus. Eukaryotes also contain multiple, membraneenclosed sub-compartments called organelles.

44. (a) What is the diagram? (b) Describe what it illustrates about the relationship between animals, halophiles, slime molds, and gram-positive bacteria. (c) Explain the evidence used in construction of this diagram.

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Chapter 01: Biochemistry Concepts and Themes

ANSWER: (a) This diagram is a phylogenetic tree. (b) It illustrates the evolutionary distance between different organisms in the three domains of life. Of the organisms listed, animals are most closely related to slime molds, next closest to halophilic archaea. Archaea and Eukarya are more similar to each other than either is to Bacteria, making gram-positive bacteria the most distant relative to all of the organisms listed in this question. (c) The distances between the branches of the tree are calculated by comparing sequence similarity between homologous proteins in different organisms.

45. Why is the use of the expression ‘Mr = 18,000 daltons’ incorrect?

ANSWER: Mr is a ratio and therefore dimensionless. Daltons are only used when describing molecular mass, not molecular weight or relative molecular mass.

46. An idea or experiment is within the realm of science if it adheres to which key elements?

ANSWER: An idea or experiment is within the realm of science if it focuses on the natural universe, is carried out within a scientific context, and if it has reproducible results that are reported honestly.

47. What is a hypothesis?

ANSWER: A hypothesis is a proposal that provides a reasonable explanation for observations. It is not yet substantiated by sufficient experimental tests.

48. Scientific endeavors can be understood by applying a systematic process of inquiry called: a. an assumption of objectivity. b. a hypothesis. c. a postulate of objectivity. d. the scientific method.

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Chapter 01: Biochemistry Concepts and Themes e. a scientific theory.

ANSWER: d 49. Scientific hypotheses and theories are often presented in peer-reviewed scientific literature. According to the Publishers Association of the United Kingdom, there are more than _____ peer-reviewed scientific journals worldwide. a. 200 b. 1,900 c. 10,000 d. 16,000 e. 1.4 million ANSWER: d 50. The practice of science as we know it is about _______ years old. a. 200 b. 1,900 c. 10,000 d. 16,000 e. 1.4 million ANSWER: a 51. Science advances in a variety of ways; not just by the implementation of the scientific method. Discuss two other productive pathways towards scientific advancement.

ANSWER: There are 4 alternative approaches to scientific knowledge in Table 1.1 of the text. If the student discusses the ‘Hypothesis→experiment→answer’ approach, this answer is incorrect; this is the classical scientific method. The 4 correct approaches are: The Serendipity approach occurs when an experiment designed to distinguish between outcome X and outcome Y, results in unexpected outcome Z. The Inspiration approach occurs when simply moments of pondering about a hypothesis lend to the rise of new technologies and major advances. The Model-building and deduction approach can occur at a computer or a desk where modeling tools and technologies can reveal scientific knowledge; experiments are usually done to confirm the predictions of model-building.

The Exploration and observation approach occurs when simply moments of observation, even without a hypothesis, lend to the attainment of new scientific knowledge.

52. A scientist is any person: a. who applies some version of the scientific method to solve a problem in the natural world. b. who makes an observation. c. who reproduces a scientific advancement. d. who works in a laboratory. e. with an advanced science degree.

ANSWER: a 53. Which approach is considered the classic scientific method?

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Chapter 01: Biochemistry Concepts and Themes a. Exploration and observation b. Hypothesis → experiment → answer c. Inspiration d. Model-building and deduction e. Serendipity ANSWER: b 54. If a result suggests a particularly exciting idea, honesty demands that: a. the idea can be observed and measured. b. the idea can be tested. c. the next experiment conducted be the one most likely to disprove that idea. d. the result be reproduceable. e. the result can be produced in a vacuum.

ANSWER: c 55. Scientific endeavors are grounded in the assumption that the forces and phenomena that make up the universe are unchanging. This is related to: a. honesty. b. observation. c. objectivity. d. reproducibility. e. serendipity.

ANSWER: d 56. The MOST precise modern definition of a gene is a segment of genetic material that: a. codes for one polypeptide. b. codes for a final functional polypeptide or RNA product. c. determines one phenotype. d. determines one trait. e. codes for one protein.

ANSWER: b 57. Functional DNA is NOT found in: a. bacterial nucleoids. b. chloroplasts. c. lysosomes. d. mitochondria. e. nuclei.

ANSWER: c 58. Which list ranks the organisms in order of increasing number of genes?

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Chapter 01: Biochemistry Concepts and Themes a. S. cerevisiae, humans, nematodes b. S. cerevisiae, nematodes, humans c. nematodes, humans, S. cerevisiae d. nematodes, S. cerevisiae, humans e. humans, S. cerevisiae, nematodes ANSWER: b 59. What are the minimal components of a virus? a. DNA and a lipid membrane b. genome (DNA or RNA), a lipid membrane, and a viral coat c. genome (DNA or RNA) and a viral coat d. protein genome and a lipid membrane ANSWER: c 60. Which item is NOT electrophilic? a. proton b. sulfhydryl c. protonated imine d. carbonyl group e. phosphoryl group ANSWER: b 61. Which atom is a common electrophile? a. hydrogen atom b. oxygen atom of a carbonyl group c. oxygen atom of a phosphate group d. oxygen atom of a hydroxide group e. phosphorus of a phosphate group ANSWER: e 62. Which atom is a possible nucleophile? a. oxygen atom in a hydroxide ion b. oxygen atom of a carboxylate group c. nitrogen atom of a deprotonated amine d. All of the answers are correct. e. None of the answers are correct.

ANSWER: d 63. Why is carbon a main element used in living organisms?

ANSWER: Carbon has a great capacity to make substances of widely different sizes, shapes, and composition.

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Chapter 01: Biochemistry Concepts and Themes 64. The reaction A + B → C has a ΔG'° of –20 kJ/mol at 25°C. Starting under standard conditions, one can predict that: a. at equilibrium, the concentration of B will exceed the concentration of A. b. at equilibrium, the concentration of C will be less than the concentration of A. c. at equilibrium, the concentration of C will be much greater than the concentration of A or B. d. C will rapidly break down to A + B. e. when A and B are mixed, the reaction will proceed rapidly toward formation of C. ANSWER: c 65. For the reaction A → B, the Keq' is 104. If a reaction mixture originally is 1 mM in A and contains no B, which statement must be true? a. At equilibrium, there will be far more B than A. b. The rate of the reaction is very slow. c. The reaction requires coupling to an exergonic reaction in order to proceed. d. The reaction will proceed toward B at a very high rate. e. ΔG'° for the reaction will be large and positive.

ANSWER: a 66. For the reaction A → B, the Keq' is 10–6. If a reaction mixture originally is 1 mM in both A and B, which statement must be true? a. At equilibrium, there will still be equal levels of A and B. b. The rate of the reaction is very slow. c. At equilibrium, the amount of A will greatly exceed the amount of B. d. The reaction will proceed toward B at a very high rate. e. ΔG'° for the reaction will be large and negative.

ANSWER: c 67. Which statement is true for the reaction A + B → C + D if its Keq for the reaction is 104? a. When the reaction has reached equilibrium, there will be significantly more A and B than C and D in the solution. b. The formation of A and B will proceed at a high rate. c. The value of ΔG'° will be large and positive. d. All of the statements are true. e. None of the statements are true.

ANSWER: e 68. The products of scientific research are widely available in: a. international data banks. b. literature search engines. c. open-source programs. d. protein data banks.

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Chapter 01: Biochemistry Concepts and Themes e. All of the above are correct.

ANSWER: e 69. Why is a laboratory notebook not a widely useful type of data bank?

ANSWER: Widely useful data banks are online resources that curate and allow the interrogation of scientific information that has been acquired over many decades. Laboratory notebooks are typically used by scientists to track their daily methods and observations while conducting experiments. They are typically not widely available online in their raw form; but are instead used to write publicly available literature. Widely useful data banks include data banks that curate genomic sequence information, literature search engines, and the protein data bank.

70. Which is not a literature search engine? a. Google Scholar b. PubMed Central c. National Center for Biotechnology Information d. Web of Science e. None of the above ANSWER: c 71. What are Jmol and PyMOL? What can they be used for?

ANSWER: Jmol and PyMOL are open-source programs available to anyone. They allow one to examine structural files and visualize structures of proteins and other macromolecules.

72. To find related DNA sequence of interest in several organisms, you can do a broad search using a program such as: a. BLAST. b. Jmol. c. PyMOL. d. NCBI. e. NMR. ANSWER: a 73. The Worldwide Protein Data Bank (PDB) has over __________ macromolecular structures. a. 50 b. 100 c. 2,000 d. 10,000 e. 200,000 ANSWER: e 74. Which data bank is a free depository of articles funded by the NIH and other U.S. government agencies? a. Google Scholar Copyright Macmillan Learning. Powered by Cognero.

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Chapter 01: Biochemistry Concepts and Themes b. PubMed Central c. National Center for Biotechnology Information d. Web of Science e. Worldwide Protein Data Bank ANSWER: b 75. Which data bank is a powerful subscription-based service accessible at academic and many commercial libraries by students, faculty, and employees? a. Google Scholar b. PubMed Central c. National Center for Biotechnology Information d. Web of Science e. Worldwide Protein Data Bank ANSWER: d 76. Which data bank curates protein and other macromolecular structures derived from x-ray crystallography, NMR, and cryo-electron microscopy? a. Google Scholar b. PubMed Central c. National Center for Biotechnology Information d. Web of Science e. Worldwide Protein Data Bank ANSWER: e 77. Which organization offers a wide range of resources, including the stored genomic sequence information from thousands of different organisms? a. Google Scholar b. PubMed Central c. National Center for Biotechnology Information d. Web of Science e. Worldwide Protein Data Bank ANSWER: c

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Chapter 02: Water: The Chemistry of Life 1. Which statement about hydrogen bonds is NOT true? a. Hydrogen bonds account for the high boiling point of water compared to molecules of similar size. b. In liquid water, the average water molecule forms hydrogen bonds with three to four other water molecules. c. Individual hydrogen bonds are much weaker than covalent bonds. d. Individual hydrogen bonds in liquid water exist for many seconds and sometimes for minutes. e. The strength of a hydrogen bond depends on the linearity of the three atoms involved in the bond.

ANSWER: d 2. Which contributes to water's unusual properties? a. the geometry of water molecules b. the polarity of water c. the ability of water molecules to form hydrogen bonds d. the electronegativity of the oxygen in water e. All of the answers are correct.

ANSWER: e 3. Ice is _____ dense than water because _____. a. less; frozen water maintains more hydrogen bonds than liquid water b. less; liquid water maintains more hydrogen bonds than frozen water c. more; frozen water maintains more hydrogen bonds than liquid water d. more; liquid water maintains more hydrogen bonds than frozen water e. more; frozen water cannot hydrogen bond ANSWER: a 4. Which image correctly represents hydrogen bonding?

a. A and D b. A and C c. B and C d. C and D e. B and D ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 02: Water: The Chemistry of Life 5. Oxygen and carbon dioxide are both biologically important gases. Which statement about these gases is true? a. O2 and CO2 are both nonpolar and are very soluble in water. b. O2 and CO2 are both polar and are very soluble in water. c. O2 and CO2 are both nonpolar and are poorly soluble in water. d. CO2 contains polar bonds and is very soluble in water, but O2 is nonpolar and is poorly soluble in water. e. O2 contains polar bonds and is very soluble in water, but CO2 is nonpolar and is poorly soluble in water.

ANSWER: c 6. Which statement about hydrogen bonds is false? a. Hydrogen bonds are highly directional. b. Hydrogen bonds are capable of holding molecules in a specific geometric arrangement. c. Hydrogen bonds are strongest when the three atoms in the bond are in a straight line. d. Hydrogen bonds are strongest when the oxygen atom is perpendicular to the hydrogen donor. e. Hydrogen bonds occur when a hydrogen with a partial positive charge is positioned between two atoms with partial negative charges.

ANSWER: d 7. In water, which substance would most force surrounding water molecules to become highly ordered? a. CH2O b. CH4 c. NH3 d. CH3CH2OH e. None of the answers is correct.

ANSWER: b 8. Which diagram correctly illustrates the clustering of lipids in the formation of a micelle?

a. A b. B c. both A and B d. neither A nor B Copyright Macmillan Learning. Powered by Cognero.

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Chapter 02: Water: The Chemistry of Life ANSWER: b 9. When a nonpolar solute is dispersed in water, which will occur? a. The entropy of the water decreases. b. The entropy of the water increases. c. The enthalpy of solution is negative. d. The system reaches equilibrium quickly. e. All of the answers are correct.

ANSWER: a 10. Which statement regarding long-chain fatty acids in aqueous solution is NOT true? a. Fatty acids in small concentrations are surrounded by highly ordered water molecules in a cage-like structure. b. Fatty acids will cluster together to minimize the lipid surface area. c. Fatty acids will form micelles to sequester hydrophobic groups from water. d. The driving force of solubilizing fatty acids is increasing entropy of the fatty acid. e. All of the statements are true.

ANSWER: d 11. Which statement correctly describes the forces that drive the formation of micelles by amphipathic molecules, such as fatty acids, dissolved in water? a. Head groups of the molecules are sequestered in the interior of the micelle, maximizing hydrogen bonding of the hydrophobic tail with surrounding solvent molecules. b. Head groups of the molecules are exposed on the outer surface of the micelle, maximizing hydrogen bonding between hydrophobic tails. c. Head groups of the molecules are exposed on the outer surface of the micelle, minimizing the order of the surrounding water molecules. d. Hydrophilic tails are exposed on the outer surface of the micelle, maximizing hydrogen bonding between the tails and surrounding water molecules. e. Hydrophobic tails are exposed on the outer surface of the micelle, maximizing hydrogen bonding between head groups.

ANSWER: c 12. Which force plays the greatest role in stabilizing biological membranes? a. hydrogen bonding because it is a strong noncovalent bond b. hydrophobic interactions that increase solvent entropy c. covalent interactions because they are very stable interactions d. electrostatic interactions between oppositely charged ions e. van der Waals interactions because of the attraction between transient dipoles ANSWER: b 13. When two atoms are joined together covalently, the van der Waals radius of the atoms in the covalent bond Copyright Macmillan Learning. Powered by Cognero.

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Chapter 02: Water: The Chemistry of Life are _____ than the radius of the atoms alone because the joined atoms are _____. a. shorter; pulled together by the shared electron pair b. shorter; pulled together by the attraction of the nucleus to the bonded atom c. shorter; pulled together by hydrogen bonding d. longer; repelled due to the shared electron pair e. longer; repelled due to the nuclear repulsion between bonded atoms ANSWER: a 14. Which diagram illustrates an amphipathic molecule?

a. A b. B c. C d. D e. E ANSWER: d 15. Distilled white vinegar has a pH of 2.4. What is the [H+] of distilled white vinegar? a. 2.4 M b. 2.5 × 10–12 M c. 2.5 × 10–3 M d. 3.98 × 10–3 M e. 3.98 × 10–6 M ANSWER: d 16. If the pH of a solution is 5.5, what is the pOH? a. 8.5 b. –5.5 c. –8.5 d. 14 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 02: Water: The Chemistry of Life e. 6.5 ANSWER: a 17. Milk of magnesia has a pH of 10.2. What is the [OH–] of milk of magnesia? a. 6.31 × 10–11 M b. 1.58 × 10–4 M c. 1.58 × 10–5 M d. 1.02 × 10–3 M e. 6.31 × 10–4 M ANSWER: b 18. The H+ concentration of a solution is 5.6 × 10–5 M. What is the pH? a. –4.25 b. 4.25 c. 5.65 d. –9.75 e. 9.75 ANSWER: b 19. The OH– concentration of a solution is 4.3 × 10–10 M. What is the pH? a. –9.37 b. 9.37 c. –4.63 d. 4.63 e. 13.4 ANSWER: d 20. If the Ka of an acid is 1.38 × 10–7, what is the pKa? a. 6.86 b. 7.14 c. 4.37 d. 10.7 e. 1.38 ANSWER: a 21. Which is the conjugate base of H2PO41–? a. H3PO4 b. H2PO42– Copyright Macmillan Learning. Powered by Cognero.

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Chapter 02: Water: The Chemistry of Life c. HPO42– d. HPO43– e. PO43– ANSWER: c 22. According to the titration curve shown, acid A is _____ because the pH _____.

a. weak; resists change when 50% titrated b. strong; resists change when 50% titrated c. weak; changes dramatically when 50% titrated d. strong; changes dramatically when 50% titrated e. It cannot be determined from the information given.

ANSWER: a 23. According to the titration curve shown, what is the approximate pKa of acid A?

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Chapter 02: Water: The Chemistry of Life

a. 0.2 b. 0.5 c. 3.2 d. 3.8 e. 4.8 ANSWER: d 24. List the acids in increasing order of strength (weakest to strongest): nitrous acid (Ka = 4.0 × 10–4), carbonic acid (Ka = 4.4 × 10–7), acetic acid (Ka = 1.7 × 10–5), phosphoric acid (Ka = 7.3 × 10–3). a. acetic acid, carbonic acid, nitrous acid, phosphoric acid b. carbonic acid, acetic acid, nitrous acid, phosphoric acid c. acetic acid, nitrous acid, carbonic acid, phosphoric acid d. phosphoric acid, nitrous acid, acetic acid, carbonic acid e. carbonic acid, phosphoric acid, nitrous acid, acetic acid ANSWER: b 25. Ibuprofen is a weak acid with a pKa of 4.9 (structure is shown with the ionizable hydrogen marked with a star). Ibuprofen is absorbed through the stomach and the small intestine. In these tissues, absorption is a function of polarity. Charged and very polar molecules are absorbed slowly; neutral hydrophobic molecules absorb quickly. If the stomach pH is about 1.5 and the small intestine pH is about 6, where (and why) will more ibuprofen be absorbed into the bloodstream?

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Chapter 02: Water: The Chemistry of Life

a. More ibuprofen will be absorbed in the small intestine because it will be uncharged due to the pH being greater than the pKa. b. More ibuprofen will be absorbed in the stomach because it will be uncharged due to the pH being lower than the pKa. c. More ibuprofen will be absorbed in the small intestine because it will be charged due to the pH being greater than the pKa. d. More ibuprofen will be absorbed in the stomach because it will be charged due to the pH being lower than the pKa. e. Ibuprofen will be absorbed equally well in both the stomach and small intestine.

ANSWER: b 26. Which compound is a diprotic acid? a. CH3COOH b. NH4+ c. H2CO3 d. H3PO4 e. CH3CH2OH ANSWER: c 27. Polar molecules cannot easily pass through the cell membrane, but hydrophobic molecules can easily pass through the membrane. Both molecules shown below can raise blood pressure. Compare the molecules and select the statement that is true.

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Chapter 02: Water: The Chemistry of Life

a. Ephedrine can more easily pass through the cell membrane than epinephrine. b. Epinephrine can more easily pass through the cell membrane than ephedrine. c. Both epinephrine and ephedrine can pass through the cell membrane equally well. d. Neither epinephrine nor ephedrine can pass through the cell membrane. e. None of the statements is true.

ANSWER: a 28. Which answer choice is involved in important biological buffering systems? a. histidine b. bicarbonate c. phosphate d. bicarbonate and phosphate e. histidine, bicarbonate, and phosphate ANSWER: e 29. Of the seven the steps listed, which four (in the correct order) are needed to prepare 1 L of a 0.02 M Tris buffer solution at pH 7.6? A lab station has a 0.1 M solution of Tris in its protonated form, 0.1 M solution of HCl and NaOH, and plentiful distilled water. The Ka of Tris is 8.32 × 10–9.

1. Calculate the volume of 0.1 M Tris to use (C1V1 = C2V2).

2. Calculate the volume of 0.1 M HCl to use (C1V1 = C2V2).

3. Calculate the volume of 0.1 M NaOH to use (C1V1 = C2V2).

4. Use the Henderson-Hasselbalch equation to calculate the ratio of Tris base to protonated Tris.

5. Adjust the pH to 7.6 by adding 0.1 M NaOH. 6. Adjust the pH to 7.6 by adding 0.1 M HCl. 7. Fill to 1 L with distilled water. a. 1, 4, 6, 7 b. 1, 4, 5, 7 c. 1, 2, 4, 7 d. 1, 3, 4, 7 e. 1, 7, 4, 6 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 02: Water: The Chemistry of Life ANSWER: b 30. Name and briefly define four types of noncovalent interactions that occur between biological molecules.

ANSWER: (1) Hydrogen bonds: weak electrostatic attractions between one electronegative atom (such as oxygen or nitrogen) and a hydrogen atom covalently linked to a second electronegative atom; (2) electrostatic interactions: relatively weak charge-charge interactions (attractions of opposite charges, repulsions of like charges) between two ionized groups; (3) hydrophobic interactions: the forces that tend to bring two hydrophobic groups together, reducing the total area of the two groups that is exposed to surrounding molecules of the polar solvent (water); (4) van der Waals interactions: weak interactions between the electric dipoles that two close-spaced atoms induce in each other.

31. Explain the fact that ethanol (CH3CH2OH) is more soluble in water than is ethane (CH3CH3).

ANSWER: Ethanol can form hydrogen bonds with water molecules, but ethane cannot. When ethanol dissolves, there is some decrease in the system's entropy due to the formation of clathrate around the alkyl group of ethanol. However, this decrease in entropy is more than compensated for by the favorable interactions (hydrogen bonds) of the hydroxyl group of ethanol with water molecules. Ethane cannot form such hydrogen bonds and is therefore insoluble in water.

32. Explain the fact that triethylammonium chloride ((CH3CH2)3N-HCl) is more soluble in water than is triethylamine ((CH3CH2)3N).

ANSWER: Triethylammonium chloride is an ionic substance. The positive charge of the triethylammonium ion makes it hydrophilic. This leads to stronger favorable interactions with water, leading to increased solubility. Triethylamine is uncharged and forms weaker interactions with water molecules.

33. Explain why amphipathic molecules tend to form micelles in water. What force drives micelle formation?

ANSWER: Micelle formation minimizes the surface area of the hydrophobic part of amphipathic molecules that contacts the polar solvent, water. Hydrophobic interactions between hydrophobic moieties are the driving force for micelle formation. When amphipathic molecules form micelles in water, the entropy decrease, due to the formation of ordered arrays of water molecules around the hydrophobic moieties, is compensated by an overall increase in entropy due to the ordering of water molecules.

34. For each pair, choose the conjugate base.

RCOOH RCOO– RNH2

RNH3+

H2PO4–

H3PO4

H2CO3 HCO3– ANSWER: RCOO–, RNH2, H2PO4–, HCO3– 35. Phosphoric acid (H3PO4) has three dissociable protons with the pKa values shown below. Which form of phosphoric acid predominates in a solution at pH 4? Explain.

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Chapter 02: Water: The Chemistry of Life H3PO4

2.14

H2PO4–

6.86

HPO42–

12.4

ANSWER: At pH 4, the first dissociable proton (pKa = 2.14) has been titrated nearly completely, and the second (pKa = 6.86) has just started to be titrated. The dominant form at pH 4 is therefore H2PO4–, the form with one dissociated proton (see Fig. 2.9 and 2.12).

36. Define pKa for a weak acid in two ways: (1) in relation to its acid dissociation constant, Ka, and (2) by reference to a titration curve for the weak acid.

ANSWER: (1) pKa = –log Ka. (2) See Fig. 2.10; pKa is the value of pH at the inflection point in a plot of pH versus extent of titration of the weak acid. At the pKa, the concentration of ionized acid equals the concentration of nonionized acid.

37. Describe a titration curve for a monoprotic weak acid and indicate the region in which the buffering capacity of the system is greatest.

ANSWER: The inflection point, which occurs when the weak acid has been exactly one-half titrated with NaOH, occurs at a pH equal to the pKa of the weak acid. The region of greatest buffering capacity (where the titration curve is flattest) occurs at pH values of pKa ± 1 (see Fig. 2.12).

38. What is the pH of a solution containing 0.2 M acetic acid (pKa = 4.7) and 0.1 M sodium acetate?

ANSWER: pH = pKa + log [conjugate base]/[acid] pH = 4.7 + log (0.1/0.2) pH = 4.7 – 0.3 pH = 4.4 39. A solution is made by combining 50 mL of a 0.1 M sodium acetate solution with 150 mL of 1 M acetic acid (pKa = 4.7). What is the pH of the resulting solution?

ANSWER:

40. For a weak acid with a pKa of 6.0, show how to calculate the ratio of acid to base at pH 5.

ANSWER:

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Chapter 02: Water: The Chemistry of Life 41. After adding 100 mL of a solution containing 0.5 mol of acetic acid per liter to 400 mL of 0.5 M NaOH, what is the final pH? (The pKa of acetic acid is 4.7.)

ANSWER: Addition of 200 mM of NaOH (400 mL×0.5 M) to 50 mM of acetic acid (100 mL×0.5 mM) completely titrates the acid so that it can no longer act as a buffer. This leaves 150 mM of NaOH dissolved in 500 mL of total solution, an [OH–] of 0.3 M. Given [OH–], [H+] can be calculated from the equilibrium constant for the dissociation of water (the ion product of water): [H+][OH–] = 10–14 M2 [H+] = 10–14 M2/0.3 M pH is, by definition, log (1/[H+]) pH = log (0.3 M/10–14 M2) = 13.48 42. A weak acid HA, has a pKa of 5.0. If 1.0 mol of this acid and 0.1 mol of NaOH were dissolved in one liter of water, what is the final pH? ANSWER: Combining 1 mol of weak acid with 0.1 mol of NaOH yields 0.9 mol of weak acid and 0.1 mol of salt.

43. In proteins, the amino acid histidine (His) plays an important role in many biological reactions. The pKa for the protonation of His to form HisH+ = 6.0. When pH = 7.0, what is the fraction of total histidine that will be in the HisH+ form?

ANSWER: Use the Henderson-Hasselbalch equation to determine the ratio of [His] to [HisH+]. pH = pKa + log ([His]/[HisH+]) 7.0 = 6.0 + log ([His]/[HisH+]) 1.0 = log ([His]/[HisH+]) [His]/[HisH+] = antilog (1) = 10 To determine the fraction of the total in the HisH+ form, [His]total = [His] + [HisH+], fraction = [HisH+]/[His]total = [HisH+]/([His] + [HisH+]) substitute from ratio calculated above + + + = [HisH ]/(10[HisH ] + [HisH ]) = 1/11, or 0.09 44. As two atoms get nearer to each other, do van der Waals attractive forces always increase?

ANSWER: No, the attractive forces created by the transient dipole-induced dipole interaction increases until the van der Waals radius is reached, where the net attraction is maximal. At this point, the electron clouds begin to repel each other.

45. Speculate why weak forces, not strong forces, are the basis for molecular recognition among biomolecules.

ANSWER: Weak forces make biological interactions reversible within physiological constraints. The Copyright Macmillan Learning. Powered by Cognero.

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Chapter 02: Water: The Chemistry of Life cumulative effects of many weak interactions can be very significant. Strong forces would require large amounts of energy to reverse any interaction.

46. Adding a solute to water lowers the freezing point of water. The solute changes this colligative property of water by _____ the water. a. lowering the concentration of b. increasing hydrogen bonding in c. altering the ionic bonding within d. changing the pH of e. changing the temperature of ANSWER: a 47. A hydronium ion is: a. H3O+. b. immediately formed when hydrogen ions are released in water. c. a hydrated proton. d. formed by the dissociation of water. e. All of the answers are correct.

ANSWER: e 48. What is the pH of a 1 M HCl solution? a. 0 b. 0.1 c. 1 d. 10 e. –1 ANSWER: a 49. What is the pH of a 0.1 M NaOH solution? a. 0.1 b. 1.0 c. 12.8 d. 13 e. 14 ANSWER: d 50. Which statement is true about the properties of aqueous solutions? a. A pH change from 5.0 to 6.0 reflects an increase in the hydroxide ion concentration, [OH–], of 20%. b. A pH change from 8.0 to 6.0 reflects a decrease in the proton concentration, [H+], by a factor of 100. c. Charged molecules are generally insoluble in water. d. Hydrogen bonds form readily in aqueous solutions.

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Chapter 02: Water: The Chemistry of Life e. The pH can be calculated by adding 7 to the value of the pOH. ANSWER: d 51. The pH of a sample of blood is 7.4, while gastric juice is pH 1.4. The blood sample has _____ [H+] than the gastric juice. a. 0.189 times the b. 5.29 times lower c. 6 times lower d. 6,000 times lower e. one million times lower ANSWER: e 52. The aqueous solution with the lowest pH is: a. 0.01 M HCl. b. 0.1 M acetic acid (pKa = 4.76). c. 0.1 M formic acid (pKa = 3.75). d. 0.1 M HCl. e. 10–12 M NaOH. ANSWER: d 53. The aqueous solution with the highest pH is: a. 1 M HCl. b. 1 M NH3 (pKa = 9.25). c. 0.5 M NaHCO3 (pKa = 3.77). d. 0.1 M NaOH. e. 0.001 M NaOH. ANSWER: d 54. Phosphoric acid is tribasic, with three pKa values: 2.14, 6.86, and 12.4. Which ionic form predominates at pH 4.5? a. H3PO4 b. H2PO4– c. HPO42– d. PO43– e. None of the answers are correct.

ANSWER: b 55. When water is found bound in the internal structure of a biomolecule, which statement is false? a. The properties of the bound water molecules are different from those of the "bulk" water of the solvent.

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Chapter 02: Water: The Chemistry of Life b. The bound water molecules may provide a path for proton hopping. c. The bound water molecules may form an essential part of the protein's ligand-binding site. d. The orientation of bound water molecules is precise. e. None of the above; all of the statements are true.

ANSWER: e 56. "Proton hopping" essentially means that: a. an individual proton jumps from one electronegative group to the next. b. a free proton moves from one hydroxyl group of ionized water to the next. c. several protons move between hydrogen-bonded water molecules, causing the net movement of a proton over a long distance in a short time. d. individual protons are freer to move among and between water molecules in solution. e. hydronium ions are freer to move among and between water molecules in solution.

ANSWER: c 57. Which statement about buffers is true? a. A buffer composed of a weak acid of pKa = 5 has greater buffering capacity at pH 4 than at pH 6. b. At pH values lower than the pKa, the conjugate base concentration is higher than that of the conjugate acid. c. The pH of a buffered solution remains constant no matter how much acid or base is added to the solution. d. The best buffers are those composed of strong acids and strong bases. e. When pH = pKa, the weak acid and conjugate base concentrations in a buffer are equal.

ANSWER: e 58. Substance X has a pKa of 7.4. What is the pH of the final solution formed by mixing 100 mL of a 1.0 M solution of X that is initially at pH 8.0 and 30 mL of 1.0 M hydrochloric acid? a. 6.5 b. 6.8 c. 7.2 d. 7.4 e. 7.5 ANSWER: d 59. The Henderson-Hasselbalch equation: a. allows the graphic determination of the molecular weight of a weak acid from its pH alone. b. does not explain the behavior of di- or tribasic weak acids. c. employs the same value for pKa for all weak acids. d. is equally useful with solutions of acetic acid and of hydrochloric acid. e. relates the pH of a solution to the pKa and the concentrations of acid and conjugate base.

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Chapter 02: Water: The Chemistry of Life ANSWER: e 60. Consider an acetate buffer, initially at the same pH as its pKa (4.76). When sodium hydroxide (NaOH) is mixed with this buffer, the: a. pH remains constant. b. pH rises more than if the same amount of NaOH is added to an acetate buffer initially at pH 6.76. c. pH rises more than if the same amount of NaOH is added to unbuffered water at pH 4.76. d. ratio of acetic acid to acetate ion in the buffer decreases. e. sodium acetate formed precipitates because it is less soluble than acetic acid.

ANSWER: d 61. A compound is known to have two ionizable groups: a free amino group with a pKa of 8.8, and another group with a pKa between 5 and 7. A mixture of 100 mL of a 0.2 M solution of this compound at pH 8.2 and 40 mL of 0.2 M hydrochloric acid has a final pH of 6.2. What is the pKa of the second ionizable group? a. The pH cannot be determined from this information. b. 5.4 c. 5.6 d. 6.0 e. 6.2 ANSWER: c 62. Three buffers are made by combining a 1 M solution of acetic acid with a 1 M solution of sodium acetate in the ratios shown.

Buffer 1: Buffer 2: Buffer 3:

1 M acetic acid 10 mL 50 mL 90 mL

1 M sodium acetate 90 mL 50 mL 10 mL

Which statement is true of the resulting buffers? a. pH of buffer 1 < pH of buffer 2 < pH of buffer 3 b. pH of buffer 1 = pH of buffer 2 = pH of buffer 3 c. pH of buffer 1 > pH of buffer 2 > pH of buffer 3 d. The problem cannot be solved without knowing the value of pKa. e. None of the statements are true.

ANSWER: c 63. A 1.0 M solution of a compound with two ionizable groups (pKa values = 6.2 and 9.5; 100 mL total) has an initial pH of 6.8. If a biochemist adds 60 mL of 1.0 M HCl to this solution, what is the final pH? a. 5.60 b. 8.90 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 02: Water: The Chemistry of Life c. 9.13 d. 9.32 e. The pH cannot be determined from this information.

ANSWER: a 64. The objective is to maintain a pH = 7.0 for an enzyme-catalyzed reaction that will produce hydrogen ions along with the desired product. At equal concentrations, which weak acid, if any, will serve as the better buffer for the reaction: acid A with pKa = 6.5, or acid B with pKa = 7.5? a. acid A b. Water is as good as either of the acids available. c. acid B d. Both are equally effective.

ANSWER: a 65. According to the titration curve shown, what is a good buffering range for acid A?

a. 0.3 M to 0.8 M OH– b. 0.9 M to 1.0 M OH– c. pH 3.5 to pH 4.2 d. pH 2.8 to pH 3.8 e. pH 2.8 to pH 4.8 ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 02: Water: The Chemistry of Life 66. According to the Henderson-Hasselbalch equation, when is the pH equal to the pKa? a. when the concentration of acid is close to zero b. when the pH approaches 7 c. when the concentration of the conjugate base is equal to the ionization constant for water d. when the concentration of the conjugate base is equal to the concentration of the acid e. None of the answers are correct.

ANSWER: d

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Chapter 03: Amino Acids, Peptides, and Proteins 1. The chirality of an amino acid results from the fact that its α carbon: a. has no net charge. b. is within a carboxylic acid. c. is bonded to four different chemical groups. d. is in the L absolute configuration in naturally occurring proteins. e. is symmetric.

ANSWER: c 2. Of the 20 standard amino acids, only _____ is not optically active. The reason is that its side chain _____. a. alanine; is a simple methyl group b. glycine; is a hydrogen atom c. glycine; is unbranched d. lysine; contains only nitrogen e. proline; forms a covalent bond with the amino group ANSWER: b 3. Two amino acids of the standard 20 contain sulfur atoms. They are: a. cysteine and serine. b. cysteine and threonine. c. methionine and cysteine. d. methionine and serine. e. threonine and serine.

ANSWER: c 4. All of the amino acids that are found in proteins, except for proline, contain a(n) _____ group. a. amino b. carbonyl c. carboxyl d. ester e. thiol ANSWER: a 5. Which statement about aromatic amino acids is correct? a. All are strongly hydrophilic. b. Histidine's ring structure results in its being categorized as aromatic or basic, depending on pH. c. On a molar basis, tryptophan absorbs more ultraviolet light than tyrosine. d. The major contribution to the characteristic absorption of light at 280 nm by proteins is the phenylalanine R group. e. The presence of a ring structure in its R group determines whether or not an amino acid is aromatic.

ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 03: Amino Acids, Peptides, and Proteins 6. What are the name and abbreviations of the amino acid shown?

a. aspartate; asp; D b. aspartic acid; asp; P c. asparagine; asn; N d. alanine; ala; A e. arginine; arg; R ANSWER: c 7. The uncommon amino acid selenocysteine has an R group with the structure —CH2—SeH (pKa ≈ 5). In an aqueous solution, pH = 7.0, the R group of selenocysteine would be: a. deprotonated. b. uncharged. c. protonated. d. dimerized. e. aromatic.

ANSWER: a 8. Which two amino acids differ from each other by only one atom? a. Ser and Thr b. Leu and Ile c. Ala and Ser d. Asp and Asn e. Ser and Cys ANSWER: e 9. Titration of valine by a strong base, for example NaOH, reveals two pK values. Which is the titration reaction Copyright Macmillan Learning. Powered by Cognero.

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Chapter 03: Amino Acids, Peptides, and Proteins occurring at pK2 (pK2 = 9.62)? a. —COOH + OH– → —COO– + H2O b. —COOH + —NH2 → —COO– + —NH2+ c. —COO– + —NH2+ → —COOH + —NH2 d. —NH3+ + OH– → —NH2 + H2O e. —NH2 + OH– → —NH– + H2O ANSWER: d 10. In a highly basic solution, pH = 13, which is the dominant form of glycine? a. NH2—CH2—COOH b. NH2—CH2—COO– c. NH2—CH3+—COO– d. NH3+—CH2—COOH e. NH3+—CH2—COO– ANSWER: b 11. For amino acids with neutral R groups, at any pH below the pI of the amino acid, the population of amino acids in solution will have: a. a net negative charge. b. a net positive charge. c. no charged groups. d. no net charge. e. positive and negative charges in equal concentration.

ANSWER: b 12. At pH 7.0, converting a glutamic acid to γ-carboxyglutamate will have what effect on the overall charge of the protein containing it? a. It will become more negative. b. It will become more positive. c. It will stay the same. d. There is not enough information to answer the question. e. The answer depends on the salt concentration.

ANSWER: a 13. At pH 7.0, converting a proline to hydroxyproline will have what effect on the overall charge of the protein containing it? a. It will become more negative. b. It will become more positive. c. It will stay the same.

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Chapter 03: Amino Acids, Peptides, and Proteins d. There is not enough information to answer the question. e. The answer depends on the salt concentration.

ANSWER: c 14. The peptide alanylglutamylglycylalanylleucine has: a. a disulfide bridge. b. five peptide bonds. c. four peptide bonds. d. no free carboxyl group. e. two free amino groups.

ANSWER: c 15. An octapeptide composed of four repeating glycylalanyl units has: a. one free amino group on an alanyl residue. b. one free amino group on an alanyl residue and one free carboxyl group on a glycyl residue. c. one free amino group on a glycyl residue and one free carboxyl group on an alanyl residue. d. two free amino and two free carboxyl groups. e. two free carboxyl groups, both on glycyl residues.

ANSWER: c 16. In a conjugated protein, a prosthetic group is: a. a fibrous region of a globular protein. b. a nonidentical subunit of a protein with many identical subunits. c. a part of the protein that is not composed of amino acids. d. a subunit of an oligomeric protein. e. synonymous with "protomer."

ANSWER: c 17. Prosthetic groups in the class of proteins known as glycoproteins are composed of: a. carbohydrates. b. flavin nucleotides. c. lipids. d. metals. e. phosphates.

ANSWER: a 18. The term "specific activity" differs from the term "activity" in that specific activity: a. is measured only under optimal conditions. b. is the activity (enzyme units) in a milligram of protein. c. is the activity (enzyme units) of a specific protein. d. refers only to a purified protein.

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Chapter 03: Amino Acids, Peptides, and Proteins e. refers to proteins other than enzymes.

ANSWER: b 19. Which type of structure refers to particularly stable arrangements of amino acid residues in a protein that give rise to recurring patterns? a. primary structure b. secondary structure c. tertiary structure d. quaternary structure e. None of the answers is correct.

ANSWER: b 20. Which type of structure describes the overall three-dimensional folding of a polypeptide? a. primary structure b. secondary structure c. tertiary structure d. quaternary structure e. None of the answers is correct.

ANSWER: c 21. The functional differences, as well as differences in three-dimensional structures, between two different enzymes from E. coli result directly from their different: a. affinities for ATP. b. amino acid sequences. c. roles in DNA metabolism. d. roles in the metabolism of E. coli. e. secondary structures.

ANSWER: b 22. One method used to prevent disulfide bond interference with protein sequencing procedures is: a. cleaving proteins with proteases that specifically recognize disulfide bonds. b. protecting the disulfide bridge against spontaneous reduction to cysteinyl sulfhydryl groups. c. reducing disulfide bridges and preventing their re-formation by further modifying the —SH groups. d. removing cystines from protein sequences by proteolytic cleavage. e. sequencing proteins that do not contain cysteinyl residues.

ANSWER: c 23. Even when a gene is available and its sequence of nucleotides is known, chemical studies of the protein are still required to determine: a. molecular weight of the protein. b. the amino-terminal amino acid.

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Chapter 03: Amino Acids, Peptides, and Proteins c. the location of disulfide bonds. d. the number of amino acids in the protein. e. whether the protein has the amino acid methionine in its sequence.

ANSWER: c 24. The term "proteome" has been used to describe: a. regions (domains) within proteins. b. regularities in protein structures. c. the complement of proteins expressed by an organism's genome. d. the structure of a protein-synthesizing ribosome. e. the tertiary structure of a protein.

ANSWER: c 25. Compare the sequences taken from four different proteins, and select the answer that BEST characterizes their relationships.

Protein 1 2 3 4

A DVEKGKKIDIMKCS DVQRALKIDNNLGQ LVTRPLYIFPNEGQ FFMNEDALVARSSN

B C HTVEKGGKHKTGPNLH GLFGRKTGQAPGYSYT HTVEKGAKHKTAPNVH GLADRIAYQAKATNEE HTLEKAAKHKTGPNLH ALKSSKDLMFTVINDD HQFAASSIHKNAPQFH NLKDSKTYLKPVISET

a. Based only on sequences in column B, protein 4 reveals the greatest evolutionary divergence. b. Comparing proteins 1 and 2 in column A reveals that these two proteins have diverged the most throughout evolution. c. Protein 4 is the protein that shows the greatest overall homology to protein 1. d. Proteins 2 and 3 show a greater evolutionary distance than proteins 1 and 4. e. The portions of amino acid sequence shown suggest that these proteins are completely unrelated.

ANSWER: a 26. All of the 20 common amino acids contain an R group that is attached to the: a. carboxyl group. b. amino group. c. α carbon. d. β carbon. e. It depends on the amino acid.

ANSWER: c 27. The amino acid proline is unique because the R group: a. has a phosphate attached. b. is cyclical.

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Chapter 03: Amino Acids, Peptides, and Proteins c. is attached to the carboxylic acid carbon. d. is positively charged. e. None of the answers is correct.

ANSWER: b 28. Which correctly matches the amino acid name with its one-letter abbreviation? a. tyrosine, T b. lysine, L c. phenylalanine, P d. aspartic acid, D e. proline, R ANSWER: d 29. On the molecule, the arrow is pointing to the _____ carbon.

a. amino b. α c. β d. γ e. carboxylic acid ANSWER: c 30. Which pair of amino acids can be used to measure the concentration of proteins based on absorption of UV light? a. proline and valine b. serine and threonine c. aspartate and glutamate d. lysine and histidine e. tyrosine and tryptophan ANSWER: e 31. Why can only some amino acids be used to measure protein concentration based on absorption of UV light? a. Only some amino acids are aromatic. b. Only some amino acids are positively charged.

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Chapter 03: Amino Acids, Peptides, and Proteins c. Only some amino acids are strongly hydrophobic. d. Only some amino acids are strongly hydrophilic. e. Only some amino acids are negatively charged.

ANSWER: a 32. The MOST useful way to classify amino acids is by: a. molecular weight. b. pKa. c. polarity. d. propensity in proteins. e. alphabetical order.

ANSWER: c 33. The protein bovine serum albumin (BSA), has a molecular weight of about 66,400. Using the Lambert-Beer Law, calculate the concentration of a sample of BSA with an absorbance of 1.3 at 280 nm, assuming the molar extinction coefficient at 280 nm is 43,824 M–1 cm–1. a. 0.51 mg/mL b. 0.86 mg/mL c. 1.97 mg/mL d. 2.56 mg/mL e. 3.78 mg/mL ANSWER: c 34. Amino acids without ionizable R groups can act as a zwitterion in a(n) _____ solution. a. nonpolar b. boiling c. acidic d. basic e. neutral ANSWER: e 35. According to the diagram, which is the BEST estimate of this amino acid's isoelectric point?

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Chapter 03: Amino Acids, Peptides, and Proteins

a. 0.5 b. 1 c. 1.5 d. 6 e. 11 ANSWER: d 36. According to the diagram, at which pH will this amino acid have a net charge of 0?

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Chapter 03: Amino Acids, Peptides, and Proteins

a. 11 b. 6 c. 1.5 d. 1 e. 0.5 ANSWER: b 37. According to the diagram, how many titratable groups does this amino acid have?

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Chapter 03: Amino Acids, Peptides, and Proteins

a. 0 b. 1 c. 2 d. 3 e. It cannot be determined by this diagram.

ANSWER: c 38. According to the diagram, what is the average net charge of this amino acid at pH = 1?

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Chapter 03: Amino Acids, Peptides, and Proteins

a. +1 b. +0.5 c. 0 d. –0.5 e. –1 ANSWER: a 39. According to the diagram, does this amino acid have a titratable R group?

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Chapter 03: Amino Acids, Peptides, and Proteins

a. yes, as evidenced by the steep center of the curve b. yes, as evidenced by the two relatively flat parts of the curve c. no, as evidenced by only two relatively flat parts of the curve d. no, as evidenced by the steepness of the center of the curve e. It cannot be determined from this graph.

ANSWER: c 40. When the ribosome forms a peptide bond, which two atoms form a covalent bond? a. the α carbon and the amino nitrogen b. the amino nitrogen and carboxylic acid carbon c. the α carbon and the carboxylic acid carbon d. the amino nitrogen and the β carbon e. It depends on the amino acids being joined.

ANSWER: b 41. Which compound is a side product of a peptide bond formation? a. water b. phosphate c. ammonium d. carboxylic acid e. None of the answers is correct.

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Chapter 03: Amino Acids, Peptides, and Proteins ANSWER: a 42. By convention, polypeptides are read in which order? a. N to C terminus b. C to N terminus c. 5′ to 3′ d. 3′ to 5′ e. smallest to largest amino acid by molecular weight ANSWER: a 43. The structure below is a molecule used in the sweeteners Equal and NutraSweet. When heated, the peptide bond will hydrolyze into methanol and which two amino acids?

a. asparagine and tyrosine b. aspartic acid and phenylalanine c. glutamic acid and phenylalanine d. aspartic acid and tryptophan e. glutamic acid and histidine ANSWER: b 44. According to the table, ion-exchange chromatography would be the LEAST effective method to separate which protein mixture?

Protein Molecular weight pI A 12,500 3.2 B 13,200 5.7 C 13,500 10.5 D 36,900 5.8 E 78,400 6.1 a. proteins A, B, and C b. proteins A, C, and D c. proteins B, D, and E d. proteins B, C, and D e. All of the answers are correct.

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Chapter 03: Amino Acids, Peptides, and Proteins 45. According to the table, size-exclusion (gel-filtration) chromatography could NOT be used to separate which protein mixture?

Protein Molecular weight pI A 12,500 3.2 B 13,200 5.7 C 13,500 10.5 D 36,900 5.8 E 78,400 6.1 a. proteins A, D, and F b. proteins C, E, and F c. proteins B, E, and F d. proteins A, B, and C e. All of the answers are correct.

ANSWER: d 46. The gel illustrates the purification of the RecA protein from E. coli. This gel was loaded from the top.

Which statement BEST describes how proteins "run" in an SDS gel?

a. The proteins are separated primarily by molecular weight—the largest molecular weight proteins are on the bottom, and the smallest are on the top. b. The proteins are separated primarily by molecular weight—the smallest molecular weight proteins are on the bottom, and the largest are on the top. c. The proteins are separated primarily by charge—the protein with the smallest pI is on the top, and the protein with the largest pI is on the bottom. d. The proteins are separated primarily by charge—the protein with the largest pI is on the top, and the protein with the smallest pI is on the bottom. e. The proteins are separated primarily by their propensity to precipitate—as electrophoresis progresses, the heat generated by the current denatures proteins faster or slower. The slowest to denature is on the bottom, and the fastest to denature is on the top.

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Chapter 03: Amino Acids, Peptides, and Proteins 47. Which of the lanes in the gel would represent an "uninduced" sample?

a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: b 48. What does the molecule SDS do to proteins? a. It neutralizes any charged residues on the protein. b. It denatures large portions of proteins. c. It distributes a large negative charge throughout the protein. d. It neutralizes any charged residues on the protein and denatures large portions of proteins. e. It distributes a large negative charge throughout the protein after it has been denatured.

ANSWER: e 49. What are the structural characteristics common to all amino acids found in naturally occurring proteins?

ANSWER: All amino acids found in naturally occurring proteins have an α carbon to which are attached a carboxylic acid, an amine, a hydrogen, and a variable side chain. All the amino acids are also in the L configuration.

50. Briefly describe the five major groupings of amino acids.

ANSWER: Amino acids may be categorized by the chemistry of their R groups: (1) nonpolar aliphatics; (2) polar, uncharged; (3) aromatic; (4) positively charged; (5) negatively charged. (See Fig. 3-4.)

51.

A Tyr-Lys-Met

B Gly-Pro-Arg

C Asp-Trp-Tyr

D Asp-His-Glu

E Leu-Val-Phe

Which of the tripeptides: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 03: Amino Acids, Peptides, and Proteins ____ (a) is most negatively charged at pH 7? ____ (b) contains the amino acid with the simplest structure? ____ (c) contains the largest number of nonpolar R groups? ____ (d) contains sulfur? ____ (e) will have the greatest light absorbance at 280 nm?

ANSWER: (a) D; (b) B; (c) E; (d) A; (e) C 52. Describe the structures of the amino acids phenylalanine and aspartate in the ionization state expected at pH 7.0. Why is aspartate very soluble in water, whereas phenylalanine is much less soluble?

ANSWER: Aspartate has a polar (hydrophilic) side chain, which forms hydrogen bonds with water. In contrast, phenylalanine has a nonpolar (hydrophobic) side chain. (See Fig. 3-5.)

53. Why do amino acids, when dissolved in water, become zwitterions?

ANSWER: Near pH = 7, the carboxylic acid group (—COOH) will dissociate to become a negatively charged —COO– group, and the —NH2 amino group will attract a proton to become a positively charged — NH3+ group.

54. As more OH– equivalents (base) are added to an amino acid solution, what titration reaction will occur around pH = 9.5?

ANSWER: Around pH = 9.5, the —NH3+ group will be titrated according to the reaction —NH3+ + OH–→ — NH2 + H2O. 55. What is the pI, and how is it determined for amino acids that have nonionizable R groups?

ANSWER: The pI is the isoelectric point. It occurs at a characteristic pH when a molecule has an equal number of positive and negative charges, or no net charge. For amino acids with nonionizable R groups, pI is the arithmetic mean of a molecule's two pKa values: pI =

(pK1 + pK2)

56. The amino acid histidine has a side chain for which the pKa is 6.0. Calculate what fraction of the histidine side chains will carry a positive charge at pH 5.4. Be sure to show your work.

ANSWER: pH = pKa + log([acid]/[conjugate base]) pKa – pH = log([acid]/[conjugate base]) antilog (pKa – pH) = [acid]/[conjugate base] antilog (6.0 – 5.4) = [acid]/[conjugate base] 4 = [acid]/[conjugate base], or 4[conjugate base] = [acid] Therefore, at pH 5.4, four-fifths (80%) of the histidine will be in the protonated (acid) form.

57. How can a polypeptide have only one free amino group and one free carboxyl group?

ANSWER: This is possible only if the peptide has no side chains with carboxyl or amino groups. Then, with the exception of the single amino-terminal amino acid and the single carboxyl-terminal amino acid, all Copyright Macmillan Learning. Powered by Cognero.

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Chapter 03: Amino Acids, Peptides, and Proteins the other α-amino and carboxyl groups are covalently condensed into peptide bonds.

58. Hydrolysis of peptide bonds is an exergonic reaction. Why, then, are peptide bonds quite stable?

ANSWER: Peptide bonds are stable because hydrolysis of peptide (or amide) bonds has a high activation energy and as a result occurs very slowly.

59. What ionic form of Gly–Ala–Glu predominates at pH 7? Explain in terms of the structural component of the tripeptide.

ANSWER: The peptide must have an amino-terminal Gly residue, a carboxyl-terminal Glu residue, and ionized amino and carboxyl groups.

60. If the average molecular weight of the 20 standard amino acids is 138, why do biochemists divide a protein's molecular weight by 110 to estimate its number of amino acid residues?

ANSWER: For each peptide bond formed, a molecule of water is lost, bringing the average molecular weight down to 120. To reflect the preponderance of low molecular weight amino acids, the average molecular weight is lowered further to 110.

61. Lys residues make up 10.5% of the weight of ribonuclease. The ribonuclease molecule contains 10 Lys residues. Calculate the molecular weight of ribonuclease.

ANSWER: From the structure of lysine, we can calculate its molecular weight (146); when it condenses (loses H2O, Mr = 18 daltons) to form a peptide bond, the resulting residue contributes 146 – 18 = 128 daltons to the protein's molecular weight. If 10 Lys residues contribute 10.5% of the protein's molecular weight, each Lys residue is 1.05%. To calculate the total molecular weight, divide 128 daltons by 1.05% (0.0105); the result is 12,190 daltons. (The actual value is 13,700 daltons.)

62. Why do smaller molecules elute after large molecules when a mixture of proteins is passed through a sizeexclusion (gel-filtration) column?

ANSWER: The column matrix is composed of cross-linked polymers with pores of selected sizes. Smaller molecules can enter pores in the polymer beads from which larger molecules would be excluded.

Smaller molecules therefore have a larger three-dimensional space in which to diffuse, making their path through the column longer. Larger molecules migrate faster because they pass directly through the column, unhindered by the bead pores.

63. For each of these methods of separating proteins, describe the principle of the method, and tell what property of proteins allows their separation by this technique. (a) ion-exchange chromatography (b) size-exclusion (gel-filtration) chromatography (c) affinity chromatography ANSWER: (a) Ion-exchange chromatography separates proteins on the basis of their charges. (b) Size-exclusion or gel-filtration chromatography separates on the basis of size. (c) Affinity chromatography separates proteins with specific, high affinity for some ligand (attached to an inert support) from other proteins with no such affinity. (See Fig. 3-14.)

64. A biochemist is attempting to separate a DNA-binding protein (protein X) from other proteins in a solution.

Only three other proteins (A, B, and C) are present. The proteins have the properties: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 03: Amino Acids, Peptides, and Proteins pI (isoelectric point)

Size

Protein A B C X

Bind to DNA?

Mr 7.4 3.8 7.9 7.8

82,000 21,500 23,000 22,000

yes yes no yes

What type of protein separation techniques might the biochemist use to separate: (a) protein X from protein A? (b) protein X from protein B? (c) protein X from protein C? ANSWER: (a) Size-exclusion (gel-filtration) chromatography to separate on the basis of size; (b) ion-exchange chromatography or isoelectric focusing to separate on the basis of charge; (c) specific affinity chromatography, using immobilized DNA. 65. What factors would make it difficult to interpret the results of a gel electrophoresis of proteins in the absence of sodium dodecyl sulfate (SDS)?

ANSWER: SDS denatures a protein and distributes a large negative charge throughout the protein after it has been denatured. Without SDS, protein migration through a gel would be influenced by the protein's intrinsic net charge—which could be positive or negative—and its unique three-dimensional shape, in addition to its molecular weight. Thus, it would be difficult to ascertain the difference between proteins based upon a comparison of their mobilities in gel electrophoresis.

66. You are given a solution containing an enzyme that converts B into A. Describe what you would do to determine the specific activity of this enzyme solution.

ANSWER: First, add a known volume of the enzyme solution (say, 0.01 mL) to a solution of its substrate B and measure the initial rate at which product A is formed, expressed as μmol/mL of enzyme solution/min. Then measure the total protein concentration, expressed as mg/mL. Finally, divide the enzyme activity (μmol/min/mL) by the protein concentration (mg/mL); the quotient is the specific activity.

67. As a protein is purified, both the amount of total protein and the activity of the purified protein decrease.

Why, then, does the specific activity of the purified protein increase?

ANSWER: Specific activity is the units of enzyme activity (μmol of product/min) divided by the amount of protein (mg). As the protein is purified, some of it is lost in each step, resulting in a drop in activity.

However, other contaminating proteins are lost to a much greater extent. Therefore, with each purification step, the purified protein constitutes a greater proportion of the total, resulting in an increase in specific activity. (See also Table 3-3.)

68. Define the primary structure of a protein.

ANSWER: The primary structure of a protein is its unique sequence of amino acids and any disulfide bridges present in the native structure, that is, its covalent bond structure.

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Chapter 03: Amino Acids, Peptides, and Proteins 69. In one or two sentences, describe the usefulness of each of these reagents or reactions in the analysis of protein structure: (a) oxidation agent (performic acid) (b) protease (c) reducing agent (dithiothreitol or β-mercaptoethanol) ANSWER: (a) Used to break disulfide bonds or "bridges" and produce two cysteic acid residues; (b) used to produce specific peptide fragments from a polypeptide; (c) used to break disulfide bonds or "bridges" or to keep them from forming and to keep Cys residues in their reduced form.

70. A polypeptide is hydrolyzed, and it is determined that there are 3 Lys residues and 2 Arg residues (as well as other residues). How many peptide fragments can be expected when the native polypeptide is incubated with the proteolytic enzyme trypsin?

ANSWER: Six fragments would be expected, unless the carboxyl-terminal residue is Lys or Arg; in which case there would be five.

71. Conjugated proteins contain chemical substituents in addition to amino acids. List three classes of conjugated proteins and identify the type of prosthetic group associated with each one.

ANSWER: Any of these are acceptable answers: Lipoproteins, with lipid groups Glycoproteins, with carbohydrate groups Phosphoproteins, with phosphoryl groups Hemoproteins, with heme groups Flavoproteins, with flavin nucleotide groups Metalloproteins, with metal ions (zinc, iron, calcium, etc.)

72. Describe two major differences between chemical synthesis of polypeptides and synthesis of polypeptides in the living cell.

ANSWER: There are many such differences; here are a few. Chemical synthesis proceeds from carboxyl terminus to amino terminus; in the living cell, the process starts at the amino terminus and ends at the carboxyl terminus. In the living cell, synthesis occurs under physiological conditions; chemical synthesis does not. Chemical synthesis is only capable of synthesizing short polypeptides; cells can produce proteins of several thousand amino acids.

73. For the study of a protein in detail, an effort is usually made to first: a. conjugate the protein to a known molecule. b. determine its amino acid composition. c. determine its amino acid sequence. d. determine its molecular weight. e. purify the protein.

ANSWER: e 74. In a mixture of the five proteins listed, which should elute second in size-exclusion (gel-filtration) chromatography? a. cytochrome c, Mr = 13,000 b. immunoglobulin G, Mr = 145,000 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 03: Amino Acids, Peptides, and Proteins c. ribonuclease A, Mr = 13,700 d. RNA polymerase, Mr = 450,000 e. serum albumin, Mr = 68,500 ANSWER: b 75. By adding SDS (sodium dodecyl sulfate) during the electrophoresis of proteins, it is possible to: a. determine a protein's isoelectric point. b. determine an enzyme's specific activity. c. determine the amino acid composition of the protein. d. preserve a protein's native structure and biological activity. e. separate proteins exclusively on the basis of molecular weight.

ANSWER: e 76. How can isoelectric focusing be used in conjunction with SDS gel electrophoresis?

ANSWER: Isoelectric focusing can separate proteins of the same molecular weight on the basis of differing isoelectric points. SDS gel electrophoresis can then separate proteins with the same isoelectric points on the basis of differing molecular weights. When combined in two-dimensional electrophoresis, a great resolution of large numbers of proteins can be achieved.

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Chapter 04: Protein Structure 1. Which interactions are NOT considered to be weak in proteins? a. hydrogen bonds b. hydrophobic interactions c. ionic bonds d. peptide bonds e. van der Waals forces ANSWER: d 2. The MOST important contribution to the stability of a protein's conformation appears to be the: a. entropy increase from the decrease of ordered water molecules forming a solvent shell around it. b. maximum entropy increase from ionic interactions between the ionized amino acids in a protein. c. sum of free energies of formation of many weak interactions among the hundreds of amino acids in a protein. d. sum of free energies of formation of many weak interactions between its polar amino acids and surrounding water. e. stabilizing effect of hydrogen bonding between the carbonyl group of one peptide bond and the amino group of another.

ANSWER: a 3. In an aqueous solution, protein conformation is determined by two major factors. One is the formation of the maximum number of hydrogen bonds. The other is the: a. formation of the maximum number of hydrophilic interactions. b. maximization of ionic interactions. c. minimization of entropy by the formation of a water solvent shell around the protein. d. placement of hydrophobic amino acid residues within the interior of the protein. e. placement of polar amino acid residues around the exterior of the protein.

ANSWER: d 4. Which statement is NOT an appropriate description for van der Waals interactions? a. They involve dipole-dipole interactions. b. Their strength depends on the distance between the two interacting atoms. c. They are highly specific. d. An individual van der Waals interaction does not contribute significantly to the stability of a protein. e. They can involve hydrophobic amino acids.

ANSWER: c 5. Which statement about intrinsically disordered proteins is true? a. They contain small hydrophobic cores. b. They represent misfolded conformations of cellular proteins. c. They have no stable three-dimensional structure and therefore have no cellular function. d. They are responsible for proteostasis.

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Chapter 04: Protein Structure e. They can interact with multiple protein-binding partners and are central to protein interaction networks.

ANSWER: e 6. Which backbone arrangement BEST represents that of two peptide bonds? a. Cα—N—Cα—C—Cα—N—Cα—C b. Cα—N—C—C—N—Cα c. C—N—Cα—Cα—C—N d. Cα—C—N—Cα—C—N e. Cα—Cα—C—N—Cα—Cα—C ANSWER: d 7. Which pairs of bonds within a peptide backbone show free rotation around both bonds? a. Cα—C and N—Cα b. C=O and N—C c. C=O and N—Cα d. N—C and Cα—C e. N—Cα and N—C ANSWER: a 8. Roughly how many amino acids are there in one turn of an α helix? a. 1 b. 2.8 c. 3.6 d. 4.2 e. 10 ANSWER: c 9. In an α helix, the R groups on the amino acid residues: a. alternate between the outside and the inside of the helix. b. are found on the outside of the helix spiral. c. cause only right-handed helices to form. d. generate the hydrogen bonds that stabilize the helix. e. stack within the interior of the helix.

ANSWER: b 10. An ideal α helix has a length per turn of 0.54 nm and a rise along the helix axis for each amino acid residue of 0.15 nm. About how many turns are there for a segment of an ideal α helix that contains 54 amino acids? a. 15 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 04: Protein Structure b. 8 c. 100 d. 29 e. less than 1 ANSWER: a 11. Which feature tends to prevent the formation of an α helix? a. aromatic residues spaced three or four residues apart b. a sequence with no Gly residues c. a positively charged Lys residue at the carboxyl terminal of a helix d. the presence of a Pro residue e. two Ala residues side by side ANSWER: d 12. The three-dimensional conformation of a protein may be strongly influenced by amino acid residues that are very far apart in sequence. This relationship is in contrast to secondary structure, where the amino acid residues are: a. always side by side. b. generally near each other in sequence. c. invariably restricted to about 7 of the 20 standard amino acids. d. often on different polypeptide strands. e. usually near the polypeptide chain's amino terminus or carboxyl terminus.

ANSWER: b 13. Which statement is false? a. Collagen is a protein in which the polypeptides are mainly in the α-helical conformation. b. Disulfide linkages are important for keratin structure. c. Gly residues are particularly abundant in collagen. d. Silk fibroin is a protein in which the polypeptide is almost entirely in the β conformation. e. α-Keratin is a protein in which the polypeptides are mainly in the α-helical conformation.

ANSWER: a 14. Studies of the globular myoglobin structure demonstrated that: a. "corners" between α-helical regions invariably lacked proline residue. b. highly polar or charged amino acid residues tended to be located interiorally. c. as expected, myoglobin was completely different from hemoglobin. d. the structure was very compact, with virtually no internal space available for water. e. the α helix predicted by Pauling and Corey was not found in myoglobin.

ANSWER: d 15. Proteins often have regions that can fold and function as an independent entity from the whole protein.

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Chapter 04: Protein Structure These regions are called: a. domains. b. oligomers. c. peptides. d. sites. e. subunits.

ANSWER: a 16. Which statement concerning protein domains is true? a. They are a form of secondary structure. b. They are examples of structural motifs. c. They consist of separate polypeptide chains (subunits). d. They have been found only in prokaryotic proteins. e. They may retain their correct shape even when separated from the rest of the protein.

ANSWER: e 17. The structural classification of proteins (based on motifs) is based primarily on their: a. amino acid sequence. b. evolutionary relationships. c. function. d. secondary structure content and arrangement. e. subunit content and arrangement.

ANSWER: d 18. A repeating structural unit in a multimeric protein is known as a(n): a. domain. b. motif. c. oligomer. d. protomer. e. subunit.

ANSWER: d 19. Proteostasis is the cellular process by which protein(s): a. are synthesized. b. are folded. c. are modified. d. are degraded. e. levels are maintained.

ANSWER: e 20. An average protein will NOT be denatured by: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 04: Protein Structure a. a detergent such as sodium dodecyl sulfate. b. heating to 90 °C. c. iodoacetic acid. d. pH 10. e. urea.

ANSWER: c 21. Which factor is LEAST likely to result in protein denaturation? a. altering net charge by changing pH b. changing the salt concentration c. disruption of weak interactions by boiling d. exposure to detergents e. mixing with organic solvents such as acetone ANSWER: b 22. Experiments on denaturation and renaturation after the reduction and reoxidation of the —S—S— bonds in the enzyme ribonuclease have shown that: a. folding of denatured ribonuclease into the native, active conformation, requires the input of energy in the form of heat. b. native ribonuclease does not have a unique secondary and tertiary structure. c. the completely unfolded enzyme, with all —S—S— bonds broken, is still enzymatically active. d. the enzyme, dissolved in water, is thermodynamically stable relative to the mixture of amino acids whose residues are contained in ribonuclease. e. the primary sequence of ribonuclease is sufficient to determine its specific secondary and tertiary structure.

ANSWER: e 23. Which statement concerning the process of spontaneous folding of proteins is false? a. It may be an essentially random process. b. It may be defective in some human diseases. c. It may involve a gradually decreasing range of conformational species. d. It may involve initial formation of a highly compact state. e. It may involve initial formation of local secondary structure.

ANSWER: a 24. Protein S will fold into its native conformation only when protein Q is also present in the solution. However, protein Q can fold into its native conformation without protein S. Protein Q, therefore, may function as a _____ for protein S. a. proteasome b. molecular chaperone c. protein precursor Copyright Macmillan Learning. Powered by Cognero.

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Chapter 04: Protein Structure d. structural motif e. supersecondary structural unit ANSWER: b 25. Proteins in their functional, folded conformation are called _____ proteins. a. native b. unique c. intrinsic d. inherent e. natural ANSWER: a 26. Which group of amino acids would MOST likely be found in the core of a protein that is folded into a threedimensional structure and soluble in water? a. N, Y, and K b. I, M, and V c. V, T, and R d. M, S, and Y e. F, Y, and Q ANSWER: b 27. Which residues are MOST likely to be enriched in an intrinsically disordered protein that is soluble in water? a. Y, K, W b. E, C, F c. V, I, L d. R, K, A e. H, F, W ANSWER: d 28. Using the Ramachandran plot, identify the secondary structure adopted by an amino acid with phi and psi angles of –90 and 60°, respectively.

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Chapter 04: Protein Structure

a. right-handed α helix b. left-handed α helix c. β sheet d. twisted β sheets e. Amino acids will not adopt this conformation.

ANSWER: c 29. Using the Ramachandran plot, identify the secondary structure adopted by an amino acid with phi and psi angles of –90 and –180°, respectively.

a. right-handed α helix b. left-handed α helix c. β sheet d. twisted β sheets Copyright Macmillan Learning. Powered by Cognero.

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Chapter 04: Protein Structure e. Amino acids will not adopt this conformation.

ANSWER: c 30. Which amino acid is MOST likely to be found outside the highlighted/shaded regions on a Ramachandran plot? a. Met b. Cys c. Ala d. Gly e. His ANSWER: d 31. The _____ secondary structure is formed more readily than other types of secondary structures, most likely due to _____. a. β-turn; hydrogen bonds with solvent molecules b. β-sheet; R-group contacts c. β-sheet; internal hydrogen bonds d. α-helical; R-group contacts e. α-helical; internal hydrogen bonds ANSWER: e 32. This secondary structure is an example of a(n):

a. parallel β sheet. b. antiparallel β sheet. c. right-handed α helix. d. left-handed α helix. e. β turn.

ANSWER: c 33. This secondary structure is an example of a(n):

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Chapter 04: Protein Structure

a. parallel β sheet. b. antiparallel β sheet. c. right-handed α helix. d. left-handed α helix. e. β turn.

ANSWER: b 34. Proline residues are infrequently found in _____ due to their _____. a. α helices; decreased ability to serve as hydrogen-bond donors b. α helices; large positive charge that disrupts the repeating structure c. β sheets; decreased ability to serve as hydrogen-bond donors d. β sheets; large positive charge that disrupts the repeating structure e. β turns; decreased flexibility as an amino acid ANSWER: a 35. Long-range interactions between residues on a single polypeptide chain could BEST be classified as _____ structure. a. primary b. secondary c. tertiary d. quaternary e. globular ANSWER: c 36. Interactions between residues on separate polypeptide chains could be BEST classified as _____ structure. a. primary b. secondary c. tertiary d. quaternary e. global ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 04: Protein Structure 37. A _____ protein will often have properties that allow it to be both strong and flexible. a. macromolecular b. fibrous c. globular d. helical e. membrane ANSWER: b 38. The dimensions of an α helix from a fibrous protein, such as keratin, differ slightly from those in a globular protein. What is the reason for this difference in helical dimensions? a. An α helix from a fibrous protein is left-handed, not right-handed. b. An α helix from a fibrous protein contains mainly hydrophobic residues, elongating the helix. c. An α helix from a fibrous protein contains mainly polar residues, elongating the helix. d. An α helix from a fibrous protein forms a coiled coil, distorting the helix. e. An α helix from a fibrous protein contains mainly Gly residues for more efficient packing.

ANSWER: d 39. A lack of ascorbic acid (vitamin C) will result in reduced stability of: a. keratin. b. collagen. c. fibroin. d. myoglobin. e. None of the answers are correct.

ANSWER: b 40. As humans age, their connective tissue, which is rich in collagen, becomes more rigid and brittle. What is the molecular cause of this change? a. Collagen fibrils change their degree of supercoiling over time. b. Many older adults have vitamin C deficiencies that result in collagen oxidation. c. As cells age, a variant of collagen is expressed. d. Cross-links between collagen fibrils accumulate over time. e. UV damage to collagen accumulates over time.

ANSWER: d 41. Silk fibers have a low ability to stretch. Wool fibers have a high ability to stretch. The difference between the stretching ability of silk and wool fibers is due to their protein components. Which statement correctly attributes the secondary structure contributions to this stretch ability? a. Both silk and wool fibers have roughly equal distributions of a helices and β sheets, but they are arranged in different conformations. b. Silk is mainly β sheets; wool is mainly α helices. c. Silk is mainly α helices; wool is mainly β sheets.

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Chapter 04: Protein Structure d. Silk contains mainly β barrels; wool contains mainly coiled coil α helices. e. Silk contains coiled coil α helices; wool contains mainly β barrels.

ANSWER: b 42. Which method would be MOST useful to solve the atomic resolution structure of a protein that happens to easily form a repeating structure? a. NMR b. x-ray crystallography c. mass spectrometry d. circular dichroism e. electron microscopy ANSWER: b 43. Which statement about protein folding is NOT true? a. Burial of hydrophobic residues requires at least two layers of secondary structure. b. Generally, α helices and β sheets are found in different layers of a protein structure. c. Residues close together in tertiary structure are always close together in primary structure. d. The β conformation is most stable in a right-handed twist. e. The backbone of a β sheet cannot readily hydrogen-bond to an adjacent α helix.

ANSWER: c 44. What is the correct way to classify this protein according to its secondary structural elements?

a. all β b. all α c. α / β d. β / α / β e. α / β / α ANSWER: c 45. Intrinsically disordered proteins may: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 04: Protein Structure a. wrap around their targets. b. have multiple interaction partners. c. be at the center of protein interaction networks. d. lack a hydrophobic core. e. All of the answers are correct.

ANSWER: e 46. If a protein is not folded correctly or becomes partially unfolded, what could NOT be a consequence? a. The protein may form an inactive aggregate that leads to disease. b. The protein may be remodeled by a chaperone. c. The protein may be degraded by the proteasome. d. The protein may be refolded. e. All of these consequences are possible.

ANSWER: e 47. Which disease is NOT one characterized by or associated with an unfolded protein aggregate? a. Alzheimer disease b. Huntington's disease c. Parkinson disease d. scurvy e. All of these diseases are linked to unfolded protein aggregates.

ANSWER: d 48. Any given protein is characterized by a unique amino acid sequence (primary structure) and threedimensional (tertiary) structure. How are these related?

ANSWER: The three-dimensional structure is determined by the amino acid sequence. This means that the amino acid sequence contains all of the information that is required for the polypeptide chain to fold up into a discrete three-dimensional shape.

49. Name four factors (bonds or other forces) that contribute to stabilizing the native structure of a protein, and describe one condition or reagent that interferes with each type of stabilizing force.

ANSWER: Any of the following forces stabilize native protein structures and are disrupted by the listed conditions or reagents: (a) disulfide bonds by reducing conditions or mercaptoethanol or dithiothreitol, (b) hydrogen bonds by pH extremes or urea, (c) hydrophobic interactions by detergents or urea or guanidine hydrochloride, (d) ionic interactions by changes in pH or ionic strength, and (e) van der Waals interactions by any unfolding condition.

50. When a polypeptide is in its native conformation, there are weak interactions between its R groups.

However, when it is denatured, there are similar interactions between the protein groups and water. What then accounts for the greater stability of the native conformation?

ANSWER: In the unfolded polypeptide, there are ordered solvation shells of water around the protein groups.

The number of water molecules involved in such ordered shells is reduced when the protein folds, resulting in higher entropy. Hence, the lower free energy and greater stability of the native Copyright Macmillan Learning. Powered by Cognero.

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Chapter 04: Protein Structure conformation.

51. Describe the resonance structure of a peptide bond, and explain why there is no rotation around the C—N bond.

ANSWER: The intermediate resonance structure imparts a partial double-bond characteristic to the C—N bond, thereby prohibiting rotation. (See Fig. 4-3.)

52. Provide a brief explanation for this statement: Soluble globular proteins can be distinguished from soluble intrinsically disordered proteins based on their amino acid content.

ANSWER: Soluble globular proteins contain significant percentages of hydrophobic amino acids that are usually clustered to facilitate packing into the core of the structure. Intrinsically disordered proteins are enriched in charged amino acids and small amino acids such as Gly and Ala. 53. Where are the hydrogen bonding typically found between two residues in an a helix?

ANSWER: Hydrogen bonds occur between every carbonyl oxygen in the polypeptide backbone and the peptide —NH of the fourth amino acid residue toward the amino terminus of the chain. (See Fig. 4-5.)

54. Describe three of the important features of the α-helical polypeptide structure. Provide one or two sentences for each feature.

ANSWER: The α-helical structure of a polypeptide is tightly wound around a long central axis. Each turn of the right-handed helix contains 3.6 residues and stretches 5.4 Å along the axis. The peptide NH is hydrogen-bonded to the carbonyl oxygen of the fourth amino acid along the sequence toward the amino terminus. The R groups of the amino acid residues protrude outward from the helical backbone.

55. Describe three of the important features of a β-sheet polypeptide structure. Provide one or two sentences for each feature.

ANSWER: In the β-sheet structure, several extended polypeptides, or two regions of the same polypeptide, lie side by side and are stabilized by hydrogen bonding between adjacent chains. Adjacent chains may be either parallel (with a repeat distance of about 6.5 Å) or antiparallel (7 Å repeat). The R groups are often small and alternately protrude from opposite faces of the β sheet.

56. Why is silk fibroin so strong, but at the same time so soft and flexible?

ANSWER: Unlike collagen and keratin, silk fibroin has no covalent cross-links between adjacent strands or between its stacked sheets. The lack of cross-links make fibroin very flexible. Fibroin's unusual tensile strength derives from the fact that the peptide backbone of antiparallel β strands is fully extended and that the R groups in the stacked pleated sheets interdigitate. These features prevent any longitudinal sliding of the sheets across one another.

57. What is typically found in the interior of a water-soluble globular protein?

ANSWER: Hydrophobic amino acid residues cluster away from the surface in globular proteins. So, much of the protein's interior is a tightly packed combination of hydrocarbon and aromatic ring R groups with very few water molecules.

58. Explain what is meant by motifs in protein structure.

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Chapter 04: Protein Structure ANSWER: Motifs are particularly stable arrangements of elements of secondary structure (e.g., α helix and β conformation), including the connections between them. They are found in a variety of proteins.

59. Explain (succinctly) the theoretical and/or experimental arguments in support of this statement: The primary sequence of a protein determines its three-dimensional shape and thus its function.

ANSWER: Experiments on the denaturation and renaturation of ribonuclease A (RNase A) showed that a completely denatured enzyme (ribonuclease) could fold spontaneously into its native, enzymatically active form with only the primary sequence to guide it.

60. Each of these reagents or conditions will denature a protein. For each, describe in one or two sentences what the reagent/condition does to destroy native protein structure. (a) urea (b) high temperature (c) detergent (d) low pH ANSWER: (a) Urea acts primarily by disrupting hydrophobic interactions. (b) High temperature provides thermal energy greater than the strength of the weak interactions (hydrogen bonds, electrostatic interactions, hydrophobic interactions, and van der Waals forces), breaking these interactions. (c) Detergents bind to hydrophobic regions of the protein, preventing hydrophobic interactions among several hydrophobic patches on the native protein. (d) Low pH causes protonation of the side chains of Asp, Glu, and His, preventing electrostatic interactions.

61. How can changes in pH alter the conformation of a protein?

ANSWER: Changes in pH can influence the extent to which certain amino acid side chains (or the amino and carboxyl termini) are protonated. The result is a change in net charge on the protein, which can lead to electrostatic attractions or repulsions between different regions of the protein. The final effect is a change in the protein's three-dimensional shape or even complete denaturation.

62. Once a protein has been denatured, how can it be renatured? If renaturation does not occur, what might be the explanation?

ANSWER: Because a protein may be denatured through the disruption of hydrogen bonds and hydrophobic interactions by salts or organic solvents, removal of those conditions will reestablish the original aqueous environment. The removal of these conditions will often permit the protein to fold once again into its native conformation. If the protein does not renature, it may be because the denaturing treatment removed a required prosthetic group, or because the normal folding pathway requires the presence of a polypeptide chain-binding protein or molecular chaperone. The normal folding pathway could also be mediated by a larger polypeptide, which is then cleaved.

63. What are two mechanisms by which "chaperone" proteins assist in the correct folding of polypeptides?

ANSWER: Chaperones protect unfolded polypeptides from aggregation by binding to hydrophobic regions.

They can also provide a microenvironment that promotes correct folding.

64. What important concepts regarding protein thermal denaturation can be inferred from the egg white of a boiled egg?

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Chapter 04: Protein Structure ANSWER: Denatured proteins often precipitate and/or aggregate. Denaturation is often not a reversible process.

65. Identify and quantify the secondary structures present in this protein.

ANSWER: This peptide contains four α helices and two β sheets.

66. A protein is highly unlikely to have both ϕ and ψ angles equal to 0° due to: a. steric hindrance between the side chain and the peptide backbone. b. polarity differences between adjacent side chains. c. hydrophobic interactions between adjacent side chains. d. steric hindrance between adjacent side chains. e. too few van der Waals contacts for proper folding.

ANSWER: a

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Chapter 05: Protein Function and Ligand Binding 1. The interactions of ligands with proteins: a. are relatively nonspecific. b. are relatively rare in biological systems. c. are usually irreversible. d. are usually transient. e. usually result in the inactivation of the proteins.

ANSWER: d 2. A prosthetic group of a protein is a nonprotein structure that is _____ the protein. a. a ligand of b. a part of the secondary structure of c. a substrate of d. permanently associated with e. transiently bound to ANSWER: d 3. When oxygen binds to a heme-containing protein, the two open coordination bonds of Fe2+ are occupied by: a. one O atom and one amino acid atom. b. one O2 molecule and one amino acid atom. c. one O2 molecule and one heme atom. d. two O atoms. e. two O2 molecules.

ANSWER: b 4. In the binding of oxygen to myoglobin, the relationship between the concentration of oxygen and the fraction of binding sites occupied can BEST be described as: a. hyperbolic. b. linear with a negative slope. c. linear with a positive slope. d. random. e. sigmoidal.

ANSWER: a 5. Which statement about protein-ligand binding is correct? a. The Ka is equal to the concentration of ligand when all of the binding sites are occupied. b. The Ka is independent of such conditions as salt concentration and pH. c. The larger the Ka, the weaker the affinity. d. The larger the Ka, the faster is the binding. e. The larger the Ka, the smaller the Kd (dissociation constant).

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Chapter 05: Protein Function and Ligand Binding ANSWER: e 6. Myoglobin and the subunits of hemoglobin have: a. no obvious structural relationship. b. very different primary and tertiary structures. c. very similar primary and tertiary structures. d. very similar primary structures, but different tertiary structures. e. very similar tertiary structures, but different primary structures.

ANSWER: e 7. Calculate the fraction of binding sites occupied by a ligand if the equilibrium concentrations of protein-ligand complex and free protein are 8 mM and 12 mM, respectively. a. 0.67 b. 0.40 c. 0.60 d. 0.75 e. 0.50 ANSWER: b 8. Consider a protein-ligand equilibrium where P + L ⇌ PL. What happens to the ratio of protein-ligand complex to free protein if the equilibrium concentration of free ligand is doubled? Note: P = protein; L = ligand; PL = protein-ligand complex. a. The ratio decreases by a factor of four. b. The ratio increases by a factor of four. c. The ratio is doubled. d. The ratio is halved. e. There is no effect on the ratio.

ANSWER: e 9. Examine the binding curves for two proteins (A and B) that bind the same ligand. Which statement applies?

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Chapter 05: Protein Function and Ligand Binding a. The dissociation constant for A is less than that of B. b. B has more than one type of ligand that can bind. c. B has a stronger affinity for the ligand than protein A does. d. Protein A is oligomeric. e. Protein A has multiple binding sites for the ligand.

ANSWER: a 10. An allosteric interaction between a ligand and a protein is one in which: a. the binding of a molecule to a binding site affects binding of additional molecules to the same site. b. the binding of a molecule to a binding site affects binding properties of another site on the protein. c. the binding of the ligand to the protein is covalent. d. multiple molecules of the same ligand can bind to the same binding site. e. two different ligands can bind to the same binding site.

ANSWER: b 11. In hemoglobin, the transition from the T state to R state (low to high affinity) is triggered by: a. Fe2+ binding. b. heme binding. c. oxygen binding. d. subunit association. e. subunit dissociation.

ANSWER: c 12. What is the biological advantage to the sigmoidal binding curve of hemoglobin for oxygen? a. It ensures that hemoglobin has a high affinity for oxygen. b. It allows hemoglobin to bind oxygen irreversibly. c. It ensures that hemoglobin can bind oxygen only weakly. d. It allows hemoglobin to shift between low and high affinities for oxygen. e. It allows hemoglobin to bind oxygen at varying pH. ANSWER: d 13. Carbon monoxide (CO) is toxic to humans because it: a. binds to myoglobin and causes it to denature. b. is rapidly converted to toxic CO2. c. binds to the globin portion of hemoglobin and prevents the binding of O2. d. binds to the Fe in hemoglobin and prevents the binding of O2. e. binds to the heme portion of hemoglobin and causes heme to unbind from hemoglobin.

ANSWER: d 14. The amino acid substitution of Val for Glu in hemoglobin S results in aggregation of the protein because of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 05: Protein Function and Ligand Binding _____ interactions between molecules. a. covalent b. disulfide c. hydrogen-bonding d. hydrophobic e. ionic ANSWER: d 15. The fundamental cause of sickle cell disease is a change in the structure of: a. blood. b. capillaries. c. hemoglobin. d. red cells. e. the heart.

ANSWER: c 16. The binding site of a protein is complementary to a specific ligand due to which characteristic of the binding site? a. its size b. its shape c. its charge d. its hydrophobicity e. All of the answers are correct.

ANSWER: e 17. Which molecule binds MOST strongly to the heme iron? a. N2 b. O2 c. CO2 d. CO e. H2O ANSWER: d 18. Hemoglobin is involved in the transport of which ion and/or molecule through the blood? a. CO2 b. O2 c. H+ d. both CO2 and O2 e. CO2, O2, and H+ Copyright Macmillan Learning. Powered by Cognero.

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Chapter 05: Protein Function and Ligand Binding ANSWER: e 19. A Hill coefficient (nH) of 2 indicates that there: a. is positive cooperativity in ligand binding. b. are two ligand-binding sites. c. is negative cooperativity in ligand binding. d. is positive cooperativity in ligand binding and there are two ligand-binding sites. e. is negative cooperativity in ligand binding and there are two ligand-binding sites.

ANSWER: a 20. Sickle cell anemia is caused by a single mutation in the hemoglobin protein from a Glu to a Val, which results in which alteration in the hemoglobin subunits? a. more neutral charges and formation of a nucleophilic contact point b. fewer negative charges and formation of a hydrophobic contact point c. fewer positive charges and formation of a hydrophobic contact point d. more negative charges and formation of a hydrophilic contact point e. more positive charges and formation of a hydrophilic contact point ANSWER: b 21. Two proteins bind to the same ligand, and protein A has half of its binding sites occupied when the ligand concentration is 0.5 mM, while protein B has half of its binding sites occupied when the ligand concentration is at 0.3 mM. Which protein binds the ligand more strongly, and what is the dissociation constant for that proteinligand interaction? a. Protein A binds more strongly, and Kd = 0.3 mM. b. Protein A binds more strongly, and Kd = 0.5 mM. c. Protein B binds more strongly, and Kd = 0.3 mM. d. Protein B binds more strongly, and Kd = 0.5 mM. e. Both proteins bind the ligand with the same affinity, and Kd = 0.4 mM. ANSWER: c 22. Which statement comparing hemoglobin and myoglobin is NOT correct? a. Hemoglobin binds O2 cooperatively, while myoglobin does not. b. Hemoglobin is a multisubunit protein, while myoglobin is not. c. Hemoglobin and myoglobin are both primarily composed of α helices. d. Hemoglobin has a stronger binding affinity for O2 than myoglobin does. e. Hemoglobin and myoglobin both contain the heme prosthetic group.

ANSWER: d 23. Describe the concept of induced fit in protein-ligand binding.

ANSWER: Induced fit refers to the structural adaptations that occur when a ligand binds to a protein. This often Copyright Macmillan Learning. Powered by Cognero.

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Chapter 05: Protein Function and Ligand Binding involves a conformational change in the protein that alters the binding site to make it more complementary to the ligand.

24. Explain why most multicellular organisms use an iron-containing protein for oxygen binding rather than free Fe2+. Your answer should include an explanation of (a) the role of heme and (b) the role of the protein itself.

ANSWER: (a) Binding of oxygen to free Fe2+ would result in the formation of reactive oxygen species that can damage biological structures. Heme-bound iron is less reactive in this regard. (b) Binding of oxygen to free heme can result in irreversible oxidation of the Fe2+ to Fe3+ that does not bind oxygen.

Situating the heme group in a protein environment helps to prevent this from occurring.

25. Describe how you would determine the Ka (association constant) for a ligand and a protein.

ANSWER: An experiment would be carried out in which a fixed amount of the protein is incubated with varying amounts of ligand for long enough to reach equilibrium. The fraction of protein molecules that have a molecule of ligand bound is then determined. A plot of this fraction (θ) versus ligand concentration [L] should yield a hyperbola. The value of [L] when θ= 0.5 is equal to 1/Ka. 26. Why is carbon monoxide (CO) toxic to aerobic organisms?

ANSWER: It binds to heme with a higher affinity than oxygen and thus prevents oxygen from binding to hemoglobin.

27. For the binding of a ligand to a protein, what is the relationship between the Ka (association constant), the Kd (dissociation constant), and the affinity of the protein for the ligand?

ANSWER: Ka = 1/Kd. The larger the Ka (and the smaller the Kd), the higher the affinity of the protein for the ligand.

28. What fraction of ligand binding sites are occupied (θ) when [ligand] = Kd? Show your work.

ANSWER: Half of the ligand binding sites are occupied when [ligand] = Kd. 29. Explain why the structure of myoglobin makes it function well as an oxygen-storage protein, whereas the structure of hemoglobin makes it function well as an oxygen-transport protein.

ANSWER: The geometry of O2 and CO binding to heme is slightly different. In myoglobin, there is a histidine residue that does not interact with the heme iron but can interact with a ligand that is bound to the heme. This histidine stabilizes bound O2 through hydrogen bonding but has less ability to stabilize CO binding because the CO-heme complex is less polar and sits at a slightly different angle. As a result, compared with free heme, the relative affinity of protein-bound heme for CO is greatly decreased and that for O2 is increased.

30. Explain briefly why the relative affinity of heme for oxygen and carbon monoxide is changed by the presence of the myoglobin protein.

ANSWER: The hyperbolic binding of oxygen to the single binding site of myoglobin results in a high affinity even at the relatively low partial pressures of O2 that occur in tissues. In contrast, the cooperative (sigmoidal) binding of O2 to the multiple binding sites of hemoglobin results in high affinity at high Copyright Macmillan Learning. Powered by Cognero.

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Chapter 05: Protein Function and Ligand Binding partial pressures such as occur in the lungs, but lower affinity in the tissues. This permits hemoglobin to bind O2 in the lungs and release it in the tissues.

31. Explain how the effects of sickle cell disease demonstrate that hemoglobin undergoes a conformational change upon releasing oxygen.

ANSWER: In hemoglobin S, the wild-type glutamate at residue 6 of the β chain is replaced by valine. When oxygen is bound, both hemoglobin A and hemoglobin S are soluble. However, due to conformational changes in the protein that occur when oxygen is released, the hydrophobic valine residue in hemoglobin S is exposed, and the deoxy form of the protein becomes very insoluble. This exposed hydrophobic "patch" causes aggregation of deoxyhemoglobin S into long insoluble fibrous aggregates, resulting in red blood cells with distorted shapes (and leading to the symptoms of the disease). (Fig. 5-20) 32. Describe how the cooperativity of oxygen binding in hemoglobin is advantageous for the transport of oxygen from the lungs to the tissues and why myoglobin would not be a good protein to use for oxygen transport.

ANSWER: Cooperativity of oxygen binding in hemoglobin means that, when there is not much oxygen bound to the hemoglobin protein, binding of additional oxygen will not be enhanced. This is the case in the tissues, where there are low oxygen levels. In the lungs, however, oxygen levels are high, and the binding of some oxygen to hemoglobin will enhance binding of additional oxygen molecules. In this way, at high oxygen concentrations, such as in the lungs, hemoglobin will bind several oxygen molecules. When in the tissues, however, hemoglobin will release oxygen since it is at a lower concentration, and by releasing oxygen, the affinity of hemoglobin for the other bound oxygens will decrease, leading to more effective release of oxygen in the tissues. Myoglobin, which does not exhibit cooperativity in oxygen binding, will bind oxygen very well in the lungs but will not release oxygen in the tissues, thus being a poor protein to use for oxygen transport.

33. Which equation describes the quantitative relationship between the dissociation constant and the fraction of ligand-binding sites on the protein that are occupied by ligand? a.

Ka = [PL] b.

c.

d.

e.

Ka =

ka

Y=

[PL]

Y=

[L]

Y=

[L]

ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 05: Protein Function and Ligand Binding 34. A/an ______ binds to the binding site in a protein. a. antibody b. cofactor c. globin d. ligand e. target ANSWER: d 35. The term Ka represents the: a. association equilibrium constant. b. association rate constant. c. dissociation equilibrium constant. d. dissociation rate constant. e. equilibrium expression.

ANSWER: a 36. Which compounds consist of four pyrrole rings linked by methene bridges? a. coordination bonds b. globins c. hemoglobins d. myoglobins e. porphyrins ANSWER: e 37. Which protein-ligand pair has the greatest affinity?

Protein

Ligand Kd(M)a

Avidin (egg white)

Biotin

Insulin Receptor (human)

Insulin 1×10-10

Nickel-binding protein (Escherichia coli)

Ni2+

1×10-7

Calmodulin (rat)b

Ca2+

3×10-6, 2×10-5

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Chapter 05: Protein Function and Ligand Binding

a. avidin-biotin b. insulin receptor-insulin c. nickel-binding protein-Ni2+ d. calmodulin-Ca2+ e. A pair with Kd = 2 x 10-10 M ANSWER: a 38. Which compounds are a specialized family of transport proteins containing heme? a. antigens b. antibodies c. globins d. hemes e. porphyrins ANSWER: c 39. Show your work for calculating the Kd of a reaction with Ka = 2 nM. ANSWER: Kd = 1/ Ka Kd = 1 / 2 nM 40. The T state: a. is most stable when oxygen is bound. b. is most stable when oxygen is not bound. c. is the high-affinity state for hemoglobin. d. is the predominant conformation of oxyhemoglobin. e. transitions to the R state when oxygen becomes unbound.

ANSWER: b 41. For a protein with n binding sites, the expression for ligand-binding sites on the protein that are occupied by ligand is: a.

Ka = [PL]n

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Chapter 05: Protein Function and Ligand Binding b.

c.

d.

e.

Y=

[L]n

Y=

[PL]

Y = [L]

Y = [L]n

ANSWER: e 42. Hemoglobin has: a. one globin unit and one heme prosthetic group. b. two globin units and two heme prosthetic groups. c. two globin units and four heme prosthetic groups. d. four globin units and four heme prosthetic groups. e. four globin units and eight heme prosthetic groups.

ANSWER: d 43. Deoxyhemoglobin is: a. hemoglobin with oxygen molecules bound to the globin groups. b. hemoglobin without oxygen molecules bound to the globin groups. c. hemoglobin with oxygen molecules bound to the heme groups. d. hemoglobin without oxygen molecules bound to the heme groups. e. myoglobin with oxygen molecules bound to the heme groups.

ANSWER: d 44. Describe the structural differences between myoglobin and hemoglobin.

ANSWER: Myoglobin is a monomer; it has one globin unit and one heme prosthetic group. The monomer consists of one peptide chain. Hemoglobin is a tetramer; it has four globin units and four heme prosthetic groups. Each globin unit is formed from one peptide chain, so hemoglobin has four peptide chains. Two of the peptide chains are α chains and two are β chains. Also, hemoglobin has a quaternary structure, while myoglobin, being a monomer, does not have a quaternary structure.

45. Carbon monoxide can lead to death when only ______ of the hemoglobin binding sites are occupied by CO molecules. a. 10-20% b. 20-30% c. 30-40% d. 40-50% Copyright Macmillan Learning. Powered by Cognero.

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Chapter 05: Protein Function and Ligand Binding e. 50-60% ANSWER: e 46. When CO combines with hemoglobin, the complex is referred to as: a. carboxyhemoglobin b. carbomonoxyhemoglobin c. decarboxyhemoglobin d. deoxyhemoglobin monoxide e. hemoxycarbon monoxide ANSWER: a 47. A/An _____ is an alternative form of a gene that arises by mutation. a. allele b. genome c. chromosome d. anemia e. residue ANSWER: a 48. Because the hemoglobin content of persons living with sickle cell disease is only about half the normal value, these persons present with symptoms of ______. a. anemia b. carboxyhemoglobin c. CO poisoning d. CRISPR e. deoxyHbS ANSWER: a 49. At COHb levels of 20% to 30%, a. symptoms are rarely observed. b. the individual experiences mild headaches. c. the individual experiences severe headaches that are generally accompanied by nausea, dizziness, confusion, disorientation, and some visual disturbances. d. the individual experiences severe neurological symptoms. e. the individual loses consciousness and can sink into coma.

ANSWER: c 50. Describe how CO binding of hemoglobin increases the globin’s affinity for O2.

ANSWER: CO binding facilitates hemoglobin’s transition from the T state to the R state. So, as one or two subunits of a hemoglobin tetramer bind CO, the other subunits can more efficiently bind O2.

Therefore, the affinity for O2 is increased in these remaining subunits.

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Chapter 05: Protein Function and Ligand Binding

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Chapter 06: Protein Function and Enzymes 1. One of the enzymes involved in glycolysis, aldolase, requires Zn2+ for catalysis. When the enzyme lacks zinc, it is referred to as the: a. apoenzyme. b. coenzyme. c. holoenzyme. d. prosthetic group. e. substrate.

ANSWER: a 2. Enzymes are potent catalysts because they: a. are consumed in the reactions they catalyze. b. are very specific and can prevent the conversion of products back to substrates. c. drive reactions to completion while other catalysts drive reactions to equilibrium. d. increase the equilibrium constants for the reactions they catalyze. e. lower the activation energy for the reactions they catalyze.

ANSWER: e 3. The role of an enzyme in an enzyme-catalyzed reaction is to: a. bind a reaction intermediate and prevent it from converting back to substrate. b. ensure that all of the substrate is converted to product. c. ensure that the product is more stable than the substrate. d. increase the rate at which substrate is converted into product. e. make the free-energy change for the reaction more favorable.

ANSWER: d 4. Which statement is true of enzyme catalysts? a. Their catalytic activity is independent of pH. b. They are generally equally active on D and L isomers of a given substrate. c. They can increase the equilibrium constant for a given reaction by a 1,000-fold or more. d. They can increase the reaction rate for a given reaction by many orders of magnitude. e. To be effective, they must be present at the same concentration as their substrate.

ANSWER: d 5. Which statement about enzyme catalysts is false? a. They sometimes bind covalently to substrates or intermediates during catalysis. b. They increase the equilibrium constant for a reaction, thus favoring product formation. c. They increase the stability of the transition state. d. They have a high level of specificity for their substrates. e. They can be effective, even at a concentration much lower than that of their substrate.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 06: Protein Function and Enzymes 6. Which statement is false? a. A reaction may not occur at a detectable rate even though it has a favorable equilibrium. b. After a reaction, the enzyme involved becomes available to catalyze the reaction again. c. For S→P, a catalyst shifts the reaction equilibrium to the right. d. Lowering the temperature of a reaction usually lowers the reaction rate. e. Substrate binds to an enzyme's active site.

ANSWER: c 7. Enzymes differ from other catalysts in that only enzymes: a. are not consumed in the reaction. b. display specificity toward a single reactant. c. fail to influence the equilibrium point of the reaction. d. form an activated complex with the reactants. e. lower the activation energy of the reaction catalyzed.

ANSWER: b 8. Compare the two reaction coordinate diagrams and select the answer that correctly describes their relationship. In each case, the single intermediate is the ES complex.

a. Diagram (a) describes a strict lock and key model, whereas (b) describes a transition-state complementarity model. b. The activation energy for the catalyzed reaction is arrow 5 in (a) and arrow 7 in (b). c. The activation energy for the uncatalyzed reaction is given by the sum of arrows 5 and 6 in (a) and by the sum of arrows 7 and 8 in (b). d. The contribution of binding energy is given by arrow 5 in (a) and by arrow 7 in (b). e. The ES complex is given by point 2 in (a) and point 3 in (b).

ANSWER: a 9. Which statement regarding the binding energy derived from enzyme-substrate interactions is true? a. It cannot provide enough energy to explain the large rate accelerations brought about by enzymes. b. It is sometimes used to hold two substrates in the optimal orientation for reaction. c. It is the result of covalent bonds formed between enzyme and substrate.

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Chapter 06: Protein Function and Enzymes d. Most of it is derived from covalent bonds between enzyme and substrate. e. Most of it is used up simply binding the substrate to the enzyme.

ANSWER: b 10. The concept of induced fit refers to the fact that: a. enzyme specificity is induced by enzyme-substrate binding. b. enzyme-substrate binding induces an increase in the reaction entropy, thereby catalyzing the reaction. c. enzyme-substrate binding induces movement along the reaction coordinate to the transition state. d. substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation. e. when a substrate binds to an enzyme, the enzyme induces a loss of water (desolvation) from the substrate.

ANSWER: d 11. In this diagram of the first step in the reaction catalyzed by the protease chymotrypsin, the process of general base catalysis is illustrated by the number _____, and the process of covalent catalysis is illustrated by the number _____.

a. 1; 2 b. 1; 3 c. 2; 3 d. 2; 1 e. 3; 2 ANSWER: a 12. The benefit of measuring the initial rate of a reaction (V0) is that at the beginning of a reaction: a. [ES] can be measured accurately. b. changes in [S] are negligible, so [S] can be treated as a constant. c. changes in Km are negligible, so Km can be treated as a constant. d. V0 = Vmax. e. varying [S] has no effect on V0.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 06: Protein Function and Enzymes 13. Which statement about a plot of V0 versus [S] for an enzyme that follows Michaelis-Menten kinetics is false? a. As [S] increases, the initial velocity of reaction V0 also increases. b. At very high [S], the velocity curve becomes a horizontal line intersecting the y-axis at Km. c. Km is the [S] at which V0 = 1/2Vmax. d. The shape of the curve is a hyperbola. e. The y-axis is a rate term with units such as mmol/min.

ANSWER: b 14. Michaelis and Menten assumed that the overall reaction for an enzyme-catalyzed reaction could be written as

Using this reaction, the rate of breakdown of the enzyme-substrate complex can be described by the expression: a. k1([Et] – [ES]). b. k1([Et] – [ES])[S]. c. k2[ES]. d. k–1[ES] + k2[ES]. e. k–1[ES].

ANSWER: d 15. The steady state assumption, as applied to enzyme kinetics, implies: a. Km = Ks. b. the enzyme is regulated. c. the ES complex is formed and broken down at equivalent rates. d. the Km is equivalent to the cellular substrate concentration. e. the maximum velocity occurs when the enzyme is saturated.

ANSWER: c 16. An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 μmol. If, in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (12 μmol) of product to be formed? a. 1.5 min b. 13.5 min Copyright Macmillan Learning. Powered by Cognero.

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Chapter 06: Protein Function and Enzymes c. 27 min d. 3 min e. 6 min ANSWER: c 17. These data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics.

V0 (μmol/min)

Substrate added (mmol/L)

217 325 433 488 647

0.8 2 4 6 1,000

The Km for this enzyme is approximately: a. 1 mM. b. 1,000 mM. c. 2 mM. d. 4 mM. e. 6 mM. ANSWER: c 18. For enzymes in which the slowest (rate-limiting) step is the reaction

Km becomes equivalent to: a. kcat. b. the [S], where V0 = Vmax. c. the dissociation constant, Kd, for the ES complex. d. the maximal velocity. e. the turnover number.

ANSWER: c 19. The number of substrate molecules converted to product in a given unit of time by a single enzyme molecule at saturation is referred to as the: a. dissociation constant. b. half-saturation constant. c. maximum velocity. d. Michaelis-Menten number. e. turnover number.

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Chapter 06: Protein Function and Enzymes ANSWER: e 20. In competitive inhibition, an inhibitor: a. binds at several different sites on an enzyme. b. binds covalently to the enzyme. c. binds only to the ES complex. d. binds reversibly at the active site. e. lowers the characteristic Vmax of the enzyme.

ANSWER: d 21. Enzyme X exhibits maximum activity at pH = 6.9. X shows a fairly sharp decrease in its activity when the pH goes much lower than 6.4. One likely interpretation of this pH activity is that: a. a Glu residue on the enzyme is involved in the reaction. b. a His residue on the enzyme is involved in the reaction. c. the enzyme has a metallic cofactor. d. the enzyme is found in gastric secretions. e. the reaction relies on specific acid-base catalysis.

ANSWER: b 22. Phenyl-methane-sulfonyl-fluoride inactivates serine proteases by binding covalently to the catalytic serine residue at the active site; this enzyme-inhibitor bond is NOT cleaved by the enzyme. What type of inhibition does this represent? a. irreversible b. competitive c. noncompetitive d. mixed e. pH inhibition ANSWER: a 23. Penicillin and related drugs inhibit the enzyme _____; this enzyme is produced by _____. a. β-lactamase; bacteria b. transpeptidase; human cells c. transpeptidase; bacteria d. lysozyme; human cells e. aldolase; bacteria ANSWER: c 24. Which statement about allosteric control of enzymatic activity is false? a. Allosteric effectors give rise to sigmoidal V0 versus [S] kinetic plots. b. Allosteric proteins are generally composed of several subunits. c. An effector may either inhibit or activate an enzyme.

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Chapter 06: Protein Function and Enzymes d. Binding of the effector changes the conformation of the enzyme molecule. e. Heterotropic allosteric effectors compete with substrate for binding sites.

ANSWER: e 25. Allosteric enzymes: a. are regulated primarily by covalent modification. b. usually catalyze several different reactions within a metabolic pathway. c. usually have more than one subunit. d. usually have only one active site. e. usually show strict Michaelis-Menten kinetics.

ANSWER: c 26. How is trypsinogen converted to trypsin? a. A protein kinase-catalyzed phosphorylation converts trypsinogen to trypsin. b. An increase in Ca2+ concentration promotes the conversion. c. Proteolysis of trypsinogen forms trypsin. d. Trypsinogen dimers bind an allosteric modulator, cAMP, causing dissociation into active trypsin monomers. e. Two inactive trypsinogen dimers pair to form an active trypsin tetramer.

ANSWER: c 27. Which statement regarding enzyme activity is correct? a. Enzymes bind their substrates better than the transition state. b. Enzymes bind the products better than the reaction intermediates. c. Enzymes increase the activation energy required for the reaction to take place. d. Enzymes bind their substrates better than the transition state and the transition state better than the reaction products. e. Enzymes bind the transition state better than the reaction products and reduce the activation energy required for the reaction to take place.

ANSWER: e 28. Binding energy between an enzyme and a substrate contributes to catalysis in which way? a. Binding energy provides the enzyme specificity for the substrate. b. Binding energy contributions allow for entropy reduction in the substrate-enzyme complex. c. Binding energy compensates for energy changes as a result of desolvation of the substrate. d. Binding energy contributes to the process of induced fit between the substrate and the enzyme. e. All of the answers are correct.

ANSWER: e 29. If chemical reactions will eventually reach an equilibrium state, what is the purpose of enzymes in a biological system?

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Chapter 06: Protein Function and Enzymes a. Enzymes are consumed to speed up chemical reactions. b. Enzymes speed up chemical reactions without being used up in the process. c. Enzymes slow down chemical reactions. d. Enzymes alter the equilibrium state between reactants and products. e. Enzymes prevent the formation of unstable reaction intermediates.

ANSWER: b 30. In what way does an uncompetitive inhibitor bind to an enzyme? a. It reversibly binds to the enzyme active site. b. It irreversibly binds to the enzyme active site. c. It reversibly binds to the enzyme-substrate complex but does not bind to the free enzyme. d. It reversibly binds to both the free enzyme and the enzyme-substrate complex. e. It irreversibly binds to an enzyme allosteric site.

ANSWER: c 31. Treatment of methanol poisoning by using ethanol is an example of what type of enzyme inhibition? a. mixed b. uncompetitive c. competitive d. noncompetitive e. suicide ANSWER: c 32. An enzyme's specificity constant (kcat/Km) has a maximum upper limit due to: a. substrate concentration. b. enzyme concentration. c. the enzyme turnover number. d. diffusion of the enzyme and substrate together. e. the Michaelis constant.

ANSWER: d 33. Chymotrypsin catalyzes the cleavage of proteins at peptide bonds adjacent to aromatic amino acid residues by using which catalytic mechanism? a. general acid-base catalysis b. metal ion catalysis c. covalent catalysis d. both general acid-base catalysis and metal ion catalysis e. both general acid-base catalysis and covalent catalysis ANSWER: e 34. The HIV protease enzyme uses a general acid-base catalysis mechanism to cleave viral polypeptides but Copyright Macmillan Learning. Powered by Cognero.

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Chapter 06: Protein Function and Enzymes does not use a covalent catalysis. This enzyme functions optimally in the pH range of 4 to 6. Due to the specific amino acids involved in this catalysis, HIV protease is a member of which subclass of proteases? a. metalloproteases b. serine proteases c. aspartyl proteases d. cysteine proteases e. lysine proteases ANSWER: c 35. Which amino acid may NOT be targeted for phosphorylation to modulate the activity of an enzyme? a. Tyr b. His c. Ser d. Ala e. Thr ANSWER: d 36. Which type of enzyme is responsible for attaching phosphate groups to specific amino acids? a. a protein phosphatase b. a ribosylase c. a protein kinase d. a protein glycosylase e. an ATPase ANSWER: c 37. Define the terms "cofactor" and "coenzyme."

ANSWER: A cofactor is any chemical component required for enzyme activity; it includes both organic molecules, called coenzymes, and inorganic ions.

38. Draw and label a reaction coordinate diagram for an uncatalyzed reaction, S→P, and the same reaction catalyzed by an enzyme, E. ANSWER: See Figure 6-3.

39. The difference in (standard) free-energy content, ΔG'º, between substrate S and product P may vary considerably among different reactions. What is the significance of these differences?

ANSWER: The difference in free-energy content between substrate (or reactant) and product for each reaction reflects the relative amounts of each compound present at equilibrium. The greater the difference in free energy, the greater the difference in amounts of each compound at equilibrium.

40. For a reaction that can take place with or without catalysis by an enzyme, what would be the effect of the enzyme on the: (a) standard free-energy change of the reaction? (b) activation energy of the reaction?

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Chapter 06: Protein Function and Enzymes (c) initial velocity of the reaction? (d) equilibrium constant of the reaction?

ANSWER: (a) no change; (b) decrease; (c) increase; (d) no change 41. Sometimes the difference in (standard) free-energy content (ΔG'º) between a substrate S and a product P is very large, yet the rate of chemical conversion (S→P) is quite slow. Why? ANSWER: The rate of conversion from substrate to product (or the reverse reaction, from product to substrate) does not depend on the free-energy difference between them. The rate of the reaction depends on the activation energy of the reaction (ΔG‡), which is the difference between the free-energy content of S (or P) and the reaction transition state.

42. What is the difference between general acid-base catalysis and specific acid-base catalysis? (Assume that the solvent is water.)

ANSWER: Specific acid-base catalysis refers to catalysis by the constituents of water; that is, the donation of a proton by the hydronium ion (H3O+) or the acceptance of a proton by the hydroxyl ion (OH–).

General acid-base catalysis refers to the donation or acceptance of a proton by weak acids and bases other than water.

43. Two different enzymes are able to catalyze the same reaction, A→B. They both have the same Vmax, but differ in their Km values. For enzyme 1, the Km is 1.0 mM; for enzyme 2, the Km is 10 mM. When enzyme 1 was incubated with 0.1 mM A, it was observed that B was produced at a rate of 0.0020 mmol/min. (a) What is the value of the Vmax of the enzymes? (b) What will be the rate of production of B when enzyme 2 is incubated with 0.1 mM A? (c) What will be the rate of production of B when enzyme 1 is incubated with 1 M (i.e., 1,000 mM) A? ANSWER: (a) 0.022 mmol/min; (b) 0.00022 mmol/min; (c) 0.022 mmol/min 44. Write out the equation that describes the mechanism for enzyme action used as a model by Michaelis and Menten. List the important assumptions used by Michaelis and Menten to derive a rate equation for this reaction.

ANSWER: The two equations are

One assumption is that [P] = 0 so that the rate of the reaction depends exclusively on the breakdown of ES and is not influenced by the reverse reaction; that is, k–2 can be ignored and V0 = k2[ES]. This condition is possible only if early reaction times are measured; the velocity, therefore, is an initial velocity. A second assumption is that the rate of ES formation equals the rate of ES breakdown; in other words, the reaction is at a steady state. A third assumption is [S] >> [Et], so that total [S], which equals free substrate and enzyme-bound substrate, is essentially equal to [S].

45. For the reaction E + S→ES→P the Michaelis-Menten constant, Km, is actually a combination of three terms. What are they? How is Km determined graphically?

ANSWER: Km = (k2 + k–1)/k1, where k–1 and k1 are the rate constants for the breakdown and association, Copyright Macmillan Learning. Powered by Cognero.

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Chapter 06: Protein Function and Enzymes respectively, of the ES complex and k2 is the rate constant for the breakdown of ES to form E + P. Km can be determined graphically on a plot of V0 versus [S] by finding the [S] at which V0 = 1/2Vmax. More conveniently, on a double-reciprocal plot, the intercept on the x-axis equals –1/Km. 46. An enzyme catalyzes a reaction at a velocity of 20 μmol/min when the concentration of substrate (S) is 0.01 M. The Km for this substrate is 10–5 M. Assuming that Michaelis-Menten kinetics are followed, what will the reaction velocity be when the concentration of S is (a) 10–5 M and (b) 10–6 M? ANSWER: The velocity of 20 μmol/min is essentially Vmax because it is measured at [S] >> Km. (a) When [S] = 10–5 M = Km, V = 1/2Vmax, or 10 μmol/min. (b) When [S] is 10–6 M, velocity can be calculated from the Michaelis-Menten equation V0 = Vmax [S]/(Km + [S]) = (20 μmol/min)(10–6 M)/(10–5 + 10–6) = 1.8 μmol/min 47. How does the total enzyme concentration affect turnover number and Vmax?

ANSWER: The turnover number (kcat) is the number of substrate molecules converted to product in a given time by a single enzyme molecule. The turnover number is not affected by the total enzyme concentration, [Et]. For any given reaction, however, Vmax can change because Vmax is the product of turnover number × the total enzyme concentration, or Vmax = kcat [Et].

48. Methanol (wood alcohol) is highly toxic because it is converted to formaldehyde in a reaction catalyzed by the enzyme alcohol dehydrogenase NAD+ + methanol → NADH + H+ + formaldehyde Part of the medical treatment for methanol poisoning is to administer ethanol (ethyl alcohol) in amounts large enough to cause intoxication under normal circumstances. Explain this in terms of examples of enzymatic reactions.

ANSWER: Ethanol is a structural analog of methanol. Ethanol competes with methanol for the binding site of alcohol dehydrogenase, slowing the conversion of methanol to formaldehyde, and allowing its clearance by the kidneys. The effect of ethanol is that of a competitive inhibitor.

49. The initial rate of an enzyme reaction is measured as a function of substrate concentration in the presence and absence of an inhibitor. The data obtained: [S] 0.0001 0.0002 0.0005 0.001 0.002 0.005 0.01 0.02 0.05 0.1 0.2

V0 –Inhibitor 33 50 71 83 91 96 98 99 100 100 100

+Inhibitor 17 29 50 67 80 91 95 98 99 100 100

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Chapter 06: Protein Function and Enzymes (a) What is the Vmax in the absence of an inhibitor? (b) What is the Km in the absence of an inhibitor? (c) When [S] = 0.0004, what will V0 be in the absence of an inhibitor? (d) When [S] = 0.0004, what will V0 be in the presence of an inhibitor? (e) What kind of inhibitor is it likely to be?

ANSWER: (a) 100; (b) 0.0002; (c) 66.7; (d) 40; (e) competitive 50. Chymotrypsin belongs to a group of proteolytic enzymes called the serine proteases, many of which have an Asp, His, and Ser residue that are crucial to the catalytic mechanism. The serine hydroxyl functions as a nucleophile. What do the other two amino acids do to support this nucleophilic reaction?

ANSWER: In chymotrypsin, histidine functions as a general base, accepting a proton from the serine hydroxyl, thereby increasing serine's reactivity as a nucleophile. The negatively charged Asp stabilizes the positive charge that develops on the His. 51. For serine to work effectively as a nucleophile in covalent catalysis in chymotrypsin a nearby amino acid, histidine, must serve as a general base catalyst. Briefly describe, in words, how these two amino acids work together.

ANSWER: The serine is a polar hydroxyl with the oxygen functioning as an electronegative nucleophile. A nearby histidine residue with pKa ≈ 6.0, however, functions as a base to abstract the proton from the serine hydroxyl group. The result is to substantially increase the electronegativity of the serine oxygen, making it a much stronger nucleophile. This, in turn, lowers the activation energy of the covalent catalysis between serine and the carbonyl carbon of the substrate peptide bond.

52. Why is a transition-state analog not necessarily the same as a competitive inhibitor?

ANSWER: The structure of a competitive inhibitor may be similar to the structure of the free substrate. A similar structure will mean that the competitive inhibitor can associate with the enzyme at the active site, effectively blocking the normal substrate from binding. A transition-state analog, however, is similar in structure to the transition-state of the reaction catalyzed by the enzyme. Often, a transition-state analog will bind tightly to an enzyme, and it is not easily replaced by substrate.

53. Explain how a biochemist might discover that a certain enzyme is allosterically regulated.

ANSWER: The enzyme would show kinetics that do not fit the Michaelis-Menten equation; the plot of V versus [S] would be sigmoidal, not hyperbolic. The enzyme kinetics would be affected by molecules other than the substrate(s).

54. What is a zymogen (proenzyme)? Explain briefly with an example.

ANSWER: A zymogen is an inactive form of an enzyme that is activated by one or more proteolytic cleavages in its sequence. Chymotrypsinogen, trypsinogen, and proelastase are all zymogens which become chymotrypsin, trypsin, and elastase, respectively, after proper cleavage.

55. Describe why the Michaelis constant (Km) and the dissociation constant (Kd) should not be considered to be the same. Use the rate constant definitions of these terms in the explanation, and also indicate in what case Km and Kd can be considered to be the same.

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Chapter 06: Protein Function and Enzymes ANSWER: The Km refers to the rate of product formation plus the rate of enzyme and substrate dissociation with the sum divided by the rate of substrate and enzyme association, or (k2 + k–1)/k1. This is not the same as the Kd, which is defined just as the ratio between enzyme and substrate dissociation and enzyme and substrate association, or k–1/k1. Since the Km includes k2 as an additional term, it should not be considered to be the same as the Kd. However, in the limited case where k2 is rate-limiting and is much less than k–1, then k2 can be approximated as zero, which reduces the Km term to (0 + k–1)/k1. This then makes the Km and Kd terms equal, and then they can be considered the same, although this case does not apply for most enzymes.

56. Enzyme-catalyzed reactions are characterized by the formation of a complex between: a. a coenzyme and a product. b. a coenzyme and a substrate. c. an enzyme and a product. d. an enzyme and a substrate. e. a product and a substrate.

ANSWER: d 57. Chymotrypsin is a: a. hydrolase. b. lactase. c. lyase. d. polymerase. e. protease.

ANSWER: e 58. Chymotrypsin enhances the rate of peptide bond hydrolysis by a factor of at least: a. 104. b. 106. c. 107. d. 109. e. 1010.

ANSWER: d 59. Beta-lactam antibiotics inhibit bacterial synthesis of: a. β-lactam. b. cell membranes. c. cell walls. d. peptidoglycan. e. trypsin.

ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 06: Protein Function and Enzymes 60. Which is not a HIV protease inhibitor? a. Amoxicillin b. Indinavir c. Lopinavir d. Nelfinavir e. Saquinavir ANSWER: a 61. Allosteric enzymes which use their substrates as a modulator are being subjected to _________ regulation. a. allosteric b. heterotropic c. homotropic d. kinetic e. mixed ANSWER: c 62. Allosteric enzymes which use molecules other than their substrates as a modulator are being subjected to _________ regulation. a. allosteric b. heterotropic c. homotropic d. kinetic e. mixed ANSWER: b 63. Which type of enzyme is responsible for removing phosphate groups from specific amino acids? a. a protein phosphatase b. a ribosylase c. a protein kinase d. a protein glycosylase e. an ATPase ANSWER: a

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Chapter 07: Carbohydrates 1. Which of these monosaccharides is NOT an aldose? a. erythrose b. fructose c. glucose d. glyceraldehyde e. ribose ANSWER: b 2. The reference compound for naming D and L isomers of sugars is: a. fructose. b. glucose. c. glyceraldehyde. d. ribose. e. sucrose.

ANSWER: c 3. When two carbohydrates are epimers: a. one is a pyranose, the other a furanose. b. one is an aldose, the other a ketose. c. they differ in length by one carbon. d. they differ only in the configuration around one carbon atom. e. they rotate plane-polarized light in the same direction.

ANSWER: d 4. Which pair is epimeric? a. D-glucose and D-glucosamine b. D-glucose and D-mannose c. D-glucose and L-glucose d. D-lactose and D-sucrose e. L-mannose and L-fructose ANSWER: b 5. Which pair is anomeric? a. D-glucose and D-fructose b. D-glucose and L-fructose c. D-glucose and L-glucose d. α-D-glucopyranose and β-D-glucopyranose e. α-D-glucopyranose and β-L-glucopyranose ANSWER: d 6. When the linear form of glucose cyclizes, the product is a(n): Copyright Macmillan Learning. Powered by Cognero.

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Chapter 07: Carbohydrates a. anhydride. b. glycoside. c. hemiacetal. d. lactone. e. oligosaccharide.

ANSWER: c 7. Which pair is interconverted in the process of mutarotation? a. D-glucose and D-fructose b. D-glucose and D-galactose c. D-glucose and D-glucosamine d. D-glucose and L-glucose e. α-D-glucopyranose and β-D-glucopyranose ANSWER: e 8. Which compound is NOT a reducing sugar? a. fructose b. glucose c. glyceraldehyde d. ribose e. sucrose ANSWER: e 9. Which monosaccharide is NOT a carboxylic acid? a. 6-phospho-gluconate b. gluconate c. glucose d. glucuronate e. muramic acid ANSWER: c 10. When forming the disaccharide maltose from two glucose monosaccharides: a. water is eliminated. b. a hemiacetal is converted to an acetal. c. the resulting disaccharide is no longer a reducing sugar. d. water is eliminated, and a hemiacetal is converted to an acetal. e. None of the answers are correct.

ANSWER: d 11. From the abbreviated name of the compound Gal(β1 → 4)Glc, we know that: a. C-4 of glucose is joined to C-1 of galactose by a glycosidic bond.

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Chapter 07: Carbohydrates b. the compound is a D-enantiomer. c. the galactose residue is at the reducing end. d. the glucose is in its pyranose form. e. the glucose residue is the β anomer.

ANSWER: a 12. Starch and glycogen are both polymers of: a. fructose. b. glucose-1-phosphate. c. sucrose. d. α-D-glucose. e. β-D-glucose.

ANSWER: d 13. Which statement about starch and glycogen is false? a. Amylose is unbranched; amylopectin and glycogen contain many (α1 → 6) branches. b. Both are homopolymers of glucose. c. Both serve primarily as structural elements in cell walls. d. Both starch and glycogen are stored intracellularly as insoluble granules. e. Glycogen is more extensively branched than starch.

ANSWER: c 14. Which is a heteropolysaccharide? a. cellulose b. chitin c. glycogen d. glycosaminoglycan e. starch ANSWER: d 15. The basic structure of a proteoglycan consists of a core protein and a: a. glycolipid. b. glycosaminoglycan. c. lectin. d. lipopolysaccharide. e. peptidoglycan.

ANSWER: b 16. In glycoproteins, the carbohydrate moiety is always attached through the amino acid residues: a. asparagine, serine, or threonine. b. aspartate or glutamate.

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Chapter 07: Carbohydrates c. glutamine or arginine. d. glycine, alanine, or aspartate. e. tryptophan, aspartate, or cysteine.

ANSWER: a 17. Which compound(s) is a dominant feature of the outer membrane of the cell wall of gram-negative bacteria? a. amylose b. cellulose c. glycoproteins d. lipopolysaccharides e. lipoproteins ANSWER: d 18. The biochemical property of lectins that is the basis for most of their biological effects is their ability to bind to: a. amphipathic molecules. b. hydrophobic molecules. c. specific lipids. d. specific oligosaccharides. e. specific peptides.

ANSWER: d 19. A glycoconjugate may include a carbohydrate as well as a(n): a. amino acid. b. lipid. c. saccharide. d. protein or lipid. e. amino acid, a lipid, or a saccharide.

ANSWER: d 20. In a(n) _____, the carbonyl group is at the end of the carbon chain, but in a(n) _____, the carbonyl group can be at any other position. a. aldose; ketose b. ketose; aldose c. hexose; heptose d. aldose; hexose e. ketose; tetrose ANSWER: a 21. Which monosaccharide is NOT a hexose? a. fructose Copyright Macmillan Learning. Powered by Cognero.

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Chapter 07: Carbohydrates b. ribose c. mannose d. glucose e. galactose ANSWER: b 22. These two monosaccharides could BEST be described as:

a. ketoses. b. anomers. c. rotamers. d. epimers. e. enantiomers.

ANSWER: d 23. Which arrow correctly identifies the atom that will become the anomeric carbon?

a. A b. B c. C d. D e. E ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 07: Carbohydrates 24. Based on the form of the cyclic sugar in a Haworth projection, which Fischer projection formula could have formed this structure?

a. A b. B c. C d. D e. E ANSWER: a 25. When drawing a Haworth perspective formula from a Fisher projection, which statement is true regarding anomers? a. It is α if the hydroxyl group on the anomeric carbon is up. b. It is α if the hydroxyl group on the anomeric carbon is down. c. It is α if the hydroxyl group on the anomeric carbon is on the same side of the ring as the C-6. d. It is α if the hydroxyl group on the anomeric carbon is on the opposite side of the ring as the C-6. e. None of the answers are correct.

ANSWER: d 26. During which process are chemical bonds NOT broken? a. mutarotation b. interconverting between two configurations c. interconverting between two conformations d. interconverting between two anomers Copyright Macmillan Learning. Powered by Cognero.

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Chapter 07: Carbohydrates e. None of the answers are correct: all involve breaking a chemical bond.

ANSWER: c 27. Which modification cannot be made to a sugar molecule by an organism? a. oxidation of the carbonyl carbon b. replacing a hydroxyl group with an amino group c. oxidizing a carbon atom to a carboxyl group d. replacing a hydroxyl group with a hydrogen atom e. All of the answer choices are possible sugar derivatives.

ANSWER: e 28. What is the correct linkage designation for the glycosidic bond between the two monosaccharide rings?

a. (α1→4) b. (β1→4) c. (1→4) d. (α1→β4) e. (α1→α4) ANSWER: b 29. Which statement about cellulose is true? a. Cellulose is the main storage polysaccharide of animal cells. b. Cellulose is a highly branched polysaccharide. c. Cellulose is a homopolysaccharide. d. Cellulose glucose residues contain the same configuration as amylose glucose residues. e. Cellulose cannot be broken down by invertebrate animals.

ANSWER: c 30. Which polysaccharide is NOT a homopolysaccharide? a. glycogen b. chitin c. amylose Copyright Macmillan Learning. Powered by Cognero.

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Chapter 07: Carbohydrates d. hyaluronan e. cellulose ANSWER: d 31. The polysaccharide chitin is MOST chemically similar to: a. cellulose. b. dextran. c. glycogen. d. starch. e. amylose.

ANSWER: a 32. Which answer choice is the BEST description of the benefits that are provided by the extended conformation that the heteropolysaccharide glycosaminoglycans assume in solution? a. The repulsive forces of densely packed negative charges are minimized. b. The hydrophobic portions of the molecules may pack closely together to maximize van der Waals forces. c. The patterns of sulfated and nonsulfated sugar residues are more easily recognized by their respective ligands. d. The repulsive forces of densely packed negative charges are minimized, and the hydrophobic portions of the molecules may pack closely together to maximize van der Waals forces. e. The repulsive forces of densely packed negative charges are minimized, and the patterns of sulfated and nonsulfated sugar residues are more easily recognized by their respective ligands.

ANSWER: e 33. Which glycosaminoglycan is found in the synovial fluid of joints? a. chondroitin sulfate b. dermatan sulfate c. hyaluronan d. keratin sulfate e. heparan sulfate ANSWER: c 34. Which statement is false regarding proteoglycans? a. Both syndecans and glypicans can be shed into the extracellular space. b. Proteoglycan shedding is unregulated in cells. c. A protease or phospholipase is responsible for releasing proteoglycans. d. Proteoglycans are extracellular proteins with attached sulfated glycosaminoglycans. e. Cell features such as proteoglycans can be changed quickly.

ANSWER: b 35. About what percentage of mammalian proteins are glycosylated?

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Chapter 07: Carbohydrates a. 5% b. 15% c. 50% d. 90% e. 95% ANSWER: c 36. Oligosaccharides differ from nucleic acids and proteins in that oligosaccharides are commonly: a. branched. b. polymeric. c. acidic. d. basic. e. None of the answers are correct.

ANSWER: a 37. The antiviral drugs oseltamivir (Tamiflu) and zanamivir (Relenza) do NOT work by: a. inhibiting the viral enzyme that trims the host cell's oligosaccharides. b. causing the viral particles to aggregate. c. competing with the host cell's oligosaccharides for binding. d. preventing the release of viruses from the infected cell. e. All of the answers are correct.

ANSWER: e 38. Explain why all monosaccharides and disaccharides are soluble in water.

ANSWER: These compounds have many hydroxyl groups, each of which can form hydrogen bonds with water.

39. The compound L-glyceraldehyde is shown. Describe a stereochemically correct representation of C-1 and C-2 of D-glucose.

ANSWER: In D-glucose, the positions of the —H and —OH on C-2 are the reverse of those for C-2 of Lglyceraldehyde. (See Fig. 7-2.)

40. Categorize each of these as an aldose, a ketose, or neither.

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Chapter 07: Carbohydrates

ANSWER: Molecules (b) and (d) are aldoses; (a) is a ketose; (c) and (e) are neither.

41. (a) In the pyranose ring form structure, how many asymmetric carbons (chiral centers) does each of the anomers of glucose have? (b) How many stereoisomers of the glucose are theoretically possible?

ANSWER: (a) The number of chiral centers is 5; all the carbons except C-6. (b) The number of possible stereoisomers for a compound with n chiral centers is 2n; in this case, 25 or 32 possible isomers.

42. Identify all the epimeric pairs in the structures, taking into account the possibility of mutarotation in solution.

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Chapter 07: Carbohydrates

ANSWER: Epimeric pairs should have one carbon of altered stereochemistry compared to each other, as long as that carbon is not the anomeric carbon, because of the easy mutarotation to the opposite anomer.

Therefore, C and D are anomers of each other, but are both epimers of A. Likewise, E and F are anomers of each other, but both are epimers of B. 43. In the structure, (a) how many of the monosaccharide units are furanoses, and how many are pyranoses? (b) What is the linkage between the two monosaccharide units? (c) Is this a reducing sugar? Explain.

ANSWER: (a) 2 pyranoses; (b) β1→4; (c) yes; There is a free anomeric carbon on one of the monosaccharide units that can undergo oxidation upon ring opening.

44. Match these molecules with their biological roles. (a) glycogen (b) starch (c) chitin (d) cellulose (e) peptidoglycan (f) hyaluronate (g) proteoglycan

_____ viscosity, lubrication of extracellular secretions _____ carbohydrate storage in plants _____ exoskeleton of insects _____ structural component of bacterial cell wall _____ structural component of plant cell walls _____ extracellular matrix of animal tissues _____ carbohydrate storage in animal liver

ANSWER: f; b; c; e; d; g; a 45. Describe one biological advantage of storing glucose units in branched polymers (glycogen, amylopectin) rather than in linear polymers.

ANSWER: The enzymes that act on these polymers to mobilize glucose for metabolism act only on their nonreducing ends. With extensive branching, there are more such ends for enzymatic attack than would be present in the same quantity of glucose stored in a linear polymer. In effect, branched polymers increase the substrate concentration for these enzymes.

46. Explain how it is possible that a polysaccharide molecule, such as glycogen, may have only one reducing end, and yet have many nonreducing ends.

ANSWER: The molecule is branched with each branch ending in a nonreducing end.

47. The glycosaminoglycans are negatively charged at neutral pH. What components of these polymers confer the negative charge?

ANSWER: Uronic acids, such as glucuronic acid, and sulfated hydroxyl groups confer negative charges. See Figure 7-13.

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Chapter 07: Carbohydrates 48. Describe the principal components of a typical proteoglycan, including their relationships and connections to one another.

ANSWER: A typical proteoglycan consists of a core protein with covalently attached glycosaminoglycan polysaccharides, such as chondroitin sulfate and keratin sulfate. The polysaccharides generally attach to a serine residue in the protein via a trisaccharide (Gal–Gal–Xyl). (See Fig. 7-14.)

49. Describe the differences between a proteoglycan and a glycoprotein.

ANSWER: Both are made up of proteins and polysaccharides. In proteoglycans, the carbohydrate moiety dominates, constituting 95% or more of the mass of the complex. In glycoproteins, the protein constitutes a larger fraction, generally 50% or more of the total mass.

50. Describe the structure of a proteoglycan aggregate such as is found in the extracellular matrix.

ANSWER: A proteoglycan aggregate is a supramolecular assembly of proteoglycan monomers. Each monomer consists of a core protein with multiple, covalently linked polysaccharide chains. Hundreds of these monomers can bind noncovalently to a single extended molecule of hyaluronic acid to form large structures.

51. How do oligosaccharide portions of glycoproteins change the properties of the proteins?

ANSWER: Hydrophilic carbohydrates can alter the polarity and solubility of the proteins. Steric and charge interactions may influence the conformation of polypeptide regions and protect it from proteolysis.

Carbohydrates attached to proteins can provide binding sites for lectins.

52. What are lectins? What are some biological processes that involve lectins?

ANSWER: Lectins are proteins that bind to specific oligosaccharides. They interact with specific cell surface glycoproteins thus mediating cell-cell recognition and adhesion.

53. Briefly explain how the drugs Tamiflu and Relenza work to prevent the flu.

ANSWER: A newly formed flu virus needs to get out of its host cell in order to infect new cells and make more virus. As the new virus buds off the host cell, it initially binds (via lectins) to cell surface receptors located in the membrane that are specific for sialic acid residues. Tamiflu and Relenza compete for binding and prevent the release of viruses from the infected cells. They also causes viral particles to aggregate, which further blocks another cycle of infection. (Fig. 7.21) 54. Which is not a fibrous matrix protein? a. collagen b. elastin c. fibronectin d. proteoglycans e. None of the above ANSWER: d 55. What is the correct sequence for leukocyte extravasation in the inflammatory response?

ANSWER: See figure 7.20: selectin binding→integrin binding→arrest→adhesion→extravasation 56. What are cell membrane lectins that mediate cell-cell recognition and adhesion?

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Chapter 07: Carbohydrates a. fibronectin b. glycoprotein c. glypican d. proteoglycan e. selectins ANSWER: e 57. Lectins bind carbohydrates with _______ specificity. a. no b. low c. moderate d. high e. N/A: Lectins do not bind with carbohydrates ANSWER: d 58. The association between cells and the proteoglycan of the extracellular matrix is mediated by the membrane protein: a. fibronectin b. glycoprotein c. integrin d. proteoglycan e. elastin ANSWER: c 59. Which is not a membrane component? a. glycolipids b. glypicans c. heparin d. lipopolysaccharides e. selectins ANSWER: c

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Chapter 08: Lipids, Membranes, and Membrane Proteins 1. Which molecules or substances contain, or are derived from, fatty acids? a. glycerophospholipids b. galactolipids c. sphingolipids d. sulfolipids e. All of these contain or are derived from fatty acids.

ANSWER: e 2. Which factor is NOT a consequence of the partial hydrogenation of vegetable oils? a. longer shelf life b. lower melting temperature c. reduction of carbon–carbon double bonds to single bonds d. conversion of some cis double bonds to trans double bonds e. increased risk of cardiovascular disease upon consumption by humans ANSWER: b 3. Which statement is true? a. Glycerophospholipids are found only in the membranes of plant cells. b. Glycerophospholipids contain fatty acids linked to glycerol through amide bonds. c. Phosphatidylcholine is a sphingolipid. d. Some sphingolipids include oligosaccharides in their structure. e. Triacylglycerols are the principal components of erythrocyte membranes.

ANSWER: d 4. Which compound is NOT a glycerophospholipid? a. phosphatidylcholine b. phosphatidylethanolamine c. phosphatidylserine d. cardiolipin e. ceramide ANSWER: e 5. Which statement about sterols is true? a. Sterols have four fused rings in their structure. b. Sterols are found in the membranes of all living cells. c. Sterols are soluble in water, but less so in organic solvents such as chloroform. d. Cholesterol is the principal sterol in fungi. e. The principal sterol of animal cells is ergosterol.

ANSWER: a 6. Which term BEST describes the cholesterol molecule?

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Chapter 08: Lipids, Membranes, and Membrane Proteins a. amphipathic b. nonpolar, charged c. nonpolar, uncharged d. polar, charged e. polar, uncharged ANSWER: a 7. An example of a glycerophospholipid that is involved in cell signaling is: a. arachidonic acid. b. ceramide. c. phosphatidylinositol. d. testosterone. e. vitamin A (retinol).

ANSWER: c 8. What is the abbreviated nomenclature for octadecenoic acid with two double bonds? a. 16:1 b. 14:2 c. 18:2 d. 18:0 e. 18:1 ANSWER: c 9. Which statement is true about fatty acids? a. Fatty acids with longer chains have lower melting points. b. Fatty acids with longer chains have higher solubility in water. c. Fatty acids with more double bonds have higher melting points. d. Fatty acids with longer chains have higher melting points and lower solubility in water. e. None of the statements are true.

ANSWER: d 10. What is the abbreviated nomenclature for a 14-carbon omega-3 fatty acid that contains no other double bonds? a. 14:1 (Δ11) b. 17:1 (Δ14) c. 14:1 (Δ3) d. 13:1 (Δ11) e. 14:0 (Δ11) ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 08: Lipids, Membranes, and Membrane Proteins 11. Which type of chemical linkage is used to join fatty acids to glycerol? a. ether linkage b. peptide bond c. ester linkage d. double bond e. peroxide bond ANSWER: c 12. Which term BEST describes the polarity of membrane lipids? a. ambidextrous b. completely nonpolar c. amphipathic d. completely polar e. amphibolic ANSWER: c 13. Which compound is NOT a type of membrane lipid? a. sphingolipid b. glycerophospholipid c. sulfolipid d. sterol e. triacylglycerol ANSWER: e 14. Which glycerophospholipid has a net charge of zero at pH 7? a. phosphatidylserine b. phosphatidic acid c. phosphatidylcholine d. phosphatidylglycerol e. phosphatidylinositol 4,5-bisphosphate ANSWER: c 15. Which phospholipid head group is positively charged at pH 7? a. serine b. ethanolamine c. choline d. both serine and ethanolamine e. both ethanolamine and choline ANSWER: e 16. Fatty acids are attached to sphingosine by what type of chemical linkage?

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Chapter 08: Lipids, Membranes, and Membrane Proteins a. ether b. amide c. ester d. peroxide e. alkyl ANSWER: b 17. Circle the fatty acid in each pair that has the higher melting temperature. (a) 18:1 (Δ9) 18:2 (Δ9,12) (b) 18:0 18:1 (Δ9) (c) 18:0 16:0 ANSWER: (a) 18:1 (Δ9); (b) 18:0; (c) 18:0 18. Describe the trends in the melting points of fatty acids with respect to (a) chain length and (b) unsaturation; (c) explain these trends in molecular terms.

ANSWER: All other things being equal, (a) the longer the acyl chain, the higher the melting temperature; and (b) the more unsaturation, the lower the melting temperature. (c) The melting temperature is a measure of the thermal energy needed to break the intermolecular interactions that stabilize the solid form of a lipid, which depends on how well the lipid molecules pack together. The longer the carbon chains are, the greater the surface area, which means more collective van der Waals attractions between neighboring chains. Saturated lipids can adopt a nearly straight conformation that packs easily. In unsaturated chains, a cis double bond introduces a kink into the acyl chain so that it does not pack as easily as saturated lipids.

19. Draw the structure of phosphatidylserine in ionic form at pH 7.

ANSWER: The head group and general glycerophospholipid structures are shown in Figure 8.5. At neutral pH, there is a charge on the phosphate group, and serine is in the zwitterionic form; it has a protonated amino group and an ionized carboxyl group.

20. Draw the structure of phosphatidylcholine. Circle the part of the molecule that is polar and draw an arrow to the part that is nonpolar.

ANSWER: The head group and general glycerophospholipid structures are shown in Figure 8.5. At neutral pH, there is a negative charge on the phosphate group, and the quaternary amino group of choline carries a fixed positive charge; this entire phosphorylcholine moiety is polar. The acyl chains attached to glycerol are the nonpolar part of the molecule.

21. Draw and describe the basic structure shared by all glycerophospholipids.

ANSWER: The general glycerophospholipid structure is shown in Figure 8.5. All glycerophospholipids have two fatty acids in ester linkage with C-1 and C-2 of glycerol; often, the fatty acid at C-1 is saturated and that at C-2 is unsaturated. C-3 of glycerol is joined to an alcohol-containing head group through a phosphodiester linkage, which is negatively charged at neutral pH. 22. Describe why saturated fatty acids have higher melting points than do unsaturated fatty acids of the same chain length.

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Chapter 08: Lipids, Membranes, and Membrane Proteins ANSWER: Saturated fatty acids contain no double bonds and are therefore very flexible in their orientation.

However, the fully extended conformation is the most energetically favorable, and many saturated fatty acids can pack together tightly when they are all in the extended conformation. This tight packing leads to significant hydrophobic interactions, which increases the overall melting point of the saturated fatty acids. Unsaturated fatty acids contain one or more double bonds, which form kinks in the fatty acid structure that makes them less flexible. Those kinks also interfere with the tight packing together of the fatty acids, leading to less stable aggregates of these fatty acids, with fewer hydrophobic interactions, and therefore lower melting points than saturated fatty acids that have the same chain length. (See Figure 8.1 for images.)

23. Which statement about membranes is true? a. Most plasma membranes contain more than 70% proteins. b. Sterol lipids are common in bacterial plasma membranes. c. Sterol lipids are common in human cell plasma membranes. d. Sterol lipids are common in plant cell plasma membranes. e. The plasma membranes of all cell types within a particular organism have basically the same lipid and protein composition.

ANSWER: c 24. The inner (plasma) membrane of E. coli is about 75% lipid and 25% protein by weight. How many molecules of membrane lipids are there for each molecule of protein? (Assume the average protein has a molecular weight of 50,000 and the average lipid has a molecular weight of 750.) a. 1 b. 50 c. 200 d. 10,000 e. 50,000 ANSWER: c 25. Which statement about the composition of membranes is true? a. All biological membranes contain cholesterol. b. Free fatty acids are major components of all membranes. c. The inner and outer membranes of mitochondria have different protein compositions. d. The lipid composition of all membranes of eukaryotic cells is essentially the same. e. The lipid:protein ratio is similar in most membranes.

ANSWER: c 26. Membrane proteins: a. are sometimes covalently attached to lipid moieties. b. are sometimes covalently attached to carbohydrate moieties. c. are composed of the same 20 amino acids found in soluble proteins. d. diffuse laterally in the membrane unless they are anchored.

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Chapter 08: Lipids, Membranes, and Membrane Proteins e. All of the answers are correct.

ANSWER: e 27. Peripheral membrane proteins: a. are generally noncovalently bound to membrane lipids. b. are usually denatured when released from membranes. c. can be released from membranes only by treatment with detergent(s). d. may have functional units on both sides of the membrane. e. penetrate deeply into the lipid bilayer.

ANSWER: a 28. An integral membrane protein can be extracted with: a. a buffer of alkaline or acid pH. b. a chelating agent that removes divalent cations. c. a solution containing detergent. d. a solution of high ionic strength. e. hot water.

ANSWER: c 29. Which statement describes a general feature of the lipid bilayer in all biological membranes? a. Individual lipid molecules are free to diffuse laterally on the surface of the bilayer. b. Individual lipid molecules in one face (monolayer) of the bilayer readily diffuse (flip-flop) to the other monolayer. c. Polar, uncharged compounds readily diffuse across the bilayer. d. The bilayer is stabilized by covalent bonds between neighboring phospholipid molecules. e. The polar head groups face inward toward the inside of the bilayer.

ANSWER: a 30. The fluidity of a lipid bilayer will be increased by: a. decreasing the number of unsaturated fatty acids. b. decreasing the temperature. c. increasing the length of the alkyl chains. d. increasing the temperature. e. substituting 18:0 (stearic acid) in place of 18:2 (linoleic acid).

ANSWER: d 31. Glucose transport into erythrocytes is an example of: a. active transport. b. antiport. c. electrogenic uniport. d. passive transport.

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Chapter 08: Lipids, Membranes, and Membrane Proteins e. symport.

ANSWER: d 32. Which type of membrane transport uses ion gradients as the energy source (driving force) for transport? a. facilitated diffusion b. passive transport c. primary active transport d. secondary active transport e. simple diffusion ANSWER: d 33. Consider the transport of glucose into an erythrocyte by facilitated diffusion. What is the free-energy change for glucose uptake into the cell when the glucose concentrations are 5 mM on the outside and 0.1 mM on the inside? (These values may be of use: R = 8.315 J/mol·K; T = 298 K; F (Faraday constant) = 96,480 J/V·mol; N = 6.022 × 1023/mol.) a. less than 2 kJ/mol b. about –10 kJ/mol c. about 30 kJ/mol d. about –30 kJ/mol e. It is impossible to calculate without knowledge of the membrane potential.

ANSWER: b 34. Consider the transport of K+ from the blood (where its concentration is about 4 mM) into an erythrocyte that contains 150 mM K+. The transmembrane potential is about 60 mV, inside negative relative to outside. What is the free-energy change for this transport process? (These values may be of use: R = 8.315 J/mol·K; T = 298 K; F (Faraday constant) = 96,480 J/V·mol; N = 6.022 × 1023/mol.) a. about 5 J/mol b. about 15 J/mol c. about 5 kJ/mol d. about 15 kJ/mol e. It is impossible to calculate with the information given.

ANSWER: d 35. In one catalytic cycle, the Na+/K+ ATPase transporter transports _____, and converts one _____. a. two Na+ out, three K+ in; ATP to ADP + Pi b. three Na+ out, two K+ in; ATP to ADP + Pi c. three Na+ in, two K+ out; ATP to ADP + Pi d. one Na+ out, one K+ in; ATP to ADP + Pi e. two Na+ out, three K+ in; ADP + Pi to ATP Copyright Macmillan Learning. Powered by Cognero.

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Chapter 08: Lipids, Membranes, and Membrane Proteins ANSWER: b 36. Movement of water across membranes is facilitated by proteins called: a. annexins. b. aquaporins. c. hydropermeases. d. selectins. e. transportins.

ANSWER: b 37. A ligand-gated ion channel (such as the nicotinic acetylcholine receptor): a. is a charged lipid in the membrane bilayer that allows ions to pass through. b. is a membrane protein that permits a ligand to pass through the membrane only when opened by the appropriate ion. c. is a membrane protein that allows ions to pass through the membrane only when opened by the appropriate ligand. d. is a molecule that binds to the membrane allowing ions to pass through. e. always requires a second ligand to close the channel once it is opened.

ANSWER: c 38. Which statement does NOT describe a characteristic of the fluid mosaic model of membranes? a. Hydrophobic interactions predominate between lipids in the membrane. b. Proteins and lipids are able to move laterally at different rates. c. Protein structures are able to move easily from one leaflet to another. d. Noncovalent forces predominate in intermolecular interactions. e. Proteins may traverse a membrane or be associated with one side or the other of the membrane.

ANSWER: c 39. Which phrase BEST describes the term "amphitropic" protein? a. cytosolic lipid-linked proteins b. GPI-linked proteins c. proteins that are reversibly membrane-associated or cytosolic d. proteins that are either integral or lipid-linked e. proteins that can be on either leaflet of the bilayer ANSWER: c 40. What is meant by the term "positive-inside rule"? a. The typical transmembrane potential for plasma membranes is positive on the inside. b. Integral plasma membrane proteins commonly have positively charged amino acids on the cytosolic side. c. Concentrations of cations are usually higher inside than outside of cells.

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Chapter 08: Lipids, Membranes, and Membrane Proteins d. All of the answers are correct. e. None of the answers are correct.

ANSWER: b 41. Which process does NOT require a protein facilitator? a. simple diffusion b. ionophore-mediated diffusion c. a primary active uniport d. a passive antiporter e. both simple diffusion and ionophore-mediated diffusion ANSWER: a 42. Which factor MOST significantly determines the direction in which an ion moves through an ion channel in a membrane? a. the electrochemical gradient across the membrane b. the size and shape of the channel c. the ion selectivity of the channel d. the electrochemical gradient across the membrane and the size and shape of the channel e. the size, shape, and ion selectivity of the channel ANSWER: a 43. Which statement is true for passive transport across a biological membrane? a. Passive transport is driven by a solute electrochemical gradient. b. Passive transport is driven by ATP. c. Passive transport is irreversible. d. Passive transport is driven by both a solute electrochemical gradient and ATP. e. Passive transport is driven by a solute electrochemical gradient and is irreversible.

ANSWER: a 44. The plasma membrane of an animal cell consists of 45% by weight of phospholipids and 55% proteins.

What is the mole ratio (moles of lipid/moles of protein) if the average molecular weight of phospholipids is 750 and the average molecular weight of membrane proteins is 50,000?

ANSWER: The lipid-to-protein molar ratio is approximately 55. In 100 g of membrane, there are 45 g/750 g·mol–1 = 0.06 mol phospholipid, and 55 g/50,000 g·mol–1 = 1.1 × 10–3 mol protein. So lipid/protein = 0.06/0.0011 = 55.

45. In the context of the fluid mosaic model, indicate the positions and orientations of phospholipids, cholesterol, integral and peripheral membrane proteins, and the carbohydrate moieties of glycoproteins and glycolipids.

ANSWER: Phospholipids and sterols are found in both faces of the lipid bilayer. Integral membrane proteins penetrate or span the lipid bilayer, but peripheral membrane proteins associate at the membrane surface with lipid head groups or integral membrane proteins. The carbohydrate moieties of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 08: Lipids, Membranes, and Membrane Proteins glycolipids and glycoproteins are invariably on the outside face of the plasma membrane. (See Figure 8.10.)

46. Explain the differences between integral and peripheral membrane proteins.

ANSWER: Integral membrane proteins are very firmly associated with the membrane; their hydrophobic domains are associated with the fatty acyl groups in the interior in hydrophobic interactions.

Peripheral membrane proteins are more loosely associated and usually do not penetrate the hydrophobic interior of the bilayer. Conditions that reduce ionic interactions and hydrogen bonds commonly release them.

47. A protein is found to extend through the membrane of a cell. Describe this protein in terms of the location of particular types of amino acid side chains in its structure and its ability to move within the membrane.

ANSWER: This integral membrane protein associates with the lipid bilayer through hydrophobic interactions between domains containing many hydrophobic amino acids and the fatty acyl chains of membrane lipids. Polar and charged residues are located on portions of the protein that protrude from either face of the membrane. The protein is free to diffuse laterally in the plane of the membrane, but cannot move across the lipid bilayer.

48. Distinguish between simple diffusion, facilitated diffusion, and active transport across a membrane for these questions. (More than one may be true.) (a) Which processes are energy-dependent? (b) Which processes need some kind of carrier protein(s)? (c) Which processes can be saturated by substrate? (d) Which processes can establish a concentration gradient?

ANSWER: (a) only active transport; (b) facilitated diffusion and active transport; (c) facilitated diffusion and active transport; (d) only active transport 49. Explain why nonpolar compounds are generally able to diffuse across biological membranes without the aid of a specific transport system.

ANSWER: A nonpolar compound is more soluble in nonpolar solutes, (such as benzene or chloroform), than in polar solvents, (such as water). They generally pass through biological membranes unaided because it is energetically favorable for them to move from the aqueous solvent into the nonpolar region of the bilayer interior.

50. Compare the structure and activity of a membrane transport protein that transports a polar substance across a membrane with a typical soluble enzyme. How are transporters and enzymes similar? How are they different?

ANSWER: Both proteins have domains that specifically recognize and bind to their substrates via weak interactions; they both, therefore, mediate processes that are stereospecific and saturable. The membrane transporter usually spans the membrane, having one or several hydrophobic domains that interact with the hydrophobic acyl chains of fatty acids in the interior of the lipid bilayer. Soluble enzymes tend to have their hydrophobic residues buried within their interiors, and their hydrophilic residues on the surface, where they interact with water.

51. Compare and contrast symport and antiport. Which term best describes the transport system mediated by the Na+K+ ATPase?

ANSWER: Symport and antiport are both types of cotransport systems in which two solutes move through the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 08: Lipids, Membranes, and Membrane Proteins membrane simultaneously. In symport, both move in the same direction; in antiport, one solute goes in one direction, the other in the opposite direction. The Na+K+ ATPase of the plasma membrane is an antiport system. It moves K+ in and Na+ out in a ratio of two K+ per three Na+. Neither of the two ions can be transported unless the other is present, which is characteristic of cotransport systems.

(See Figures 8.16 and 8.19.)

52. What kinds of forces or bonds anchor an integral membrane protein in a biological membrane? a. a covalently bound lipid anchors to the bilayer b. hydrophobic interactions between the hydrophobic domains of the protein and the fatty acyl chains of the bilayer interior c. ionic interactions and hydrogen bonds between charged and polar side chains of the protein and polar head groups of the bilayer lipids d. a covalently bound lipid anchors to the bilayer and hydrophobic interactions between the hydrophobic domains of the protein and the fatty acyl chains of the bilayer interior e. a covalently bound lipid anchors to the bilayer and ionic interactions and hydrogen bonds between charged and polar side chains of the protein and polar head groups of the bilayer lipids ANSWER: b 53. What forces hold a peripheral membrane protein to the membrane? a. a covalently bound lipid anchors to the bilayer b. hydrophobic interactions between the hydrophobic domains of the protein and the fatty acyl chains of the bilayer interior c. ionic interactions and hydrogen bonds between charged and polar side chains of the protein and polar head groups of the bilayer lipids d. a covalently bound lipid anchors to the bilayer and hydrophobic interactions between the hydrophobic domains of the protein and the fatty acyl chains of the bilayer interior e. a covalently bound lipid anchors to the bilayer and ionic interactions and hydrogen bonds between charged and polar side chains of the protein and polar head groups of the bilayer lipids ANSWER: c

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Chapter 09: Nucleotides and Nucleic Acids 1. The compound that consists of ribose linked by an N-glycosidic bond to N-9 of adenine is: a. a deoxyribonucleoside. b. a purine nucleotide. c. a pyrimidine nucleotide. d. adenosine monophosphate. e. adenosine.

ANSWER: e 2. Purines have ____ ring(s), with each ring containing _____ nitrogen(s); Pyrimidines have one ring, with each ring containing _____ nitrogen(s). a. one; one; one b. one; two; two c. two; one; two d. two; two; one e. two; two; two ANSWER: e 3. A major component of RNA but not of DNA is: a. adenine. b. cytosine. c. guanine. d. thymine. e. uracil.

ANSWER: e 4. The difference between thymine and uracil is one _____ group on the _____ ring. a. methylene; pyrimidine b. methyl; pyrimidine c. hydroxyl; ribose d. amine; pyrimidine e. methyl; purine ANSWER: b 5. The difference between a ribonucleotide and a deoxyribonucleotide is a: a. deoxyribonucleotide has an —H instead of an —OH at C-2. b. deoxyribonucleotide has an —H instead of an —OH at C-3. c. ribonucleotide has an extra —OH at C-4. d. ribonucleotide has less structural flexibility than deoxyribonucleotide. e. ribonucleotide is a pyranose, deoxyribonucleotide is a furanose.

ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 09: Nucleotides and Nucleic Acids 6. The phosphodiester bonds that link adjacent nucleotides in both RNA and DNA: a. always link A with T and G with C. b. are susceptible to alkaline dehydration. c. are uncharged at neutral pH. d. form between the planar rings of adjacent bases. e. join the 3' hydroxyl of one nucleotide to the 5' hydroxyl of the next.

ANSWER: e 7. The phosphodiester bond that joins adjacent nucleotides in DNA: a. associates ionically with metal ions, polyamines, and proteins. b. is positively charged. c. is susceptible to alkaline hydrolysis. d. links C-2 of one base to C-3 of the next. e. links C-3 of deoxyribose to N-1 of thymine or cytosine.

ANSWER: a 8. The alkaline hydrolysis of RNA does NOT produce: a. 2'-AMP. b. 2',3'-cGMP. c. 2'-CMP. d. 3',5'-cAMP. e. 3'-UMP. ANSWER: d 9. The DNA oligonucleotide abbreviated pATCGAC: a. has seven phosphate groups. b. has a hydroxyl at its 3' end. c. has a phosphate on its 3' end. d. has an A at its 3' end. e. violates Chargaff's rules.

ANSWER: b 10. For the oligoribonucleotide pACGUAC, the nucleotide at the: a. 3' end has a phosphate at its 3' hydroxyl. b. 3' end is a purine. c. 5' end has a 5' hydroxyl. d. 5' end has a phosphate on its 5' hydroxyl. e. 5' end is a pyrimidine.

ANSWER: d 11. The nucleic acid bases: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 09: Nucleotides and Nucleic Acids a. absorb UV light maximally at 280 nm. b. are all about the same size. c. are relatively hydrophilic. d. are roughly planar. e. can all stably base-pair with one another.

ANSWER: d 12. In a double-stranded nucleic acid, cytosine typically base-pairs with: a. adenosine. b. guanine. c. inosine. d. thymine. e. uracil.

ANSWER: b 13. In the Watson-Crick model for the DNA double helix (B form) the A=T and G≡C base pairs share which property? a. The distance between the two glycosidic (base-sugar) bonds is the same in both base pairs, within a few tenths of an angstrom. b. The molecular weights of the two base pairs are identical. c. The number of hydrogen bonds formed between the two bases of the base pair is the same. d. The plane of neither base pair is perpendicular to the axis of the helix. e. The proton-binding groups in both base pairs are in their charged or ionized form.

ANSWER: a 14. Which is true in typical DNA? a. A = G b. A = C c. A = U d. A + T = G + C e. A + G = T + C ANSWER: e 15. Which base compositions for double-stranded DNA are possible? a. 5% A, 45% G, 45% C, 5% T, 0% U b. 20% A, 20% G, 20% C, 20% T, 20% U c. 35% A, 15% G, 35% C, 15% T, 0% U d. All of the answers are correct. e. None of the answers are correct.

ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 09: Nucleotides and Nucleic Acids 16. In the Watson-Crick structure of B-form DNA, the: a. planes of the bases are nearly perpendicular to the helical axis. b. adenine content of one strand must equal the thymine content of the same strand. c. nucleotides are arranged in the A form. d. purine content (fraction of bases that are purines) must be the same in both strands. e. two strands are parallel.

ANSWER: a 17. In the Watson-Crick model for the DNA double helix, which statement is NOT true? a. The two strands run antiparallel to one another. b. The base-pairing occurs on the inside of the double helix. c. The double helix is right-handed. d. There are two equally sized grooves that run up the sides of the helix. e. The two strands have complementary sequences.

ANSWER: d 18. In the Watson-Crick model of DNA structure: a. both strands run in the same direction, 3'→5'; they are parallel. b. phosphate groups project toward the middle of the helix, where they are protected from interaction with water. c. T can form three hydrogen bonds with either G or C in the opposite strand. d. the distance between the sugar backbone of the two strands is just large enough to accommodate either two purines or two pyrimidines. e. the vertical distance between two adjacent bases in one strand is about 3.4 Å.

ANSWER: e 19. Which statement is NOT true of all naturally occurring DNA? a. Deoxyribose units are connected by 3',5'-phosphodiester bonds. b. The amount of A always equals the amount of T. c. The ratio A+T/G+C is constant for all natural DNAs. d. The two complementary strands are antiparallel. e. Two hydrogen bonds form between A and T. ANSWER: c 20. In the Watson-Crick model of DNA structure (B-form DNA): a. a purine in one strand always hydrogen bonds with a purine in the other strand. b. A=T pairs share three hydrogen bonds. c. G≡C pairs share two hydrogen bonds. d. the 5' ends of both strands are at one end of the helix. e. the bases occupy the interior of the helix.

ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 09: Nucleotides and Nucleic Acids 21. The double helix of DNA in the B form is stabilized by: a. covalent bonds between the 3' end of one strand and the 5' end of the other. b. hydrogen bonding between the phosphate groups of two side-by-side strands. c. hydrogen bonds between the riboses of each strand. d. nonspecific base-stacking interaction between two adjacent bases in the same strand. e. ribose interactions with the planar base pairs.

ANSWER: d 22. B-form DNA in vivo is a _____-handed helix, _____ Å in diameter with a rise of _____ Å per base pair. a. left; 20; 3.9 b. right; 18; 3.4 c. right; 18; 3.6 d. right; 20; 3.4 e. right; 23; 2.6 ANSWER: d 23. For the helix in double-stranded B-form DNA, the majority of the stability can be attributed to: a. base-pairing interactions via hydrogen bonds. b. hydration of the phosphate backbone. c. metal cations interacting with the phosphate backbone and base-stacking interactions. d. covalent bonds between adjacent bases in one strand. e. the presence of a dipole along each strand.

ANSWER: c 24. The sequence of one strand of double-stranded DNA is shown in the 5' → 3' direction. Which sequence is palindromic? a. AGGTCC b. CCTTCC c. GAATCC d. GGATCC e. GTATCC ANSWER: d 25. Which base compositions for single-stranded RNA are possible? a. 5% A, 45% G, 45% C, 0% T, 5% U b. 25% A, 25% G, 25% C, 0% T, 25% U c. 35% A, 10% G, 30% C, 0% T, 25% U d. All of these answers are correct. e. None of these answers are correct.

ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 09: Nucleotides and Nucleic Acids 26. Double-stranded regions of RNA: a. are less stable than double-stranded regions of DNA. b. can be observed in the laboratory, but probably have no biological relevance. c. can form between two self-complementary regions of the same single strand of RNA. d. do not occur. e. have the two strands arranged in parallel (unlike those of DNA, which are antiparallel).

ANSWER: c 27. Double-stranded regions of RNA typically take on a(n) _____-handed helix. a. A-form left b. A-form right c. B-form left d. B-form right e. Z-form left ANSWER: b 28. When double-stranded DNA is heated at neutral pH, which change does NOT occur? a. the absorption of UV light increases (260 nm) b. the covalent N-glycosidic bond between the base and the pentose breaks c. the helical structure unwinds d. the hydrogen bonds between A and T break e. the viscosity of the solution decreases ANSWER: b 29. Which deoxyoligonucleotide is complementary to the DNA sequence (5')AGACTGGTC(3')? a. (5')CTCATTGAG(3') b. (5')GACCAGTCT(3') c. (5')GAGTCAACT(3') d. (5')TCTGACCAG(3') e. (5')TCTGGATCT(3') ANSWER: b 30. The ribonucleotide polymer (5')GUGAUCAAGC(3') could only form a double-stranded structure with: a. (5')CACUAGUUCG(3'). b. (5')CACAUGUUCG(3'). c. (5')CACUUUCGCCC(3'). d. (5')GCUUGAUCAC(3'). e. (5')GCCUAGUUUG(3').

ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 09: Nucleotides and Nucleic Acids 31. In the laboratory, several factors are known to cause alteration of the chemical structure of DNA. The factor(s) likely to be important in a living cell is/are: a. heat. b. low pH. c. reactive oxygen species. d. UV light. e. both reactive oxygen species and UV light.

ANSWER: e 32. In living cells, nucleotides and their derivatives can serve as: a. carriers of metabolic energy. b. enzyme cofactors. c. intracellular signals. d. precursors for nucleic acid synthesis. e. All of the answers are correct.

ANSWER: e 33. The energy carrier ATP is an example of a: a. deoxyribonucleoside triphosphate. b. dinucleotide. c. peptide. d. ribonucleotide. e. ribonucleoside triphosphate.

ANSWER: e 34. The chemical difference between a nucleotide and a nucleoside is that: a. a different nitrogenous base is added to each. b. nucleotides do not contain phosphate groups. c. nucleosides do not contain phosphate groups. d. nucleotides do not contain pentose sugars. e. nucleosides do not contain pentose sugars.

ANSWER: c 35. A pyrimidine is joined covalently to the pentose sugar through the _____ of the base to the _____ carbon of the pentose. a. N-1; 1' b. N-9; 1' c. N-1; 5' d. N-9; 5' e. None of the answers are correct.

ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 09: Nucleotides and Nucleic Acids 36. The correct name for the molecule is:

a. 5-methylcytidine. b. C5-methylcytidine. c. 5-methyluridine. d. C5-methylthymidine. e. 5-methylthymidine.

ANSWER: a 37. The covalent backbone of DNA and RNA consists of alternating: a. phosphate groups and nitrogenous bases. b. phosphate groups and pentose residues. c. pentose residues and nitrogenous bases. d. pentose residues and cyclic nucleosides. e. None of the answers are correct.

ANSWER: b 38. DNA is _____ chemically stable than RNA due to the _____-hydroxyl groups on _____. a. more; 2'; RNA b. more; 2'; DNA c. more; 3'; RNA d. less; 2'; RNA e. less; 2'; DNA ANSWER: a 39. Which statement is false about base stacking when nucleic acids are in a double-stranded conformation? a. Stacking interactions occur when two or more bases are positioned with their rings parallel to each other. b. Stacking interactions are hydrophilic in nature. c. Stacking involves a combination of van der Waals and hydrophobic interactions between bases. d. Base stacking helps to minimize contact of the bases with water. e. Base stacking interactions are important in stabilizing the three-dimensional structure of nucleic acids.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 09: Nucleotides and Nucleic Acids 40. The MOST striking difference between A-, B-, and Z-form DNA is that: a. A-form does not contain a major and minor groove. b. B-form has the smallest helical diameter. c. Z-form is left-handed. d. B-form has the greatest number of base pairs per helical turn. e. A-form maximizes base-stacking interactions.

ANSWER: c 41. Which single-stranded nucleic acid could form a hairpin structure? a. (5')TTTGCGATACTCATCGCATT(3') b. (5')TTTGCGATACTCACGCTATT(3') c. (5')TTTGCGATACTCTGCGATTT(3') d. All of the answers are correct. e. None of the answers are correct.

ANSWER: a 42. A tetraplex of DNA is MOST likely to form with which residue? a. adenosine b. thymidine c. guanosine d. cytidine e. A tetraplex is equally likely with all bases.

ANSWER: c 43. These structures would MOST likely be formed by:

a. W: RNA; X: RNA; Y: DNA b. W: RNA; X: DNA; Y: DNA c. W: DNA; X: DNA; Y: RNA d. W: DNA; X: RNA; Y: RNA e. W: RNA; X: RNA; Y: RNA Copyright Macmillan Learning. Powered by Cognero.

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Chapter 09: Nucleotides and Nucleic Acids ANSWER: a 44. Which DNA primer would have the highest melting temperature? a. GCATCGGC b. AATCGGAT c. ATACAGATCGGC d. ACCGGCAGGTCGGC e. ATACGCAGATCGGC ANSWER: d 45. Which type of damage to DNA structure is MOST likely to be caused by UV light? a. deamination b. pyrimidine dimers c. depurination d. depyrimidination e. hydrolysis of the phosphodiester bond ANSWER: b 46. Which molecule contains the greatest amount of stored chemical energy? a. dGTP b. GDP c. dGDP d. GMP e. cGMP ANSWER: a 47. One of the MOST common regulatory second messengers in cells is: a. FAD. b. NAD+. c. coenzyme A. d. ATP. e. cAMP. ANSWER: e 48. Discuss the similarities and differences between the structures of nucleosides and nucleotides.

ANSWER: Both have a nitrogenous base and a pentose; nucleotides also have a phosphate group, which nucleosides lack.

49. Match the type of bond with its role. (a) phosphodiester _____ links base to pentose in nucleotide (b) N-glycosidic _____ joins adjacent nucleotides in one strand (c) phosphate ester _____ joins complementary nucleotides in two strands Copyright Macmillan Learning. Powered by Cognero.

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Chapter 09: Nucleotides and Nucleic Acids (d) hydrogen ANSWER: b; a; d; c

_____ difference between a nucleoside and a nucleotide

50. (a) Compounds that contain a nitrogenous base, a sugar, and a phosphate group are called _____. (b) Two purines found in DNA are _____ and _____. (c) A pyrimidine found in all DNA but in only some RNA is _____. (d) In DNA, the base pair _____ is held together by three hydrogen bonds; (e) the base pair _____ has only two such bonds.

ANSWER: (a) nucleotides; (b) adenine; guanine; (c) thymidine; (d) G≡C; (e) A=T 51. The composition (mole fraction) of one of the strands of a double-helical DNA is [A] = 0.3, and [G] = 0.24.

Calculate the composition, if possible. If impossible, write I. For the same strand: (a) [T] = ____ (b) [C] = ____ (c) [T] + [C] = ____ For the other strand: (d) [A] = ____ (e) [T] = ____ (f) [A] + [T] = ____ (g) [G] = ____ (h) [C] = ____ (i) [G] + [C] = ____ ANSWER: (a) I; (b) I; (c) 0.46; (d) I; (e) 0.3; (f) I; (g) I; (h) 0.24; (i) I 52. What is the approximate length in micrometers of a DNA molecule (in the B-form) containing 10,000 base pairs?

ANSWER: 3.4 Å × 10,000 = 34,000 Å = 3.4 μm 53. Describe briefly what is meant by saying that two DNA strands are complementary.

ANSWER: The nucleotide sequences of complementary strands are such that: Wherever an A occurs in one strand, there is a T in the other strand with which it can form a hydrogen-bonded base pair.

Wherever a C occurs in one strand, a G occurs in the other. A is the base complementary to T, and C is the base complementary to G. 54. In one sentence, identify the most obvious structural difference between A-form (Watson-Crick) DNA and Z-form DNA. ANSWER: A-form DNA is a right-handed helix; Z-form DNA is a left-handed helix. (See Fig. 9.14.)

55. Write a double-stranded DNA sequence containing a six-nucleotide palindrome.

ANSWER: Any double-stranded sequence that has the form: 1-2-3-4-5-6-6'-5'-4'-3'-2'-1' 1'-2'-3'-4'-5'-6'-6-5-4-3-2-1 where each number and its prime represent correctly paired bases (A with T, C with G) all along the double-stranded molecule.

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Chapter 09: Nucleotides and Nucleic Acids 56. Describe briefly how noncovalent interactions contribute to the three-dimensional shapes of RNA molecules.

ANSWER: Hydrogen bonding in regions of complementarity within an RNA chain can result in regions of double helix that are stabilized by base stacking. Breaks in complementary regions can result in loops and bulges that, together with the helical regions, can generate a precise three-dimensional structure.

57. Describe qualitatively how the Tm for a double-stranded DNA depends on its nucleotide composition.

ANSWER: In general, the higher the proportion of G and C, the higher the melting temperature, Tm. More thermal energy is required to break the three hydrogen bonds holding G≡C pairs than to break the two hydrogen bonds holding A=T pairs. Also, G≡C pairs stack better than A=T pairs, thereby stabilizing the double helix better.

58. A solution of DNA is heated slowly until the Tm is reached. What is the likely structure of the DNA molecules at this temperature?

ANSWER: The DNA molecules are partially denatured; in each molecule approximately 50% of the DNA is single-stranded and 50% is double-helical. The single-stranded regions, which appear as bubbles within the molecules, are those that denatured at lower temperatures because of their higher content of A=T base pairs.

59. Based on the spontaneous deamination of cytosine, explain why it is advantageous for DNA to contain thymine, not uracil.

ANSWER: Cytosine undergoes spontaneous deamination to form uracil. If uracil were normally present in DNA, a deamination event could not be recognized as damaged goods. Thus, C→U changes would not be repaired by normal cellular processes and these new U's would end up pairing with A, leading to a point mutation. The extra methyl group in thymine is sufficient for differentiating between U and T. 60. What is the principal effect of UV radiation on DNA? ANSWER: UV radiation causes the formation of a dimer between adjacent thymine bases on the same DNA strand. This results in a kink in the double helix at that site.

61. Speculate why adenosine, rather than another large molecule, is used often as a cofactor in a nucleotidebinding fold of different enzymes.

ANSWER: Although adenosine does not contribute more than other nucleotides in binding energy, once ATP was selected through evolution as a source of chemical energy, its abundance made it a logical choice for incorporation into enzymes. Additionally, a single protein domain could be used in multiple enzymes to bind adenosine.

62. The second messenger ______ forms from ATP in a reaction catalyzed by adenylyl cyclase. a. cADP b. cAMP c. dAMP d. cATP Copyright Macmillan Learning. Powered by Cognero.

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Chapter 09: Nucleotides and Nucleic Acids e. dATP ANSWER: b 63. The α, β and β, γ linkages of nucleoside triphosphates are: a. O-glycosidic bonds. b. N-β-glycosyl bonds. c. phosphate bonds. d. phosphoanhydride bonds. e. phosphodiester bonds.

ANSWER: d 64. The nucleotides or nucleosides in enzymatic cofactors all include a _______ base. a. adenine b. cytosine c. guanine d. thymine e. uracil ANSWER: a 65. Why is it speculated that adenine is always found in enzymatic cofactors (including nucleosides or nucleotides)?

ANSWER: In the structures of enzymatic cofactors, adenine is used by various enzymes to bind to them. When these cofactors first formed in Earth’s early atmosphere, adenine was likely abundant in the primordial soup from which life arose and additional adenine bases can be generated from five molecules of HCN. Also, the purine ring of adenine is readily bound via base stacking in a ribozyme active site.

66. Complete the following table on the abbreviations of ribonucleoside 5’-phosphates. How do the abbreviations differ for deoxyribonucleoside 5’-phosphates?

Base Mono- DiTriAdenine AMP a ATP Guanine b GDP GTP Cytosine c d CTP ANSWER: a) ADP b) GMP c) CMP d) CDP The abbreviations for deoxyribonucleoside 5’-phosphates would have a ‘d’ in front of them to indicate the ‘deoxy’.

67. Describe the role of some nucleotides in regulation and signaling.

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Chapter 09: Nucleotides and Nucleic Acids ANSWER: Nucleotides are often second messengers that are produced after interactions between receptors and their hormones (or other extracellular chemical signaling molecules). The nucleotides signal and cause changes intracellularly. Adenosine 3’,5’-cyclic monophosphate (cAMP), guanosine 3’,5’cyclic monophosphate (cGMP), and guanosine 5’-diphosphate,3’-diphosphate (ppGpp) are examples of regulatory nucleotides.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism 1. When a DNA molecule is described as replicating bidirectionally, that means that it has two: a. chains. b. polymerases that synthesize 3′→5′ and 5′→3′. c. origins. d. replication forks. e. termination points.

ANSWER: d 2. An Okazaki fragment is a: a. fragment of DNA resulting from endonuclease action. b. fragment of RNA that is a subunit of the 30S ribosome. c. piece of DNA that is synthesized in the 3′→5′ direction. d. segment of DNA that is an intermediate in the synthesis of the lagging strand. e. segment of mRNA synthesized by RNA polymerase.

ANSWER: d 3. Which statement about enzymes that interact with DNA is true? a. E. coli DNA polymerase I is unusual in that it possesses only a 5′→3′ exonucleolytic activity. b. Endonucleases degrade circular but not linear DNA molecules. c. Exonucleases degrade DNA at a free end. d. Many DNA polymerases have a proofreading 5′→3′ exonuclease. e. Primases synthesize a short stretch of DNA to prime further synthesis.

ANSWER: c 4. E. coli DNA polymerase III: a. can initiate replication without a primer. b. is efficient at nick translation. c. is the principal DNA polymerase in chromosomal DNA replication. d. represents over 90% of the DNA polymerase activity in E. coli extracts. e. requires a free 5′-hydroxyl group as a primer.

ANSWER: c 5. The proofreading function of DNA polymerase does NOT involve: a. a 3′→5′ exonuclease. b. base pairing. c. detection of mismatched base pairs. d. phosphodiester bond hydrolysis. e. reversal of the polymerization reaction.

ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism 6. Prokaryotic DNA polymerase III: a. contains a 5′→3′ proofreading activity to improve the fidelity of replication. b. does not require a primer molecule to initiate replication. c. has subunits that act as a clamp to improve the processivity of DNA synthesis. d. synthesizes DNA in the 3′→5′ direction. e. synthesizes only the leading strand; DNA polymerase I synthesizes the lagging strand.

ANSWER: c 7. The role of the Dam methylase is to: a. add a methyl group to uracil, converting it to thymine. b. modify the template strand for recognition by repair systems. c. remove a methyl group from thymine. d. remove a mismatched nucleotide from the template strand. e. replace a mismatched nucleotide with the correct one.

ANSWER: b 8. When bacterial DNA replication introduces a mismatch in double-stranded DNA, the methyl-directed repair system: a. cannot distinguish the template strand from the newly replicated strand. b. changes both the template strand and the newly replicated strand. c. corrects the DNA strand that is methylated. d. corrects the mismatch by changing the newly replicated strand. e. corrects the mismatch by changing the template strand.

ANSWER: d 9. In base-excision repair, the first enzyme to act is: a. AP endonuclease. b. Dam methylase. c. DNA glycosylase. d. DNA ligase. e. DNA polymerase.

ANSWER: c 10. Which mechanism is used to repair a chemically modified base in DNA? a. mismatch repair b. base-excision repair c. nucleotide-excision repair d. direct repair e. More than one mechanism is used for this type of lesion.

ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism 11. Which statement does NOT accurately describe a difference between E. coli DNA polymerases Pol I and Pol III? a. Pol I has a 3′→5′ exonuclease, and DNA Pol III does not. b. Pol I has a slower polymerization rate than Pol III. c. Pol I is less processive than Pol III. d. Pol I is a single subunit, while Pol III is a multisubunit protein. e. Both polymerases synthesize in the 5′→3′ direction.

ANSWER: a 12. In DNA replication in E. coli: a. the leading strand requires multiple primers. b. polymerization is mostly 3′→5′. c. the process always finishes at a conserved terminator sequence. d. the process proceeds bidirectionally. e. the action of topoisomerases is not required.

ANSWER: d 13. Which statement is false regarding the DNA ligase mechanism? a. The mechanism is an example of covalent catalysis. b. Either ATP or NAD+ may contribute an adenyl group to the reaction. c. DNA ligase makes a phosphodiester bond between two nucleic acids. d. One of the products of the reaction is ADP. e. Okazaki fragments are joined together.

ANSWER: d 14. The entire complex of enzymes and proteins that carries out replication is called the: a. polymerase. b. replicatase. c. replisome. d. Okazaki complex. e. duplicasome.

ANSWER: c 15. Why does DNA replication in E. coli have a leading and lagging strand? a. Because the replication is 5′→3′ on one strand and 3′→5′ on the other strand. b. Because of the difference in supercoiling between the two strands. c. Because the replication of each strand is at a different replication fork. d. Because the DNA strands are antiparallel.

ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism 16. What does it mean to say that DNA replication is semiconservative?

ANSWER: Semiconservative means that after replication, each double-stranded DNA contains one newly synthesized strand and one strand from the original template DNA. The template DNA was thus conserved (saved), but it only comprises half (semi) of a newly synthesized duplex DNA. 17. The DNA shown is replicated from left to right. Label the templates for leading strand and lagging strand synthesis.

(5′)ACTTCGGATCGTTAAGGCCGCTTTCTGT(3′) (3′)TGAAGCCTAGCAATTCCGGCGAAAGACA(5′) ANSWER: The polarity of the strands indicates that the top strand is the template for lagging strand synthesis, and the bottom strand is the template for leading strand synthesis. (See Fig. 10.2.)

18. All known DNA polymerases catalyze synthesis only in the 5′→3′direction. Nevertheless, during semiconservative DNA replication in the cell, they are able to catalyze the synthesis of both daughter chains, which would appear to require synthesis in the 3′→5′direction. Explain the process by which DNA polymerase synthesizes both daughter strands during DNA replication.

ANSWER: During DNA replication, one strand is synthesized continuously and the other is synthesized by a discontinuous mechanism. The daughter chain, which appears to be growing in the 3′→5′ direction (the lagging strand), is actually being synthesized by the continual initiation of new chains and their elongation in the 5′→3′ direction.

19. What is an Okazaki fragment? What enzyme(s) is required for its formation in E. coli?

ANSWER: An Okazaki fragment is an intermediate in DNA replication in E. coli. It is a short fragment of newly synthesized DNA, attached to the 3′ end of a short RNA primer. Such fragments are produced by the combined action of primase (part of the primosome) and DNA polymerase III during replication of the lagging strand. (See Fig. 10.7.)

20. Diagram the reaction catalyzed by DNA polymerase that occurs between deoxyribose at the end of a DNA chain and the 5′ phosphates of a deoxyribonucleoside triphosphate. Include the chemical structure of the phosphate group, indicate the locations of the sugar and base, and show the rearrangements of electrons that occur.

ANSWER: See Figure 10.3.

21. A suitable substrate for DNA polymerase is shown. Label the primer and template strand, and indicate which end of each strand must be 3′ or 5′.

To observe DNA synthesis on this substrate in vitro, what additional reaction components must be added?

ANSWER: The top strand (the primer) has its 5′ end to the left; the bottom (template) strand has the opposite polarity. For DNA synthesis with this substrate in vitro, one would have to add DNA polymerase, the four deoxynucleoside triphosphates, Mg2+, and a suitable buffer.

22. All known DNA polymerases can only elongate a preexisting DNA chain (i.e., require a template and a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism primer) but cannot initiate a new DNA chain. Nevertheless, during semiconservative DNA replication in the cell, entirely new daughter DNA chains are synthesized. Explain the process that occurs in the cell that allows for the synthesis of daughter chains by DNA polymerase.

ANSWER: In the cell, initiation of DNA chains occurs via the synthesis of an RNA primer by an RNA polymerase type of enzyme (primase). This primer is elongated by DNA polymerase to produce the daughter DNA chain. The RNA is removed by 5' exonucleolytic hydrolysis before replication is completed.

23. DNA replication in E. coli begins at a site in the DNA called the (a) _____. At the replication fork the (b) _____ strand is synthesized continuously while the (c) _____ strand is synthesized discontinuously. On the strand synthesized discontinuously, the short pieces are called (d) _____ fragments. An RNA primer for each of the fragments is synthesized by an enzyme called (e) _____, and this RNA primer is removed after the fragment is synthesized by the enzyme (f) _____, using its (g) _____ activity. The nicks left behind in this process are sealed by the enzyme (h) _____.

ANSWER: (a) origin; (b) leading; (c) lagging; (d) Okazaki; (e) primase; (f) DNA Pol I; (g) 5′→3′ exonuclease; (h) DNA ligase 24. DNA synthesis on the lagging strand in E. coli is a complex process known to involve several proteins.

Initiation of a new chain is catalyzed by the enzyme (a) _____, and elongation is catalyzed by the enzyme (b) _____. Synthesis is discontinuous, yielding short segments called (c) _____, which are eventually joined by the enzyme (d) _____, which requires the cofactor (e) _____.

ANSWER: (a) primase; (b) DNA Pol III; (c) Okazaki fragments; (d) DNA ligase; (e) NAD+ 25. The high fidelity of DNA replication is due primarily to immediate error correction by the 3′→5′ exonuclease (proofreading) activity of the DNA polymerase. Some incorrectly paired bases escape this proofreading, and further errors can arise from challenges to the chemical integrity of the DNA. List the four classes of repair mechanisms that the cell can use to help correct such errors.

ANSWER: The four classes consist of (1) mismatch repair, (2) base-excision repair, (3) nucleotide-excision repair, and (4) direct repair.

26. List three types of DNA damage that require repair.

ANSWER: Defects in DNA that require repair include (1) mismatches that occur during replication; (2) abnormal bases; (3) pyrimidine dimers produced by UV irradiation. Other answers are possible.

27. Explain how inheriting mutations in genes encoding DNA repair enzymes could lead to increased cancer risk.

ANSWER: Mutations in DNA repair genes impair the cell’s ability to fix DNA damage, leading to an increased mutation rate. Since cancers begin with mutations in oncogenes, an increased mutation rate would lead to an increased risk of cancer.

28. RNA polymerase: a. binds tightly to a region of DNA thousands of base pairs away from the DNA to be transcribed. b. can synthesize RNA chains without a primer. c. has a λ subunit, which acts as a proofreading ribonuclease.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism d. separates DNA strands throughout a long region of DNA (up to thousands of base pairs), then copies one of them. e. synthesizes RNA chains in the 3′→5′ direction.

ANSWER: b 29. Which statement about E. coli RNA polymerase is false? a. Elongation by RNA polymerase is primarily done by the σ factor. b. RNA polymerase holoenzyme has several subunits. c. RNA produced by this enzyme will be complementary to the DNA template. d. The enzyme adds nucleotides to the 3′ end of the growing RNA chain. e. The enzyme cannot synthesize RNA in the absence of DNA. ANSWER: a 30. Which statement about E. coli RNA polymerase (core enzyme) is false? a. In the absence of the σ subunit, core polymerase has little specificity for where initiation begins. b. The core enzyme contains several different subunits. c. The core enzyme has no polymerizing activity until the σ subunit is bound. d. The RNA chain grows in a 5′→3′ direction. e. The RNA product is complementary to the DNA template.

ANSWER: c 31. Which property does NOT apply to RNA polymerase from E. coli (core enzyme alone)? a. It can extend an RNA chain and initiate a new chain. b. It is required for the synthesis of mRNA, rRNA, and tRNA in E. coli. c. It produces an RNA polymer that begins with a 5′-triphosphate. d. It recognizes specific start signals in DNA. e. It requires all four ribonucleoside triphosphates and a DNA template.

ANSWER: d 32. The σ factor of E. coli RNA polymerase: a. associates with the promoter before binding core enzyme. b. combines with the core enzyme to confer specific binding to a promoter. c. is inseparable from the core enzyme. d. is required for termination of an RNA chain. e. will catalyze synthesis of RNA from both DNA template strands in the absence of the core enzyme.

ANSWER: b 33. Which characteristic does NOT apply to the ρ terminator in E. coli? a. The ρ terminator has an ATP-dependent RNA-DNA helicase activity.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism b. The ρ terminator migrates in the 3′→5′ direction along RNA. c. The ρ terminator causes RNA polymerase release at a CA-rich sequence near the end of the transcript. d. ATP is hydrolyzed by the ρ terminator during the termination process. e. All of these characteristics apply to the ρ terminator.

ANSWER: b 34. Processing of a primary mRNA transcript in a eukaryotic cell does NOT normally involve: a. attachment of a long poly(A) sequence at the 3′ end. b. conversion of normal bases to modified bases, such as inosine and pseudouridine. c. excision of intervening sequences (introns). d. joining of exons. e. methylation of one or more guanine nucleotides at the 5′ end.

ANSWER: b 35. The 5'-terminal cap structure of eukaryotic mRNAs is a(n) _____ nucleotide joined to the mRNA via a _____. a. 7-methylcytosine; 2′,3′-cyclic linkage b. 7-methylguanosine; 5′,3′-diphosphate linkage c. 7-methylguanosine; 5′,5′-triphosphate linkage d. N6-methyladenosine; 5′,5′-phosphodiester bond e. O6-methylguanosine; 5′,5′-triphosphate linkage ANSWER: c 36. Eukaryotic nuclear pre-mRNA splicing by the spliceosome requires: a. a guanine nucleoside or nucleotide. b. endoribonucleases. c. polynucleotide phosphorylase. d. RNA polymerase II. e. small nuclear ribonucleoproteins (snRNPs).

ANSWER: e 37. Which statement is false about eukaryotic mRNA and its processing? a. Exons are used for polypeptide synthesis. b. Introns are complementary to their adjacent exons and will form hybrids with them. c. The mature mRNA is substantially shorter than the corresponding region on the DNA. d. The mRNA is originally synthesized in the nucleus but ends up in the cytoplasm. e. The splicing that yields a mature mRNA occurs at very specific sites in the RNA primary transcript.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism ANSWER: b 38. Which statement about the synthesis of rRNA and tRNA in E. coli is true? a. Both rRNA and some tRNAs are part of the same primary transcript. b. Each rRNA sequence (16S, 23S, 5S) is transcribed as a separate primary transcript. c. Primary tRNA transcripts undergo methylation, but rRNA sequences are not methylated. d. The tRNA sequences all lie at the 3′ end of the rRNA transcripts. e. There is a single copy of the rRNA genes.

ANSWER: a 39. One process the reverse transcriptase of an animal RNA virus catalyzes is: a. degradation of the RNA strand in a DNA-RNA hybrid. b. insertion of the viral genome into a chromosome of the host (animal) cell. c. RNA formation in the 3′→5′ direction. d. RNA synthesis but not DNA synthesis. e. synthesis of an antisense RNA transcript.

ANSWER: a 40. Reverse transcriptase: a. can utilize only RNA templates. b. has a 3′→5′ proofreading exonuclease but not a 5′→3′ exonuclease. c. is activated by AZT. d. is encoded by retroviruses. e. synthesizes DNA with the same fidelity as a typical DNA polymerase.

ANSWER: d 41. What acts as the nucleophile in the mechanism of nucleotide addition by RNA polymerase? a. 5′ phosphate of an incoming nucleotide b. water molecule c. 5′ hydroxyl of the template DNA d. 3′ hydroxyl from the RNA being extended e. aspartate in the active site ANSWER: d 42. What metal ion cofactors are required to bind nucleotide triphosphate substrates by E. coli RNA polymerase? a. Fe2+ b. Zn2+ c. Mg2+ Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism d. Mn2+ e. None of these ions are required.

ANSWER: c 43. Which statement is true regarding the RNA product synthesized by E. coli RNA polymerase? a. Its sequence is identical to that of the template strand. b. Its sequence is identical to that of the nontemplate strand. c. It is able to form a duplex with the template strand. d. It is able to form a duplex with the nontemplate strand e. None of the statements are true.

ANSWER: c 44. What statement is true regarding the synthesis of nucleic acids by E. coli RNA polymerase? a. The RNA-DNA hybrid assumes a B-form DNA conformation. b. The RNA-DNA hybrid is antiparallel. c. The RNA-DNA hybrid is between 20 and 30 base pairs long. d. Unwinding the DNA duplex creates negative supercoils in the template DNA. e. None of the statements are true.

ANSWER: b 45. Which statement is false with respect to different σ subunits in bacterial RNA polymerase? a. Specific amino acids in σ subunits will interact with the polymerase. b. DNA sequences spanning approximately three turns of the helix will specifically interact with σ subunits. c. Both DNA and RNA polymerases interact with σ subunits. d. Multiple σ subunits exist in a single E. coli RNA polymerase. e. The binding of NusA is competitive with the binding of σ subunits in E. coli.

ANSWER: d 46. Which activity is NOT associated with reverse transcriptases? a. RNA-dependent DNA synthesis b. synthesis of RNA primers c. RNA degradation d. DNA-dependent DNA synthesis e. All of these are associated with reverse transcriptase.

ANSWER: b 47. Consider the small hypothetical gene. The DNA base pairs corresponding to the first and last nucleotides transcribed are indicated by a + and an *, respectively. Give the sequence of the mRNA strand produced by this gene. Label the 5′ and 3′ends of the mRNA. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism

ANSWER: (5')GCUAUGUACGUAGCUACGU(3') 48. Write the sequence of the messenger RNA molecule synthesized from a DNA template strand having the sequence (5′)ATCGTACCGTTA(3′).

ANSWER: (3')UAGCAUGGCAAU(5'); also acceptable is (5')UAACGGUACGAU(3') 49. List one important property that distinguishes RNA polymerases from DNA polymerases, and list one important property they share.

ANSWER: Among the distinguishing characteristics: RNA polymerase does not require a primer, but DNA polymerase does; RNA polymerase lacks the 3'→5' proofreading exonuclease activity present in DNA polymerase. Among the shared properties: both enzymes use nucleoside triphosphates as substrates, require Mg2+ to produce an antiparallel complement to the template, and synthesize nucleic acids in the 5'→3' direction.

50. Shown is an RNA molecule being transcribed from a strand of DNA. Indicate the 5′ and 3′ ends of the RNA molecule and the strand of DNA that is complementary to the RNA molecule. In which direction is synthesis occurring?

ANSWER:

51. Describe the sequence of events in the initiation of transcription by E. coli RNA polymerase.

ANSWER: The core enzyme plus σ subunit, called holoenzyme, binds to the promoter region forming a closed complex (i.e., in which the DNA double helix is not unwound). This is converted to an open complex by the unwinding of a short region of the promoter. Synthesis of the RNA chain begins within the complex. The complex then moves along the DNA away from the promoter region and the σ subunit dissociates.

52. Compare and contrast ρ-dependent and ρ-independent termination of transcription in prokaryotes.

ANSWER: In both, the transcription complex pauses at specific sequences in the DNA template. In ρCopyright Macmillan Learning. Powered by Cognero.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism independent termination, the formation of hairpin structures causes pausing, which leads to dissociation of the product RNA and the polymerase from the template DNA and hence to termination. In ρ-dependent termination, the features that cause pausing are unclear, and dissociation requires the presence of the ρ protein.

53. Rifampin blocks RNA polymerase; why would this be lethal to a cell that was sensitive to this drug?

ANSWER: Without RNA synthesis, there could be no protein synthesis, and the cell would be unable to replenish proteins that had been turned over or synthesize new proteins. Since both processes are necessary to life, the sensitive cell would stop growing and die.

54. Compare reverse transcription and transcription in terms of the listed characteristics. a. Direction of polynucleotide synthesis b. Nature of template c. Nature of primer d. Incorporated nucleotides ANSWER: Reverse Transcription Transcription (a) Direction of polynucleotide synthesis (b) Nature of template (c) Nature of primer (d) Incorporated nucleotides

5'→3' RNA or DNA tRNA dNTPs

5'→3' DNA none NTPs

55. Housekeeping genes in bacteria are commonly expressed constitutively, but not all of these genes are expressed at the same level (the same number of molecules per cell). The primary mechanism responsible for variations in the level of constitutive enzymes from different genes is that: a. all constitutive enzymes are synthesized at the same rate, but are not degraded equally. b. their promoters have different affinities for RNA polymerase holoenzyme. c. some constitutively expressed genes are more inducible than others. d. some constitutively expressed genes are more repressible than others. e. the same number of mRNA copies are made from each gene but are translated at different rates.

ANSWER: b 56. Which normally binds to the operator region? a. attenuators b. inducers c. mRNAs d. repressors e. suppressor tRNAs ANSWER: d 57. The diagram represents a hypothetical operon in the bacterium E. coli. The operon consists of two structural genes (A and B), which code for the enzymes A-ase and B-ase, respectively, and also includes P (promoter) and O (operator) regions as shown.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism

For this arrangement of an operon, which statement is true about mRNA synthesis? a. All four regions (A, B, O, and P) will be transcribed into an mRNA that will then be translated into four different proteins. b. Gene B will never be transcribed from this operon. If B-ase is produced, it is from another site where B has its own promoter and operator. c. Gene B will be transcribed before gene A. d. An activator must bind to the O region before transcription can begin. e. When RNA polymerase makes mRNA from this operon, it begins RNA synthesis just to the left of gene A. ANSWER: e 58. The diagram represents a hypothetical operon in the bacterium E. coli. The operon consists of two structural genes (A and B) that code for the enzymes A-ase and B-ase, respectively, and also includes P (promoter) and O (operator) regions as shown.

When a certain compound (X) is added to the growth medium of E. coli, the separate enzymes A-ase and B-ase are both synthesized at a 50-fold higher rate than in the absence of X (which has a molecular weight of about 200). Which statement is true of such an operon? a. Adding X to the growth medium causes a repressor protein to be released from the O region. b. Adding X to the growth medium causes a repressor protein to bind tightly to the O region. c. Synthesis of the mRNA from this operon is not changed by the addition of compound X. d. X must be an activator that binds at the operator. e. Adding X causes an activator to be released from the promoter.

ANSWER: a 59. Transcription of the lactose (lac) operon in E. coli is stimulated by: a. a mutation in the repressor gene that strengthens the affinity of the repressor for the operator. b. a mutation in the repressor gene that weakens the affinity of the repressor for the operator. c. a mutation in the repressor gene that weakens the affinity of the repressor for the inducer. d. binding of the repressor to the operator. e. the presence of glucose in the growth medium.

ANSWER: b 60. Which is correct about operators and operons? a. An operator is a repressor binding site. An operon consists of the gene cluster, promotor, and regulatory sequences. b. An operator is a molecule that binds to an operon to facilitate the binding of architectural regulators. c. An operon is a protein that binds to an operator sequence to repress transcription. d. An operon is a clustering of four or more operators.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism e. There is no difference; the terms are interchangeable (synonyms).

ANSWER: a 61. Which statement about eukaryotic gene regulation is correct? a. Large polycistronic transcripts are common. b. Most regulation is positive, involving activators rather than repressors. c. Eukaryotic regulatory proteins are similar in complexity and size to prokaryotic ones. d. Transcription does not involve promoters. e. Transcription occurs without major changes in chromosomal organization.

ANSWER: b 62. Which statement about eukaryotic gene regulation is NOT correct? a. Access to eukaryotic promoters is restricted by the structure of chromatin. b. Most regulation is positive, involving activators rather than repressors. c. Larger and more multimeric proteins are involved in regulation of eukaryotic transcription. d. Transcription and translation are separated in both space and time. e. Strong promoters in eukaryotes are generally fully active in the absence of regulatory proteins.

ANSWER: e 63. A(n) _____ is a DNA sequence. a. coactivator b. corepressor c. enhancer d. inducer e. transactivator ANSWER: c 64. Which type of eukaryotic regulatory proteins interacts with enhancers? a. basal transcription factors b. coactivators c. repressors d. TATA-binding proteins e. transcription activators ANSWER: e 65. Consider the lac operon of E. coli. When there is neither glucose nor lactose in the growth medium: a. CAP protein binds to the lac operator. b. CAP protein displaces the Lac repressor from the lac promoter. c. the repressor is bound to the lac operator. d. RNA polymerase binds the lac promoter and transcribes the lac operon. e. the operon is fully induced.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism ANSWER: c 66. The three types of regulatory proteins that regulate transcription initiation by RNA polymerase are: a. effectors, specificity factors, and enhancers. b. activators, effectors, and repressors. c. effectors, enhancers, and repressors. d. specificity factors, repressors, and activators. e. enhancers, repressors, and activators.

ANSWER: d 67. If the binding of a small molecule to a regulatory protein lowers the Kd of the protein/operator-binding interaction, what can reasonably be stated? a. The small molecule activates transcription by decreasing inhibition by the protein. b. The small molecule inhibits transcription by increasing inhibition by the protein. c. The small molecule activates transcription by increasing activation by the protein. d. The small molecule inhibits transcription by decreasing activation by the protein. e. None of these statements are reasonable.

ANSWER: b 68. Allolactose binding to the I gene product in the regulation of the lac operon is an example of: a. transcriptional activation by reducing repressor binding. b. transcriptional inhibition by increasing repressor binding. c. transcriptional activation by increasing activator binding. d. transcriptional inhibition by decreasing activator binding. e. architectural regulation.

ANSWER: a 69. Which outcome will be true when E. coli is grown in the absence of glucose? a. The level of cyclic AMP will be low. b. The CRP (cAMP receptor protein) cannot bind to its target DNA. c. If lactose is present, synthesis of lac operon mRNA will occur at a high level. d. A repressor is activated by cyclic AMP and will prevent the binding of RNA polymerase to the lac operon promoter. e. None of the statements are true.

ANSWER: c 70. What kinds of regulation exist for the lac operon in E. coli? a. negative regulation and transcription attenuation b. negative regulation and positive regulation c. positive regulation and transcription attenuation d. only negative regulation Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism e. only transcription attenuation ANSWER: b 71. Describe and contrast positive regulation and negative regulation of gene expression.

ANSWER: Positive regulation involves an activator protein that, when bound in the vicinity of the regulated gene, facilitates the binding of RNA polymerase to its promoter. The affinity of the activator for the DNA may be either increased or decreased by a signal molecule. Negative regulation involves a repressor protein that, when bound near the regulated gene, hinders access of RNA polymerase to its promoter or inhibits its activity. Repressor affinity for its binding site (the operator) may be increased or decreased by a signal molecule. (See Figure 10.26.)

72. Define operon and polycistronic mRNA. ANSWER: An operon is several contiguous structural genes that are coordinately regulated by promoter and operator regions, as depicted in Figure 10.27. All of the genes in an operon are transcribed in a polycistronic mRNA. That is, a single mRNA molecule encoding more than one structural gene.

Note: Operons are not found in eukaryotic organisms.

73. Draw a simple map of the lactose operon indicating the relative positions of promoter, operator, cap-binding site, repressor gene (I), and the structural lacZ gene of the operon.

ANSWER: See Figure 10.28.

74. Briefly explain (a) why there is a lag in cell growth when bacteria are switched from a medium containing glucose to one containing lactose. (b) When the growth medium contains both lactose and glucose, what proteins will be bound to the lac operon regulatory region? (c) If only lactose is in the growth medium, what proteins will be bound to the lac operon regulatory region?

ANSWER: (a) The lag represents the time required for lactose to enter the cell, be converted into the inducer allolactose, and turn on the synthesis of the genes of the lac operon that are essential for growth on lactose. (b) When both glucose and lactose are present, neither Lac repressor nor CAP will be bound. (c) If only lactose is present, CAP will be bound. (See Figure 10.28.)

75. Describe briefly the relationship between chromatin structure and transcription in eukaryotes.

ANSWER: Heterochromatin is highly condensed and transcriptionally inert because the chromatin proteins make promoters inaccessible. The less condensed euchromatin has undergone alteration of its structure ("remodeling"), allowing some regions to be transcribed. These alterations include covalent modification (such as acetylation) of histones and displacement of nucleosomes, creating exposed regions of DNA (hypersensitive sites) that are probably binding sites for regulatory proteins.

76. Define in one to two sentences (a) heterochromatin; (b) euchromatin; (c) chromatin remodeling.

ANSWER: (a) Heterochromatin is highly condensed chromatin that is transcriptionally inactive, often associated with structures such as centromeres. (b) Euchromatin is most of the chromatin, not nearly as highly condensed as heterochromatin, and able to be activated for transcription. (c) Chromatin remodeling reflects structural changes resulting from enzymes that covalently modify the core histone complex in nucleosomes (such as histone acetyltransferases) and those that move or displace nucleosomes using the energy of ATP hydrolysis.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism 77. "A protein with a gene that is actively being transcribed will be present in high concentrations." Why is this statement not accurate?

ANSWER: Protein concentration does not exclusively depend on the rate of transcription. It will also depend on the stability of the RNA, the rate of translation, the rate of protein processing (e.g., posttranslational modifications), and the rate of protein degradation.

78. Northern blotting is a technique used to detect RNA content. When analyzing transcriptional regulation with Northern blots, the RNA of housekeeping genes is typically examined in parallel with the gene of interest. Why would this be important?

ANSWER: Since housekeeping genes are expressed constitutively, their RNA concentration should not vary. In this way, these genes act as internal controls for RNA content.

79. The DNA in the E. coli chromosome is BEST described as: a. a single circular double-helical molecule. b. a single linear double-helical molecule. c. a single linear single-stranded molecule. d. multiple linear double-helical molecules. e. multiple linear single-stranded molecules.

ANSWER: a 80. The DNA in a eukaryotic chromosome is BEST described as: a. a single circular double-helical molecule. b. a single linear double-helical molecule. c. a single linear single-stranded molecule. d. multiple linear double-helical molecules. e. multiple linear single-stranded molecules.

ANSWER: b 81. The genome of organism A is 4.6 million base pairs long, while that of organism B is 12.1 million base pairs long. Given this information, what can be stated regarding these two organisms? a. Organism B will contain approximately three times the number of genes as organism A. b. Organism B will contain approximately 7.5 million more base pairs in introns compared with organism A. c. Organism B will contain three times as many chromosomes as organism A. d. The genome of each has its own unique number of genes.

ANSWER: d 82. Histones are _____ that are usually associated with _____. a. acidic proteins; DNA b. acidic proteins; RNA c. basic proteins; DNA d. basic proteins; RNA Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism e. coenzymes derived from histidine; enzymes ANSWER: c 83. Which protein does NOT contribute to the octameric histone core? a. H1 b. H2A c. H2B d. H3 e. H4 ANSWER: a 84. The fundamental repeating unit of organization in a eukaryotic chromosome is the: a. centrosome. b. lysosome. c. microsome. d. nucleosome. e. polysome.

ANSWER: d 85. Nucleosomes: a. are important features of chromosome organization in eukaryotes and bacteria. b. are composed of proteins rich in acidic amino acids, such as Asp and Glu. c. are composed of protein and RNA. d. bind DNA and alter its supercoiling. e. occur in chromatin at irregular intervals along the DNA molecule.

ANSWER: d 86. Which statement is false regarding histones? a. Overall, histones are positively charged proteins. b. Lys and Arg residues are abundant in histones relative to other proteins. c. Histone H1 is found in the nucleosome core. d. The nucleosome core contains eight histone proteins.

ANSWER: c 87. What kind of posttranslational modifications can occur in histones? a. only methylation b. only acetylation c. only phosphorylation d. methylation and acetylation e. methylation, acetylation, and phosphorylation ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism 88. How many nucleosomes would be expected to be found in a 1,000 bp piece of linear DNA? a. zero to one b. two to three c. four to five d. six to seven e. eight to nine ANSWER: c 89. Heterochromatin is comprised of: a. linked chromatin between two or more chromosomes. b. DNA at the centromere. c. DNA-protein regions that are transcriptionally active. d. DNA-RNA hybrid regions that form during transcription. e. highly condensed transcriptionally inactive regions of DNA. ANSWER: e 90. Describe the composition and structure of a nucleosome.

ANSWER: A nucleosome consists of 146 base pairs of double-stranded DNA wrapped around a core of histones (small, basic proteins). This core contains two copies each of histones H2A, H2B, H3, and H4. See Figures 10.13 and 10.14.

91. Introns: a. are frequently present in prokaryotic genes but are rare in eukaryotic genes. b. are spliced out before transcription. c. are translated but not transcribed. d. can occur many times within a single gene. e. encode unusual amino acids in proteins.

ANSWER: d 92. What are introns?

ANSWER: Introns are regions of genes (primarily eukaryotic) that are transcribed into pre-mRNA but are not translated into protein. They do not code for amino acid sequences.. As a result, they interrupt the colinearity between the nucleotide sequence of the gene and the amino acid sequence of the encoded protein.

93. Which statement correctly describes promoters in E. coli? a. A promoter may be present on either side of a gene or in the middle of it. b. All promoters have the same sequence that is recognized by RNA polymerase holoenzyme. c. Every promoter has a different sequence, with little or no resemblance to other promoters. d. Many promoters are similar and resemble a consensus sequence, which has the highest affinity for RNA polymerase holoenzyme.

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Chapter 10: Biological Information Part 1: DNA and RNA Metabolism e. Promoters are not essential for gene transcription, but can increase its rate by two- to threefold.

ANSWER: d 94. Which statement is false for euchromatin? a. Euchromatin is a form of chromatin. b. Euchromatin is being more actively transcribed than heterochromatin. c. Euchromatin is more densely packed than heterochromatin. d. None of these statements are false.

ANSWER: d

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Chapter 11: Biological Information Part 2: Protein Metabolism 1. A certain bacterial mRNA is known to represent only one gene and to contain about 800 nucleotides.

Assuming that the average amino acid residue contributes 110 Da to the peptide molecular weight, the largest polypeptide that this mRNA could code for would have a molecular weight (Da) of about: a. 800. b. 5,000. c. 30,000. d. 80,000. e. An upper limit cannot be determined from the data given.

ANSWER: c 2. Assuming that the average amino acid residue contributes 110 Da to the molecular weight of a peptide, what will be the approximate minimum length of the mRNA encoding a protein with a molecular weight of 50,000 Da? a. 133 nucleotides b. 460 nucleotides c. 1,400 nucleotides d. 5,000 nucleotides e. A minimum length cannot be determined from the data given.

ANSWER: c 3. Which statement describes a feature of the wobble hypothesis? a. A naturally occurring tRNA exists in yeast that can read both arginine and lysine codons. b. A tRNA can recognize only one codon. c. Some tRNAs can recognize codons that specify two different amino acids if both are nonpolar. d. The third base of the codon pairs loosely with the corresponding base of its anticodon. e. The middle position of an anticodon is not important to the amino acid specified.

ANSWER: d 4. Which statement is true about the genetic code? a. All codons recognized by a given tRNA encode different amino acids. b. There is wide variation in the genetic code among all organisms. c. Several different codons may encode the same amino acid. d. The base in the middle position of the tRNA anticodon sometimes permits wobble base pairing with two or three different codons. e. The first position of the tRNA anticodon is always adenosine.

ANSWER: c 5. Which step is NOT one of the steps of protein synthesis? a. termination and complete proteolytic degradation of the ribosome b. elongation of the peptide chain c. initiation of peptide synthesis Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11: Biological Information Part 2: Protein Metabolism d. activation of the amino acids e. protein folding and posttranslational processing ANSWER: a 6. Which statement about ribosomes is true? a. The large subunit contains rRNA molecules; the small subunit does not. b. The RNA in ribosomes plays a structural, not a catalytic, role. c. There are about 25 ribosomes in an E. coli cell. d. There are two major ribosomal subunits, each compromising multiple proteins and RNA. e. Ribosomes are relatively small, with molecular weights of less than 10,000.

ANSWER: d 7. Which statement about tRNA molecules is false? a. A, C, G, and U are the only bases present in the molecule. b. Although composed of a single strand of RNA, each molecule contains several short, double-helical regions. c. Any given tRNA will accept only one specific amino acid. d. The amino acid attachment is always to an A nucleotide at the 3′ end of the molecule. e. There is at least one tRNA for each of the 20 amino acids.

ANSWER: a 8. Which statement about the tRNA that normally accepts phenylalanine is false? (mRNA codons for phenylalanine are UUU and UUC.) a. It may be represented as tRNAPhe. b. It will accept only the amino acid phenylalanine. c. The tRNA likely contains 73 to 93 nucleotides. d. Phenylalanine can be specifically attached to an —OH group at the 3′ end. e. The tRNA must contain the sequence UUU. ANSWER: e 9. Which statement is false about tRNA molecules? a. The 3′-terminal sequence is CCA. b. Their anticodons are complementary to the triplet codon in the mRNA. c. They contain more than four different bases. d. They contain several double-helical regions. e. With the right enzyme, any given tRNA molecule will accept any of the 20 amino acids.

ANSWER: e 10. Aminoacyl-tRNA synthetases (amino acid activating enzymes): a. "recognize" specific tRNA molecules and specific amino acids.

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Chapter 11: Biological Information Part 2: Protein Metabolism b. attach an amino acid to the appropriate tRNA in conjunction with another enzyme. c. interact directly with free ribosomes. d. are specific for one or two amino acids. e. require GTP to activate the amino acid.

ANSWER: a 11. Which statement about aminoacyl-tRNA synthetases is false? a. Some aminoacyl-tRNA synthetases have an editing/proofreading capability. b. Aminoacyl-tRNA synthetases attach an amino acid to the 3′ end of a tRNA. c. Aminoacyl-tRNA synthetases cleave ATP to AMP + PPi. d. Aminoacyl-tRNA synthetases will use any tRNA species but are highly specific for a given amino acid. e. There is a different aminoacyl-tRNA synthetase for every amino acid.

ANSWER: d 12. The aminoacyl-tRNA synthetase that attaches an amino acid to a tRNA: a. only recognizes one specific tRNA. b. attaches a specific amino acid to any available tRNA species. c. attaches the amino acid at the 5′ end of the tRNA. d. catalyzes the formation of an ester bond. e. hydrolyzes ATP to ADP + Pi. ANSWER: d 13. In the activation of an amino acid for protein synthesis: a. leucine can be attached to tRNAPhe by the aminoacyl-tRNA synthetase specific for leucine. b. methionine is first formylated, then attached to tRNAMet. c. the amino acid is attached to the 5′ end of the tRNA through a phosphodiester bond. d. there is at least one specific activating enzyme and one specific tRNA for each amino acid. e. two separate enzymes are required, one to form the aminoacyl adenylate and the other to attach the amino acid to the tRNA. ANSWER: d 14. Which statement is true for protein synthesis in eukaryotes? a. All proteins are initially synthesized with methionine at their C terminus. b. All proteins are initially synthesized with methionine at their N terminus. c. All proteins are initially synthesized with tryptophan at their C terminus. d. All proteins are initially synthesized with a repeat of five amino acids at their carboxyl terminal. e. None of the statements are true.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11: Biological Information Part 2: Protein Metabolism 15. Which statement about bacterial mRNA translation is true? a. A ribosome initiates translation near the end of the mRNA that is synthesized last. b. An mRNA is rarely degraded and is usually passed on to the daughter cells after cell division. c. During protein synthesis, ribosomes move along the mRNA in the 5′→3′ direction. d. Ribosome subunits stay together without dissociation for the synthesis of hundreds of proteins. e. The codon signaling peptide termination is located in the mRNA near its 5′ end.

ANSWER: c 16. Formation of the ribosomal initiation complex for bacterial protein synthesis does NOT require: a. EF-Tu. b. N-formylmethionyl-tRNAfMet. c. GTP. d. initiation factor 2 (IF2). e. mRNA. ANSWER: a 17. In bacteria, the elongation stage of protein synthesis does NOT involve: a. aminoacyl-tRNAs. b. EF-Tu. c. GTP. d. IF2. e. peptidyl transferase.

ANSWER: d 18. Bacterial ribosomes: a. bind tightly to specific regions of DNA, forming polysomes. b. contain at least one catalytic RNA molecule (ribozyme). c. contain three types of RNAs and five different proteins. d. have specific, different binding sites for each of the 20 tRNAs. e. require puromycin for normal function.

ANSWER: b 19. On average approximately how many NTPs must be converted to NDPs to incorporate one amino acid into a protein? a. zero b. one c. two d. four e. twelve ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11: Biological Information Part 2: Protein Metabolism 20. In eukaryotes, which molecule is covalently linked to proteins headed for degradation? a. mannose b. ubiquitin c. guanosine d. acetyl-CoA e. methionine ANSWER: b 21. What is an open reading frame (ORF)? a. a DNA sequence that can be transcribed into five or more different mRNAs b. a DNA sequence that contains no initiation codon c. a stretch of at least 50 codons that does not contain a termination codon d. an mRNA sequence that can be translated into more than one protein e. "junk" DNA that has no known function ANSWER: c 22. The 5′→3′ initiation codon is _____, and it codes for the amino acid _____. a. AUG; Met b. UUA; Met c. CCG; Glu d. GAA; Glu e. GGG; Gly ANSWER: a 23. Which amino acids are coded by a single codon? a. only methionine b. only isoleucine c. methionine and isoleucine d. methionine and tryptophan e. methionine, isoleucine, and tryptophan ANSWER: d 24. What does it mean to say that the genetic code is degenerate? a. More than one tRNA can carry a given amino acid. b. The genetic code is universal throughout living organisms. c. There is more than one codon for most amino acids. d. Only the first two nucleotides in a codon are needed to specify the amino acid. e. The genetic code cannot be described as being degenerate.

ANSWER: c 25. If only the nucleotides C and G are present in an mRNA strand beyond the start codon, which of the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11: Biological Information Part 2: Protein Metabolism following amino acids would NOT be incorporated into the resulting polypeptide?

5′end ↓ U C A G U UUU Phe UCU Ser UAU Tyr UGU Cys UUC Phe UCC Ser UAC Tyr UGC Cys UUA Leu UCA Ser UAA Stop UGA Stop UUG Leu UCG Ser UAG Stop UGG Trp C CUU Leu CCU Pro CAU His CGU Arg CUC Leu CCC Pro CAC His CGC Arg CUA Leu CCA Pro CAA Gln CGA Arg CUG Leu CCG Pro CAG Gln CGG Arg A AUU Ile ACU Thr AAU Asn AGU Ser AUC Ile ACC Thr AAC Asn AGC Ser AUA Ile ACA Thr AAA Lys AGA Arg AUG Met ACG Thr AAG Lys AGG Arg G GUU Val GCU Ala GAU Asp GGU Gly GUC Val GCC Ala GAC Asp GGC Gly GUA Val GCA Ala GAA Glu GGA Gly GUG Val GCG Ala GAG Glu GGG Gly a. Ala b. Arg c. Glu d. Gly e. Pro ANSWER: c 26. Attachment of an amino acid to a tRNA: a. consumes the equivalent of at least two phosphate bonds. b. requires ATP to activate the carboxylate of the amino acid. c. has as a final product an ester linkage between an adenosine and the amino acid. d. All of the answers are correct.

ANSWER: d 27. Which statement is false regarding the role of EF-G in prokaryotic translation? a. EF-G can bind to the A site of the ribosome. b. EF-G is associated with the translocation step of translation. c. GTP is hydrolyzed while bound to EF-G. d. EF-G is structurally similar to the EF-Tu–tRNA complex. e. None of the statements is false.

ANSWER: e 28. The Shine-Dalgarno sequence: a. is found on the 5′ side of the transcription initiation codon in mRNA. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11: Biological Information Part 2: Protein Metabolism b. forms a duplex with sequences in prokaryotic mRNA. c. is important in eukaryotic gene translation. d. is complementary to the initiator tRNA anticodon. e. is a nontranslated sequence after the stop codon in mRNA. ANSWER: a 29. Which is an extremely toxic protein that inactivates ribosomes by depurination of an adenosine? a. ricin b. cyanide c. chloramphenicol d. puromycin e. arsenic ANSWER: a 30. Which reaction in protein synthesis leads to the formation of an ester bond in one of its products? a. tRNA charging with amino acids b. binding of aminoacyl-tRNA to the ribosome c. reaction of acylated tRNA in the A site with acylated tRNA in the P site d. the translocation step in protein synthesis e. release of the final peptide product ANSWER: a 31. If fMet-tRNAfMet was able to bind to EF-Tu, it would be: a. delivered to the ribosomal A site. b. delivered to the ribosomal P site. c. delivered to the ribosomal E site. d. unable to bind to the ribosome at all. e. degraded.

ANSWER: a 32. If a glycosylated protein is purified from the ER, what would likely be true about this protein? a. The glycosyl groups will contain very little mannose. b. The glycosylation is likely to be N-linked. c. The protein will still contain a signal sequence. d. The protein will be targeted to lysosomes. e. The protein will be an integral membrane protein.

ANSWER: b 33. Ubiquitin is added to proteins targeted for degradation by formation of a(n) _____ bond between _____ of ubiquitin and a _____ the target protein. a. ester; the C terminus; serine of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11: Biological Information Part 2: Protein Metabolism b. thioester; the C terminus; cysteine of c. amide; the C terminus; lysine of d. ester; a glutamate; serine in e. thioester; a glutamate; cysteine in ANSWER: c 34. Which statement is false concerning ubiquitin? a. It is a prokaryotic protein and is not present in eukaryotes. b. This protein is covalently added to proteins targeted for degradation. c. This protein is small (<100 amino acids). d. Its sequence is highly conserved among the organisms in which it is present. e. It forms an amide bond to a target protein.

ANSWER: a 35. Consider the hypothetical short mRNA shown. Where does translation begin, and what would be the sequence of the protein produced if this were translated in an E. coli cell?

5′-AUAGGAGGUUUGACCUAUGCCUCGUUUAUAGCC-3′ C A G 5′ end ↓ U U UUU Phe UCU Ser UAU Tyr UGU Cys UUC Phe UCC Ser UAC Tyr UGC Cys UUA Leu UCA Ser UAA Stop UGA Stop UUG Leu UCG Ser UAG Stop UGG Trp C CUU Leu CCU Pro CAU His CGU Arg CUC Leu CCC Pro CAC His CGC Arg CUA Leu CCA Pro CAA Gln CGA Arg CUG Leu CCG Pro CAG Gln CGG Arg A AUU Ile ACU Thr AAU Asn AGU Ser AUC Ile ACC Thr AAC Asn AGC Ser AUA Ile ACA Thr AAA Lys AGA Arg AUG Met ACG Thr AAG Lys AGG Arg G GUU Val GCU Ala GAU Asp GGU Gly GUC Val GCC Ala GAC Asp GGC Gly GUA Val GCA Ala GAA Glu GGA Gly GUG Val GCG Ala GAG Glu GGG Gly ANSWER: Begin translation at the start codon, AUG. Sequence: N–Met–Pro–Arg–Leu–C 36. Beyond the start codon, a segment of the template strand of a double-stranded DNA contains the sequence (5′)-CTT TGA TAA GGA TAG CCC TTC-(3′).

U C A G 5′ end ↓ U UUU Phe UCU Ser UAU Tyr UGU Cys UUC Phe UCC Ser UAC Tyr UGC Cys UUA Leu UCA Ser UAA Stop UGA Stop Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11: Biological Information Part 2: Protein Metabolism C

A

G

UUG Leu CUU Leu CUC Leu CUA Leu CUG Leu AUU Ile AUC Ile AUA Ile AUG Met GUU Val GUC Val GUA Val GUG Val

UCG Ser CCU Pro CCC Pro CCA Pro CCG Pro ACU Thr ACC Thr ACA Thr ACG Thr GCU Ala GCC Ala GCA Ala GCG Ala

UAG Stop CAU His CAC His CAA Gln CAG Gln AAU Asn AAC Asn AAA Lys AAG Lys GAU Asp GAC Asp GAA Glu GAG Glu

UGG Trp CGU Arg CGC Arg CGA Arg CGG Arg AGU Ser AGC Ser AGA Arg AGG Arg GGU Gly GGC Gly GGA Gly GGG Gly

a. What is the base sequence of the mRNA that can be transcribed from this strand segment? b. What amino acid sequence could be coded by the mRNA base sequence in (a), using only the first reading frame starting at the 5′ end? c. Suppose the other (complementary) strand segment is used as a template for transcription. What is the amino acid sequence of the resulting peptide, again starting from the 5′ end and using only the first reading frame?

ANSWER: (a) (5')GAA GGG CUA UCC UUA UCA AAG(3') (b) Glu–Gly–Leu–Ser–Leu–Ser–Lys (c) The codons translate to Leu–Stop–Stop. No peptide would be produced because of the stop codons.

37. Describe the possible outcomes that could occur as a result of a single base change in an mRNA. ANSWER: The most likely result is a single amino acid change in the encoded protein, when a codon is altered to one of another amino acid. However, some nucleotide changes will be "silent" and not change the protein if the altered codon still specifies the original amino acid. Conversion to a nonsense (stop) codon will result in a truncated polypeptide, while alteration of the normal stop codon to a "sense" codon will result in a lengthened protein. Alteration of the initiation codon may result in a total failure to translate, and result in no protein product. Changes in mRNA sequence outside the protein-coding region may affect translational efficiency, splicing, or mRNA turnover rates.

38. In protein synthesis, 61 codons specify the 20 amino acids. Base pairing between the codon and the tRNA anticodon assures that the correct amino acid will be inserted into the nascent polypeptide chain. Why then does the cell require only 32 different tRNAs to recognize 61 different codons?

ANSWER: Certain tRNAs have the unusual nucleotide inosinate in the first anticodon position, which base pairs with the third nucleotide of the codon, referred to as the wobble base. Because inosinate can base pair with A, U, or C, a tRNA containing inosinate can recognize three different codons. In each recognized codon, there is a standard anticodon-codon base pair with the first two bases of the codon; "wobble" in the third base pair allows one tRNA to read three different codons. Similarly, tRNAs with U or G in the first anticodon position also exhibit a wobble effect that permits pairing with two different codons.

39. Indicate whether the statements are true (T) or false (F). ___ A ribosome is the complex within which protein synthesis occurs.

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Chapter 11: Biological Information Part 2: Protein Metabolism ___ Ribosomes contain many separate proteins. ___ Ribosomes contain several different RNAs. ___ The three ribosomal RNAs in a bacterial ribosome are distributed in three separate, large ribosomal subunits. ___ The catalytic machinery of the ribosome is protein-based. ___ There are four binding sites for aminoacyl-tRNAs on a ribosome.

ANSWER: T; T; T; F; F; F 40. A given mRNA sequence might be translated in any of three reading frames. Describe how prokaryotes and eukaryotes determine the correct reading frame.

ANSWER: In prokaryotes, the Shine-Dalgarno sequence in the mRNA base pairs with a complementary sequence in the 16S RNA of the ribosome. This positions the correct start codon (AUG) on the 30S ribosomal subunit. Thus, the initiating AUG codon is distinguished by its proximity to the ShineDalgarno sequence. In eukaryotes, the initiating AUG codon is the first AUG that the ribosome encounters as it scans the mRNA from its 5' end. In both cases, the initiating AUG sets the correct reading frame.

41. Match the factor or enzyme at the right with the stage(s) of prokaryotic synthesis at which it acts. If a factor or enzyme participates in two stages of protein synthesis, indicate both of them. ___ Amino acid activation (a) release factor ___ Initiation (b) EF-Tu ___ Elongation (c) aminoacyl-tRNA ___ Termination (d) Shine-Dalgarno sequence ANSWER: c; d; b; a 42. Indicate whether the statements are true (T) or false (F). ___ Assembly of a complete ribosome onto an mRNA requires ATP hydrolysis. ___ Aminoacylation, or "charging," of tRNA requires the formation of an aminoacyl-AMP intermediate. ___ Aminoacyl-tRNA binding to the A site of the ribosome requires the accessory factor EF-G and GTP hydrolysis. ___ Translocation of a growing polypeptide from the A to the P site on the ribosome requires EF-G and GTP hydrolysis. ___ Release of the completed peptide requires release factors, but no NTP hydrolysis.

ANSWER: F; T; F; T; T 43. Number the steps in the proper order with regard to protein synthesis in prokaryotes. Note that not all steps are listed. ___ Aminoacyl-tRNA binds to the A site. ___ Deacylated tRNA is released from the ribosome. ___ Peptide bond formation shifts the growing peptide from the A to the P site. ___ The 50S subunit binds to the 30S subunit and mRNA, forming the initiation complex.

ANSWER: 2; 4; 3; 1 44. Polypeptide chain elongation in E. coli occurs by the cyclical repetition of three steps. What are these steps and what cellular components are necessary for each of them to occur?

ANSWER: The three steps: (1) An aminoacyl-tRNA is brought to the A site by EF-Tu with bound GTP; (2) the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11: Biological Information Part 2: Protein Metabolism peptidyl transferase activity of the ribosome, which acts as a ribozyme, catalyzes peptide bond formation; (3) the ribosome translocates three nucleotides down the mRNA, helped by EF-G (translocase) and hydrolysis of GTP. This shifts the peptidyl-tRNA from the A site to the P site and the deacylated tRNA from the P site to the E site, after which the tRNA is released from the E site, and once EF-G is released, another aminoacyl-tRNA can bind to the now-empty A site.

45. Following the synthesis of their polypeptide chain, many proteins require further posttranslational modifications before they attain their full biological activity or function. List and describe briefly at least four examples of specific modification that can occur.

ANSWER: (1) Amino-terminal modification (N-acetylation or deacetylation), (2) proteolytic cleavage, such as removal of signal sequences used for targeting, (3) side-chain modification (phosphorylation, carboxylation, methylation, etc.), (4) attachment of N-linked (to Asn) or O-linked (to Ser or Thr) oligosaccharide moieties, (5) isoprenylation/farnesylation of Cys side chains, (6) incorporation of prosthetic groups (heme, biotin, etc.), (7) processing of proenzymes or zymogens, and (8) formation of disulfide crosslinks.

46. When first synthesized, proinsulin has an additional leader or signal sequence at its amino terminus. This complete molecule is called preproinsulin and the signal sequence is cleaved off to give proinsulin. Briefly, what is the likely function of the signal sequence?

ANSWER: The leader peptide in preproinsulin is a signal sequence directing it to the endoplasmic reticulum, from which proinsulin (following cleavage of the signal sequence) enters the Golgi complex and is packaged into a secretory vesicle for secretion via exocytosis.

47. Describe the role of ubiquitin in mediating intracellular protein breakdown.

ANSWER: Ubiquitin, a protein found in all eukaryotic cells, is covalently attached to proteins targeted for degradation. The carboxyl-terminal residue of ubiquitin is first activated by the formation of a thioester with the first enzyme (E1) in the pathway in an ATP-dependent reaction. After transfer of the activated ubiquitin from the first enzyme to a second enzyme (E2), to which it is also connected via a thioester linkage, ubiquitin is finally joined through its carboxyl terminus to a lysine ε-amino group in the target protein by a third enzyme (E3). Multiple ubiquitin molecules can be attached to a target protein, labeling the protein for degradation by the cellular proteasome.

48. The codon for Trp is UGG, while UGA is a stop codon. Briefly explain why it would not be possible to have the reverse (UGG = stop, UGA = Trp).

ANSWER: In order to recognize the codon UGA, the anticodon for tRNATrp would have to be UCA, but the anticodon U is in the wobble position, so it might also recognize UGG (the anticodon UCA recognizes UG (A/G)), potentially causing a premature stop of translation. With UGG as the Trp codon, the anticodon is CCA, and C is in the wobble base position. However, even in this position, C will only pair with G, thus maintaining the integrity of the code. Inosine in the first position of the anticodon would recognize UG (A/U/C).

49. Name three GTP-binding factors that function in bacterial protein translation, and describe their function.

ANSWER: IF2, an initiation factor that is involved in binding of the initiator tRNA; EF-Tu, the elongation factor associated with binding of aminoacyl-tRNA molecules into the A site; EF-G that is associated with translocation of the ribosome (and also facilitates disassembly of the ribosome after release of the completed polypeptide).

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Chapter 11: Biological Information Part 2: Protein Metabolism

50. Which statement is false regarding riboswitches? a. Riboswitches are a type of cis-regulatory element. b. Riboswitches are natural ligand-binding RNA structures found in prokaryotes and some eukaryotes. c. Riboswitches are often found at the 5' end of an untranslated region of genes. d. Each riboswitch can bind an assortment of small metabolites. e. Riboswitches can affect either transcription or translation.

ANSWER: d 51. Gene silencing by miRNA interference acts by _____ of the target gene. a. only inhibiting transcription b. only inhibiting translation c. only inhibiting splicing d. degrading the mRNA or inhibiting translation e. only inhibiting polyadenylation ANSWER: d 52. Which method is NOT a way of regulating the concentration of a protein in a cell? a. transcriptional regulation b. translational regulation c. degradation of proteins d. posttranslational modification of proteins e. All of these methods are ways of regulating protein concentrations.

ANSWER: e 53. Which effect is NOT expected from changing the sequence in the –10 region of an E. coli promoter for a protein-encoding gene? a. decreased binding of σ subunits b. decreased transcription of the mRNA c. decreased binding of ribosomes to the mRNA d. lower concentrations of the protein product in the cell e. All of these effects would be expected.

ANSWER: c 54. What kinds of regulation exist for the trp operon in E. coli? a. negative regulation and transcription attenuation b. negative regulation and positive regulation c. positive regulation and transcription attenuation d. only negative regulation e. only transcription attenuation Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11: Biological Information Part 2: Protein Metabolism ANSWER: a 55. Which factor is NOT associated with translational regulation? a. attenuation in the trp operon b. ribosomal protein/mRNA interactions c. riboswitches d. DsrA interaction with rpoS e. All of these are examples of translational regulation.

ANSWER: a 56. What is a role of the riboswitches that regulate mRNA function in cis? a. to bind to a specific ligand b. to form a complex with the operator sequence c. to bind to the ribosome and inhibit its ribozyme activity d. to prevent ribosome binding to the Shine-Dalgarno sequence e. All of the answers are correct.

ANSWER: a 57. Describe two different mechanisms by which riboswitches can modulate mRNA. ANSWER: Upon binding an effector, riboswitches can (1) stabilize a termination structure, or (2) block the ribosomal binding site.

58. Large numbers of microRNAs (miRNAs), also known as small temporal RNAs (stRNAs), have now been discovered in higher eukaryotes. Describe their characteristics and general function.

ANSWER: Primary miRNA transcripts are about 70 nucleotides, with self-complementary internal sequences that can form hairpin structures. These are cleaved into short duplexes of 20 to 25 nucleotides. One strand is transferred to and binds the target mRNA. This can then inhibit gene expression by blocking translation of the target mRNA or facilitating its degradation.

59. The Hsp70 family and elaborate protein complexes called chaperonins are two types of ________, which help proteins fold into their native conformations. a. anticodons b. chaperones c. clients d. proteosomes e. riboswitches ANSWER: b 60. Which statement is true? a. Chaperonins denature folded proteins. b. The lifetime of proteins is highly regulated. c. Posttranslational processing of proteins is uncommon.

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Chapter 11: Biological Information Part 2: Protein Metabolism d. Proteins are folded by proteosomes. e. Proteins may need to have segments added by proteases.

ANSWER: b 61. A ________ is a supramolecular assembly that functions in the degradation of damaged or unneeded cellular proteins. a. anticodon b. client c. proteasome d. proteosome e. ubiquitin ANSWER: c 62. Regarding translation in eukaryotes versus that in prokaryotes (bacteria), indicate whether the statements are true (T) or false (F). ___ In eukaryotes, the 3′end of the mRNA is associated with the 5′end during initiation, whereas in prokaryotes it is not. ___ In prokaryotes, translation is initiated at the AUG codon nearest a Shine-Dalgarno sequence in the mRNA, whereas in eukaryotes it is initiated at an AUG codon near the 3′end of the mRNA. ___ In prokaryotes, translation is initiated with Met, whereas in eukaryotes it is initiated with fMet. ___ In prokaryotes, translation and transcription of an mRNA molecule can occur concurrently, whereas in eukaryotes they cannot.

ANSWER: T; F; F; T

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Chapter 12: Nucleic Acid Technologies 1. Restriction enzymes: a. act at the membrane to restrict the passage of certain molecules into the cell. b. are highly specialized ribonucleases that degrade mRNA soon after its synthesis. c. are sequence-specific DNA endonucleases. d. are proteases that cleave peptides only at recognized sequences. e. catalyze the addition of a certain amino acid to a specific tRNA. ANSWER: c 2. The E. coli recombinant plasmid pBR322 has been widely utilized in genetic engineering experiments.

Which feature does pBR322 NOT have? a. a number of conveniently located recognition sites for restriction enzymes b. a number of palindromic sequences near the EcoRI site, which permit the plasmid to assume a conformation that protects newly inserted DNA from nuclease degradation c. a replication origin, which permits it to replicate autonomously d. resistance to two different antibiotics, which permits rapid screening for recombinant plasmids containing foreign DNA e. small overall size, which facilitates entry of the plasmid into host cells ANSWER: b 3. A convenient cloning vector with which to introduce foreign DNA into E. coli is a(n): a. E. coli chromosome. b. messenger RNA. c. plasmid. d. yeast "ARS" sequence. e. yeast transposable element.

ANSWER: c 4. Common features found in a cloning plasmid used for protein expression do NOT include: a. a multiple cloning site insert. b. an origin of replication. c. an antibiotic-resistance marker. d. a ribosome-binding site. e. telomeric ends.

ANSWER: e 5. Which is NOT used to study protein function? a. green fluorescent protein b. label to identify conserved regions c. affinity chromatography d. mutant versions of proteins e. fusion proteins Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12: Nucleic Acid Technologies ANSWER: b 6. Current estimates indicate that humans have about _____ protein-coding genes. a. 2,000 b. 10,000 c. 20,000 d. 100,000 e. 200,000 ANSWER: c 7. Current estimates indicate that _____ of the human genome is translated into protein. a. less than 0.5% b. roughly 1.5% c. roughly 10% d. roughly 25% e. more than 50% ANSWER: b 8. Which method is NOT used in linkage analysis? a. comparing densely spaced polymorphisms b. collecting DNA from a family affected by the disease of interest c. sequencing selected parts of the genome d. introducing retroviruses at the mutated locus e. looking for SNP variants ANSWER: d 9. The sequence of one strand of dsDNA is shown in the 5'→3' direction. Which sequence is a palindrome? a. AATGCC b. GGATCC c. GATATG d. CCCGCG e. AGTAGT ANSWER: b 10. When cloning a gene into a plasmid, which enzyme would be used to covalently attach the gene to the plasmid DNA? a. DNA polymerase b. exonuclease III c. DNA ligase d. alkaline phosphatase e. restriction endonuclease Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12: Nucleic Acid Technologies ANSWER: c 11. A good cloning vector should NOT have: a. an origin of replication. b. an antibiotic resistance gene. c. several unique restriction sites. d. a small size. e. genes allowing its insertion into the host chromosome.

ANSWER: e 12. Which amino acid when repeated six to ten times at the N- or C-terminal ends of a protein allows that protein to bind to nickel and cobalt? a. Glu b. His c. Ala d. Tyr e. Asp ANSWER: b 13. What type of insert is referred to as a linker? a. type of cloning vector b. DNA fragment with multiple recognition sequences for restriction endonucleases c. type of expression vector d. enzyme that joins DNA molecules together e. method of getting DNA into a cell ANSWER: b 14. A good expression vector for protein expression in E. coli should have which feature(s)? a. an origin of replication b. a selectable marker c. a promoter, operator, and ribosome-binding site d. All of the answers are correct. e. None of the answers are correct.

ANSWER: d 15. The gene for green fluorescent protein was originally isolated from which species? a. panda b. firefly c. mushroom d. jellyfish e. glowworm Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12: Nucleic Acid Technologies ANSWER: d 16. What common protein tertiary structure does green fluorescent protein contain? a. α-helix b. β-barrel c. coiled-coil d. Greek key e. four-helix bundle ANSWER: b 17. Which protein forms a complex with a guide RNA to target DNA for cleavage and destruction? a. RecA b. Cas9 c. EcoRI d. Gal4p e. GST ANSWER: b 18. To study the effects of inactivating a specific gene on the development of a mouse, which technique or system would be used to inactivate that gene? a. transfection b. CRISPR/Cas9 c. tandem affinity purification d. yeast two-hybrid analysis e. oligonucleotide-directed mutagenesis ANSWER: b 19. Which makes up the largest portion of human DNA sequence types (25.9%)? a. introns b. exons c. transposons d. SINEs e. simple-sequence repeats ANSWER: a 20. Due primarily to single-nucleotide polymorphisms, each human differs from every other human, on average, by what percentage of the human genome? a. 6% b. 15% c. 0.1% d. 98% Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12: Nucleic Acid Technologies e. 1.5% ANSWER: c 21. Which repeating sequences in the human genome are the targets of technologies used in forensic DNA analysis? a. simple-sequence repeats b. single-nucleotide polymorphisms c. introns d. short tandem repeats e. long repetitive sequences ANSWER: d 22. The process of linkage analysis does NOT rely on which technique to determine the genetic causes of diseases? a. analysis of family pedigrees b. identification of SNPs that are most often inherited with the disease-causing gene c. the sequence of the entire chromosome where the disease-causing gene is located to directly identify it d. comparing the DNA of people who have the disease with the DNA of people who don't have the disease e. obtaining many sequences of a region of the chromosome that is associated with the disease ANSWER: c 23. Which statement about transposons is NOT correct? a. They are a form of molecular parasite. b. They have played a major role in human evolution. c. All transposons are active and move around the human genome. d. Some transposons move to another genomic location via an RNA intermediate. e. They often contain genes that encode proteins that catalyze transposon movement.

ANSWER: c 24. A plasmid that encodes resistance to ampicillin and tetracycline is digested with the restriction enzyme PstI, which cuts the plasmid at a single site in the ampicillin-resistance gene. The DNA is then annealed with a PstI digest of human DNA, ligated, and used to transform E. coli cells. (a) What antibiotic would be put in an agar plate to ensure that the cells of a bacterial colony contain the plasmid? (b) What antibiotic-resistance phenotypes will be found on the plate? (c) Which phenotype will indicate the presence of plasmids that contain human DNA fragments?

ANSWER: (a) tetracycline; (b) resistant to both antibiotics and resistant to only tetracycline; (c) The resistance to only tetracycline phenotype indicates that the gene for ampicillin resistance has been interrupted by the insertion of a human DNA fragment.

25. Match each feature of the plasmid pBR322 (at left) with one appropriate description presented (at right).

Descriptions may be used more than once.

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Chapter 12: Nucleic Acid Technologies

_____ ampR _____ ori _____ tetR _____ BamHI _____ PstI

(a) Permits selection of bacteria containing the plasmid. (b) A sequence required for packaging recombinant plasmids into bacteriophage. (c) Origin of replication. (d) Cleavage of the plasmid here does not affect antibiotic sequence-resistance genes. (e) Insertion of foreign DNA here permits identification of bacteria-containing recombinant plasmids.

ANSWER: a; c; a; e; e 26. Explain briefly the properties of the plasmid pBR322 that make it so convenient as a vector for cloning fragments of foreign DNA. ANSWER: pBR322 has two antibiotic-resistance markers, so that its presence in a bacterium can be detected, bacteria that carry it can be selected, and insertion of cloned DNA into one of the resistance markers can be detected by loss of antibiotic resistance. The plasmid also has an origin of replication so that it replicates autonomously in E. coli. Several convenient restriction sites allow easy insertion of restricted DNA fragments. Its small size facilitates its entry into E. coli by standard transformation protocols.

27. What sequences are required in an expression vector (for use with E. coli) that are NOT essential in a cloning plasmid?

ANSWER: Regulated expression of the cloned gene requires: (1) a bacterial promoter and (2) its associated operator, (3) a transcription-termination sequence, and (4) a ribosome-binding site.

28. Which would be expected to be larger, the percentage of the human genome that is translated into protein or the percentage of the genome of a bacterium that is translated into protein? Why? ANSWER: Because bacteria do not have introns, the percentage of their genome that is translated into protein would be larger than that of humans, who do have introns.

29. As the cost of sequencing personal human genome rapidly declines, consider the benefits and risks of having every fetus's DNA sequenced before birth.

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Chapter 12: Nucleic Acid Technologies ANSWER: Answers are expected to vary, but key points that should be brought up include: (1) identification of treatable diseases before they become a problem; (2) identification of untreatable diseases that lead to questions like abortion; (3) identification of trending mutations that may or may not be causative for some diseases; (4) too much data (remember the human genome has 3 billion bp) without sufficient understanding of what it means regarding individual human health.

30. Describe how the CRISPR/Cas9 system is used to inactivate a gene during the process of genomic engineering.

ANSWER: First, a plasmid with the CRISPR/Cas9 system is inserted into a target cell. This plasmid encodes both the Cas9 protein and the CRISPR single guide RNA (sgRNA) that is used to target a region of genomic DNA for cleavage. The sgRNA forms a complex with Cas9, and due to specific base pairing between the sgRNA and the genomic DNA target, the Cas9 protein is bound to the DNA target. The nuclease domains of the Cas9 protein can then cleave both of the DNA strands in the target gene, resulting in a double-strand break. This double-strand break is typically repaired by nonhomologous end joining, which often deletes or alters base pairs when the break is repaired. The alteration or deletion of base pairs may then inactivate the target gene on the cell's genome.

31. The image shows a DNA sequencing gel using the Sanger method. Migration during electrophoresis was from the top to the bottom. What is the sequence of the template DNA?

a. (5')ACTAGTGACCGT(3') b. (5')TGCCAGTGATCA(3') c. (5')ACGGTCACTAGT(3') d. (5')TGATCACTGGCA(3') e. It cannot be determined from this information.

ANSWER: d 32. The image shows a DNA sequencing gel using the Sanger method. Migration during electrophoresis was Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12: Nucleic Acid Technologies from the top to the bottom. Which band represents the shortest and longest fragments of DNA?

a. shortest = Z, longest = W b. shortest = X, longest = Z c. shortest = W, longest = Y d. shortest = Y, longest = X e. shortest = W, longest = Z ANSWER: a 33. Define short tandem repeat (STR) and explain briefly how an STR locus can be used to convict or acquit criminal suspects or provide paternity information.

ANSWER: A short tandem repeat is a location on a chromosome where a short (usually 4 bp) DNA sequence is repeated many times. The precise number of times the locus is repeated varies in the human population. The length of the STR can be determined using PCR, followed by capillary gel electrophoresis. Analysis of multiple STR loci can yield a profile unique to an individual, providing evidence for a criminal case or paternity test.

34. Which technique would be used to locate gene alterations that define inherited diseases? a. Comparative genomics b. CRISPR/Cas c. DNA cloning d. Mass spectrometry e. RNA-Seq ANSWER: a 35. Which technique would be used to study the evolution and migration of human ancestors?

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Chapter 12: Nucleic Acid Technologies a. CRISPR/Cas b. DNA cloning c. Human genomics d. Linkage analysis e. RNA-Seq ANSWER: c 36. Which technique defines cellular proteomes? a. CRISPR/Cas b. DNA cloning c. Linkage analysis d. Mass spectrometry e. RNA-Seq ANSWER: d 37. Genomic information can be used to: a. identify the perpetrator of a crime. b. identify sources of genetic disease. c. identify unknown victims of a crime. d. All of the above options are true. e. None of the above options are true.

ANSWER: d 38. The transcriptome consists of: a. all the cellular DNA present at a given moment. b. all the cellular RNA present at a given moment. c. all the cellular introns present at a given moment. d. all the mutations identified in an offspring. e. all the proteins present in a cell at a given moment.

ANSWER: b 39. In DNA sequencing by the Sanger (dideoxy) method: a. radioactive dideoxy ATP is included in each of four reaction mixtures before the enzymatic synthesis of complementary strands. b. specific enzymes are used to cut the newly synthesized DNA into small pieces, which are then separated by electrophoresis. c. the dideoxynucleotides must be present at high levels to obtain long stretches of DNA sequence. d. the role of the dideoxy CTP is to occasionally terminate the enzymatic synthesis of DNA where a G occurs in the template strands. e. the template DNA strand is radioactive.

ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12: Nucleic Acid Technologies 40. In the chemical synthesis of DNA the: a. dimethoxytrityl group catalyzes the formation of the phosphodiester bond. b. direction of synthesis is 5' to 3'. c. maximum length of oligonucleotide that can be synthesized is 8 to 10 nucleotides. d. only components that are needed are a DNA sample and a pair of oligonucleotide primers. e. protecting cyanoethyl groups are removed after each step.

ANSWER: b 41. When performing a polymerase chain reaction (PCR), what is the advantage of using a DNA polymerase isolated from a thermophilic organism? a. The PCR does not need to be cooled below the denaturation step. b. The PCR does not need primers to proceed. c. The PCR does not need to have fresh enzyme added after each cycle. d. The PCR does not need to have fresh dNTPs added after each cycle. e. The PCR does not need to have fresh template DNA added after each cycle.

ANSWER: c 42. The polymerase chain reaction mixture does NOT include: a. all four deoxynucleoside triphosphates. b. DNA containing the sequence to be amplified. c. DNA ligase. d. heat-stable DNA polymerase. e. oligonucleotide primers.

ANSWER: c 43. Which statement about the polymerase chain reaction (PCR) is false? a. DNA amplified by PCR can be cloned. b. DNA amplification is linear in magnitude. c. Newly synthesized DNA must be heat-denatured before the next round of DNA synthesis begins. d. The boundaries of the amplified DNA segment are determined by the synthetic oligonucleotides used to prime DNA synthesis. e. The technique is sufficiently sensitive that DNA sequences can be amplified from a single animal or human hair.

ANSWER: b 44. When two genes share detectable sequence similarity, those genes or their gene products, are said to be: a. homologs b. symbiologs. c. complementary sequences. d. co-genes.

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Chapter 12: Nucleic Acid Technologies e. housekeeping genes.

ANSWER: a 45. Which statement is true about genetic mutations? a. Mutations may arise from an unrepaired mistake in DNA replication. b. Mutations may arise from incorrectly repaired damage to one of the DNA strands. c. Mutations in reproductive cells can be passed to offspring. d. Mutations may better equip an organism or cell to survive in its environment. e. All of the statements are true.

ANSWER: e 46. How did the addition of different-colored fluorescent tags on each ddNTP improve the Sanger DNA sequencing method? a. Scientists did not need to use a heat-stable DNA polymerase. b. A DNA primer was no longer needed. c. All four ddNTPs could be incorporated into a single reaction. d. Scientists could now use short tandem repeats to assist their sequencing. e. The template DNA could now be double-stranded.

ANSWER: c 47. An example of a next-generation method of DNA sequencing is _____ sequencing. a. capillary array b. reverse Sanger c. tandem PCR d. reversible terminator e. MALDI-MS ANSWER: d 48. What happens in automated Sanger DNA sequencing reaction if one forgets to add (a) all fluorescent ddNTPs? (b) fluorescent ddGTP? (c) all dNTPs? (d) dGTP? ANSWER: (a) The polymerase will synthesize the whole strand without any breaks, but there will not be any detectable product at all because there are no short fragments and there is no detectable molecule. (b) The polymerase will never stop at a complementary C when incorporating a G. All other bases will work normally. Thus, it will be interpreted that the sequence does not contain any C. However, on carefully inspection, there should be an extra space (gap) in the sequence, where a normal nontermination dGTP was incorporated. (c) Essentially, no reaction will occur, with the exception of incorporating the first ddNTP. Thus, nothing will be seen. (d) The polymerase reaction will proceed up to the first complementary C in the sequence and then terminate. Thus, at best a very short strand of DNA sequence will be seen.

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Chapter 12: Nucleic Acid Technologies 49. (a) List and explain the function of the ingredients required for a polymerase chain reaction (PCR). (b) Describe the steps of each cycle of PCR. ANSWER: (a) PCR reactions need a template DNA molecule containing a region to be amplified, synthetic DNA oligonucleotide primers to serve as the starting point for DNA polymerase to extend the template, dNTPs that serve as the raw materials to be added to the growing nucleotide chains and the energy for the reaction, and a thermostable DNA polymerase to catalyze 5' to 3' DNA synthesis. (b) Step 1: Template denaturation is done at a high temperature; Step 2: the temperature of the reaction is lowered for primers to anneal to the template DNA; Step 3: the temperature is raised to the optimum temperature for the Taq (or other thermostable) DNA polymerase to catalyze DNA synthesis.

50. Certain restriction enzymes produce cohesive (sticky) ends. This means that they: a. cut both DNA strands at the same base pair. b. cut in regions of high G≡C content, leaving ends that can form more hydrogen bonds than ends of high A=T content. c. make a staggered double-strand cut, leaving ends with a few nucleotides of single-stranded DNA protruding. d. make ends that can anneal to cohesive ends generated by any other restriction enzyme. e. stick tightly to the ends of the DNA they have cut.

ANSWER: c

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Chapter 13: Introduction to Intermediary Metabolism 1. For the reaction A → B, ΔG'° = –60 kJ/mol. The reaction is started with 10 mM A; no B is initially present.

After 24 hours, analysis reveals the presence of 2 mM of B, 8 mM of A. Which statement is the MOST likely explanation? a. A and B have reached equilibrium concentrations. b. An enzyme has shifted the equilibrium toward A. c. B formation is kinetically slow; equilibrium has not been reached in 24 hours. d. Formation of B is thermodynamically unfavorable. e. The result described is impossible, given the fact that ΔG'° is –60 kJ/mol.

ANSWER: c 2. For glucose 1-phosphate to enter glycolysis, it must first be converted to glucose 6-phosphate, which is then converted to fructose 6-phosphate as follows: glucose 1-phosphate → glucose 6-phosphate glucose 6-phosphate → fructose 6-phosphate

ΔG'° = –7.1 kJ/mol ΔG'° = +1.7 kJ/mol

ΔG'° for the overall reaction is: a. –8.8 kJ/mol. b. –7.1 kJ/mol. c. –5.4 kJ/mol. d. +5.4 kJ/mol. e. +8.8 kJ/mol.

ANSWER: c 3. The standard free-energy changes for the reactions are given. ΔG'° = –43.0 kJ/mol phosphocreatine → creatine + Pi ΔG'° = –30.5 kJ/mol ATP → ADP + Pi What is the overall ΔG'° for this reaction? phosphocreatine + ADP → creatine + ATP a. –73.5 kJ/mol b. –12.5 kJ/mol c. +12.5 kJ/mol d. +73.5 kJ/mol e. ΔG'° cannot be calculated without Keq'.

ANSWER: b 4. If the ΔG'° of the reaction A → B is –40 kJ/mol, under standard conditions the reaction: a. is at equilibrium. b. will never reach equilibrium. c. will not occur spontaneously.

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Chapter 13: Introduction to Intermediary Metabolism d. will proceed at a rapid rate. e. will proceed spontaneously from A to B. ANSWER: e 5. The ΔG'° values for the two reactions are given. oxaloacetate + acetyl-CoA + H2O → citrate + CoA–SH + H+ citrate synthase

ΔG'° = –32.2 kJ/mol

oxaloacetate + acetate → citrate citrate lyase

ΔG'° = –1.9 kJ/mol

What is the ΔG'° for the hydrolysis of acetyl-CoA? acetyl-CoA + H2O → acetate + CoASH + H+ a. –34.1 kJ/mol b. –32.2 kJ/mol c. –30.3 kJ/mol d. +61.9 kJ/mol e. +34.1 kJ/mol ANSWER: c 6. Which item is NOT nucleophilic? a. proton b. oxygen of alcohol c. oxygen of carboxylate d. nitrogen of creatine e. side chain of arginine ANSWER: a 7. Which statement is NOT true? a. The carbon adjacent to a carbonyl can be resonance stabilized to form a carbanion. b. A carbonyl carbon can be made more electrophilic by a nearby metal ion. c. The carbon adjacent to an imine can be resonance stabilized to form a carbanion. d. Decarboxylation of an α-keto acid goes through a carbocation intermediate. e. A Claisen ester condensation reaction goes through a carbanion intermediate.

ANSWER: d 8. The reaction ATP → ADP + Pi is an example of a(n) _____ reaction. a. homolytic cleavage b. rearrangement Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13: Introduction to Intermediary Metabolism c. free radical d. group transfer e. oxidation-reduction ANSWER: d 9. Which statement is true about biochemical oxidation-reduction reactions? a. They usually proceed through homolytic cleavage. b. During oxidation a compound gains electrons. c. Dehydrogenases typically remove two electrons and two hydrides. d. There are two commonly accessed oxidation states of carbon. e. Every oxidation must be accompanied by a reduction.

ANSWER: e 10. Which factor does NOT contribute to the large, negative, free-energy change upon hydrolysis of highenergy compounds? a. electrostatic repulsion in the reactant b. low activation energy of forward reaction c. stabilization of products by extra resonance forms d. stabilization of products by ionization e. stabilization of products by solvation ANSWER: b 11. The hydrolysis of ATP has a large negative ΔG'°; nevertheless, it is stable in solution due to: a. entropy stabilization. b. ionization of the phosphates. c. resonance stabilization. d. the hydrolysis reaction being endergonic. e. the hydrolysis reaction having a large activation energy.

ANSWER: e 12. The hydrolysis of phosphoenolpyruvate proceeds with a ΔG'° of about –62 kJ/mol. The greatest contributing factors to this reaction are the destabilization of the reactant by electrostatic repulsion and stabilization of the product pyruvate by: a. electrostatic attraction. b. ionization. c. polarization. d. resonance. e. tautomerization.

ANSWER: e 13. Muscle contraction involves the conversion of _____ energy to _____energy.

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Chapter 13: Introduction to Intermediary Metabolism a. chemical; kinetic b. chemical; potential c. kinetic; chemical d. potential; chemical e. potential; kinetic ANSWER: a 14. Biological oxidation-reduction reactions always involve: a. direct participation of oxygen. b. formation of water. c. mitochondria. d. transfer of electrons. e. transfer of hydrogens.

ANSWER: d 15. Biological oxidation-reduction reactions never involve: a. transfer of e– from one molecule to another. b. formation of free e–. c. transfer of :H- (hydride ion) from one molecule to another. d. formation of H+ atoms. e. None of the answers is correct.

ANSWER: b 16. The standard reduction potentials (E'°) for the half-reactions are given.

E'° = +0.031 V fumarate + 2H+ + 2e– → succinate E'° = –0.219 V FAD + 2H+ + 2e– → FADH2 If succinate, fumarate, FAD, and FADH2, all at l M concentrations and in the presence of succinate dehydrogenase, were mixed together, what would happen initially? a. Fumarate and succinate would become oxidized; FAD and FADH2 would become reduced. b. Fumarate would be reduced to form succinate; FADH2 would be oxidized and form FAD. c. No reaction would occur because all reactants and products are already at their standard concentrations. d. Succinate would be oxidized to form fumarate; FAD would be reduced and form FADH2. e. Succinate would be oxidized to form fumarate; FADH2 would be unchanged because it is a cofactor.

ANSWER: b 17. The structure of NAD+ does NOT include: a. a flavin nucleotide. b. a phosphoanhydride bond.

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Chapter 13: Introduction to Intermediary Metabolism c. an adenine nucleotide. d. nicotinamide. e. two ribose residues.

ANSWER: a 18. Which compound is classified as high energy? a. phosphocreatine b. ATP c. phosphoenolpyruvate d. acetyl-CoA e. All of the answers are correct.

ANSWER: e 19. In this reaction, which molecule is the reductant (reducing agent)?

1,3-bisphosphoglycerate + NADH + H+ → glyceraldehyde-3-phosphate + Pi + NAD+ a. 1,3-bisphosphoglycerate b. NADH and NAD+ c. H+ d. Pi e. NADH ANSWER: e 20. Which molecule is NOT considered a universal electron carrier? a. NAD b. NADP c. FMN d. FAD e. AMP ANSWER: e 21. Which statement is false regarding biological oxidations? a. The redox potential is the affinity of a substance to accept electrons. b. Oxidation can occur by addition of oxygen to a molecule. c. Electrons are transferred from substances with a high reduction potential to substances with a lower reduction potential. d. Oxidases, oxygenases, and dehydrogenases are all enzymes that catalyze biological oxidationreduction reactions. e. NAD+ is a cosubstrate of certain dehydrogenases and always accepts electrons in the form of hydride ions.

ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13: Introduction to Intermediary Metabolism 22. Which statement is false regarding carbonyl groups? a. The carbon atom in a carbonyl group is more oxidized than the carbon atom in carbon dioxide. b. The carbon atom of a carbonyl group is an electrophile. c. Electron delocalization by a carbonyl group stabilizes an adjoining carbanion, thus facilitating its formation. d. Positively charged metal ions may enhance the ability of a carbonyl group to act as an electron sink. e. The functional group of aldehydes and ketones is a carbonyl group.

ANSWER: a 23. Which statement is true regarding coupled reactions? a. The standard free-energy changes of the coupled reactions are multiplicative, and their equilibrium constants are additive. b. The coupled reactions do not need to share a common intermediate. c. For the sum of the coupled reactions to occur spontaneously, the free energy of the endergonic reaction must be greater in absolute magnitude than the free energy of the exergonic reaction. d. The standard free-energy changes of the coupled reactions are additive, their equilibrium constants are multiplicative, and the coupled reactions must share a common intermediate. e. None of the statements is true.

ANSWER: d 24. Which statement is true regarding the actual free-energy change for a reaction? a. The actual free-energy change varies depending on the pathway by which the reaction occurs. b. When the standard free energy is large and positive, the reaction will never occur. c. The actual free energy varies depending on the standard free energy and the concentration of reactants and products. d. The actual free energy depends on the rate of the reaction and the presence of catalysts. e. None of the statements are true.

ANSWER: c 25. Which conjugate redox pair participates in biological redox reactions? a. Fe2+ and Fe3+ b. NADP+ and NAD+ c. FMN and FADH2 d. succinate and fumarate e. both Fe2+ and Fe3+ and succinate and fumarate ANSWER: e 26. What symbol represents the standard transformed free energy used by biochemists? a. ΔG b. ΔG' Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13: Introduction to Intermediary Metabolism c. ΔG'° d. ΔG° ANSWER: c 27. Acyl phosphates, phosphoesters, and phosphoanhydrides are compounds of significantly different phosphoryl group transfer potentials. Which choice correctly ranks these phosphorylated intermediates with respect to their phosphoryl group transfer potential from highest to lowest? a. 1,3-bisphosphoglycerate > adenosine triphosphate > glucose 6-phosphate b. glucose 6-phosphate > adenosine triphosphate > 1,3-bisphosphoglycerate c. adenosine triphosphate > 1,3-bisphosphoglycerate > glucose 6-phosphate d. adenosine triphosphate > glucose 6-phosphate > 1,3-bisphosphoglycerate e. 1,3-bisphosphoglycerate > glucose 6-phosphate > adenosine triphosphate ANSWER: a 28. Explain in quantitative terms the circumstances under which this reaction can proceed. citrate → isocitrate ΔG'° = +13.3 kJ/mol ANSWER: Students may explain that an endergonic reaction can proceed if it is coupled to another reaction that is exergonic, as such the net free-energy change of the coupled reactions is negative.

29. Explain what is meant by the statement: "Standard free-energy changes are additive." Give an example of the usefulness of this additive property in understanding how cells carry out thermodynamically unfavorable chemical reactions.

ANSWER: When two chemical reactions can be summed to give a third reaction, the standard free-energy change for the third reaction is the arithmetic sum of the standard free-energy changes of the other two reactions. Thus, cells can link thermodynamically unfavorable reactions with thermodynamically favorable reactions to produce a reaction that is thermodynamically favorable overall. Many possible examples may be given.

30. Classify each of the starred * atoms as an electrophile or a nucleophile.

ANSWER: (a) nucleophile (b) electrophile (c) electrophile (d) nucleophile Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13: Introduction to Intermediary Metabolism (e) nucleophile 31. Why is the actual free energy (ΔG) of hydrolysis of ATP in the cell different from the standard free energy (ΔG'°)?

ANSWER: The concentrations of the reactant (ATP) and the products (ADP and Pi) are not equal to 1 M. The pH is not exactly equal to 7.0 (the standard pH) and the temperature is usually higher than the standard temperature of 25 ∘C. 32. The free energy of hydrolysis of phosphoenolpyruvate is –61.9 kJ/mol. Explain in chemical terms why this compound has a large, negative value for ΔG'°.

ANSWER: The product of hydrolysis is enol pyruvate, which quickly tautomerizes to pyruvate. This keto-enol tautomerization stabilizes the products of hydrolysis relative to phosphoenolpyruvate, which cannot tautomerize.

33. In general, when ATP hydrolysis is coupled to an energy-requiring reaction, the actual reaction often consists of the transfer of a phosphate group from ATP to another substrate, rather than direct hydrolysis of the ATP. Explain.

ANSWER: Hydrolysis of the ATP would result in the loss of most of the free energy as heat. In a transfer reaction, the gamma (third) phosphate of ATP is transferred to the reaction substrate to produce a high-energy reactive phosphorylated intermediate. This activates the substrate by increasing its free energy, ensuring an exergonic reaction.

34. What is oxidation? What is reduction? Can oxidation occur without a simultaneous reduction? Why or why not?

ANSWER: Oxidation is a loss of electrons; reduction is a gain of electrons. Free electrons are unstable (do not occur), so whenever an electron is released by the oxidation of one species, an electron must be accepted by the reduction of another species.

35. For each pair of ions or compounds, indicate which is the more highly reduced species. (a) Co2+/Co+ (b) lactate/pyruvate (c) Fe3+/Fe2+ (d) NAD+/NADH ANSWER: (a) Co+; (b) lactate; (c) Fe2+; (d) NADH 36. Which of the following statements is true regarding the metabolic pathways surrounding fatty acids? a. Fatty acid synthesis and degradation processes are regulated in a reciprocal manner. b. Fatty acids are simultaneously synthesized and degraded within cells. c. Fatty acid synthesis and degradation are regulated randomly. d. Throughout fatty acid synthesis and degradation, the same enzymes are regulated. e. Fatty acid synthesis and degradation occurs in the same cellular compartments to maintain homeostasis.

ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13: Introduction to Intermediary Metabolism 37. Which of the following is an example of a catabolic reaction pathway? a. glycolysis b. DNA replication c. protein synthesis d. photosynthesis e. muscle growth ANSWER: a 38. What is an alternative term commonly used to describe anabolic reactions? a. enzyme-catalyzed reactions b. degradation c. metabolism d. biosynthesis e. physiology ANSWER: d 39. Which of the following is a common end-product of catabolic reactions? a. proteins b. nucleic acids c. ATP d. ADP e. CO2 ANSWER: e 40. Which of the following is a common end-product of anabolic reactions? a. lactic acid b. lipids c. ATP d. ADP e. H20 ANSWER: b 41. Besides the degradation of organic nutrient molecules, catabolic reactions are also responsible for: a. protein synthesis b. obtaining carbon from the environment c. reducing electron carriers d. polymerization of precursor molecules e. elimination of excess nutrients ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13: Introduction to Intermediary Metabolism 42. What is the advantage of reciprocally regulating anabolic and catabolic reaction sequences? a. To minimize the need for certain enzymes that participate in both catabolic and anabolic reaction sequences b. To allow for reactions to take place in only one compartment c. To maximize the reaction rate for either anabolic or catabolic reactions d. To prevent the degradation of newly synthesized biomolecules e. There is no advantage because this regulation occurs randomly ANSWER: d 43. Describe two ways in which enzymes in anabolic or catabolic reaction pathways can be regulated inside the cell.

ANSWER: A key enzyme in a pathway may be activated or inhibited allosterically, or its amount may be changed by the rates of translation and degradation of the enzyme.

44. Explain why anabolic and catabolic pathways are not catalyzed by exactly the same set of enzymes, operating in one direction for anabolism and the opposite direction for catabolism.

ANSWER: If anabolic and catabolic pathways are catalyzed by exactly the same set of enzymes, inhibition of an enzyme involved in catabolism would also inhibit the reaction sequence in anabolism, preventing the cell from upregulating one pathway while downregulating the reciprocal pathway.

45. Classify each as an energy-containing nutrient or energy-depleted end product.

1. carbohydrates 2. CO2 3. H2O 4. fats 5. proteins ANSWER: (a) energy-containing nutrient (b) energy-depleted end product (c) energy-depleted end product (d) energy-containing nutrient (e) energy-containing nutrient 46. Among the common enzyme reactions that participate in metabolism, which reaction is associated with the release of a carbon dioxide molecule? a. condensation b. decarboxylation c. rearrangement d. elimination e. free radical ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13: Introduction to Intermediary Metabolism 47. Among the common enzyme reactions involved in metabolism, which reaction is associated with the formation of an intermediate compound containing an –OH group on each carbon of a C=C double bond? a. condensation b. decarboxylation c. rearrangement d. elimination e. free radical ANSWER: c 48. Among the common enzyme reactions involved in metabolism, which reaction is associated with the formation of an “activated” metabolic intermediate? a. decarboxylation b. elimination c. free radical d. group transfer e. oxidation-reduction ANSWER: d 49. Describe how an isomerization reaction can proceed without a change in the overall oxidation state of a molecule.

ANSWER: The overall oxidation state of a molecule that undergoes isomerization does not change because an initial reduction of one carbon (C1) is coupled to an oxidation of a nearby carbon (C2). Thus, the overall oxidation state of the molecule remains the same.

50. Explain why catabolic and anabolic pathways frequently take place in different cellular compartments.

ANSWER: Paired catabolic and anabolic pathways will compete for many of the same enzymes to function.

Placing catabolic and anabolic pathways in different cellular compartments prevents this competition, allowing the process to take place smoothly. Moreover, the separate compartments prevent newly synthesized molecules from being immediately degraded by the reciprocal pathway, thereby avoiding wasteful, energy-consuming reactions and facilitating regulation of each pathway.

51. Which phosphorylated compound has the largest negative value for the standard free-energy change (ΔG'°) upon hydrolysis? a. phosphoenolpyruvate b. glucose 6-phosphate c. phosphocreatine d. ATP e. acetyl-CoA ANSWER: a 52. Explain why oxygen is the preferred molecule used by animals as the provider of the thermodynamic driving force for mitochondrial ATP production by oxidative phosphorylation.

ANSWER: Oxygen’s very high reduction potential (E′° = 0.816 V) allows for the spontaneous transfer of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13: Introduction to Intermediary Metabolism electrons from the electron transport chain to molecular oxygen, generating the power needed for the mitochondrial ATP production machinery.

53. Which of the following is not a factor affecting the overall activity of enzymes? a. mRNA degradation b. phosphorylation-dephosphorylation events c. enzyme size d. protein degradation e. enzyme-substrate binding ANSWER: c 54. Which of the following statements best describes homeostasis? a. Regulation of the dynamic steady state conditions inside the cell in response to rapidly changing external circumstances. b. The ensemble of metabolic intermediates, regulatory signals, and secondary metabolites present in a given cell or tissue under specific conditions. c. The relative affinity of the electron acceptor of each redox pair for electrons. d. The exergonic release of energy as a result of ATP hydrolysis. e. The process of transforming a compound into any of its isomeric forms, which involves changing stereochemistry.

ANSWER: a 55. Which of the following statements best describes the metabolome? a. Regulation of the dynamic steady state conditions inside the cell in response to rapidly changing external circumstances. b. The ensemble of metabolic intermediates, regulatory signals, and secondary metabolites present in a given cell or tissue under specific conditions. c. The relative affinity of the electron acceptor of each redox pair for electrons. d. The exergonic release of energy as a result of ATP hydrolysis. e. The process of transforming a compound into any of its isomeric forms, which involves changing stereochemistry.

ANSWER: b 56. Which of the following statements best describes the phrase “dynamic steady state”? a. The change in the gross composition of cells due to the relative proportions of carbohydrates, fat, and protein in the diet that vary from meal to meal. b. The collection of regulatory mechanisms for maintaining homeostasis at the molecular, cellular, and organismal levels. c. The movement of metabolites through each pathway in the appropriate direction to meet the cell’s, or the organism’s, needs under changing circumstances. d. The gradual changes in regulation of catabolic and anabolic reactions in response to the varying proportions of nutrients ingested from meal to meal.

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Chapter 13: Introduction to Intermediary Metabolism e. Maintenance of the overall concentration of specific metabolites despite consistent and sometimes rapid concentration fluctuations in response to external circumstances.

ANSWER: e 57. Covalent modifications of enzymes or other proteins, such as phosphorylation, occur in the span of: a. minutes or hours b. days c. weeks d. seconds or minutes e. from milliseconds to many hours ANSWER: d 58. Changes in gene expression that alter the number of enzymes occur in the span of: a. minutes or hours b. days c. weeks d. seconds or minutes e. from milliseconds to many hours ANSWER: a 59. Explain why covalent modifications of enzymes or other proteins occur in the span of seconds to minutes while a change in gene expression that alters the number of enzymes occurs in the span of many minutes or hours.

ANSWER: The number of chemical reactions that occur when transcribing and translating a gene greatly surpasses the number of chemical reactions required for the covalent modification of enzymes.

Depending on the length of the coding region, hundreds or thousands of reactions are performed to transcribe a gene. This is also true for the process of translation. In comparison, the covalent modification of enzymes, such as in the case of phosphorylation, which occurs after a single chemical reaction with ATP, can take seconds.

60. For each factor affecting the activity of enzymes, indicate whether it affects the number of molecules or modulates its activity. (a) transcription (b) extracellular signals (c) phosphorylation-dephosphorylation events (d) substrate-enzyme binding (e) allosteric effector binding ANSWER: (a) number of molecules, (b) number of molecules, (c) modulates activity, (d) modulates activity, (e) modulates activity.

61. Explain how glucose homeostasis is maintained in the blood during both normal and diseased states.

ANSWER: During normal conditions, glucose is maintained at a blood concentration of 5 mM. At this time, glucose's entry rate from various sources into the blood is balanced by changes in glucose uptake Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13: Introduction to Intermediary Metabolism from the blood into various tissues. In diabetes mellitus, for example, this balance is disrupted, starting with the failure to remove glucose from the blood due to a lack of or insensitivity to insulin.

62. Which statement is true regarding the actual free-energy change of a reaction? a. If the actual free-energy change is negative, the reaction is spontaneous. b. If the actual free energy is zero, the rate of the forward and reverse reactions are equal. c. The actual free energy becomes less negative as the reaction proceeds spontaneously. d. All of the statements are true. e. None of the statements is true.

ANSWER: d

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis 1. The conversion of one molecule of fructose 1,6-bisphosphate to two molecules of pyruvate by the glycolytic pathway results in a net formation of _____ molecule(s) of _____ and _____ molecule(s) of ATP. a. one; NAD+; two b. one; NADH; one c. two; NAD+; four d. two; NADH; two e. two; NADH; four ANSWER: e 2. Which of the following enzymes catalyzes a reaction in glycolysis that requires ATP as a substrate? a. hexokinase b. glyceraldehyde 3-phosphate dehydrogenase c. pyruvate kinase d. aldolase e. phosphoglycerate kinase ANSWER: a 3. Which of the following enzymes catalyzes a reaction in glycolysis that produces ATP as a product? a. hexokinase b. glyceraldehyde 3-phosphate dehydrogenase c. pyruvate kinase d. aldolase e. phosphofructokinase-1 ANSWER: c 4. The steps of glycolysis between glyceraldehyde 3-phosphate and 3-phosphoglycerate do NOT involve: a. ATP synthesis. b. catalysis by phosphoglycerate kinase. c. oxidation of NADH to NAD+. d. the formation of 1,3-bisphosphoglycerate. e. utilization of Pi. ANSWER: c 5. The first reaction in glycolysis that results in the formation of an energy-rich compound is catalyzed by: a. glyceraldehyde 3-phosphate dehydrogenase. b. hexokinase. c. phosphofructokinase-1. d. phosphoglycerate kinase. e. triose phosphate isomerase.

ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis 6. Which item is a cofactor in the reaction catalyzed by glyceraldehyde 3-phosphate dehydrogenase? a. ATP b. Cu2+ c. heme d. NAD+ e. NADP+ ANSWER: d 7. Inorganic fluoride inhibits enolase. In an anaerobic system that is metabolizing glucose as a substrate, which compound would be expected to increase in concentration following the addition of fluoride? a. 2-phosphoglycerate b. glucose c. glyoxylate d. phosphoenolpyruvate e. pyruvate ANSWER: a 8. Glycogen is converted to monosaccharide units by: a. glucokinase. b. glucose 6-phosphatase. c. glycogen phosphorylase. d. glycogen synthase. e. glycogenase.

ANSWER: c 9. Which cofactor participates directly in MOST of the oxidation-reduction reactions in the fermentation of glucose to lactate? a. ADP b. ATP c. FAD/FADH2 d. glyceraldehyde 3-phosphate e. NAD+/NADH ANSWER: e 10. In comparison with the resting state, actively contracting human muscle tissue has a: a. higher concentration of ATP. b. higher rate of lactate formation. c. lower consumption of glucose. d. lower rate of oxygen consumption.

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis e. lower ratio of NADH to NAD+.

ANSWER: b 11. Which statement is NOT true concerning glycolysis in anaerobic muscle? a. Fructose 1,6-bisphosphatase is one of the enzymes of the pathway. b. It is an endergonic process. c. It results in net synthesis of ATP. d. It results in synthesis of NADH. e. Its rate is slowed by a high [ATP]/[ADP] ratio.

ANSWER: b 12. Which statement is NOT correct? a. Aerobically, oxidative decarboxylation of pyruvate forms acetate (acetyl-CoA) that enters the citric acid cycle. b. In anaerobic muscle, pyruvate is converted to lactate. c. In yeast growing anaerobically, pyruvate is converted to ethanol. d. Reduction of pyruvate to lactate regenerates a cofactor essential for glycolysis. e. Under anaerobic conditions, pyruvate does not form because glycolysis does not occur.

ANSWER: e 13. The ultimate electron acceptor in the fermentation of glucose to ethanol is: a. acetaldehyde. b. acetate. c. ethanol. d. NAD+. e. pyruvate.

ANSWER: a 14. In the alcoholic fermentation of glucose by yeast, thiamine pyrophosphate is a coenzyme required by: a. aldolase. b. hexokinase. c. lactate dehydrogenase. d. pyruvate decarboxylase. e. transaldolase.

ANSWER: d 15. Which statement about the pentose phosphate pathway is correct? a. It generates 36 molecules of ATP per molecule of glucose consumed. b. It generates 6 molecules of CO2 for each molecule of glucose consumed. c. It is a reductive pathway; it consumes NADH. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis d. It is present in plants, but not in animals. e. It provides precursors for the synthesis of nucleotides.

ANSWER: e 16. The metabolic function of the pentose phosphate pathway is to: a. act as a source of ADP biosynthesis. b. generate NADPH and pentoses for the biosynthesis of fatty acids and nucleic acids. c. participate in oxidation-reduction reactions during the formation of H2O. d. provide intermediates for the citric acid cycle. e. synthesize phosphorus pentoxide.

ANSWER: b 17. Glucose breakdown in certain mammalian and bacterial cells can occur by mechanisms other than classic glycolysis. In MOST of these, glucose 6-phosphate is oxidized to 6-phosphogluconate, which is then further metabolized by: a. a type of aldolase that splits glucose 6-phosphate to form glyceric acid and glyceraldehyde 3phosphate. b. a type of aldolase that splits glucose 6-phosphate to form glycolic acid and erythrose 4-phosphate. c. conversion to 1,6-bisphosphogluconate. d. decarboxylation to produce keto- and aldopentoses. e. oxidation to a six-carbon dicarboxylic acid.

ANSWER: d 18. Which enzyme acts in the pentose phosphate pathway? a. 6-phosphogluconate dehydrogenase b. aldolase c. glycogen phosphorylase d. phosphofructokinase-1 e. pyruvate kinase ANSWER: a 19. The oxidation of three molecules of glucose by the pentose phosphate pathway may result in the production of _____ molecules of pentose, _____ molecules of NADPH, and _____ molecules of CO2. a. two; four; eight b. three; four; three c. three; six; three d. four; three; three e. four; six; six ANSWER: c 20. Which enzyme or enzymes catalyze an energy-conserving reaction in glycolysis?

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis a. hexokinase b. pyruvate kinase c. glyceraldehyde 3-phosphate dehydrogenase d. pyruvate kinase and glyceraldehyde 3-phosphate dehydrogenase e. hexokinase, pyruvate kinase, and glyceraldehyde 3-phosphate dehydrogenase ANSWER: d 21. How much ATP and NADH is produced from the conversion of one molecule of dihydroxyacetone phosphate (DHAP) to one molecule of pyruvate? a. two molecules of ATP and two molecules of NADH b. two molecules of ATP and one molecule of NADH c. two molecules of ATP and zero molecules of NADH d. one molecule of ATP and two molecules of NADH e. one molecule of ATP and zero molecules of NADH ANSWER: b 22. Which products of glycolysis store energy that can later be used by the cell? a. fructose 1,6-bisphosphate b. ATP and NADH c. 1,3-bisphosphoglycerate and phosphoenolpyruvate d. fructose 1,6-bisphosphate, 1,3-bisphosphoglycerate, and phosphoenolpyruvate e. ATP, NADH, 1,3-bisphosphoglycerate, and phosphoenolpyruvate ANSWER: b 23. Why is an anaerobic fate of pyruvate necessary in skeletal muscle? a. It is required to maintain the reducing environment of the cytosol. b. It allows for the reoxidation of NADH to NAD+, as NAD+ is in limited supply during anaerobic conditions. c. It decreases the pH of the skeletal muscle cell, thereby increasing release of oxygen from myoglobin. d. The anaerobic process directly generates additional ATP for working muscle tissue. e. None of the answers is correct.

ANSWER: b 24. What is the general name for an enzyme that catalyzes an intramolecular phosphoryl group shift? a. phosphatase b. mutase c. kinase d. phosphorylase e. phosphate translocase ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis 25. Ethanol is produced by yeast under anaerobic conditions. This is accomplished by 2 reactions beyond the 10 reactions of glycolysis. What is the purpose served by these 2 reactions in the metabolism of yeast cells? a. They are required to reoxidize NADH. b. They are required to ensure a net oxidation of glucose. c. They are required to sustain gluconeogenesis. d. They are required to produce NADPH. ANSWER: a 26. Which statement does NOT describe a general function of the pentose phosphate pathway? a. The pentose phosphate pathway is used to produce NADPH for reductive biosynthesis in adipocytes. b. The pentose phosphate pathway is used to facilitate the elimination of dangerous peroxides in erythrocytes. c. The pentose phosphate pathway allows for the entry of dietary pentose intermediates into the glycolytic pathway. d. The pentose phosphate pathway allows for the conversion of hexoses into pentoses that may be used as precursors in the synthesis of nucleosides. e. The pentose phosphate pathway produces reduced molecules whose electrons may be shuttled to the mitochondria for oxidative phosphorylation.

ANSWER: e 27. Which compound links glycolysis and nucleotide synthesis? a. acetyl-CoA b. oxaloacetate c. citrate d. glucose 6-phosphate e. glucose 3-phosphate ANSWER: d 28. There are a variety of fairly common human genetic diseases in which enzymes required for the breakdown of fructose, lactose, or sucrose are defective. However, there are very few cases of people having a genetic disease in which one of the enzymes of glycolysis is severely affected. Why are such mutations seen so rarely?

ANSWER: The glycolytic pathway is so central to all of cellular metabolism that mutations in glycolytic enzymes are lethal; embryos with such mutations would not survive.

29. Define "fermentation" and explain, by describing relevant reactions, how it differs from glycolysis. Include in the explanation a discussion of the role of NADH in the reaction(s).

ANSWER: Fermentation is the operation of the glycolytic pathway under anaerobic conditions. Under aerobic conditions, the pyruvate produced by glycolysis is oxidized to acetyl-CoA, which is then passed through the citric acid cycle. NADH produced in the oxidation of glyceraldehyde 3-phosphate passes electrons to O2, and is thus recycled to NAD+, allowing continuation of the glycolytic reactions. When no O2 is available to reoxidize the NADH produced by the glyceraldehyde 3phosphate dehydrogenase reaction, electrons from NADH must be passed to the product of glycolysis, pyruvate, forming lactate or ethanol.

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis 30. Describe how NADH is recycled to NAD+ under anaerobic conditions. Why is it important to recycle NADH produced during glycolysis to NAD+?

ANSWER: Cells contain a limited supply of NAD+ and NADH. The oxidation of glyceraldehyde 3-phosphate requires NAD+ as an electron acceptor—it converts NAD+ to NADH. Unless this NADH is recycled to NAD+, oxidative metabolism in this cell will cease for lack of an electron acceptor. Under aerobic conditions, NADH passes electrons to O2; under anaerobic conditions, NADH reduces pyruvate to lactate, and is thereby recycled to NAD+.

31. All of the intermediates in the glycolytic pathway are phosphorylated. Give two plausible reasons why this might be advantageous to the cell.

ANSWER: There are at least three reasons: (1) so they will not cross the plasma membrane and be lost; (2) the resulting high-energy phosphoryl groups can be transferred directly to ADP to make ATP; and (3) the high binding energy of the ionic interactions between the (–) charged phosphoryl groups and (+) charged enzyme side chains helps to stabilize the enzyme-substrate interaction, thereby lowering the reaction activation energy.

32. Describe the glycolytic pathway from glucose to glyceraldehyde 3-phosphate. Show structures of intermediates, enzyme names, and indicate where any cofactors participate.

ANSWER: This part of the pathway involves the reactions catalyzed by hexokinase (with ATP as a cofactor), phosphohexose isomerase, phosphofructokinase-1 (with ATP as a cofactor), aldolase, and triose phosphate isomerase. (See Fig. 14-4.)

33. Describe the glycolytic pathway from glyceraldehyde 3-phosphate to pyruvate, showing structures of intermediates and names of enzymes. Indicate where any cofactors participate.

ANSWER: The answer should show the reactions catalyzed by glyceraldehyde 3-phosphate dehydrogenase (with NAD+ and Pi as cofactors), phosphoglycerate kinase (produces ATP from ADP), phosphoglycerate mutase, enolase, and pyruvate kinase (produces ATP from ADP). (See Fig. 14-6.)

34. In the conversion of glucose to pyruvate via glycolysis, all of the enzymes listed participate. Indicate the order in which they function by numbering them. _____ hexokinase _____ triose phosphate isomerase _____ phosphohexose isomerase _____ enolase _____ glyceraldehyde 3-phosphate dehydrogenase _____ pyruvate kinase _____ phosphofructokinase-1 Which of the enzymes represents the most important regulation point in glycolysis?

Which catalyzes a reaction in which ATP is produced?

Which catalyzes a reaction in which NADH is produced?

ANSWER: 1, 4, 2, 6, 5, 7, 3; phosphofructokinase-1; pyruvate kinase; glyceraldehyde 3-phosphate dehydrogenase 35. The yeast used in brewing the alcoholic beverage beer can break down glucose either aerobically or Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis anaerobically using alcoholic fermentation. Explain why beer is brewed under anaerobic conditions.

ANSWER: Since aerobic breakdown of glucose yields more energy than fermentation, in the presence of oxygen, yeast cells would not carry out fermentation and no alcohol would be produced. Under anaerobic conditions, the yeast ferment the glucose to carbon dioxide and ethanol, both of which are key ingredients in beer.

36. Rat liver is able to metabolize glucose by both the glycolytic and the pentose phosphate pathways. Indicate in the blanks if the properties are of the glycolytic pathway (G), pentose phosphate pathway (P), both (G + P), or neither (0). _____ NAD+ is involved. _____ CO2 is liberated. _____ Phosphate esters are intermediates. _____ Glyceraldehyde 3-phosphate is an intermediate. _____ Fructose 6-phosphate is an intermediate.

ANSWER: G, P, G + P, G, G 37. The bacterium E. coli can grow in a medium in which the only carbon source is glucose. How does this organism obtain ribose 5-phosphate for the synthesis of ATP? (Do not describe ATP synthesis, just the origin of ribose 5-phosphate.) Show structures and indicate where cofactors participate.

ANSWER: Ribose 5-phosphate is produced from glucose by the reactions of the pentose phosphate pathway.

Glucose is converted to glucose 6-phosphate by hexokinase, at the expense of one ATP molecule, and then enters the pentose phosphate pathway, where it undergoes three reactions in the oxidative stage, two of which require NADP+ as a cofactor, and is converted to ribulose 5-phosphate. In the nonoxidative stage, ribulose 5-phosphate is converted to ribose 5-phosphate. (See Fig. 14.14.)

38. An extract of adipose (fat) tissue can metabolize glucose to CO2. When glucose labeled with 14C in either C-1 or C-6 was added to the extract, 14CO2 was released with the time courses shown. What is the major path of glucose oxidation in this extract? Explain how this conclusion is reached.

ANSWER: The tissue must be oxidizing glucose primarily by the pentose phosphate pathway in which C-1 is released (as CO2) earlier than C-6. During glycolysis, C-1 and C-6 become equivalent at the level of glyceraldehyde 3-phosphate, and C-1 and C-6 are thus released at the same rate (during passage of acetate through the citric acid cycle).

39. Which of the following is not included in the four major metabolic pathways that determines the fate of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis glucose when glucose is abundant? a. glycolysis b. glycogen synthesis c. pyruvate fermentation d. fatty acid degradation e. pentose phosphate pathway ANSWER: d 40. After the first phase of glycolysis, how many ATP molecules have been used and how many have been produced? a. 0; 2 b. 2; 2 c. 2; 4 d. 4; 2 e. 2; 0 ANSWER: e 41. After the first and second phase of glycolysis, how many ATP molecules have been used and how many have been produced? a. 0; 2 b. 2; 2 c. 2; 4 d. 4; 2 e. 2; 0 ANSWER: c 42. What is the purpose of phosphorylation of glycolytic intermediates by ATP in phase 1 of glycolysis? a. to prevent excess glucose from entering the cell b. to release free energy in the form of heat to power the conversion of ADP to ATP c. to reduce the overall ATP concentration, which might inhibit the upcoming ATP payoff phase d. to prevent free glucose from leaving the cell, to partially conserve the energy in phosphoanydride bonds for use in subsequent reactions, and to lower the activation energy and increase the specificity of enzymatic reactions e. to increase the ADP concentration in the cell by hydrolysis ANSWER: d 43. In which cellular compartment does glycolysis occur? a. nucleus b. mitochondria c. cytoplasm d. endoplasmic reticulum Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis e. Golgi complex ANSWER: c 44. How many pyruvate molecules are produced from 5 molecules of glucose after glycolysis? a. 2 b. 4 c. 6 d. 8 e. 10 ANSWER: e 45. How many net ATP molecules are produced from 3 molecules of glucose after glycolysis? a. 2 b. 6 c. 9 d. 12 e. 15 ANSWER: b 46. Based on the nomenclature, explain the structural difference between fructose 1,6-bisphosphate and adenosine diphosphate.

ANSWER: Compounds with two phosphates linked together as a pyrophosphoryl group are diphosphates, while compounds with two phosphates located at different atoms often contain the term bisphosphate.

47. Considering the change in free energy during glycolysis, explain why the conversion of glucose to pyruvate is essentially an irreversible process.

ANSWER: The entire process of glycolysis results in a large negative free-energy change of -85 kJ/mol.

Specifically, the phosphorylated molecule inside the cell that releases the highest amount of free energy when hydrolyzed is phosphoenolpyruvate (-61.9 kJ/mol; Ch. 13). Since there is no molecule inside the cell that can overcome the energy required to reverse this reaction and the overall freeenergy change for glycolysis is large and negative, glycolysis is essentially irreversible.

48. Which statement is true about β-granules? a. The granules in the liver are essentially absent after a 24-hour fast. b. Muscle and liver granules release glucose at similar rates. c. In the liver, granules cluster together in groups of several thousand. d. Granules are generally found in cellular liposomes or embedded in the endoplasmic reticulum. e. The outermost granular tiers consist mainly of highly branched β-chains.

ANSWER: a 49. In glycogen _____ glycosidic bonds occur at branch points, while all others are _____ glycosidic bonds. a. (β1 → 4); (α1 → 4) Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis b. (α1 → 6); (α1 → 4) c. (α1 → 4); (α1 → 6) d. (β1 → 6); (α1 → 4) e. All glycosidic bonds in glycogen are (α1 → 4).

ANSWER: b 50. Which statement about mammalian glycogen synthase is NOT correct? a. It is especially predominant in the liver and muscle. b. The donor molecule is a sugar nucleotide. c. This enzyme creates a new (α1 → 4) linkage between the UDP-glucose and the nonreducing end of a glycogen branch. d. This enzyme adds glucose units to the nonreducing end of glycogen branches. e. This enzyme adds the initial glucose unit to a tyrosine residue in glycogenin.

ANSWER: e 51. The glycogen-branching enzyme catalyzes: a. degradation of (α1 → 4) linkages in glycogen. b. formation of (α1 → 4) linkages in glycogen. c. formation of (α1 → 6) linkages during glycogen synthesis. d. glycogen degradation in tree branches. e. removal of unneeded glucose residues at the ends of branches.

ANSWER: c 52. Glycogenin: a. catalyzes the conversion of starch into glycogen. b. is the enzyme responsible for forming branches in glycogen. c. is the gene that encodes glycogen synthase. d. is the primer on which new glycogen chains are initiated. e. regulates the synthesis of glycogen.

ANSWER: d 53. Which statement is true of glycogen synthase? a. Activation of the enzyme involves a phosphorylation. b. It catalyzes the addition of glucose residues to the nonreducing end of a glycogen chain by formation of (α1 → 4) bonds. c. It uses glucose 6-phosphate as the donor of glucose units. d. It catalyzes the addition of glucose residues at branch points by formation of (α1 → 6) bonds. e. The enzyme has measurable activity only in the liver.

ANSWER: b 54. Which statement is true of glycogen synthase?

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis a. It is activated by at least 11 different protein kinases. b. It is activated by Ca2+ ions in skeletal muscle. c. It catalyzes the formation of (α1 → 4) glycosidic bonds. d. It uses glucose 1-phosphate as a substrate. e. All of the statements are true.

ANSWER: c 55. In terms of the reaction mechanism, to which enzyme in glycolysis is phosphoglucomutase MOST similar? a. phosphohexose isomerase b. triose phosphate isomerase c. phosphoglycerate mutase d. enolase e. phosphoglycerate kinase ANSWER: c 56. Starting from glucose, what is the net number of nucleotide triphosphate molecules used in the synthesis of each unit of glycogen? a. zero b. one c. two d. three e. four ANSWER: c 57. Which statement is true for glycogen granules in liver cells? a. Liver glycogen granules account for up to 10% of the weight of the liver. b. Glycogen granules in the liver are much larger than those in muscle. c. Glycogen is stored in the cytosol as granules. d. All of the statements are true.

ANSWER: d 58. Sugar nucleotides are substrates in the biosynthesis of glycogen. Describe the properties of sugar nucleotides that make them suitable for this role.

ANSWER: 1. Since their formation is metabolically irreversible, they help keep the pathways in which they are involved also irreversible. 2. During a catalyzed reaction, binding energy contributions may be made by noncovalent interactions between the enzyme and the nucleotide portion of the sugar nucleotide.

3. The nucleotidyl group makes a good leaving group in reactions. 4. The nucleotidyl portions can serve as "tags" to target molecules for a particular cellular use.

59. Diagram the pathway from glucose to glycogen; show the participation of cofactors and name the enzymes involved.

ANSWER: (1) Glucose + ATP → glucose 6-phosphate + ADP; hexokinase Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14: Carbohydrate Metabolism Part 1: Glycolysis and Glycogen Synthesis (2) Glucose 6-phosphate → glucose 1-phosphate; phosphoglucomutase (3) Glucose 1-phosphate + UTP → UDP-glucose + PPi; UDP-glucose pyrophosphorylase (4) UDP-glucose → glycogen + UDP; glycogen synthase 60. What is the biological advantage of synthesizing glycogen with many branches?

ANSWER: Both glycogen synthase and glycogen phosphorylase act at the nonreducing ends of glycogen chains. Branched glycogen has far more nonreducing ends for these enzymes to work on than would the equivalent amount of linear glycogen chains. Having more ends effectively increases the concentration of substrate for the enzymes, thereby increasing the rate of glycogen synthesis and breakdown.

61. Explain the role of glycogenin.

ANSWER: Glycogenin is a protein that acts as the "primer" for the initiation of new glycogen molecules. It catalyzes the transfer of a glucose residue from UDP-glucose to a tyrosine hydroxyl group in glycogenin and then catalyzes the addition of seven more glucose residues to start the formation of a glycogen chain. Once the chain is started, glycogen synthase can extend the chain, and the branching enzyme can create branches. As more glucose residues are added, glycogenin becomes the core of the glycogen granule.

62. Name three GTP-binding factors that function in bacterial protein translation, and describe their function.

ANSWER: IF2, an initiation factor that is involved in binding of the initiator tRNA; EF-Tu, the elongation factor associated with binding of aminoacyl-tRNA molecules into the A site; EF-G that is associated with translocation of the ribosome (and also facilitates disassembly of the ribosome after release of the completed polypeptide).

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Chapter 15: Carbohydrate Metabolism Part 2: Gluconeogenesis and Glycogen Degradation 1. The action of which three enzymes are needed to release glucose from the outer branches of glycogen so that it may be used in glycolysis? a. protein kinase A; glycogen debranching enzyme; phosphoglucomutase b. protein kinase A; phosphoglucomutase; phosphoprotein phosphatase 1 c. glycogen phosphorylase; glycogen synthase; hexokinase IV d. glycogen synthase; glycogen synthase kinase 3; glycogen debranching enzyme e. glycogen phosphorylase; glycogen debranching enzyme; phosphoglucomutase ANSWER: e 2. The breakdown of glycogen to glucose 1-phosphate is called: a. neoglycogenesis. b. deglycogenization. c. glycogen anabolism. d. glycogenolysis.

ANSWER: d 3. The enzyme glycogen phosphorylase: a. catalyzes a cleavage of (β1 → 4) bonds. b. catalyzes a cleavage of (α1 → 4) bonds. c. is a substrate for a kinase. d. uses glucose 6-phosphate as a substrate. e. uses glucose as a substrate.

ANSWER: b 4. Glycogen is converted to monosaccharide units by: a. glucokinase. b. glucose 6-phosphatase c. glycogen phosphorylase. d. glycogen synthase. e. glycogenase.

ANSWER: c 5. Glycogen phosphorylase removes glucose (as glucose 1-phosphate) from the _____ ends of glycogen chains until it is _____ a branch point. a. reducing; four glucose units from b. nonreducing; at c. nonreducing; four glucose units from d. reducing; at e. nonreducing; one glucose unit from ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15: Carbohydrate Metabolism Part 2: Gluconeogenesis and Glycogen Degradation 6. Which statement is true of muscle glycogen phosphorylase? a. It catalyzes cleavage of the (α1 → 6) bonds at the branch points of glycogen. b. It catalyzes the degradation of glycogen by hydrolysis of glycosidic bonds. c. It degrades glycogen to form glucose 6-phosphate. d. It removes glucose residues from the nonreducing ends of glycogen chains by phosphorolysis. e. It adds glucose residues to the reducing ends of a growing glycogen chain.

ANSWER: d 7. The _____ tissues do NOT contribute glucose to the blood because they lack the enzyme _____. a. muscle and adipose; glucose 6-phosphatase b. kidney and adipose; glucose 6-phosphatase c. muscle and adipose; glycogen phosphorylase b d. muscle and liver; glycogen phosphorylase b ANSWER: a 8. Which statement is true of glycogen synthesis and breakdown? a. Phosphorylation activates the enzyme responsible for breakdown, and inactivates the enzyme responsible for synthesis. b. Synthesis is catalyzed by the same enzyme that catalyzes breakdown. c. The glycogen molecule "grows" at its reducing end. d. The immediate product of glycogen breakdown is free glucose. e. Under normal circumstances, glycogen synthesis and glycogen breakdown occur simultaneously and at high rates.

ANSWER: a 9. Glycogen phosphorylase a can be inhibited at an allosteric site by: a. AMP b. calcium c. GDP d. glucagon e. glucose ANSWER: e 10. Is glycogen phosphorylase activated, inhibited, or unaffected in response to a glucagon signal in liver tissue? a. activated b. inhibited c. unaffected ANSWER: a 11. Is glycogen phosphorylase activated, inhibited, or unaffected in response to a glucagon signal in skeletal muscle?

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Chapter 15: Carbohydrate Metabolism Part 2: Gluconeogenesis and Glycogen Degradation a. activated b. inhibited c. unaffected ANSWER: c 12. Is glycogen phosphorylase activated, inhibited, or unaffected in response to an insulin signal in skeletal muscle? a. activated b. inhibited c. unaffected ANSWER: b 13. Which statement is true for glycogen phosphorylase b? a. It catalyzes a phosphorolytic cleavage of (α1 → 6) glycosidic bonds. b. It catalyzes a hydrolytic cleavage of (α1 → 4) glycosidic bonds. c. It is regulated both allosterically and via covalent modification. d. It uses UDP-glucose as a substrate. e. It is an important blood glucose sensor in myocytes.

ANSWER: c 14. Glycogen storage disease type 1a is associated with mutations in: a. glucose 6-phosphatase. b. phosphoglucomutase. c. pyruvate kinase. d. glycogen phosphorylase.

ANSWER: a 15. Epinephrine binding to β-adrenergic receptors on hepatocytes causes the eventual activation of protein kinase A (PKA). Which enzyme is a substrate for phosphorylation by PKA? a. phosphorylase a b. phosphorylase b kinase c. glycogen synthase b d. phosphoprotein phosphatase I e. pyruvate kinase ANSWER: b 16. Protein kinase A (PKA) stimulates glycogen breakdown. Which molecular signal activates PKA? a. decrease in [AMP] b. increase in [insulin] c. increase in [cAMP] d. high blood glucose Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15: Carbohydrate Metabolism Part 2: Gluconeogenesis and Glycogen Degradation e. low [UDP] ANSWER: c 17. Describe the process of glycogen breakdown in muscle. Include a description of the structure of glycogen, the nature of the breakdown reaction and the breakdown product, and the required enzyme(s).

ANSWER: Muscle glycogen consists of linear polymers of (α1 → 4)-linked D-glucose with many branches formed by (α1 → 6) glycosidic linkages to D-glucose. Glycogen phosphorylase in muscle catalyzes phosphorolytic cleavage of the terminal residue at the nonreducing ends, producing glucose 1phosphate. When phosphorylase approaches (α1 → 6) branch points, a second enzyme (the "debranching enzyme") removes the four glucose residues nearest the branch point and reattaches them in (α1 → 4) linkage at a nonreducing end. Now phosphorylase can continue to degrade the molecule.

18. In mammalian liver, glucose 1-phosphate, the product of glycogen phosphorylase, can enter glycolysis or replenish blood glucose. Describe the first reaction necessary by which these two fates may proceed.

ANSWER: To enter glycolysis, glucose 1-phosphate must undergo isomerization to glucose 6-phosphate by phosphoglucomutase. To replenish glucose in the bloodstream, glucose 1-phosphate must be hydrolyzed to free glucose by glucose 1-phosphatase.

19. Order the steps leading to glycogen breakdown resulting from the stimulation of liver cells by glucagon.

1) Activation of protein kinase A 2) cAMP levels rise 3) Phosphorylation of phosphorylase b 4) Phosphorylation of phosphorylase b kinase ANSWER: The correct temporal order is 2, 1, 4, 3.

20. Although most of the glycogen stored in a human body is found in skeletal muscle, this tissue does not play a role in maintaining blood sugar levels between meals. How are muscle cells (myocytes) precluded from doing so?

ANSWER: Myocytes do not express glucose 6-phosphatase, they cannot catalyze the hydrolysis of glucose 6phosphate released from glycogen stores to glucose that can be transported out of the cell.

21. During exercise, glycogen in both liver and muscle cells can be converted to glucose metabolites for ATP generation in skeletal muscle. Do liver glycogen and muscle glycogen supply the same amount of ATP to skeletal muscles? Explain.

ANSWER: No, there is one fewer ATP produced from the glucose stored in liver glycogen compared with the glucose stored in muscle glycogen. The glycogen in the liver releases glucose, which must be phosphorylated to glucose 6-phosphate in skeletal muscle; this uses an ATP molecule. Muscle glycogen uses Pi to release glucose as glucose 1-phosphate; this step bypasses the hexokinase reaction and saves using an ATP in the activation of glucose from muscle glycogen.

22. An enzyme used in both glycolysis and gluconeogenesis is: a. 3-phosphoglycerate kinase. b. glucose 6-phosphatase. c. hexokinase.

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Chapter 15: Carbohydrate Metabolism Part 2: Gluconeogenesis and Glycogen Degradation d. phosphofructokinase-1. e. pyruvate kinase.

ANSWER: a 23. Which statement about gluconeogenesis is false? a. For starting materials, it can use carbon skeletons derived from certain amino acids. b. It consists entirely of the reactions of glycolysis, operating in the reverse direction. c. It employs the enzyme glucose 6-phosphatase. d. It is one of the ways that mammals maintain normal blood glucose levels between meals. e. It requires metabolic energy (ATP or GTP).

ANSWER: b 24. Which enzyme involved in glycolysis is NOT involved in gluconeogenesis? a. 3-phosphoglycerate kinase b. aldolase c. enolase d. phosphofructokinase-1 e. phosphoglucose isomerase ANSWER: d 25. Gluconeogenesis must use "bypass reactions" to circumvent three reactions in the glycolytic pathway that are highly exergonic and essentially irreversible. Which of the three enzymes listed below must be bypassed in the gluconeogenic pathway?

1) Hexokinase 2) Phosphoglycerate kinase 3) Phosphofructokinase-1 4) Pyruvate kinase 5) Triosephosphate isomerase a. 1, 2, 3 b. 1, 2, 4 c. 1, 4, 5 d. 1, 3, 4 e. 2, 3, 4 ANSWER: d 26. Which statement about gluconeogenesis in animal cells is TRUE? a. A rise in the cellular level of fructose 2,6-bisphosphate stimulates the rate of gluconeogenesis. b. An animal fed a large excess of fat in the diet will convert any fat not needed for energy production into glycogen to be stored for later use. c. The conversion of fructose 1,6-bisphosphate to fructose 6-phosphate is not catalyzed by phosphofructokinase-1, the enzyme involved in glycolysis.

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Chapter 15: Carbohydrate Metabolism Part 2: Gluconeogenesis and Glycogen Degradation d. The conversion of glucose 6-phosphate to glucose is catalyzed by hexokinase, the same enzyme involved in glycolysis. e. The conversion of phosphoenolpyruvate to 2-phosphoglycerate occurs in two steps, including a carboxylation.

ANSWER: c 27. Which enzyme is NOT used in both glycolysis and gluconeogenesis? a. glucose 6-phosphatase b. aldolase c. phosphoglycerate mutase d. phosphofructokinase-2 e. glyceraldehyde 3-phosphate dehydrogenase ANSWER: a 28. Which compound cannot serve as the starting material for the synthesis of glucose via gluconeogenesis? a. acetate b. glycerol c. lactate d. oxaloacetate e. pyruvate ANSWER: a 29. In humans, gluconeogenesis: a. can result in the conversion of some amino acids into blood glucose. b. helps to reduce blood glucose after a carbohydrate-rich meal c. is activated by the hormone insulin. d. is essential in the conversion of fatty acids to glucose. e. requires the enzyme hexokinase.

ANSWER: a 30. Which organ does gluconeogenesis mainly take place? a. muscle b. liver c. brain d. intestines e. equally in all organs ANSWER: b 31. What is gluconeogenesis, and what useful purposes does it serve in people?

ANSWER: Gluconeogenesis is the biosynthesis of glucose from simpler, noncarbohydrate precursors such as oxaloacetate or pyruvate. During periods of fasting, when carbohydrate reserves have been Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15: Carbohydrate Metabolism Part 2: Gluconeogenesis and Glycogen Degradation exhausted, gluconeogenesis provides glucose for metabolism in tissues (brain, erythrocytes) that derive their energy primarily from glucose metabolism.

32. In gluconeogenesis, how do animals convert pyruvate to phosphoenolpyruvate? Show structures, enzymes, and cofactors.

ANSWER: Pyruvate is converted into phosphoenolpyruvate in two steps: (1) Pyruvate + HCO-3 + ATP → oxaloacetate + ADP + Pi (2) Oxaloacetate + GTP ⇋ CO2 +GDP + phosphoenolpyruvate The first reaction is catalyzed by pyruvate carboxylase, which requires biotin as a cofactor; the second, by phosphoenolpyruvate carboxykinase. (See Fig 15.5.)

33. Compare the energy requirements between the metabolic breakdown of glucose by glycolysis and formation of new glucose by gluconeogenesis. Provide the exact number and type of high energy molecules used in each process.

ANSWER: Metabolic breakdown of glucose by glycolysis requires two equivalents of ATP for formation of glucose 6-phosphate and fructose 1,6-bisphosphate. Formation of glucose by gluconeogenesis involves a much higher energy investment: two equivalents of ATP are needed to produce two molecules of oxaloacetate, two equivalents of GTP are needed to convert the oxaloacetate to PEP, and two equivalents of ATP are needed to form two molecules of 1,3-bisphosphoglycerate.

34. Hexokinase _____ predominates in the liver, and it requires a much _____ concentration of glucose to reach half-saturation as compared to hexokinases in myocytes (muscle cells). a. III; lower b. II; higher c. IV; higher d. I; lower e. V; higher ANSWER: c 35. What role(s) does ATP have with respect to PFK-1? a. only substrate b. substrate and allosteric activator c. only allosteric inhibitor d. substrate and allosteric inhibitor e. only allosteric activator ANSWER: d 36. Glucokinase: a. acts in the conversion of liver glycogen to glucose 1-phosphate. b. converts fructose 6-phosphate to glucose 6-phosphate. c. converts glucose 6-phosphate to fructose 6-phosphate. d. is a hexokinase isozyme found in liver hepatocytes.

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Chapter 15: Carbohydrate Metabolism Part 2: Gluconeogenesis and Glycogen Degradation e. is found in all mammalian tissues.

ANSWER: d 37. There is reciprocal regulation of glycolytic and gluconeogenic reactions interconverting fructose-6phosphate and fructose-1,6-bisphosphate. Which statement about this regulation is NOT correct? a. Fructose-2,6-bisphosphate activates phosphofructokinase-1. b. Fructose-2,6-bisphosphate inhibits fructose-1,6-bisphosphatase. c. The fructose-1,6-bisphosphatase reaction is exergonic. d. The phosphofructokinase-1 reaction is endergonic. e. This regulation allows control of the direction of net metabolite flow through the pathway.

ANSWER: d 38. An increase in activity which leads to the dephosphorylation and activation of phosphofructokinase-2? a. glucagon b. insulin c. pyruvate d. citrate e. ADP ANSWER: b 39. Which method directly regulates PFK-1? a. feedback inhibition by fructose 2,6-bisphosphate b. feed forward activation by phosphoenolpyruvate c. allosteric activation by AMP d. its phosphorylation in response to glucagon signaling e. allosteric activation by citrate ANSWER: c 40. Glucokinase (hexokinase IV) differs from hexokinase (hexokinase I): a. by having a higher affinity for glucose and not being inhibited by high levels of glucose. b. by requiring larger concentrations of glucose before reaching maximal activity. c. to allow the liver to export free glucose to the bloodstream when blood sugar levels drop. d. so that the liver does not compete with other tissues for glucose when glucose levels are high. e. to allow the pancreas to trigger insulin release when blood sugar levels are low.

ANSWER: b 41. When blood glucose is abnormally high, the pancreas releases: a. epinephrine b. glucagon c. glucose d. insulin Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15: Carbohydrate Metabolism Part 2: Gluconeogenesis and Glycogen Degradation e. trypsin ANSWER: d 42. When blood glucose is abnormally low, the pancreas releases: a. epinephrine b. glucagon c. glucose d. insulin e. trypsin ANSWER: b 43. What is the primary role of gluconeogenesis in mammals? a. provide glucose used in the liver b. provide energy to the liver to send the appropriate chemical signal elsewhere c. import noncarbohydrate precursors to synthesize glucose in the liver d. export glucose to be used in other tissues e. ensure that glycogen stores are completely exhausted in other tissues ANSWER: d 44. Which of the following is true regarding the relationship between PFK-1 and PFK-2? a. As glucagon increases, PFK-2 activity increases, causing PFK-1 activity to increase. b. As glucagon increases, PFK-2 activity decreases, causing PFK-1 activity to increase. c. As insulin increases, PFK-2 activity decreases, causing PFK-1 activity to decrease. d. As insulin increases, PFK-2 activity decreases, causing PFK-1 activity to increase. e. As insulin increases, PFK-2 activity increases, causing PFK-1 activity to increase.

ANSWER: e 45. Which of the following is true regarding the relationship between FBPase-1 and FBPase-2? a. As glucagon increases, FBPase-2 activity increases, causing FBPase-1 activity to increase. b. As glucagon increases, FBPase-2 activity increases, causing FBPase-1 activity to decrease. c. As glucagon increases, FBPase-2 activity decreases, causing FBPase-1 activity to increase. d. As insulin increases, FBPase-2 activity increases, causing FBPase-1 activity to increase. e. As insulin increases, FBPase-2 activity decreases, causing FBPase-1 activity to increase.

ANSWER: a 46. Under what circumstances does the bifunctional protein phosphofructokinase-2/fructose 2,6-bisphosphatase (PFK-2/FBPase-2) become phosphorylated, and what are the consequences of its phosphorylation to the glycolytic and gluconeogenic pathways?

ANSWER: Glucagon, signaling low blood sugar, stimulates cAMP synthesis, which leads to the phosphorylation of PFK-2/FBPase-2 (among other proteins). This phosphorylation enhances FBPase-2 activity and inhibits PFK-2 activity of the enzyme, resulting in lower levels of fructose 2,6-bisphosphate (F26BP). In the absence of F26BP as an allosteric effector, the activity of PFK-1 is Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15: Carbohydrate Metabolism Part 2: Gluconeogenesis and Glycogen Degradation reduced (inhibiting glycolysis) and the activity of FBPase-1 is enhanced (stimulating gluconeogenesis), thus enabling the liver to replenish blood glucose. See Figure 15.12.

47. Does the regulation of PFK-1 affect the activity of hexokinase I in muscle tissue? Explain.

ANSWER: Yes. Inhibition of PFK-1 will lead to a buildup of glucose 6-phosphate. Hexokinase I, which is found in skeletal muscle, is inhibited by high concentrations of glucose 6-phosphate.

48. Explain the two mechanisms responsible for regulating the activity of PFK-1 and FBPase-1.

ANSWER: One mechanism that regulates PFK-1 and FBPase-1 activity is the allosteric binding of ATP or AMP. High levels of ATP inhibit PFK-1 activity (while increasing FBPase-1 activity) to promote gluconeogenesis. When ATP levels are low, AMP relieves ATP inhibition of PFK-1, causing higher PFK-1 enzyme activity (while simultaneously decreasing FBPase-1 activity) to promote glycolysis.

PFK-1 and FBPase-1 activity can also be regulated hormonally by insulin and glucagon. Insulin increases the activity of PFK-2 by dephosphorylating it to produce more fructose 2, 6-bisphosphate, a potent allosteric activator of PFK-1 and allosteric inhibitor of FBPase-1. Glucagon produces the opposite effect by phosphorylating the same bifunctional protein responsible for PFK-2 activity to increase its FBPase-2 activity, leading to the decrease in fructose 2, 6-bisphosphate (inhibiting PFK1 and activating FBPase-1).

49. Describe the differences in affinity among hexokinase isozymes that allow glucose to be immediately consumed in muscle tissues while glucose is more readily imported or exported in the liver.

ANSWER: The predominant hexokinase isozyme of muscle cells (hexokinase II) has a high affinity for glucose — it is half-saturated at about 0.1 mM. On the other hand, the predominant hexokinase isozyme of liver cells (hepatocytes) is hexokinase IV (also called glucokinase), which is half-saturated (about 10 mM). Given that blood glucose concentration is 4 to 5 mM, hexokinase II isozyme normally acts at or near its Vmax (maximum rate of reaction) in muscle cells, leading to the immediate phosphorylation of glucose upon entering the cell. In contrast, the rate of phosphorylation of glucose by hexokinase IV in the liver allows some glucose molecules to remain free of phosphorylation.

Under conditions of low blood glucose, the glucose generated by gluconeogenesis leaves the cell before being trapped by phosphorylation.

50. In anaerobic skeletal muscle, breakdown of glycogen yields _____ ATP per glucose equivalent. a. 1 b. 2 c. 3 d. 4 e. 32 ANSWER: c 51. Compare in general terms the effects of epinephrine, glucagon, and insulin on glucose metabolism.

ANSWER: Epinephrine and glucagon cause an increase in the blood glucose level. In muscle cells, epinephrine acts when a higher than normal level of glucose is required. In the liver cells, glucagon acts when the blood glucose level is unusually low. Both stimulate gluconeogenesis and glycogen breakdown and decrease glycolysis and glycogen synthesis. Insulin causes a decrease in blood glucose levels; it Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15: Carbohydrate Metabolism Part 2: Gluconeogenesis and Glycogen Degradation acts by increasing glycogen synthesis, glycolysis, and glucose uptake by cells as well as by decreasing glycogen breakdown.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle 1. Which statement is NOT true of the reaction catalyzed by the pyruvate dehydrogenase complex? a. Biotin participates in the decarboxylation. b. Both NAD+ and a flavin nucleotide act as electron carriers. c. The reaction occurs in the mitochondrial matrix. d. The substrate is held by the lipoyllysine "swinging arm." e. Two different cofactors containing —SH groups participate.

ANSWER: a 2. Which compound is NOT required for the oxidative decarboxylation of pyruvate to form acetyl-CoA? a. ATP b. CoA-SH c. FAD d. lipoate e. NAD+ ANSWER: a 3. Which combination of cofactors is involved in the conversion of pyruvate to acetyl-CoA? a. biotin, FAD, and TPP b. biotin, NAD+, and FAD c. NAD+, biotin, and TPP d. pyridoxal phosphate, FAD, and lipoate e. TPP, lipoate, and NAD+ ANSWER: e 4. Which statement about the oxidative decarboxylation of pyruvate in aerobic conditions in animal cells is correct? a. One of the products of the reactions of the pyruvate dehydrogenase complex is a thioester of acetate. b. The methyl (—CH3) group is eliminated as CO2. c. The process occurs in the cytosolic compartment of the cell. d. The pyruvate dehydrogenase complex uses cofactors: NAD+, lipoic acid, pyridoxal phosphate, and FAD. ANSWER: a 5. Which compound is NOT an intermediate of the citric acid cycle? a. acetyl-CoA b. citrate c. oxaloacetate d. succinyl-CoA e. α-ketoglutarate ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle 6. Oxaloacetate uniformly labeled with 14C (i.e., with equal amounts of 14C in each of its carbon atoms) is condensed with unlabeled acetyl-CoA. After a single pass through the citric acid cycle back to oxaloacetate, what fraction of the original radioactivity will be found in the oxaloacetate? a. all b. one-half c. one-third d. one-fourth e. three-fourths ANSWER: b 7. Each molecule of acetyl-CoA oxidized by the citric acid cycle results in the net production of _____ molecule(s) of _____. a. one; citrate b. one; FADH2 c. one; NADH d. one; oxaloacetate e. seven; ATP ANSWER: b 8. Which factor is NOT associated with the oxidation of substrates by the citric acid cycle? a. CO2 production b. flavin reduction c. the presence of lipoic acid in some of the enzyme systems d. pyridine nucleotide oxidation e. All of these factors are associated with the oxidation of substrates by the citric acid cycle.

ANSWER: d 9. The oxidative decarboxylation of α-ketoglutarate proceeds by means of multistep reactions. Which cofactor is NOT required? a. ATP b. coenzyme A c. lipoic acid d. NAD+ e. thiamine pyrophosphate ANSWER: a 10. The reaction of the citric acid cycle that is MOST similar to the pyruvate dehydrogenase complex-catalyzed conversion of pyruvate to acetyl-CoA is the conversion of: a. citrate to isocitrate. b. fumarate to malate. c. malate to oxaloacetate.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle d. succinyl-CoA to succinate. e. α-ketoglutarate to succinyl-CoA. ANSWER: e 11. Which enzymatic activity would be decreased by thiamine deficiency? a. fumarase b. isocitrate dehydrogenase c. malate dehydrogenase d. succinate dehydrogenase e. α-ketoglutarate dehydrogenase complex ANSWER: e 12. The reaction of the citric acid cycle that produces an ATP equivalent (in the form of GTP) by substrate level phosphorylation is the conversion of: a. citrate to isocitrate. b. fumarate to malate. c. malate to oxaloacetate. d. succinate to fumarate. e. succinyl-CoA to succinate.

ANSWER: e 13. The reaction catalyzed by which enzyme is NOT linked to the reduction of NAD+? a. isocitrate dehydrogenase b. malate dehydrogenase c. pyruvate dehydrogenase d. succinate dehydrogenase e. the α-ketoglutarate dehydrogenase complex ANSWER: d 14. Which cofactor is required for the conversion of succinate to fumarate in the citric acid cycle? a. ATP b. biotin c. FAD d. NAD+ e. NADP+ ANSWER: c 15. Anaplerotic reactions: a. produce oxaloacetate and malate to maintain constant levels of citric acid cycle intermediates. b. produce biotin needed by pyruvate carboxylase. c. recycle pantothenate used to make CoA. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle d. produce pyruvate and citrate to maintain constant levels of citric acid cycle intermediates. e. All of the answers are correct.

ANSWER: a 16. Intermediates in the citric acid cycle are used as precursors in the biosynthesis of: a. amino acids. b. nucleotides. c. fatty acids. d. sterols. e. All of the answers are correct.

ANSWER: e 17. The conversion of one molecule of pyruvate to three molecules of CO2 via pyruvate dehydrogenase and the citric acid cycle also yields _____ molecules of NADH, _____ molecule(s) of FADH2, and _____ molecule(s) of ATP (or GTP). a. two; two; two b. three; one; one c. three; two; zero d. four; one; one e. four; two; one ANSWER: d 18. Entry of acetyl-CoA into the citric acid cycle is decreased when: a. [AMP] is high. b. NADH is rapidly oxidized through the respiratory chain. c. The ratio of [ATP]/[ADP] is low. d. The ratio of [ATP]/[ADP] is high. e. The ratio of [NAD+]/[NADH] is high.

ANSWER: d 19. Citrate synthase and the NAD+-specific isocitrate dehydrogenase are two key regulatory enzymes of the citric acid cycle. These enzymes are inhibited by: a. acetyl-CoA and fructose 6-phosphate. b. AMP and/or NAD+. c. AMP and/or NADH. d. ATP and/or NAD+. e. ATP and/or NADH. ANSWER: e 20. The glyoxylate pathway is important in bacteria, plants, fungi, and protists because it enables them to: a. synthesize glucose from fatty acids.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle b. form acetyl-CoA from malate. c. get rid of isocitrate formed from the aconitase reaction. d. obtain glyoxylate for cholesterol biosynthesis. e. obtain glyoxylate for pyrimidine synthesis.

ANSWER: a 21. The pyruvate dehydrogenase complex and the -ketoglutarate dehydrogenase complex are homologous enzyme assemblies that require several different coenzymes to carry out their reactions. Which compound does NOT participate directly in the function of these complexes? a. flavin mononucleotide b. coenzyme A c. lipoic acid d. thiamine pyrophosphate e. nicotinamide adenine dinucleotide ANSWER: a 22. Pyruvate dehydrogenase complex is regulated by: a. allosteric regulation. b. covalent modification. c. proteolytic cleavage. d. allosteric regulation and covalent modification. e. allosteric regulation, covalent modification, and proteolytic cleavage.

ANSWER: d 23. Which statement is false regarding the reaction catalyzed by the α-ketoglutarate dehydrogenase complex? a. A long lipoyllysine arm transfers intermediates from one active site to another within the complex. b. This reaction would be expected to be abnormally low in a person suffering from beriberi, a dietary deficiency of thiamine. c. Two different coenzymes containing sulfhydryl groups participate in the overall reaction. d. Both this complex and the pyruvate dehydrogenase complex contain a common subunit. e. The proximity of the catalytic enzymes allows for side reactions to take place in the cluster.

ANSWER: e 24. Which statement is false regarding the citric acid cycle? a. Succinate dehydrogenase is a flavoprotein associated with the electron-transport chain. b. The rate of the citric acid cycle increases when the [NAD+]/[NADH] ratio is high. c. In eukaryotes, the citric acid cycle occurs in the mitochondrial matrix. d. Acetyl-CoA inhibits the α-ketoglutarate dehydrogenase complex. e. The rate of the citric acid cycle decreases when the [ATP]/[ADP] ratio is high.

ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle 25. When is the consumption of acetyl-CoA by the citric acid cycle decreased? a. when the concentration of NAD+ is high b. when the concentration of oxaloacetate is high c. when NADH is rapidly reoxidized via the electron transport chain d. when the demand for ATP is low e. when the concentrations of NAD+ and oxaloacetate are both high ANSWER: d 26. Which statement is false with respect to the pyruvate dehydrogenase complex? a. It is located in the mitochondrial matrix. b. It consists of multiple E1, E2, and E3 enzymes. c. Its activity is increased in the presence of coenzyme A and ADP. d. The proximity of the catalytic enzymes in the cluster slows down the overall rate. e. The overall reaction is essentially irreversible.

ANSWER: d 27. Arsenic inhibits the pyruvate dehydrogenase complex by inactivating the dihydrolipoamide component of dihydrolipoyl transacetylase. Which enzyme is also MOST likely to be inhibited by arsenic? a. GAP dehydrogenase b. isocitrate dehydrogenase c. succinyl-CoA synthetase d. malate dehydrogenase e. α-ketoglutarate dehydrogenase ANSWER: e 28. Which reaction in the citric acid cycle will NOT proceed in the absence of inorganic phosphate ions? a. the reaction catalyzed by α-ketoglutarate dehydrogenase b. the reaction catalyzed by succinate dehydrogenase c. the reaction catalyzed by succinyl-CoA synthetase d. the reaction catalyzed by aconitase e. the reaction catalyzed by fumarase ANSWER: c 29. Which reaction of the citric acid cycle produces a net of one water molecule? a. citrate to isocitrate b. fumarate to malate c. succinyl-CoA to succinate d. succinate to succinyl-CoA e. None of the answers is correct.

ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle 30. Which reaction of the citric acid cycle requires a water molecule? a. formation of citrate b. succinyl-CoA to succinate c. fumarate to malate d. All of the answers are correct.

ANSWER: d 31. Which cofactors associated with the pyruvate dehydrogenase complex stay bound to the complex and can be classified as prosthetic groups?

1. TPP 2. lipoate 3. CoA-SH 4. NAD+ 5. FAD a. 1 and 5 b. 1, 2, and 4 c. 1, 2, and 5 d. 3 and 4 e. 4 and 5 ANSWER: c 32. Which step is NOT catalyzed by the correct subunit of the pyruvate dehydrogenase complex? a. E1 catalyzes the decarboxylation of pyruvate. b. E1 catalyzes the oxidation of a hydroxyethyl group to an acetyl group. c. E2 catalyzes the transfer of an acetyl group to CoA-SH. d. E3 catalyzes the reduction of oxidized lipoate. e. E3 catalyzes the transfer of electrons from FADH2 to NAD+.

ANSWER: d 33. Which does NOT inhibit the activity of the pyruvate dehydrogenase (PDH) complex? a. dichloroacetate (DCA) b. NAD+ c. ATP d. acetyl-CoA e. phosphorylation of PDH ANSWER: b 34. Which vitamin is NOT a precursor for a cofactor in the pyruvate dehydrogenase complex? a. thiamine for TPP b. riboflavin for FAD Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle c. biotin for lipoate d. niacin for NAD+ e. pantothenate for CoA ANSWER: c 35. Which type of reaction is found in the citric acid cycle? a. isomerization b. hydration c. condensation d. redox e. All of the answers are correct.

ANSWER: e 36. The citric acid cycle has an important anabolic function by: a. allowing for the conversion of certain carbon atoms from even-chain fatty acids into glucose. b. producing lactate for gluconeogenesis. c. providing metabolites for the synthesis of porphyrin rings, amino acids, and glucose. d. completing the oxidation of acetyl-CoA and storing electrons in the form of NADH. ANSWER: c 37. Although ADP is not directly involved in the isocitrate dehydrogenase reaction, the activity of this enzyme is dependent on the concentration of ADP in the matrix. Briefly explain how.

ANSWER: ADP is an allosteric activator of isocitrate dehydrogenase. Furthermore, when ADP levels are high in the matrix, ATP synthase and the electron transport chain are active. Enhanced activity of the electron transport chain leads to increased levels of NAD+ in the matrix. The availability of NAD+ enhances the activity of isocitrate dehydrogenase.

38. Briefly describe the relationship of the pyruvate dehydrogenase complex reaction to glycolysis and the citric acid cycle.

ANSWER: The pyruvate dehydrogenase complex converts pyruvate, the product of glycolysis, into acetyl-CoA, the starting material for the citric acid cycle.

39. Describe the enzymes, cofactors, intermediates, and products of the pyruvate dehydrogenase complex.

ANSWER: The pyruvate dehydrogenase complex consists of multiple copies of each of three enzymes. The first enzyme to act is pyruvate dehydrogenase (E1), which converts pyruvate to CO2 and the hydroxyethyl derivative of thiamine pyrophosphate (TPP). The same enzyme then oxidizes the hydroxyethyl group to an acetyl group attached to enzyme-bound lipoate through a thioester linkage.

The second enzyme, dihydrolipoyl transacetylase (E2), transfers the acetyl group to coenzyme A, forming acetyl-CoA. The third enzyme, dihydrolipoyl dehydrogenase (E3), oxidizes the dihydrolipoate to its disulfide form, passing the electrons through FAD to NAD+. (See Fig. 16-6.)

40. There is an overly high level of pyruvate in a patient's blood and urine. One possible cause is a genetic Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle defect in the enzyme pyruvate dehydrogenase, but another plausible cause is a specific vitamin deficiency.

Explain what vitamin might be deficient in the diet, and why that would account for high levels of pyruvate to be excreted in the urine. How would the correct explanation be determined?

ANSWER: The most likely explanation is that the patient has a deficiency of thiamine, without which the cell cannot make thiamine pyrophosphate, the cofactor for pyruvate dehydrogenase. The inability to oxidize pyruvate produced by glycolysis to acetyl-CoA would lead to accumulation of pyruvate in blood and urine. The most direct test for this deficiency is to feed a diet supplemented with thiamine and determine whether urinary pyruvate levels fall.

41. Match the cofactors with their roles in the pyruvate dehydrogenase complex reaction.

Cofactors: A. coenzyme A (CoA-SH) B. NAD+ C. thiamine pyrophosphate (TPP) D. FAD E. lipoic acid in oxidized form Roles: _____ attacks and attaches to the central carbon in pyruvate _____ oxidizes FADH2 _____ accepts the acetyl group from reduced lipoic acid _____ oxidizes the reduced form of lipoic acid _____ initial electron acceptor in oxidation of pyruvate ANSWER: C, B, A, D, E 42. What is the function of FAD in the pyruvate dehydrogenase complex? How is it regenerated?

ANSWER: FAD serves as the electron acceptor in the reoxidation of the cofactor dihydrolipoate. It is converted to FADH2 by this reaction and is regenerated by the passage of electrons to NAD+.

43. The human disease beriberi is caused by a deficiency of thiamine in the diet. People with severe beriberi have higher than normal levels of pyruvate in their blood and urine. Explain this observation in terms of specific enzymatic reaction(s).

ANSWER: Thiamine is essential for the synthesis of the cofactor thiamine pyrophosphate. Without this cofactor the pyruvate dehydrogenase complex cannot convert pyruvate into acetyl-CoA, so the pyruvate produced by glycolysis accumulates.

44. There are few, if any, humans with defects in the enzymes of the citric acid cycle. Explain this observation in terms of the role of the citric acid cycle.

ANSWER: The citric acid cycle is central to all aerobic energy-yielding metabolisms and also plays a critical role in biosynthetic reactions by providing precursors. Mutations in the enzymes of the citric acid cycle are likely to be lethal during fetal development.

45. Preparation of an extract of muscle results in a dramatic decrease in the concentration of citric acid cycle intermediates compared to their concentrations in the tissue. However, in 1935, Szent-Gyorgi showed that the production of CO2 by the extract increased when succinate was added. In fact, for every molecule of succinate added, many extra molecules of CO2 were produced. Explain this effect in terms of the known catabolic Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle pathways.

ANSWER: Succinate is an intermediate in the citric acid cycle that is not consumed but is regenerated by the operation of the cycle. By adding succinate to an extract that is depleted in citric acid cycle intermediates, these intermediates are replenished, and the cycle can resume operating, oxidizing acetyl-CoA to CO2.

46. Which part of the citric acid cycle from only isocitrate to fumarate are the high-energy phosphate compounds or reduced electron carriers produced or consumed, and name the enzyme that catalyzes each step.

ANSWER: This part of the citric acid cycle includes the reactions catalyzed by isocitrate dehydrogenase, the αketoglutarate dehydrogenase complex, succinyl-CoA synthetase, and succinate dehydrogenase.

GTP, NADH, and FADH2 are produced. (See Fig. 16.15.)

47. Show the reactions by which α-ketoglutarate is converted to malate in the citric acid cycle.

ANSWER: The reactions are those catalyzed by the α-ketoglutarate dehydrogenase complex, succinyl-CoA synthetase, succinate dehydrogenase, and fumarase. (See Fig. 16.15.)

48. Show the steps of the citric acid cycle in which a six-carbon compound is converted into the first fourcarbon intermediate in the path. For each step, show structures of substrate and product, name the enzyme responsible and show where cofactors participate.

ANSWER: The reactions are those catalyzed by isocitrate dehydrogenase, the α-ketoglutarate dehydrogenase complex, and succinyl-CoA synthetase. (See Fig. 16.15.)

49. Show the steps of the citric acid cycle from only succinyl-CoA to oxaloacetate. For each step, show structures of substrate and product, name the enzyme responsible, and show where cofactors participate.

ANSWER: These are the steps catalyzed by succinyl-CoA synthetase, succinate dehydrogenase, fumarase, and malate dehydrogenase. (See Fig. 16.15.)

50. Carbon dioxide is produced in two reactions in the citric acid cycle. For each of these reactions, name and show the structures of reactant and product, name the enzyme, and show how cofactors participate.

ANSWER: See the isocitrate dehydrogenase and α-ketoglutarate dehydrogenase reactions. (See also Fig. 16.15.)

51. In which reaction of the citric acid cycle does substrate-level phosphorylation occur?

ANSWER: Substrate-level phosphorylation of GDP to GTP occurs in the succinyl-CoA synthetase reaction in which succinyl-CoA is converted to succinate during the citric acid cycle.

52. A researcher is in charge of genetically engineering a new bacterium that will derive all of its ATP from sunlight by photosynthesis. Will the researcher put the enzymes of the citric acid cycle in this organism? Briefly explain why or why not.

ANSWER: Yes; even though the citric acid cycle is not needed for catabolic reactions in this organism, the enzymes of the cycle are still essential. They produce precursors of amino acids (such as αketoglutarate and oxaloacetate), of heme (succinyl-CoA), and of a variety of other essential products.

53. Match the cofactor with its function in the citric acid cycle. A given function may be used more than once or not at all.

Cofactor Function Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle (a) NAD+/NADH 1. carries O2 2. carries small carbon-containing molecules (b) FAD/FADH2 (c) CoA 3. carries e– (d) thiamine 4. carries small nitrogen-containing molecules (e) biotin ANSWER: (a) 3; (b) 3; (c) 2; (d) 2; (e) 2 54. The reactions that occur in the pyruvate dehydrogenase (PDH) complex to convert pyruvate to acetyl-CoA happen in the following order: a. decarboxylation, oxidation of lipoate, acetyl transfer to coenzyme A, reduction of lipoate. b. reduction of lipoate, oxidation of lipoate, decarboxylation, acetyl transfer to coenzyme A. c. acetyl transfer to coenzyme A, decarboxylation, reduction of lipoate, oxidation of lipoate. d. decarboxylation, reduction of lipoate, acetyl transfer to coenzyme A, oxidation of lipoate. e. oxidation of lipoate, decarboxylation, reduction of lipoate, acetyl transfer to coenzyme A. ANSWER: d 55. Which intermediate found in the citric acid cycle is reduced by the cytosolic malate dehydrogenase isozyme? a. malate b. fumarate c. aspartate d. oxaloacetate e. citrate ANSWER: d 56. Surprisingly, there is no transporter to bring NADH generated by glycolysis, fatty acid oxidation, and other processes into the mitochondrion; however, NADH is a significant cofactor that donates its electron to power the electron transport chain found in the inner mitochondrial membrane. How do the electrons extracted from cytosol find their way to the electron transport chain in the mitochondria? a. Electrons extracted from the cytosol do not provide a significant source of electrons for the electron transport chain and are not transported across the mitochondrial membrane. b. NADH is oxidized in the cytosol, and free electrons diffuse through the mitochondrial membrane. c. NADH is converted to FADH2 and is transported across the mitochondrial membrane. d. Oxaloacetate is oxidized to form malate that transports electrons across the mitochondrial membrane. e. Malate acts as an electron carrier that transports electrons across the mitochondrial membrane.

ANSWER: e 57. Which citric acid cycle intermediate can be withdrawn to be used in the synthesis of amino acids? a. citrate b. oxaloacetate c. malate d. fumarate Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle e. iso-citrate ANSWER: b 58. The citric acid cycle has an important anabolic function by: a. allowing for the conversion of certain carbon atoms from even-chain fatty acids into glucose. b. producing lactate for gluconeogenesis. c. providing metabolites for the synthesis of cholesterol, amino acids, and glucose. d. completing the oxidation of acetyl-CoA and storing electrons in the form of NADH. ANSWER: c 59. Which citric acid intermediate is transported out of the mitochondria in the noncanonical citric acid cycle? a. citrate b. oxaloacetate c. malate d. fumarate e. iso-citrate ANSWER: a 60. How can a disease such as cancer primarily impact flux through the citric acid cycle? a. Citrase produced in the cycle of the mitochondria is retained for increased ATP production. b. By shuttling malate back to the cytosol to synthesize NADH. c. Altering the amount of carbon used for mitochondrial energy production. d. By reducing the number of carbon used for other biosynthetic pathways to power glycolysis. e. By increasing NADH production in the mitochondria to power ATP synthesis.

ANSWER: c 61. Which condition is associated with niacin deficiency (NAD+/NADH)? a. Scurvy b. Rickets c. Beriberi d. Gliomas and blastoglioma e. Pellagra ANSWER: e 62. In humans, there are genes that encode three forms of isocitrate dehydrogenase (IDH). Which form of isocitrate dehydrogenase is the predominant form found in the cytosol? a. IDH1 b. IDH2 c. IDH3 d. IDH2 and IDH3 e. IDH1, IDH2, and IDH3 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle ANSWER: a 63. The citric acid cycle plays a central role in the generation of cellular energy by funneling NADH and FADH2 into the: a. noncanonical citric acid cycle b. pentose phosphate pathway c. gluconeogenesis d. electron transport chain e. fatty acid oxidation ANSWER: a 64. Which cellular event causes a major change in a cell’s metabolic needs? a. cell death b. differentiation c. hormonal signaling d. migration e. quiescence ANSWER: b 65. Which citric acid intermediate is not needed to maintain the function of the noncanonical citric acid cycle? a. citrate b. malate c. isocitrate d. oxaloacetate e. none of the above ANSWER: c 66. Describe how the noncanonical citric acid cycle reduces the overall number of NADH generated in the mitochondria.

ANSWER: Recall that NADH is generated in steps 3, 4, and 8 of the citric acid cycle (See Fig. 16.15). Since the noncanonical citric acid cycle sequesters citrate away from the mitochondria, this prevents the generation of two NADH molecules (from steps 3 and 4) while maintaining NADH generated in step 8 (compare Fig. 16.15 and Fig. 16.20) by converting malate to oxaloacetate. In other words, a citrate molecule that enters the citric acid cycle can generate 3 NADH molecules in the mitochondria compared to the noncanonical citric acid cycle that produces only 1 NADH molecule per citrate in the mitochondria, resulting in an overall decrease of 2 NADH molecules generated in the mitochondria.

67. Explain why tumor cells divert more of the carbon derived from glucose oxidation and glycolysis is channeled into various biosynthetic pathways (like synthesis of amino acids and nucleotides) in the cytosol.

Consider the most common hallmarks of cancer.

ANSWER: Cell proliferation is a hallmark of cancer. It is associated with a higher rate of DNA replication, leading to an increase in demand for nucleotides. At the same time, cell proliferation also depends Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16: Pyruvate Oxidation and the Citric Acid Cycle on cell cycle proteins (along with other proteins) that also leads to an increase in demand for amino acids. Although the number of ATP generated per glucose molecule is decreased, the cell can compensate by using more glucose molecules to produce the necessary ATP. 68. Show the steps of the noncanonical citric acid cycle, starting from citrate to mitochondrial oxaloacetate. For each step, show structures of substrate and product, name the enzyme responsible and show where cofactors participate.

ANSWER: These steps are catalyzed by ATP citrate lyase, malate dehydrogenase 1, malate dehydrogenase 2 in that order. (See Fig. 16.20.)

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Chapter 17: Lipid Catabolism and Anabolism 1. In β oxidation, which cofactor is required for the conversion of a β-ketoacyl-CoA to an acyl-CoA and an acetyl-CoA molecule? a. NAD+ b. FAD c. H2O d. CoA-SH ANSWER: d 2. In β oxidation, which cofactor is required for the conversion of β-hydroxyacyl-CoA to β-ketoacyl-CoA? a. NAD+ b. FAD c. H2O d. CoA-SH ANSWER: a 3. In the equation, which substance is being reduced?

a. β-ketoacyl-CoA Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17: Lipid Catabolism and Anabolism b. L-β-hydroxyacyl-CoA c. NADH d. NAD+ e. H+ ANSWER: d 4. Which statement is true about perilipins? a. They are proteins on the surface of lipid droplets that help prevent untimely lipid mobilization. b. They are proteins in the interior of lipid droplets that stabilize the pH. c. They are proteins that aid in the transport of lipids in the bloodstream. d. They are modulators of hormone-sensitive lipase.

ANSWER: a 5. Which is NOT a symptom of ketosis? a. reduced blood pH b. high levels of malonyl-CoA c. presence of ketone bodies in the urine d. smell of acetone in the breath ANSWER: b 6. Which statement is true regarding fatty acids in adipocytes? a. They are primarily bound to albumin and stored in lipid droplets. b. They cannot be oxidized for energy in this tissue. c. In adipocytes, fatty acids from the bloodstream are assembled into triglycerides by esterifying them to glycerol. d. There is constant movement of fatty acids in and out of adipose tissues that is independent of hormone signals.

ANSWER: c 7. What is the major site of the formation of β-hydroxybutyrate from fatty acids? a. intestinal mucosa b. kidney c. liver d. adipose tissue e. skeletal muscle ANSWER: c 8. Which statement is false regarding the catabolic oxidation of one molecule of palmitate (16:0)? a. Eight molecules of acetyl-CoA are formed. b. AMP and two Pi are formed in the cytosol. c. The oxidation reactions occur in the mitochondria.

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Chapter 17: Lipid Catabolism and Anabolism d. Eight molecules of H2O are required. e. Carnitine is required for the transport of fatty acids into the mitochondrial matrix.

ANSWER: d 9. Which statement is false regarding ketone bodies? a. During prolonged starvation, the brain adapts to using ketone bodies as a fuel. b. Excess acetone production during ketoacidosis is exhaled through the lungs. c. Ketone bodies can be used as a carbon source for fatty acid synthesis. d. Ketone bodies can be converted to glucose during ketosis. e. Ketone bodies are formed when the glucagon/insulin ratio is high.

ANSWER: d 10. Which compound is NOT considered a ketone body? a. acetoacetate b. β-hydroxybutyrate c. acetone d. dihydroxyacetone ANSWER: d 11. Some athletes believe that dietary carnitine supplements will enhance their ability to perform aerobic exercise. Which statement BEST describes the scientific basis of this belief? a. Carnitine is required for the transport of medium- and long-chain fatty acids into the mitochondrial matrix, so increasing the amount of carnitine will increase the rate of transport. b. Carnitine activates carnitine palmitoyltransferase 1 thereby increasing the rate of transport of fatty acids into the mitochondrial matrix. c. Carnitine levels in the cytoplasm are reduced when large amounts of fatty acyl-carnitine are transported into the mitochondrial matrix. d. Carnitine enhances the activation of fatty acids to fatty acyl-CoA, thereby increasing the rate of their transport into the mitochondrial matrix. e. Carnitine is an allosteric activator of carnitine palmitoyltransferase 2.

ANSWER: a 12. Which statement regarding ketone bodies is false? a. The production of excess ketone bodies can cause a condition known as ketosis, which is associated with acidosis. b. In hepatocytes, the production of ketone bodies is an anaplerotic process because the products of βhydroxybutyrate oxidation can enter the citric acid cycle. c. Ketone bodies are produced by the liver during starvation or in cases when insulin production is insufficient. d. The production of ketone bodies increases when blood glucagon levels are elevated.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17: Lipid Catabolism and Anabolism 13. What role does serum albumin play in fat metabolism? a. It is a component of the chylomicron lipoprotein. b. It acts as a carrier for free fatty acids in the bloodstream. c. It binds hormone-sensitive lipase that is responsible for the release of triglycerides from adipocytes. d. It is involved in the transport of short-chain fatty acids across the plasma membrane. e. It is important in the transport of ketone bodies from the liver to the brain.

ANSWER: b 14. Which tissue cannot use ketone bodies as a fuel? a. liver b. heart c. skeletal muscle d. brain ANSWER: a 15. In which cellular compartment are ketone bodies synthesized? a. the cytosol b. the mitochondrial matrix c. the mitochondrial intermembrane space d. the endoplasmic reticulum e. peroxisomes ANSWER: b 16. Which statement(s) about acetoacetate and β-hydroxybutyrate is false?

1. They can be used as a fuel by skeletal muscle.

2. They are synthesized in the liver.

3. They give rise to acetone.

4. They require acetoacetyl-CoA for their synthesis.

5. They may be regarded as water-soluble, transportable forms of citrate in the blood. a. 1 and 2 b. only 3 c. 1 and 4 d. 4 and 5 e. only 5 ANSWER: e 17. Which enzyme is involved in activation of fatty acids in the cytosol? a. acyl-CoA synthetase b. inorganic pyrophosphatase c. carnitine palmitoyltransferase 1 d. both acyl-CoA synthetase and inorganic pyrophosphatase Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17: Lipid Catabolism and Anabolism e. both inorganic pyrophosphatase and carnitine palmitoyltransferase 1 ANSWER: d 18. Which statement is true regarding the regulation of β oxidation? a. AMPK activates acetyl-CoA carboxylase, which leads to the inhibition of β oxidation. b. A buildup in the levels of acetyl-CoA inhibits the thiolase reaction in β oxidation. c. High levels of NADP+ activate β-hydroxyacyl-CoA dehydrogenase. d. Malonyl-CoA inhibits carnitine palmitoyltransferase 2, which leads to the inhibition of β oxidation. e. β oxidation is enhanced in erythrocytes during starvation.

ANSWER: b 19. In hepatocytes (liver cells), what is a possible fate of acetyl-CoA produced from the oxidation of a fatty acid? a. oxidation via the citric acid cycle b. use in the synthesis of ketone bodies c. use in gluconeogenesis d. both oxidation via the citric acid cycle and use in the synthesis of ketone bodies e. both oxidation via the citric acid cycle and use in gluconeogenesis ANSWER: d 20. In myocytes (muscle cells), what is a possible fate of acetyl-CoA produced from the oxidation of a fatty acid? a. oxidation via the citric acid cycle b. use in the synthesis of ketone bodies c. use in gluconeogenesis d. both oxidation via the citric acid cycle and use in the synthesis of ketone bodies e. both oxidation via the citric acid cycle and use in gluconeogenesis ANSWER: a 21. Which form of a fatty acid is NOT found in the blood? a. acyl-CoA b. free fatty acid bound to albumin c. a component of a chylomicron d. a component of a triacylglycerol ANSWER: a 22. The role of hormone-sensitive triacylglycerol lipase is to: a. hydrolyze lipids stored in the liver. b. hydrolyze membrane phospholipids in hormone-producing cells. c. hydrolyze triacylglycerols stored in adipose tissue. d. synthesize lipids in adipose tissue.

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Chapter 17: Lipid Catabolism and Anabolism e. synthesize triacylglycerols in the liver.

ANSWER: c 23. Which compound is able to cross the inner mitochondrial membrane? a. acetyl-CoA b. fatty acyl-carnitine c. fatty acyl-CoA d. malonyl-CoA ANSWER: b 24. What is the correct order of function of the enzymes of β oxidation?

1. β-hydroxyacyl-CoA dehydrogenase 2. thiolase 3. enoyl-CoA hydratase 4. acyl-CoA dehydrogenase a. 1, 2, 3, 4 b. 3, 1, 4, 2 c. 4, 3, 1, 2 d. 1, 4, 3, 2 ANSWER: c 25. Which statement(s) is true of the oxidation of one molecule of palmitate (a 16-carbon saturated fatty acid; 16:0) by the β-oxidation pathway, beginning with the free fatty acid in the cytoplasm?

1. Activation of the free fatty acid requires the equivalent of two ATPs. 2. Inorganic pyrophosphate is produced.

3. Carnitine functions as an electron acceptor.

4. Eight molecules of FADH2 are formed.

5. Eight molecules of acetyl-CoA are formed.

6. There is no direct involvement of NAD+. a. only 1 and 5 b. 1, 2, and 5 c. 1, 2, and 6 d. 1, 3, and 5 e. only 5 ANSWER: b 26. Which statement applies to the β oxidation of fatty acids? a. The process takes place in the cytosol of mammalian cells. b. Carbon atoms are removed from the acyl chain one at a time. c. Before oxidation, fatty acids must be converted to their CoA derivatives. d. NADP+ is the electron acceptor.

ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17: Lipid Catabolism and Anabolism 27. The conversion of palmitoyl-CoA (16:0) to myristoyl-CoA (14:0) and one molecule of acetyl-CoA by the βoxidation pathway results in the net formation of: a. one FADH2 and one NADH. b. one FADH2 and one NADPH. c. one FADH2, one NADH, and one ATP. d. two FADH2 and two NADH. e. two FADH2, two NADH, and one ATP. ANSWER: a 28. The metabolite that regulates the activity of carnitine palmitoyltransferase 1 is: a. acetyl-CoA. b. carnitine-CoA. c. malonyl-CoA. d. NADH. e. CoA. ANSWER: c 29. In the first step of fatty acid oxidation, the fatty acid (R—COOH) is converted to its coenzyme A derivative in the reaction: R—COOH + ATP + CoA-SH → R—CO—S-CoA + AMP + PPi The standard free-energy change (ΔG'º) for this reaction is –15 kJ/mol.

What will tend to make the reaction even more favorable when it takes place within a cell?

ANSWER: The hydrolysis of PPi by inorganic pyrophosphatase, for which ΔG'º is –19 kJ/mol, makes the overall ΔG'º more negative and drives the reaction forward by ensuring a low concentration of the product PPi 30. Adding carnitine to a culture of isolated, intact mitochondria stimulates the oxidation of acetyl-CoA. Why? ANSWER: Carnitine is essential in the transport of fatty acyl groups into the mitochondrial matrix, where fatty acid oxidation occurs. Acetyl-CoA produced by β oxidation of fatty acids can then be oxidized via the citric acid cycle in the mitochondrial matrix.

31. Describe the first two steps of β oxidation of saturated fatty acids. Indicate where any cofactors participate.

ANSWER: The first two steps of β oxidation are catalyzed by fatty acyl-CoA dehydrogenase and enoyl hydratase. Note the formation of FADH2 in the first step and the addition of water across the double bond in the second step. (See Figure 17.22) 32. Describe the third and fourth steps in the β oxidation of a saturated fatty acid. Name the enzymes involved and indicate where any cofactors participate.

ANSWER: The third and fourth steps in β oxidation are catalyzed by β-hydroxyl-CoA dehydrogenase and acylCoA acetyltransferase (thiolase). Note the formation of NADH in the third step and the cleavage of the carbon–carbon bond in a reverse Claisen condensation that involves an additional molecule of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17: Lipid Catabolism and Anabolism free CoA in the fourth step. (See Figure 17.22) 33. One of the steps in fatty acid oxidation in mitochondria involves the addition of water across a double bond.

What is the next step in the process? Show structures and indicate where any cofactor(s) participates.

ANSWER: The next step is the reaction catalyzed by β-hydroxyacyl-CoA dehydrogenase, for which NAD+ is cofactor. (See Figure 17.22) 34. Write a balanced equation for the β oxidation of palmitoyl-CoA, a 16-carbon, fully saturated fatty acid, and indicate how much of each product is formed.

ANSWER: The overall reaction is: Palmitoyl-CoA + 7CoA-SH + 7FAD + 7NAD+ + 7H2O → 8 acetyl-CoA + 7FADH2 + 7NADH + 7H+ 35. A laboratory report shows the presence of a high concentration of ketone bodies in the urine of a patient, what disease would be suspected? Why do ketone bodies accumulate in such patients?

ANSWER: The patient is probably an untreated diabetic, but the condition might also result from fasting. In either case, the unavailability of glucose from the blood stimulates gluconeogenesis in the liver. As the substrate for glucose formation, oxaloacetate is withdrawn from the citric acid cycle, bringing that cycle to a near halt. The fatty acids being oxidized in the liver yield acetyl-CoA, which now cannot be oxidized via the citric acid cycle. The enzyme thiolase catalyzes the production of acetoacetyl-CoA, which is then converted into ketone bodies that are exported from the liver. (See Figure 17.24.)

36. Indicate whether the statement is likely true or false. Explain.

After a breakfast rich in fat and very low in carbohydrates, acetyl-CoA levels rise and ketone body synthesis decreases.

ANSWER: This statement is likely false. Following an overnight fast, the glycogen stores in the liver are significantly reduced. Therefore, if one has a breakfast very low in carbohydrates, the gluconeogenic pathway will likely be activated to maintain blood glucose levels. If an individual is catabolizing fatty acids under gluconeogenic conditions, the excess acetyl-CoA that cannot be oxidized by the citric acid cycle will be converted to ketones. Therefore, after a breakfast rich in fat and very low in carbohydrates, acetyl-CoA levels will rise and ketone body synthesis will likely increase.

37. Which compound is NOT required in the synthesis of fatty acids? a. acetyl-CoA b. biotin c. bicarbonate d. malonyl-CoA e. NADH ANSWER: e 38. Which statement is NOT true of the reaction producing malonyl-CoA during fatty acid synthesis? a. It is stimulated by citrate. b. It requires acyl carrier protein (ACP).

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Chapter 17: Lipid Catabolism and Anabolism c. It requires CO2 (or bicarbonate). d. One molecule of ATP is converted to ADP + Pi for each malonyl-CoA synthesized. e. The cofactor is biotin.

ANSWER: b 39. If malonyl-CoA is synthesized from 14CO2 and unlabeled acetyl-CoA, and the labeled malonate is then used for fatty acid synthesis, the final product (fatty acid) will have radioactive carbon in: a. every C. b. every even-numbered C atom. c. every odd-numbered C atom. d. no part of the molecule. e. only the omega-carbon atom (farthest carbon from C-1).

ANSWER: d 40. Which statement applies to synthesis of malonyl-CoA? a. It is synthesized from acetyl-CoA and bicarbonate. b. The reaction is catalyzed by acetyl-CoA carboxylase. c. The two-step synthesis is irreversible. d. In animal cells the synthesis takes place in the cytoplasm. e. All of the statements apply to the synthesis of malonyl-CoA. ANSWER: e 41. Which statement is true about precursors required for fatty acid synthesis in animal cells? a. NADPH is produced in the cytosol by the pentose phosphate pathway. b. NADPH is produced in the nucleus by malic enzyme. c. Acetyl-CoA is transported out of the mitochondrion via the aspartate shuttle. d. CoA is transported across the mitochondrial membrane. e. Malonyl-CoA is formed in the mitochondrion.

ANSWER: a 42. Which hormones stimulate the mobilization/release of fatty acids from adipose tissue to meet energy demands? a. insulin and epinephrine b. glucagon and epinephrine c. insulin and glucagon d. cortisol and insulin ANSWER: b 43. Which is the precursor for the synthesis of both glycerophospholipids and triacylglycerols? a. CDP-choline b. CDP-diacylglycerol Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17: Lipid Catabolism and Anabolism c. phosphatidate phosphatase d. phosphatidic acid e. phosphoethanolamine ANSWER: d 44. Which statement about triacylglycerol synthesis is correct? a. Humans can store more energy in glycogen than in triacylglycerols. b. Insulin stimulates conversion of dietary carbohydrates into triacylglycerols. c. It is not a hormone-sensitive process. d. Mammals are unable to convert carbohydrates into triacylglycerols. e. Phosphatidate is not on the pathway of triacylglycerol synthesis.

ANSWER: b 45. What cofactor is required for acetyl-CoA carboxylase activity? a. NADPH b. thiamine pyrophosphate c. lipoamide d. biotin ANSWER: d 46. Which statement is false concerning acetyl-CoA carboxylase? a. It catalyzes the rate-limiting step in fatty acid synthesis. b. This enzyme is inhibited by palmitate in mammals. c. This enzyme is allosterically activated by citrate in mammals. d. In mammals, this enzyme is inactivated under conditions when glycolysis is active.

ANSWER: d 47. During fatty acid synthesis, malonyl-CoA becomes covalently attached to acyl carrier protein via _____ formation to a _____. a. ester-bond; serine b. thioester-bond; cysteine c. amide-bond; lysine d. ether-bond; serine e. thioester-bond; prosthetic group ANSWER: e 48. After four rounds of fatty acid synthase activity, what size is the fatty acyl intermediate? a. 4 carbons long b. 5 carbons long c. 8 carbons long d. 9 carbons long Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17: Lipid Catabolism and Anabolism e. 10 carbons long ANSWER: e 49. Which is another name for lipin? a. acyl-CoA synthetase b. acyl transferase c. PEP carboxykinase d. phosphatidic acid phosphatase e. CDP-diacylglycerol ANSWER: d 50. The reaction sequence that leads to fatty acid synthesis includes (1) condensation, (2) first reduction reaction, (3) dehydration, and (4) second reduction. Which fatty acid synthase domain is associated with step (3) dehydration? a. ER b. MAT c. KS d. KR e. DH ANSWER: e 51. The reaction sequence that leads to fatty acid synthesis includes (1) condensation, (2) first reduction reaction, (3) dehydration, and (4) second reduction. Which fatty acid synthase domain is associated with step (2) first reduction reaction? a. ER b. MAT c. KS d. KR e. DH ANSWER: d 52. If performing fatty acid biosynthesis with CO2 labeled with 14C, where does the label end up in a fatty acid?

ANSWER: There will not be any radioactivity in any fatty acid because the carboxyl group that is added to form malonyl-CoA is lost in the first condensation reaction catalyzed by FAS. 53. The reaction sequence that leads to fatty acid synthesis includes (1) condensation, (2) first reduction reaction, (3) dehydration, and (4) second reduction. Show the first reduction reaction, with any required cofactors.

ANSWER: β-ketobutyryl-ACP + NADPH + H+ → β-hydroxybutyryl-ACP + NADP+ (See Figure 17.3) 54. Show the structure of each intermediate in the conversion of β-hydroxybutyryl-ACP to butyryl-ACP by the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17: Lipid Catabolism and Anabolism fatty acid synthase complex. Show where cofactors participate. In the first intermediate, circle the carbon atoms that are derived from malonyl-CoA. ANSWER: β-Hydroxybutyryl-ACP is first dehydrated, yielding trans-Δ2-butenoyl-ACP, which is then reduced to butyryl-ACP, with NADPH as the reducing agent. (See Figure 17.3) The first two carbon atoms bound to the ACP are derived from malonyl-CoA. 55. Describe how fatty acid synthesis is regulated through compartmentalization and energy availability in mammals.

ANSWER: Fatty acid synthesis occurs in the cytosol and ER using cytoplasmic acetyl-CoA generated by the enzyme citrate lyase (using citrate as a substrate). This cytoplasmic pool of acetyl-CoA is distinct from the mitochondrial pool since it is regulated by insulin. Additionally, insulin promotes the synthesis of fatty acids and increases lipid storage through triacylglycerol. Hormones like glucagon and epinephrine have the opposite effect by stimulating the release of triacylglycerol from adipose tissue and simultaneously decreasing the rate of glycolysis and increasing the rate of gluconeogenesis in the liver. Moreover, glucagon inhibits acetyl-CoA carboxylase and decreases malonyl-CoA production, effectively preventing the recharging of ACP with another molecule of malonyl-CoA. Energetically, fatty acid synthesis requires significant ATP and NADPH (2 NADPH and 1 ATP per round of FAS reactions or 14 NADPH and 7 ATP per palmitate). Therefore, a high NADPH/NADP+ ratio is needed for the fatty acid synthesis to proceed.

56. Describe how free fatty acids are incorporated into glycerol backbones to form triacylglycerols.

ANSWER: The process begins with the acylation of the two free hydroxyl groups of L-glycerol 3-phosphate by two fatty acyl-CoA molecules, yielding phosphatidic acid. Lipin hydrolyzes phosphatidic acid to form a 1,2-diacylglycerol, followed by its transesterification with a third fatty acyl-CoA. 57. Cholesterol is synthesized from: a. acetyl-CoA. b. choline. c. lipoic acid. d. malate. e. oxalate.

ANSWER: a 58. A 30-carbon precursor of the steroid nucleus is: a. farnesyl pyrophosphate. b. geranyl pyrophosphate. c. isopentenyl pyrophosphate. d. lysolecithin. e. squalene.

ANSWER: e 59. Chylomicrons carry _____ in the _____.

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Chapter 17: Lipid Catabolism and Anabolism a. triacylglycerols; cell b. triacylglycerols; blood c. cholesterols; blood d. fatty acids; blood e. fatty acids; cell ANSWER: b 60. Lipoprotein particles in human blood do NOT contain: a. an apolipoprotein B isoform. b. cholesterol. c. cholesteryl esters. d. albumin. e. triglycerides.

ANSWER: d 61. Which lipoprotein particle is rich in cholesteryl esters? a. chylomicrons b. VLDL c. LDL d. HDL ANSWER: c 62. Which statement is a step in the uptake of cholesterol from the bloodstream into cells? a. LDL containing ApoB-100 is recognized by the LDL receptor. b. LDL binding to the LDL receptor leads to endocytosis of the complex. c. The cholesteryl esters in the LDL are hydrolyzed to release free cholesterol. d. The ApoB-100 protein is degraded. e. All of the above are steps in the uptake of cholesterol.

ANSWER: e 63. Which factor does NOT contribute to lowering the risk of atherosclerosis? a. statin drugs b. HDL c. reverse cholesterol transport d. apoA-I e. foam cells ANSWER: e 64. What is the net cost, in terms of high energy molecules, in the synthesis of squalene from acetyl-CoA? a. 2 NADPH and 3 ATP b. 6 NADPH and 9 ATP Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17: Lipid Catabolism and Anabolism c. 6 NAPDH and 18 ATP d. 12 NADPH and 18 ATP e. 13 NADPH and 18 ATP ANSWER: d 65. Which apolipoprotein is found in chylomicrons but not in other lipoproteins? a. apoB-48 b. apoB-100 c. apoC-II d. apoC-III e. apoE ANSWER: a 66. Reverse cholesterol transport is primarily associated with which lipoprotein? a. chylomicrons b. VLDL c. IDL d. LDL e. HDL ANSWER: e 67. Conversion of VLDL to LDL is associated with what changes? a. depletion of cholesteryl esters b. an increase in particle diameter c. apoC-II-mediated hydrolysis of triglycerides by lipoprotein lipase d. apoE-mediated receptor interactions ANSWER: c 68. Which apolipoprotein binds to the LDL receptor? a. apoB-100 b. apoC-I c. apoC-II d. apoD e. apoE ANSWER: a 69. Which lipoproteins do NOT originate from the liver? a. chylomicrons and VLDL b. VLDL and HDL c. LDL and HDL d. chylomicrons Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17: Lipid Catabolism and Anabolism e. VLDL and LDL ANSWER: d 70. Which protein is NOT typically found in HDL? a. apoA-1 b. apoE c. apoH d. apoC-I e. apoD ANSWER: c 71. What statement is false regarding the regulation of HMG-CoA reductase? a. HMG-CoA reductase is inhibited by phosphorylation. b. Elevated levels of ATP will inhibit HMG-CoA reductase. c. Elevated blood glucose will increase the activity of HMG-CoA reductase. d. Statins act as competitive inhibitors for this enzyme. e. Low cellular concentrations of cholesterol will increase the expression of this enzyme and the LDL receptor.

ANSWER: b 72. Show the reaction that limits the rate of cholesterol synthesis from acetate, indicating the role of any cofactors that participate.

ANSWER: This is the reaction catalyzed by HMG-CoA reductase. Two NADPH molecules serve as cofactors.

(See Figure 17.10) 73. Show the pathway from acetyl-CoA to mevalonate, indicating the roles of any cofactors.

ANSWER: First, two molecules of acetyl-CoA are condensed to form acetoacetyl-CoA. Addition of a third acetyl group via the enzyme HMG-CoA synthase results in the formation of HMG-CoA. HMG-CoA is then reduced to mevalonate by HMG-CoA reductase, requiring two NADPH molecules. (See Figure 17.10-34) 74. A molecule of squalene is shown; it is composed of six isoprene units. Draw lines to indicate the junctions between the six isoprene units.

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Chapter 17: Lipid Catabolism and Anabolism ANSWER:

75. What are plasma lipoproteins? What is their general role in mammalian metabolism?

ANSWER: Plasma lipoproteins are protein-lipid aggregates that circulate in the blood, carrying phospholipids, triacylglycerols, cholesterol, and cholesteryl esters from their points of synthesis or absorption to the tissues in which they will be used. The lipid and protein content, and therefore the density, of plasma lipoproteins varies, giving rise to particles separable by ultracentrifugation: HDL, LDL, VLDL, and chylomicrons, for example.

76. Describe the process by which cholesterol esters in the bloodstream enter cells.

ANSWER: The process is receptor-mediated endocytosis (described in Figure 17.16). LDL particles containing cholesterol esters interact via apoB-100 with specific LDL receptors on the cell surface, initiating endocytosis that brings the LDL into the cell within endosomes. Endosomes fuse with lysosomes, and lysosomal enzymes degrade the apoproteins and breakdown the LDL contents; cholesterol, triacylglycerol, and amino acids are released into the cytosol, and LDL receptors recirculate, reappearing on the cell surface.

77. In comparing fatty acid biosynthesis with β oxidation of fatty acids, which statement is incorrect? a. A thioester derivative of crotonic acid (trans-2-butenoic acid) is an intermediate in the synthetic path and in the degradative path. b. A thioester derivative of D-β-hydroxybutyrate is an intermediate in the synthetic path but not in the degradative path. c. Fatty acid biosynthesis uses NADPH exclusively, whereas β oxidation uses NAD+ exclusively. d. Fatty acid degradation is catalyzed by cytosolic enzymes; fatty acid synthesis by mitochondrial enzymes.

ANSWER: d 78. Which statement is true regarding the entry of fatty acyl groups into the mitochondrial matrix? a. It is a rate-limiting step in the oxidation of these fatty acids. b. It is regulated by the concentration of chylomicrons. c. Entry is mediated by perilipin. d. Once these fatty acyl molecules enter the matrix, they may either be transported back out of the mitochondrion or proceed to β oxidation.

ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17: Lipid Catabolism and Anabolism 79. Neutral lipids that lack a charge are stored in adipocytes as lipid droplets. Describe the first five steps of free fatty acid mobilization from triacylglycerols, starting with epinephrine and ending with perilipin binding.

ANSWER: In step 1, the hormone epinephrine is secreted in response to low blood glucose levels.

In step 2, the hormone stimulates the enzyme adenylyl cyclase in the adipocyte plasma membrane via a G protein. Adenylyl cyclase produces the intracellular second messenger cyclic adenosine monophosphate (cyclic AMP, or cAMP).

In step 3, cyclic AMP–dependent protein kinase (PKA) triggers phosphorylation of (step 4) perilipin to bind to (step 5) CGI-58 (comparative gene identification-58). (See Figure 17.18) 80. Neutral lipids that lack a charge are stored in adipocytes as lipid droplets. Describe steps six (6) through eleven (11) of free fatty acid mobilization from triacylglycerols, starting with perilipin-bound CGI-58 and ending with fatty acid oxidation to oxygen.

ANSWER: Step 6: Perilipin-CGI-58 activates adipose triglyceride lipase (ATGL) to hydrolyze triacylglycerol, producing diacylglycerol and a free fatty acid.

Step 7: PKA also phosphorylates hormone-sensitive lipase to hydrolyze a free fatty acid from diacylglycerol.

Step 8: Monoacylglycerol lipase (MGL) hydrolyzes monoacylglycerol to a fatty acid and glycerol.

Step 9: Fatty acids leave the adipocyte and are transported in the blood bound to serum albumin.

Step 10: The fatty acids are released from albumin and enter a myocyte (muscle cell) via a specific fatty acid transporter.

Step 11: Fatty acids are oxidized to CO2 in the myocyte.

(See Figure 17.18)

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Chapter 18: Amino Acid Catabolism and Anabolism 1. Which cofactor is typically required in transamination reactions? a. pyridoxal phosphate b. coenzyme Q c. biotin d. thiamine pyrophosphate e. nicotinamide adenine dinucleotide ANSWER: a 2. Which amino acid can be considered to be both glucogenic and ketogenic? a. threonine b. aspartate c. leucine d. serine ANSWER: a 3. In which pair of compounds is the second molecule produced by the deamination of the first molecule? a. aspartic acid and oxaloacetate b. glutamine and glutamate c. alanine and pyruvate d. All of the answers are correct.

ANSWER: d 4. Which amino acid(s) can be transported into mitochondria from the cytosol in conjunction with the urea cycle? a. glutamine b. glutamate c. ornithine d. All of the answers are correct.

ANSWER: d 5. Which reaction associated with the urea cycle occurs in the cytosol? a. the synthesis of carbamoyl phosphate b. the synthesis of citrulline c. the synthesis of arginosuccinate d. transamination between glutamate and oxaloacetate to yield aspartate ANSWER: c 6. Classify the reaction type for the conversion of arginine into urea and ornithine by arginase. a. hydrolysis b. condensation c. transamination Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18: Amino Acid Catabolism and Anabolism d. proteolytic cleavage ANSWER: a 7. Which amino acid can be catabolized to either acetyl-CoA or succinyl-CoA? a. asparagine b. serine c. alanine d. isoleucine e. leucine ANSWER: d 8. Which amino acid is not catabolized to acetyl-CoA? a. tryptophan b. lysine c. tyrosine d. valine e. leucine ANSWER: d 9. Conversion of ammonia to carbamoyl phosphate in mitochondria consumes how many ATP molecules? a. none b. one c. two d. three ANSWER: c 10. Which statement is true regarding the conversion of arginine to ornithine in the urea cycle? a. The enzyme catalyzing this reaction is a hydrolase. b. This enzyme-catalyzed reaction occurs in the matrix. c. One of the products of this reaction is fumarate. d. The enzyme catalyzing this reaction is a ligase. e. The enzyme catalyzing this reaction is a hydrolase, and the reaction occurs in the matrix.

ANSWER: a 11. If 15N-labeled glutamate is introduced into liver cells, what compounds in the urea cycle will NOT be rapidly labeled with the nitrogen? a. ornithine b. aspartate c. arginine d. urea e. both ornithine and aspartate Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18: Amino Acid Catabolism and Anabolism ANSWER: a 12. What enzyme converts arginine into urea and ornithine? a. arginase b. urease c. ornithine synthase d. argininosuccinase ANSWER: a 13. Phenylketonuria is associated with disorders in the catabolism of which amino acid? a. phenylalanine b. tyrosine c. tryptophan d. histidine ANSWER: a 14. Which compound is NOT transported across the inner mitochondrial membrane as part of either the urea cycle or the aspartate-argininosuccinate shunt? a. fumarate b. aspartate c. ornithine d. citrulline ANSWER: a 15. Alanine and pyruvate are interconverted by which enzyme? a. alanine-pyruvate hydrolase b. pyruvate transferase c. pyruvate kinase d. alanine racemase e. alanine aminotransferase ANSWER: e 16. Serotonin is produced from the metabolism of which amino acid? a. valine b. serine c. cysteine d. tyrosine e. tryptophan ANSWER: e 17. Which compound can be described as glucogenic and ketogenic?

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Chapter 18: Amino Acid Catabolism and Anabolism a. leucine b. phenylalanine c. lysine d. aspartate e. glutamine ANSWER: b 18. Which amino acid is catabolized to succinyl-CoA but not acetyl-CoA? a. threonine b. isoleucine c. proline d. valine ANSWER: d 19. Which two amino acids are exclusively ketogenic? a. leucine and lysine b. isoleucine and leucine c. alanine and glutamine d. proline and leucine e. lysine and arginine ANSWER: a 20. Which statement is false regarding ketogenic and glucogenic amino acids? a. Exclusively glucogenic amino acids cannot be used to make ketone bodies. b. Ketogenic amino acids can be used to synthesize fatty acids. c. Leucine and isoleucine are both exclusively ketogenic amino acids. d. Both valine and threonine catabolism produce succinyl-CoA. ANSWER: c 21. Methionine synthase uses what substrate to synthesize methionine? a. cysteine b. S-adenosylmethionine c. ATP d. homocysteine ANSWER: d 22. Which amino acid enters the citric acid cycle as α-ketoglutarate during their catabolism? a. proline b. threonine c. aspartate Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18: Amino Acid Catabolism and Anabolism d. leucine e. tyrosine ANSWER: a 23. Which statement is true regarding lysine? a. This is a nonessential amino acid. b. Catabolism of lysine shares multiple steps with that of tryptophan. c. Carbons of catabolized lysine enter the citric acid cycle at succinyl-CoA. d. Lysine is both a glucogenic and ketogenic enzyme.

ANSWER: b 24. What defines an essential amino acid? a. It is strictly glucogenic. b. It is ketogenic and glucogenic. c. It is strictly ketogenic. d. It is synthesized by the body. e. It must be obtained in the diet.

ANSWER: e 25. Which statement is true regarding the branched-chain α-keto acid dehydrogenase complex? a. This enzyme is activated by phosphorylation. b. This enzyme catalyzes a reaction homologous to succinate dehydrogenase. c. This enzyme catalyzes an oxidative decarboxylation. d. Defects in this complex are a cause of phenylketonuria.

ANSWER: c 26. The coenzyme required for all transaminations is derived from: a. niacin. b. vitamin B6. c. riboflavin. d. thiamin. e. vitamin B12.

ANSWER: b 27. Pyridoxal phosphate participates in a variety of reactions in amino acid metabolism, EXCEPT for: a. hydrolysis. b. decarboxylation. c. racemization. d. transamination.

ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18: Amino Acid Catabolism and Anabolism 28. Which statement is NOT true of the reaction catalyzed by glutamate dehydrogenase? a. It is similar to transamination in that it involves the coenzyme pyridoxal phosphate. b. NH4+ is produced. c. The enzyme can use either NAD+ or NADP+ as a cofactor. d. The enzyme is glutamate-specific, but the reaction plays a role in the deamination of other amino acids. e. α-Ketoglutarate is produced from an amino acid.

ANSWER: a 29. Glutamate is metabolically converted to α-ketoglutarate and NH4+ by a process described as: a. deamination. b. hydrolysis. c. oxidative deamination. d. reductive deamination. e. transamination.

ANSWER: c 30. Urea synthesis in mammals takes place primarily in tissues of the: a. brain. b. kidney. c. liver. d. skeletal muscle. e. small intestine.

ANSWER: c 31. Which substance is NOT involved in the production of urea from NH4+ via the urea cycle? a. aspartate b. ATP c. carbamoyl phosphate d. malate e. ornithine ANSWER: d 32. Which substance directly donates a nitrogen atom for the formation of urea during the urea cycle? a. adenine b. aspartate c. creatine d. glutamate e. ornithine ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18: Amino Acid Catabolism and Anabolism 33. In the urea cycle, ornithine transcarbamoylase catalyzes: a. cleavage of urea to ammonia. b. formation of citrulline from ornithine and another reactant. c. formation of ornithine from citrulline and another reactant. d. formation of urea from arginine. e. transamination of arginine.

ANSWER: b 34. Which statement is false in reference to the mammalian synthesis of urea? a. Krebs was a major contributor to the elucidation of the pathway involved. b. The amino acid arginine is the immediate precursor to urea. c. The carbon atom of urea is derived from mitochondrial HCO3-. d. One of the nitrogens of urea is derived from aspartate. e. The process of urea production is an energy-yielding series of reactions.

ANSWER: e 35. Which amino acid can be directly synthesized from a citric acid cycle intermediate via transamination? a. glutamate b. serine c. threonine d. tyrosine e. proline ANSWER: a 36. Outline two paths by which 14N-labeled glutamate can lead to labeled products in the urea cycle.

ANSWER: Via either deamination by glutamate dehydrogenase with subsequent incorporation of 14NH+4 into carbamoyl phosphate or via transamination by aspartate aminotransferase to form 14N-labeled aspartate (label at α-amino group).

37. Transamination reactions are catalyzed by a family of enzymes, all of which require _____ as a coenzyme.

Prior to transamination, the coenzyme in the aldehyde form condenses with the _____ group of lysine to form a(n) _____.

ANSWER: pyridoxal phosphate (PLP); ε-amino; Schiff base (or imine or aldimine) 38. Give the name of the α-keto acid resulting when these amino acids undergo transamination with αketoglutarate: (a) glutamate, (b) aspartate, (c) alanine.

ANSWER: (a) α-ketoglutarate, (b) oxaloacetate, (c) pyruvate

39. Describe the roles of glutamine synthetase and glutaminase in the metabolism of amino groups in mammals.

ANSWER: Ammonia, which is toxic, is produced from the breakdown of nitrogen-containing molecules. The Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18: Amino Acid Catabolism and Anabolism ammonia is combined with glutamate to form glutamine in a reaction that requires ATP and is catalyzed by glutamine synthetase. Glutamine is moved from extrahepatic tissues to the liver, where the amino group is released from glutamine by glutaminase; the products are glutamate and ammonia. The ammonia delivered in this way to the liver is converted to urea and then excreted.

40. Show the reaction in which ammonia is formed from glutamate, include any required cofactors and/or intermediates.

ANSWER: This is the reaction catalyzed by glutamate dehydrogenase; either NAD+ or NADP+ act as cofactors in the first step, and water is used in the second step; see Figure 18.10.

41. During starvation, more urea production occurs. Explain this observation.

ANSWER: During starvation, cellular proteins are degraded to amino acids and their carbon skeletons are oxidized for energy. Specific and PLP-dependent transaminases catalyze the transfer of amino groups from various amino acids to α-ketoglutarate to form glutamate. Glutamate dehydrogenase removes the amino groups, which are ultimately excreted as urea.

42. Explain why a doctor might provide high concentrations of the amino acid Arg to patients who have a deficiency in one of the first three enzymes of the urea cycle.

ANSWER: The Arg is converted efficiently to urea and ornithine since the last enzyme of the urea cycle is still functioning normally. Thus, the patient produces higher than normal concentrations of ornithine, which is required for the rest of the cycle to work. That is, the flux through the pathway is increased by providing higher concentrations of the intermediates.

43. Name four amino acids that can be converted into pyruvate or a citric acid cycle intermediate, and name the intermediate formed from each.

ANSWER: (1) aspartate, oxaloacetate; (2) glutamate, α-ketoglutarate; (3) alanine, pyruvate; (4) glutamine, αketoglutarate. (Order is not important.)

44. Name one amino acid whose catabolism results in the intermediate shown: (a) pyruvate, (b) oxaloacetate, (c) α-ketoglutarate, (d) succinyl-CoA, (e) fumarate.

ANSWER: Possible answers: (a) alanine, tryptophan, glycine, serine, cysteine, threonine; (b) aspartate, asparagine; (c) glutamate, glutamine, arginine, histidine, proline; (d) isoleucine, threonine, methionine, valine; (e) phenylalanine, tyrosine. (See Fig. 18.13) 45. Degradation of amino acids yields compounds that are common intermediates in the major metabolic pathways. Explain the distinction between glucogenic and ketogenic amino acids in terms of their metabolic fates.

ANSWER: The glucogenic amino acids are those that are catabolized to intermediates that can serve as substrates for gluconeogenesis: pyruvate and any of the four- or five-carbon intermediates of the citric acid cycle. Ketogenic amino acids are catabolized to yield acetyl-CoA or acetoacetyl-CoA, the precursors for ketone body formation.

46. A laboratory report shows the presence of a high concentration of phenylalanine and its metabolites in the urine of a patient. What disease would be suspected? What defect(s) in metabolism accounts for the accumulation of phenylalanine in such patients?

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Chapter 18: Amino Acid Catabolism and Anabolism ANSWER: Phenylketonuria, the disease that results from a defect in either phenylalanine hydroxylase or the enzyme that catalyzes regeneration of tetrahydrobiopterin.

47. Diagram the degradative pathway from proline to an intermediate of either glycolysis or the citric acid cycle. Show structures of intermediates and indicate where cofactors are involved.

ANSWER: The shortest pathway is from proline to Δ1-pyrroline-5-carboxylate, γ-glutamate semialdehyde, glutamate, and α-ketoglutarate (See Figure 18.20 in reverse order). NAD+ serves as a cofactor 48. Which statement about the fixation of atmospheric nitrogen (N2) into NH3 by living cells is false? a. It involves the transfer of eight electrons per molecule of N2. b. It occurs in certain microorganisms, but not in humans. c. It requires a source of electrons, which can be supplied by ferredoxin. d. It requires one ATP per molecule of fixed N2. e. It requires two key protein components, each containing iron.

ANSWER: d 49. The enzymatic machinery to fix atmospheric N2 into NH4+ is: a. a means of producing ATP when excess N2 is available. b. composed of two key proteins, each containing iron. c. relatively stable when exposed to O2. d. specific to plant cells. e. unaffected by the supply of electrons.

ANSWER: b 50. Glutamine synthetase converts _____ to glutamine, whereas glutamate synthase converts _____ to glutamate. a. formate; ammonia b. asparagine; α-ketoglutarate c. α-ketoglutarate; oxaloacetic acid d. α-ketoglutarate; α-ketoglutarate e. glutamate; α-ketoglutarate ANSWER: e 51. Which list correctly arranges nitrogen-containing species from most to least oxidized? a. NO3- > NO2- > NH4+ > N2 b. NO3- > NO2- > N2 > NH4+ c. NO2- > NO3- > NH4+ > N2 d. N2 > NO2- > NO3- > NH4+ e. NO2- > NO3- > N2 > NH4+ ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18: Amino Acid Catabolism and Anabolism 52. Which statement is false regarding glutamine synthetase? a. It is found in all organisms. b. Many end products of pathways that use glutamine are negative effectors for this enzyme. c. It is regulated by multiple feedback inhibitors and covalent modification. d. ATP is an inhibitor of this enzyme. e. In mammals, it converts NH3 to glutamine.

ANSWER: d 53. Which substrate cannot contribute to net gluconeogenesis in mammalian liver? a. alanine b. glutamate c. palmitate d. pyruvate e. α-ketoglutarate ANSWER: c 54. The amino acid alanine can be transaminated to yield: a. fumarate. b. pyruvate. c. succinate. d. α-ketoglutarate.

ANSWER: b 55. Describe the two reactions, one catalyzed by glutamine synthetase and the other by glutamine synthase, that assimilate NH4+ into glutamate.

ANSWER: In the reaction catalyzed by glutamine synthetase, NH4+ is incorporated into glutamate to produce glutamine via a two-step reaction, starting with the activation of glutamate using ATP to produce gamma-glutamyl phosphate and ADP. Then, gamma-glutamyl phosphate is subjected to a nucleophilic attack by NH4+ to produce glutamine, Pi, and H+.

In the reaction catalyzed by glutamate synthase, which only occurs in bacteria and plants, the citric acid cycle intermediate α-ketoglutarate is reductively aminated, with glutamine acting as the amino group donor and NADPH as the electron donor. This reaction produces two glutamate molecules and NADP+.

56. Describe the path nitrogen takes from tissues, such as muscle, to the liver.

ANSWER: In the muscle, NH4+ is added to glutamate, converting it to glutamine via glutamine synthetase. In addition, alanine is produced via alanine aminotransferase when the amino group of glutamate is transferred to the excess pyruvate produced in muscle. Alanine is exported to the blood and transported to the liver, where alanine aminotransferase regenerates pyruvate and glutamate by transferring the amino group of alanine to α-ketoglutarate. Glutamine and glutamate can then be broken down to release NH4+, which subsequently enters the urea cycle.

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Chapter 18: Amino Acid Catabolism and Anabolism 57. The amino acid _____ is an intermediate in the biosynthesis of _____. a. histidine; purines b. glycine; heme c. serine; heme d. serine; sphingosine e. glutamine; glutathione ANSWER: b 58. Glutathione is a(n): a. enzyme essential in the synthesis of glutamate. b. isomer of oxidized glutamic acid. c. methyl-group donor in many biosynthetic pathways. d. product of glutamate and methionine. e. tripeptide of glycine, glutamate, and cysteine.

ANSWER: e 59. L-Dopa is an intermediate in the conversion of: a. phenylalanine to homogentisic acid. b. phenylalanine to tyrosine. c. tyrosine to epinephrine. d. tyrosine to phenylalanine. e. tyrosine to phenylpyruvate.

ANSWER: c 60. Parkinson disease is associated with an underproduction of: a. dopamine. b. ATP. c. epinephrine. d. insulin. e. bile salts.

ANSWER: a 61. Four molecules of porphobilinogen undergo a series of complex reactions to generate one molecule of heme. Which of the following reactions do NOT belong to this sequence? a. dehydration b. deamination c. transamination d. decarboxylation e. addition of iron ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18: Amino Acid Catabolism and Anabolism 62. Which of the following reactions is not an example of pyridoxal phosphate-mediated decarboxylation? a. tyrosine to dopamine b. histidine to histamine c. glutamate to γ-aminobutyrate d. alanine to pyruvate e. tryptophan to serotonin ANSWER: d 63. Show how to form δ-aminolevulinate from either glycine and succinyl-CoA or from glutamate.

ANSWER: In most animals, glycine and succinyl-CoA condense to form α-amino-β-ketoadipate, which is decarboxylated to form δ-aminolevulinate (Fig. 18.22a). Glutamate can be converted to αketoglutarate, which is decarboxylated in the citric acid cycle to form succinyl-CoA. Succinyl-CoA can then participate in the above reaction.

64. Match the signaling molecule with its amino acid precursor (a given precursor may be used more than once or not at all).

Signal Molecule Amino Acid Precursor a) epinephrine 1. histidine (b) γ-aminobutyrate 2. glutamic acid (c) histamine 3. tyrosine (d) serotonin 4. tryptophan ANSWER: (a) 3; (b) 2; (c) 1; (d) 4 65. Arrange the reactants and products according to the order in which they are used or produced to describe the complex series of reactions required for heme biosynthesis, starting with δ-aminolevulinate. There are six total reactants and products; you may use any reaction more than once.

Reactions (a) deamination (b) dehydration (c) decarboxylation (d) addition of iron (e) condensation ANSWER: e, a, b, c, c, d 66. One amino acid directly involved in the purine biosynthetic pathway is: a. alanine. b. glycine. c. glutamate. d. leucine. e. tryptophan.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18: Amino Acid Catabolism and Anabolism 67. De novo purine biosynthesis is distinguished from de novo pyrimidine biosynthesis by: a. condensation of the completed purine ring with ribose phosphate. b. incorporation of CO2. c. participation of PRPP (phosphoribosyl pyrophosphate). d. participation of aspartate.

ANSWER: b 68. The ribosyl phosphate moiety needed for the synthesis of orotidylate, inosinate, and guanylate is directly provided by: a. 5-phosphoribosyl-1-pyrophosphate. b. cytosine 5'-phosphate. c. guanosine 5'-phosphate. d. ribose 5-phosphate. e. ribulose 5-phosphate.

ANSWER: a 69. The synthesis of purine and pyrimidine nucleotides differ in that: a. ATP is required in the synthesis of purines but not in the synthesis of pyrimidines. b. Purine biosynthesis starts with the formation of PRPP, whereas pyrimidines incorporate PRPP near the end of the pathway. c. Purine formation requires a tetrahydrofolate (THF) derivative, whereas pyrimidine formation does not. d. Pyrimidine biosynthesis is tightly regulated in the cell, whereas purine biosynthesis is not. e. Pyrimidines go through many steps, adding a single carbon or nitrogen each time, whereas the basic skeleton for purines is formed by two main precursors.

ANSWER: b 70. Precursors for the biosynthesis of the pyrimidine ring system include: a. carbamoyl phosphate and aspartate. b. glutamate, NH3, and CO2. c. glycine and succinyl-CoA. d. glycine, glutamine, CO2, and aspartate. e. inosine and aspartate.

ANSWER: a 71. The MOST direct precursors of the nitrogens of UMP are: a. aspartate and carbamoyl phosphate. b. glutamate and aspartate. c. glutamate and carbamoyl phosphate. d. glutamine and aspartate. e. glutamine and carbamoyl phosphate.

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Chapter 18: Amino Acid Catabolism and Anabolism ANSWER: a 72. CMP, UMP, and TMP all have _____ as a common precursor. a. adenosine b. aspartate c. glutamine d. inosine e. S-adenosylmethionine ANSWER: b 73. Which statement is NOT true of the reaction catalyzed by ribonucleotide reductase? a. Glutathione is part of the path of electron transfer. b. It acts on nucleoside diphosphates. c. Its mechanism involves formation of a free radical. d. There is a separate enzyme for each nucleotide (ADP, CDP, GDP, UDP). e. Thioredoxin acts as an essential electron carrier.

ANSWER: d 74. Which statement correctly describes the biosynthetic pathway for purine nucleotides? a. Purine deoxynucleotides are made via the same path as ribonucleotides, followed by reduction of the ribose moiety. b. The first enzyme in the path is aspartate transcarbamoylase (ATCase). c. The nitrogen in the purine base that is bonded to ribose in the nucleotide is derived originally from glycine. d. The pathway occurs only in plants and bacteria, not in animals. e. The purine rings are first synthesized, then condensed with ribose phosphate.

ANSWER: a 75. Which molecule does NOT require 5-phosphoribosyl-1-pyrophosphate (PRPP) as a substrate for biosynthesis? a. adenosine monophosphate b. cytidine monophosphate c. guanosine monophosphate d. uridine monophosphate e. All of these molecules require PRPP for biosynthetic pathways.

ANSWER: e 76. In purine biosynthesis, which molecule contributes two of the ring nitrogens? a. urea b. formate c. aspartate Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18: Amino Acid Catabolism and Anabolism d. glycine e. glutamine ANSWER: e 77. Which is the immediate precursor of thymidylate (dTMP)? a. dGMP b. dCTP c. UTP d. dATP e. dUMP ANSWER: e 78. Which statement is true regarding the conversion of UTP into CTP by cytidylate synthetase? a. Glutamate is also a substrate for this reaction. b. This reaction consumes one phosphoanhydride bond. c. This enzyme is an isomerase. d. Ammonia is a substrate for the overall reaction. e. Folate is a required cofactor for the reaction.

ANSWER: b 79. Draw the structure of 5'-IMP. Indicate with arrows those carbon atoms donated with the aid of tetrahydrofolate and circle the atoms derived from glycine.

ANSWER: The structure of 5'-IMP with the origin of each atom is shown in Figure 18.27. Incorporation of the atoms derived from folate requires tetrahydrofolate as a cofactor.

80. Draw the structure of 5'-UMP (uridylic acid). Circle the carbon atoms donated by atoms derived from aspartate.

ANSWER: The structure of UMP is shown in Figure 18.26. The atoms contributed to UMP by aspartate in the first step of the pathway include all carbons in the nucleotide ring except the attached amide carbon in N-carbamoylaspartate.

81. Carbamoyl phosphate synthetase II is activated by ATP and PRPP but inhibited by UMP. How does this make metabolic sense? How does it compare to the regulation of ATCase?

ANSWER: ATP is a signal that energy levels are high, and synthesis of nucleotides increases when adequate energy is available to ensure comparable amounts of different nucleotides. UMP is one of the end products of the pyrimidine synthesis pathways, in which carbamoyl phosphate is an initial substrate (in the ATCase reaction), so UMP is acting as a feedback inhibitor. This is analogous to ATCase regulation, where ATP (sufficient energy) is an activator and CTP (the final product) acts as an inhibitor.

82. Cytidylate synthetase (CTP synthetase) is allosterically regulated by both ATP and UTP. Predict how they will affect the activity of this enzyme.

ANSWER: CTP synthetase catalyzes the ATP-dependent conversion of UTP into CTP. As substrates, both ATP Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18: Amino Acid Catabolism and Anabolism and UTP likely act as activators for the enzyme, maintaining a balance of the two pyrimidine nucleotides.

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Chapter 19: Electron Transfer and Oxidative Phosphorylation 1. Which statement regarding oxidative phosphorylation is true? a. The rate of electron transfer is independent of the magnitude of the proton electrochemical gradient. b. ATP synthase utilizes oxidized cofactors. c. Proton transfer through the Fo subunit of ATP synthase is required in order for ATP synthesis to occur. d. The addition of an uncoupler decreases the rate of reoxidation of NADH and FADH2, therefore decreases the rate of the citric acid cycle.

ANSWER: c 2. The consumption of acetyl-CoA by the citric acid cycle is decreased under which condition? a. high concentration of NAD+ b. high concentration of oxaloacetate c. NADH being rapidly reoxidized via the electron-transfer chain d. the mitochondrial ATP synthase being inactive e. high concentrations of both NAD+ and oxaloacetate ANSWER: d 3. The electron-transfer chain and ATP synthase are effectively coupled. Which statement BEST describes how this occurs? a. The electron-transfer chain reoxidizes reduced cofactors. b. The electron-transfer chain generates an electrochemical gradient that provides energy for the production of ATP. c. ATP synthase utilizes oxidized cofactors. d. Through a series of conformational changes, ATP synthase generates ATP. ANSWER: b 4. Which statement is NOT true of electron transfer? a. The mobile electron carrier coenzyme A is located in the inner mitochondrial membrane. b. Electrons flow from redox centers with a lower reduction potential to redox centers with a higher reduction potential. c. The electron-transfer chain oxidizes reduced cofactors. d. Cytochrome c has a higher reduction potential than FADH2.

ANSWER: a 5. During oxidative phosphorylation, the proton-motive force (electrochemical gradient) that is generated by electron transfer is used to: a. create a pore in the inner mitochondrial membrane. b. generate the substrates (ADP and Pi) for ATP synthase. c. provide free energy needed for ATP synthesis. d. reduce NAD+ to NADH. e. reduce O2 to H2O. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19: Electron Transfer and Oxidative Phosphorylation ANSWER: c 6. Coenzyme Q is a prosthetic group for which complex in the electron-transfer chain? a. Complex I b. Complex II c. Complex III d. None of the above; Q is in the lipid bilayer of the inner mitochondrial membrane.

ANSWER: d 7. Which statement does NOT describe a difference between the inner and outer mitochondrial membranes? a. The outer mitochondrial membrane contains porins, whereas the inner mitochondrial membrane does not. b. The outer mitochondrial membrane is permeable to large proteins (Mr > 10,000), whereas the inner mitochondrial membrane is not. c. A pH gradient exists across the inner mitochondrial membrane but not the outer mitochondrial membrane. d. The inner mitochondrial membrane contains a smaller percentage of lipid relative to protein compared with the outer mitochondrial membrane. e. The outer mitochondrial membrane has a smaller surface area than the inner mitochondrial membrane.

ANSWER: b 8. What factor makes the outer mitochondrial membrane permeable to protons? a. proton symports b. proton antiports c. porins d. the presence of proton ionophores e. the presence of negatively charged lipids ANSWER: c 9. Complex III contains which cofactor as a prosthetic group? a. iron-sulfur clusters (centers) b. FAD c. FMN d. coenzyme Q e. copper ANSWER: a 10. Which component of the electron-transfer chain is NOT an integral membrane protein? a. NADH dehydrogenase b. cytochrome c c. cytochrome oxidase Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19: Electron Transfer and Oxidative Phosphorylation d. ubiquinone:cytochrome c oxidoreductase e. succinate dehydrogenase ANSWER: b 11. If a drug that specifically prevented the interaction of cytochrome c with other proteins was added to respiring mitochondria in a test tube, what effects would be observed? a. ATP synthesis would immediately stop. b. Oxygen consumption would increase. c. Reduced cofactors (NADH/FADH2) would accumulate. d. Coenzyme Q would become oxidized. e. Proton export from the matrix would increase.

ANSWER: c 12. How many cytochrome c need to be reoxidized for every oxygen molecule (O2) converted to water in the electron-transfer chain? a. one b. two c. three d. four e. eight ANSWER: d 13. What compound is a product of the reaction involving the superoxide free radical catalyzed by superoxide dismutase? a. water b. hydrogen peroxide c. reduced glutathione d. oxidized glutathione e. NADPH ANSWER: b 14. Which statement is true regarding the reduction potential of mitochondrial glycerol 3-phosphate dehydrogenase? a. It will be higher than the reduction potentials of both NADH and Complex III. b. It will be lower than the reduction potentials of both NADH and Complex III. c. It will be higher than that of NADH but lower than that of Complex III. d. It will be lower than that of NADH but higher than that of Complex III. ANSWER: c 15. Which statement is false regarding 2,4-dinitrophenol? a. It is a proton carrier.

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Chapter 19: Electron Transfer and Oxidative Phosphorylation b. It may be either neutral or negatively charged. c. It is soluble in the lipid core of membranes. d. It requires a specific transporter to enter mitochondria. e. It decreases ATP synthesis while increasing oxygen consumption.

ANSWER: d 16. Which statement is true regarding the gamma (γ) subunit of ATP synthase? a. It contains mostly α-helical regular secondary structure. b. It exists as a dimer in the F1 portion of ATP synthase. c. It prevents the F1 portion from rotating relative to the Fo portion. d. It interacts with the a and b subunits of the Fo portion. e. It binds protons as part of the proton-transfer process.

ANSWER: a 17. Which statement explains why the [ATP]/[ADP][Pi] ratio is relatively stable in a cell? a. [ATP] is always relatively low. b. ATP consumption is balanced by ATP synthase activity. c. ATP is used to regulate protein kinases. d. ATPases ensure that ATP concentration does not become too large. e. Energy consumption is minimized to ensure ATP is not consumed.

ANSWER: b 18. Which statement does NOT describe characteristics of the protein inhibitor IF1? a. This protein is relatively small (< 100 amino acids). b. This protein exists as a monomer at pH 7.0. c. This protein will bind to the Fo portion of ATP synthase. d. This protein functions as a dimer when it is inhibiting ATP synthesis. e. This protein inhibits ATP hydrolysis by ATP synthase.

ANSWER: c 19. What will NOT occur when 2,4-dinitrophenol (an uncoupler of oxidative phosphorylation) is added to actively respiring mitochondria? a. The rate of ATP synthesis becomes more efficient. b. The rate of NADH reoxidation at Complex I will increase. c. The rate of oxidation of intermediates in the citric acid cycle will increase. d. Proton export from the matrix will occur. e. Heat will be released.

ANSWER: a 20. A new compound isolated from mitochondria is claimed to represent a previously unrecognized carrier in Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19: Electron Transfer and Oxidative Phosphorylation the electron-transfer chain. It is given the name coenzyme Z. Which line of evidence is the LEAST conclusive in assigning this compound a position in the electron-transfer chain? a. Alternate oxidation and reduction of the mitochondrion-bound coenzyme Z can be readily demonstrated. b. Removal of coenzyme Z from the mitochondria results in a decreased rate of oxygen consumption. c. The rate of oxidation and reduction of mitochondrion-bound coenzyme is of the same order of magnitude as the overall rate of electron transfer in mitochondria as measured by oxygen consumption. d. The reduction potential of Z is between that of two compounds known to participate in the electrontransfer chain. e. When added to a mitochondrial suspension, coenzyme Z is taken up very rapidly and specifically by the mitochondria.

ANSWER: e 21. Reduced QH2 is NOT formed by: a. Complex I and NADH. b. Complex II and succinate. c. Complex III and cytochrome c. d. fatty acid oxidation. e. oxidation of glycerol 3-phosphate.

ANSWER: c 22. In the reoxidation of QH2 by purified ubiquinone:cytochrome c oxidoreductase (Complex III) from heart muscle, the overall stoichiometry of the reaction requires two molecules of cytochrome c per molecule of QH2 because: a. cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor. b. cytochrome c is a two-electron acceptor, whereas QH2 is a one-electron donor. c. cytochrome c is water soluble and operates between the inner and outer mitochondrial membranes. d. heart muscle has a high rate of oxidative metabolism, and therefore requires twice as much cytochrome c as QH2 for electron transfer to proceed normally. e. two molecules of cytochrome c must first combine physically before they are catalytically active.

ANSWER: a 23. Which statement is NOT a feature of Complex IV? a. Cytochrome c is a one-electron donor. b. Oxygen is a substrate. c. Copper is an essential metal for the reaction. d. For every electron passed to Complex IV, two protons are consumed from the matrix (n side). e. In order to generate two water molecules, Complex IV must go through the catalytic cycle two times.

ANSWER: e 24. Which statement is NOT true of the proton-motive force?

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Chapter 19: Electron Transfer and Oxidative Phosphorylation a. One component of the proton-motive force is the chemical gradient of protons. b. One component of the proton-motive force is the charge gradient of protons. c. Generation of the proton-motive force in mitochondria requires succinate. d. The proton-motive force is generated by the electron-transfer chain in mitochondria. e. The proton-motive force drives ATP synthesis in mitochondria.

ANSWER: c 25. Cyanide, oligomycin, and 2,4-dinitrophenol (DNP) are inhibitors of mitochondrial aerobic phosphorylation.

Which statement correctly describes the mode of action of the three inhibitors? a. Cyanide and DNP inhibit electron transfer, and oligomycin inhibits the synthesis of ATP. b. Cyanide inhibits electron transfer, whereas oligomycin and DNP inhibit the synthesis of ATP. c. Cyanide, oligomycin, and DNP compete with O2 for cytochrome oxidase (Complex IV). d. Oligomycin and cyanide inhibit synthesis of ATP; DNP inhibits electron transfer. e. Oligomycin inhibits electron transfer, whereas cyanide and DNP prevent the synthesis of ATP. ANSWER: b 26. In normal mitochondria, the rate of NADH consumption (oxidation) will: a. be increased in active muscle, decreased in inactive muscle. b. be very low if the ATP synthase is inhibited, but increase when an uncoupler is added. c. decrease if mitochondrial ADP is depleted. d. decrease when cyanide is used to prevent electron transfer. e. All of the answers are correct.

ANSWER: e 27. Which statement about the chemiosmotic theory is correct? a. Electron transfer in mitochondria is accompanied by an asymmetric release of protons on one side of the inner mitochondrial membrane. b. It predicts that oxidative phosphorylation can occur, even in the absence of an intact inner mitochondrial membrane. c. The effect of uncoupling reagents is a consequence of their ability to carry electrons through membranes. d. The membrane ATP synthase has no significant role in the chemiosmotic theory.

ANSWER: a 28. Upon the addition of 2,4-dinitrophenol to a suspension of mitochondria carrying out oxidative phosphorylation linked to the oxidation of malate, what does NOT occur? a. Oxygen consumption decreases. b. Oxygen consumption increases. c. The ability of ATP synthase to efficiently produce ATP drops. d. The proton gradient dissipates. e. The rate of transport of electrons from NADH to O2 becomes maximal.

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Chapter 19: Electron Transfer and Oxidative Phosphorylation ANSWER: a 29. Uncoupling of mitochondrial oxidative phosphorylation: a. allows continued mitochondrial ATP formation but halts O2 consumption. b. halts all mitochondrial metabolism. c. halts mitochondrial ATP formation but allows continued O2 consumption. d. slows down the citric acid cycle. e. slows the conversion of glucose to pyruvate by glycolysis.

ANSWER: c 30. 2,4-Dinitrophenol (DNP) and oligomycin inhibit mitochondrial oxidative phosphorylation. DNP is an uncoupling agent; oligomycin blocks the ATP synthesis reaction itself. Therefore, DNP will: a. allow electron transfer in the presence of oligomycin. b. allow oxidative phosphorylation in the presence of oligomycin. c. block electron transfer in the presence of oligomycin. d. diminish O2 consumption in the presence of oligomycin.

ANSWER: a 31. Which statement about energy conservation in the mitochondrion is false? a. Drugs that inhibit the ATP synthase will also inhibit the flow of electrons down the chain of carriers. b. For oxidative phosphorylation to occur, it is essential to have a closed membranous structure with an inside and an outside. c. The yield of ATP per molecule of oxidizable substrate depends on the substrate. d. Uncouplers (such as dinitrophenol) have exactly the same effect on electron transfer as inhibitors such as cyanide; both block further electron transfer to oxygen. e. Uncouplers "short circuit" the proton gradient, thereby dissipating the proton-motive force as heat.

ANSWER: d 32. The relative concentrations of ATP and ADP control the cellular rates of: a. glycolysis. b. oxidative phosphorylation. c. pyruvate oxidation. d. the citric acid cycle. e. All of the answers are correct.

ANSWER: e 33. The rate of oxidative phosphorylation in mitochondria is controlled primarily by: a. feedback inhibition by CO2. b. the availability of NADH from the TCA cycle. c. the concentration of citrate or the glycerol 3-phosphate shuttle. d. the mass-action ratio of the ATP-ADP system.

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Chapter 19: Electron Transfer and Oxidative Phosphorylation e. the presence of uncoupling protein 1.

ANSWER: d 34. Which factor is NOT controlled by hypoxia-inducible factor? a. glucose transport b. glycolysis c. citric acid cycle d. Complex I of the respiratory chain e. Complex IV of the respiratory chain ANSWER: d 35. Mammals produce heat by using which endogenous uncoupling agent? a. the small molecule 2,4-dinitrophenol synthesized by the cell b. uncoupling protein 1 c. the protein thioredoxin d. the protein cytochrome c e. a modified form of the FoF1 ATPase ANSWER: b 36. Which statement about human mitochondria is true? a. Most mitochondrial proteins are encoded by nuclear genes. b. Mitochondrial genes are inherited from both maternal and paternal sources. c. rRNA and tRNA are imported from the cytoplasm and used in mitochondrial protein synthesis. d. The mitochondrial genome codes for all proteins found in mitochondria. e. The mitochondrial genome is not subject to mutations.

ANSWER: a 37. As readers read and answer this question, they are (presumably) consuming oxygen. What single reaction accounts for most of their oxygen consumption?

ANSWER: O2 is converted to H2O by electrons from the respiratory chain. The final step is the one catalyzed by cytochrome oxidase (Complex IV).

38. Diagram the path of electron flow from NADH to the final electron acceptor during electron transfer in mitochondria. For each electron carrier, indicate whether only electrons, or both electrons and protons, are accepted/donated by that carrier. Also, indicate where electrons from succinate oxidation enter the chain of carriers.

ANSWER: NADH (both) → Complex I → QH2 (both) → Complex III → Cyt c (only e–) → Complex IV → O2 (both) (See Fig. 19.7) Electrons from succinate enter at Complex II. 39. Compound X is an inhibitor of mitochondrial ATP synthesis. It was observed that when compound X was Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19: Electron Transfer and Oxidative Phosphorylation added to cells, the NAD+/NADH ratio decreased. Would X be expected to be an uncoupling agent or an inhibitor of respiratory electron transfer? Explain.

ANSWER: It is an inhibitor of electron transfer; its addition lowers the NAD+/NADH ratio because NADH produced by oxidative reactions in mitochondria can no longer be reoxidized by electron flow to O2.

If it had been an uncoupling agent, the NADH could be reoxidized, the ratio of NAD+/NADH would be unchanged, but there would be no buildup of a proton gradient (i.e., no synthesis of ATP).

40. Cyanide ion (CN–) blocks electron transfer in mitochondria involving cytochrome. 2,4-Dinitrophenol (DNP) is an uncoupler of mitochondrial oxidative phosphorylation. For each inhibitor, draw a graph of ATP synthesized vs. time and O2 consumed vs. time with labeled curves to show the effects of adding each of these compounds separately to a suspension of mitochondria supplied with O2, succinate, ADP, and Pi. Use an arrow to indicate the time of drug addition.

ANSWER:

41. Give an example of (a) an uncoupler of oxidative phosphorylation, and (b) an inhibitor of respiration. (c) Describe the difference in the effects of such uncouplers and inhibitors on mitochondrial function.

ANSWER: (a) Uncouplers include DNP, valinomycin, and FCCP. (b) Respiration inhibitors include antimycin A, piericidin A, CN–, rotenone, and amytal. (c) Uncouplers stop formation of ATP while allowing electron transfer to continue. Inhibitors of respiration block both electron transfer and phosphorylation. (See Table 19-4.)

42. Mitochondria carrying out oxidative phosphorylation consume oxygen. Explain what happens to this oxygen, and describe the effect of an uncoupling agent such as 2,4-dinitrophenol (DNP) on the rate of oxygen consumption. Assume there is a sufficient supply of oxidizable substrate, ADP, and Pi. ANSWER: O2 is reduced to H2O by electrons passed through the respiratory chain. Addition of DNP, which uncouples phosphorylation (ATP synthesis) from electron flow, actually stimulates O2 consumption slightly by removing the "drag" of ATP synthesis.

43. The skunk cabbage (Symplocarpus foetidus) can maintain a temperature of 10 to 25 ºC higher than the temperature of the surrounding air. Suggest a mechanism for this.

ANSWER: One possible heat-generating mechanism is an uncoupled mitochondria in which some of the energy of electron flow is dissipated as heat. This is most easily achieved in brown adipocytes through uncoupling protein 1 (UCP1), which bypasses some of the mitochondrial complexes that normally Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19: Electron Transfer and Oxidative Phosphorylation capture favorable electron flow toward making a proton gradient that then drives ATP synthesis.

44. In his studies of alcoholic fermentation by yeast, Louis Pasteur noted that the sudden addition of oxygen (O2) to a previously anaerobic culture of fermenting grape juice resulted in a dramatic decrease in the rate of glucose consumption. This Pasteur effect can be counteracted by the addition of 2,4-dinitrophenol (DNP), an uncoupler of oxidative phosphorylation. (a) Why would the yeast cells consume less glucose in the presence of oxygen? Estimate how much less glucose they would use. (b) Why would DNP counteract or prevent the Pasteur effect?

ANSWER: (a) During aerobic oxidation of glucose, the yield of ATP is about 38 molecules per molecule of glucose. Under anaerobic conditions, only 2 ATP are produced per glucose. Thus, to get the same amount of ATP, a cell needs to use 19 times more glucose anaerobically. (b) Because DNP blocks the principal path of aerobic ATP production, aerobic cells in the presence of DNP need about the same amount of glucose as anaerobic cells.

45. Describe the role of hypoxia-inducible factor (HIF-1) in reducing reactive oxygen species (ROS).

ANSWER: Under conditions of low oxygen, the electron-transfer chain may accidentally generate ROS because of the low levels of the final electron acceptor (oxygen). Therefore, HIF-1 is induced and serves as a transcription factor to increase glycolysis (i.e., anaerobic consumption of glucose). Some of the proteins whose transcription are turned on to yield this response are the glucose transporter, glycolytic enzymes, and lactate dehydrogenase. Additionally, the activity of pyruvate dehydrogenase is reduced (via phosphorylation) to slow the citric acid cycle and the flux of electrons into the electron-transfer chain.

46. Mutations in mitochondrial genes frequently produce diseases that affect the brain and skeletal muscle (mitochondrial encephalomyopathies). Why are these two tissues particularly sensitive to mitochondrial mutations?

ANSWER: Both the brain and skeletal muscles are heavily dependent on an abundant supply of ATP for normal functioning.

47. Which of the following cellular processes is energetically unfavorable? a. Transfer of electrons across the electron transport chain. b. Formation of superoxide. c. Movement of small molecules and ions across the outer mitochondrial membrane. d. Movement of protons from the mitochondrial matrix to the intermembrane space. e. Diffusion of CO2 out of the mitochondria.

ANSWER: d 48. Describe the electron transport chain's three complexes that lead to the transfer of protons from the matrix to the intermembrane space. Include the number of protons transferred in each complex and the electron carrier involved.

ANSWER: The first complex in the electron transport chain transfers four protons from the matrix to the intermembrane space, which occurs at complex I, a proton pump driven by the energy from the electron transfer originating from NADH. The second complex (complex III) also transfers four protons to the intermembrane space, two from Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19: Electron Transfer and Oxidative Phosphorylation QH2 and another two from the mitochondrial matrix, driven by the energy from electron transfer originating from QH2.

Finally, the third complex (complex IV) uses electron transfer from four Cyt c to pump protons into the intermembrane space. Complex IV moves four protons for every one molecule of O2 reduced.

49. Describe the roles of the enzymes glutathione reductase, glutathione peroxidase, and superoxide dismutase in producing or neutralizing reactive oxygen species.

ANSWER: Superoxides come from the unwanted transfer of electrons from ubiquinol to oxygen in the electron transport chain. These superoxides can spontaneously form hydroxyl free radicals or convert to hydrogen peroxide mediated by the enzyme superoxide dismutase. Hydrogen peroxide can harm the cell when overproduced, so it is converted to a water molecule via glutathione peroxidase and two glutathione molecules. The conversion of hydrogen peroxide to water simultaneously removes electrons from glutathione, covalently bonding the two molecules together to form glutathione disulfide (GSSG). GSSG is reduced by glutathione reductase in an NADPH-driven reaction to recycle glutathione and convert more hydrogen peroxide to water. (See Fig. 19.15) 50. Describe the three principles of mitochondrial energy production.

ANSWER: The first principle states that high-energy electrons enter the electron transport chain in a series of redox reactions that ultimately lead to oxygen. The second principle states that the energetically favorable flow of electrons across the electron transport chain is coupled to an energetically unfavorable transport of protons against its concentration gradient. Finally, the third principle states that the energetically favorable flow of protons down their concentration gradient provides the free energy needed for ATP synthesis.

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Chapter 20: Metabolism and Biosignaling 1. One distinction between peptide and steroid hormones is that peptide hormones: a. act through nonspecific receptors, whereas steroid hormones act through specific receptors. b. are generally water insoluble, whereas steroid hormones are water soluble. c. are more stable than steroid hormones. d. bind to cell surface receptors, whereas steroid hormones often bind to nuclear receptors. e. bind to their receptors with high affinity, whereas steroid hormones bind with low affinity.

ANSWER: d 2. Insulin is an example of a(n) _____ hormone. a. catecholamine b. eicosanoid c. paracrine d. peptide e. steroid ANSWER: d 3. The maturation of insulin from its precursor (preproinsulin) involves: a. acetylation. b. oxidation. c. phosphorylation. d. proteolysis. e. reduction.

ANSWER: d 4. Epinephrine is an example of a(n) _____ hormone. a. catecholamine b. eicosanoid c. paracrine d. peptide e. steroid ANSWER: a 5. An example of a steroid hormone is: a. epinephrine. b. retinoic acid. c. testosterone. d. thromboxane. e. thyroxine.

ANSWER: c 6. _____ hormones are released into the blood and carried to target cells throughout the body, while _____ Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20: Metabolism and Biosignaling hormones are released into the extracellular space and diffuse to neighboring target cells. a. Endocrine; paracrine b. Paracrine; endocrine c. Peptide; steroid d. Steroid; peptide e. Autocrine; paracrine ANSWER: a 7. Which statement about metabolism in the mammalian liver is false? a. Most plasma lipoproteins are synthesized in the liver. b. The enzymatic complement of liver tissue changes in response to changes in the diet. c. The liver synthesizes most of the urea produced in the body. d. The presence of glucose 6-phosphatase makes the liver one of the few organs able to release glucose into the bloodstream. e. Under certain conditions, most of the functions of the liver can be performed by other organs.

ANSWER: e 8. Which statement does NOT describe the fate of amino acids in the liver? a. Amino acids are incorporated into new proteins. b. Nitrogen equivalents from amino acids are converted to urea. c. Carbon equivalents from amino acids are converted to glucose or citric acid cycle intermediates. d. Amino acids are stored for later use. e. Amino acids serve as precursors to nucleotide biosynthesis.

ANSWER: d 9. Which statement does NOT describe the fate of fatty acids in the liver? a. Fatty acids are oxidized to generate ATP. b. Fatty acids are converted to ketone bodies. c. Fatty acids can be converted to cholesterol. d. Fatty acids are stored for later use. e. Fatty acids are converted to glucose.

ANSWER: e 10. In skeletal muscle, phosphocreatine functions as a(n): a. reservoir of Pi for mitochondria. b. reservoir of high-energy phosphate to replenish ATP. c. reservoir of amino acids for protein synthesis. d. electron acceptor under anaerobic conditions. e. None of the answers is correct.

ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20: Metabolism and Biosignaling 11. The Cori cycle is the: a. conversion of lactate to pyruvate in skeletal muscle to drive glycogen synthesis. b. interconversion between glycogen and glucose 1-phosphate. c. production of lactate from glucose in peripheral tissues, and the resynthesis of glucose from lactate in liver. d. synthesis of alanine from pyruvate in skeletal muscle and the synthesis of pyruvate from alanine in the liver. e. synthesis of urea in liver and degradation of urea to carbon dioxide and ammonia by bacteria in the gut.

ANSWER: c 12. Which statement is true? a. The brain prefers glucose as an energy source but can use ketone bodies. b. Muscle cannot use fatty acids as an energy source. c. In a well-fed human, about equal amounts of energy are stored as glycogen and as triacylglycerol. d. Fatty acids cannot be used as an energy source in humans because humans lack the enzymes of the glyoxylate cycle. e. Amino acids are a preferable energy source over fatty acids.

ANSWER: a 13. Which factor is NOT involved in triggering insulin release from β cells in response to glucose? a. reduced K+ efflux through ATP-gated K+ channels b. increased concentrations of ATP c. cAMP-dependent phosphorylation of lipase d. phosphorylation of glucose to glucose 6-phosphate e. increased Ca2+ levels ANSWER: c 14. The largest energy store in a well-nourished human is: a. ATP in all tissues. b. blood glucose. c. liver glycogen. d. muscle glycogen. e. triacylglycerols in adipose tissue.

ANSWER: e 15. Long-term maintenance of body weight is regulated by the hormone: a. adiposin. b. hypothalmin. c. leptin. d. obesin.

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Chapter 20: Metabolism and Biosignaling e. testosterone.

ANSWER: c 16. The hormone leptin _____ appetite, and the hormone insulin _____ appetite. a. increases; increases b. increases; decreases c. decreases; increases d. decreases; decreases e. has no effect on; has no effect on ANSWER: d 17. The peptide hormone adiponectin, produced in adipose tissue, circulates in the blood and: a. enhances fatty acid synthesis in liver cells. b. increases the rate of β oxidation of fatty acids in muscle cells. c. inhibits glucose uptake and catabolism in muscle and liver cells. d. reduces the transport of fatty acids into muscle cells. e. stimulates gluconeogenesis in liver cells.

ANSWER: b 18. Under conditions of prolonged starvation, which fuel is MOST catabolized by the brain in humans? a. ketone bodies b. glucose c. amino acids d. odd-chain fatty acids e. glycerol ANSWER: a 19. Which statement is true regarding the late stages of starvation? a. The brain has decreased its need for glucose by switching to lactate as a major energy source. b. High levels of acetyl-CoA derived from lipid catabolism lead to the production of acetoacetate. c. Ketone bodies are produced from amino acids liberated from noncritical proteins. d. Gluconeogenesis in the liver is reduced because tissues have begun to rely on ketones for fuel. e. Catabolism of triacylglycerols provides the major source of a gluconeogenic metabolite in the form of glycerol.

ANSWER: b 20. Which statement is true regarding the islet cells of the pancreas? a. α cells secrete glucagon to stimulate glycogenesis in skeletal muscle. b. α cells secrete glucagon to stimulate glycolysis in the liver. c. β cells secrete glucagon to stimulate glycogenolysis in the liver. d. β cells secrete insulin to stimulate glucose uptake by cells, including in skeletal muscle, the liver, Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20: Metabolism and Biosignaling and adipose tissue. e. α cells secrete insulin to stimulate glycolysis in skeletal muscle.

ANSWER: d 21. In an attempt to control type 2 diabetes, some people follow a high-fat, low-carbohydrate diet. Which substance(s) would NOT have an increased concentration in the blood after two weeks on this diet? a. fatty acids b. glucagon c. acetone d. glucose e. amino acids ANSWER: d 22. Which statement is true regarding diabetes mellitus? a. Diabetes mellitus is characterized by the abnormal depletion of blood glucose by the liver. b. Diabetes mellitus is characterized by reduced levels of gluconeogenesis in the liver. c. Diabetes mellitus is characterized by the destruction of glucagon-secreting cells of the pancreas. d. Diabetes mellitus is characterized by the secretion of glucagon in the urine leading to "sweet urine." e. Diabetes mellitus is characterized by ketonemia, metabolic acidosis, and ketonuria.

ANSWER: e 23. Which hormone triggers the release of α-melanocyte-stimulating hormone from anorexigenic neurons? a. leptin b. neuropeptide Y c. ghrelin d. both leptin and neuropeptide Y e. both leptin and ghrelin ANSWER: a 24. Which hormone acts on orexigenic neurons in the arcuate nucleus to produce hunger before a meal? a. leptin b. insulin c. ghrelin d. cortisol e. PPARδ ANSWER: c 25. Which is closest to a normal blood glucose range? a. 40–65 mg/100 mL whole blood b. 50–150 mg/100 mL whole blood c. 100–120 mg/100 mL whole blood Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20: Metabolism and Biosignaling d. 120–125 mg/100 mL whole blood e. 60–90 mg/100 mL whole blood ANSWER: e 26. Which statement is true regarding the protein leptin? a. Leptin is a hormone made primarily by hepatocytes that functions to regulate energy balance by curtailing appetite. b. Leptin is an adipokine released when adipose cells are well-filled with triacylglycerols. c. Leptin is involved in regulating the sympathetic nervous system, resulting in a decrease in blood pressure, heart rate, and thermogenesis. d. The amount of leptin released by adipose tissue is independent of the number and size of adipocytes; rather, it is strictly correlated with ingestion of food.

ANSWER: b 27. Which statement is false regarding the protein ghrelin? a. Levels of ghrelin are lowest in the blood just before a meal and highest just after a meal. b. Ghrelin is a peptide hormone produced by endocrine cells lining the gastrointestinal tract. c. The receptors for ghrelin are found in the same location as the receptors for leptin. d. Regulation of gastric ghrelin secretion may be influenced by specific microbes in the gut. e. Ghrelin stimulates the release of orexigenic peptides from arcuate neurons in the hypothalamus.

ANSWER: a 28. Name four general classes of hormones and give an example of each.

ANSWER: (1) Peptide (e.g., insulin, glucagon); (2) amine (e.g., norepinephrine); (3) steroid (e.g., testosterone, progesterone); (4) lipid (e.g., prostaglandins). (See Table 20.1.)

29. Which class(es) of hormone acts via nuclear receptors? Give an example of this type of hormone and briefly describe its mode of action.

ANSWER: Steroid hormones typically act via nuclear receptors. Examples are the sex hormones testosterone and progesterone. Since steroids are nonpolar molecules, they pass through the plasma membrane and interact with receptor proteins in the nucleus. The hormone-receptor complex interacts with DNA and alters the expression of specific genes.

30. Briefly explain the role of phosphocreatine in muscle.

ANSWER: During maximal muscle activity, muscle cells are in danger of depleting their ATP levels. Oxygen cannot be provided rapidly enough to support the electron-transfer chain, fermentation is not nearly as efficient as aerobic metabolism, and the decrease in pH caused by lactate accumulation is detrimental to the efficiency of the muscles. Phosphocreatine provides an additional source of energy that can convert ADP back to ATP via the creatine kinase reaction to help maintain muscle activity.

31. Explain the signaling pathway by which an increase in blood glucose causes the release of insulin from pancreatic β cells.

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Chapter 20: Metabolism and Biosignaling ANSWER: Glucose in the blood is transported into pancreatic β cells via the GLUT2 transporter, where it is phosphorylated and then metabolized via glycolysis and the citric acid cycle, ultimately leading to higher ATP concentrations via oxidative phosphorylation. Higher ATP levels close ATP-gated K+ channels, depolarizing the membrane, which leads to the opening of voltage-gated Ca2+ channels, an increase in the intracellular Ca2+ concentration, and the subsequent release of insulin via exocytosis.

32. Explain the response and the consequences of prolonged fasting on fuel metabolism by the liver.

ANSWER: Under conditions of prolonged fasting (i.e., absence of glucose in the diet), the liver is responsible for providing energy, primarily to the brain. Fatty acids from adipocytes are broken down and converted to ketone bodies, which can be used as an alternative fuel source by the brain. Proteins from muscle cells are broken down to provide an alternative route to glucose via citric acid cycle intermediates and gluconeogenesis. The consequences include muscle wastage, loss of weight, acidic blood, and excess ketone bodies and urea in the urine.

33. What is leptin? How does it function in the long-term maintenance of body mass?

ANSWER: Leptin is a small polypeptide that is produced in adipocytes and is carried by the blood to the brain.

It is produced when there are adequate lipid stores in adipocytes and interacts with a receptor in the hypothalamus. This interaction produces a cascade of effects that tend to suppress appetite and increase energy expenditure.

34. Briefly describe the effects of the hormone adiponectin.

ANSWER: Adiponectin is a peptide hormone produced by adipose tissues that triggers increased uptake and oxidation of fatty acids in liver and muscle cells. It also blocks gluconeogenesis and fatty acid biosynthesis in the liver and sensitizes the liver to insulin.

35. How do very low concentrations of hormones result in large amplification of the original signal?

ANSWER: The high affinity of the hormone-receptor interaction allows cells to respond to very low concentrations of hormone. All hormone receptor subtypes share three major features: receptor binding, transduction, and response. At each step during signal transduction, a small quantity of the signal molecule activates a larger number of molecules at the next step. When this occurs over several steps, the multiplicative effect can result in a several millionfold amplification of the original signal.

36. Which of the following is NOT a response of pancreatic β cells to increased blood glucose levels? a. increased conversion of glucose to glucose 6-phosphate b. increased glycogen breakdown c. increased insulin secretion d. increased intracellular calcium concentrations e. increased oxidative phosphorylation ANSWER: b 37. In humans, the Cori cycle shifts part of the metabolic burden of active muscle to the liver. Which two metabolic pathways are involved in this cycle? a. glycolysis and gluconeogenesis Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20: Metabolism and Biosignaling b. pentose phosphate pathway and gluconeogenesis c. glycolysis and the pentose phosphate pathway d. gluconeogenesis and the Calvin cycle e. None of the answers is correct.

ANSWER: a 38. What is a typical response of hepatocytes hours after a shift to a high carbohydrate diet? a. increase synthesis of enzymes essential for amino acid catabolism. b. increase synthesis of enzymes essential for gluconeogenesis. c. decrease in available enzymes essential to carbohydrate metabolism d. decrease in liver enzyme turnover. e. increase synthesis of enzymes essential for fat synthesis.

ANSWER: e 39. The presence of hexokinase IV allows hepatocytes to continue phosphorylating glucose when the glucose concentration rises well above levels that would overwhelm other hexokinases. Once the glycogen storage reaches capacity, how is hexokinase IV regulated to prevent further production of glycogen in the liver? a. targeted to the lysosome for degradation b. post-translationally modified for inactivation c. translocated into the nucleus d. allosterically inhibited by glucose 6-phosphate e. none of the above ANSWER: c 40. In the Cori cycle, actively contracting muscle cells convert pyruvate to lactate. The lactate is transported in the blood to the liver, where it is "recycled" by gluconeogenesis to glucose, which is transported back to the muscle for additional ATP production. Why do active muscle cells simply not export pyruvate, which can also be converted in the liver to glucose?

ANSWER: Muscle contracting anaerobically must retain pyruvate so that it can be reduced to lactate and regenerate NAD+, which is required for continued glycolysis and the production of ATP during low oxygen conditions.

41. Describe the role of lipases when the demand for stored fuel rises and the roles of epinephrine and insulin in regulating lipase activity.

ANSWER: Lipases in adipocytes hydrolyze stored triacylglycerols to release free fatty acids. The release of fatty acids from adipocytes is greatly accelerated by epinephrine, which stimulates the cAMPdependent phosphorylation of perilipin. This gives lipases that are specific for tri-, di-, and monoacylglycerols access to triacylglycerols in lipid droplets. Insulin counterbalances this effect of epinephrine, decreasing the activity of the lipase.

42. Which hormone stimulates appetite? a. ghrelin b. insulin Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20: Metabolism and Biosignaling c. adiponectin d. both ghrelin and insulin e. both ghrelin and adiponectin ANSWER: e 43. Leptin and leptin receptors work together to regulate the signal for hunger and energy expenditure. Which cell type produces leptin, and where are leptin receptors typically found? a. Adipocytes produce leptin, while leptin receptors are predominantly located in the hypothalamus. b. Cells of the arcuate nucleus in the hypothalamus produce leptin and contain leptin receptors. c. Leptins are produced in the hypothalamus, while leptin receptors are found in adipocytes. d. Leptin and leptin receptors are produced and located in adipocytes. e. Leptin and leptin receptors are found in all cell types.

ANSWER: a 44. Since the blood level of the hormone neuropeptide Y (NPY) rises during starvation and is elevated in mice with either two defective copies of the leptin gene (ob/ob) or two defective copies of the leptin receptor gene (db/db). What can you conclude regarding the overall role of leptin in regulating feeding behavior? a. Leptin stimulates the production and release of GLP-1. b. Leptin stimulates the production and release of α-MSH. c. Leptin inhibits the production and release of GLP-1. d. Leptin inhibits the production and release of GLP-1 and stimulates the production and release of αMSH. e. Leptin stimulates the production and release of GLP-1 and inhibits the production and release of αMSH. ANSWER: d 45. AMPK shifts metabolism toward energy generation and away from energy-requiring biosynthesis. Which of the following DOES NOT regulate AMPK activity? a. adiponectin receptor-mediated phosphorylation b. elevated levels of intracellular AMP c. an increase in intracellular calcium ion concentration d. decreased levels of ATP e. increased activity of liver kinase B1 ANSWER: c 46. Which of the following DOES NOT influence the amount of leptin released by adipocytes? a. weight loss b. the number of leptin receptors c. the size of adipocytes d. the number of adipocytes e. the prolonged energy expenditure Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20: Metabolism and Biosignaling ANSWER: b 47. Which signal originating from the hypothalamus triggers the mobilization of lipids in adipose tissue and generation of heat, increasing energy homeostasis in response to leptin? a. α-MSH b. GLP-1 c. thyroid stimulating hormone d. neuropeptide Y (NPY) e. norepinephrine ANSWER: e 48. One example of research that advances the treatment of obesity and associated diseases, including type 2 diabetes, cardiovascular disease, and cancer, is the discovery and characterization of glucagon-like peptide 1 (GLP-1). GLP-1 binds to the GLP-1 receptor, a G protein-coupled receptor found in which of the following cell types? a. cardiovascular system b. pancreatic α cells c. kidney d. hypothalamus e. all of these cell types ANSWER: e 49. Leptin is an example of a(n) _____ hormone. a. amine b. lipid c. autocrine d. peptide e. steroid ANSWER: d 50. What is the difference between orexigenic and anorexigenic neurons?

ANSWER: Orexigenic neurons stimulate eating by producing and releasing the hormone glucagon-like peptide 1 (GLP-1), which causes the next neuron in the circuit to send the signal to the brain: Eat! Anorexigenic neurons are appetite-suppressing and produce α-melanocyte–stimulating hormone, which causes the next neuron in the circuit to send the signal to the brain: Stop eating!

51. Although adipokines are a major regulator of body weight, hormones from numerous other tissue types also exert a role. Given the list of hormones below, describe where each hormone is produced and which tissue it acts on. (a) ghrelin (b) incretins (c) peptide YY (PYY3-36) Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20: Metabolism and Biosignaling (d) thyroid-stimulating hormone (TSH) ANSWER: (a) produced in the gastrointestinal tract (primarily the stomach); acts in the hypothalamus (b) produced in the gut after ingestion of a meal; acts on the pancreas to increase insulin secretion and decrease glucagon secretion (c) produced in the intestine, signals satiety in the brain (d) produced in the pituitary, activates the thyroid to produce thyroxine 52. Type 1 diabetes, sometimes referred to as insulin-dependent diabetes mellitus (IDDM), is caused by the autoimmune destruction of pancreatic β cells. Given insulin's role in regulating eating behaviors, predict how individuals with type 1 diabetes might respond to the presence of food in the absence of insulin.

ANSWER: In regulating eating behaviors, insulin plays a role in inhibiting the production of the hormone neuropeptide Y (NPY) in the hypothalamus; thus, the signal sent by NPY to other neurons to "eat more, metabolize less" is reduced. However, in the absence of insulin and the presence of food, NPY can be regulated in different ways. For example, in the presence of food in the stomach, ghrelin (a positive regulator of NPY) ceases to be released. Additionally, peptide YY is produced in the intestine in response to food, signaling satiety in the brain and inhibiting NPY release. In addition to inhibiting the production of NPY, insulin also triggers the release of α-MSH, signaling to eat less and metabolize more fuel. However, leptin can also trigger the release of α-MSH. Since eating behaviors can be regulated in various ways, the absence of insulin in the presence of food can only slightly influence eating behaviors in individuals with type 1 diabetes.

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