Test Bank for Biochemistry A Short Course 5th Edition 2025. By Justin Hines, Catherine Reinke, John

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Chapter 1 1. What underlies the unity of biochemistry? a. Some organisms are highly uniform at the molecular level. b. All living beings are highly uniform at the organismal level. c. Organisms are highly uniform at the cellular level. d. All organisms are highly uniform at the molecular level. e. Some living beings are highly uniform at the cellular level.

ANSWER: d 2. How many times more carbon is contained in organisms than in Earth's crust? a. 60 times b. 50 times c. 9.5 times d. 10 times e. 100 times ANSWER: b 3. Why is carbon preferable to be the chemical base for life? a. Carbon dioxide is essentially insoluble in water. b. It escapes the biochemical circulation once it is utilized. c. The stability and strength of the carbon–carbon bonds make it a good base for larger molecular complexes. d. By reacting with hydrogen, it can undergo combustion, which provides organisms with energy. e. Carbon dioxide is a strong construction material.

ANSWER: c 4. What is NOT a primary function of proteins? a. signal receptors b. energy storage c. signal molecules d. structural units e. defensive agents ANSWER: b 5. What does being a catalyst mean? a. slowing down the rate of reactions to increase control over them b. being fully utilized in the course of a reaction; causing no unnecessary change in the reaction components c. repeatedly enhancing the rate of chemical reactions without being consumed d. repeatedly decreasing the rate of chemical reactions without being affected by enzymes e. enhancing the rate of chemical reactions being permanently affected during their course to control the environmental changes .. Powered by Cognero.

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Chapter 1 ANSWER: c 6. Which structure CANNOT be a nucleotide? a. a five-carbon sugar attached to a base and at least one phosphoryl group b. a five-carbon sugar attached to a heterocyclic ring structure and one phosphoryl group c. a five-carbon sugar attached to a base and two phosphoryl groups d. a five-carbon sugar attached to a base and four phosphoryl groups e. a five-carbon sugar attached to a heterocyclic ring structure and three phosphoryl groups ANSWER: d 7. What determines the nature of an organism? a. the information stored in proteins b. the information contained in deoxyribonucleotides c. the information stored in the deoxyribonucleic acid d. the information stored in biological polymers e. the information contained in the ribonucleic acid ANSWER: c 8. What is NOT a nucleotide? a. adenine b. thymine c. uracil d. cytosine e. adenosine ANSWER: e 9. Which statement about compositional differences between DNA and RNA is TRUE? a. Ribonucleotides contain an additional hydroxyl group. b. Ribonucleotides exploit three different bases, while deoxyribonucleotides exploit four. c. Deoxyribonucleotides contain an additional hydroxyl group. d. Deoxyribonucleotides contain a five-carbon sugar, whereas RNA bases contain a six-carbon sugar. e. Ribonucleotides exploit four different bases, while deoxyribonucleotides exploit three.

ANSWER: a 10. What is the molecular basis for establishing intracellular compartments? a. the formation of membranous barriers due to ability of proteins to form extended linear structures b. the ability of lipids to form barriers due to their dual properties regarding solubility in water c. the formation of lipid barriers due to the small size of lipids d. the development of lipid barriers due to hydrophilic interactions within lipid molecules e. the formation of lipid barriers due to the dual nature of lipids regarding their interactions with proteins .. Powered by Cognero.

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Chapter 1 ANSWER: b 11. What can serve as sites of cell-to-cell interactions? a. molecules of glycogen b. branched chains of carbohydrates c. branched proteins d. hydrophobic tails of lipids e. hydrophilic heads of lipids ANSWER: b 12. Choose the CORRECT statement about functions of lipids and proteins. a. They both can be involved in signal transduction. b. They both form a cell skeleton. c. They both act as receptors for signal molecules. d. They both can provide large amounts of cellular energy upon combustion. e. They both are involved in the transfer of biological information.

ANSWER: a 13. What determines the function of a cell or tissue? a. information contained in the genome b. selective transcription of RNA c. translation of RNA d. replication of DNA e. selective expression of genes ANSWER: e 14. What is the central dogma? a. the flow of information across all biomolecules of a cell b. the scheme proposed by Francis Crick to describe complex cell-to-cell interactions c. the basic scheme of the flow of information in the cell for eventual rendering of the genetic information into a functional form d. the scheme proposed in 1958 to describe the genome e. the basic scheme of selective replication of DNA ANSWER: c 15. What is the basic unit of life? a. DNA b. protein c. RNA d. cell e. biomolecular complexes .. Powered by Cognero.

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Chapter 1 ANSWER: d 16. What constitutes the cell membrane? a. a lipid bilayer organized through interaction of lipid hydrophobic heads with each other and lipid hydrophilic tails with the environment b. the outer layer of a cytoplasm c. a lipid bilayer organized through interaction of lipid hydrophobic tails with each other and lipid hydrophilic heads with an aqueous environment d. the periplasmic space e. a lipid bilayer organized by interaction with membrane proteins ANSWER: c 17. How can one distinguish eukaryotic and prokaryotic cells? a. Eukaryotic cells do not have any membranous compartments. b. The cytoplasm is present only in eukaryotic cells. c. The DNA is free in the cytoplasm of eukaryotic cells. d. Membrane-enclosed compartments are absent in prokaryotes. e. Eukaryotic cells are surrounded by two membranes with the periplasmic space in between.

ANSWER: d 18. What are two biochemical features that are obligatory for a cell to exist? a. plasma membrane and nucleus b. nucleus and cytoplasm c. storage for genetic information and a lipid bilayer d. plasma membrane and nucleoid e. plasma membrane and cytoplasm ANSWER: e 19. What is an example of the structural function of carbohydrates? a. formation of branched complexes involved in cell-to-cell interaction b. construction of a cell wall c. formation of glycogen polymers d. modification of proteins e. construction of starch in plants ANSWER: b 20. Which is an organelle? a. ribosome b. cytoskeleton c. plasma membrane d. secretory granule .. Powered by Cognero.

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Chapter 1 e. cell wall ANSWER: d 21. Which sequence of steps preceding DNA replication is TRUE? a. The DNA polymerase gene is transcribed in the nucleus → the DNA is translated into the enzyme on ER → the enzyme is translocated through nuclear pores into the nucleus. b. mRNA of the DNA polymerase gene is synthesized in the nucleus → the mRNA is transferred to the cytoplasm → the DNA polymerase is synthesized on the ribosome → the enzyme is translocated through nuclear pores into the nucleus. c. The DNA polymerase gene is translated in the nucleus → the mRNA is transferred to the cytoplasm → transcription takes place in the ribosome → the enzyme is translocated through nuclear pores into the nucleus. d. mRNA of the DNA polymerase gene is translated in the cytoplasm→ the enzyme is transferred to the nucleus → transcription takes place in the ribosome. e. mRNA of the DNA polymerase gene is synthesized in the cytoplasm → the mRNA is transferred to the nucleus → the DNA polymerase is synthesized on the ribosome → the enzyme is translocated through nuclear pores out of the nucleus.

ANSWER: b 22. In which organelles is chemical energy generated? a. nucleus and mitochondrion b. chloroplast and nucleus c. mitochondrion and endosome d. mitochondrion and chloroplast e. nucleus and endosome ANSWER: d 23. What is the source of cellular energy? a. glucose b. mitochondrion c. adenosine triphosphate d. glycogen e. lipids ANSWER: c 24. The quantity of which organelle in liver cells can be markedly changed by increasing the ingestion of drugs by an organism? a. smooth ER b. zymogen granules c. rough ER d. the Golgi complex e. ER lumen .. Powered by Cognero.

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Chapter 1 ANSWER: a 25. What organelles is NOT surrounded by two membranes? a. nucleus b. chloroplast c. mitochondrion d. the Golgi complex e. chromoplast ANSWER: d 26. Which statement about a plant cell is TRUE? a. The conversion of sunlight into a form of energy useful for a cell takes place in the plant vacuole. b. The plant cell wall is constructed from linear polymers of amino acids. c. The conversion of sunlight into a form of energy useful for a cell takes place in the mitochondrion. d. Chloroplasts power the whole living world. e. The plant cell wall is constructed mostly from branched polymers of carbohydrates.

ANSWER: d 27. How big is a human microbiome in relation to the number of human cells? a. relates as 0:1 b. relates as 2:1 c. relates as 13:1 d. relates as 1:2 e. approximately equal ANSWER: e 28. What is the role of DNA polymerase? a. It catalyzes the transcription of DNA. b. It structures the collection of genes. c. It is the catalyst of synthesis of new strands of DNA. d. It selectively expresses heritable information of an organism. e. It is the catalyst of separation of DNA strands.

ANSWER: c 29. Where are secreted proteins modified? a. smooth ER and transport vesicle b. the Golgi complex and smooth ER c. rough ER and transport vesicle d. the Golgi complex and transport vesicle e. rough ER and the Golgi complex ANSWER: e .. Powered by Cognero.

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Chapter 1 30. What regions of the endoplasmic reticulum can lack ribosomes? a. rough ER and budding-off secretory vesicles b. smooth ER and transport vesicles budding off the rough ER c. smooth ER and secretory vesicles budding off the rough ER d. rough ER and transport vesicles budding off the smooth ER e. smooth ER and sequestered plasma membrane ANSWER: b 31. Which statement about the difference between endocytosis and phagocytosis is TRUE? a. Endosomes formed only in phagocytosis can be fused with lysosomes. b. Small amounts of material can be taken into the cell by phagocytosis but only large ones by endocytosis. c. Endosomes formed only in endocytosis can be fused with lysosomes. d. Large amounts of material can be taken into the cell by phagocytosis but only small ones by endocytosis. e. Only immune cells can endocytize cholesterol.

ANSWER: d 32. Which process is similar to the formation of lysosomes? a. formation of secretory granules b. endocytosis c. phagocytosis d. formation of transport vesicles e. stacking of membranes ANSWER: a 33. Where does sorting of proteins take place? a. rough ER b. transport vesicle c. the Golgi complex d. smooth ER e. secretory vesicle ANSWER: c 34. What components can assist in protein folding? a. stacked membranes b. chaperons c. zymogens d. ribosomes e. lipids .. Powered by Cognero.

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Chapter 1 ANSWER: b 35. Which organelles are unique for plant cells? a. chloroplasts and plant vacuole b. chloroplasts c. cell wall and plant vacuole d. chloroplasts and filaments ANSWER: a 36. What characterizes the matrix of the mitochondria? a. membrane that is in touch with the cytoplasm b. highly invaginated membrane c. combustion of fuel molecules d. intermembrane space e. actin filaments ANSWER: b 37. Which is common between the biomolecules that contribute to the transfer of biological information? a. diverse three-dimensional structure b. dimerization c. dual chemical nature d. repeating units in the linear structure e. repeating units in the branched structure ANSWER: d 38. In higher organisms, what is a polymer with double-stranded phosphodiester-linked monomers? a. RNA b. DNA c. protein d. carbohydrate e. lipid ANSWER: b 39. What determines folding of a protein? a. hydrophobic or hydrophilic environment b. the location of a protein in the cell c. the pH of the cytoplasm d. the sequence of amino acids that constitute the protein e. interactions of proteins with lipids ANSWER: d .. Powered by Cognero.

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Chapter 1 40. The half-life of what molecule is likely to be the shortest? a. protein b. lipid c. carbohydrate d. DNA e. mRNA ANSWER: e 41. What is another function of a plasma membrane apart from its barrier function? a. providing to the interior of the cell an enclosed environment that no molecules may cross b. providing a selectively permeable barrier with the aid of transport proteins c. giving eukaryote and prokaryote cells structural strength d. allowing only a free passage of water in and out of the cell e. facilitated entrance of hormones ANSWER: b 42. Poisons that kill an organism as a result of a loss of high-energy ATP molecules are MOST likely to target which organelle? a. mitochondria b. cytoskeleton c. cytoplasm d. endoplasmic reticulum e. nucleus ANSWER: a 43. A secreted protein would be processed through organelles in what order? a. nucleus, secretory vesicle, Golgi complex b. cytoplasm, Golgi complex, cytosol, secretory vesicle c. endoplasmic reticulum, secretory vesicle, Golgi complex d. nucleus, cytoplasm, endoplasmic reticulum, Golgi complex, secretory vesicle e. endoplasmic reticulum, Golgi complex, secretory granule ANSWER: e 44. Extracellular material is taken into the eukaryotic cell via: a. exocytosis. b. phagocytosis. c. formation of zymogen granules. d. secretory pathway. e. diffusion.

ANSWER: b .. Powered by Cognero.

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Chapter 1 45. If substance X inhibits secretion of the labeled protein but fully synthesized, folded, and glycosylated protein is still found in the cell, where is the most likely site for substance X to act? a. nucleus during translation b. budding off the secretory granule c. translation on the ribosome d. enzyme modification in the Golgi complex e. nucleus during transcription ANSWER: b 46. Which match between the loss of a particular organelle with the associated disease is TRUE? a. hypercholesterolemia – smooth endoplasmic reticulum b. diabetes – endosome c. Tay-Sachs disease – lysosome d. muscle degeneration – mitochondria e. stroke – Golgi body ANSWER: c 47. What experiments would you perform to verify the existence of the endoplasmic reticulum in a strain of bacteria? a. determine whether the bacterium can synthesize ATP in the presence of fuel molecules and O2 b. determine whether the bacterium can synthesize proteins c. determine whether the bacterium generates CO2 in the presence of fuel molecules d. determine whether the bacterium has an internal membrane-enclosed compartment e. determine whether lipids can form a bilayer structure in the bacterial cell ANSWER: d 48. What is/are the rigid material(s) that provide(s) structural support to a plant cell? a. plant cytoskeleton b. plasma membrane c. cell wall d. outer membrane e. microfilaments and microtubules ANSWER: c 49. Filaments and microtubules are components of a network called the: a. chloroplast. b. cytoplasm. c. cytoskeleton. d. cell wall. e. mitochondrion. .. Powered by Cognero.

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Chapter 1 ANSWER: c 50. Translation takes place on/in the: a. ribosomes. b. smooth endoplasmic reticulum. c. nucleus. d. DNA polymerases. e. DNA parent strand.

ANSWER: a 51. Which three elements make up the majority of organisms, listed in descending order of abundance (as a percentage) in humans? a. carbon, oxygen, hydrogen b. hydrogen, oxygen, carbon c. hydrogen, carbon, nitrogen d. nitrogen, hydrogen, oxygen e. hydrogen, oxygen, nitrogen ANSWER: b 52. The abundance of which element makes living organisms unique? a. hydrogen b. oxygen c. phosphorous d. nitrogen e. carbon ANSWER: e 53. When comparing carbon–carbon and silicon–silicon bonds, which is FALSE? a. Carbon–carbon bonds are stronger than silicon–silicon bonds and provide greater stability to large molecules. b. Carbon–carbon bonds release more energy when undergoing combustion, compared to silicon– silicon bonds. c. Molecules with carbon–carbon bonds are better construction materials compared to molecules with silicon–silicon bonds d. Carbon–carbon bonding leads to greater solubility in water compared to silicon bonding. e. Carbon-based molecules are better fuel sources than silicon-based molecules.

ANSWER: d 54. The linear structure of proteins is constructed by linking ____ with _____ bonds. a. nucleotides; covalent b. amino acids; peptide .. Powered by Cognero.

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Chapter 1 c. amino acids; phosphodiester d. polypeptides; hydrogen e. amino acids; hydrogen ANSWER: b 55. Which is NOT a functional role of proteins? a. catalysis b. signal transduction c. cell-cell recognition d. cell mobility e. storage of genetic information ANSWER: e 56. Which three components make up the monomers that are put together to form DNA? a. ribose sugar, nitrogenous base, and three hydroxyl groups b. ribose sugar, nitrogenous base, and three phosphoryl groups c. deoxyribose sugar, nitrogenous base, and three phosphoryl groups d. deoxyribose sugar, nitrogenous base, and three amino groups e. deoxyribose sugar, nitrogenous base, and three hydroxyl groups ANSWER: c 57. Which statement regarding DNA and RNA is FALSE? a. DNA and RNA both contain pentose sugar backbones. b. DNA and RNA both consist of nucleotides connected by phosphodiester linkages. c. DNA and RNA both consist of four unique nucleotides, differing only in the nitrogenous base. d. DNA and RNA are both directional molecules. e. DNA and RNA are both helical structures composed of strands connected by hydrogen bonding.

ANSWER: e 58. Which is a result of the hydrophobic and hydrophilic dual chemical nature of lipids? a. the ability to perform enzymatic functions and speed up biochemical reactions b. the ability to form barriers that delineate an “inside” and “outside” at a biochemical level c. the ability to act as signal molecules d. the ability to combust to provide large amounts of cellular energy e. the ability to store and transfer genetic information ANSWER: b 59. A particular mutation renders DNA polymerase nonfunctional. Which is MOST LIKELY to be affected by this mutation? a. replication of DNA b. transcription of DNA molecules to functional RNAs .. Powered by Cognero.

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Chapter 1 c. transcription of DNA molecules to mRNA d. translation of mRNA molecules to proteins e. folding of functional RNAs ANSWER: a 60. Liver cells and heart cells contain the exact same genetic information. What is responsible for liver and heart cells having drastic structural and functional differences? a. DNA replication b. mutations c. protein sorting d. selective gene expression e. ribosomes ANSWER: d 61. An unusual accumulation of mRNA molecules is observed in the cytoplasm of a cell with low levels of protein. Which enzyme or cell structure is likely malfunctioning in this cell? a. DNA polymerase b. plasma membrane c. ribosomes d. RNA polymerase e. secretory granule ANSWER: c 62. You are a researcher studying the function of a protein called mastermind. This protein functions in neuronal cells to assist in the transmission of electrical signals. In one of your experiments, you notice that mastermind is present in the neuronal cells, but not functioning properly. Other cellular functions appear normal. Select the likely explanation for your observations. a. A mutation in the mastermind mRNA has occurred. b. Ribosomes are not functioning properly. c. RNA polymerase is not functioning properly. d. The mastermind gene is missing from the genome. e. DNA polymerase is not functioning properly.

ANSWER: a 63. A nonfunctional RNA polymerase would MOST directly affect which biological process? a. folding of polypeptides into functional proteins b. conversion of mRNA into protein c. transfer of genetic information from DNA to RNA d. replication of genetic information e. exportation of mRNA from the nucleus to the cytoplasm ANSWER: c .. Powered by Cognero.

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Chapter 1 64. What is meant when scientists say DNA replication is semi-conservative? a. Half of an offspring’s genetic content comes from one parent and the other half comes from the other parent. b. DNA replication consumes energy to synthesize the first strand and synthesizes the second strand without expending additional energy. c. Each strand of a parent DNA molecule acts as a template strand to make two new daughter DNA molecules. d. DNA polymerase is only needed to synthesize one of the strands. e. Each strand of DNA is synthesized without a template.

ANSWER: c 65. Which is an example of the selective permeability of cell membranes? a. oxygen diffusing freely across the plasma membrane b. proteins embedded in the plasma membrane facilitating entry of glucose c. free movement of water across a cell membrane d. phospholipids containing hydrophobic and hydrophilic regions. e. proteins associated with the plasma membrane transducing information ANSWER: b 66. Which of the following is NOT a biochemical process that occurs in the cytoplasm? a. mRNA synthesis b. glucose metabolism c. fatty acid synthesis d. protein synthesis e. endosome-lysosome fusion ANSWER: a 67. Which biological process occurs on rough endoplasmic reticulum (ER) but not on smooth ER? a. processing of exogenous chemicals b. transcription c. translation d. phagocytosis e. DNA replication ANSWER: c 68. What is a consequence of mutations in chaperone proteins in the lumen of the ER? a. improperly sorted proteins b. misfolded proteins c. untranslated proteins d. hyperactive proteins e. unmodified proteins .. Powered by Cognero.

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Chapter 1 ANSWER: b 69. The primary function of lysosomes is to: a. bring sorted proteins to the correct organelle or plasma membrane. b. engulf foreign microorganisms or exogenous agents. c. fuse with secretory granules to dump cargo into the extracellular environment. d. fuse with endosomes to digest extracellular material and damaged intracellular organelles. e. store water, ions, and various nutrients.

ANSWER: d 70. Which is TRUE of both familial hypercholesterolemia and Tay–Sachs disease? a. Both diseases are the result of improper gene expression. b. Both diseases are the result of disruptions in the microbiome. c. Both diseases are the result of improper combustion of fuels. d. Both diseases are the result of malfunctions in specific cellular organelles. e. Both diseases are the result of an inability to store glucose molecules as glycogen.

ANSWER: d

.. Powered by Cognero.

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Chapter 2 1. Molecules that are readily soluble in water are considered: a. nonpolar. b. polar. c. zwitterionic. d. volatile. e. dielectric.

ANSWER: b 2. The interaction that is described by Coulomb's law is called: a. hydrophobic. b. weak. c. electrostatic. d. hydrogen. e. physical.

ANSWER: c 3. What would be the solubility of a salt in a solvent with a low dielectric constant? a. excellent b. good c. suitable d. poor e. exclusive ANSWER: d 4. Hydrophobic molecules are driven together by: a. entropy. b. enthalpy. c. van der Waals interactions. d. affinity. e. hydrogen bonds.

ANSWER: a 5. The protein that interacts with both water and the hydrophobic regions of the membrane is considered: a. polar. b. nonpolar. c. amphibious. d. anabolic. e. amphipathic.

ANSWER: e 6. Protonation of a base yields its: .. Powered by Cognero.

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Chapter 2 a. conjugate molecule. b. conjugate base. c. conjugate acid. d. ionized derivative. e. ionized base.

ANSWER: c 7. An organic acid ionizes to form a conjugate base and: a. hydrogen. b. water. c. a hydroxyl ion. d. a proton. e. hydrogen peroxide.

ANSWER: d 8. Gastroesophageal reflux disease is a common example of a pathological change in: a. pKa of acetic acid. b. hydrogen bonds in a key digestive enzyme. c. protein structure. d. water content in cells. e. pH level.

ANSWER: e 9. What is the pH of human blood? a. 5.5 b. 7.0 c. 9.4 d. 6.4 e. 7.4 ANSWER: e 10. Choose the functional group that CANNOT form a hydrogen bond with water. a. carbonyl group b. amino group c. aromatic ring d. alcohol group e. sulfhydryl group ANSWER: c 11. Which combination can associate to form a hydrogen bond? .. Powered by Cognero.

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Chapter 2 a. N–H and O b. N–H and S c. O–H and P d. C=O and S e. C=O and P ANSWER: a 12. What is the typical length of noncovalent bonds? a. 0.4 angstroms b. 4 angstroms c. 40 angstroms d. 4 nm e. 0.04 nm ANSWER: b 13. In a typical cell, the water content is about: a. 60%. b. 70%. c. 80%. d. 90%. e. 99%.

ANSWER: b 14. Oxygen is an electronegative atom. That means that in a molecule of water the: a. electrons of covalent bonds spend more time near the oxygen atom than near the hydrogen atoms. b. electrons of covalent bonds spend more time near the hydrogen atoms than near the oxygen atom. c. electrons of bonds are located closely to the hydrogen atoms. d. electrons of bonds are located closely to the oxygen atom. e. bonds joining the hydrogen atoms to the oxygen atom are noncovalent.

ANSWER: a 15. What does it mean that liquid water has a partly ordered structure? a. There are van der Waals interactions between water molecules. b. Hydrogen-bonded clusters of molecules are continually being formed and broken apart in liquid water. c. In the physiological range of temperatures liquid water can undergo phase transition to gas that is completely unordered and to ice that is completely ordered. d. There is Brownian motion of water molecules. e. Water molecules can form hydrogen bonds with polar molecules but not with nonpolar molecules.

ANSWER: b .. Powered by Cognero.

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Chapter 2 16. How many neighboring water molecules are hydrogen-bonded on average to one water molecule in a sample of pure water? a. 1.4 b. 2.4 c. 3.4 d. 4.4 e. 5.4 ANSWER: c 17. What type of interactions is NOT a weak interaction? a. nuclear interactions b. salt bridges c. van der Waals interactions d. ionic bonds e. hydrogen bonds ANSWER: a 18. What is the amount of energy needed to apply a 1-newton force over a distance of 1 angstrom? a. 1010 J b. 10–10 J c. 1 J d. 10 J e. 1 kcal ANSWER: b 19. What is the amount of energy needed to raise the temperature of 2 kilograms of water from 14.5°C to 15.5°C? a. 2 J b. 2 kJ c. 2 cal d. 20 cal e. 2 kcal ANSWER: e 20. What factor does NOT affect electrostatic interaction according to Coulomb's law? a. the charges on the two interacting atoms b. a change in entropy if the interaction occurs c. the distance between the two interacting atoms d. the dielectric constant of the medium e. presence and concentration of other charged corpuscles in the medium .. Powered by Cognero.

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Chapter 2 ANSWER: b 21. How much weaker are hydrogen bonds as compared to covalent bonds? a. from 2 to 10 times b. from 20 to 50 times c. from 100 to 1000 times d. from 1000 to 10,000 times e. from 10,000 to a million times ANSWER: b 22. What is the difference in length between a typical hydrogen bond and a covalent bond? a. They are equal in length. b. A hydrogen bond is somewhat shorter than a covalent bond. c. A hydrogen bond is much shorter than a covalent bond. d. A hydrogen bond is somewhat longer than a covalent bond. e. A hydrogen bond is much longer than a covalent bond.

ANSWER: d 23. What group in nucleotide bases is a hydrogen-bond donor? a. N–H b. S–H c. P–O d. C=O e. C–H ANSWER: a 24. Van der Waals interactions do not depend on: a. the distance between the two interacting atoms. b. the distribution of electrons around the nuclei. c. the presence of other charged corpuscles in the medium. d. a transient asymmetry in electrical charge of atoms. e. the geometry of the large molecules.

ANSWER: c 25. What amino acids are most likely to be found in the core of a water-soluble globular protein? a. nonpolar b. polar but uncharged c. positively charged d. negatively charged e. random ANSWER: a .. Powered by Cognero.

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Chapter 2 26. Choose the molecule around which in an aqueous solution water molecules are most ordered. a. alcohol b. aliphatic amino acid c. aromatic amino acid d. hydrocarbon chain e. carboxylic acid ANSWER: d 27. HCl is a strong acid that is easily and completely hydrolyzed in an aqueous solution. What is the concentration of hydroxyl ions in a 50 mM aqueous HCL solution? a. 2 × 10–7 M b. 5 × 10–6 M c. 2 × 10–12 M d. 2 × 10–13 M e. 5 × 10–4 M ANSWER: d 28. What is the net charge of a glycine molecule in human blood at pH 7.4, for an amino group of glycine pKa = 9.6 and for a carboxyl group pKa = 2.3? a. –2 b. –1 c. 0 d. +1 e. +2 ANSWER: c 29. Choose the correct name for the plot of pH changes in a weak acid solution under stepwise addition of a strong acid or base. a. van der Waals curve b. Michaelis–Menten curve c. Henderson–Hasselbalch curve d. compensatory respiratory alkalosis curve e. titration curve ANSWER: e 30. Tris buffers are commonly used in biochemistry because they buffer within the physiological range of pH due to a pKa of 8.1. What is the [A–]/[HA] ratio in a 0.1 M tris solution with pH 9.1? a. 1:100 b. 1:10 .. Powered by Cognero.

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Chapter 2 c. 10:1 d. 2:1 e. 1:2 ANSWER: c 31. What organ in the human body is crucial for blood pH regulation by compensatory respiratory alkalosis? a. heart b. kidneys c. liver d. lungs e. muscles ANSWER: d 32. Tris buffers are commonly used in biochemistry because they buffer within the physiological range of pH due to a pKa of 8.1. What is the concentration of the conjugate base in a 0.1 M tris solution with pH 5.1? a. 0.1 M b. 0.01 M c. 0.05 M d. 0.5 nM e. 0.1 mM ANSWER: e 33. What is the concentration of acetate in a 0.1 M acetic acid solution at pH near to pKa = 4.76? a. 0.5 M b. nearly 0.1 M c. 0.05 M d. 0.01 M e. almost 0 ANSWER: b 34. What is the [A–]/[HA] ratio when a weak acid is in a solution two pH units below its pKa? a. 1:100 b. 1:10 c. 10:1 d. 2:1 e. 1:2 ANSWER: a 35. What is the hydroxyl ion concentration in a urine sample that has a pH of 6? a. 10–6 M .. Powered by Cognero.

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Chapter 2 b. 10–8 M c. 106 M d. 10–14 M e. 6 M ANSWER: b 36. What is the H+ concentration in a urine sample that has a pH of 6? a. 10–6 M b. 10–8 M c. 106 M d. 10–14 M e. 8 M ANSWER: a 37. Typical van der Waals energies are about: a. 4–20 kJ/mol. b. 2–4 kJ/mol. c. 200–400 kJ/mol. d. 2–4 J/mol. e. 200–400 MJ/mol.

ANSWER: b 38. Nonpolar molecules in water: a. dissolve independently. b. aggregate together. c. precipitate. d. dissociate to ions. e. form hydrogen bonds with water molecules.

ANSWER: b 39. What is the [A–]/[HA] ratio when a weak acid is in a solution one pH unit above its pKa? a. 1:1 b. 1:10 c. 10:1 d. 2:1 e. 1:2 ANSWER: c 40. What are the primary chemical components present in a phosphate buffer at pH 7.4? .. Powered by Cognero.

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Chapter 2 a. H3PO4 and PO43– b. H2PO4– and PO43– c. HPO42– and PO43– d. H2PO4– and HPO42– e. H3PO4 and HPO42– ANSWER: d 41. What is the concentration of acetic acid in 250 ml of a 100 mM acetate buffer at pH 4.76? a. 250 mM b. 100 mM c. 50 mM d. 75 mM e. 25 mM ANSWER: c 42. Citric acid is an important intermediate in glucose metabolism and is synthesized in the mitochondrial matrix. The pKa values for each of the three carboxylic groups of the citric acid are 3.1, 4.8, and 6.4. What is the charge on a citrate molecule formed in the mitochondrial matrix where the pH is 7.8? a. +3 b. +2 c. –3 d. –2 e. +1 ANSWER: c 43. A student observes that when an unknown molecule is added to water, it forms micelles, which under the right conditions can form membranes. What can this student infer about this phenomenon? a. The unknown molecule is amphipathic. b. The micelle formation is driven by the resulting decrease in entropy of water. c. The unknown molecule forms many van der Waals interactions with water. d. The micelle formation is driven by the hydrophilic effect. e. The unknown molecule dissociates to ions in water.

ANSWER: a 44. What is the term for the movement of particles due to the random fluctuations of energy content of the environment? a. dissociation b. Brownian motion c. hydrophobic interaction d. van der Waals interaction .. Powered by Cognero.

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Chapter 2 e. entropy ANSWER: b 45. What is the term for the electrostatic interactions between atoms with opposite electrical charges? a. salt bridges b. Brownian motion c. hydrophobic interactions d. hydrogen bonds e. van der Waals interactions ANSWER: a 46. Water weakens the electrostatic interaction of ions due to its: a. ionic bonds or salt bridges. b. Brownian motion. c. entropy. d. ion product of water. e. dielectric constant.

ANSWER: e 47. What is the contact distance when two atoms no longer repulse each other yet have the strongest attraction? a. amphipathic b. Brownian c. hydrophobic d. hydrogen e. van der Waals ANSWER: e 48. Which type of amino acid is responsible for increasing entropy as a protein folds? a. nonpolar b. polar but uncharged c. charged positively d. charged negatively e. amphipathic ANSWER: a 49. The biochemical time scale is on the order of: a. 10−24 s to 10−18 s. b. 10−18 s to 10−12 s. c. 10−12 s to 10−6 s. d. 10−6 s to 10−3 s. e. 10−3 s to 10−1 s. .. Powered by Cognero.

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Chapter 2 ANSWER: c 50. Which of these molecules is polar? a. H2 b. O2 c. N2 d. C6H14 e. H2S ANSWER: e 51. Which type of interactions maintains the column of water in a tree trunk? a. covalent bonding b. ionic interactions c. hydrogen bonding d. van der Waals interactions e. salt bridges ANSWER: c 52. In which environment are electrostatic interactions the strongest? a. in a vacuum b. in water c. in ammonia d. in oil e. in acetic acid ANSWER: a 53. Which of these electric dipole orientations does NOT occur? a. the negative end of a dipole pointing toward a positively charged ion b. the negative end of one dipole pointing toward the positive end of another dipole c. two dipoles positioned so that both the positive and negative ends can interact d. the positive end of a dipole pointing toward a negatively charged ion e. the positive end of one dipole pointing toward the positive end of another dipole ANSWER: e 54. When sodium chloride dissolves in water: a. the sodium ions form noncovalent interactions with the hydrogen atoms of water molecules. b. the chloride ions form noncovalent interactions with the oxygen atoms of water molecules. c. the sodium ions form noncovalent interactions with each other. d. the sodium ions form noncovalent interactions with the chloride ions. e. the chloride ions form noncovalent interactions with the hydrogen atoms of water molecules.

ANSWER: e .. Powered by Cognero.

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Chapter 2 55. Which type of interactions is principally responsible for the stability of a DNA double helix? a. covalent bonding b. ionic interactions c. hydrogen bonding d. van der Waals interactions e. salt bridges ANSWER: c 56. What would be the most likely consequence if complementary base pairs in DNA interacted through covalent bonding instead of through hydrogen bonding? a. The DNA double helix would be less stable. b. The DNA base sequences would be more vulnerable to harmful reactions. c. The DNA could undergo transcription, but not replication. d. The genes on the DNA could not be expressed. e. The DNA could undergo replication, but not transcription.

ANSWER: d 57. What group in nucleotide bases is a hydrogen bond acceptor? a. N–H b. S–H c. P–O d. C=O e. C–H ANSWER: d 58. Which type of interactions allows geckos to walk up walls? a. covalent bonding b. ionic interactions c. hydrogen bonding d. van der Waals interactions e. salt bridges ANSWER: d 59. What motto can be used to describe van der Waals interactions? a. stability in distance b. stability in strength c. stability in numbers d. stability in charge e. stability in smallness ANSWER: c .. Powered by Cognero.

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Chapter 2 60. What is the Second Law of Thermodynamics? a. Energy cannot be created or destroyed. b. The total entropy of a system and its surroundings always increases in a spontaneous process. c. Energy can be created, but not destroyed. d. The total entropy of a system and its surroundings always decreases in a spontaneous process. e. Energy can be destroyed, but not created.

ANSWER: b 61. A student adds an unknown solute to water and observes that the solute forms globules within the water.

The unknown solute molecules most likely: a. form hydrogen bonds with water. b. dissociate into ions. c. form hydrogen bonds with one another. d. decrease the order of the water molecules. e. increase the order of the water molecules.

ANSWER: d 62. What is the primary driver of protein folding? a. hydrogen bonding b. hydrophobic effect c. ionic interactions d. van der Waals interactions e. dipole-dipole interactions ANSWER: b 63. Which of these functional groups CANNOT form hydrogen bonds? a. hydrocarbon chains b. carboxylic acids c. amines d. thiols e. ketones ANSWER: a 64. Compensatory respiratory alkalosis: a. decreases blood pH by increasing the production of CO2. b. increases blood pH by increasing the release of CO2. c. increases blood pH by decreasing the production of CO2. d. decreases blood pH by decreasing the release of CO2. e. increases blood pH by increasing the production of CO2.

ANSWER: b .. Powered by Cognero.

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Chapter 2 65. What is the pH of gastric (stomach) juice? a. 1 to 2 b. 3 to 4 c. 5 to 6 d. 7 to 8 e. 9 to 10 ANSWER: a 66. Which expression represents the pKa of an acid? a. b.

c. d.

e.

ANSWER: e 67. Which expression is true when pH = pKa? a. b. c.

d.

e.

ANSWER: a 68. Which scientist’s discovery laid the groundwork for the development of techniques for tracking individual atoms and molecules in biochemical reactions? a. Lawrence Joseph Henderson .. Powered by Cognero.

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Chapter 2 b. Charles-Augustin de Coulomb c. Marie Curie d. Robert Brown e. Karl Albert Hasselbalch ANSWER: c 69. What is the pH of a lactic acid solution (pKa = 3.86) when the concentration of lactic acid is 0.1 M and the concentration of lactate is 0.01 M? a. 1.86 b. 2.86 c. 3.86 d. 4.86 e. 5.86 ANSWER: b 70. Uncontrolled diabetes can lead to the accumulation of acidic compounds called ketone bodies in the blood.

What happens to the H2CO3/HCO3⁻ buffer system in a person with this condition? a. HCO3⁻ is converted into H2CO3, which is then converted into CO2. b. H2CO3 is converted into HCO3⁻, which is then converted into CO2. c. CO2 is converted into H2CO3, which is then converted into HCO3⁻. d. CO2 is converted into HCO3⁻, which is then converted into H2CO3. e. HCO3⁻ is converted into CO2, which is then converted into H2CO3.

ANSWER: a

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Chapter 3 1. What is NOT a function of amino acids? a. building blocks of proteins b. neurotransmitters c. precursors of nucleic acids d. building blocks of lipids e. precursors of hormones ANSWER: d 2. Recall the structural formula of an amino acid in the dipolar form. Choose the group that is coming out of the plane of the page toward the viewer. a. COO– b. NH3+ c. R d. C e. H ANSWER: c 3. Recall the structural formula of an amino acid in the dipolar form. Choose the group that is projected behind the page away from the viewer. a. COO– b. NH3+ c. R d. H e. C ANSWER: d 4. What are the CORRECT constituents of the α-amino acid? a. central carbon with hydrogen atom, amino group, carboxyl group, R group b. α hydrogen, amino group, carboxyl group, D group c. α carbon, amino group, carboxyl group, R group d. central carbon with hydrogen atom, amide group, carboxyl group, L group e. α carbon, amino group with hydrogen atom, carboxyl group, L group ANSWER: a 5. What makes α-amino acids chiral to the largest extent? a. D isomers are more dominant in α solution. b. The α-carbon atom is tetrahedral. c. α-amino acids are able to exist in both mirror-image forms. d. L isomers are more dominant in a solution. e. Four different groups of α-amino acids are connected to the tetrahedral α-carbon atom. .. Powered by Cognero.

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Chapter 3 ANSWER: e 6. What does NOT vary between amino acids? a. shape b. charge c. hydrophobicity d. L form in proteins e. hydrogen-bonding capacity ANSWER: d 7. Which amino acid contains a terminal carboxamide? a. glycine b. glutamine c. lysine d. glutamate e. phenylalanine ANSWER: b 8. Which statement about the role of functional groups of amino acids is FALSE? a. A narrow range of functional groups provides for the broad spectrum of protein function. b. Most of the properties of amino acids are conferred by functional groups. c. Functional groups are easier to identify in stereochemical renderings. d. A wide range of functional groups provides for the broad spectrum of protein function. e. Under physiological conditions, at least two functional groups of amino acids are charged.

ANSWER: a 9. Which amino acid is more conformationally restricted and why? a. glycine because it is achiral b. proline because its side chain is bonded to both the nitrogen and the α-carbon atoms c. glycine because two hydrogen atoms are bonded to the α-atom d. proline because its side chain is aliphatic e. proline because its side chain is bonded to both the carboxyl carbon and the α-carbon atom ANSWER: b 10. Choose only the aromatic amino acids. a. Phe, Trp, Ser b. Trp, Tyr, Ile c. Tyr, Phe, His d. Phe, Tyr, Leu e. Trp, Phe, Val ANSWER: c .. Powered by Cognero.

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Chapter 3 11. Which amino acids have a branched side chain? a. Val, Leu, Met b. Ile, Leu, Ala c. Ala, Ile, Pro d. Leu, Gly, Val e. Ile, Leu, Val ANSWER: e 12. Which statement about serine, threonine, and tyrosine is FALSE? a. All of them have a hydroxyl group. b. All of them have nonpolar counterpart amino acids. c. They are all aliphatic. d. They are all polar. e. All of them form zwitterions at physiological pH. ANSWER: c 13. Interactions between these amino acids are parts of the driving force for protein folding. Choose all correct answers. a. Ile and Trp b. Ser and Tyr c. Trp and Met d. Phe and Gly e. Val and Thr ANSWER: a, c 14. The side chain of which amino acids can completely lose a proton at a slightly basic pH? a. methionine b. cysteine c. serine d. threonine e. tyrosine ANSWER: b 15. What would be the base-to-acid ratio of histidine at pH 3? a. 100 b. 1000 c. 0.003 d. 300 e. 0.001 ANSWER: e .. Powered by Cognero.

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Chapter 3 16. Which amino acid contains a guanidinium functional group? a. alanine b. lysine c. histidine d. arginine e. asparagine ANSWER: d 17. Which amino acid contains an imidazole ring? a. proline b. arginine c. asparagine d. tryptophan e. histidine ANSWER: e 18. Which amino acid contains a group uncharged or positively charged near a neutral pH? a. histidine b. proline c. cysteine d. serine e. aspartate ANSWER: a 19. What amino acid contains a thioether bond? a. serine b. threonine c. cysteine d. isoleucine e. methionine ANSWER: e 20. The tripeptide that should have the highest solubility in water is: a. HRK. b. QRH. c. KDG. d. WET. e. RKM. ANSWER: a .. Powered by Cognero.

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Chapter 3 21. Which functional group distinguishes asparagine from aspartic acid? a. hydroxyl b. carboxamine c. sulfhydryl d. carboxamide e. guanidinium ANSWER: d 22. Which amino acid can be used as a taste enhancer? a. glutamine b. asparagine c. glutamate d. aspartate e. glycine ANSWER: c 23. How many amino acids can human beings synthesize? a. 20 b. 11 c. 5 d. 18 e. 3 ANSWER: b 24. What is the reason for the edema in kwashiorkor? a. insufficient protein in the blood plasma b. distorted hydrophilicity of the protein c. distorted distribution of water within capillaries d. disrupted hydrophobic effect e. excessive protein in the blood plasma ANSWER: a 25. Which amino acids with ionizable groups have a typical pKa value higher than 10? Select all that apply. a. lysine b. cysteine c. tyrosine d. arginine e. histidine ANSWER: a, c, d 26. Which amino acid has a thiol group? .. Powered by Cognero.

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Chapter 3 a. histidine b. serine c. cysteine d. tyrosine e. asparagine ANSWER: c 27. Which ionizable group has the lowest affinity for protons? a. asparagine side chain group b. aspartic acid side chain group c. terminal α-amino group d. glutamic acid side chain group e. terminal α-carboxyl group ANSWER: e 28. What tripeptide is able to accept the largest number of protons at pH 7.4? a. RHN b. HKE c. DQE d. EHD e. DKE ANSWER: d 29. What amino acids can be involved in acid–base catalysis? a. glutamic acid and asparagine b. tyrosine and serine c. asparagine and arginine d. glutamic acid and cysteine e. threonine and lysine ANSWER: d 30. Which tripeptide is able to accept the least number of protons at pH 11? a. RHN b. HKE c. DQE d. EHD e. DKE ANSWER: a 31. Which amino acid can be part of a protein? a. L glycine .. Powered by Cognero.

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Chapter 3 b. D glycine c. L alanine d. D alanine e. L imidazole ANSWER: c 32. Which amino acid side chains are capable of ionization? Select all that apply. a. aspartate b. histidine c. serine d. alanine e. tyrosine ANSWER: a, b, e 33. The amino acids that contain sulfur are: a. cysteine and serine. b. cysteine and threonine. c. cysteine and methionine. d. threonine and methionine. e. serine and methionine.

ANSWER: c 34. Which amino acid is often present in the active site of enzymes because of its ability to both donate and accept a proton at a physiological pH? a. arginine b. aspartic acid c. histidine d. glutamine e. lysine ANSWER: c 35. Which statement about the compactness of the protein interior is TRUE? a. A small number of sizes and shapes of hydrocarbon side chains enable their compact packing. b. A wide variety of aliphatic side chains complicate their packing with little empty space. c. Large and different polar side chains restrict protein folding. d. A wide variety of sizes and shapes of hydrocarbon side chains enable their packing with little empty space. e. A small variety of aromatic side chains are used to direct protein folding to a more compact form.

ANSWER: d 36. What is the driving force for formation of the unique spatial structure of water-soluble proteins? .. Powered by Cognero.

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Chapter 3 a. the tendency of hydrophilic groups to cluster together b. the tendency of hydrophobic groups to repel each other c. the tendency of polar groups to interact with water molecules d. the tendency of hydrophobic groups to cluster together e. the tendency of hydrophilic groups to repel each other ANSWER: d 37. Which is a chiral type of amino acids found in proteins? a. L amino acids b. D amino acids c. zwitterions d. α-amino acids e. polar amino acids ANSWER: a 38. The amino acid with the smallest-size side chain allowing the greatest flexibility in a protein is _______.

ANSWER: glycine 39. In what pH range is alanine present predominantly in its zwitterionic form? a. 0–2 b. 9–14 c. 8–10 d. 2–4 e. 2–9 ANSWER: e 40. What would be the overall charge of the peptide Asp-Gly-Arg-His at pH 1? a. –1 b. 0 c. 1 d. 2 e. 3 ANSWER: e 41. Which amino acids would be most soluble in a nonpolar solvent such as benzene? a. valine b. histidine c. glutamine d. glycine e. threonine ANSWER: a .. Powered by Cognero.

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Chapter 3 42. Below is a list of five tripeptides identified by their single letter codes. Which tripeptide contains an amino acid capable of forming covalent disulfide bonds? a. FNC b. RGK c. VIL d. MDE e. SYT ANSWER: a 43. In which part of a globular protein are Trp and Phe located and why? a. exterior due to the hydrophilic effect b. interior due to the hydrophobic effect c. exterior forming polar H bonds with water d. interior forming ionic bonds with other amino acids e. exterior forming ionic polar bonds with water ANSWER: b 44. Amino acids can contain all of the following functional groups EXCEPT: a. indole. b. thioester. c. phenyl. d. sulfhydryl. e. amine.

ANSWER: b 45. Which amino acids contain reactive aliphatic hydroxyl groups? Select all that apply. a. serine b. threonine c. methionine d. cysteine e. alanine ANSWER: a, b 46. Which amino acids are positively charged at a neutral pH? a. lysine and arginine b. histidine and arginine c. cysteine and arginine d. lysine and proline e. glutamine and histidine ANSWER: a .. Powered by Cognero.

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Chapter 3 47. What would be the interactions between side chains of aspartate and arginine at a neutral pH? a. hydrophobic b. ionic c. hydrogen bonding d. steric e. covalent ANSWER: b 48. Below is a list of five tripeptides identified by their single letter codes. Which tripeptide has the most polar side chains? a. FNC b. RGK c. VIL d. MDE e. SYT ANSWER: e 49. Below is a list of five tripeptides identified by their single letter codes. Which tripeptide is negatively charged at a physiological pH? a. FNC b. RGK c. VIL d. MDE e. SYT ANSWER: d 50. Which of these amino acids is considered a polar amino acid with a neutral R group? a. valine b. lysine c. serine d. arginine e. aspartic acid ANSWER: c 51. Which amino acid has a side chain that can form a disulfide bond? a. alanine b. serine c. cysteine d. glutamine e. leucine .. Powered by Cognero.

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Chapter 3 ANSWER: c 52. Which of these amino acids has a negatively charged side chain at neutral pH? a. glycine b. histidine c. methionine d. aspartic acid e. arginine ANSWER: d 53. Which amino acid has a ring structure that makes it more conformationally restricted than other amino acids? a. proline b. tryptophan c. phenylalanine d. tyrosine e. isoleucine ANSWER: a 54. What is the significance of the inflection points in the regions of an amino acid titration curve where pH changes are minimized? a. They indicate the pH values at which nearly all molecules of the amino acid are electrically neutral (uncharged). b. They represent the pKa values of the ionizable groups in the amino acid. c. They reflect regions in which the ionizable groups in the amino acid have minimal buffering capacity. d. They correspond to the pH values at which the ionizable groups in the amino acid are fully ionized.

ANSWER: b 55. Which form of amino acids was selected during evolution to be a constituent of proteins? a. D isomer b. L isomer c. zwitterion d. neutral e. charged ANSWER: b 56. Which amino acid is most likely to be found in the active site of an enzyme whose reaction mechanism requires the transfer of a proton from one molecule to another? a. serine b. threonine c. histidine .. Powered by Cognero.

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Chapter 3 d. glutamine e. alanine ANSWER: c 57. Which statement best describes the isoelectric point (pI) of an amino acid with an ionizable side chain? a. It is the pH at which the side chain is fully ionized. b. It is the pH at which the amino acid carries no net charge. c. It is the arithmetic mean of all the pKa values of the amino acid. d. It is the pH at which the amino acid is most soluble in water.

ANSWER: b 58. Which amino acid has a side chain that contains a thioether group? a. valine b. methionine c. cysteine d. leucine e. isoleucine ANSWER: b 59. How many different proteogenic amino acids can a species use to construct proteins? a. 10 b. 15 c. 20 d. 25 e. 30 ANSWER: c 60. What is the significance of a pKa value in the context of amino acids? a. It indicates the concentration of the amino acid in solution. b. It represents the pH at which an amino acid is at its isoelectric point. c. It is the pH at which an amino acid changes from a polar to a nonpolar molecule. d. It is the pH at which a functional group in an amino acid can gain or lose a proton. e. It determines the overall charge of the protein at physiological pH. ANSWER: d 61. How many inflection points does the titration curve of glycine have? a. 0 b. 1 c. 2 d. 3 e. 4 .. Powered by Cognero.

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Chapter 3 ANSWER: d 62. The isoelectric point (pI) of glutamic acid can be calculated by averaging the pKa values of: a. the α-carboxyl group and the R group. b. the α-amino group and the R group. c. the two ionizable groups with positive charges. d. all ionizable groups present. e. the α-amino group and the α-carboxyl group.

ANSWER: a 63. Selenocysteine is unique among amino acids because it: a. contains a sulfur atom instead of selenium. b. is the only amino acid that is not chiral. c. is considered the 21st proteogenic amino acid. d. can only be found in human proteins. e. is not found in any organisms.

ANSWER: c 64. Kwashiorkor is a condition that highlights the importance of which type of nutrient in the diet? a. fats b. carbohydrates c. vitamins d. proteins e. minerals ANSWER: d 65. What is the primary reason that only L isomers of amino acids are used in protein construction? a. L isomers are more stable than D isomers. b. L isomers are easier to synthesize biologically. c. L isomers were likely selected during evolution by chance. d. D isomers are not found in nature.

ANSWER: c 66. Which amino acid is unique in being achiral? a. alanine b. valine c. glycine d. leucine ANSWER: c 67. What is the significance of the one-letter code for amino acids developed by Margaret Oakley Dayhoff? .. Powered by Cognero.

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Chapter 3 a. It facilitated the early comparison of protein sequences by computers. b. It represents the three-dimensional structure of amino acids. c. It is used to identify amino acids that are essential in the diet. d. It classified amino acids based on their polarity and charge at neutral pH. ANSWER: a 68. Which amino acid has a side chain that can form covalent bonds that are particularly important in stabilizing some proteins? a. serine b. threonine c. cysteine d. methionine ANSWER: c 69. Histidine's side chain has a pKa near 6.0, making it functionally significant in enzymes because: a. it can significantly acidify the active site environment. b. it can easily bind and release protons at physiological pH. c. it can contribute to the hydrophobic effect that drives the formation of the unique three-dimensional architecture of the enzyme. d. it can form covalent bonds that stabilize the enzyme’s structure.

ANSWER: b 70. What is the primary role of disulfide bonds in the structure of proteins? a. They increase the solubility of proteins in water. b. They help in the transport of proteins across cell membranes. c. They maintain the stability of a protein's tertiary structure. d. They act as catalysts for biochemical reactions within the protein.

ANSWER: c

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Chapter 4 1. What will be the weight of a protein consisting of 250 amino acid residues? a. 27,500 g mol–1 b. 22,000 g mol–1 c. 5500 g mol–1 d. 25,000 g mol–1 e. 250 kDa ANSWER: a 2. What groups of amino acids are linked in the primary sequence? a. β-amino group and α-carboxyl group b. α-amino group and β-carboxyl group c. α-amino group and α-carboxyl group d. α-nitro group and α-carboxyl group e. β-nitro group and β-carboxyl group ANSWER: c 3. What does polarity mean? a. ends of a polypeptide chain have different charges of the carboxyl group b. ends of a polypeptide chain always have different amino acid residues c. ends of an amino acid residue can link only the same amino acid residue d. a polypeptide chain has the amino-terminal residue and the carboxyl-terminal residue e. a polypeptide chain has the amino-terminal residue with a different charge ANSWER: d 4. Which statements about the polypeptide Gly-Tyr-Gly-Phe-Met-Ser are CORRECT? Select all that apply. a. Glycine is the C-terminal residue. b. Serine is the N-terminal residue. c. Serine is the C-terminal residue. d. Methionine is the N-terminal residue. e. Glycine is the N-terminal residue.

ANSWER: c, e 5. What is the regularly repeating part of a polypeptide called? a. the distinctive side chain b. a variable part c. a peptide d. the backbone e. an oligopeptide ANSWER: d .. Powered by Cognero.

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Chapter 4 6. What is the distance between the nitrogen and carbon atoms in a peptide bond? a. 1.45 Å b. 1.27 Å c. 1.51 Å d. 5.4 Å e. 1.32 Å ANSWER: e 7. Why are peptide bonds of proteins found mainly in the trans configuration in nature? a. In the cis configuration, side chains of amino acids are most distant from each other but steric clashes between the groups are necessary. b. In the trans configuration, side chains of amino acids are most distant from each other and steric clashes between the groups are excluded. c. In the trans configuration, side chains of amino acids are least distant from each other and steric clashes between the groups are excluded. d. In the trans configuration, the two adjacent rigid peptide units may rotate about amino acid bonds. e. The trans configuration contributes to the rotation of radicals in a clockwise direction.

ANSWER: b 8. What is the difference between a peptide and a protein? a. the number of amino acid residues b. the configuration of peptide bonds c. the structure of amino acid residues d. the charge of functional groups e. torsion angles ANSWER: a 9. What structures of polypeptide chains were proposed by Linus Pauling and Robert Corey? a. primary structures b. turns and loops c. secondary structures d. the alpha helix and the beta pleated sheet e. cis and trans configurations ANSWER: d 10. In the formation of the alpha helix, the carbonyl group is connected by: a. a hydrogen bond not with a neighboring NH group, but with a NH group located at a distance of four amino acid residues from it. b. a hydrogen bond with a neighboring NH group. c. a disulfide bond not with a neighboring NH group, but with a NH group located at a distance of four amino acid residues from it. .. Powered by Cognero.

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Chapter 4 d. a peptide bond with a neighboring NO group. e. a hydrogen bond with a single amino acid on an adjacent strand.

ANSWER: a 11. What helices are energetically more favorable? a. right-handed b. counterclockwise c. left-handed d. twisted ribbons e. disulfide-bonded ANSWER: a 12. Aspartate can't form the α-helix because it: a. has a ring structure. b. has closely located radicals that have a tendency to steric clashes. c. has hydrogen-bond acceptors that compete for the main chain groups. d. is a too short amino acid to form the α-helix. e. does not have NH groups.

ANSWER: c 13. What is the distance between adjacent amino acids along the α helix and along the β strand (respectively)? a. 3.5 Å and 1.5 Å b. 5.4 Å and 45 Å c. 5.4 Å and 3.6 Å d. 1.5 Å and 36 Å e. 1.5 Å and 3.5 Å ANSWER: e 14. Name the secondary structure shown in the figure

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Chapter 4

a. α helix b. reverse turns c. β sheet d. parallel β sheet e. mixed β sheet ANSWER: e 15. Collagen has: a. pyrrolidine rings, which stabilize helices and glycine residues at every third position (because only glycine residues can fit inside of the super-helical cable). b. pyrrolidine rings, which can push off and destabilize the helix and glycine residues at every second position (because only glycine residues can fit inside of the super-helical cable). c. rings of proline and glutamine residues at every second position. d. rings of glycine, which can stabilize helices and proline residues at every third position (because only proline residues can fit inside of the super-helical cable). e. rings of proline and glycine residues at every second position.

ANSWER: a 16. What tertiary structure does myoglobin have? a. eight β sheets and α helices, turns and loops b. eight α helices, turns and loops between helices, and heme group c. eight β sheets, turns and loops between helices, and heme group d. seven β sheets, turns and loops between helices e. seven α helices, turns and loops between helices ANSWER: b 17. Where are the majority of nonpolar residues located in myoglobin? a. The outside of myoglobin. .. Powered by Cognero.

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Chapter 4 b. The interior of myoglobin. c. Myoglobin consists of only nonpolar residues. d. The outside and the interior of myoglobin. e. The outside of the super helix.

ANSWER: b 18. How can you protect side chains of the main chain from interaction with water? a. pair all of the NO and CO groups by peptide bonding b. convert all nonpolar residues to polar ones c. form α helices and β sheets d. form quaternary structure e. form superhelices ANSWER: c 19. What is a helix-turn-helix motif? a. supersecondary structure in which an α helix is separated from another α helix by a turn b. supersecondary structure in which a β sheet is separated from a helix by a turn c. secondary structure in which two right-handed α helices are intertwined d. secondary structure in which two left-handed α helices are intertwined e. secondary structure in which an α helix is separated from a β sheet by a loop ANSWER: a 20. What types of interactions are in the quaternary structure? Select all that apply. a. disulfide bonds b. ionic bonds c. hydrogen bonds d. peptide bonds e. van der Waals forces ANSWER: b, c, e 21. A denatured protein is a protein: a. with high enzymatic activity. b. with destroyed noncovalent bonds. c. in which the disulfides are fully converted into sulfhydryls. d. that was treated with β-mercaptoethanol. e. that has been transformed into a random coiled peptide incapable of normal activities.

ANSWER: e 22. What led to the conclusion that information about catalytic activity is contained in the primary sequence? a. Anfinsen's experiment, which showed that denatured protein can spontaneously restore its structure after exposure to denaturing agents .. Powered by Cognero.

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Chapter 4 b. Sanger's experiment, which showed that a protein has a precisely defined amino acid sequence c. Sanger's experiment, which showed that denatured protein can spontaneously restore its structure after exposure to denaturing agents d. Ramachandran's experiment about the φ and ψ angles e. Anfinsen's experiment, which showed that a protein has a precisely defined amino acid sequence ANSWER: a 23. What is the difference between calculated and actual folding time called? a. Dawkins' paradox b. folding funnel c. Ramachandran plot d. Levinthal's paradox e. Anfinsen's paradox ANSWER: d 24. What proteins have multiple conformations? a. ribonuclease and α-keratin b. collagen and prions c. synuclein and lymphotactin d. myoglobin and lymphotactin e. synuclein and collagen ANSWER: c 25. Proteins that are exceptions to the paradigm that a given protein amino acid sequence will fold into a particular three-dimensional structure are those that: a. perform different functions in different conditions and with a different partner. b. are only hydrophobic. c. are only hydrophilic. d. perform different functions with the same partners. e. have a different structure.

ANSWER: a 26. Prions are: a. proteins that are normally present in the brain. b. agents that are similar in size to viruses but consist only of protein. c. agents causing viral brain diseases. d. uncommon for brain proteins with a complex structure. e. proteins contributing to the correct folding of proteins.

ANSWER: a 27. In a Ramachandran plot: .. Powered by Cognero.

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Chapter 4 a. the values of 0 to +180 for the angle of the N–αC bond and of 0 to –180 for the angle of the αC–C bond are strongly disfavored. b. the φ angle is positive if the amino group is rotated to the left from the α carbon. c. only a small part of conformations is sterically excluded because two atoms cannot be in the same place at the same time. d. clockwise rotations of the torsion angles correspond to the negative values. e. the ψ angle corresponds to the bond between the amino group and α carbon.

ANSWER: a 28. Which amino acids of the portion of the polypeptide AIGHTSM will be hydrogen-bonded to each other in the α helix? a. A-T, I-S, G-M b. A-H, I-T, G-S c. A-G, I-H, G-T d. A-I, G-H, T-S e. A-M, I-S, G-T ANSWER: a 29. Choose the CORRECT statement concerning the folding funnel. a. As the percentage of protein residues in native conformation increases, the total energy of the protein decreases, which increases the entropy of the protein. b. The energy of the protein grows as more hydrophobic interactions occur upon protein folding. c. In the nucleation-condensation model, both local and long-range interactions take place to lead to the formation of the native state. d. The molten globule model assumes that local interactions facilitate long-range hydrophobic interactions upon protein folding. e. The molten globule state has a lower energy than the partially correct intermediate.

ANSWER: c 30. In the study of a new protein, you have observed the existence of several distinct stable conformations of the protein depending on presence or absence of other proteins in the solution. One form is 100% α helical, another contains about 75% α helices, and for the third one half of the protein had no distinctive secondary elements.

Explain the results of your study. a. Most if not all of the forms are intermediates of the native protein. b. The protein should belong to the IDP class and is most likely involved in some signaling pathway. c. An intrinsically disordered form of the protein should have been present in the absence of any other protein, while in the presence of some other protein the folding should have proceeded as implied by the model of molten globule. d. The protein doesn't adhere to the paradigm of the folding funnel, and the existence of several forms strongly implies that the protein belongs to the metamorphic proteins. e. The protein doesn't adhere to the paradigm of the folding funnel, and the intrinsically disordered structure implies that this protein is metamorphic. .. Powered by Cognero.

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Chapter 4 ANSWER: b 31. Choose the CORRECT statement concerning prions. a. The prion precursor contains mostly extended structural elements rather than tightly coiled ones. b. A large number of β strands prevents extensive contacts between different proteins. c. The difference in the free energy between PrP and PrPSC is small. d. Prions usually consist of two subunits of the normal PrP protein. e. The decrease in the content of α-helical structural elements is unique for prions and always leads to pathological conditions.

ANSWER: c 32. Choose the CORRECT statement about the secondary structure of protein. a. Adjacent strands in the parallel β sheet can be formed from the amino acids that are neighbors in the protein sequence. b. Loops and turns are required only to connect β strands while α helices are taking the whole length of the protein. c. Neighboring amino acids can form any possible secondary structure. d. Adjacent strands in the antiparallel β sheet can be formed from the amino acids that are neighbors in the protein sequence. e. One β strand is stable enough to exist in the protein structure.

ANSWER: d 33. How many hydrogen bonds should be present within the antiparallel β sheet that curls into a closed barrellike structure if overall it has 8 β strands each 5 amino acids long? Do not consider the bonds that should be present in turns between the strands. a. 10 b. 30 c. 36 d. 44 e. 80 ANSWER: c 34. Compact globular units, called _______, are polypeptide chains folded into two or more compact regions.

ANSWER: domains 35. The secondary structure that is stabilized by CO and NH hydrogen bonding within the peptide chain is called a(n): a. turn. b. loop. c. α helix. d. β sheet. e. β strand. .. Powered by Cognero.

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Chapter 4 ANSWER: c 36. Every third residue in the protein collagen is: a. leucine. b. glycine. c. proline. d. tyrosine. e. titin.

ANSWER: b 37. Select all that apply. Human hemoglobin: a. is the oxygen-carrying protein in mitochondria. b. exists as an α2β2 tetramer. c. consists of α helices and β sheets. d. contains heme groups. e. has a tertiary structure.

ANSWER: b, d 38. The amino acid residue that would MOST likely be buried in the interior of a water-soluble globular protein is: a. aspartate. b. serine. c. phenylalanine. d. lysine. e. glutamine.

ANSWER: c 39. The folding of a protein into its native shape can best be described as a(n): a. random event. b. random event catalyzed by ribosome proteins to maintain a low energy structure. c. series of controlled folds with a few random-shaped structures. d. series of repeatable random events where the lowest energy structure is maintained. e. event where the highest possible energy state is stabilized with discrete folding intermediates.

ANSWER: d 40. What structures did Pauling and Corey predict in 1951? Select all that apply. a. α helix b. β sheet c. turns between β sheets d. loops between α helices e. β strand .. Powered by Cognero.

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Chapter 4 ANSWER: a, b 41. Peptides differ from proteins in: a. the direction of β strands. b. the number of amino acid residues. c. the three-dimensional structure. d. the primary structure. e. polarity.

ANSWER: b 42. Lymphotactin is an example of: a. metamorphic proteins. b. peptides. c. protein hormones. d. fatty-acid-binding proteins. e. fibrous proteins.

ANSWER: a 43. Two amino acids undergo oxidation to form a dimer called: a. proline. b. glycine. c. cystine. d. tyrosine. e. leucine.

ANSWER: c 44. What determines a protein's function? a. its structure b. its gene sequence c. N-terminal amino acids d. time of its formation e. C-terminal amino acids ANSWER: a 45. Why is the peptide bond planar? a. Bulky side chains prevent free rotation around the bond. b. It exhibits partial double-bond character, preventing rotation. c. Hydrogen bonding between the NH and C=O groups limits movement. d. The bond can have either cis conformation or trans conformation. e. It has different peptide bonds.

ANSWER: b .. Powered by Cognero.

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Chapter 4 46. Where are β turns and loops often found? a. in a hydrophobic pocket b. on the interior cleft c. at the protein interface with a ligand d. on the surface of proteins e. between β sheets ANSWER: d 47. Your study group is trying to identify differences in the four levels of protein structure. The secondary structure but not the tertiary structure is stabilized by: a. ionic attractions between oppositely charged side chains. b. H-bonding between polar side chains. c. hydrophobic interactions between nonpolar side chains. d. H-bonding between the oxygen of the backbone carbonyl and the hydrogen of the backbone amine. e. van der Waals forces.

ANSWER: d 48. A clinician friend comes to you and tells you she has a patient that she thinks has some sort of defect in the collagen structure. She wants to know what kinds of structural differences there might be. Which of the following is FALSE for defects leading to scurvy or brittle bone disease? a. Proline residues are not hydroxylated. b. Glycine is replaced by other amino acids. c. Prolyl hydroxylase activity is deficient. d. Accumulation of defective collagen causes cell death. e. There is violation of the quaternary structure.

ANSWER: e 49. Which of the following structures is lost when a peptide bond is formed between two amino acids? a. amino group b. water c. carboxyl group d. carbonyl group e. amino-terminal residue ANSWER: b 50. Which of the following proteins contains examples of the α-helical character? Select all that apply. a. keratin b. ferritin c. myosin d. tropomyosin .. Powered by Cognero.

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Chapter 4 e. CD4 ANSWER: a, b, c, d 51. The three dimensional structure of proteins depends on its ____. a. primary structure b. secondary structure c. tertiary structure d. quaternary structure e. polypeptide structure ANSWER: a 52. Which statement about peptide bonds is TRUE? a. Peptide bonds join amino acids by linking the α-carboxyl group of one amino acid residue to the side chain of another amino acid residue. b. Peptide bonds join amino acids by linking the α-carboxyl group of one amino acid residue to the αcarboxyl group of another amino acid residue. c. Peptide bonds formed between amino acids result in the loss of a water molecule. d. The equilibrium of peptide bonds between amino acids lies on the side of synthesis rather than hydrolysis under most conditions. e. Peptide bonds are quite stable kinetically because their rate of hydrolysis is extremely fast.

ANSWER: c 53. Which of the following amino acid residues is the carboxyl-terminal residue in the polypeptide Met-SerTyr-Phe-Val-Leu? a. methionine b. serine c. tyrosine d. phenylalanine e. leucine ANSWER: e 54. The regularly repeating part of a polypeptide is called the ____. a. side chain b. alpha chain c. amino chain d. main chain e. beta chain ANSWER: d 55. A protein with a molecular weight of 76,000 g mol-1 has a mass of ____. a. 76,000 kilodaltons .. Powered by Cognero.

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Chapter 4 b. 760,000 kilodaltons c. 76 kilodaltons d. 76 daltons e. 760,000 daltons ANSWER: c 56. The constrained conformation of peptide backbones are a result of what? a. the planar nature of peptide bonds b. the uncharged nature of peptide bonds c. the partial double-bond character of peptide bonds d. the trans configuration of peptide bonds e. the cis configuration of peptide bonds ANSWER: c 57. A Ramachandran plot shows that the majority of possible ϕ and ψ combinations are prevented due to ____. a. steric collisions between atoms b. a preference for the cis configuration of the peptide bond c. the length of bond between the nitrogen and the α-carbon atoms d. the planar nature of peptide bonds ANSWER: a 58. Hydrogen-bonding between the repeating functional groups of the polypeptide backbone is responsible for which level of protein structure? a. primary structure b. secondary structure c. tertiary structure d. quaternary structure ANSWER: b 59. The α helix of a protein contains five total turns. How many total amino acid residues compose this α helix? a. 45 b. 100 c. 6 d. 18 e. 5 ANSWER: d 60. Which of the following statements about α helices is TRUE? a. In schematic representations of proteins, α helices are depicted as globular balls. b. The pitch of the α helix is the sum of the length of all the turns of the helix. c. Each residue in an α helix is related to the next one by a rise of 2.5 Å along the helix axis. .. Powered by Cognero.

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Chapter 4 d. Right-handed α helices are energetically more favorable because there are fewer steric clashes between the side chains and the backbone. e. The CO group of residue i in an α helix forms a hydrogen bond with the NH group of residue i + 3.

ANSWER: d 61. Which of the following statements about α helices, β sheets, and β strands is FALSE? a. The structural confirmation of a β strand is almost fully extended rather than tightly coiled as in the α helix. b. The α helix is the most common secondary structure found in proteins, followed by the β sheet. c. The distance between adjacent amino acids along a β strand is approximately 3.5 Å, in contrast with a distance of 1.5 Å along an α helix. d. The side chains of the amino acids composing α helices extend outward in a helical array, and the side chains of adjacent amino acids in β sheets point in opposite directions. e. None of the above.

ANSWER: e 62. In a parallel β sheet, hydrogen bonds connect each amino acid with ___ amino acid(s) on the adjacent strand. a. one b. two c. three d. four e. five ANSWER: b 63. Which level of protein structure is primarily the result of the interactions between the side chains of the peptide chain? a. primary structure b. secondary structure c. tertiary structure d. quaternary structure ANSWER: c 64. β-mercaptoethanol is a chemical that disrupts disulfide bridges. Which level of protein structure would be MOST affected by exposure of a protein to β-mercaptoethanol? a. primary structure b. secondary structure c. tertiary structure d. quaternary structure ANSWER: c 65. Which is NOT a molecular interaction found in the quaternary structure of proteins? .. Powered by Cognero.

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Chapter 4 a. Van der Waals forces b. hydrogen bonding c. disulfide bonds d. ionic bonds e. hydrophobic interactions ANSWER: c 66. Where are nonpolar residues normally located in a protein that functions in an aqueous environment, such as in the interior of a cell? a. interior of the protein, shielded from water b. exterior of the protein, exposed to water c. exterior of the protein, interacting with other nonpolar residues d. interior of the protein, interacting with other polar residues e. none of the above ANSWER: a 67. Polypeptide subunits interact to form which level of protein structure? a. primary structure b. secondary structure c. tertiary structure d. quaternary structure ANSWER: d 68. What term describes a protein converting into a randomly coiled peptide devoid of normal functional activity? a. mutation b. denaturation c. deamination d. prion e. metamorphic ANSWER: b 69. The folding of proteins inside living cells is aided by the presence of what? a. prions b. ribosomes c. chaperones d. helper T cells e. none of the above ANSWER: c 70. What statement(s) is TRUE of the diseases Alzheimer disease, Parkinson disease, Huntington disease, and .. Powered by Cognero.

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Chapter 4 transmissible spongiform encephalopathies? a. These diseases are associated with improperly folded proteins. b. These diseases result in the formation of protein aggregates called amyloid fibrils. c. A common feature of these diseases is the conversion of normally soluble proteins into insoluble fibrils rich in β sheets. d. all of the above ANSWER: d

.. Powered by Cognero.

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Chapter 5 1. The prosthetic group of myoglobin that causes its ability to bind oxygen is: a. heme. b. porphyrin. c. protoporphyrin. d. pyrrole. e. oxymyoglobin.

ANSWER: a 2. What is the term for hemoglobin in the R state? a. αβ dimer b. oxyhemoglobin c. deoxyhemoglobin d. desoxyhemoglobin e. cytoglobin ANSWER: b 3. What is the term for hemoglobin in the T state? a. αβ dimer b. oxyhemoglobin c. deoxyhemoglobin d. desoxyhemoglobin e. cytoglobin ANSWER: c 4. What is the quaternary structure of adult human hemoglobin A? a. monomer b. dimer of the α subunit and β subunit c. trimer of the α, β, and γ subunits d. tetramer of the two α subunits and two β subunits e. tetramer of the four α subunits ANSWER: d 5. How does oxygen-binding affinity of hemoglobin change after the binding of 2,3-BPG? a. increases greatly b. increases slightly c. decreases d. is strongly inhibited e. does not change ANSWER: c 6. Unlike adult forms of hemoglobin, fetal hemoglobin contains: .. Powered by Cognero.

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Chapter 5 a. two γ chains instead of α chains. b. one γ chain instead of an α chain. c. two ε chains instead of β chains. d. one ε chain instead of an α chain. e. two γ chains instead of β chains.

ANSWER: e 7. Why does fetal hemoglobin bind oxygen more tightly than adult hemoglobin? a. Fetal hemoglobin has a single mutation in the β subunit; therefore, it is called the γ chain. b. The substitution of a serine residue for histidine 143 in the γ chain of the 2,3-BPG-binding site causes reduced affinity for 2,3-bisphosphogycerate. c. The substitution of a histidine residue for serine 143 in the γ chain of the 2,3-BPG-binding site causes increased affinity for 2,3-bisphosphogycerate. d. One of the substitutions in the fetal α chain is alanine for proline, which facilitates a van der Waals interaction and depresses the formation of the R state. e. One of the changes in the fetal α chain is a substitution of proline for alanine, which disrupts a van der Waals contact and facilitates the formation of the R state.

ANSWER: b 8. With how many positively charged amino acids does 2,3-BPG interact in each chain of deoxyhemoglobin? a. 1 b. 2 c. 3 d. 4 e. 6 ANSWER: c 9. Where is the 2,3-BPG-binding site in deoxyhemoglobin? a. in the central cavity b. near the heme pocket c. in the hydrophobic core d. on the α1β1–α2β2 interface e. in the groove on the surface ANSWER: a 10. The effect of pH on oxygen binding of hemoglobin is called: a. pH effect. b. metabolic acidosis effect. c. Born effect. d. Herrick effect. e. Bohr effect. .. Powered by Cognero.

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Chapter 5 ANSWER: e 11. During the acidification inside erythrocytes, the specific way to stabilize the T state of hemoglobin is the formation of the salt bridge between: a. β1 His 146 and β2 Asp 94. b. β1 Asp 94 and N-terminal His of the β1 chain. c. α2 Lys 40 and β1 Asp 94. d. protonated β1 Asp 94 and C-terminal His of the β1 chain. e. β1 Asp 94 and protonated C-terminal His of the β1 chain.

ANSWER: e 12. Hemoglobin: a. doesn't help to transport hydrogen ions. b. doesn't help to transport carbon dioxide from actively metabolizing tissues to lungs. c. isn't the main transporter of oxygen. d. isn't the main transporter of carbon dioxide. e. doesn't bind oxygen in lungs where the partial pressure of oxygen is high.

ANSWER: d 13. In the human organism, myoglobin: a. helps to transport hydrogen ions in the body. b. helps to transport carbon dioxide from actively metabolizing tissues to lungs. c. facilitates the diffusion of oxygen to cellular sites that require oxygen d. is the main storage of carbon dioxide in muscles. e. binds oxygen in lungs where the partial pressure of oxygen is high and releases it in muscles in times of need.

ANSWER: c 14. Unlike hemoglobin A, hemoglobin S: a. tends to form large fibrous aggregates when it is deoxygenated. b. tends to form large fibrous aggregates when it is oxygenated. c. has much lower oxygen-binding affinity. d. has much higher oxygen-binding affinity. e. has modified regulation due to mutation.

ANSWER: a 15. What is the reason hemoglobin S aggregates? Choose the MOST accurate statement. a. The exposed hydrophobic valine makes a better van der Waals contact surface between Hb tetramers. b. The exposed hydrophobic valine on the β chain interacts with a complementary hydrophobic patch on the β chain of another hemoglobin molecule. .. Powered by Cognero.

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Chapter 5 c. The exposed hydrophobic valine on the β chain interacts with a hydrophobic patch on the α chain of another hemoglobin molecule. d. The hydrophobic valine, instead of glutamate, disturbs folding of the β chain; therefore, more hydrophobic side chains become exposed. e. The valine side chain produces a hydrophobic patch in oxygenated hemoglobin.

ANSWER: b 16. Which form of hemoglobin aggregates in a newborn that is heterozygous for the sickle-cell hemoglobin allele? a. oxygenated HbA b. deoxygenated HbS c. oxygenated HbF d. deoxygenated HbA e. oxygenated HbS ANSWER: b 17. What is the evidence that myoglobin binds oxygen in a noncooperative manner? a. sigmoidal shape of the oxygen-binding curve b. hyperbolic shape of the oxygen-binding curve c. saturation level of the oxygen-binding curve d. the speed of oxygen binding at high pO2 e. the speed of oxygen binding at moderate pO2 ANSWER: b 18. What is the evidence that hemoglobin binds oxygen in a cooperative manner? a. saturation level of the oxygen-binding curve b. hyperbolic shape of the oxygen-binding curve c. sigmoidal shape of the oxygen-binding curve d. the speed of oxygen binding at high pO2 e. the speed of oxygen binding at moderate pO2 ANSWER: c 19. Choose the WRONG statement describing the coordination of the iron ion in hemoglobin. a. The Fe2+ ion is coordinated to the four nitrogen atoms in the center of the protoporphyrin of the heme. b. The fifth coordination site is occupied by the "proximal histidine" of the globin chain. c. The oxygen is bound to the sixth coordination site of the iron. d. The bonding between iron and oxygen leads to the rearrangement of electrons of the iron ion such that the number of electrons is decreased and the iron ion becomes effectively smaller. e. Histidine that occupies the fifth coordination site is capable of inhibiting oxidation of the iron to the .. Powered by Cognero.

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Chapter 5 ferric state.

ANSWER: e 20. How many heme groups does a normal adult hemoglobin tetramer contain? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: d 21. What is the maximal number of oxygen molecules that can be bound to one tetramer of normal adult hemoglobin A? a. 5 b. 4 c. 3 d. 2 e. 1 ANSWER: b 22. Carbon dioxide reacts with the amino terminal groups of hemoglobin to form carbamate groups, which carry a _______ charge.

ANSWER: negative 23. Normal adult hemoglobin A: a. isn't a tetramer. b. isn't composed of two α subunits and two β subunits. c. doesn't have a structure very similar to myoglobin. d. can't be best described as a pair of nonidentical αβ dimers. e. is not able to bind an oxygen molecule by each subunit.

ANSWER: d 24. What is the Bohr effect? a. the ability of hemoglobin to retain oxygen when in competition with myoglobin b. the regulation of hemoglobin oxygen binding by hydrogen ions and carbon dioxide c. the alteration of hemoglobin conformation during low oxygen stress d. the ability of hemoglobin to bind oxygen when the iron ion moves out of the porphyrin plane e. the regulation of hemoglobin oxygen-binding affinity by pH level ANSWER: b 25. Choose a FALSE statement about how oxygen binding to hemoglobin results in the T-to-R-state transition. .. Powered by Cognero.

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Chapter 5 a. When oxygen binds to hemoglobin, the iron ion moves out of the porphyrin plane. b. The movement of the iron ion causes changes in the quaternary structure that correspond to T-to-Rstate transition. c. The histidine residue bound in the fifth coordination site moves with the iron. d. Because proximal histidine is part of an α helix, the α helix moves as well. e. The carboxyl terminal of the α helix lies between the two αβ dimers.

ANSWER: a 26. What is the partial pressure of oxygen in the alveoli of the human lungs? a. 10 torr b. 20 torr c. 100 torr d. 156 torr e. 760 torr ANSWER: c 27. Which statements concerning the following equilibrium is FALSE? CO2 + H2O H2CO3 a. An increase in the pressure of CO2 in actively metabolizing tissues will result in a decrease of pH inside red blood cells. b. This reaction is catalyzed by carbonic anhydrase in red blood cells. c. H2CO3 dissociates to H+ and a bicarbonate ion, HCO3–. d. The majority of CO2 is transported to the lungs in the form of HCO3–. e. The oxygen-binding affinity of hemoglobin increases as pH decreases ANSWER: e 28. How many oxygen-binding sites in hemoglobin are occupied when blood travels through capillaries? a. about 98% b. about 66% c. about 7% d. about 32% e. about 50% ANSWER: d 29. In the allosteric regulation of oxygen binding to Hb, _____ does NOT participate. a. carbon dioxide b. carbon monoxide c. hydrogen ion d. carbonic acid e. 2,3-bisphosphoglycerate .. Powered by Cognero.

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Chapter 5 ANSWER: b 30. What is the oxidation state of the heme iron in myoglobin and hemoglobin under normal conditions? a. Fe2+ b. Fe3+ c. ferric ion d. oxidized e. Fe0 ANSWER: a 31. What prevents oxidation of the heme iron to the ferric form? a. protoporphyrin b. proximal histidine c. distal histidine d. β 82 lysine e. 2,3-bisphosphoglycerate ANSWER: c 32. What is the magnitude of rotation of one αβ dimer with respect to the other one upon oxygenation of hemoglobin? a. 3 degrees b. 30 degrees c. 90 degrees d. 5 degrees e. 15 degrees ANSWER: e 33. What would be the expected result of substitution of aspartate 94 of the β chain with a serine residue? a. 2,3-BPG would bind tighter because of the loss of a negative charge. b. Oxygen-binding affinity would increase because of a disruption of a van der Waals contact, thus facilitating the formation of the R state. c. The formation of the salt bridge with C-terminal histidine in the β chain would be disrupted; therefore, hemoglobin sensitivity to acidification is expected to be reduced. d. 2,3-BPG would bind less tightly because of the loss of a negative charge. e. This substitution would have no effect on the allosteric properties of hemoglobin.

ANSWER: c 34. If any of the histidine residues substitute with an arginine residue in the 2,3-BPG binding site of hemoglobin, 2,3-BPG would likely be bound: a. tighter because of the loss of a positive charge. b. tighter because of the gain of a positive charge. .. Powered by Cognero.

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Chapter 5 c. less tightly because of the loss of a positive charge. d. less tightly because of the gain of a positive charge. e. in the same way.

ANSWER: e 35. Carbon dioxide forms carbamate groups in hemoglobin by reacting with: a. aspartate residues. b. cysteine residues. c. N-terminal amino groups. d. tyrosine residues. e. heme.

ANSWER: c 36. Sickle-cell anemia is caused by: a. a decreased production of α chains of hemoglobin. b. a substitution of a Glu residue for a Phe residue at the β6 position. c. the loss of the heme group because the proximal His is oxidized. d. a substitution of a Val residue for a Glu residue at the β6 position. e. a substitution of a Glu residue for His at the C-terminal of the α chain.

ANSWER: d 37. If a Lys residue is substituted with a Ser residue in the 2,3-bisphosphoglycerate (2,3-BPG) binding site of hemoglobin, 2,3-BPG would bind: a. 2 tighter because of the loss of a positive charge. b. tighter because of the gain of a positive charge. c. less tightly because of the loss of a positive charge. d. less tightly because of the gain of a positive charge. e. in the same way.

ANSWER: c 38. fMRI measures the changes in the: a. fifth coordination site upon binding of iron, revealing the amount of carbon monoxide bound in carbon monoxide poisoning. b. fifth coordination site upon binding of iron, revealing sensory brain activity. c. magnetic properties of γ-chain hemoglobin and is a noninvasive way of measuring fetal hemoglobin levels during pregnancy. d. magnetic properties of hemoglobin, revealing the relative amounts of deoxy- and oxyhemoglobin in the circulation of a specific organ. e. pO2 levels in the circulatory system in any organism.

ANSWER: d .. Powered by Cognero.

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Chapter 5 39. What is the organic portion of the heme group in hemoglobin? a. carbonic acid b. histidine c. pyrrole ring d. carbamate e. protoporphyrin ANSWER: e 40. What is the chemical form in which most of the carbon dioxide is transported in the blood? a. carbonic acid b. carbonate ion c. bicarbonate ion d. carbamate e. carbon monoxide ANSWER: c 41. What is the molecule whose function is to facilitate diffusion of oxygen in muscle cells? a. hemoglobin b. myoglobin c. protoporphyrin d. carbamate e. histidine ANSWER: b 42. What factor influences the binding of oxygen to myoglobin? a. the concentration of bicarbonate ions, HCO3– b. the partial pressure of oxygen, pO2 c. the concentration of hemoglobin present in the tissue d. the concentration of 2,3-bisphosphogycerate e. the concentration of the cooperative effect ANSWER: b 43. The iron atom in heme is bound to the fifth coordination site of which molecule? a. protoporphyrin b. oxygen c. carbonic acid d. carbamate e. histidine ANSWER: e .. Powered by Cognero.

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Chapter 5 44. What statement concerning the differences between hemoglobin and myoglobin is TRUE? a. Both hemoglobin and myoglobin are tetrameric proteins. b. Hemoglobin exhibits a hyperbolic O2 saturation curve, while myoglobin exhibits a sigmoid-shaped curve. c. Hemoglobin exhibits cooperative binding of O2, while myoglobin does not. d. Hemoglobin exhibits a higher degree of O2 saturation at all physiologically relevant partial pressures of O2 than myoglobin does. e. Both hemoglobin and myoglobin help to transport carbon dioxide from actively metabolizing tissues to lungs.

ANSWER: c 45. Which of the statements concerning myoglobin is FALSE? a. The globin chain contains an extensive α-helix structure. b. The heme group is bound to the globin chain by two disulfide bonds between cysteine residues. c. The iron of the heme group is in the Fe2+ oxidation state. d. The diameter of the iron ion decreases upon binding of oxygen. e. The function of myoglobin is oxygen storage in muscle.

ANSWER: b 46. Fetal hemoglobin: a. is composed of two α and two γ subunits. b. binds 2,3-BPG more tightly than normal adult hemoglobin. c. binds oxygen with less affinity than HbA at all pO2. d. does not exist in the T state. e. does not exist in the R state.

ANSWER: a 47. What statement about oxygen transport by hemoglobin is TRUE? a. Oxygen binds to the proximal histidine residue of the globin chain. b. Bonding of carbon dioxide to hemoglobin molecules increases the binding of oxygen. c. Hemoglobin binds more oxygen as the pH is lowered. d. Hemoglobin binds more oxygen at higher 2,3-BPG concentrations. e. The binding of each O2 molecule to hemoglobin increases its affinity for the next O2.

ANSWER: e 48. Molecular recognition is referred to as: a. the ability of specific molecules to bind to one another in the face of many alternatives. b. the chemical nature of molecules to recognize similar-sized molecules. c. the evolutionary nature of molecules to adapt to recognizing other molecules. d. the self-recognition abilities of all biochemical molecules. .. Powered by Cognero.

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Chapter 5 ANSWER: a 49. In receptor, R, and ligand, L, binding interactions, L1/2 represents: a. The time it takes for half of all receptors to be paired with a ligand b. The concentration of L at which half of the receptor amount is bound to L and half is free c. The concentration of R at which half of the ligand amount is bound to R and half is free d. The half-life of L ANSWER: b 50. Estrogen receptors have the strongest affinity for the ligand estradiol. However, estrogen receptors also bind bisphenol A, a precursor found in plastics, and tamoxifen, a cancer drug. Bisphenol A and tamoxifen can both be labelled as which of the following? a. disruptive ligands b. off-target ligands c. competitor ligands d. unnatural ligands ANSWER: c 51. For the estrogen receptor, bisphenol A and tamoxifen both have Kd values of 200nM, compared to a Kd of 1nM with estradiol. What does this say about the molecular affinities of the estrogen receptor for bisphenol A and tamoxifen compared to estradiol? a. The bisphenol A and tamoxifen ligands bind with significantly lower affinity compared to estradiol. b. The bisphenol A and tamoxifen ligands bind with significantly higher affinity compared to estradiol. c. The dissociation rates of the bisphenol A and tamoxifen ligands is lower than the dissociation rate of the estradiol ligand. d. The association rates of the bisphenol A and tamoxifen ligands is higher than the association rate of the estradiol ligand.

ANSWER: a 52. Which of the following is the most commonly used metric to measure and communicate binding affinity by biochemists? a. association constants b. equilibrium constants c. dissociation constants d. affinity constants ANSWER: c 53. Which protein-ligand interaction is one of the strongest known to biochemists and is leveraged as an affinity tag in numerous laboratory methods? a. avidin-streptavidin b. immunoglobulin-RNAseH c. horseradish peroxidase-BCIP .. Powered by Cognero.

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Chapter 5 d. alkaline phosphatase-NBT e. avidin-biotin ANSWER: e 54. Why is molecular recognition especially important to the immune system? a. Cells of the immune system have no ability to distinguish between foreign molecules and molecules from normal cells and thus require molecular recognition. b. Cells of the immune system must identify and respond to signs of invading pathogens such as bacteria and viruses in the presence of vast numbers of highly similar molecules from normal cells. c. The immune system needs to recognize molecular entities in order to transmit electrical signals. d. The immune system needs to recognize molecular entities in order to provide cells and tissues with oxygen. e. The immune system needs to recognize molecular entities in order to metabolise sugars for energy.

ANSWER: b 55. The adaptive immune response uses: a. soluble proteins called immunoglobulins that recognize foreign particles and mark them for destruction b. enzymatic barriers such as lysozyme to defend against foreign pathogens c. physical barriers such as skin to limit exposure to foreign invaders d. T-cell-mediated responses against bacteria and viruses ANSWER: a 56. Vaccines prevent serious infection without causing disease by priming the: a. innate immune system b. cell-mediated immune system c. humoral immune system d. natural immune system e. adaptive immune system ANSWER: e 57. A major function of hemoglobin, a component of red blood cells, is to: a. bind and carry CO2 from the lungs to cells and tissues b. bind and carry CO2 back to the lungs c. bind and carry O2 back to the lungs d. transport O2 from the heart to the lungs e. transport O2 from the lungs to the heart ANSWER: b 58. The bound prosthetic group that allows both hemoglobin and myoglobin to bind oxygen is called___. a. myosin .. Powered by Cognero.

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Chapter 5 b. heme c. actin d. histidine ANSWER: b 59. A mutation occurs in the distal histidine of heme. Which of the following would be the most deleterious effect on hemoglobin and myoglobin due to this mutation? a. the pickup of dangerous ligands like carbon monoxide in place of oxygen b. the reduction of the heme iron to the ferric ion c. smaller ligands will be given access to the heme iron d. movement of the iron ion into the plane of the porphyrin ANSWER: a 60. Functional magnetic resonance imaging (fMRI), a powerful method for examining brain function, reveals active parts of the brain by: a. highlighting regions of increased blood flow richer in oxyhemoglobin b. highlighting regions of decreased blood flow richer in oxyhemoglobin c. highlighting regions of increased blood flow richer in deoxyhemoglobin d. highlighting regions of decreased blood flow richer in deoxyhemoglobin ANSWER: a 61. Allosteric proteins are characterized by their more complicated properties compared to other proteins, usually arising from their: a. irregular splice sites b. requirement for chaperone assistance when folding c. multisubunit quaternary structures d. hydrophobic alpha helices ANSWER: c 62. The sigmoidal oxygen-binding curve of hemoglobin in red blood cells demonstrates a binding behavior referred to as: a. complimentary b. cooperative c. symbiotic d. direct ANSWER: b 63. What is the physiological significance of the cooperative binding of oxygen by hemoglobin? a. cooperative binding leads to efficient oxygen transport b. red blood cells export oxygen and import carbon dioxide at greater rates c. hemoglobin reaches equilibrium with myoglobin more efficiently .. Powered by Cognero.

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Chapter 5 d. cooperative binding limits the pickup of dangerous molecules besides oxygen such as carbon monoxide ANSWER: a 64. Why is hemoglobin a more efficient oxygen transporter than myoglobin? a. Myoglobin does not bind oxygen tight enough to be an efficient oxygen transporter. b. Hemoglobin binds oxygen more tightly than myoglobin, making it a better carrier of oxygen. c. Myoglobin binds oxygen too tightly to be an efficient oxygen transporter. d. Myoglobin has a higher affinity for dangerous molecules such as carbon monoxide compared to hemoglobin.

ANSWER: c 65. The deoxygenated state of hemoglobin is referred to as: a. the A state b. the R state c. the T state d. the low state ANSWER: c 66. The oxygenated state of hemoglobin is referred to as: a. the A state b. the R state c. the T state d. the high state ANSWER: b 67. Why does fetal hemoglobin bind oxygen more tightly than adult hemoglobin? a. Fetal hemoglobin has a single mutation in the β subunit; therefore, it is called the γ chain. b. The substitution of a serine residue for histidine 143 in the γ chain of the 2,3-BPG-binding site causes reduced affinity of 2,3-bisphosphogycerate for fetal hemoglobin. c. The substitution of a histidine residue for serine 143 in the γ chain of the 2,3-BPG-binding site causes increased affinity of 2,3-bisphosphogycerate for fetal hemoglobin. d. One of the substitutions in the fetal α chain is alanine for proline, which facilitates a van der Waals interaction and depresses the formation of the R state. e. One of the changes in the fetal α chain is a substitution of proline for alanine, which disrupts a van der Waals contact and facilitates the formation of the R state.

ANSWER: b

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Chapter 6 1. Sedimentation of a particle isn't affected by: a. mass. b. shape. c. the density of the solution. d. charge. e. size.

ANSWER: d 2. Genome rather than proteome: a. is a fixed characteristic of the cell. b. provides a list of gene products that are present in an organism. c. varies with the cell type and developmental stage. d. is influenced by the presence of hormones or inhibitors, or other environmental conditions. e. is the level of functional information, which encompasses the types, functions, and interactions of proteins that yield a functional unit.

ANSWER: a 3. The technique that obtains the isolation of the protein of interest from the thousands of other proteins in the cell is called: a. salting out. b. purification. c. precipitation. d. dialysis. e. exclusion.

ANSWER: b 4. What technique significantly increases the resolving power of protein purification? a. salting out b. gel electrophoresis c. ion-exchange chromatography d. high-performance liquid chromatography e. affinity chromatography ANSWER: d 5. What properties rise upon protein purification? a. enzyme activity b. amount of total protein c. specific activity d. bands on the gel electrophoresis e. total activity ANSWER: c .. Powered by Cognero.

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Chapter 6 6. Choose the CORRECT definition of the given terms. a. The total amount of protein is the measure of enzyme activity. b. Enzyme activity is the ratio of total activity to the amount of protein in the enzyme assay. c. An assay is not always based on some unique biochemical properties of the protein of interest. d. The purification level is obtained by dividing specific activity of the initial extract by the total volume of the fraction. e. The total activity is obtained by measuring specific activity of the fraction used in the assay and multiplying it by the fraction's total volume.

ANSWER: b 7. A mixture of all components of the cell that do NOT contain intact cells is called: a. a pellet. b. supernatant. c. dialysate. d. centrifugate. e. homogenate.

ANSWER: e 8. If you want to get several fractions of decreasing density, each still containing hundreds of different proteins, in a step-by-step fashion what technique would you use? a. gradient centrifugation b. differential centrifugation c. western blotting d. gel electrophoresis e. one of the chromatography methods ANSWER: b 9. Select all that apply. An intermediate product of protein purification is: a. crude supernatant extract. b. crude pellet extract. c. homogenate. d. supernatant extract. e. dialysate.

ANSWER: b, d 10. What technique is used to remove salt from a solution? a. gel electrophoresis b. salting out c. dialysis d. centrifugation .. Powered by Cognero.

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Chapter 6 e. fractioning ANSWER: c 11. Proteins with different sedimentation coefficients can be separated by: a. gradient centrifugation. b. differential centrifugation. c. two-dimensional electrophoresis. d. affinity chromatography. e. ELISA. ANSWER: b 12. The protein in a solution with a high concentration of (NH4)2SO4 doesn't bind to a column. What technique should you use to solve this problem? a. SDS-PAGE b. ELISA c. centrifugation d. dialysis e. HPLC ANSWER: d 13. In which order will proteins A (55 kDa), B (60 Da), and C (50,000 Da) emerge from the column upon molecular exclusion chromatography? a. A → B → C b. A → C → B c. B → A → C d. B → C → A e. C → A → B ANSWER: b 14. How can a protein that is positively charged at pH 7 be eluted in ion-exchange chromatography? a. displacement of the protein by another ligand b. increasing volume of the buffer c. increasing pH of the buffer d. increasing concentration of salt in the buffer e. decreasing concentration of salt in the buffer ANSWER: d 15. What technique is used to separate proteins by their net charge? a. molecular exclusion chromatography b. ion-exchange chromatography c. two-dimensional electrophoresis .. Powered by Cognero.

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Chapter 6 d. ELISA e. immunoprecipitation ANSWER: b 16. What amino acids should be prevailing in a protein at pH 7 for binding of that protein to beads containing negatively charged carboxylate groups? a. serine b. phenylalanine c. histidine d. glycine e. cysteine ANSWER: c 17. Molecular exclusion chromatography is a technique of protein separation according to: a. size. b. isoelectric point. c. binding affinity. d. charge. e. definition of individual proteins.

ANSWER: a 18. What is affinity chromatography based on? a. separation of proteins by size b. moving a molecule with a net charge in an electric field c. affinity of proteins for specific chemical groups or specific molecules d. separation of proteins by their net charge e. high affinity of the antibody for the proteins ANSWER: c 19. What is ion-exchange chromatography based on? a. separation of proteins by size b. moving a molecule with a net charge in an electric field c. affinity of proteins for specific chemical groups or specific molecules d. separation of proteins by their net charge e. high affinity of the antibody for the proteins ANSWER: d 20. The largest number of individual proteins from a mixture is separated by: a. affinity chromatography. b. molecular exclusion chromatography. c. ion-exchange chromatography. .. Powered by Cognero.

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Chapter 6 d. gel filtration. e. high-performance liquid chromatography.

ANSWER: e 21. What technique is used to separate proteins by their charge and then by their size? a. molecular exclusion chromatography b. ion-exchange chromatography c. two-dimensional electrophoresis d. ELISA e. immunoprecipitation ANSWER: c 22. Choose the FALSE statement about protein electrophoresis. a. The gel serves as a molecular sieve that enhances separation. b. The distance and speed that a protein moves depend on its shape. c. Small proteins move rapidly through the gel. d. The proteins in the gel can be visualized by staining them with silver. e. It is performed in a thin, horizontal slab of polyacrylamide.

ANSWER: e 23. Separation of proteins only by their molecular weight can be achieved by putting them in an environment where they are surrounded by negative charges. What provides these negative charges? a. the flow from the cathode to the anode b. electricity c. adjusted pH d. SDS e. β-mercaptoethanol ANSWER: d 24. Protein purification can be monitored by measuring: a. specific activity and performing SDS-PAGE. b. total activity and performing SDS-PAGE. c. total protein and performing HPLC. d. specific activity and performing two-dimensional electrophoresis. e. total activity and performing HPLC. ANSWER: a 25. For enzymes, the assay is a measure of: a. total protein. b. total enzyme. c. enzyme activity. .. Powered by Cognero.

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Chapter 6 d. specific activity. e. total activity.

ANSWER: c 26. Differential centrifugation rather than gradient centrifugation: a. is referred to as ultracentrifugation. b. doesn't depend on the shape of the particle. c. requires the formation of a density gradient in a centrifuge tube. d. can use radioactivity marks to determine the receptor from the fraction. e. applies the harvesting of the separated bands of protein, collecting them drop by drop.

ANSWER: b 27. Choose the CORRECT statement. a. The binding between Ig and antigen is a step in the immune response. b. An epitope is composed of heavy and light chains. c. Antigens are synthesized by an animal in response to the presence of foreign substances. d. The Fc domain of the Ig is formed by two heavy chains and two flexible linkers. e. Nucleic acids can't be effective antigens.

ANSWER: a 28. What reagent helps to visualize amino acids in amino acid composition analysis? a. phenyl isothiocyanate b. clostripain c. β-mercaptoethanol d. hydroxylamine e. fluorescamine ANSWER: e 29. Polyclonal antibodies rather than monoclonal antibodies: a. can be obtained by fusing of antibody-producing cells and myeloma cells. b. are generated from a large number of cells of a single kind. c. are used in purification of receptors or other proteins by immunoprecipitation. d. such as Trastuzumab are used to treat some forms of breast cancer. e. have an advantage for the detection of a protein with low abundance.

ANSWER: e 30. Select all that apply. Enzyme-linked immunosorbent assay: a. uses an enzyme, noncovalently linked to a specific antibody, that reacts with a colorless substrate to produce a colored product. b. is used in tests for HIV infection. c. comes in several forms, including indirect and sandwich. .. Powered by Cognero.

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Chapter 6 d. can detect less than 10–9 g of a protein. e. can be performed only with polyclonal antibodies.

ANSWER: a, c, e 31. Which statements describe sandwich ELISA? Select all that apply. a. The antibody to a particular antigen is adsorbed to the bottom of a well. b. The production of color indicates the amount of an antibody to a specific antigen. c. This technique is used in tests for HIV infection. d. The detection antibody with an enzyme linked to it binds to a specific antibody. e. It is a quantitative method that permits the measurement of small quantities of an antigen.

ANSWER: a, e 32. What technique allows the detection of very small quantities of a particular protein in a cell or in body fluid and makes it possible to find a protein in a complex mixture, which is useful in monitoring protein purification and in the cloning of genes? a. sandwich ELISA b. indirect ELISA c. gradient centrifugation d. immunoblotting e. MALDI-TOF ANSWER: d 33. What technique helps to determine amino acid sequence by removing one amino acid at a time from the amino end of a peptide? a. Edman degradation b. specific cleavage c. peptides overlapping d. electrospray ionization e. MALDI-TOF ANSWER: a 34. What reagent cleaves the carboxyl side of lysine and arginine residues in proteins? a. hydroxylamine b. staphylococcal protease c. clostripain d. trypsin e. carboxypeptidase A ANSWER: d 35. Which statement about mass spectrometry is FALSE? a. In electrospray ionization, the protein or peptide under study is co-precipitated with an organic .. Powered by Cognero.

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Chapter 6 compound that absorbs laser light of an appropriate wavelength. b. In the time of flight analysis, tiny amounts of biomolecules as small as a few picomoles (pmol) or femtomoles (fmol) can be analyzed. c. In tandem mass spectrometry, peptides can be fragmented by bombardment with argon to generate a family of product ions in which one or more amino acids are removed from one end of the initial peptide analyte. d. MALDI-TOF is one of the most accurate means of determining protein mass. e. In MALDI-TOF, peptide masses are matched in a database against proteins that have been "electronically cleaved" by a computer simulating the same fragmentation technique used for the experimental sample.

ANSWER: a 36. What is the most common protease used in sample preparation for a MALDI-TOF assay? a. pepsin b. papain c. trypsin d. thrombin e. carboxypeptidase A ANSWER: c 37. When enzymes are purified, the assay is often based on: a. light absorbance. b. catalytic activity. c. pH. d. temperature changes. e. mRNA levels.

ANSWER: b 38. Two-dimensional electrophoresis is a combination of what two techniques? a. isoelectric focusing and affinity chromatography b. ion-exchange chromatography and SDS-PAGE c. affinity chromatography and SDS-PAGE d. isoelectric focusing and SDS-PAGE e. isoelectric focusing and ion-exchange chromatography ANSWER: d 39. What technique can be used to determine the size of a target protein? a. Edman degradation b. affinity chromatography c. western blot d. ELISA .. Powered by Cognero.

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Chapter 6 e. isoelectric focusing gel ANSWER: c 40. Antibodies as reagents are used by: a. western blotting. b. gradient centrifugation. c. gel electrophoresis. d. ELISA. e. MALDI-TOF. ANSWER: d 41. Which technique CANNOT be used for quantitative analysis? a. gradient centrifugation b. ELISA c. enzyme assay d. SDS-PAGE e. chromatography ANSWER: a 42. What is TRUE regarding gel filtration chromatography and PAGE? a. In both, small proteins move most rapidly. b. In both, large proteins move most rapidly. c. In PAGE, large proteins move most rapidly but in gel filtration, small proteins move most rapidly. d. In gel filtration, large proteins move most rapidly but in PAGE, small proteins move most rapidly. e. In both, proteins are always in their native structure.

ANSWER: d 43. Two proteins are similar in size but differ significantly in the number of acidic and basic amino acids. What technique would be BEST suited for separating these two proteins? a. SDS-PAGE and gel-filtration chromatography b. isoelectric focusing and dialysis c. immunoprecipitation and affinity chromatography d. isoelectric focusing and ion-exchange chromatography e. SDS-PAGE and immunoprecipitation ANSWER: d 44. Two proteins are similar in the number of acidic and basic amino acids but significantly differ in size. What technique would be BEST suited for separating these two proteins? a. SDS-PAGE and gel-filtration chromatography b. isoelectric focusing and dialysis c. immunoprecipitation and affinity chromatography .. Powered by Cognero.

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Chapter 6 d. isoelectric focusing and ion-exchange chromatography e. isoelectric focusing and affinity chromatography ANSWER: a 45. Calmodulin is a calcium-binding protein expressed in eukaryotic cells. What two techniques would greatly reduce the number of steps to purify calmodulin? a. SDS-PAGE and gel-filtration chromatography b. isoelectric focusing and dialysis c. immunoprecipitation and affinity chromatography d. isoelectric focusing and ion-exchange chromatography e. immunoprecipitation and ion-exchange chromatography ANSWER: c 46. Select all that apply. In isoelectric focusing: a. proteins migrate to the location at which they have no net charge. b. the pH of a gel depends on the isoelectric point of the protein. c. a gel for isoelectric focusing contains SDS. d. a mixture of proteins with the same isoelectric point is subjected to electrophoresis. e. the isoelectric point (pI) of a protein is the pH at which its net charge is zero.

ANSWER: a, e 47. What is the term for the subset of gene products actually expressed by a cell? a. genomics b. metabolome c. genome d. Svedberg e. proteome ANSWER: e 48. You are interested in studying a powerful enzyme that is expressed in low amounts. What technique would you choose to determine how much of it is found in the tissue of interest? a. MALDI-TOF mass spectrometry to determine the protein mass b. ELISA to identify any antigenic determinants c. Amino Acid Composition Analysis because it can be done for the whole protein d. two-dimensional electrophoresis to determine charge and size data of the protein e. salting out to concentrate the protein for further study ANSWER: b 49. The sedimentation coefficient is expressed in _______ units.

ANSWER: Svedberg .. Powered by Cognero.

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Chapter 6 50. The functional level of information content, which encompasses the types, functions, and interactions of proteins, is known as the ____. a. genome b. proteome c. transcriptome d. microbiome e. proteinome ANSWER: b 51. Unlike the genome, the proteome: a. is constant across all cell types. b. does not change during developmental stages. c. is not a fixed characteristic within a cell. d. provides an accounting of all the biological information within a cell.

ANSWER: c 52. In general terms, what does it mean to purify a protein? a. separate the protein from the proteome from contaminating genetic information from the genome b. disassociate the polypeptide subunits consisting of the protein quaternary structure c. remove untranslated regions from the protein d. isolate the protein of interest from the other proteins in the cell ANSWER: d 53. In protein purification steps, an assay is: a. a comprehensive experiment to test the effect of the purified protein on cell function. b. a test to determine the quantity of purified protein. c. a test based on a unique identifying property of the protein that is used to determine if the purification steps are proceeding correctly. d. a process of determining where in the purification steps error has occurred.

ANSWER: c 54. Specific activity is___________. a. the ratio of enzyme activity to the amount of protein in the enzyme assay b. the total enzymatic activity of the specific protein of interest being purified c. the total amount of active protein isolated by the purification process d. the ratio of the amount of protein to the rate of enzyme activity in the enzyme assay ANSWER: a 55. Which step of the protein purification process grants access to the protein of interest by fractionating cellular components? a. cell lysis .. Powered by Cognero.

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Chapter 6 b. supernatal centrifugation c. homogenation d. differential centrifugation e. pelleting ANSWER: d 56. The cell fraction that is used as a source of protein for further purification is known as the ____. a. crude extract b. isolated source c. specific isolate d. protein homogenate e. protein fraction ANSWER: a 57. Which of the following is NOT a basis for protein purification? size a. solubility b. size c. charge d. binding affinity e. molecular weight ANSWER: e 58. The process of separating proteins by solubility includes: a. salting in to fractionate a mixture of proteins. b. proteins that retain their activity in the presence of high salt concentrations. c. a subsequent salt removal step using dialysis. d. constant salt concentrations at which all proteins precipitate.

ANSWER: c 59. Gel-filtration chromatography, also known as molecular exclusion chromatography, separates proteins by which characteristic? a. solubility b. size c. charge d. binding affinity e. molecular weight ANSWER: b 60. Ion-exchange chromatography separates proteins by _____. a. solubility .. Powered by Cognero.

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Chapter 6 b. size c. charge d. binding affinity e. molecular weight ANSWER: c 61. The plant protein concanavalin A binds reversibly to glucose and is purified by passing a crude extract through a column of beads containing covalently attached glucose residues. Concanavalin A binds to the beads whereas other proteins do not. Which property of proteins does this purification use to isolate Concanavalin A? a. solubility b. size c. charge d. binding affinity e. molecular weight ANSWER: d 62. The plant protein concanavalin A binds reversibly to glucose and is purified by passing a crude extract through a column of beads containing covalently attached glucose residues. Concanavalin A binds to the beads whereas other proteins do not. Which type of chromatography does this process incorporate? a. ion-exchange chromatography b. molecular exclusion chromatography c. affinity chromatography d. gel-filtration chromatography e. high-performance liquid chromatography ANSWER: c 63. To evaluate the success of each purification step, scientists can visualise the amount of protein using the properties of electrical charge. This process is known as: a. specific activity b. chromatography c. gel electrophoresis d. hybridization ANSWER: c 64. What is the purpose of sodium dodecyl sulfate (SDS) during SDS-PAGE? a. SDS makes all proteins in the reaction positively charged. b. SDS causes all proteins in the mixture to have the same charge-to-mass ratio. c. SDS reduces disulfide bonds. d. SDS completely linearizes the proteins.

ANSWER: b 65. Which technique isolates proteins based on their relative contents of acidic and basic residues? .. Powered by Cognero.

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Chapter 6 a. thin-layer chromatography b. SDS-PAGE c. isoelectric focusing d. gel electrophoresis e. in situ hybridization ANSWER: c 66. The sedimentation coefficients (Svedberg units) for myoglobin and carbonic anhydrase are reported to be 1.97 and 3.23, respectively. What do these values say about the speed in which these two molecules move in a centrifugal field? a. Myoglobin moves faster in a centrifugal field than carbonic anhydrase. b. Myoglobin moves slower in a centrifugal field than carbonic anhydrase. c. Myoglobin moves at the same rate as carbonic anhydrase in a centrifugal field. d. Sedimentation coefficients do not provide insight as to the speed at which molecules move in a centrifugal field.

ANSWER: b 67. Immunological techniques employ _____ in order to specifically target and isolate proteins of interest. a. antigens b. epitopes c. T-cells d. antibodies e. macrophages ANSWER: d 68. Polyclonal and monoclonal antibodies differ in the number of ____ they recognize. a. antigens b. epitopes c. individuals d. antiserum e. constant regions ANSWER: b 69. During immunoprecipitation, the supernatant containing proteins of interest is passed over a matrix containing antibodies that bind to the proteins. This technique is analogous to which process? a. gel electrophoresis b. ion-exchange chromatography c. ultracentrifugation d. affinity-chromatography e. SDS-PAGE ANSWER: d .. Powered by Cognero.

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Chapter 6

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Chapter 7 1. How many molecules can carbonic anhydrase hydrate per second and what are these molecules? a. 10 molecules of water b. 105 molecules of oxygen c. 106 molecules of carbon monoxide d. 106 molecules of carbon dioxide e. 10 molecules of carbon dioxide ANSWER: d 2. What is proteolysis? a. reaction of functional groups' removal b. oxidation–reduction reactions c. reaction of DNA synthesis d. reaction of DNA replication e. reaction of hydrolysis of a peptide bond ANSWER: e 3. How many major classes of enzymes are known? a. more than 1000 b. 7 c. 21 d. 65 e. 5 ANSWER: b 4. What ensures the accuracy of interaction between enzyme and substrate? a. the intricate three-dimensional structure of the enzyme protein b. the intricate structure of the substrate c. the presence of coenzymes d. the small size of molecules e. the two-dimensional structure of the enzyme protein ANSWER: a 5. What bonds does thrombin hydrolyze? a. bonds only on the carboxyl side of lysine and arginine residues b. lysine–arginine bonds in particular peptide sequences c. arginine–glycine bonds in particular peptide sequences d. lysine–glycine bonds in particular peptide sequences e. any peptide bond ANSWER: c .. Powered by Cognero.

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Chapter 7 6. The class of enzymes called ligases: a. catalyzes the formation of double bonds. b. transfers electrons between molecules. c. transfers functional groups between molecules. d. cleaves molecules by the addition of water. e. joins two molecules in a reaction powered by ATP hydrolysis.

ANSWER: e 7. What is the mechanism of action of a hydrolase? a. shuffling of amine groups between donor and acceptor molecules b. removal of a water molecule c. addition of atoms or functional groups d. transfer of a functional group to water e. addition of oxygen ANSWER: d 8. Which statement about cofactors is CORRECT? a. A cofactor is a small molecule that is formed as a result of a reaction catalyzed by enzymes. b. Cofactors can be subdivided into two groups: small organic molecules, derived from vitamins, and metals. c. Cofactors can be subdivided into two groups: small organic molecules, derived from vitamins, and prosthetic (helper) groups. d. An enzyme without its cofactor is referred to as a holoenzyme. e. The catalytic activity of many substrates depends on the presence of small molecules called cofactors.

ANSWER: b 9. What is a holoenzyme? a. an enzyme without its cofactor b. an enzyme with its cofactor c. an enzyme with an active site d. an enzyme without prosthetic groups e. an enzyme with the transition state ANSWER: b 10. What are small organic molecules that contribute to the catalytic activity of many enzymes and are derived from vitamins? a. substrates b. coenzymes c. helper groups d. holoenzymes .. Powered by Cognero.

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Chapter 7 e. apoenzymes ANSWER: b 11. What is the distinction between coenzymes and a normal substrate? a. Coenzymes are often derived from vitamins. b. Coenzymes can be subdivided into two groups, but substrates form only one group. c. Normal substrates are used by a variety of enzymes. d. An enzyme without its coenzyme is referred to as an apoenzyme; an enzyme with its substrate is called a holoenzyme. e. A coenzyme is a large molecule, and a substrate is a small molecule.

ANSWER: a 12. Whether the reaction will take place or not depends mainly on: a. amount of available substrate. b. intricate structure of the enzyme. c. Gibbs free energy difference. d. presence of coenzyme. e. the rate of reaction.

ANSWER: c 13. What TWO thermodynamic properties does one need to consider in order to understand how enzymes operate? a. the change in temperature as the reaction proceeds b. the gas constants for the products and reactants c. the free-energy difference between the reactants and the cofactors d. the free-energy difference between the products and the reactants e. the free energy required to initiate the conversion of reactants into products ANSWER: d, e 14. Which statement about free energy is INCORRECT? a. A reaction can take place spontaneously only if the free-energy change is positive. b. A reaction cannot take place spontaneously if the free-energy change is positive. c. In a system at equilibrium, there is no net change in the concentrations of the products and reactants, and the free-energy change is zero. d. The free-energy change of a reaction is independent of the path (or molecular mechanism) of the transformation. e. The free-energy change does not signify whether a reaction will proceed at a perceptible rate.

ANSWER: a 15. What does "spontaneously" mean in the context of thermodynamics? a. The reaction will take place with the input of free energy. .. Powered by Cognero.

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Chapter 7 b. The reaction does not release energy. c. The reaction will take place when ΔG is zero. d. The reaction depends only on the free energy of the products. e. The reaction will take place without the input of energy.

ANSWER: e 16. What does the free-energy change (ΔG) depend on? Select all that apply. a. concentration of the reactants b. gas constant and absolute temperature c. concentration of water d. nature of the reactants e. concentration of enzymes ANSWER: a, d 17. How much energy (in kcal) is released during the oxidation of 1 mol of glucose, if it is known that ΔG = – 17.2 kJ? a. –4.11 kcal b. –0.239 kcal c. –298 kcal d. –4110 kcal e. –1720 kcal ANSWER: a 18. Will a reaction take place spontaneously, if ΔG°' is –5.46 kcal mol–1? a. No, because ΔG°' is negative. b. Yes, because ΔG°' is negative. c. The criterion of spontaneity for a reaction is ΔG, not ΔG°'. d. The criterion of spontaneity for a reaction is K'eq, not ΔG°'. e. No, because the reaction has stopped.

ANSWER: c 19. What would be the equilibrium constant (approximately) if ΔG°'' is 20 kJ mol–1? a. 3 × 104 b. 10 c. 5.69 d. 1.36 e. 3 × 10–4 ANSWER: e 20. What can enzymes alter in reactions? .. Powered by Cognero.

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Chapter 7 a. Gibbs free energy b. equilibrium constant c. reaction rate d. reaction equilibrium e. substrate concentration ANSWER: c 21. A reaction can occur spontaneously only if ΔG is _______.

ANSWER: negative 22. What determines the rate of reaction? a. free-energy change b. formation of the transition state c. concentration of the enzyme d. free energy required to initiate the conversion of reactants into products e. free energy required to initiate the conversion of products into reactants ANSWER: d 23. The active site of an enzyme: a. is a series of amino acids that bind the enzyme. b. is a linear sequence of amino acids that react with each other. c. binds covalently to the substrate. d. takes up most of the total volume of the enzyme. e. is responsible for the specificity of some enzymes.

ANSWER: e 24. Which reactions proceed with an input of energy? a. exergonic b. isomerization c. oxidation–reduction d. enzyme-catalyzed e. endergonic ANSWER: e 25. The _____ is the least stable and most seldom-occurring species along the reaction pathway, and has the highest free energy. a. activation energy b. transition state c. cofactor d. coenzyme e. apoenzyme .. Powered by Cognero.

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Chapter 7 ANSWER: b 26. What is the free energy of activation? a. the difference in concentration between the product and the substrate b. the difference in free energy between the product and the substrate c. a thermodynamic property that is a measure of useful energy, or energy that is capable of doing work d. the difference in free energy between the product and the transition state e. the difference in free energy between the transition state and the substrate ANSWER: e 27. Which process lowers the activation energy? a. formation of the product from a reactant b. catalysis of the reaction by enzymes c. achieving reaction equilibrium d. formation of the product from the transition state e. formation of the active site ANSWER: b 28. If the activation energy is lowered: a. more product will be formed faster. b. less product will be formed overall. c. fewer substrate molecules will have the energy required to reach the transition state. d. the same amount of product will be formed but at a slower rate. e. more molecules of the enzyme will be needed.

ANSWER: a 29. The first step in catalysis by enzymes is the formation of the: a. transition state. b. active sites. c. cofactor–substrate complex. d. enzyme–substrate complex. e. binding energy between the enzyme and the substrate.

ANSWER: d 30. Which statement about the active site of an enzyme is INCORRECT? a. The active site is a three-dimensional cleft. b. Active sites are unique enzymes. c. The active site is the region that binds the substrates. d. The active site takes up a small part of the total volume of an enzyme. e. Active sites may include distant residues.

ANSWER: b .. Powered by Cognero.

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Chapter 7 31. Which weak reversible interactions bind the substrate to the enzyme? Select all that apply. a. hydrogen bonds b. hydrophobic interactions c. covalent bonds d. van der Waals forces e. ionic interactions ANSWER: a, b, d, e 32. Which statement describing the interactions between a substrate and an enzyme is FALSE? a. The enzyme and the substrate interact by means of short-range forces. b. The substrate may bind only to certain conformations of the enzyme. c. The mechanism of catalysis is dynamic, involving structural changes with multiple intermediates. d. Enzymes are flexible and the shapes of active sites can be markedly modified by the binding of a substrate. e. The lock-and-key model is the most accurate model of enzyme–substrate interactions.

ANSWER: e 33. What did Emil Fischer suggest? a. induced-fit model b. term "induced fit" c. lock-and-key model d. conformation selection e. term "active site" ANSWER: c 34. What does conformation selection mean? a. the mechanism of catalysis b. binding of a substrate only to certain conformations of the enzyme c. interaction of an enzyme and the substrate by means of short-range forces d. a process of dynamic recognition e. the general rule that polar residues are located on the surface of proteins ANSWER: b 35. The maximal binding energy is released when the: a. Gibbs free energy is negative. b. Gibbs free energy is positive. c. energy of the substrate and that of the product are different. d. α-carbon atom of the inhibitor is trigonal. e. substrate is in the transition state.

ANSWER: e .. Powered by Cognero.

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Chapter 7 36. What fact is confirmed by the effects of transition-state analogs? a. selective binding of the transition state b. selective binding of the substrate c. dynamic recognition of the substrate d. flexibility of enzymes e. lack of enzyme specificity ANSWER: a 37. Why does the interaction of a transition-state analog with the enzyme take place? a. Compounds are more tightly bound to the transition state. b. The structure of some atoms of a transition-state analog is similar to the structure of transition-state atoms. c. The energy between the substrate and the product is different. d. The full complement of interactions is formed only when the substrate is in the transition state. e. The substrate may bind only to certain conformations of the transition-state analogue.

ANSWER: b 38. What is the common strategy by which catalysis occurs? a. increasing the probability of product formation b. shifting the reaction equilibrium c. stabilizing the transition state d. increasing the free-energy difference between the transition state and the product e. increasing the activation energy ANSWER: c 39. Which term BEST describes an enzyme binding to its substrate? a. nonspecific b. static recognition c. induced fit d. irreversible e. lock and key ANSWER: c 40. An enzyme will specifically bind its substrate because of: a. a tight lock-and-key binding mechanism. b. a high number of hydrophobic amino acids in the center of the protein. c. a large number of weak interactions at the active site. d. additional nonprotein cofactors. e. a lower activation energy upon binding.

ANSWER: c .. Powered by Cognero.

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Chapter 7 41. Enzymes that transfer electrons between substrates are called: a. apoenzymes. b. hydrolyases. c. lyases. d. isomerases. e. oxidoreductases.

ANSWER: e 42. A tightly bound cofactor might be called a(n): a. apoenzyme. b. holoenzyme. c. transition state. d. prosthetic group. e. cosubstrate.

ANSWER: d 43. Riboflavin is a water-soluble organic substance that is not synthesized by humans. Metabolically, it is converted into a substance called flavin adenine dinucleotide, which is required by succinate dehydrogenase.

Choose the correct statement. a. Riboflavin is a coenzyme. b. Flavin adenine dinucleotide is a vitamin. c. Succinate dehydrogenase is a coenzyme. d. Flavin adenine dinucleotide is a coenzyme. e. Riboflavin is an enzyme catalyzing formation of succinate dehydrogenase.

ANSWER: d 44. The molecular structure that is short-lived and is neither a substrate nor a product is known as: a. substrate analog. b. transition state. c. enzyme–substrate complex. d. cofactor. e. stable intermediate.

ANSWER: b 45. How can a reaction be described if it is exergonic? a. at equilibrium b. spontaneous c. with a positive free-energy change d. oxidation–reduction e. with a cofactor .. Powered by Cognero.

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Chapter 7 ANSWER: b 46. At equilibrium, the Gibbs free energy is/has: a. a positive value. b. neutral. c. a negative value. d. zero. e. equal to one.

ANSWER: d 47. Which of the following is NOT a cofactor? a. Zn2+ b. thiamine pyrophosphate c. coenzyme A d. tetrahydrofolate e. carbonic anhydrase ANSWER: e 48. Choose the CORRECT statement. Enzymes: a. force reactions to proceed in only one direction. b. alter the equilibrium of the reaction. c. alter the standard free energy of the reaction. d. without their cofactors are referred to as coenzymes. e. are very specific.

ANSWER: e 49. Which factor is the basis for the difference between ΔG°' and ΔG values? a. the temperature b. whether the reaction is at equilibrium c. the concentrations of the reactants and products d. the mechanism of the reaction e. whether the reaction is spontaneous ANSWER: c 50. Many transition-state analogs bind more tightly than the native substrate. This fact reinforces the concept that: a. transition-state analogs are planar structures. b. transition-state analogs are highly charged at a physiological pH. c. binding to the transition state is done through a lock-and-key mechanism. d. transition-state analogs are hydrophobic. e. binding to the transition state is done through an induced-fit mechanism. .. Powered by Cognero.

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Chapter 7 ANSWER: e 51. What are catalytically active RNA molecules called? a. ribosomes b. enzymes c. ribozymes d. proteases e. ligases ANSWER: c 52. Which statement comparing papain and thrombin is CORRECT? a. Papain has much greater substrate specificity than thrombin. b. Thrombin has much greater substrate specificity than papain. c. Papain and thrombin are both highly specific proteases. d. Papain and thrombin are both undiscriminating proteases. e. Papain cleaves peptide bonds, but thrombin does not.

ANSWER: b 53. Which of these enzymes is a hydrolase? a. lactate dehydrogenase b. fumarase c. trypsin d. aminoacyl-tRNA synthetase e. triose phosphate isomerase ANSWER: c 54. In the tissues, carbonic anhydrase catalyzes the addition of water to carbon dioxide to form carbonic acid. In the lungs, carbonic anhydrase catalyzes this reaction in the reverse direction, breaking down carbonic acid into carbon dioxide in water. What class of enzyme is carbonic anhydrase? a. oxidoreductase b. transferase c. lyase d. isomerase e. ligase ANSWER: c 55. The citric acid cycle enzyme succinyl CoA synthetase catalyzes the reversible conversion of succinyl CoA to succinate and produces one molecule of GTP from GDP and inorganic phosphate (Pi). Note that the name of this enzyme denotes the reaction in the reverse direction. What class of enzyme is succinyl CoA synthetase? a. oxidoreductase b. hydrolase .. Powered by Cognero.

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Chapter 7 c. lyase d. isomerase e. ligase ANSWER: e 56. Which statement comparing prosthetic groups and stoichiometric coenzymes is CORRECT? a. Both must be present in stoichiometric ratios with other substrates. b. Both are unchanged in the overall chemical reaction. c. Both are tightly bound to the enzyme. d. Both are part of an apoenzyme. e. Both are small organic molecules.

ANSWER: e 57. A magnesium ion (Mg2+) plays an essential role in the synthesis of ATP by ATP synthase. Which term describes the magnesium ion in ATP synthase? a. cofactor b. coenzyme c. prosthetic group d. cosubstrate e. stoichiometric coenzyme ANSWER: a 58. Nicotinamide adenine dinucleotide exists in an oxidized form (NAD+) and a reduced form (NADH). The enzyme lactate dehydrogenase oxidizes one molecule of NADH to NAD+ for each molecule of pyruvate that it reduces to lactate. The NAD+ produced in this reaction can then be used by other oxidoreductase enzymes.

Which term describes nicotinamide adenine dinucleotide? a. substrate b. product c. prosthetic group d. stoichiometric coenzyme e. enzyme ANSWER: d 59. Which equation is TRUE for a system at equilibrium? a. ΔG = 0 b. ΔG = 1 c. K'eq = 0 d. K'eq = 1 e. ΔG = ΔG°' ANSWER: a .. Powered by Cognero.

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Chapter 7 60. What is the standard free-energy change (ΔG°') for a reaction if the equilibrium constant under standard conditions (K'eq) is 100? a. −11,400 kJ mol⁻1 b. −11.4 kJ mol⁻1 c. −5.69 kJ mol⁻1 d. +11.4 kJ mol⁻1 e. +11,400 kJ mol⁻1 ANSWER: b 61. The conversion of 3-phosphoglycerate to 2-phosphoglycerate has a standard free-energy change (ΔG°') of +4.6 kJ mol⁻1. Calculate the free-energy change (ΔG) of this reaction if the concentration of 3-phosphoglycerate is 6.3 × 10⁻5 M, the concentration of 2-phosphoglycerate is 4.6 × 10⁻5 M, and the temperature is 25 °C (298 K). a. −5.4 kJ mol⁻1 b. −3.1 kJ mol⁻1 c. +1.3 kJ mol⁻1 d. +3.1 kJ mol⁻1 e. +5.4 kJ mol⁻1 ANSWER: d 62. The isomerization of glucose 6-phosphate to fructose 6-phosphate has a standard free-energy change (ΔG°') of +1.7 kJ mol⁻1. In the cell, the Gibbs free-energy change (ΔG) for this reaction is −2.7 kJ mol⁻1 at 25 °C. Which statement about this reaction is CORRECT? a. The concentration of glucose 6-phosphate is equal to that of fructose 6-phosphate in the cell. b. The concentration of fructose 6-phosphate is higher than that of glucose 6-phosphate in the cell. c. The concentration of glucose 6-phosphate is higher than that of fructose 6-phosphate in the cell. d. The concentration of fructose 6-phosphate is higher than that of glucose 6-phosphate at equilibrium. e. The concentration of glucose 6-phosphate is equal to that of fructose 6-phosphate at equilibrium.

ANSWER: c 63. Which statement describing enzymes is FALSE? a. All enzymes lower the activation energy of a reaction. b. All enzymes speed up a chemical reaction. c. All enzymes are proteins. d. All enzymes are biological catalysts. e. All enzymes facilitate the formation of a transition state.

ANSWER: c 64. A chemical reaction of substrate S to form product P goes through a transition state (X‡) that has: a. a higher free energy than does either S or P. b. a lower free energy than does either S or P. c. a higher free energy than S but a lower free energy than P. .. Powered by Cognero.

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Chapter 7 d. a lower free energy than S but a higher free energy than P. e. the same free energy as S but a higher free energy than P. ANSWER: a 65. How do enzymes affect the free energy of a reaction? a. They decrease both the free energy of activation (ΔG‡) and the free-energy change (ΔG). b. They increase both the free energy of activation (ΔG‡) and the free-energy change (ΔG). c. They decrease the free energy of activation (ΔG‡) but increase the free-energy change (ΔG). d. They increase the free energy of activation (ΔG‡) but do not affect the free-energy change (ΔG). e. They decrease the free energy of activation (ΔG‡) but do not affect the free-energy change (ΔG).

ANSWER: e 66. In enzymes, the active site is generally a: a. nonpolar microenvironment that may contain polar residues. b. nonpolar microenvironment that only contains hydrophobic residues. c. polar microenvironment that only contains polar residues. d. polar microenvironment that may contain hydrophobic residues. e. polar microenvironment that largely excludes water.

ANSWER: a 67. Which statement comparing the models of enzyme binding is CORRECT? a. The dynamic recognition model has been replaced by the induced fit model. b. The induced fit model has been replaced by the dynamic recognition model. c. The lock-and-key model has been replaced by the induced fit model. d. The induced fit model has been replaced by the lock-and-key model. e. The lock-and-key model and the induced fit model are considered to be equally valid.

ANSWER: c 68. The formation of a large number of weak interactions between a complementary enzyme and substrate releases free energy. The released free energy is called the: a. Gibbs free energy b. binding energy c. activation energy d. standard free energy e. free energy of catalysts ANSWER: b 69. Which compound is a potent inhibitor of proline racemase? a. L-proline b. D-proline c. phenylalanine .. Powered by Cognero.

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Chapter 7 d. cysteine e. pyrrole 2-carboxylic acid ANSWER: e 70. Why is pyrrole 2-carboxylic acid a potent inhibitor of proline racemase? a. It mimics L-proline, preventing L-proline from binding to the active site. b. It mimics D-proline, preventing D-proline from binding to the active site. c. Its binding affinity for the active site is lower than that of the substrate. d. Its trigonal geometry mimics that of the transition state. e. Its tetrahedral geometry mimics that of the transition state.

ANSWER: d

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Chapter 8 1. The velocity of a reaction is the quantity of the: a. reactant that appears in a specified unit of time. b. product that disappears in a specified unit of time. c. solution color that disappears in a specified unit of time. d. reactant that disappears in a specified unit of time. e. solution color that appears in a specified unit of time.

ANSWER: d 2. What is the reaction order if the reaction rate was doubled by doubling the reactant concentration? a. zero order b. first order c. second order d. pseudo-first order e. pseudo-second order ANSWER: b 3. What is the reaction order if the reaction rate was increased by a factor of 4 by doubling the reactant concentration? a. zero order b. first order c. second order d. pseudo-first order e. pseudo-second order ANSWER: c 4. Which rate equation is TRUE if doubling the concentration of reactant A doubles the reaction rate and doubling the concentration of reactant B has no effect on the rate? a. V = k[A] b. V = k[A][B] c. V = k[A]2[B]1 d. V = k[A]2[B] e. V = k[A]2 ANSWER: a 5. Which statement about pseudo-first-order reactions is FALSE? a. Bimolecular reactions can be pseudo-first-order reactions. b. The reaction rate will not appear to depend on the concentration of reactant A, whose concentration greatly exceeds that of reactant B. c. The reaction has a pseudo-first order if the concentration of one reactant greatly exceeds that of the second one and if the second reactant is present at low concentrations. d. The reaction rate will not appear to depend on the concentration of reactant A, whose concentration .. Powered by Cognero.

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Chapter 8 is significantly less than that of reactant B. e. Unimolecular reactions cannot be pseudo-first-order reactions.

ANSWER: d 6. Choose the CORRECT statement concerning the reaction order. a. First-order rate constants have units M–1 s–1. b. Biochemical reactions cannot have a zero order. c. Bimolecular reactions always behave as second-order reactions. d. Second-order reactions are usually unimolecular. e. First-order rate constants have unit s–1.

ANSWER: e 7. Which statement about enzyme kinetics is TRUE? a. Michaelis–Menten enzymes have sophisticated regulatory properties. b. The kinetic description of enzymatic activity is required to understand the structure of enzymes. c. The action of allosteric enzymes increases the metabolic chaos. d. Allosteric enzymes allow the efficient integration of metabolism. e. All enzymes obey Michaelis–Menten kinetics.

ANSWER: d 8. Which characteristics are usually the first ones to be determined when studying enzyme activity? a. Vmax and KM b. V0 and KM c. Vmax and V0 d. [S] and KM e. Vmax and k2 ANSWER: a 9. In the Michaelis–Menten model: a. the reaction equilibrium is attained with time, and the change in the concentration of the product allows measuring of the reaction rate. b. a unique characteristic of the enzyme is investigated when the enzyme concentration is relatively constant and the product formation is negligible. c. enzyme concentration is relatively constant only near the end of the reaction, so only in the beginning of the reaction can the velocity be measured precisely. d. only in the beginning of the reaction, there is no net change in concentrations of S and P, so the reaction velocity can be measured precisely. e. the initial velocity depends on [S] only when the entire enzyme is bound to the substrate but the product has not accumulated yet.

ANSWER: b .. Powered by Cognero.

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Chapter 8 10. The Michaelis constant (KM): a. depends on the concentration of the enzyme. b. is always the same for an enzyme, regardless of the substrate. c. describes the properties of the enzyme-substrate reaction. d. can be attained only when all of the enzyme is bound to substrate. e. is the maximal velocity possible for an enzyme-catalyzed reaction.

ANSWER: c 11. What is the consequence of a high KM value of an enzyme? a. Moderate concentration of the substrate will be enough for maximal enzyme activity. b. This enzyme has a high maximal velocity. c. The enzyme is present in a large concentration in the cell. d. A high concentration of the substrate is required for maximal enzyme activity. e. The velocity will be directly proportional to the substrate concentration.

ANSWER: d 12. Which statement about Vmax is TRUE? a. Vmax is independent of enzyme concentration. b. For Michaelis–Menten enzymes, the maximal velocity is approached asymptotically. c. Vmax can be attained only when half of the enzyme is bound to the substrate. d. The reaction velocity is maximal when the concentration of the substrate equals KM. e. The reaction velocity is half Vmax when the entire enzyme is bound to the substrate.

ANSWER: b 13. Which description of the reciprocal of initial velocity is FALSE? a. When it is half the maximal speed, the substrate concentration equals KM. b. It gives a way to define KM when it is plotted against the reciprocal of the substrate concentration. c. When it is plotted against the reciprocal of the substrate concentration, it allows KM/Vmax to be defined as a slope of the graph. d. It allows definition of the reciprocal of the maximal velocity in a double-reciprocal plot. e. It is part of the Lineweaver–Burk equation.

ANSWER: a 14. Which condition must be met to determine the turnover number of an enzyme? a. The reaction velocity is half of Vmax. b. The enzyme is fully saturated with substrate. c. The substrate concentration is equal to [E]T. d. Half of the enzyme is saturated with substrate. .. Powered by Cognero.

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Chapter 8 e. The reaction velocity is at zero.

ANSWER: b 15. Calculate the turnover of an enzyme if the maximal velocity is 0.2 M sec–1 and the product concentration after the reaction is 1 M. a. 0.2 sec b. 0.2 sec–1 c. 5 sec d. 0.5 sec e. 0.5 sec–1 ANSWER: b 16. An enzyme has specificity constant values of 1.0 × 103, 1.1 × 104, and 4.0 × 10–1 for compounds A, B, and C, respectively. For which compound does the enzyme have a higher efficiency and why? a. Compound C; a low kcat value accounts for the high and specific rate of catalysis, while a high KM value lowers the required concentration of the compound. b. Compound B; a high KM value accounts for the high and specific rate of catalysis, while a low kcat value lowers the required concentration of the compound. c. Compound A; higher kcat accounts for the high and specific rate of catalysis, while a low KM value lowers the required concentration of the compound. d. Compound C; the least amount of the compound is required to saturate the enzyme, while having the highest rate of catalysis. e. Compound B; the least amount of the compound is required to saturate the enzyme, while having the highest rate of catalysis.

ANSWER: e 17. Which constant represents the nature of the enzyme–substrate interaction? a. specificity constant b. constant of the catalysis rate c. Michaelis constant d. rate constant k1 e. allosteric constant ANSWER: c 18. The KM of the cytosolic isomer of fumarase is equal to 5 µM, whereas KM of the mitochondrial isomer is equal to 50 µM. What should the value of k2 be for a less active isomer of the enzyme to achieve kinetic perfection? a. between 5 × 108 and 5 × 109 s⁻1 b. between 2 × 107 and 2 × 108 s⁻1 c. between 1 × 108 and 1 × 109 s⁻1 .. Powered by Cognero.

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Chapter 8 d. between 5 × 102 and 5 × 103 s⁻1 e. between 2 × 106 and 2 × 107 s⁻1 ANSWER: d 19. What ternary complex must form for catalysis to occur by the enzyme (E) that converts pyruvate to lactate using NADH as a cosubstrate? a. E (lactate) (NAD+) b. E (NADH) (pyruvate) c. E (pyruvate) (lactate) d. E (NADH) (NAD+) e. E (lactate) (NADH) ANSWER: b 20. What is the reaction class if all substrates must bind to the enzyme in an arranged manner before any product is released? a. random sequential b. ping-pong c. ordered sequential d. second order e. first order ANSWER: c 21. An example of a reaction of the double-displacement class is the: a. conversion of pyruvate and NADH to lactate and NAD+ by lactate dehydrogenase with formation of a ternary complex. b. conversion of creatine and ATP to phosphocreatine and ADP by creatine kinase with ATP being released before both substrates bind the enzyme. c. conversion of aspartate and α-ketoglutarate to oxaloacetate and glutamate with formation of a ternary complex. d. cleavage of a peptide bond by chymotrypsin with the formation of a substituted intermediate on a serine residue. e. isomerization of dihydroxyacetone phosphate to glyceraldehyde 3-phosphate.

ANSWER: d 22. What is NOT a feature of allosteric enzymes? a. Their activity can be modified by environmental signals. b. They have a sigmoidal dependence of the reaction velocity on the substrate concentration. c. Each allosteric enzyme is capable of conducting multiple reactions. d. Their activity isn't regulated by a threshold effect. e. They depend on alterations in the quaternary structure.

ANSWER: d .. Powered by Cognero.

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Chapter 8 23. Consider the pathways shown, in which en denotes different enzymes and capital letters denote different compounds.

Which regulatory scheme would produce the appropriate amount of K? a. K inhibits e1 and e10 while F inhibits e1, and I inhibits e10. b. F and I inhibit e1 and e10, respectively, and stimulate e10 and e1, respectively, whereas K inhibits e1 and e10. c. F stimulates e10 and I stimulates e1 while K inhibits e1 and e10. d. K stimulates e1 and e10 while F inhibits e1, and I inhibits e10. e. F and I stimulate e1 and e10, respectively, and inhibit e10 and e1, respectively, whereas K inhibits e1 and e10.

ANSWER: b 24. According to the concerted model, it is NOT characteristic for allosteric enzymes to: a. catalyze multiple reactions within one enzyme. b. maintain two distinct conformations of the enzyme that can be spontaneously interconverted. c. bind the substrate more readily while in the more stable conformation. d. follow the symmetry rule by all of the enzyme active sites. e. disrupt the equilibrium between the conformations in favor of the least stable form after substrate binding.

ANSWER: c 25. What is TRUE regarding the action of allosteric effectors? a. Negative effectors inhibit the less stable form of the enzyme. b. Positive effectors stimulate the more stable form of the enzyme. c. Negative effectors stabilize the less stable form of the enzyme. d. Negative effectors inhibit the more stable form of the enzyme. e. Negative effectors stabilize the more stable form of the enzyme.

ANSWER: e 26. What is TRUE for the sequential model for allosteric enzymes? a. The binding of a substrate disrupts the equilibrium between two states of the enzyme. b. The initial molecule of the substrate prevents unproductive colliding between other substrate .. Powered by Cognero.

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Chapter 8 molecules and the tense form of the enzyme. c. The rate constant for the binding of a substrate is lower for the last subunit of the enzyme than for the first one. d. The affinity to the substrate increases due to the influence of neighboring subunits. e. Cooperativity in the sequential model can only be positive.

ANSWER: d 27. Which enzyme of the given metabolic pathway is most likely to be an allosteric enzyme that controls the synthesis of metabolite H?

a. e1 b. e2 c. e3 d. e5 e. e7 ANSWER: e 28. Which enzyme of the given metabolic pathway is most likely to catalyze a committed step for the synthesis of metabolite D?

a. e1 b. e2 c. e3 d. e4 e. e6 ANSWER: c 29. What is the name for allosteric regulation when a molecule blocks the committed step of its own synthesis pathway? a. committed inhibition b. feedback inhibition c. concerted model d. sequential model .. Powered by Cognero.

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Chapter 8 e. cooperativity ANSWER: b 30. Which type of experiment allows simultaneous studying of millions of enzyme molecules? a. ensemble studies b. cooperative studies c. in singulo d. homotropic e. heterotropic ANSWER: a 31. Which type of method allows investigation of molecular heterogeneity? a. homotropic b. ensemble studies c. heterotropic d. in singulo e. cooperative studies ANSWER: d 32. An enzyme will be MOST sensitive to changes in the substrate concentration when: a. [S] is near KM. b. [S] is below KM. c. KM is below Vmax. d. [S] is above KM. e. KM is Vmax/2.

ANSWER: b 33. Judging by the given Lineweaver–Burk plot, identify the concentration of substrate that is required for enzyme A to operate at half of its maximal speed.

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Chapter 8

a. 0.5 b. −0.5 c. −2 d. 2.5 e. 0.4 ANSWER: a 34. Judging by the given Lineweaver–Burk plot, identify the turnover number of enzyme B if its total concentration is 0.2 M.

a. 0.1 b. 10 c. 5 .. Powered by Cognero.

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Chapter 8 d. −1 e. 2.5 ANSWER: b 35. Which description of the concerted model for allosteric enzymes is TRUE? a. At a low substrate concentration, L0 is small. Upon increasing the substrate concentration, in accordance with the symmetry rule, L0 increases. The disrupted equilibrium of the enzyme states results in a sharp increase of the reaction velocity. As the concentration reaches the threshold, the enzyme becomes more active. b. Upon increasing the substrate concentration, due to cooperativity, the number of enzyme molecules in the T state decreases. The equilibrium between enzyme states is disturbed due to a lower stability of the R state. As the concentration reaches the threshold, the enzyme becomes more active. c. At a low substrate concentration, L0 is large. Increasing the substrate concentration will result in accumulation of the enzyme in the R state, which increases L0. The disrupted equilibrium of the enzyme states results in a sharp increase of the reaction velocity. When the substrate concentration is way above KM, the enzyme activity is very sensitive to the changes in the concentration, which results in a sharp increase of the reaction velocity. d. Upon increasing the substrate concentration due to the symmetry rule, the number of enzyme molecules in the T state increases. The equilibrium between enzyme states is disturbed due to a lower stability of the T state. When the substrate concentration is way above KM, the enzyme activity is very sensitive to the changes in the concentration, which results in a sharp increase of the reaction velocity. e. At a low substrate concentration, L0 is large. Upon increasing the substrate concentration, L0 markedly decreases. The disrupted equilibrium of the enzyme states results in a sharp increase of the reaction velocity. As the concentration reaches the threshold, the enzyme becomes more active.

ANSWER: e 36. What is the critical feature of the Michaelis–Menten model of enzyme catalysis? a. increasing the probability of product formation b. shifting the reaction equilibrium c. formation of an ES complex d. reaching the reaction equilibrium e. accumulation of the product ANSWER: c 37. What value of [S] as a fraction of KM is required to obtain 20% Vmax? a. 0.2 KM b. 0.25 KM c. 0.5 KM d. 0.75 KM e. 0.8 KM .. Powered by Cognero.

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Chapter 8 ANSWER: b 38. Allosteric effectors alter the equilibrium between the: a. ES state. b. R and T forms of a protein. c. forward and reverse reaction rate. d. formation of the product and its reverse reaction. e. concentration thresholds.

ANSWER: b 39. A loss of allosteric regulation in the production of purine nucleotides may result in: a. excess nucleotides for DNA. b. a loss of PRS catalytic activity. c. decreased urate degradation. d. a decrease in urate concentration. e. an accumulation of excess urate.

ANSWER: e 40. The formula V0 = Vmax ([S]/([S] + KM)) indicates the relationship between the: a. enzyme activity and the equilibrium constant. b. rate of a catalyzed reaction and the equilibrium constant. c. enzyme activity and substrate concentration. d. probable substrate concentration and the normal concentration in vivo. e. turnover number and the rate of catalysis.

ANSWER: c 41. When the substrate concentration is much greater than KM, the rate of catalysis is almost equal to: a. [S] + KM. b. kcat. c. Vmax. d. Vmax/2. e. (Vmax/KM)[S].

ANSWER: c 42. If the enzyme concentration is 5 nM, the substrate concentration is 5 mM, and KM is 5 μM: a. the enzyme will be fully saturated with the substrate. b. most of the enzyme won't have the substrate bound. c. there will be more enzyme than substrate. d. the enzyme will display first-order kinetics. .. Powered by Cognero.

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Chapter 8 e. the rate of the reaction will be half its maximal value.

ANSWER: a 43. Homotropic effects of allosteric enzymes: a. are due to the effects of substrates. b. are due to the effects of allosteric activators. c. shift the kinetics curve to the left. d. shift the kinetics curve to the right. e. account for the linear nature of the kinetics curve.

ANSWER: a 44. When [S] << KM, the enzymatic velocity depends on: a. the values of kcat/KM, [S], and [E]T. b. Vmax of the reaction. c. the affinity of the substrate for the catalytic site. d. the values of kcat, [S], and [E]T. e. the formation of the ES complex.

ANSWER: a 45. Allosteric effectors: a. can cause large changes in enzymatic activity. b. can lead to a decrease in the availability of a protein. c. don't alter the sensitivity of a metabolic pathway. d. decrease the sensitivity of the enzyme at nearly all concentrations of the substrate. e. alter enzyme activity by binding to the active site of an enzyme.

ANSWER: a 46. Which statement is TRUE when the amount of substrate is far greater than the amount of enzyme present? a. [S] is much less than KM. b. V0 is half of Vmax. c. The enzyme displays second-order kinetics. d. The enzyme displays first-order kinetics. e. The enzyme displays zero-order kinetics.

ANSWER: e 47. During the early stages of an enzyme purification protocol, when cells have been lysed but cytosolic components have not been separated, the reaction velocity-versus-substrate concentration is sigmoidal. As you continue to purify the enzyme, the curve shifts to the right. Explain your results. a. This is an enzyme that displays Michaelis–Menten kinetics and you purify away a homotropic inhibitor. .. Powered by Cognero.

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Chapter 8 b. This is an enzyme that displays Michaelis–Menten kinetics, and you must use a Lineweaver–Burk plot to determine KM and Vmax correctly. c. This is an allosteric enzyme and you must use a Lineweaver–Burk plot to determine KM and Vmax correctly. d. This is an allosteric enzyme and during purification you purify away a heterotropic activator. e. This is an allosteric enzyme displaying a double-displacement mechanism and during purification you purify away one of the substrates.

ANSWER: d 48. Calculate the turnover number for the enzyme if its Mr is 75,000 and if by assaying 5 μg of the enzyme under saturating [S] concentrations, you determine that Vmax is approached asymptotically and is equal to 1.68 μmol s⁻1. a. 2.25 × 106 s⁻1 b. 1.50 × 105 s⁻1 c. 2.50 × 105 s⁻1 d. 2.52 × 104 s⁻1 e. You also need to know KM for this enzyme to calculate the turnover number.

ANSWER: d 49. Which term describes a reaction if doubling the concentration of reactants A and B does not affect the rate? a. zero order b. first order c. second order d. pseudo-first order e. pseudo-second order ANSWER: a 50. Which rate equation describes a zero-order reaction? a. V = k b. V = k[A] c. V = k[B] d. V = k[A][B] e. V = k[A]2[B]2 ANSWER: a 51. Which rate equation BEST describes a pseudo-first-order reaction? a. V = k b. V = k[A] c. V = k[A]2 d. V = k[A][B] .. Powered by Cognero.

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Chapter 8 e. V = k[A]2[B]2 ANSWER: b 52. Which rate constant represents the dissociation of the enzyme–substrate (ES) complex into enzyme (E) and substrate (S)? a. k−2 b. k−1 c. k0 d. k1 e. k2 ANSWER: b 53. Which expression gives the Michaelis constant (KM)? a. (k−2 + k1)/k2 b. (k−1 + k1)/k2 c. (k−2 + k2)/k1 d. (k−1 + k2)/k1 e. (k1 + k2)/k−1 ANSWER: d 54. A Lineweaver–Burk plot displays _____ on the horizontal axis and _____ on the vertical axis. a. V0; [S] b. 1/[S]; 1/V0 c. [S]; V0 d. 1/V0; 1/[S] e. Vmax; KM ANSWER: b 55. Judging by the given Lineweaver–Burk plot, identify the maximal velocity of the reaction catalyzed by

.. Powered by Cognero.

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Chapter 8

enzyme A.

a. −2 b. 0.4 c. 0.5 d. 1.25 e. 2.5 ANSWER: b 56. Judging by the given Lineweaver–Burk plot, identify the value of KM/Vmax for enzyme B.

a. −1 b. −0.5 c. 0.5 d. 1 .. Powered by Cognero.

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Chapter 8 e. 2 ANSWER: c 57. Experimental evidence suggests that substrate concentrations in vivo are _____ the KM value of enzymes. a. approximately equal to b. approximately half c. approximately twice d. several orders of magnitude less than e. several orders of magnitude greater than ANSWER: a 58. The KM value for an enzyme does NOT depend on: a. the particular substrate. b. the pH of the environment. c. the temperature of the environment. d. the ionic strength of the environment. e. the substrate concentration.

ANSWER: e 59. Which term describes an enzyme’s sensitivity to changes in substrate concentration? a. allostericity b. binding affinity c. catalycity d. elasticity e. specificity ANSWER: d 60. At substrate values well above KM, enzymes display: a. the same catalytic activity but are less sensitive to changes in substrate concentration. b. the same catalytic activity but are more sensitive to changes in substrate concentration. c. lower catalytic activity but are more sensitive to changes in substrate concentration. d. greater catalytic activity and are more sensitive to changes in substrate concentration. e. greater catalytic activity but are less sensitive to changes in substrate concentration.

ANSWER: e 61. Which expression gives the specificity constant? a. Vmax/[E]T b. 1/kcat c. kcat/KM d. KM/Vmax e. k−1/k1 .. Powered by Cognero.

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Chapter 8 ANSWER: c 62. The range of KM values for most enzymes is _____, and the range of kcat values for most enzymes is _____. a. 10 to 107 M; 1 to 106 s–1 b. 10–1 to 10–7 M; 10–6 to 1 s–1 c. 10–1 to 10–7 M; 1 to 106 s–1 d. 1 to 106 M; 10–1 to 10–7 s–1 e. 10 to 107 M; 10–6 to 1 s–1 ANSWER: c 63. The enzyme chymotrypsin has a kcat value of 100 s⁻1 and a KM value of 5000 μM with the substrate acetyl-Ltryptophanamide. The enzyme penicillinase has a kcat value of 2000 s⁻1 and a KM value of 50 μM with the substrate benzylpenicillin. How do these two enzymes with their respective substrates compare to one another? a. Both enzymes have attained kinetic perfection. b. Chymotrypsin has attained kinetic perfection, but penicillinase has not. c. Penicillinase has attained kinetic perfection, but chymotrypsin has not. d. Chymotrypsin is more efficient than penicillinase, but neither enzyme has attained kinetic perfection. e. Penicillinase is more efficient than chymotrypsin, but neither enzyme has attained kinetic perfection.

ANSWER: e 64. The enzyme creatine kinase catalyzes the transfer of a phosphate group from ATP to creatine to form phosphocreatine and ADP. In this reaction, either ATP or creatine can bind to the enzyme first. After the reaction occurs, phosphocreatine is released from the enzyme followed by ADP. The reaction catalyzed by creatine kinase is an example of a _____ reaction. a. ordered sequential b. random sequential c. double-displacement d. monosubstrate e. substituted enzyme ANSWER: b 65. Which statement about double-displacement reactions is FALSE? a. One or more products are released before all substrates bind the enzyme. b. The enzyme is temporarily modified during the reaction. c. This type of reaction is used to shuttle amino groups between amino acids and ɑ-ketoacids. d. The defining feature of this type of reaction is a ternary complex. e. This type of reaction starts with two substrates and yields two products.

ANSWER: d 66. Which expression gives the allosteric constant (L0)? a. T/R b. V0/Vmax .. Powered by Cognero.

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Chapter 8 c. kcat/KM d. KM/Vmax e. k−1/k1 ANSWER: a 67. Which statement CORRECTLY describes an allosteric activator? a. It increases the T/R ratio and shifts the sigmoidal kinetics curve to the right. b. It increases the T/R ratio and shifts the sigmoidal kinetics curve to the left. c. It decreases the T/R ratio and shifts the sigmoidal kinetics curve to the right. d. It decreases the T/R ratio and shifts the sigmoidal kinetics curve to the left. e. It does not change the T/R ratio or shift the sigmoidal kinetics curve.

ANSWER: d 68. The glycolytic enzyme phosphofructokinase-1 catalyzes the transfer of a phosphate group from ATP to fructose 6-phosphate (F6P) to form fructose 1,6-bisphosphate (F-1,6-BP) and ADP. This enzyme displays sigmoidal kinetics, and the presence of fructose 2,6-bisphosphate (F-2,6-BP) shifts the sigmoidal curve to the left. What is F-2,6-BP in relation to phosphofructokinase-1? a. a substrate b. a positive homotropic effector c. a negative homotropic effector d. a positive heterotropic effector e. a negative heterotropic effector ANSWER: d 69. Researchers are investigating an enzyme that catalyzes the conversion of compound A to compound B. They find that 60% of the enzyme molecules are in the T state in the absence of compound X, whereas 90% of the enzyme molecules are in the T state in the presence of compound X. What is compound X in relation to this enzyme? a. a substrate b. a positive homotropic effector c. a negative homotropic effector d. a positive heterotropic effector e. a negative heterotropic effector ANSWER: e 70. Ensemble studies of an enzyme reveal a kcat value of 25 s–1. Single-molecule experiments reveal that the enzyme exists in two distinct conformational states with kcat values of 10 s–1 and 30 s–1. Which statement BEST explains these results? a. Half of the enzyme molecules exist in the conformational state with a kcat value of 10 s–1, and the other half exist in the conformational state with a kcat value of 30 s–1. b. A quarter of the enzyme molecules exist in the conformational state with a kcat value of 10 s–1, and the rest exist in the conformational state with a kcat value of 30 s–1. .. Powered by Cognero.

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Chapter 8 c. Three-quarters of the enzyme molecules exist in the conformational state with a kcat value of 10 s–1, and the rest exist in the conformational state with a kcat value of 10 s–1. d. Nearly all the enzyme molecules exist in the conformational state with a kcat value of 25 s–1. e. The results of the ensemble studies are likely erroneous.

ANSWER: b

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Chapter 9 1. What is a term for an electron-rich chemical that is attracted to the regions of positive charge in another molecule? a. ligand b. electrophile c. nucleophile d. inhibitor e. radical ANSWER: c 2. What is the term for an electron-deficient chemical that is attracted to the regions of negative charge in another molecule? a. ligand b. electrophile c. nucleophile d. inhibitor e. radical ANSWER: b 3. Choose the term for the catalytic mechanism that brings two substrates into proximity and in the proper orientation on a single binding surface of an enzyme. a. covalent catalysis b. general acid–base catalysis c. metal ion catalysis d. catalysis by approximation and orientation e. irreversible catalysis ANSWER: d 4. The antibiotic penicillin is an example of a(n) _____ inhibitor. a. irreversible b. competitive c. noncompetitive d. uncompetitive e. reversible ANSWER: a 5. What is the key difference between an enzyme that uses a covalent catalysis mechanism and the other catalytic strategies? a. Enzymes that utilize a covalent catalysis strategy facilitate reactions by donating electrons, whereas the other types facilitate reactions by accepting electrons. b. Enzymes that utilize a covalent catalysis strategy don't use metal ions as a cofactor. c. Enzymes that utilize a covalent catalysis strategy use water as a proton donor or acceptor, whereas the other types use other molecules for that role. .. Powered by Cognero.

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Chapter 9 d. Enzymes that utilize a covalent catalysis strategy facilitate the formation of the intermediate compound, whereas the other types facilitate the formation of the transition state. e. Enzymes that utilize a covalent catalysis strategy covalently bind the substrate to one or more of the amino acids in the active site, whereas the other types bind substrates to the site by noncovalent interactions.

ANSWER: e 6. A(n) _______ inhibitor most often has a structure similar to the substrate and reversibly binds to the active site of the enzyme.

ANSWER: competitive 7. What is the typical form of the curve describing dependence of enzymatic activity on temperature? a. bell-shaped curve with one maximum b. curve of two bells with two peaks c. sigmoidal d. hyperbolic e. growing curve with saturation ANSWER: a 8. What is the typical form of the curve describing pH dependence of enzymatic activity? a. sigmoidal b. hyperbolic c. growing curve with saturation d. bell-shaped curve with one maximum e. curve of two bells with two peaks ANSWER: d 9. What is the molecular basis for the (general) dependence of enzymatic activity on hydrogen ion concentration? a. metal ions b. special cofactors c. ionizable groups d. denaturation e. Brownian motion ANSWER: c 10. _____ is an antibiotic that acts as a competitive inhibitor. a. sulfanilamide b. p-aminobenzoic acid (PABA) c. penicillin d. doxycycline e. diisopropylphosphofluoridate (DIPF) .. Powered by Cognero.

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Chapter 9 ANSWER: a 11. _____ is an antibiotic that functions at low concentrations as a noncompetitive inhibitor. a. sulfanilamide b. p-aminobenzoic acid (PABA) c. penicillin d. doxycycline e. diisopropylphosphofluoridate (DIPF) ANSWER: d 12. What compound is an irreversible group-specific inhibitor? a. sulfanilamide b. p-aminobenzoic acid (PABA) c. penicillin d. doxycycline e. diisopropylphosphofluoridate (DIPF) ANSWER: e 13. What is the kinetic hallmark of reversible competitive inhibition? a. Vmax cannot be attained, even at high substrate concentrations. b. Vmax can be attained at sufficiently high concentrations of substrate. c. The apparent value for KM decreases as more inhibitor is added. d. The apparent value for KM remains unchanged as more inhibitor is added. e. Vmax decreases as more inhibitor is added.

ANSWER: b 14. What is the kinetic hallmark of reversible uncompetitive inhibition? a. Vmax increases as more inhibitor is added. b. The apparent value for KM increases as more inhibitor is added. c. The apparent value for KM decreases as more inhibitor is added. d. The apparent value for KM remains unchanged as more inhibitor is added. e. Vmax remains unchanged as more inhibitor is added.

ANSWER: c 15. What is a useful instrument to determine the reversible inhibition type? Choose the BEST answer. a. Michaelis–Menten equation b. determination of the maximal rate of catalysis (Vmax) upon addition of the inhibitor c. determination of the apparent value for KM upon addition of the inhibitor d. Henderson–Hasselbalch equation .. Powered by Cognero.

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Chapter 9 e. double-reciprocal plots of the rate of catalysis at different concentrations of substrate upon addition of the inhibitor and without it ANSWER: e 16. In a Lineweaver–Burk double-reciprocal plot, the intercept on the y axis is the same in the presence and in the absence of the inhibitor. Choose the correct type of inhibitor. a. irreversible b. competitive c. uncompetitive d. noncompetitive e. The data are not enough to come to a conclusion.

ANSWER: b 17. What type of inhibitor binds irreversibly to the active site of an enzyme? a. suicide b. competitive c. uncompetitive d. pure noncompetitive e. mixed noncompetitive ANSWER: a 18. What is the term used for a substrate analog that is structurally similar to the substrate and chemically modifies a residue in the active site to mark it? a. group-specific reagent b. mechanism-based reagent c. transition-state analog d. affinity label e. suicide inhibitor ANSWER: d 19. What compound is NOT an irreversible inhibitor? a. aspirin b. tosyl-L-phenylalanine chloromethyl ketone c. glyphosate d. diisopropylphosphofluoridate e. penicillin ANSWER: c 20. Choose a compound that acts by covalent modification of the serine residues in proteins. a. aspirin b. tosyl-L-phenylalanine chloromethyl ketone .. Powered by Cognero.

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Chapter 9 c. glyphosate d. diisopropylphosphofluoridate e. p-aminobenzoic acid ANSWER: d 21. Where is the reactive bond of penicillin located? a. side chain b. thiazolidine ring c. pentaglycine bridge d. peptidoglycan e. β-lactam ring ANSWER: e 22. Where is chymotrypsin MOST likely to cleave proteins? a. carboxyl-terminal side of glutamate and aspartate b. carboxyl-terminal side of tryptophan, tyrosine, phenylalanine, methionine, and isoleucine c. amino-terminal side of tryptophan, tyrosine, phenylalanine, methionine, and isoleucine d. amino-terminal side of glutamate and aspartate e. carboxyl-terminal side of lysine, arginine and histidine ANSWER: b 23. What is the form of the dependency graph of 1/V0 versus 1/S? a. logarithmic b. linear c. sigmoidal d. hyperbolic e. bell-shaped ANSWER: b 24. What stabilizes the tetrahedral intermediate of hydrolysis of a peptide bond by chymotrypsin? a. oxyanion hole b. tetrahedral pocket c. catalytic triad d. hydrophobic pocket e. serine 195 ANSWER: a 25. Withdrawal of the proton from the hydroxyl group of serine 195 in chymotrypsin by histidine 57 generates a(n): a. alkoxide ion. b. alkoxide radical. .. Powered by Cognero.

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Chapter 9 c. serine ion. d. oxyanion hole. e. hydroxyl ion.

ANSWER: a 26. What is the function of the aspartate 102 residue in the catalytic triad of chymotrypsin? a. It helps to stabilize the tetrahedral intermediate through hydrogen bonding with it. b. It helps to collapse the tetrahedral intermediate by withdrawal of a proton from it. c. It acts as a donor of a hydroxyl group in the deacylation process. d. It helps to orient the histidine residue and make it a better proton donor through hydrogen bonding and electrostatic effects. e. It helps to orient the histidine residue and make it a better proton acceptor through hydrogen bonding and electrostatic effects.

ANSWER: e 27. How many times is/are the tetrahedral intermediate(s) formed during the complete enzymatic cycle of chymotrypsin? a. 0 b. 1 c. 2 d. 3 e. 4 ANSWER: c 28. What is the reason for using substrate analogs that form colored products when studying kinetics and mechanisms of enzyme action? a. All substrate analogs are enzyme inhibitors. b. Substrate analogs covalently modify reactive groups in active sites of enzymes. c. Substrate analogs can be used as competitive inhibitors in a reaction with natural substrates. d. Formation of colored products can be readily monitored by spectrophotometers. e. Colored intermediates are formed of the substrate analogs.

ANSWER: d 29. When chymotrypsin activity is monitored with a chromogenic substrate, the kinetics within milliseconds show a burst phase and a steady-state phase. What does this tell us about chymotrypsin's mechanism of catalysis? a. The chromogenic substrate is an uncompetitive inhibitor. b. Hydrolysis proceeds in a single step. c. The burst is observed because the first step is more rapid than the second step. d. The cleavage of the substrate does not obey Michaelis–Menten kinetics. e. The chromogenic substrate is not suitable for studying chymotrypsin's mechanism of catalysis. .. Powered by Cognero.

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Chapter 9 ANSWER: c 30. What term is used for a compound that does not necessarily mimic a native substrate but is used to modify specific functional groups of amino acids in enzymes? a. group-specific reagent b. mechanism-based reagent c. transition-state analog d. affinity label e. suicide inhibitor ANSWER: a 31. Which statement about the binding of a substrate to chymotrypsin is FALSE? a. The substrate binds a deep hydrophobic pocket within chymotrypsin. b. The binding region locks the substrate into an appropriate conformation for the active site to do its job. c. The specificity of chymotrypsin depends almost entirely on which carboxylic acid is on the carboxylterminal side of the peptide bond to be cleaved. d. Few polar groups are present on the surface of the binding region. e. The specificity site of chymotrypsin is called the S1 pocket.

ANSWER: c 32. In a Lineweaver–Burk double-reciprocal plot, the slope of the line is the same in the presence and in the absence of the inhibitor. Choose the correct type of inhibitor. a. reversible b. competitive c. uncompetitive d. noncompetitive e. The data are not enough to determine the type.

ANSWER: c 33. In the double-reciprocal plot of the rate of catalysis, the intercept with the x axis is unchanged in the presence and in the absence of the inhibitor. Choose the correct type of inhibitor. a. irreversible b. competitive c. uncompetitive d. noncompetitive e. The data are not enough to determine the type.

ANSWER: e 34. What is the kinetical hallmark of pure noncompetitive inhibition? a. Vmax increases as more inhibitor is added. .. Powered by Cognero.

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Chapter 9 b. The apparent value for KM increases as more inhibitor is added. c. The apparent value for KM decreases as more inhibitor is added. d. The apparent value for KM remains unchanged as more inhibitor is added. e. Vmax remains unchanged as more inhibitor is added.

ANSWER: d 35. Which active site residues are required for chymotrypsin activity? Select all that apply. a. His b. Asn c. Lys d. Ser e. Asp ANSWER: a, d, e 36. Choose a catalytic mechanism where a molecule other than water plays the role of a proton donor or acceptor. a. covalent catalysis b. general acid–base catalysis c. metal ion catalysis d. catalysis by approximation and orientation e. irreversible catalysis ANSWER: b 37. In what type of inhibition can the inhibitor bind only to the ES complex to form an ESI complex? a. competitive b. pure noncompetitive c. irreversible d. uncompetitive e. mixed noncompetitive ANSWER: d 38. Which irreversible inhibitor is also known as a substrate analog? a. group-specific reagent b. affinity label c. suicide inhibitor d. transition-state analog e. competitive inhibitor ANSWER: b 39. An enzyme is optimally active at a neutral pH, but its activity drops off sharply if the pH is changed. What residue(s) is/are most likely in the active site of this enzyme? .. Powered by Cognero.

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Chapter 9 a. side chains of lysine and glutamate b. two histidine amino acid side chains c. a glycine amino acid d. polar side chains e. nonpolar side chains ANSWER: a 40. Choose the catalytic strategy where a cofactor serves as an electrophile to stabilize a negative charge on a reaction intermediate. a. covalent catalysis b. general acid–base catalysis c. metal ion catalysis d. catalysis by approximation and orientation e. irreversible catalysis ANSWER: c 41. In designing a drug to inhibit an enzyme specific to a new pathogenic strain of E. coli, would you choose a group-specific inhibitor or a mechanism-based inhibitor? Why? a. mechanism-based, because it mimics the transition-state intermediate b. mechanism-based, because it modifies a catalytically active group on the enzyme c. group-specific, because it will react to specific R groups in the enzyme d. group-specific, because its activity can be enhanced with an allosteric inhibitor e. group-specific, because it is structurally similar to the enzyme's substrate ANSWER: b 42. What are the two biochemical principles that explain the enzyme activity versus temperature curve? a. The rising portion of the curve is due to an increase in Brownian motion of molecules, and the decrease is due to activation of inhibitor molecules. b. An increase in temperature increases the interactions with allosteric activators, and a decrease in temperature increases the interactions with allosteric inhibitors. c. The rising portion of the curve is due to an increase in Brownian motion of molecules, and the decrease is due to enzyme denaturation. d. The rising portion of the curve is due to an increase in enzyme synthesis, and the decrease is due to reduction in Brownian motion of the molecules. e. The rising portion of the curve is due to an increase in enzyme synthesis, and the decrease is due to activation of inhibitor molecules.

ANSWER: c 43. What determines the specificity of chymotrypsin? a. interaction of the polar residues in the active site with the substrate b. binding of the N-terminus of the substrate to the active site c. covalent binding of a His residue in the active site to the substrate .. Powered by Cognero.

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Chapter 9 d. conformational change upon binding of the substrate e. binding of a hydrophobic side chain into the S1 pocket of the enzyme ANSWER: e 44. Double-reciprocal plots are not used today to determine KM and Vmax for enzymes; however, they appear to be quite useful in studies of inhibition. Why? a. The slope of the line with the inhibitor relative to the slope of the line without the inhibitor gives information about whether the enzyme is regulated allosterically. b. The x axis determines kcat for the reaction with the inhibitor more accurately. c. The slope of the line provides information about the transition-state intermediate. d. The slope of the line with the inhibitor relative to the slope of the line without the inhibitor gives information about the mechanism of inhibition. e. The y axis can be used to determine binding constants for the substrate versus the inhibitor.

ANSWER: d 45. What is the type of catalysis in which the proton donor is NOT water? a. acid–base b. metal ion c. approximation and orientation d. covalent e. noncompetitive ANSWER: a 46. What enzyme participates in pigment formation and has a low-temperature optimum? a. cyclooxygenase b. collagenase c. tyrosinase d. chymotrypsin e. pepsin ANSWER: c 47. What is a protease enzyme with an optimal pH between 1 and 2? a. phosphofructokinase b. collagenase c. tyrosinase d. chymotrypsin e. pepsin ANSWER: e 48. What is NOT a way in which metal ions function catalytically? a. stabilizing a negative charge on a reaction intermediate .. Powered by Cognero.

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Chapter 9 b. increasing the acidity of a nearby molecule c. increasing the binding energy d. generating a nucleophile e. becoming temporarily covalently modified ANSWER: e 49. Which amino acid residue does chymotrypsin use as a general base catalyst? a. aspartate b. serine c. histidine d. phenylalanine e. leucine ANSWER: c 50. How do neutrophiles participate in chemical reactions? a. by donating electrons to electrophiles b. by accepting electrons from electrophiles c. by donating protons to electrophiles d. by accepting protons from electrophiles e. by forming noncovalent bonds with electrophiles ANSWER: a 51. Why are the ears, tail, and legs of Siamese cats darker than the rest of their body? a. Tyrosinase is only active at normal body temperature. b. Tyrosinase is inactive at normal body temperature. c. Tyrosine is only active in acidic environments. d. Tyrosinase is inactive in acidic environments. e. Tyrosinase is inhibited by molecules present in the darker areas.

ANSWER: b 52. In what type of inhibition does the inhibitor bind to an area other than the active site? a. competitive b. noncompetitive c. irreversible d. uncompetitive e. All inhibitors bind the active site.

ANSWER: b 53. What type of inhibition always increases the apparent value of KM? a. competitive b. pure noncompetitive .. Powered by Cognero.

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Chapter 9 c. mixed noncompetitive d. uncompetitive e. irreversible ANSWER: a 54. Which statement about mixed noncompetitive inhibition is TRUE? a. The inhibitor binds the enzyme–substrate complex but not the free enzyme. b. The inhibitor binds equally well to the free enzyme and enzyme–substrate complex. c. The inhibitor binds preferentially to either the free enzyme or enzyme–substrate complex, but still binds both. d. The inhibitor binds the free enzyme but not the enzyme–substrate complex. e. The inhibitor binds the substrate but not the free enzyme or enzyme–substrate complex.

ANSWER: c 55. Which statement about pure noncompetitive inhibition is TRUE? a. The inhibitor binds the enzyme–substrate complex but not the free enzyme. b. The inhibitor binds equally well to the free enzyme and enzyme–substrate complex. c. The inhibitor binds preferentially to either the free enzyme or enzyme–substrate complex, but still binds both. d. The inhibitor binds the free enzyme but not the enzyme–substrate complex. e. The inhibitor binds the substrate but not the free enzyme or enzyme–substrate complex.

ANSWER: b 56. How do the kinetic effects of pure noncompetitive inhibition compare with those of mixed noncompetitive inhibition? a. Pure noncompetitive inhibition decreases the apparent value of KM, whereas mixed noncompetitive inhibition increases the apparent value of KM. b. Pure noncompetitive inhibition increases the apparent value of KM, whereas mixed noncompetitive inhibition decreases the apparent value of KM. c. Pure noncompetitive inhibition does not change the apparent value of KM, whereas mixed noncompetitive inhibition increases or decreases the apparent value of KM. d. Pure noncompetitive inhibition increases or decreases the apparent value of KM, whereas mixed noncompetitive inhibition does not change the apparent value of KM. e. Neither pure nor mixed noncompetitive inhibition changes the apparent value of KM. ANSWER: c 57. How are the kinetic effects of competitive inhibition unique? a. Vmax only increases with competitive inhibition. b. Vmax only remains unchanged with competitive inhibition. c. Vmax only decreases with competitive inhibition. d. The apparent value of KM only remains unchanged with competitive inhibition. e. The apparent value of KM only decreases with competitive inhibition. .. Powered by Cognero.

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Chapter 9 ANSWER: b 58. With respect to the observed changes to the apparent values of KM and Vmax, all irreversible inhibitors give rise to the same pattern as a _____ inhibitor. a. competitive b. pure noncompetitive c. mixed noncompetitive d. pure or mixed noncompetitive e. uncompetitive ANSWER: b 59. How does the antibiotic penicillin kill bacteria? a. by inhibiting the synthesis of folic acid b. by inhibiting protein synthesis c. by inhibiting DNA replication d. by disrupting the cell membrane e. by inhibiting cell wall synthesis ANSWER: e 60. Penicillin mimics the _____ moiety of the normal substrate for glycopeptide transpeptidase. a. D-Ala D-Ala b. L-Ala L-Ala c. D-Gly D-Gly d. L-Ser L-Ser e. D-Ser D-Ser ANSWER: a 61. What enzyme inhibitor prevents folic acid synthesis in bacteria? a. penicillin b. p-aminobenzoic acid (PABA) c. diisopropylphosphofluoridate (DIPF) d. doxycycline e. sulfanilamide ANSWER: e 62. Which of these inhibitor types reversibly inhibits enzymes? a. group-specific reagents b. substrate analogs c. mechanism-based reagents d. transition state analogs e. affinity labels .. Powered by Cognero.

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Chapter 9 ANSWER: d 63. Which irreversible inhibitor is also known as a suicide inhibitor? a. group-specific reagent b. affinity label c. mechanism-based reagent d. substrate analog e. transition state analog ANSWER: c 64. Which of these amino acid residues is MOST likely to be targeted by chymotrypsin for peptide bond hydrolysis? a. serine b. glutamate c. glycine d. lysine e. leucine ANSWER: e 65. Tosyl-L-phenylalanine chloromethyl ketone (TPCK) binds to chymotrypsin as a(n): a. affinity label. b. group-specific reagent. c. transition state analog. d. mechanism-based inhibitor. e. suicide inhibitor.

ANSWER: a 66. Which functional group gives rise to the oxyanion hole in the chymotrypsin mechanism? a. the imidazole group of the histidine 57 b. the carboxyl group of aspartate 102 c. the hydroxyl group of serine 195 d. the amino group of the target peptide bond e. the carbonyl group of the target peptide bond ANSWER: e 67. Which functional group acts as a general acid catalyst in the chymotrypsin mechanism? a. the imidazole group of the histidine 57 b. the carboxyl group of aspartate 102 c. the hydroxyl group of serine 195 d. the amino group of the target peptide bond e. the carbonyl group of the target peptide bond .. Powered by Cognero.

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Chapter 9 ANSWER: a 68. Which sequence correctly orders these five steps in the mechanism of peptide bond hydrolysis?

1. water binding 2. nucleophilic attack on the peptide bond carbonyl group 3. nucleophilic attack on the acyl-enzyme intermediate 4. release of the amine component 5. release of the carboxylic acid component a. 1, 2, 4, 3, 5 b. 1, 3, 5, 2, 4 c. 2, 5, 1, 3, 4 d. 2, 4, 1, 3, 5 e. 3, 4, 1, 2, 5 ANSWER: d 69. How does sarin exert its toxic effects? a. It prevents the breakdown of acetylcholine in nerve synapses by binding acetylcholinesterase as a substrate and being converted into a chemically reactive intermediate. b. It prevents the synthesis of the neurotransmitter acetylcholine by binding acetylcholinesterase as a substrate and being converted into a chemically reactive intermediate. c. It prevents the breakdown of acetylcholine in nerve synapses by covalently modifying a residue in the active site of acetylcholinesterase. d. It prevents the synthesis of the neurotransmitter acetylcholine by covalently modifying a residue in the active site of acetylcholinesterase. e. It prevents the breakdown of acetylcholine in nerve synapses by acting as a transition state analog of acetylcholinesterase.

ANSWER: c 70. How does atropine mitigate the symptoms of sarin poisoning? a. by preventing sarin from permanently inhibiting acetylcholinesterase b. by preventing acetylcholine from binding to the acetylcholine receptor c. by binding to and activating the acetylcholine receptor d. by stimulating the breakdown of acetylcholine in nerve synapses e. by covalently modifying the active site of acetylcholinesterase ANSWER: b

.. Powered by Cognero.

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Chapter 10 1. (CH2O)n is the basic empirical formula for _______.

ANSWER: carbohydrates 2. What are the smallest carbohydrates? Select all that apply. a. aldose b. altrose c. dihydroxyacetone d. ketose e. triose ANSWER: c, e 3. The simplest polysaccharide is: a. a monosaccharide. b. a disaccharide. c. UDP. d. maltose. e. acetylglucosamine.

ANSWER: b 4. What is one of the sweetest chemicals known? a. β-D-glucose b. D-ribose c. D-deoxyribose d. β-D-fructopyranose e. β-D-fructofuranose ANSWER: d 5. What provides carbohydrates with the ability to form a vast array of three-dimensional structures? a. They function in parallel with polysaccharides. b. They are just as amino acids are for proteins. c. The nature of covalent bonds of monosaccharides varies highly. d. A large variety of monosaccharides can be linked together in many different ways. e. Monosaccharides are basically hydrates of carbon.

ANSWER: d 6. A tetrose has asymmetric carbon atoms in the D and L configuration in the second and third positions, respectively. What is the respective configuration of the tetrose? a. D b. L c. DL d. α .. Powered by Cognero.

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Chapter 10 e. β ANSWER: b 7. How many asymmetric carbon atoms are there in the open-chain aldoheptose molecule? a. 4 b. 5 c. 6 d. 7 e. 8 ANSWER: b 8. What is the chemical basis for cyclization of monosaccharides? a. The hydroxyl group farthest from the carbonyl carbon attacks the hydroxyl group closest to the carbonyl carbon. b. The carbonyl group attacks the hydroxyl group closest to the carbonyl carbon. c. The carbonyl group attacks the hydroxyl group farthest from the carbonyl carbon. d. The hydroxyl group closest to the carbonyl carbon attacks the carbonyl carbon. e. The hydroxyl group farthest from the carbonyl carbon attacks the carbonyl carbon.

ANSWER: e 9. Which carbon atoms are involved in the formation of α-D-fructofuranose from the open-chain form of this monosaccharide? a. C-1 aldehyde group and C-5 hydroxyl group b. C-2 aldehyde group and C-6 hydroxyl group c. C-1 keto group and C-5 hydroxyl group d. C-2 keto group and C-6 hydroxyl group e. C-2 keto group and C-5 hydroxyl group ANSWER: e 10. Which carbon atoms are involved in the formation of α-D-fructopyranose from the open-chain form of this monosaccharide? a. C-1 aldehyde group and C-5 hydroxyl group b. C-2 aldehyde group and C-6 hydroxyl group c. C-1 keto group and C-5 hydroxyl group d. C-2 keto group and C-6 hydroxyl group e. C-2 keto group and C-5 hydroxyl group ANSWER: d 11. How many asymmetric carbon atoms are there in the β-D-glucopyranose molecule? a. 2 b. 3 .. Powered by Cognero.

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Chapter 10 c. 4 d. 5 e. 6 ANSWER: d 12. Which carbon atom will always be outside of the ring in the cyclized form of a ketohexose? a. C-1 b. C-2 c. C-β d. C-5 e. C-6 ANSWER: a 13. Which carbon atom of an aldohexose CANNOT be a part of a pyranose ring? a. C-1 b. C-2 c. C-β d. C-5 e. C-6 ANSWER: e 14. Which carbon atom is anomeric in the furanose form of fructose? a. C-1 b. C-2 c. C-β d. C-5 e. C-6 ANSWER: b 15. In cyclic carbohydrates: a. steric hindrance between hydroxyl groups makes the "boat" conformation more stable than the "chair" conformation. b. hydrogen atoms provide less steric hindrance than hydroxyl groups in the axial positions. c. hydroxyl groups in the equatorial positions provide more steric hindrance than hydroxyl groups in axial positions. d. axial bonds provide more steric hindrance for functional groups since they are nearly parallel to the plane of the ring. e. axial bonds provide less steric hindrance for functional groups since they are nearly perpendicular to the plane of the ring.

ANSWER: b .. Powered by Cognero.

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Chapter 10 16. Choose the FALSE statement about ketoses. a. The anomeric carbon exists only in the cyclic form of the monosaccharide such as fructose. b. The pyranose form of fructose predominates in the free solution. c. The C-1 keto group in the open-chain form of some ketoses can form an intramolecular hemiketal. d. The positioning of the keto group in the chain leaves ketoses with one fewer asymmetric center. e. Dihydroxyacetone is one of the smallest ketoses.

ANSWER: c 17. Choose the TRUE statement about glycoproteins. a. About 80% of the simple glycoprotein molecule can be a carbohydrate. b. In proteoglycans, carbohydrate components make up about 95% of the molecule by weight. c. Half of the proteome is proteoglycans. d. In mucins, moieties of N-acetylgalactosamine can make up about 40% of the molecule by weight. e. Only 1%–2% of the proteome is glycosylated.

ANSWER: b 18. Reducing sugars are called so due to a(n): a. ability of the cyclic form of the carbohydrate to reduce substances. b. ability of the open-chain form of the carbohydrate to oxidize cupric ions. c. ability of the open-chain form of the carbohydrate to be oxidized. d. ability of the cyclic form of the carbohydrate to be oxidized. e. ability of the open-chain form of the carbohydrate to be reduced.

ANSWER: c 19. Which bonds link monosaccharides in oligosaccharides? a. N-glycosidic bonds b. O-glycosidic bonds c. ester bonds d. N-linkages e. noncovalent interactions ANSWER: b 20. What are the products of the reaction of invertase? a. α-D-glucopyranose and β-D-glucopyranose b. α-D-glucopyranose and β-L-fructopyranose c. β-D-glucopyranose and α-L-galactopyranose d. α-D-glucopyranose and β-D-fructofuranose e. β-D-fructofuranose and α-L-glucopyranose ANSWER: d 21. Which disaccharide comes from the hydrolysis of starch? .. Powered by Cognero.

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Chapter 10 a. maltase b. sucrose c. maltose d. amylose e. lactose ANSWER: c 22. What is NOT a homopolymeric polysaccharide? a. amylose b. amylopectin c. glycogen d. aggrecan e. cellulose ANSWER: d 23. What type of glycosidic bond is there in this disaccharide?

a. β-1,5 linkage b. α-1,4 linkage c. β-1,3 linkage d. α-1,3 linkage e. β-1,4 linkage ANSWER: e 24. What statement about polysaccharides is TRUE? a. The hollow helix is formed when monosaccharides are N-linked. b. Soluble O-linked polysaccharides can improve the absorption of nutrients. c. The α-1,4 linkages favor straight structures, which are more suitable for storage. d. Insoluble β-linked polysaccharides can improve the absorption of nutrients. e. Pectin can increase the rate at which digestion products pass through the large intestine.

ANSWER: b 25. During the synthesis of sucrose by a glycosyltransferase, the oxygen atom of the glycosidic bond comes from the hydroxyl group on the: a. first carbon atom of α-D-glucopyranose since in UDP-glucose, UDP is connected to this oxygen. b. second carbon atom of α-D-fructofuranose since in UDP-fructose, UDP is connected to this oxygen. c. second carbon atom of β-D-fructofuranose since oxygen of α-D-glucopyranose is left as part of UDP .. Powered by Cognero.

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Chapter 10 after the transfer. d. fifth carbon atom of β-D-fructofuranose since oxygen of α-D-glucopyranose is left as part of UDP after the transfer. e. second carbon atom of α-D-fructofuranose since oxygen of β-D-glucopyranose is left as part of UDP after the transfer.

ANSWER: c 26. How many nucleotides are required for a glycosyltransferase to synthesize a disaccharide? a. 1 b. 2 c. 3 d. 0 e. 1%–2% of the total nucleotides ANSWER: a 27. Which statement is FALSE about N-linked oligosaccharides? a. They contain N-acetylglucosamine. b. They have a trisaccharide core. c. They contain several mannoses. d. They have a core that can bind additional sugars. e. Sugars are attached to the side chain of asparagine.

ANSWER: b 28. In nutrient sensing, what does the acetyl group of GlcNAc indicate? a. carbohydrate availability b. fatty acid availability c. protein availability d. diabetes e. oligosaccharide structure ANSWER: b 29. What is the minimal number of negatively charged carboxylate or sulfate groups in one unit of glycosaminoglycan? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: a 30. What is the cause of I-cell disease (mucolipidosis II)? .. Powered by Cognero.

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Chapter 10 a. inability to degrade amino sugars b. overproduction of carbohydrates c. inability to degrade glycosaminoglycans d. dysregulation of GlcNAc transferase e. inability to synthesize amino sugars ANSWER: c 31. Hyaluronate in the cartilage acts as a: a. repeating unit of the aggrecan molecule. b. central filament for the aggrecan molecule. c. monomer of chondroitin sulfate. d. globular domain of the aggrecan molecule. e. central filament between glycosaminoglycans.

ANSWER: b 32. What property is the basis for the proteoglycans of the cartilage to function as a shock absorber? a. reversible nature of the bonds between negatively charged glycosaminoglycans and water b. reversible nature of the glycosidic bonds within proteoglycans c. irreversible nature of the glycosidic bonds within proteoglycans d. irreversible nature of the bonds between glycosaminoglycans and water e. reversible nature of the bonds between negatively charged carboxylate or sulfate groups and the proteins ANSWER: a 33. Which class of glycoproteins is characterized by the extensive O-glycosylation by N-acetylgalactosamine? a. chitins b. proteoglycans c. mucins d. simple glycoproteins e. glycosaminoglycans ANSWER: c 34. Which carbohydrate defines the B blood type? a. galactose linked to another galactose by an α-1,2 glycosidic bond b. N-acetylgalactosamine linked to galactose by an α-1,2 glycosidic bond c. N-acetylglucosamine linked to another N-acetylglucosamine by an α-1,3 glycosidic bond d. galactose linked to another galactose by an α-1,3 glycosidic bond e. N-acetylgalactosamine linked to galactose by an α-1,3 glycosidic bond ANSWER: d 35. What is the main function of lectins? .. Powered by Cognero.

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Chapter 10 a. inflammatory response b. glycosylation c. protein transport from the Golgi complex d. lubricants e. cellular interactions ANSWER: e 36. What lectin selectively binds to the endothelium? a. L-selectin b. E-selectin c. P-selectin d. hemagglutinin e. erythropoietin ANSWER: b 37. Which statement regarding isomeric forms of carbohydrates is TRUE? a. Enantiomers of glyceraldehyde differ in the configuration of two asymmetric centers. b. Epimers appear due to a new asymmetric carbon that forms upon cyclization of a carbohydrate. c. The existence of the stereoisomers does not preclude the existence of constitutional isomers for a particular carbohydrate. d. Diastereoisomers differ in the same asymmetric carbon. e. Among glucose anomers, the α form is the most abundant.

ANSWER: c 38. What is an example of a glycosaminoglycan? a. UDP b. glycogen c. cellulose d. lectins e. heparin ANSWER: e 39. Select all that apply. Carbohydrates are _____ with two or more hydroxyl groups. a. aldehydes b. ketones c. acids d. alcohols e. phenols ANSWER: a, b 40. Which amino acid residues in glycoproteins are the sugars commonly linked to? .. Powered by Cognero.

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Chapter 10 a. tyrosine and asparagine b. serine, threonine, and asparagine c. serine, tyrosine, and asparagine d. cysteine, threonine, and asparagine e. tyrosine, serine, threonine, and asparagine ANSWER: b 41. The anomer of α-D-glucopyranose is a.

b.

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d.

e.

ANSWER: c 42. Inhibitors against which viral enzyme have potential as anti-influenza agents? Select all that apply. a. calnexin b. neuraminidase c. selectin d. glycosyltransferase e. sialidase ANSWER: b, e .. Powered by Cognero.

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Chapter 10 43. All of the following are repeating units of glycosaminoglycans EXCEPT: a. chondroitin 6-sulfate. b. keratin sulfate. c. hyaluronate. d. heparin. e. hemoglobin.

ANSWER: e 44. Which enzyme digests amylopectin? a. α-amylase b. amylose c. cellulose d. lactase e. invertase ANSWER: a 45. An aldehyde and alcohol can react to form a(n): a. hemialkyl. b. hemiketal. c. hemiacetal. d. amidyl. e. carbonyl.

ANSWER: c 46. Select all that apply. The simplest carbohydrates are: a. D-glyceraldehyde. b. glycerate. c. dihydroxyacetone. d. L-glyceraldehyde. e. D-mannose.

ANSWER: a, c, d 47. Select all that apply. Glycoproteins normally: a. are found on membranes. b. are secreted as extracellular proteins. c. are found inside organelles. d. are found inside the nucleus. e. function as structural components.

ANSWER: a, b, e 48. Select all that apply. The nutritional storage forms of glucose in plants are: .. Powered by Cognero.

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Chapter 10 a. glycogen. b. amylose. c. amylopectin. d. starch. e. cellulose.

ANSWER: b, c, d 49. Some viruses gain entry into specific cells by: a. attaching to ion channels. b. cleaving the glycosidic bonds and altering protein shapes. c. binding to specific glycoproteins on the cell surface. d. adding moieties of sialic acids to membrane proteins. e. changing the permeability of membranes.

ANSWER: c 50. Select all that apply. Lectins are proteins that: a. bind carbohydrates on glycoproteins and other macromolecules. b. promote cell–cell interaction. c. are found in animals and plants. d. exploit noncovalent interactions for binding carbohydrates. e. function as lubricants.

ANSWER: a, b, c, d 51. Glyceraldehyde is a(n) _____, whereas dihydroxyacetone is a(n) _____. a. ketose; aldose b. aldose; ketose c. triose; hexose d. hexose; triose e. sugar; sugar derivative ANSWER: b 52. What is a pyranose? a. a sugar with a five-membered ring consisting of four carbon atoms and one oxygen atom b. a sugar with a five-membered ring consisting of five carbon atoms c. a sugar with a six-membered ring consisting of five carbon atoms and one oxygen atom d. a sugar with a six-membered ring consisting of six carbon atoms e. a six-carbon open-chain sugar ANSWER: c 53. Which carbon atom is anomeric in glucopyranose? a. C-1 .. Powered by Cognero.

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Chapter 10 b. C-2 c. C-3 d. C-5 e. C-6 ANSWER: a 54. Heating high-fructose corn syrup converts: a. β-D-fructopyranose into β-D-glucopyranose. b. β-D-fructofuranose into ɑ-D-fructofuranose. c. β-D-fructopyranose into ɑ-D-fructopyranose. d. β-D-fructofuranose into β-D-fructopyranose. e. β-D-fructopyranose into β-D-fructofuranose.

ANSWER: e 55. Which statement about D-glucose is FALSE? a. It is commonly called blood sugar. b. It is the only fuel used by red blood cells. c. It is more reactive than most other monosaccharides. d. It is the only fuel the brain uses under non-starvation conditions. e. It is formed from formaldehyde under prebiotic conditions.

ANSWER: c 56. Which statement about hemoglobin A1c is FALSE? a. It is a widely used biomarker. b. It is a glycosylated form of hemoglobin. c. Its level in the blood depends on blood glucose levels over a long period of time. d. It is present at elevated levels in poorly treated or untreated diabetes mellitus. e. It is a product of the nonenzymatic addition of glucose to hemoglobin.

ANSWER: b 57. Which statement is FALSE about sugar phosphorylation? a. It prevents sugars from spontaneously leaving the cell. b. It prevents sugars from interacting with transporters of the unmodified sugar. c. It facilitates the interconversion of pyranose and furanose forms. d. It facilitates sugar metabolism. e. It makes the sugars anionic.

ANSWER: c 58. What is the composition of the disaccharide maltose? a. two D-glucose residues linked by an α-1,4-glycosidic bond b. ɑ-glucose and β-fructose linked by an α-1,4-glycosidic bond .. Powered by Cognero.

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Chapter 10 c. two D-glucose residues linked by an α-1,6-glycosidic bond d. ɑ-glucose and β-fructose linked by a β-1,4-glycosidic bond e. two D-glucose residues linked by a β-1,4-glycosidic bond ANSWER: a 59. Most of the glucose units in glycogen are connected by _____-glycosidic linkages. The branches are formed by _____-glycosidic linkages. a. α-1,4; β-1,4 b. β-1,4; α-1,4 c. β-1,4; β-1,6 d. α-1,6; α-1,4 e. α-1,4; α-1,6 ANSWER: e 60. The drugs acarbose (Precose) and miglitol (Glyset) treat type 2 diabetes by: a. competitively inhibiting ɑ-amylase. b. promoting the release of the hormone insulin. c. competitively inhibiting ɑ-glucosidase. d. promoting the storage of glucose as glycogen. e. competitively inhibiting β-galactosidase.

ANSWER: c 61. Which carbohydrate is one of the most abundant organic compounds in the biosphere? a. glycogen b. amylose c. cellulose d. chitin e. pectin ANSWER: c 62. Choose the CORRECT statement about the molecular architecture of cellulose. a. The ɑ linkages produce a hollow helix. b. The β linkages produce a hollow helix. c. The ɑ linkages produce a branched polymer. d. The β linkages produce a straight chain. e. The ɑ linkages produce a straight chain.

ANSWER: d 63. Which class(es) of glycoproteins are predominantly carbohydrate by weight? a. glycoproteins and proteoglycans b. proteoglycans and mucins .. Powered by Cognero.

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Chapter 10 c. only glycoproteins d. glycoproteins and mucins e. only mucins ANSWER: b 64. Which amino acids can be linked to carbohydrates by an O-glycosidic linkage? a. asparagine and serine b. serine and threonine c. asparagine and threonine d. asparagine, serine, and threonine e. cysteine, serine, threonine ANSWER: b 65. Which marker directs proteins from the Golgi complex to lysosomes? a. mannose 6-phosphate b. N-acetylgalactosamine c. glucosamine d. galactosamine e. N-acetylglucosamine ANSWER: a 66. Which statement about erythropoietin (EPO) is FALSE? a. EPO stimulates the production of red blood cells by bone marrow. b. Some athletes have used recombinant EPO to improve their endurance. c. The glycosylated form of EPO is rapidly removed from the blood by the kidney. d. EPO is 40% carbohydrate by weight. e. Recombinant EPO is used to treat anemias.

ANSWER: c 67. The variable number of tandem repeats (VNTR) regions of mucins are rich in: a. O-glycosylated asparagine residues. b. N-glycosylated asparagine residues. c. O-glycosylated cysteine residues. d. N-glycosylated serine and threonine residues. e. O-glycosylated serine and threonine residues.

ANSWER: e 68. A blood donor produces a transferase that adds N-acetylgalactosamine to the O antigen but lacks a transferase that adds galactose to the O antigen. The blood from this donor can be safely transfused into people with which blood types? a. only type A and type O .. Powered by Cognero.

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Chapter 10 b. only type B and type O c. only type A and type AB d. only type B and type AB e. only type A and type B ANSWER: c 69. Choose the statement that CORRECTLY relates cholera to blood type. a. Individuals with type O blood are less likely to contract cholera. b. Individuals with type O blood are more likely to contract cholera. c. Individuals with type O blood who contract cholera are less likely to have severe disease. d. Individuals with type O blood who contract cholera are more likely to have severe disease. e. Individuals with type O blood are essentially immune to cholera.

ANSWER: d 70. Which molecular interaction allows an embryo to attach to the endometrium of the mother’s uterus? a. Oligosaccharides produced by the embryo bind L-selectin on the cell surface of endometrial cells. b. L-Selectin produced by the embryo binds oligosaccharides on the cell surface of endometrial cells. c. The mannose 6-phosphate receptor produced by the embryo binds mannose 6-phosphate on the cell surface of endometrial cells. d. Oligosaccharides produced by the embryo bind E-selectin on the cell surface of endometrial cells. e. E-Selectin produced by the embryo binds oligosaccharides on the cell surface of endometrial cells.

ANSWER: b

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Chapter 11 1. Which membrane lipid is composed of sphingosine, fatty acid, and a simple sugar? a. sphingoside b. cerebroside c. sphingolipid d. sphingomyelin e. cardiolipin ANSWER: b 2. Which lipids contain a branched chain of seven sugar residues? a. gangliosides b. sphingolipids c. glycolipid d. cerebrosides e. sugar-containing lipids ANSWER: a 3. What is an alcohol component in cerebroside? a. glycerol b. ethanolamine c. inositol d. sphingosine e. choline ANSWER: d 4. What is the common name of hexadecanoic acid? a. oleic acid b. stearic acid c. palmitic acid d. linoleic acid e. linolenic acid ANSWER: c 5. Which of these fatty acids has the shorthand notation 18:3? a. oleic acid b. stearic acid c. palmitic acid d. linoleic acid e. linolenic acid ANSWER: e 6. The systematic name of linoleic acid is _____ acid. .. Powered by Cognero.

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Chapter 11 a. n-hexadecanoic b. n-octadecanoic c. cis-Δ9-octadecenoic d. cis-cis-Δ9,Δ12-octadecadienoic e. all-cis-Δ9,Δ12,Δ15-octadecatrienoic ANSWER: d 7. Counting from the carboxyl terminus of a fatty acid, which carbon atom is often referred to as β? a. 1 b. 2 c. 3 d. 4 e. the last carbon atom in the chain ANSWER: c 8. What lipids are referred to as polycyclic hydrocarbons? a. phospholipids b. steroids c. sphingolipids d. triacylglycerols with highly unsaturated fatty acids e. glycolipids ANSWER: b 9. Choose the compound that is composed of a chain of hydrogen-bearing carbon atoms terminated with a carboxyl group. a. fatty acid b. triacylglycerol c. glycolipid d. phospholipid e. steroid ANSWER: a 10. Choose the notation for the double bond position and configuration in natural oleic acid. a. trans-Δ2 b. cis-Δ2 c. cis, cis-Δ9,Δ12 d. trans-Δ9 e. cis-Δ9 ANSWER: e .. Powered by Cognero.

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Chapter 11 11. In what form does a fatty acid exist at physiological pH? a. anion b. cation c. protonated molecule d. acid e. zwitterion ANSWER: a 12. The configuration of most fatty acids in biological systems is predominantly: a. cis. b. trans. c. all-trans. d. Δ12. e. ω-3.

ANSWER: a 13. A(n) _____ in the chain length will result in the formation of _____ van der Waals interactions between the chains and a(n) _____ of the melting point. a. decrease; more; increase b. increase; more; increase c. increase; less; decrease d. decrease; more; decrease e. increase; less; increase ANSWER: b 14. What linkages join fatty acids to the glycerol backbone in phosphoglycerides? a. ether linkages b. thioether linkages c. phosphate linkages d. methylene linkages e. ester linkages ANSWER: e 15. What feature do all ⍵-3 fatty acids have in common? a. a methyl group at the carboxyl end of the fatty acid b. a methyl group on the third carbon of the chain c. a double bond between carbon atoms 2 and 3 d. a triple bond at the carboxyl end of the fatty acid e. several double bonds in the chain ANSWER: e .. Powered by Cognero.

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Chapter 11 16. What is a term for hydrophobic groups that are covalently attached to proteins and tether them to the membrane? a. lipid linkages b. membrane modifications c. membrane linkages d. membrane anchors e. membrane bonds ANSWER: d 17. The glycosylphosphatidylinositol (GPI) anchor is attached to _____ of a membrane protein. a. a cysteine residue b. a cysteine residue at the carboxyl terminus c. the carboxyl terminus d. the amino terminus e. a serine residue ANSWER: c 18. A precursor to estradiol is: a. sitosterol. b. eicosapentaenoate. c. cholesterol. d. cardiolipin. e. arachidonate.

ANSWER: c 19. How many rings are in the steroid nucleus? a. 0 b. 1 c. 2 d. 3 e. 4 ANSWER: e 20. Which organisms have ether linkages rather than ester linkages in their membrane lipids? a. fungi b. archaea c. bacteria d. mycoplasma e. eukaryotes ANSWER: b .. Powered by Cognero.

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Chapter 11 21. What is the main storage form of fatty acids? a. micelles b. liposomes c. triacylglycerols d. phospholipids e. glycolipids ANSWER: c 22. What is the major site of accumulation of triacylglycerols in mammals? a. adipose tissue b. liver c. mitochondria d. endoplasmic reticulum e. muscle tissue ANSWER: a 23. What compounds are derived from the treatment of triacylglycerols with strong bases? a. salts of triacylglycerol b. diacylglycerols c. phosphatidic acids d. salts of fatty acids e. free sphingosines ANSWER: d 24. Which organisms do NOT have sterols in their membranes? a. bacteria b. fungi c. plants d. insects e. protozoa ANSWER: a 25. The glycosylphosphatidylinositol (GPI) anchor does NOT contain a(n): a. double bond. b. amino group. c. cyclopentane ring. d. ester linkage. e. mannose.

ANSWER: c 26. Derivatives of cholesterol with an alcohol functional group are: .. Powered by Cognero.

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Chapter 11 a. steroids. b. sterols. c. hopanoids. d. hopanols. e. sterones.

ANSWER: b 27. What is one of the defects that lead to Hutchinson–Gilford progeria syndrome? a. The farnesyl group is not attached to the lamin A precursor. b. The farnesyl group is not removed from the lamin A precursor. c. The glycosylphosphatidylinositol anchor is not attached to the lamin A precursor. d. The glycosylphosphatidylinositol anchor is not removed from the lamin A precursor. e. Lamin A is subject to hypermethylation.

ANSWER: b 28. The presence of cis double bonds in fatty acids limits tight packaging and the number of _____ interactions. a. hydrophobic b. ionic. c. nuclear d. hydrogen-bonding e. van der Waals ANSWER: e 29. Which phospholipid is abundant in the myelin sheath of neural cells? a. phosphatidic acid b. diacylglycerol c. diphophatidylglycerol d. sphingomyelin e. phosphatidylinositol ANSWER: d 30. Unsaturation in fatty acids _______ the melting points of lipids.

ANSWER: decreases 31. What corresponds to the polar head group of phosphatidylcholine? a. choline b. phosphorylcholine c. phosphate d. glycerophosphocholine e. glycerol ANSWER: b .. Powered by Cognero.

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Chapter 11 32. Phosphatidylethanolamine does NOT contain a(n) _____ group. a. amino b. methylene c. carboxyl d. carbonyl e. methyl ANSWER: c 33. Phosphatidylinositol does NOT contain a(n): a. amino group. b. methylene group. c. ester linkage. d. carbonyl group. e. hydroxyl group.

ANSWER: a 34. Phosphatidylcholine does NOT contain a(n): a. carbonyl group. b. hydroxyl group. c. ester linkage. d. substituted amino group. e. methylene group.

ANSWER: b 35. Cardiolipin does NOT contain a(n): a. phosphate group. b. methyl group. c. ester linkage. d. amino group. e. hydroxyl group.

ANSWER: d 36. Phosphatidate does NOT contain a(n): a. phosphate. b. glycerol. c. ester linkage. d. acyl chain. e. glucose.

ANSWER: e .. Powered by Cognero.

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Chapter 11 37. Eukaryotic cell membranes usually do NOT contain: a. phospholipids. b. glycolipids. c. steroids. d. triacylglycerols. e. sphingolipids.

ANSWER: d 38. Which is NOT a function of lipids? a. intercellular signaling b. fuel source c. structural rigidity of the cytoskeleton d. membrane component e. protection ANSWER: c 39. How many double bonds does octadecatrienoic acid have? a. 0 b. 1 c. 2 d. 3 e. 4 ANSWER: d 40. Which fact about a fatty acid is indicated by the notation 12:2? a. The hydrocarbon chain has 12 carbon atoms and two double bonds. b. The 12-carbon hydrocarbon chain is saturated. c. There is a double bond between the ɑ and β carbon atoms. d. There is a double bond involving the methyl carbon atom. e. There are two trans double bonds in the hydrocarbon chain of this 12-carbon fatty acid.

ANSWER: a 41. The following compound is NOT:

a. a component of biological membranes. b. a sphingolipid. c. a phosphoglyceride. .. Powered by Cognero.

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Chapter 11 d. amphipathic. e. phosphatidylcholine.

ANSWER: b 42. The longer is the fatty acid, the _____ the fatty acid is/has. a. more oxidized b. lower the melting point of c. more double bonds d. more reduced e. higher the melting point of ANSWER: e 43. How many carbons does palmitate have in its hydrocarbon chain? a. 12 b. 14 c. 16 d. 20 e. 24 ANSWER: c 44. The backbone of a phospholipid is: a. either glycerol or sphingosine. b. always glycerol. c. always sphingosine. d. a fatty acid. e. a steroid nucleus.

ANSWER: a 45. At which carbon of glycerol is the polar head group of phospholipids found? a. C-1 b. C-2 c. C-3 d. ⍵-carbon atom e. ɑ-carbon atom ANSWER: c 46. What functional group are polar head groups of phospholipids esterified to? a. methyl b. phosphate c. ketone d. thiol .. Powered by Cognero.

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Chapter 11 e. aldehyde ANSWER: b 47. Identify the differences in archaea membrane lipids compared to those of eukaryotes or bacteria and how these differences help them withstand extreme environmental conditions. a. The ether linkages are more readily hydrolyzed by enzymes, allowing the membrane to serve as an energy reserve. b. The glycerol moiety is esterified to multiple complex carbohydrate chains, making them more soluble in low-pH environments. c. The fatty acid chains are branched, allowing them to pack more tightly, thereby protecting membrane integrity. d. There are two phosphate esters instead of only one, giving the archaea better solubility in high-salt environments. e. Omega-3 fatty acids are common in membranes and may act as important precursors as they do in eukaryotes.

ANSWER: c 48. Cholesterol and other steroids are not soluble in blood and therefore must be transported. Predict what chemical modifications must occur for cholesterol to move through the circulatory system. a. Cholesterol forms micelles in blood, the surface of which is hydrophilic and the interior hydrophobic. b. Cholesterol cannot be transported in blood, so it is broken down and resynthesized in all cells. c. Cholesterol moves through cell membranes from tissue to tissue. d. Cholesterol forms glycolipids with large carbohydrate complexes in order to increase solubility. e. Cholesterol is esterified to a fatty acid for transport by lipoprotein particles, the surface of which is hydrophilic and the interior hydrophobic.

ANSWER: e 49. The shorthand notation indicating that there are two cis double bonds between carbons 9 and 10 and again between 12 and 13 is _____.

ANSWER: Suggested Answer: cis, cis-Δ9, Δ12 50. Olive oil is rich in monounsaturated fatty acids (MUFAs). Which free fatty acid is MOST abundant in olive oil? a. stearate (n-octadecenoate) b. oleate (cis-Δ9-octadecenoate) c. linoleate (cis, cis-Δ9,Δ12-octadecadienoate) d. linolenate (all-cis-Δ9,Δ12,Δ15-octadecadienoate) e. arachidonate (all-cis-Δ5,Δ8,Δ11,Δ14-eicosapentaenoic acid) ANSWER: b 51. Which of these compounds is an omega-3 (⍵-3) fatty acid? .. Powered by Cognero.

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Chapter 11 a. palmitate (n-hexadecenoate) b. palmitoleate (cis-Δ9-hexadecenoate) c. linoleate (cis, cis-Δ9,Δ12-octadecadienoate) d. linolenate (all-cis-Δ9,Δ12,Δ15-octadecadienoate) e. arachidonate (all-cis-Δ5,Δ8,Δ11,Δ14-eicosapentaenoic acid) ANSWER: d 52. Which of these fatty acids has the highest melting point? a. stearate (18:0) b. laurate (12:0) c. arachidate (20:0) d. myristate (14:0) e. palmitate (16:0) ANSWER: c 53. Which of these fatty acids has the lowest melting point? a. linoleate (18:2) b. oleate (18:1) c. linolenate (18:3) d. stearidonate (18:4) e. stearate (18:0) ANSWER: d 54. Eating a diet that is rich in _____ fatty acids and low in _____ fatty acids helps protect against cardiovascular disease. a. trans-unsaturated; omega-3 b. saturated; trans-unsaturated c. omega-3; trans-unsaturated d. saturated; omega-3 e. trans-unsaturated; saturated ANSWER: c 55. Which statement about triacylglycerols is FALSE? a. They are amphipathic. b. They are stored in a nearly anhydrous form. c. They are richer in energy than carbohydrates. d. They contain ester linkages. e. The fatty acid components are uncharged.

ANSWER: a 56. Which statement about waxes is FALSE? .. Powered by Cognero.

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Chapter 11 a. They contain small alcohol moieties. b. They serve many biological functions. c. They contain long-chain fatty acids. d. They are a storage form of fatty acids. e. They contain ester linkages.

ANSWER: a 57. Triacylglycerols are stored in the: a. smooth endoplasmic reticulum of adipocytes. b. cytosol of endothelial cells. c. cytosol of adipocytes. d. plasma membrane of endothelial cells. e. plasma membrane of adipocytes.

ANSWER: c 58. What do you need to make soap? a. glycerol and a strong acid b. glycerol and a strong base c. wax and a strong acid d. triacylglycerols and a strong base e. triacylglycerols and a strong acid ANSWER: d 59. What is NOT a component of phosphoglycerides? a. glycerol b. two fatty acid chains c. phosphate d. an alcohol e. sphingosine ANSWER: e 60. Which amino acid is a common component of phospholipids? a. alanine b. asparagine c. cysteine d. glutamate e. serine ANSWER: e 61. The steroid nucleus consists of _____ joined together. a. a cyclopentane ring and three cyclohexane rings .. Powered by Cognero.

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Chapter 11 b. two cyclopentane rings and two cyclohexane rings c. three cyclopentane rings and a cyclohexane ring d. two cyclopentane rings and three cyclohexane rings e. three cyclopentane rings and two cyclohexane rings ANSWER: a 62. Which statement about cholesterol is FALSE? a. It is constructed from a steroid nucleus. b. It has a hydrocarbon tail and a hydroxyl group. c. It constitutes almost 25% of intracellular membranes. d. It modulates membrane fluidity. e. It is an amphipathic molecule.

ANSWER: c 63. Compared to other phospholipids, sphingolipids tend to: a. be shorter and less saturated. b. be longer and less saturated. c. interact less with cholesterol. d. be shorter and more saturated. e. have a higher melting point.

ANSWER: e 64. Membrane lipids are _____ molecules. a. amphipathic b. hydrophobic c. positively charged d. water-insoluble e. electrically neutral ANSWER: a 65. Prostaglandin E2 is a hormone molecule that induces labor. How many carbon atoms does prostaglandin E2 have? a. 14 b. 16 c. 18 d. 20 e. 22 ANSWER: d 66. Which of these signaling molecules is NOT a derivative of arachidonate? a. prostaglandins .. Powered by Cognero.

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Chapter 11 b. steroids c. prostacyclins d. thromboxanes e. leukotrienes ANSWER: b 67. Which statement about eicosanoids is FALSE? a. They can have different effects on different cell types. b. They are derived from arachidonate. c. They bind receptors in the cell membrane. d. They travel long distances through the blood. e. They have relatively short half-lives.

ANSWER: d 68. Lipoxygenases catalyze the initial step in a pathway leading to: a. prostaglandins. b. phospholipids. c. prostacyclins. d. thromboxanes. e. leukotrienes.

ANSWER: e 69. Prostaglandin synthase catalyzes the conversion of: a. arachidonate to prostaglandin H2. b. prostaglandin H2 to prostacyclin. c. arachidonate to leukotrienes. d. prostaglandin H2 to thromboxanes. e. phospholipids to arachidonate.

ANSWER: a 70. Proteins can be anchored to the membrane by a palmitoyl or farnesyl group attached to a(n) _____ residue. a. asparagine b. methionine c. cysteine d. serine e. threonine ANSWER: c

.. Powered by Cognero.

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Chapter 12 1. Which role does the membrane play by separating the cell interior from the exterior and cell compartments from the cell interior? a. an open boundary b. lipid vesicle c. self-assembled structure d. an interface e. hydrophobic interior ANSWER: d 2. What should be the approximate length of a phospholipid molecule, considering the average thickness of membranes? a. 10 Å b. 30–50 Å c. 60–100 Å d. 50 nm e. 8–12 nm ANSWER: c 3. The process of _____ has rarely been observed. a. lateral diffusion of proteins b. transverse diffusion of proteins c. lateral diffusion of lipids d. transverse diffusion of lipids e. passive diffusion of molecules through the membrane ANSWER: b 4. Which property is NOT associated with lipid membranes? a. asymmetry b. solvent for proteins c. self-assembled leaflets d. solvent for carbohydrates e. electric polarization ANSWER: d 5. Which membrane component provides impermeability to polar molecules and ions? a. hydrophilic heads of phospholipids b. cholesterol c. hydrophobic tails of phospholipids d. integral proteins e. peripheral proteins ANSWER: c .. Powered by Cognero.

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Chapter 12 6. Which component of membranes is key to controlling membrane fluidity in prokaryotes? a. hydrophilic heads of phospholipids b. cholesterol c. hydrophobic tails of phospholipids d. integral proteins e. peripheral proteins ANSWER: c 7. The formation of lipid bilayers is determined by: a. van der Waals attractive forces. b. hydrophobic interactions. c. electrostatic interactions. d. hydrogen bonds. e. van der Waals repulsive forces.

ANSWER: b 8. What role do van der Waals interactions play in membranes? a. preventing the self-assembly of the lipid bilayer b. decreasing the melting temperature c. increasing membrane fluidity d. favoring loose packing of the hydrophobic interior e. decreasing membrane fluidity ANSWER: e 9. Choose the CORRECT statement about membrane permeability: a. Tryptophan crosses membranes 103 times more slowly than water. b. If a molecule easily crosses the solvation shell of the membrane, it can easily cross the whole membrane. c. Water can easily traverse through the membrane because it is amphipathic. d. Entry of sodium ions in the cell can increase membrane permeability for sodium ions. e. Nonpolar interactions with the core of the membrane are more favorable for Na+ than for K+.

ANSWER: d 10. The Tm of the membrane thelowest if it is composed of: a. octadecanoate without cholesterol. b. octadecanoate and a few cholesterol molecules. c. hexadecanoate and a few cholesterol molecules. d. cis-Δ9-hexadecenoate and a few cholesterol molecules. e. cis-Δ9-hexadecenoate and a lot of cholesterol molecules. .. Powered by Cognero.

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Chapter 12 ANSWER: e 11. Lipid rafts: a. are complexes of cholesterol with saturated fatty acids and proteins that decrease overall Tm of the membrane. b. are small and dynamic regions rich in cholesterol that make phase transitions of the membrane less likely. c. can change membrane fluidity since they are large. d. increase membrane fluidity and at the same time make membranes less prone to phase transitions. e. are complexes of cholesterol with unsaturated fatty acids and proteins that increase overall Tm of the membrane.

ANSWER: b 12. Why do liposomes have high clinical importance? a. They act as systemic drugs. b. Only hydrophilic molecules can be trapped inside such vesicles. c. They are quite uniform in size. d. Different drugs or biochemicals can be trapped inside liposomes. e. They increase the amount of drug delivered across the body.

ANSWER: d 13. What is the typical protein content of mitochondrial membranes? a. 18% b. 25% c. 50% d. 75% e. 100% ANSWER: d 14. Associations of proteins with membranes that do NOT exist are: a. integral membrane-spanning β sheet with nonpolar side groups outside. b. integral membrane-spanning α helices with polar side groups outside. c. integral membrane-spanning nonpolar α helices. d. peripheral polar interactions with phospholipid heads. e. partial embedding in the membrane of relatively small hydrophobic regions.

ANSWER: b 15. What property of membranes makes the transverse diffusion of phospholipids across the membrane energetically unfavorable? a. lateral diffusion b. hydrophobic interior of the membrane .. Powered by Cognero.

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Chapter 12 c. protein immobilization d. electric polarization e. membrane fluidity ANSWER: b 16. What type of transport can proceed against a concentration gradient? a. simple diffusion b. passive transport c. active transport d. facilitated diffusion e. lateral diffusion ANSWER: c 17. What is the origin of the free energy of the Na+ ion gradient? a. free energy of the K+ ion gradient b. electric polarization c. Na+-K+ ATPase d. free energy of the solvation shell e. free energy of AMP ANSWER: c 18. What will happen upon hydrolysis of 3 ATP molecules by the Na+-K+ pump? a. 3 Na+ will be transported outside the cell, and 2 K+ will be transported inside the cell. b. 6 Na+ will be transported outside the cell, and 9 K+ will be transported inside the cell. c. 3 Na+ will be transported inside the cell, and 2 K+ will be transported outside the cell. d. 9 Na+ will be transported inside the cell, and 6 K+ will be transported outside the cell. e. 9 Na+ will be transported outside the cell, and 6 K+ will be transported inside the cell.

ANSWER: e 19. Through which type of ion channels do neurotransmitters usually act? a. voltage-gated ion channels b. antiporters c. ligand-gated ion channels d. ABC transporters e. symporters ANSWER: c 20. Which statement about P-type ATPases is FALSE? a. Ca2+ ATPase is the key enzyme in muscle contraction because it transports Ca2+ ions out of the .. Powered by Cognero.

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Chapter 12 cytoplasm. b. Na+-K+ ATPase utilizes about a third of the organism's total ATP. c. Release of Ca2+ from the sarcoplasmic reticulum is required for muscle contraction. d. Na+-K+ ATPase binds two ATP molecules to exchange Na+ and K+ ions. e. H+-K+ ATPase transports protons from the cells of the stomach wall.

ANSWER: d 21. The slowest type of molecule transport is: a. transverse diffusion of proteins. b. active transport. c. facilitated diffusion. d. lateral diffusion. e. transverse diffusion of lipids.

ANSWER: a 22. Potassium ion channels facilitate diffusion because: a. the K+ ion doesn't have to lose its solvation shell upon entering the cell. b. the channel decreases desolvation energy of the potassium ion by the selectivity filter. c. resolvation of the potassium ion within the channel costs more than its desolvation. d. the K+ ion loses its solvation shell upon entering the cell. e. the channel provides a concentration gradient.

ANSWER: e 23. Sodium ions can't pass through the K+ ion channel because they: a. are too small to react with the selectivity filter. b. require hydrolysis of more than one ATP molecule. c. are too big to be efficiently rehydrated by the channel. d. are taken up by respective Na+ ion channels so they don't need to pass through K+ ion channels. e. need more binding sites in the selectivity filter.

ANSWER: a 24. What is the driving force for ion movement in the Knock-On model, which suggests that the ions never occupy adjacent sites? a. charge attraction b. patch-clamp c. interaction with carbonyl groups d. charge repulsion e. interaction with water ANSWER: d .. Powered by Cognero.

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Chapter 12 25. Some tumor cells become more sensitive to a range of drugs as they become resistant to only one drug because they: a. obtain P-glycoprotein that transports different drugs inside the cell. b. obtain the multidrug-resistance protein that transports different drugs from the cell. c. have more ATP-binding cassettes than other cells. d. have mutated the ATP-regulated chloride channel. e. obtain the MDR ion channel that transports different drugs from the cell.

ANSWER: b 26. What does NOT determine whether a small molecule will cross a membrane? a. concentration gradient b. molecule solubility in the hydrophilic environment c. replacement of the solvation shell of polar water molecules by nonpolar interactions with the membrane interior d. molecule solubility in the hydrophobic environment e. difference in the molecule concentration on both sides of the membrane ANSWER: c 27. Valinomycin is an amphipathic antibiotic capable of facilitating diffusion of potassium ions because it can effectively replace the solvation shell of the ion while diffusing through the plasma membrane. What would happen if all of the potassium ion channels were blocked and valinomycin was added to the cells immediately after? Consider that the experiment is performed on neurons in the resting state and other ion channels are intact. a. K+ would be transported inside the neuron, which would increase its resting potential. b. K+ wouldn't be transported in any direction; thus, no change in the resting potential would be observed. c. K+ would be transported from the neuron, which would increase the resting potential. d. K+ would be transported inside the neuron, which would decrease its resting potential. e. K+ would be transported from the neuron, which would decrease the resting potential.

ANSWER: c 28. How would the action potential on a neuron in the resting state change if all of the sodium ion channels were blocked? a. action potential wouldn't be generated or propagated b. no change in the action potential c. threshold for the action potential would decrease d. action potential would only be propagated e. threshold for the action potential would increase ANSWER: a .. Powered by Cognero.

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Chapter 12 29. What would happen to the neuron if the activity of Na+-K+ ATPase was completely inhibited? a. The action potential couldn't be generated. b. The resting potential would be restored too fast after the action potential. c. Repolarization of the membrane wouldn't appear. d. The resting potential wouldn't be restored after the action potential. e. Depolarization of the membrane wouldn't appear.

ANSWER: d 30. What is the first step of the action potential? a. a high conductance for Na+ b. the flow of Na+ out of the cell c. the flow of K+ into the cell d. the flow of Na+ into the cell e. a low conductance for K+ ANSWER: d 31. What would happen to the resting potential if neurons were treated with tetrodotoxin, a toxin that acts as a sodium channel blocker? a. significant decrease b. it would disappear c. significant increase d. it would propagate e. no significant changes ANSWER: e 32. What is the main cause of repolarization? a. The flow of K+ out of the cell. b. The flow of Na+ out of the cell. c. The flow of K+ into the cell. d. The flow of Na+ into the cell. e. A low conductance for K+.

ANSWER: a 33. According to the given scheme, what type of transport should be used to transport sodium ions to cells?

.. Powered by Cognero.

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Chapter 12

a. simple diffusion b. facilitated diffusion c. primary active transport d. coupled to K+ symport e. coupled to Cl– antiport ANSWER: c 34. According to the given scheme, what type of transport should be used to transport potassium ions to cells?

a. simple diffusion b. facilitated diffusion c. primary active transport d. coupled to Na+ antiport e. coupled to Cl– antiport ANSWER: c 35. According to the given scheme, choose the most appropriate types of transport and, if applicable, transporters that can be used to transport glucose from the lumen to blood if the initial concentration of glucose in the lumen is higher than in the cells.

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Chapter 12

a. simple diffusion along the glucose concentration gradient inside cells and primary active transport against the concentration gradient to blood b. symport to cells dependent on the gradient of sodium ions concentration and facilitated diffusion down the glucose concentration gradient to blood c. simple diffusion along the glucose concentration gradient inside cells and secondary active transport up the glucose concentration gradient to blood d. antiport inside cells dependent on the gradient of sodium ions concentration and symport to blood dependent on the gradient of potassium ions e. facilitated diffusion along the glucose concentration gradient to cells and active transport by the ATP-dependent antiporter to blood ANSWER: b 36. The _______ of small molecules is correlated with their relative solubilities in water and nonpolar solvents.

ANSWER: permeability 37. Fluorescence recovery after photobleaching has been used to study: a. membrane composition. b. lateral diffusion in membranes. c. protein structural motifs in membranes. d. lipid rafts in membranes. e. ion channels activity.

ANSWER: b 38. Which lipid bilayer composition would be the MOST fluid? a. polyunsaturated 18-carbon fatty acids b. saturated 18-carbon fatty acids c. saturated 16-carbon fatty acids d. polyunsaturated 16-carbon fatty acids e. one double bond in 18-carbon fatty acids ANSWER: d 39. What describes the role of a protein in the selective permeability of a membrane? .. Powered by Cognero.

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Chapter 12 a. the polar heads of phospholipids, which allow polar molecules to cross membranes b. the binding of steroids to an extracellular peripheral membrane protein c. the insertion of a porin with polar amino acids forming the β strands d. Na+-K+ ATPase activity that sets up an electrical potential across the membrane e. passive diffusion according to the second law of thermodynamics ANSWER: d 40. Which statement is consistent with the structure of biological membranes? a. All membrane proteins are integral and associate with the hydrophobic region of the membrane. b. Both proteins and lipids readily undergo transverse ("flip-flop") diffusion from the inside to the outside of the membrane. c. Membranes are symmetric. d. The membrane lipids self-assemble to form the lipid bilayer. e. A biological membrane consists of proteins sandwiched between two layers of lipids, which is referred to as a lipid bilayer.

ANSWER: d 41. Digitalis is a cardiotonic steroid that does not act like a steroid. Instead, its function is to inhibit: a. the Na+-K+ pump. b. fatty acid intake. c. calcium channels. d. potassium and sodium pores. e. the dropsy channel.

ANSWER: a 42. Antiporters and symporters are examples of: a. passive diffusion. b. primary active transporters. c. entropy driven pores. d. ABC transporters. e. secondary transporters.

ANSWER: e 43. The MOST common motif found in membrane-spanning proteins is: a. α helices of nonpolar amino acids that pass through the membrane. b. α helices of charged amino acids that form channels via extensive hydrogen bonding. c. the triple helix of α helices. d. a helix-turn-helix arrangement of the peptide strands. e. a bent β sheet.

ANSWER: a .. Powered by Cognero.

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Chapter 12 44. Would you agree that a peripheral membrane protein can act as Na+-K+ ATPase? Why? a. Yes, because Na+ and K+ are available intracellularly and extracellularly. b. Yes, because the hydrophobic amino side chains of the α helices firmly anchor the protein to the membrane. c. No, because a peripheral protein does not form a transporter across the entire membrane. d. No, because only secondary active transport proteins can act as Na+-K+ ATPase. e. No, because P-glycoprotein must also be present for Na+-K+ ATPase activity.

ANSWER: c 45. What evidence exists to show that membrane asymmetry can be preserved for long periods? a. Lateral diffusion from fluorescence recovery after photobleaching experiments. b. Transverse diffusion of proteins is rarely observed. c. Facilitated diffusion can take place in either direction depending on the concentration gradient. d. Tumor cells often become resistant to drugs. e. The α helices in membrane-spanning proteins are hydrophobic and tightly packed.

ANSWER: b 46. If you want to study how membranes regulate their curvature and budding processes, what membrane structures would you study? a. presence of lipid rafts b. degree of fatty acid unsaturation c. presence of active transport proteins d. presence of prostaglandin H2 synthase-1 e. concentration of cholesterol ANSWER: a 47. What force(s) stabilize(s) lipid bilayers? Select all that apply. a. van der Waals interactions b. electrostatic bonding between the polar heads and surrounding water c. covalent bonds between the lipid tails d. covalent bonds between the lipids and membrane proteins e. hydrogen bonding between the polar heads and surrounding water ANSWER: a, b, e 48. Choose the proper order of permeability of molecules across a membrane, from the most to the least permeable molecules. a. water, glucose, urea, sodium ion b. water, indole, glucose, sodium ion c. water, indole, sodium ion, glucose d. indole, glucose, urea, sodium ion .. Powered by Cognero.

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Chapter 12 e. indole, water, glucose, sodium ion ANSWER: b 49. The main functions of biological membranes are to separate the interior of the cell from the exterior environment and to form intracellular compartments known as ______, which are shielded from the cytoplasm. a. organelles b. ribosomes c. plasma d. cell storage ANSWER: a 50. Which of the following does NOT describe a carbohydrate typically found linked to a lipid or protein in the membranes of biological cells? a. glycolipid b. glycoprotein c. peptidoglycan d. lipidoglycan e. none of the above ANSWER: d 51. The lipids in cells often assemble into bilayers, which form membranes that are important to cell function.

Which of the following correctly describes the molecular affinities and orientations of lipid bilayers in cells? a. The hydrophobic regions of lipids face the interior of the membrane, and the hydrophilic regions of lipids face the exterior extracellular and intracellular environments. b. The positively-charged regions of lipids face the interior of the membrane, and the negativelycharged regions of lipids face the exterior extracellular and intracellular environments. c. The hydrophilic regions of lipids face the interior of the membrane, and the hydrophobic regions of lipids face the exterior extracellular and intracellular environments. d. The positively-charged regions of lipids face the exterior of membranes, and the negatively-charged regions of lipids face the interior.

ANSWER: a 52. Which of the following molecules are amphipathic? a. glycoproteins b. peptidoglycans c. carbohydrates d. phospholipids ANSWER: d 53. What is an advantage of using liposomes for drug delivery? a. Liposomes are nearly spherical in shape and are smaller than naturally occurring lipid-containing vesicles. .. Powered by Cognero.

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Chapter 12 b. The preparation of liposomes through the use of sonication results in various-sized liposomes that are suitable for delivery to numerous cell types. c. Liposomes can be engineered to target specific cells, thereby minimizing the exposure of other cells in the body to potentially toxic chemicals. d. Nucleic acids that are not incorporated into liposomes during their formation are eventually removed by filtration and are not exposed to the body.

ANSWER: c 54. Which of the following is the correct order of events that occur when a small molecule traverses a lipid bilayer? a. 1 - shed the water with which it is associated, 2 - dissolve in hydrocarbon core of the membrane, 3 diffuse to other side of the membrane, 4 - resolvation b. 1 - dissolve in hydrocarbon core of the membrane, 2 - diffuse to other side of the membrane, 3 - shed the water with which it is associated, 4 - resolvation c. 1 - diffuse to other side of the membrane, 2 - dissolve in hydrocarbon core of the membrane, 3 - shed the water with which it is associated, 4 - resolvation d. 1 - shed the water with which it is associated, 2 - diffuse to other side of the membrane, 3 resolvation, 4 - dissolve in hydrocarbon core of the membrane ANSWER: a 55. The fluidity of membranes depends on the a. boiling point of the fatty acid chains that make up the lipid bilayer. b. the pH of the extracellular environment. c. different molecular affinities for water by phospholipids. d. saturation of the fatty acid chains within phospholipids, determined by the number of double bonds.

ANSWER: d 56. How does cholesterol affect the fluidity of biological membranes? a. Cholesterol occupies the exterior of membranes and acts to displace the hydrophilic head groups of phospholipids. b. Cholesterol inserts into the membrane and disrupts the regular interactions of the fatty acid chains of phospholipids. c. Cholesterol does not have a significant effect on the fluidity of membranes. d. Cholesterol lowers the melting point of fatty acids located inside the membrane.

ANSWER: b 57. Which types of membranes are more suitable for insulation? a. protein-rich membranes b. carbohydrate-rich membranes c. lipid-rich membranes d. nucleic-acid-rich membranes ANSWER: c .. Powered by Cognero.

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Chapter 12 58. A biochemist is studying a novel cell receptor that transmits cell growth signals and determines that this receptor contains several alpha helices that span the length of the cell membrane and interact extensively with the hydrophobic region of the lipid bilayer. This cell receptor can accurately be classified by the biochemist as which type of membrane protein? a. peripheral protein b. integral protein c. channel protein d. protein pump ANSWER: b 59. Which of the following cannot be studied using the biochemical technique known as fluorescence recovery after photobleaching (FRAP)? a. membrane fluidity b. lipid rafts c. lateral movement of membrane components d. ion channel activity ANSWER: d 60. Steroids are lipophilic molecules that are capable of dissolving in the lipid bilayer. The diffusion of steroids down their concentration gradient is an example of which type of diffusion? a. facilitated diffusion b. passive transport c. simple diffusion d. simple transport ANSWER: c 61. Action potentials found in neurons are electrical signals that require the passage of charged sodium ions down their concentration gradient through protein channels. This type of movement of ions across a membrane is an example of which type of diffusion? a. facilitated diffusion b. active transport c. simple diffusion d. simple transport ANSWER: a 62. The sodium-potassium pump is a protein complex that maintains the negative resting potential of neurons by moving 3 positively charged sodium ions into the cell and 2 positively charged potassium ions out of the cell, against their respective concentration gradients. This type of movement across a membrane is an example of which type of diffusion? a. facilitated diffusion b. active transport .. Powered by Cognero.

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Chapter 12 c. passive transport d. simple diffusion ANSWER: b 63. Which of the following is NOT a P-type ATPase? a. sodium-potassium pump b. calcium ATPase c. gastric ATPase d. tyrosine ATPase ANSWER: d 64. How do mutations in the gene for the ATP-regulated protein channel CFTR cause cystic fibrosis? a. Mutations in the CFTR gene cause a decrease in fluid and salt secretion. b. Mutations in the CFTR gene cause a deaccumulation of mucus in the lungs. c. Mutations in the CFTR gene result in multi-drug resistance. d. Mutations in the CFTR gene result in nonfunctional ATPases that regulate action potentials in neurons.

ANSWER: a 65. How does secondary active transport differ from primary active transport? a. Primary active transport requires the use of membrane proteins, while secondary active transport does not. b. Primary active transport requires the direct use of ATP, while secondary active transport does not. c. Primary active transport is the coupling of an unfavorable reaction to a thermodynamically favorable reaction, while secondary active transport is the diffusion of molecules against their concentration gradient. d. Primary active transport does not require the use of a transmembrane protein, while secondary active transport does require the use of a transmembrane protein.

ANSWER: b 66. In animal cells, the movement of glucose into cells is often powered by the simultaneous entry of sodium.

Which of the following cotransporters facilitates this exchange? a. antiporter b. symporter c. primary transporter d. glucose-Na ATPase ANSWER: b 67. Neurotransmitters released by presynaptic neurons into synapses diffuse through which type of ion channel? a. voltage-gated b. ligand-gated c. chemical-gated .. Powered by Cognero.

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Chapter 12 d. ATPases ANSWER: b 68. Some bacterial cells become insensitive over time to certain antibiotic chemicals and as a result become resistant to elimination. Which of the following best explains why some of these bacterial cells are resistant? a. Some types of bacteria have P-glycoproteins that transport different chemicals into the cell. b. Some bacteria have developed a multidrug resistance protein that traverses the membrane and expels chemicals that enter the cell. c. Over time, certain types of bacteria have developed the ability to trap and degrade antibiotics in intracellular compartments. d. Certain types of antibiotics are never effective.

ANSWER: b

.. Powered by Cognero.

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Chapter 13 1. Which mechanism allows an organism to sense its environment and formulate a proper biochemical response? a. oncogene b. proto-oncogene c. receptors d. primary messenger e. signal transduction ANSWER: e 2. Choose the CORRECT order of signal transferring in the basic signal-transduction pathway. a. membrane receptor → primary messenger → second messenger → termination of the signal → activation/inhibition of the pumps, enzymes, gene-transcription factors b. primary messenger → activation/inhibition of the pumps, enzymes, gene-transcription factors → membrane receptor → second messenger → termination of the signal c. second messenger → membrane receptor → activation/inhibition of the pumps, enzymes, genetranscription factors → primary messenger → termination of the signal d. membrane receptor → activation/inhibition of the pumps, enzymes, gene-transcription factors → primary messenger → second messenger → termination of the signal e. primary messenger → membrane receptor → second messenger → activation/inhibition of the pumps, enzymes, gene-transcription factors → termination of the signal ANSWER: e 3. Which of the following signal molecules does NOT pass through the cell membrane to transfer the environmental signal? a. primary messenger b. second messenger c. membrane receptor d. activator e. ligand ANSWER: a 4. Select all that apply. Most signal molecules can't pass through the cell membrane or through transporters because of their: a. large size. b. polarity c. high concentration. d. low concentration. e. small size.

ANSWER: a, b 5. What component of the signal-transduction pathway transfers information from the environment to a cell's interior? .. Powered by Cognero.

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Chapter 13 a. primary messenger b. second messenger c. membrane receptor d. activator e. ligand ANSWER: c 6. What component of the signal-transduction pathway transduces received information into other forms of information that can alter the cell's biochemistry? a. primary messenger b. second messenger c. membrane receptor d. activator e. ligand ANSWER: c 7. What is NOT a second messenger? a. cyclic adenosine monophosphate b. cyclic guanosine monophosphate c. inositol 1,4,5-trisphosphate d. phosphatidylinositol 4,5-bisphosphate e. diacylglycerol ANSWER: d 8. What is NOT a feature of second messengers? a. They are able to diffuse easily to other compartments of the cell. b. They are able to diffuse easily to the nucleus. c. They alter the tertiary or quaternary structure of the receptor. d. They change in concentration in response to environmental signals. e. The signal may be amplified significantly in the generation of second messengers.

ANSWER: c 9. Most membrane receptor proteins of the signal-transduction pathway do NOT belong to: a. seven-transmembrane-helix receptors. b. G-protein-coupled receptors. c. dimeric receptors that are protein kinases. d. dimeric receptors that recruit protein kinases. e. GTPase-activating proteins.

ANSWER: e 10. Choose the CORRECT combination of type of membrane receptor and example of the receptor. .. Powered by Cognero.

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Chapter 13 a. seven-transmembrane-helix receptor; growth-hormone receptor b. dimeric receptors that recruit protein kinases; insulin receptor c. dimeric receptors that are protein kinases; insulin receptor d. dimeric receptors that recruit protein kinases; β-adrenergic receptor e. dimeric receptors that are protein kinases; growth-hormone receptor ANSWER: c 11. Choose the CORRECT statement about activation of G proteins. a. The ligand-binding site of the 7TM receptor lies near the cytoplasmic surface. b. The conformational change in the extracellular domain of the receptor activates a GTP-binding protein. c. GDP is bound in a pocket close to the surface at which the α subunit interacts with the βγ dimer. d. G proteins do not have the ability for self-deactivation. e. Gαi stimulates adenylate cyclase activity, thus increasing cAMP production.

ANSWER: c 12. What transmits the signal that a receptor has bound its ligand? a. the conformational change in the cytoplasmic domain of the 7TM receptor b. the dissociation of the G-protein into Gα and Gβγ subunits c. the exchange of the bound GTP for GDP in the G-protein d. the dissociation of the adenylate cyclase e. synthesis of the second messenger ANSWER: b 13. What is the function of cAMP phosphodiesterase? a. It activates protein kinase A. b. It inactivates protein kinase B. c. It joins two R and two C subunits of the PKA. d. It converts cAMP into AMP. e. It phosphorylates other proteins.

ANSWER: d 14. The R2C2 complex of PKA catalytically is inactive: a. at low concentrations of cAMP. b. in the absence of cAMP. c. at high concentrations of cAMP. d. in the absence of cAMP phosphodiesterase. e. after the dissociation of the R2C2 holoenzyme.

ANSWER: b .. Powered by Cognero.

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Chapter 13 15. Choose the CORRECT pathway of the adenylate cyclase cascade. a. ligand → 7TM receptor → Gαs → adenylate cyclase → cAMP → dissociation of the R2C2 of the PKA → proteins phosphorylation → turning off by the cAMP phosphodiesterase b. ligand → 7TM receptor → Gαq → adenylate cyclase → cAMP → dissociation of the R2C2 of the PKA → proteins phosphorylation → turning off by the cAMP phosphodiesterase c. ligand → 7TM receptor → Gαq → phospholipase C → PIP2 → IP3 releases Ca2+, while DAG activates PKC → proteins phosphorylation d. ligand → tyrosine kinase receptor → Gαs → adenylate cyclase → cAMP → dissociation of the R2C2 of the PKA → proteins phosphorylation → turning off by the cAMP phosphodiesterase e. ligand → 7TM receptor → Gαs → adenylate cyclase → cAMP → joining of the R2C2 of the PKA into holoenzyme → proteins phosphorylation ANSWER: a 16. Choose the CORRECT pathway of the phosphoinositide cascade. a. ligand → 7TM receptor → Gαq → adenylate cyclase → cAMP → dissociation of the R2C2 of the PKA → proteins phosphorylation → turning off by the cAMP phosphodiesterase b. ligand → tyrosine kinase receptor → Gαq → phospholipase C → PIP2 → IP3 releases Ca2+, while DAG activates PKC → proteins phosphorylation c. ligand → 7TM receptor → Gαq → phospholipase C → PIP2 → DAG releases Ca2+, while IP3 activates PKC → proteins phosphorylation d. ligand → 7TM receptor → Gαs → adenylate cyclase → cAMP → dissociation of the R2C2 of the PKA → proteins phosphorylation → turning off by the cAMP phosphodiesterase e. ligand → 7TM receptor → Gαq → phospholipase C → PIP2 → IP3 releases Ca2+, while DAG activates PKC → proteins phosphorylation ANSWER: e 17. What process does NOT terminate the adenylate cyclase cascade? a. dissociation of the ligand–receptor complex b. conformation changes in the receptor c. self-deactivation of the G-protein by conversion of GTP into GDP d. conversion of cAMP into AMP by the cAMP phosphodiesterase e. dissociation of the PKA to subunits ANSWER: e 18. What is the mechanism of action of the pertussis toxin secreted by Bordetella pertussis, the bacterium responsible for whooping cough? a. Protein kinase A stays active even in the absence of cAMP. b. The toxin modifies a Gαs-protein and stabilizes its active GTP-bound form. c. The genetic material is translocated d. The Gαi-protein is inactivated, which normally closes Ca2+ channels and opens K+ channels. .. Powered by Cognero.

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Chapter 13 e. Protein kinase A opens a chloride channel and inhibits the Na+–H+ exchanger.

ANSWER: d 19. What second messenger stays in the membrane after cleavage of PIP2? a. phosphatidylinositol 4,5-bisphosphate b. 1,4,5-trisphosphate c. diacylglycerol d. phospholipase C e. cyclic adenosine monophosphate ANSWER: c 20. What second messenger does NOT cause a cascade of phosphorylation to elicit a response from the cell? a. phosphatidylinositol 4,5-bisphosphate b. 1,4,5-trisphosphate c. diacylglycerol d. cyclic guanosine monophosphate e. cyclic adenosine monophosphate ANSWER: b 21. Which ion is initially present in the active site of inositol monophosphatase? Which ion replaces it and is also used to treat bipolar affective disorder? a. Na+ and K+, respectively b. Mg2+ and Ca2+, respectively c. Mg2+ and Li+, respectively d. Zn2+ and Pb2+, respectively e. Zn2+ and Mg2+, respectively ANSWER: c 22. What receptors initiate signal-transduction pathways through changes in quaternary structures after ligand binding? Select all that apply. a. seven-transmembrane-helix receptors b. G-protein-coupled receptors c. dimeric receptors that are protein kinases d. dimeric receptors that recruit protein kinases e. protein kinases ANSWER: c, d 23. Choose the CORRECT statement about receptor dimerization and JAK2. a. The insulin receptor dimerizes after ligand binding. b. Activation of JAK2 brings together intracellular domains of the receptor. .. Powered by Cognero.

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Chapter 13 c. Janus kinase 2 is associated with each intracellular domain of the receptor in an activated form. d. Self-phosphorylation of Janus kinase 2 activates it catalytically. e. Janus kinases have two nearly identical active sites, but only one site is catalytically functional.

ANSWER: e 24. Choose the CORRECT sequence of the EGF signaling pathway. a. ligand binding → dimerization of the EGF receptor → self-phosphorylation of the JAK2 proteins → Sos protein binds to the phosphotyrosine residues of the receptor → recruiting of Grb-2 protein → binding and activation of Ras protein → activation of the specific protein kinases → cell growth promotion b. ligand binding → dimerization of the EGF receptor → cross-phosphorylation of the JAK2 proteins → Grb-2 protein binds to the phosphotyrosine residues of the receptor → recruiting of Sos protein → binding and activation of Ras protein → activation of the specific protein kinases → cell growth promotion c. ligand binding to the dimeric EGF receptor → self-phosphorylation of the JAK2 proteins → Sos protein binds to the phosphotyrosine residues of the receptor → recruiting of Grb-2 protein → binding and activation of Ras protein → activation of the specific protein kinases → cell growth promotion d. ligand binding to the dimeric EGF receptor → cross-phosphorylation of the JAK2 proteins → Grb-2 protein binds to the phosphotyrosine residues of the receptor → recruiting of Sos protein → binding and activation of Ras protein → activation of the specific protein kinases → cell growth promotion e. ligand binding → dimerization of the EGF receptor → cross-phosphorylation of the JAK2 proteins → Grb-2 protein binds to the phosphotyrosine residues of the receptor → activation of the specific protein kinases → recruiting of Sos protein → binding and activation of Ras protein → cell growth promotion ANSWER: b 25. Unlike the other members of the receptor tyrosine kinase class, the insulin receptor: a. has only one catalytically functional site out of two. b. undergoes cross-phosphorylation after ligand binding. c. activates the signal-transduction pathway. d. exists as a dimer even in the absence of insulin. e. acts in gene-transcription regulation.

ANSWER: d 26. What protein is referred to as a guanine-nucleotide exchange factor? a. Sos protein b. Ras protein c. Grb-2 protein d. small G proteins e. Arf protein ANSWER: a .. Powered by Cognero.

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Chapter 13 27. What statement about small G-proteins is TRUE? a. They are heterotrimeric. b. Their molecular weight is 40–45 kDa. c. They cycle between an inactive GTP-bound form and an active GDP-bound form. d. Unlike G-proteins, they don't possess an intrinsic GTPase activity. e. The signal-transduction protein Ras belongs to the small G-proteins.

ANSWER: e 28. What disease is NOT a result of a mutation in the receptor tyrosine kinases? a. arteriosclerosis b. type 2 diabetes c. inflammation d. Cushing's syndrome e. cancer ANSWER: d 29. What is the term for the genes that normally regulate cell growth, but often cause cancer when mutated? a. oncogenes b. proto-oncogenes c. paragenes d. adaptor genes e. tumor-suppressor genes ANSWER: b 30. Choose the CORRECT sequence of the insulin signaling pathway. a. insulin binding to the receptor → dimerization of the receptor → cross-phosphorylation and activation of the receptor → binding and phosphorylation of the IRS → the lipid kinase phosphoinositide 3-kinase binds to phosphorylated sites on IRS and then converts PIP2 into PIP3 → activation of the PIP3-dependent protein kinase → phosphorylation and activation of Akt b. insulin binding to the receptor → dimerization of the receptor → self-phosphorylation and activation of the receptor → binding and phosphorylation of the IRS → the lipid kinase phosphoinositide 3kinase binds to phosphorylated sites on IRS and then converts PIP3 into PIP2→ activation of the PIP2-dependent protein kinase → phosphorylation and activation of Akt c. insulin binding to the receptor → cross-phosphorylation and activation of the receptor → binding and phosphorylation of the IRS → the lipid kinase phosphoinositide 3-kinase binds to phosphorylated sites on IRS and then converts PIP2 into PIP3 → activation of the PIP3-dependent protein kinase → phosphorylation and activation of Akt d. insulin binding to the receptor → self-phosphorylation and activation of the receptor → binding and phosphorylation of the IRS → the lipid kinase phosphoinositide 3-kinase binds to phosphorylated sites on IRS and then converts PIP2 into PIP3 → activation of the PIP3-dependent protein kinase → phosphorylation and activation of Akt .. Powered by Cognero.

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Chapter 13 e. insulin binding to the receptor → binding and phosphorylation of the IRS → cross-phosphorylation and activation of the receptor → the lipid kinase phosphoinositide 3-kinase binds to phosphorylated sites on IRS and then converts PIP2 into PIP3 → activation of the PIP3-dependent protein kinase → phosphorylation and activation of Akt ANSWER: c 31. What statement about the termination of insulin signaling is FALSE? a. Lipid phosphatases are required to remove phosphoryl groups. b. Lipid phosphatases hydrolyze PIP3 to PIP2. c. Tyrosine phosphatases remove phosphoryl groups from tyrosine residues on the insulin receptor. d. Serine phosphatases remove phosphoryl groups from activated protein kinases such as Akt. e. Proteins can promote their own inactivation by the release of a phosphoryl group.

ANSWER: e 32. Why can each activated insulin receptor phosphorylate multiple IRS molecules? a. An insulin receptor is dimeric even without ligand binding. b. An insulin receptor binds adaptor proteins. c. A phosphorylated insulin receptor doesn't hydrolyze spontaneously. d. The activated insulin receptor itself is a protein kinase. e. The activated insulin receptor recruits protein kinases.

ANSWER: d 33. Choose the CORRECT sequence of the olfactory signal-transduction cascade. a. odorant binds to the olfactory receptor → cross-phosphorylation and activation of the receptor → binding and phosphorylation of the adaptor proteins → the lipid kinase phosphoinositide 3-kinase converts PIP2 into PIP3 → activation of the PIP3-dependent protein kinase → phosphorylation and activation of Akt → initiation of an action potential b. odorant binds to the olfactory receptor → Gαs → adenylate cyclase → cAMP → dissociation of the R2C2 of the PKA → proteins phosphorylation → initiation of an action potential c. odorant binds to the olfactory receptor → Gαq → phospholipase C → PIP2 → IP3 releases Ca2+, while DAG activates PKC → proteins phosphorylation and initiation of an action potential d. odorant binds to the olfactory receptor → dimerization of the EGF receptor → cross-phosphorylation of the JAK2 proteins → Grb-2 protein binds to the phosphotyrosine residues of the receptor → recruiting of Sos protein → binding and activation of Ras protein → activation of the specific protein kinases → initiation of an action potential e. odorant binds to the olfactory receptor → Gα → adenylate cyclase → cAMP → activation of the nonspecific cation channel → flow of Ca2+ and other molecules into the cell → initiation of an action potential ANSWER: e 34. What is the composition of the Ca2+-binding motif of the EF-hand protein family? .. Powered by Cognero.

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Chapter 13 a. helix-turn-helix b. helix-loop-helix c. β sheet d. leucine zipper e. zinc-fingers ANSWER: b 35. What is the term for the genes that encode proteins that either inhibit cell growth by turning off growthpromoting pathways or trigger the death of tumor cells? a. oncogenes b. proto-oncogenes c. paragenes d. adaptor genes e. tumor-suppressor genes ANSWER: e 36. What is the target for the Ca2+–calmodulin complex that regulates the metabolism of fuel, ionic permeability, neurotransmitter synthesis, and neurotransmitter release? a. Ca2+–ATPase pump b. CaM c. calmodulin-dependent protein kinase d. EF-hand protein family e. small G-proteins ANSWER: c 37. Most signal molecules: a. easily diffuse through the membrane and bind to a receptor in the cytoplasm. b. bind to membrane receptors and transmit information across a membrane without traversing the membrane. c. carry out functions in the nucleus after binding to a receptor in the cell membrane. d. easily diffuse through the membrane, bind to a receptor in the cytoplasm, and carry out functions in the cytoplasm. e. easily diffuse through the membrane and bind to a receptor in the nucleus.

ANSWER: b 38. What amino acids can be phosphorylated? a. tyrosine, serine, threonine b. tyrosine, serine, tryptophan c. serine, threonine, asparagine d. histidine, serine, phenylalanine e. tyrosine, methionine, tryptophan .. Powered by Cognero.

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Chapter 13 ANSWER: a 39. What form of the guanyl nucleotide is bound in the unactivated state of the G-protein? a. GTP b. GDP c. GMP d. dGTP e. cGMP ANSWER: b 40. The mechanism by which insulin-signaling processes might be terminated includes: a. a change in temperature. b. the aggregation of all protein subunits. c. protein dephosphorylation by phosphatases. d. protein degradation by proteases. e. self-inhibition.

ANSWER: c 41. How does the binding of a hormone to a receptor activate a G-protein? a. It causes an exchange of GTP for bound GDP. b. γ subunit is released from binding to the β subunit. c. It causes an exchange of GDP for bound GTP. d. Second messengers activate the G-protein. e. The hormone directly activates the G-protein.

ANSWER: a 42. The enzyme responsible for induction of the phosphoinositide cascade is: a. phospholipase C. b. phospholipase A. c. C-dependent protein (CDP). d. inositol triphosphate. e. adenylate cyclase.

ANSWER: a 43. Select all that apply. Examples of diseases caused by altered G-protein activity include: a. whooping cough. b. arteriosclerosis. c. type 2 diabetes. d. cholera. e. Cushing's disease.

ANSWER: a, d .. Powered by Cognero.

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Chapter 13 44. What step of the generalized scheme of signal-transduction pathways is defective in Cushing's syndrome? a. termination b. release of the primary messenger c. relay of information by the second messenger d. reception of the primary messenger e. activation of effectors ANSWER: e 45. Intrinsic GTPase activity of Gα subunits is important in what step of the generalized scheme of signaltransduction pathways? a. termination b. release of the primary messenger c. relay of information by the second messenger d. reception of the primary messenger e. activation of effectors ANSWER: a 46. The cleavage of PIP2 is important in what step of the generalized scheme of signal-transduction pathways? a. termination b. release of the primary messenger c. relay of information by the second messenger d. reception of the primary messenger e. activation of effectors ANSWER: c 47. The action of 7TM proteins is important in what step of the generalized scheme of signal-transduction pathways? a. termination b. release of the primary messenger c. relay of information by the second messenger d. reception of the primary messenger e. activation of effectors ANSWER: d 48. The α and γ subunits of heterotrimeric G-proteins are anchored to the cell membrane by being covalently linked to: a. protein kinase A. b. small G proteins. c. fatty acids. d. protein kinase C. .. Powered by Cognero.

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Chapter 13 e. phospholipase C. ANSWER: c 49. What is the term for a gene that leads to the transformation of susceptible cell types into cell types with cancer-like characteristics? a. protein kinase A b. calmodulin c. fatty acids d. proto-oncogene e. oncogene ANSWER: e 50. What is the primary messenger responsible for the “fight or flight” response? a. insulin b. epinephrine c. dopamine d. cortisol e. serotonin ANSWER: b 51. What enzyme becomes active when bound to cyclic adenosine monophosphate (cAMP)? a. amylase b. lipase c. adenylate cyclase d. protein kinase A e. phosphatidylinositide-3-kinase ANSWER: d 52. What protein binds to calcium ions and serves as a Ca2+ sensor in eukaryotic cells? a. calmodulin b. hemoglobin c. ß-adrenergic receptor d. protein kinase A e. small G-protein ANSWER: a 53. Seven-transmembrane-helix receptors interact with a variety of G-proteins to activate downstream signaling pathways. Which G-protein stimulates adenylate cyclase enzymatic activity? a. Gαs b. Gαi c. Gαq .. Powered by Cognero.

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Chapter 13 d. Gα12/13 e. Gα15 ANSWER: a 54. Which enzyme becomes active when bound to Ca2+ and diacylglycerol? a. phospholipase C b. cyclic adenosine monophosphate c. protein kinase C d. calmodulin e. protein kinase A ANSWER: c 55. When IP3 binds to the IP3 receptor, what secondary messenger is released from the endoplasmic reticulum? a. ROS b. Ca2+ c. cAMP d. cGMP e. diacylglycerol ANSWER: b 56. What unique characteristic does the insulin receptor have that most other members of the receptor tyrosine kinase class do not? a. The insulin receptor exists as a dimer even in the absence of insulin binding b. The insulin receptor does not undergo conformational changes upon insulin binding c. Insulin subunits are joined together by a peptide bond d. The insulin receptor lacks an extracellular ligand-binding domain e. Insulin binding to the insulin receptor does not lead to phosphorylation ANSWER: a 57. Activation of the insulin receptor results in the mobilization of which transporters to the cell surface? a. Ca2+ b. glucose c. monophosphate d. epinephrine e. oxytocin ANSWER: b 58. What calcium-binding unit is found in many proteins and is characterized by a helix-loop-helix structure? a. calmodulin b. PKA regulatory domain c. EF hand .. Powered by Cognero.

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Chapter 13 d. Excalibur-like motif e. Regulatory linker domains ANSWER: c 59. Advantages of second messengers include: a. A signal can be amplified by producing many second messengers b. Second messengers can freely diffuse to other sites within the cell c. A few common second messengers can be used in multiple signaling pathways d. All of the above e. None of the above ANSWER: d 60. Which protein directly binds to the phosphorylated EGF receptor and facilitates the subsequent binding of other proteins, ultimately leading to the activation of Ras in the cell growth signaling pathway? a. Sos b. Akt c. JAK d. P13K e. Grb-2 ANSWER: e 61. How do the small G-proteins differ from the heterotrimeric G-proteins (select all that apply)? a. They are smaller b. They are not signal proteins c. They are monomeric d. They do not cycle between an active GTP-bound form and an inactive GDP-bound form e. They are not involved in signal transduction ANSWER: a, c 62. What is the difference between a proto-oncogene and an oncogene? a. Proto-oncogenes are located in the nucleus of cells, while oncogenes are found in the cytoplasm b. Proto-oncogenes are inactive genes, while oncogenes are actively transcribed c. Proto-oncogenes are involved in tumor suppression, whereas oncogenes promote normal cell growth d. Proto-oncogenes are mutated forms of oncogenes, while oncogenes are normal cellular genes e. Proto-oncogenes are normally expressed, unmutated forms of a gene, while oncogenes have suffered a mutation that leads to unrestrained growth ANSWER: e 63. Consider this pathway: epinephrine →receptor→cAMP→protein kinase→transcription factor. Identify the secondary messenger: a. epinephrine .. Powered by Cognero.

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Chapter 13 b. protein kinase c. receptor d. cAMP e. transcription factor ANSWER: d 64. Which of the following steps occurs immediately after epinephrine binds to the ß-adrenergic receptor? a. Gαs diffuses to adenylate cyclase b. GTP displaces GDP in Gα, forming the activated protein Gαs c. Adenylate cyclase is phosphorylated by protein kinase A d. cAMP binds to protein kinase A e. Phosphodiesterase breaks down cAMP ANSWER: b 65. Which step in signal transduction involves the binding of a ligand to a receptor, resulting in a conformational change in the receptor? a. Release of the primary messenger b. Signal amplification c. Signal termination and response inhibition d. Reception of the primary messenger e. Transduction and amplification of information by second messengers ANSWER: d 66. How do receptor tyrosine kinases function in signal transduction? a. They bind directly to DNA to initiate gene transcription. b. They release second messengers into the cytoplasm. c. They serve as passive channels for ion transport across the cell membrane. d. They activate downstream signaling pathways by phosphorylating tyrosine residues. e. They degrade ligands upon binding.

ANSWER: d 67. How does the drug imatinib mesylate (Gleevec) primarily function as a therapeutic agent in treating leukemia? a. It promotes overexpression of the Bcr-Abl fusion protein. b. It induces mutations in cancer cells, leading to apoptosis. c. It inhibits the activity of phosphodiesterase enzymes. d. It enhances the production of tyrosine kinase enzymes. e. It specifically inhibits the activity of the Bcr-Abl tyrosine kinase protein.

ANSWER: e 68. An inhibitor of phosphodiesterase activity would have which of the following effects? .. Powered by Cognero.

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Chapter 13 a. Blocking the response to epinephrine b. Decreasing the amount of cAMP in the cytoplasm c. Blocking the activation of G-proteins in response to epinephrine binding to its receptor d. Prolonging the effect of epinephrine by maintaining elevated cAMP levels in the cytoplasm e. Blocking the activation of protein kinase A ANSWER: d 69. Which of the following would be inhibited by a drug that specifically blocks the addition of phosphate groups to proteins? a. G-protein coupled receptor signaling b. Ligand-gated ion channel signaling c. Adenylate cyclase activity d. Phosphatase activity e. Receptor tyrosine kinase activity ANSWER: e

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Chapter 14 1. What is the name for a set of biochemical reactions by which large molecules in food are converted into smaller units? a. metabolism b. catabolism c. digestion d. enzymatic cleavage e. proteolytic destruction ANSWER: c 2. What degrades the components of the meal? a. symbiotic bacteria b. monosaccharides c. peptides d. zymogens e. digestive enzymes ANSWER: e 3. What statement about digestion is CORRECT? a. Digestion begins in the stomach. b. Almost all digestive enzymes are secreted as zymogens, except pepsin. c. Polysaccharides are cleaved into monosaccharides. d. Lipids are converted into fatty acids by α-amylase. e. All of the digestive enzymes are oxidoreductases.

ANSWER: c 4. What statement about zymogens is FALSE? a. All digestive enzymes, except amylase, are synthesized in the inactive forms. b. Before secretion, zymogens exist in granules near the cell membrane. c. Zymogens are inactivated when a part of the precursor is proteolytically cleaved. d. A pancreatic zymogen trypsinogen can be cleaved to form an enzyme such as trypsin. e. Pepsinogen has a small amount of enzyme activity.

ANSWER: c 5. What ions are present in saliva? a. Na+, Cl–, K+, HCO3– b. H+, Cl–, Ca2+ c. H2PO4–, H+, OH– d. Ca2+, K+ e. H+, OH–, K+ ANSWER: a .. Powered by Cognero.

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Chapter 14 6. How do intestinal cells protect themselves from the acidic environment? a. Pancreatic cells release secretin, which promotes the release of bicarbonates. b. Intestinal cells release secretin, which promotes the release of sodium bicarbonate. c. Intestinal cells release sodium bicarbonate, which promotes the release of potassium carbonate. d. Intestinal cells release carbonates as a response to a low pH, which promotes the release of arrestin. e. Intestinal cells release sodium bicarbonate, which promotes the release of secretin.

ANSWER: b 7. Choose the CORRECT sequence of protein digestion: a. Action of high pH and pepsinogen in the stomach → activation of CCK in the pancreas and release of digestive enzymes → in the intestine lumen, oligopeptides are formed by pancreatic proteases → on the internal surface of intestine cells, amino acids, di- and tripeptides are formed by peptidases → transport into the blood. b. Activation of CCK in the pancreas and release of digestive enzymes → in the intestine lumen, diand tripeptides are formed by pancreatic proteases → in the cytoplasm of intestine cells, amino acids are formed by specific peptidases. c. Action of low pH and pepsin in the stomach → activation of CCK in the gallbladder and pancreas leads to the release of digestive enzymes → in the small intestine lumen, oligopeptides are formed by pancreatic proteases → on the external surface of intestine cells, amino acids, di- and tripeptides are formed by specific peptidases. d. Release of digestive enzymes in the small intestine → in the intestine lumen, di- and tripeptides are formed by proteases → in the cytoplasm of intestine cells, amino acids are formed by specific peptidases. e. Action of low pH and pepsin in the stomach → activation of enteropeptidase and release of pancreatic digestive enzymes in the small intestine → in the intestine lumen, oligopeptides are formed by pancreatic proteases → on the external surface of intestine cells, di- and tripeptides are formed by peptidases → in the cytoplasm of intestine cells, amino acids are formed by specific peptidases.

ANSWER: c 8. Which statements about protein digestion are CORRECT? Select all that apply. a. The proteins are denatured by the alkaline environment of the stomach. b. The proteins are not digested at the neutral environment of the intestine. c. The denaturation is caused by the breakage of sulfate bonds due to the low pH. d. The polypeptide products of pepsin digestion stimulate the release of the cholecystokinin. e. The pancreatic proteases only hydrolyze proteins into oligopeptides.

ANSWER: d, e 9. What is gluten enteropathy? a. celiac disease that is triggered in susceptible individuals by hordein in wheat, secalin in rye, and gliadin in barley b. a pancreatic inflammatory disorder that is triggered in susceptible individuals by proteins in wheat, .. Powered by Cognero.

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Chapter 14 rye, and barley c. celiac disease that is triggered by intolerance to animal proteins d. a pancreatic inflammatory disorder that is triggered in susceptible individuals by protein in rice, soya bean, and barley e. an intestinal inflammatory disorder that is triggered in susceptible individuals by gliadin in wheat, secalin in rye, and hordein in barley ANSWER: e 10. What compounds are monosaccharides? a. glucose and lactose b. galactose and maltose c. sucrose and glucose d. galactose and fructose e. lactose and sucrose ANSWER: d 11. α-Amylase cleaves the _____ of a polysaccharide such as starch and does not digest a limit dextrin because of _____. a. α-1,4 bonds; α-1,6 bonds b. α-1,6 bonds; α-1,4 bonds c. α-1,2 bonds; α-1,4 bonds d. α-1,2 bonds; α-1,6 bonds e. α-1,6 bonds; α-1,2 bonds ANSWER: a 12. The limit dextrin is: a. polysaccharide, digested by α-amylase. b. oligosaccharide, which has α-1,4 glycosidic bonds and is not digested by amylase. c. polysaccharide, which has α-1,4 and 1,6 glycosidic bonds. d. oligosaccharide which has α-1,6 bonds and is not digested by amylase. e. oligosaccharide, which has α-1,4 glycosidic bonds and is digested by amylase.

ANSWER: d 13. Choose the CORRECT sequence of starch digestion in the intestine. a. α-Amylase cleaves the α-1,6 bonds of starch to maltose, maltotriose, and limit dextrin → inside of the intestinal cells, α-glucosidase digests maltose, maltotriose, also α-dextrinase digests the limit dextrin to simple sugars. b. α-Amylase cleaves the α-1,4 bonds of starch to maltose, maltotriose and limit dextrin → on the surface of the intestinal cells, α-glucosidase digests maltose, maltotriose, also α-dextrinase digests the limit dextrin to simple sugars. c. α-Amylase cleaves the α-1,4 bonds of starch to maltose, galactose and limit dextrin → on the surface of the intestinal cells, α-glucosidase digests glucose, galactose, also α-dextrinase digests the limit .. Powered by Cognero.

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Chapter 14 dextrin to simple sugars. d. α-Amylase cleaves the α-1,6 bonds of dextrin to maltose → on the surface of the intestinal cells, αglucosidase digests glucose, galactose, also α-dextrinase digests maltose to simple sugars. e. α-Amylase cleaves the α-1,4 bonds of cellulose to maltose, maltotriose and limit dextrin → on the surface of the blood cells, α-glucosidase digests maltose, maltotriose, also α-dextrinase digests the limit dextrin to simple sugars.

ANSWER: b 14. What transporter allows entrance of fructose into the intestinal cells? a. proton pump b. SGLT c. GLUT5 d. GLUT2 e. FABP ANSWER: c 15. What are the products of sucrose hydrolysis? a. glucose and galactose b. mannose and glucose c. galactose and fructose d. glucose and fructose e. two galactose molecules ANSWER: d 16. A newborn baby has symptoms such as intestinal colic, diarrhea, and regurgitations after the feeding. The activity of _____ is reduced. a. α-amylase b. α-dextrinase c. α-glucosidase d. sucrase e. lactase ANSWER: e 17. What would happen first if an enzyme mutation occurred that would seriously inhibit the activity of pancreatic amylase? a. The carbohydrates could not be transported into intestinal cells. b. The carbohydrates would be digested more quickly. c. The carbohydrates would not be shortened to smaller simple sugars. d. The polysaccharides would not be transported into the blood. e. Simple sugars would be resynthesized to di- or oligosaccharides.

ANSWER: c .. Powered by Cognero.

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Chapter 14 18. A young child with normal development had diarrhea, vomiting, and abdominal pain after eating. Exclusion of milk from the diet did not give a positive result. After feeding with sucrose, the level of glucose in the blood barely increased. Why could such symptoms occur? a. hereditary lactase deficiency b. secondary lactase deficiency c. disorders of transporters d. deficiency of amylase e. hereditary insufficiency of sucrase ANSWER: e 19. What is involved in the transport of monosaccharides into the intestinal epithelial cells? Select all that apply. a. a secondary active-transport process b. osmosis c. facilitated diffusion d. filtration e. diffusion without transporters ANSWER: a, c 20. Since lipids are not soluble in water, what is the problem with their digestion? a. Lipids are not exposed to bile salt. b. Lipids are not digestible. c. Lipases cannot degrade large lipid droplets. d. Lipid droplets cannot be converted into an emulsion. e. Amylases cannot degrade large lipid droplets.

ANSWER: c 21. What statement describes lipids rather than proteins and carbohydrates? a. They are not digested in the small intestine. b. They are converted into an emulsion in the intestine. c. They are not fuel for cells and tissues. d. They form globular structures called monoacylglycerols. e. They are not soluble in water.

ANSWER: e 22. What is the structure of micelles? a. The polar head groups of fatty acids are outside the micelle and the carbon chains are inside. b. The polar head groups of proteins are inside the micelle and the carbon chains are outside. c. Bile salts are inside the micelle and the carbon chains are enclosed outside. d. Phospholipids are inside and cholesterol is outside of the micelle. e. Globular structure formed by small proteins in aqueous solution.

ANSWER: a .. Powered by Cognero.

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Chapter 14 23. What is the role of cholesterol in lipid digestion? a. Cholesterol cleaves bonds of lipids. b. Cholesterol transports free fatty acids through the intestinal cell membrane. c. Bile salts necessary for lipid digestion are synthesized from cholesterol. d. Cholesterol emulsifies micelles. e. Cholesterol activates lipases.

ANSWER: c 24. The patient has a delayed outflow of bile from the gallbladder. What may this condition lead to? a. Micelles would be formed correctly. b. Bile salts wouldn't be able to function normally. c. Bile salts wouldn't be formed from cholesterol. d. Chylomicrons wouldn't be able to function normally. e. Emulsification would be impaired or less efficient.

ANSWER: b 25. Why does steatorrhea occur? a. Emulsification takes place faster because of a lack of bile salts. b. Bile salts are not formed and cannot digest lipids. c. Bile salts are synthesized from cholecystokinin and cannot form micelles. d. Production of bile salts is insufficient; micelles cannot be formed. e. Production of bile salts is excessive; micelles cannot be formed.

ANSWER: d 26. What process takes place on the cytoplasmic face of the smooth endoplasmic reticulum? a. resynthesis of the triacylglycerols from fatty acids and monoacylglycerol b. synthesis of chylomicrons from triacylglycerols, proteins, cholesterol c. resynthesis of the triacylglycerols from amino acids and diacylglycerol d. synthesis of chylomicrons from monoacylglycerols e. synthesis of the monoacylglycerol from triacylglycerols ANSWER: a 27. Chylomicrons are: a. transport particles for the transfer of lipids into the intestinal cells. b. transport particles consisting of monoacylglycerols, proteins, and cholesterol. c. specific transport proteins for transport of carbohydrates into cells. d. transport particles consisting of triacylglycerols, proteins, phospholipids and cholesterol. e. globular structures formed by small lipids in aqueous solutions.

ANSWER: d .. Powered by Cognero.

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Chapter 14 28. What is the diameter of chylomicrons? a. 250 nm b. 90 nm c. 200 nm d. 2000 nm e. 98 nm ANSWER: c 29. Choose the correct steps of lipid digestion. a. Emulsification in the stomach due to grinding and mixing → emulsification is enhanced by bile salts in the intestine → triacylglycerols of lipids are degraded to free fatty acids and monoacylglycerol by lipases in the lumen of the small intestine. b. Emulsification in the intestine due to the grinding and mixing → emulsification is enhanced by bile salts in the stomach → triacylglycerols of lipids are degraded to free fatty acids and monoacylglycerol by lipases in the lumen of the small intestine. c. Formation of micelles in the stomach due to the grinding and mixing → triacylglycerols of lipids are degraded to free fatty acids and monoacylglycerol by lipases in the lumen of the small intestine. d. Emulsification by bile salts in the stomach → emulsification is enhanced by zymogens after leaving the stomach → triacylglycerols of lipids are degraded to free fatty acids and monoacylglycerol by lipases in the pancreas. e. Emulsification in the stomach due to the grinding and mixing → emulsification is enhanced by bile salts → triacylglycerols of lipids are degraded to water and monoacylglycerol by lipases in the lumen of the small intestine.

ANSWER: a 30. A person has not eaten fats for a long time, but received a sufficient amount of carbohydrates and proteins.

After that, he was diagnosed with symptoms of a lack of vitamins, especially vitamins A, D, E, and K. Suggest the possible reason of hypovitaminosis. a. Water-soluble vitamins couldn't be absorbed. b. Micelles couldn't be formed to transport vitamins. c. Fat-soluble vitamins couldn't be absorbed. d. The diet, rich in proteins and carbohydrates, didn't contain vitamins. e. Fats take part in catabolism of vitamins.

ANSWER: c 31. Which statement about snake venom is CORRECT? a. Snake venom is a highly modified form of pancreatic juice. b. The composition of the venom does not differ among snakes of the same species. c. Various proteolytic enzymes in the venom degrade intestinal cell membranes. d. Components of snake venoms may be used as drugs to prevent heart attack. e. Neurotoxins in the snake venom kill the prey.

ANSWER: d .. Powered by Cognero.

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Chapter 14 32. Bothropasin is the proteolytic enzyme of venom of the South African pit viper Bothrops jararaca, which is used for treatment of thrombosis. Suggest the possible action of this enzyme. Select all that apply. a. digestion of phospholipids of in the cell membrane b. degradation of fibrous proteins found in membranes c. a neurotoxic effect d. hydrolysis of peptide bonds e. depletion of clotting factors ANSWER: b, d, e 33. What is the significance of the fact that amino acids are moved from the lumen of the small intestine into intestinal cells by transporters but then they are moved into blood by antiporters? a. The transporters are driven by K+-H+ ATPase, which provides energy to completely deplete the intestine of all amino acids. b. The sequential action of transporters and antiporters enables the movement of amino acids against their concentration gradient, allowing for the absorption of amino acids into the bloodstream. c. Peptidases in the intestinal cell membrane also act as amino acid transport channels driving amino acids into intestinal cells. d. The transporters on the lumen side of the intestinal cell will let oligopeptides into the cell, and the anti-porter ensures that these oligopeptides can be utilized by serum proteins. e. The anti-porters can move only polar amino acids.

ANSWER: b 34. The set of reactions that extract biologically useful energy from environmental sources is _______.

ANSWER: catabolism 35. The pancreas releases _____ to buffer the pH of the stomach juices. a. a strong base b. HCl c. NaHCO3 d. pepsin e. ATPase activating protein ANSWER: c 36. Pepsin requires _____ to achieve activity. Select all that apply. a. low pH b. partial denaturation of protein c. zymogen activation d. cholecystokinin activation e. formation of the emulsion ANSWER: a, b, c .. Powered by Cognero.

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Chapter 14 37. What enzyme activates trypsin? a. zymogenase b. pepsin c. trypsinase d. enteropeptidase e. carboxypeptidase ANSWER: d 38. What is NOT a pancreatic protease? a. pepsin b. chymotrypsin c. trypsin d. carboxypeptidase e. elastase ANSWER: a 39. Complex carbohydrates are absorbed as: a. acetyl CoA. b. starch. c. glucose and simple sugars. d. short four-glucose residues. e. glycogen.

ANSWER: c 40. Glucose transport into intestinal epithelial cells takes place by: a. active transport using the sodium sucrose gradient. b. active transport using ATP as the energy source. c. passive transport using the sodium glucose gradient. d. secondary active transport using the potassium glucose gradient. e. secondary active transport using the sodium-glucose linked cotransporter.

ANSWER: e 41. The result of lipase activity in digestion is: a. phospholipid head group hydrolysis. b. monoacylglycerol and free fatty acids. c. bile salt formation. d. hydrolysis of membrane proteins. e. emulsion formation.

ANSWER: b 42. The transport of lipids in the lymph system is carried out by: .. Powered by Cognero.

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Chapter 14 a. lipases. b. micelles. c. cholesterol. d. chylomicrons. e. free fatty acids ANSWER: d 43. What is released by the upper intestine and increases zymogen secretion? a. secretin b. pepsin c. cholecystokinin d. bile salts e. enteropeptidase ANSWER: c 44. What is the advantage of synthesizing enzymes as zymogens? a. It allows them to be packaged in chylomicrons more easily. b. Zymogens can be stored for rapid release when needed. c. It can be activated by the low pH in the stomach. d. Zymogens are resistant to inactivation by kinases. e. Diet pills can be designed to inhibit zymogen hydrolysis.

ANSWER: b 45. Snake venom contains _____, which cause(s) hydrolysis of red blood cells. a. pepsin b. zymogen c. phospholipases d. lipids e. enteropeptidase ANSWER: c 46. The low pH of partially digested food moving from the stomach to the beginning of the small intestine stimulates the release of what enzyme? a. pepsin b. secretin c. chymotrypsin d. lipase e. amylase ANSWER: b 47. What generates stomach acid? .. Powered by Cognero.

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Chapter 14 a. the N+-K+-ATPase pump b. the H+-K+-ATPase pump c. gastric lipase d. pepsinogen e. secretin ANSWER: b 48. Why is digestion important for catabolism even though no useful energy is gained in the process? a. Digestion regulates blood pH by converting lipids into electrolytes. b. Digestion allows for the breakdown of carbohydrates into nucleotides. c. Digestion renders the macromolecules in our meals into more biochemically manageable fragments. d. Digestion aids in the synthesis of ATP from fatty acids. e. Digestion hydrolyzes fats to polysaccharides that will be used in catabolism.

ANSWER: c 49. In digestion, proteins are hydrolyzed to _____, polysaccharides are hydrolyzed to _____, and fats are hydrolyzed to _____ and _____. a. monosaccharides, amino acids, fatty acids, and glycerol b. amino acids, fatty acids, glucose, and triglycerides c. nucleotides, monosaccharides, glycerol, and phospholipids d. amino acids, simple sugars, fatty acids, and glycerol e. simple sugars, nucleotides, fructose, and cholesterol ANSWER: d 50. What is the major molecular product of the second stage of digestion? a. acetyl CoA b. ATP c. glucose d. lipase e. bile salts ANSWER: a 51. How is acetyl CoA related to ATP? a. Acetyl CoA inhibits the synthesis of ATP. b. Acetyl CoA is a precursor molecule for ATP synthesis. c. Acetyl CoA competes with ATP for binding to enzymes. d. Acetyl CoA is converted directly into ATP. e. Acetyl CoA serves as a cofactor for ATP breakdown.

ANSWER: b 52. In digestion, proteins are converted into amino acids by _____, polysaccharides are cleaved into .. Powered by Cognero.

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Chapter 14 monosaccharides by _____, and lipids are converted into fatty acids by _____. a. α-amylase, proteases, lipases b. nucleases, lipases, proteases c. proteases, lipases, nucleases d. lipases, peptidases, α-amylase e. proteases, α-amylase, lipases ANSWER: e 53. What is the main cause of gastroesophageal reflux disease (GERD)? a. an overabundance of pepsin in the stomach b. an underactive gastric proton pump c. an overactive gastric proton pump d. elevated pH in the stomach e. a lack of membrane proteins in the stomach lining ANSWER: c 54. Hyponatremia is a condition that occurs when the level of sodium in the blood is too low. How could hyponatremia potentially affect the functioning of the sodium-glucose linked transporter? a. Hyponatremia could enhance the activity of the sodium-glucose linked transporter, improving glucose absorption in the intestines. b. Hyponatremia could reduce the activity of the sodium-glucose linked transporter, resulting in increased glucose uptake in the kidneys. c. Hyponatremia could disrupt the sodium gradient necessary for the function of the sodium-glucose linked transporter, leading to decreased glucose absorption in the intestines. d. Hyponatremia could increase the expression of the sodium-glucose linked transporter, resulting in excessive sodium reabsorption in the kidneys. e. Hyponatremia would have no impact on the functioning of the sodium-glucose linked transporter, as it primarily affects water balance in the body.

ANSWER: c 55. Why is the digestion of pasta more efficient after the pasta is heated in water? a. Cooking in water increases the protein content of the pasta, facilitating digestion by protease enzymes. b. Cooking in water enhances the absorption of lipids from the pasta, aiding in their breakdown by lipase enzymes. c. Cooking in water disrupts the structural integrity of the pasta, allowing α-amylase easier access to the α-1,4 linkage in starch. d. Cooking in water decreases the fiber content of the pasta, promoting faster transit through the digestive tract. e. Cooking in water increases the mineral content of the pasta, facilitating the action of mineraldependent enzymes.

ANSWER: c .. Powered by Cognero.

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Chapter 14 56. What might happen if digestive enzymes were released in their active form in the pancreas instead of being stored as zymogens? a. The digestive enzymes could prematurely digest and destroy pancreatic tissue. b. The digestive enzymes could become inactive due to the acidic environment of the pancreas. c. The digestive enzymes could be unable to reach the small intestine and would accumulate in the pancreas. d. The digestive enzymes could bypass the stomach and directly enter the bloodstream, causing systemic inflammation. e. The digestive enzymes would be unaffected and continue to function normally in the small intestine.

ANSWER: a 57. Why do triacylglycerols need to be degraded to fatty acids to be absorbed across the intestinal epithelium? a. Triacylglycerols are too large to pass through the tight junctions between intestinal epithelial cells. b. Fatty acids exhibit greater solubility in the lipid bilayer of intestinal epithelial cells compared to intact triacylglycerols. c. Conversion of triacylglycerols to fatty acids enhances the binding affinity of the lipids to transport proteins on the apical surface of intestinal epithelial cells. d. Fatty acids undergo facilitated diffusion across the intestinal epithelium, while triacylglycerols require active transport mechanisms. e. Triacylglycerols are enzymatically broken down into fatty acids to prevent their aggregation and enhance solubility for transport.

ANSWER: b 58. Where are triacylglycerols resynthesized in cells? a. the pancreas b. the esophagus c. the GLUT5 transporter d. the smooth endoplasmic reticulum e. the mitochondria ANSWER: d 59. How could trypsinogen deficiency lead to digestive issues? a. Trypsinogen deficiency results in decreased secretion of bile salts, leading to impaired fat digestion. b. Trypsinogen deficiency causes excessive production of gastric acid, resulting in acid reflux and gastric ulcers. c. Trypsinogen deficiency leads to insufficient activation of pancreatic enzymes, impairing the digestion of proteins. d. Trypsinogen deficiency increases the activity of intestinal peptidases, leading to accelerated protein digestion. e. Trypsinogen deficiency disrupts the release of intestinal hormones, affecting the regulation of digestive processes. .. Powered by Cognero.

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Chapter 14 ANSWER: c 60. What are the functions of bile salts in digestion? Select all that apply. a. Emulsification of dietary lipids to increase their surface area for enzymatic action. b. Facilitation of the absorption of fat-soluble vitamins (A, D, E, and K). c. Promotion of protein digestion by activating pancreatic proteases. d. Neutralization of gastric acid to create an optimal pH environment for enzyme activity in the small intestine. e. Formation of micelles to solubilize lipids and aid in their absorption by intestinal cells.

ANSWER: a, b, e 61. What is the main component of chylomicrons? a. triacylglycerols b. glucose c. micelles d. bile salts e. ATP ANSWER: a 62. You are investigating a murder mystery party, attempting to identify the culprit of a crime based on attendees' meal choices and the fact that the perpetrator’s blood appeared milky. Which suspect below is most likely to have committed this crime? a. Mary, who opted for a light spring mix salad for dinner b. Jon, who indulged in a ribeye steak with cream sauce and deep-fried potatoes c. Alex, who had a bowl of cereal with skim milk d. Lori, who forgot to eat dinner altogether e. Tasha, who enjoyed a bowl of chicken noodle soup ANSWER: b 63. What is the role of saliva in the digestive system? a. Saliva contains α-amylase, which starts the chemical digestion in the mouth by breaking down proteins. b. Saliva contains lipase, which starts chemical digestion in the mouth by breaking down proteins. c. Saliva contains maltase, which starts chemical digestion in the mouth by breaking down carbohydrates. d. Saliva contains α-amylase, which starts chemical digestion in the mouth by breaking down carbohydrates. e. Saliva contains sucrase, which starts the chemical digestion in the mouth by breaking down lipids.

ANSWER: d 64. 75% of the world’s population is lactose intolerant. Why is this? a. They have an insufficient amount of sucrase. .. Powered by Cognero.

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Chapter 14 b. They are allergic to lactose. c. They are lacking in lactase. d. They can break down lactose but can’t metabolize the resulting products. e. Their digestive enzymes haven’t been activated efficiently.

ANSWER: c 65. A patient presents to the hospital with the complaint of pain in his abdomen. Imaging reveals ulcers in his upper small intestine. Further tests reveal that the ulcers are caused by acid entering the small intestine not being properly neutralized. What organ in the digestive system is malfunctioning, causing these acidic secretions to harm the small intestine? a. the stomach b. the liver c. the pancreas d. the gallbladder e. the small intestine ANSWER: c

.. Powered by Cognero.

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Chapter 15 1. What purpose is NOT a major one for a continual input of free energy for mammals? a. muscle contraction b. cellular movement c. biosynthetic processes d. facilitated transport e. ATP-mediated transport ANSWER: d 2. Which statement about principles of energy that flows in living systems is FALSE? a. The formation of energy-rich molecules relies on the energy obtained from the environment. b. The number of reaction types used in metabolic pathways is unlimited. c. Metabolic pathways are predominantly controlled by allosteric regulation. d. Organization of enzymes into large complexes facilitates metabolic processes. e. Energy-releasing and energy-requiring processes are linked by molecules with a high-phosphoryltransfer potential.

ANSWER: b 3. Which statement about the control of metabolism is TRUE? a. Each metabolic pathway has unique reactions through which the whole pathway is controlled. b. Only reversible reactions can be used to regulate the whole pathway. c. Since metabolic pathways are interdependent, the regulated reactions can be identical between pathways. d. Biosynthetic and degradative pathways have nothing in common. e. In an amphibolic pathway, the regulated reaction can be reversible.

ANSWER: a 4. What is the common energy currency of the cell? a. ΔG°´ b. ΔG c. NAD+ d. acetyl CoA e. ATP ANSWER: e 5. Which statement about donors of free energy is TRUE? a. ATP is the only free-energy donor in the cell. b. ATP is the predominant free-energy donor because unlike other nucleotide triphosphates, its hydrolysis is exergonic. c. UTP can sometimes be the free-energy donor. d. Linkages between phosphoryl groups of UTP contain more energy than those of ATP. e. Linkages between phosphoryl groups of ATP contain more energy than those of UTP. .. Powered by Cognero.

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Chapter 15 ANSWER: c 6. Which statement about ATP is TRUE? a. The energy is stored in three phosphoanhydride linkages of ATP. b. The hydrolysis of α or β bonds of ATP releases a large amount of energy. c. Under typical cellular conditions, the hydrolysis of both energy-rich bonds of ATP together releases about 12 kcal mol-1. d. The hydrolysis of β or γ bonds of ATP releases a large amount of energy. e. The free energy is liberated when ATP is hydrolyzed to AMP and Pi. ANSWER: d 7. The standard free energy for hydrolysis of the phosphoanhydride bond of ATP is –30.5 kJ mol–1 (–7.3 kcal mol–1) and for hydrolysis of the phosphoester bond of fructose 1,6-bisphosphate is –16.7 kJ mol–1 (–4.0 kcal mol–1). Which of the molecules has a greater phosphoryl transfer potential and why? a. ATP; because it is spontaneous b. fructose 1,6-bisphosphate; because phosphoester bonds are higher in energy c. ATP; because only phosphoanhydride bonds provide energy that is high enough d. fructose 1,6-bisphosphate; because the overall energy difference between reactants and products is favorable e. ATP; because the overall energy difference between reactants and products is favorable ANSWER: e 8. What molecule has a phosphoryl-transfer potential lower than that of ATP? a. UTP b. PEP c. creatine-phosphate d. glucose-1-phosphate e. 1,3-BPG ANSWER: d 9. What process can increase the phosphoryl-transfer potential of ATP? a. decreased solvation of products of ATP hydrolysis b. increase in the number of molecules upon hydrolysis c. decreased net charge of products due to resonance d. decrease in the number of molecules upon hydrolysis e. increased net charge of reactants due to resonance ANSWER: b 10. Choose the CORRECT statements about ATP. a. The high phosphoryl-transfer potential of ATP can be explained by features of the ATP structure. .. Powered by Cognero.

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Chapter 15 b. Five factors differentiate stability of the reactants and products in ATP hydrolysis. c. The triphosphate unit of ATP carries about three negative charges at pH 7. d. The entropy of the reactants is greater than the entropy of the products in ATP hydrolysis. e. ΔG°´ depends on the difference in free energies of the products.

ANSWER: a 11. Choose the CORRECT statement explaining importance of hydration for the phosphoryl-transfer potential of ATP. a. Effective hydrogen bonding with water molecules increases the free energy for ATP synthesis from the products of its hydrolysis. b. High resonance stabilization of orthophosphate makes the phosphoryl transfer to ADP less favorable. c. Effective hydrogen bonding with water molecules decreases the free energy for ATP synthesis from the products of its hydrolysis. d. A high concentration of water allows obtaining more energy from the hydrolysis of ATP, which increases the phosphoryl-transfer potential. e. A high concentration of water allows obtaining more energy from the hydrolysis of ATP, which decreases the phosphoryl-transfer potential.

ANSWER: a 12. Which statement explaining the importance of electrostatic repulsion for the phosphoryl-transfer potential of ATP is TRUE? a. At a physiological pH, hydrolysis of ATP decreases the number of negative charges. b. Juxtaposition of negative charges of β and γ phosphoryl groups gives a lower free energy than that for free orthophosphate. c. A close proximity between phosphoryl groups of ATP gives a higher energy than does free orthophosphate in a solution. d. Juxtaposition of negative charges of α and β phosphoryl groups gives a lower free energy than that for free orthophosphate. e. Water more readily stabilizes negative charges of products of ATP hydrolysis than of ATP itself.

ANSWER: c 13. What is the source of energy for muscle contraction during seconds of intense sprints? a. ATP b. creatine phosphate c. aerobic metabolism d. anaerobic metabolism e. glycogen ANSWER: b 14. What is the main source of energy for muscle contraction during minutes of exercise? a. ATP b. creatine phosphate .. Powered by Cognero.

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Chapter 15 c. aerobic metabolism d. anaerobic metabolism e. glycogen ANSWER: d 15. Judging by the given reactions and their standard free energies, choose the CORRECT statement.

A + B →C ΔG°´ = –36 kJ mol–1 C+D E ΔG°´ = + 12 kJ mol–1 E + F → A ΔG°´ = +5 kJ mol–1 a. The ΔG°´ of the reaction B + E + F → C should be positive. b. Formation of compound E is coupled to the formation of A. c. E is a shared intermediate that couples formation of C and A. d. Overall free energy of the pathway should be positive. e. B is a shared intermediate that couples formation of C and A. ANSWER: c 16. How would the equilibrium constant change if the reaction X with ΔG°´= +12.2 kJ mol–1 was coupled to the hydrolysis of ATP (ΔG°´ = –30.5 kJ mol–1)? Consider that R × T is equal to 2.47. a. would increase by a factor of 105 b. would increase by 5.15 c. would decrease by a factor of 106 d. would decrease by 5.15 e. would increase by a factor of 106 ANSWER: e 17. Which statement about the role of phosphoryl groups in the cell is FALSE? a. Phosphoryl groups are used as a tag for the enzymes of reductive biosynthesis. b. Kinetic stability of the phosphate esters allows their use for regulatory purposes. c. Thermodynamic stability of the phosphate esters facilitates their manipulation by enzymes. d. Phosphorylation prevents diffusion of molecules through the plasma membrane. e. Negative charges of phosphate esters decrease the chance of nonspecific hydrolysis in aqueous environment.

ANSWER: c 18. What is the origin of energy for anabolic reactions in phototrophs? a. oxidation of carbon to CO2 b. electrochemical potential of stored glycogen c. reduction of pyruvate to lactate d. ATP .. Powered by Cognero.

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Chapter 15 e. sunlight ANSWER: e 19. Why is the cell concentration of ATP 10 to 100 times higher than KM for most enzymes that use ATP as a substrate? a. to increase protein solubility b. long-term storage of energy c. to decrease protein solubility d. for unexpected activity e. for ATP-mediated transport ANSWER: a 20. What molecule should be giving more energy upon catabolic conversion? a.

b.

c.

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e.

ANSWER: a 21. Which step for trapping of energy by the cell is TRUE? a. oxidation of fuel molecules → synthesis of molecules with a high-phosphoryl transfer potential b. energy of light → synthesis of molecules with a high-phosphoryl transfer potential c. oxidation of fuel molecules → synthesis of compounds with a high-phosphoryl transfer potential → synthesis of ATP d. energy of light → synthesis of ADP → synthesis of ATP e. oxidation of fuel molecules → synthesis of ATP → synthesis of compounds with a high-phosphoryl transfer potential ANSWER: c .. Powered by Cognero.

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Chapter 15 22. Choose the most precise description of an activated carrier. a. It is ATP. b. It is a compound that can transfer a functional group with negative ΔG of the transfer. c. It is a compound that transfers a phosphoryl group in an exergonic process. d. It is a compound that transfers a phosphoryl group in an endergonic process. e. It transfers electrons.

ANSWER: b 23. Choose the coenzyme involved in the transfer of electrons in catabolic processes. a. NADP+ b. FAD+ c. ATP d. biotin e. CoA ANSWER: b 24. How many protons can bind to FAD+ upon dehydrogenation of a substrate? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: b 25. What difference between NAD+ and FAD+ is FALSE? a. FAD+ has two reactive sites. b. NAD+ is synthesized from niacin. c. NAD+ has two reactive sites. d. FAD+ can accept two electrons. e. NAD+ can accept two electrons.

ANSWER: c 26. How many reactive sites are there in NADP+? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: a .. Powered by Cognero.

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Chapter 15 27. How many reactive sites are there in coenzyme A? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: a 28. What is the main reason hydrolysis of the thioester bond gives more energy than the oxygen ester bond? a. due to charge repulsion b. due to a favorable entropy change c. due to better solvation d. due to resonance stabilization e. due to charge attraction ANSWER: d 29. What criterion is necessary for the metabolic pathway to exist? a. All enzymes should have features that allow precise recognition of regulatory molecules. b. All individual reactions should be thermodynamically stable. c. The overall pathway should be exergonic. d. All individual reactions should be kinetically stable. e. The overall pathway should be endergonic.

ANSWER: c 30. What factor is NOT used to regulate metabolism? a. concentration of the enzyme b. concentration of the substrate c. concentration of hormones d. availability of vitamins e. accessibility of the substrate ANSWER: d 31. What is the energy charge of resting muscle if concentrations of ATP, ADP, AMP, and Pi are 4 mM, 13 µM, 3 µM, and 10 mM, respectively? a. 0 b. 0.25 c. 0.5 d. 0.75 e. 1 ANSWER: e .. Powered by Cognero.

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Chapter 15 32. Which pathways should be stimulated in a muscle cell in resting state? a. catabolic pathways because energy charge is low b. catabolic pathways because energy charge is high c. amphibolic pathways because energy charge is 0 d. anabolic pathways because energy charge is low e. anabolic pathways because energy charge is high ANSWER: e 33. What is the phosphorylation potential of resting muscle if concentrations of ATP, ADP, AMP, and Pi are 4 mM, 13 µM, 3 µM, and 10 mM, respectively? a. 0.1 b. 0.4 c. 0.5 d. 1 e. 2.5 ANSWER: b 34. Calculate the ΔG of the hexokinase, which catalyzes the following reaction: glucose + ATP → glucose 6-phosphate + ADP Consider that concentrations of glucose and glucose 6-phsphate (G6P) are 2 mM and 83 µM, and of ATP and ADP are 4 mM and 13 µM; ΔG°´ of G6P hydrolysis is –13.8 kJ mol–1, and ΔG°´of ATP hydrolysis is –30.5 kJ mol–1. a. –5.78 kJ mol–1 b. –10.92 kJ mol–1 c. –38.7 kJ mol–1 d. –27.63 kJ mol–1 e. –55.22 kJ mol–1 ANSWER: c 35. Which vitamin is important for the transport of electrons during catabolic reactions? a. panthotenate b. biotin c. niacin d. thiamine e. adenine ANSWER: c 36. Which vitamin is important for the transport of electrons during reductive biosynthesis? a. panthotenate b. biotin .. Powered by Cognero.

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Chapter 15 c. niacin d. thiamine e. adenine ANSWER: c 37. Which vitamin is important for synthesis of lipids? a. pantothenate b. biotin c. niacin d. thiamine e. adenine ANSWER: a 38. Reaction pathways that only transform fuels into cellular energy are: a. anabolic. b. catabolic. c. allosteric. d. amphibolic. e. catalytic.

ANSWER: b 39. Metabolic pathways that require an input of energy and are often biosynthetic processes are: a. anabolic. b. catabolic. c. allosteric. d. amphibolic. e. catalytic.

ANSWER: a 40. A thermodynamically unfavorable reaction can be driven by a thermodynamically favorable reaction it is _______ to.

ANSWER: coupled 41. Choose the electron donor used for reductive biosynthesis. a. NADH b. NADPH c. FADH2 d. CoA-SH e. ATP ANSWER: b .. Powered by Cognero.

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Chapter 15 42. Choose the correct coenzyme – transferring group pair? a. NADH – acyl b. tetrahydrofolate – electrons c. coenzyme A – acyl d. lipoamide – aldehyde e. thiamine pyrophosphate – glucose ANSWER: c 43. Phosphorylation of fructose 6-phosphate is an endergonic reaction with a ΔG° of 16.3 kJ/mol. How do cells overcome this thermodynamic barrier for this reaction under standard conditions? a. The enzyme that catalyzes this reaction couples it with the condensation of ADP and inorganic phosphate, resulting in an overall ΔG° of –46.8 kJ/mol. b. The enzyme that catalyzes this reaction couples it with the hydrolysis of ATP to ADP and inorganic phosphate, resulting in an overall ΔG° of –14.2 kJ/mol. c. This reaction will proceed to the right because Keq is small. d. This reaction will proceed to the right because Keq is negative. e. The reaction can be driven forward by uncoupling the reaction from the hydrolysis of ATP. ANSWER: b 44. You are interested in studying bacteria found in peat swamps and you identify a new bacterium that you believe is a chemotroph. Which of the following would you use to verify your belief? a. screen for ATP synthesis b. screen for enzymes that oxidize carbon c. screen for light gathering structures d. screen for digestive enzymes e. screen for linked reactions ANSWER: c 45. The hydrolysis of a phosphate group from ATP releases 30.5 kJ/mol, whereas the hydrolysis of a phosphate from glucose 6-phosphate releases only 13.82 kJ/mol. In that the product is the same, what accounts for the difference? a. ATP has greater resonance stabilization than the product orthophosphate. b. There is a greater increase in entropy when ATP is hydrolyzed. c. Water hydrates ATP to a greater degree than glucose 6-phosphate. d. ATP has a larger phosphoryl-transfer potential. e. The phosphate ester in ATP is more thermodynamically stable than in glucose 6-phosphate.

ANSWER: d 46. This energy source is used to regenerate ATP from ADP and Pi. a. oxidation of carbon to CO2 .. Powered by Cognero.

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Chapter 15 b. electrochemical potential of stored glycogen c. reduction of pyruvate to lactate d. reduction of pyruvate to acetyl CoA e. hydrolysis of glycerol 3-phosphate ANSWER: a 47. FAD is an electron carrier that is derived from the vitamin: a. pantothenate. b. niacin. c. riboflavin. d. thiamine. e. folate.

ANSWER: c 48. Select all that apply. Pantothenate kinase–associated degeneration: a. is a predominantly neurological disorder. b. can cause anemia. c. affects tissues that are dependent on aerobic metabolism. d. is a pathological condition characterized by iron accumulation in the muscles. e. is a disorder when a patient lacks the vitamin pantothenate.

ANSWER: a, c 49. Which activated carriers contain adenosine phosphate units? a. NADH b. FADH2 c. coenzyme A d. FMNH2 e. biotin ANSWER: a, b, c 50. Some of the mechanisms by which enzyme catalytic activity is controlled are: a. allosteric control. b. rate of transcription. c. covalent modification. d. availability of substrates. e. energy status of the cell.

ANSWER: a, c 51. Which of the following is NOT a reason for why living organisms require continual input of free energy? a. Continual input of free energy is necessary for mechanical work in muscle contraction and cellular movements. .. Powered by Cognero.

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Chapter 15 b. Active transport of molecules and ions c. Osmosis and simple diffusion of gasses down their concentration gradient d. Synthesis of macromolecules and other biomolecules from simple precursors ANSWER: c 52. _______ capture energy from sunlight, whereas _________ capture energy from the oxidation of chemicals a. Phototrophs, heterotrophs b. Phototrophs, autotrophs c. Autotrophs, chemotrophs d. Autotrophs, heterotrophs e. Phototrophs, chemotrophs ANSWER: e 53. Which of the following is NOT a unifying theme common to all metabolic reactions? a. Metabolic reactions are nonspontaneous and do not take place unless biochemical catalysts are present. b. Metabolism is a coherent network containing many common motifs. c. Metabolism involves numerous reactions but employs only a few different types of mechanisms. d. Metabolic reactions are highly regulated. e. Many of the enzymes used in metabolism are organized into large complexes.

ANSWER: a 54. What molecule has a phosphoryl-transfer potential lower than that of ATP? a. UTP b. 1,3 BFG c. Glucose-1-phosphate d. PEP e. Pyruvate ANSWER: c 55. A biochemist is investigating metabolism in kidney cells and discovers that, dependent on the energy conditions, glucose in the cell is either synthesized or broken down. This is an example of which kind of metabolic pathway? a. Anabolic b. Catabolic c. Heterobolic d. Amphipathic e. Amphibolic ANSWER: e 56. During spurts of intense exercise, energy is derived from the breakdown of which molecule in muscle? .. Powered by Cognero.

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Chapter 15 a. glucose b. glycogen c. ATP d. creatine phosphate e. UTP ANSWER: d 57. Which of the following is a requirement for a reaction to occur spontaneously? a. an enzyme must be involved b. the change in free energy is negative c. the heat of reaction is negative d. the activation energy for the reaction to occur is achieved ANSWER: b 58. Which of the following is not a factor in the high phosphoryl-transfer potential of ATP? a. at a pH of 7, the triphosphate unit of ATP carries about four negative charges b. orthophosphate has greater resonance stability than any of the phosphates in ATP c. water binds to ADP and Pi thereby rendering the synthesis of ATP less favorable d. the entropy of the products is less than that of the starting ATP molecule ANSWER: d 59. Which of the following is the predominant electron donor used for reductive biosynthesis in living cells? a. NADH b. ATP c. cAMP d. FADH2 e. NADPH ANSWER: e 60. From which vitamin is the electron carrier FADH2 derived? a. niacin b. riboflavin c. thiamine d. cobalamin e. folic acid ANSWER: b 61. During periods of strenuous exercise, reservoirs of high-potential phosphoryl groups are readily transferred from creatine phosphate to ADP by which enzyme? a. ADP phosphatase b. creatine kinase .. Powered by Cognero.

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Chapter 15 c. creatine phosphatase d. phosphoryl transferase e. ADP kinase ANSWER: c 62. What characteristic of phosphates makes them useful in biochemical systems? a. phosphate esters are thermodynamically unstable but kinetically stable in aqueous solutions b. phosphate ester formation is quick and does not require the use of an enzyme c. the addition of phosphate groups has minimal impact on target molecule conformation d. phosphate hydrolysis releases free energy but is not spontaneous ANSWER: a 63. Which of the following is NOT a factor commonly used to regulate metabolism? a. substrate accessibility and local concentration b. enzyme concentration c. enzyme catalytic activity d. substrate cleavage ANSWER: c 64. In animals, 90% of ATP is generated by the oxidation of carbon fuels through the use of proton gradients.

This process is known as: a. phosphoryl generation b. oxidative phosphorylation c. ATP oxidation d. none of the above ANSWER: b 65. What is the origin of the energy needed for anabolic reactions in chemotrophs? a. sunlight b. reduction of pyruvate to lactate c. degradation of ATP d. oxidation of ingested carbon-containing molecules to CO2 ANSWER: d 66. Which of the following statements are true? a. when the covalent bond linking the phosphates in ATP is broken, energy is released because new bonds and noncovalent interactions are formed b. ATP is the cellular energy currency because it has the highest phosphoryl-transfer potential of any biological molecule c. ATP has two phosphoanhydride bridges d. none of the above .. Powered by Cognero.

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Chapter 15 ANSWER: a, c 67. ATP is an activated carrier of phosphoryl groups because phosphoryl transfer from ATP is a(n) ___________ process. a. endergonic b. spontaneous c. exergonic d. energetically unfavorable ANSWER: c 68. An acyl group can be linked to the sulfhydryl group in CoA by which type of bond? a. phosphodiester b. hydrogen c. thioester d. disulfide ANSWER: c 69. Why is the hydrolysis of a thioester bond more thermodynamically favorable than that of an oxygen ester? a. The electrons in the carbon-oxygen double bond form less stable resonance structures with the carbon-sulfur bond than with the carbon-oxygen bond b. The electrons in the carbon-oxygen double bond form more stable resonance structures with the carbon-sulfur bond than with the carbon-oxygen bond c. The hydrolysis of a thioester occurs spontaneously, without the requirement of an enzyme d. None of the above ANSWER: a 70. Why is it biologically more efficient to ingest rather than synthesize vitamins? a. the enzymes that work in vitamin-generation pathways are energetically inefficient b. the biochemical pathways that involve vitamins are relatively uncomplex c. ingesting vitamins is energetically more favorable than synthesizing the enzymes required to form vitamins from simpler molecules d. ingestion is a quicker process than metabolism ANSWER: c

.. Powered by Cognero.

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Chapter 16 1. Glycolysis is related to the synthesis of fatty acids through the conversion of pyruvate to _______.

ANSWER: acetyl CoA 2. In which compartment of the eukaryotic cell does glycolysis occur? a. membrane b. cytoplasm c. nucleus d. mitochondria e. endoplasmic reticulum ANSWER: b 3. What is the event that ends the first stage of glycolysis? a. synthesis of ATP b. synthesis of glucose 6-phosphate c. synthesis of triose phosphates d. synthesis of pyruvate e. synthesis of NADH ANSWER: c 4. What is the event that ends the second stage of glycolysis? a. synthesis of ATP b. synthesis of glucose 6-phosphate c. synthesis of triose phosphates d. synthesis of pyruvate e. synthesis of NADH ANSWER: d 5. Which property is important for the energetic efficiency of glycolysis? a. equilibrium between triose phosphates b. hydrophobicity of the kinase cleft c. isomerization of aldoses to ketoses d. induced fit e. phosphoryl-transfer potential ANSWER: e 6. Choose the molecule that has a phosphoryl-transfer potential higher than that of ATP. a. 1,3-bisphosphoglycerate b. glyceraldehyde 3-phosphate c. dihydroxyacetone phosphate d. fructose 1,6-bisphosphate e. NAD+ .. Powered by Cognero.

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Chapter 16 ANSWER: a 7. What is the net yield of ATP upon conversion of one molecule of glucose to two molecules of pyruvate? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: b 8. What process takes part in NAD+ regeneration? a. metabolism of fructose b. synthesis of triose phosphates c. metabolism of pyruvate d. synthesis of pyruvate e. aldolase reaction ANSWER: c 9. What class of enzymes exploits NAD+ in glycolysis? a. kinase b. mutase c. decarboxylase d. dehydrogenase e. enolase ANSWER: d 10. What is the process that sustains the redox balance of glycolysis under anaerobic conditions? a. transport of lactate from the cell b. regeneration of NAD+ in the citric acid cycle c. formation of NADH in the reaction of glyceraldehyde 3-phosphate dehydrogenase d. regeneration of NAD+ in fermentation processes e. regeneration of NAD+ in the mitochondrial electron-transport chain ANSWER: d 11. What is the process that sustains the redox balance of glycolysis under aerobic conditions? a. transport of lactate from the cell b. regeneration of NAD+ in the citric acid cycle c. formation of NADH in the reaction of glyceraldehyde 3-phosphate dehydrogenase d. regeneration of NAD+ in fermentation processes .. Powered by Cognero.

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Chapter 16 e. regeneration of NAD+ in the mitochondrial electron-transport chain ANSWER: e 12. Which organism's fermentation is the only way to generate energy from the decomposition of glucose? a. prokaryote b. eukaryote c. obligate anaerobe d. yeast e. facultative anaerobe ANSWER: c 13. What disease is characterized by the development of a cataract and by the absence of a specific transferase in red blood cells? a. beriberi b. classic galactosemia c. pellagra d. Warburg effect e. lactose intolerance ANSWER: b 14. How many irreversible reactions are there in glycolysis and how many of them are involved in allosteric regulation? a. Three irreversible reactions; all of them are allosterically regulated. b. Three irreversible reactions; only the phosphofructokinase reaction is allosterically regulated. c. Three irreversible reactions; only two of them are allosterically regulated. d. Four irreversible reactions; only three of them are allosterically regulated. e. Three irreversible reactions; only the hexokinase reaction is allosterically regulated.

ANSWER: a 15. Which of the following does NOT inhibit the activity of phosphofructokinase in muscle cells? a. high level of ATP b. high pH c. low level of AMP d. low pH e. high energy charge of the cell ANSWER: b 16. The regulation of which glycolytic enzyme indirectly depends on the cell energy charge in muscle cells? a. aldolase b. hexokinase c. enolase .. Powered by Cognero.

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Chapter 16 d. phosphofructokinase e. pyruvate kinase ANSWER: b 17. Which glycolytic enzyme is the MOST important for regulation of glycolysis in mammals? a. aldolase b. hexokinase c. enolase d. phosphofructokinase e. pyruvate kinase ANSWER: d 18. Which statement about glycolytic kinases is FALSE? a. Reactions catalyzed by kinases in the first stage of glycolysis use ATP as the donor of the phosphate group, while in the second stage they use ADP as the acceptor of the phosphate group. b. Phosphofructokinase catalyzes the committed step of glycolysis in muscle cells. c. Allosteric regulation of liver pyruvate kinase depends on the energy charge of the cell. d. The expression of glycolytic kinases is controlled by the partial pressure of oxygen in tissues. e. All of the glycolytic reactions catalyzed by kinases are irreversible.

ANSWER: e 19. Which statement about β pancreatic cells is FALSE? a. Only the GLUT2 glucose transporter is present in these cells. b. The metabolism of glucose in these cells is translated into a physiological response by ion channels. c. They secrete insulin in response to a high concentration of glucose in blood. d. They are able to respond only to a high concentration of glucose because of the hexokinase isozyme that has a high KM value. e. The activity of these cells helps regulate synthesis of glycogen in the liver.

ANSWER: a 20. The actions of which glucose transporters indirectly lead to the secretion of insulin and directly lead to the increased uptake of glucose in the presence of insulin? a. GLUT1 and GLUT2 b. GLUT1 and GLUT3 c. GLUT2 and GLUT3 d. GLUT2 and GLUT4 e. GLUT1 and GLUT5 ANSWER: d 21. What reactions take place in the alcoholic fermentation of glucose starting with pyruvate? a. First, pyruvate is decarboxylated and then acetaldehyde is oxidized by NAD+. .. Powered by Cognero.

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Chapter 16 b. Pyruvate is reduced by NADH to yield lactate. c. First, pyruvate is decarboxylated and then acetaldehyde is reduced by NADH. d. Pyruvate is oxidized by NAD+ to yield lactate. e. Pyruvate is dehydrated to acetaldehyde, which is then decarboxylated to ethanol.

ANSWER: c 22. What reactions take place in the lactic acid fermentation of glucose starting with pyruvate? a. First, pyruvate is decarboxylated and then formed acetaldehyde is oxidized by NAD+. b. Pyruvate is reduced by NADH to yield lactate. c. First, pyruvate is decarboxylated and then formed acetaldehyde is reduced by NADH. d. Pyruvate is oxidized by NAD+ to yield lactate. e. Pyruvate is dehydrated to acetaldehyde, which is then decarboxylated to lactate.

ANSWER: b 23. What is the basis of maintaining the glycolytic redox balance in fermentation processes? a. production of carbon dioxide b. utilization of pyruvate c. decrease of pH in the cell d. utilization of NAD+ e. regeneration of NAD+ ANSWER: e 24. Choose the glycolytic intermediate that can be destabilized by the introduction of C2–C3 double bonds in its structure. a. 1,3-bisphosphoglycerate b. 2-phosphoglycerate c. 3-phosphoglycerate d. phosphoenolpyruvate e. pyruvate ANSWER: d 25. Why hydrophobicity of the kinases cleft upon binding of glucose is highly important? a. It favors a fast release of highly polar glucose. b. It prevents water from hydrolysis of glycopyranose. c. It attracts water to facilitate the phosphoryl group transfer. d. It prevents nonspecific energy consumption by the enzyme. e. It allows divalent ions to be used in the reaction.

ANSWER: d 26. Which of the glycolytic enzymes catalyzes the reaction that does NOT require nucleotide-based cofactors .. Powered by Cognero.

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Chapter 16 and has two products? a. aldolase b. phosphoglucomutase c. hexokinase d. glyceraldehyde 3-phosphate dehydrogenase e. phosphoglycerate mutase ANSWER: a 27. What molecule allosterically stimulates phosphorylation of fructose 6-phosphate? a. AMP b. fructose 2,6-diphosphate c. ADP d. glucose 6-phosphate e. fructose 2,6-bisphosphate ANSWER: e 28. Choose the molecule that accepts electrons in alcoholic fermentation. a. GAP b. pyruvate c. ethanol d. acetaldehyde e. NADH ANSWER: d 29. Calculate the net yield of ATP from glycolysis after consumption of 10 molecules of sucrose. Consider that sucrose is FULLY processed via glycolysis in adipocytes. a. 10 b. 15 c. 20 d. 25 e. 30 ANSWER: e 30. Why is oxidation catalyzed by glyceraldehyde 3-phosphate dehydrogenase spontaneous? a. The free energy of the thioester intermediate is lower than that of GAP, NAD+, and H2O. b. Acyl-phosphate formation is promoted by inorganic phosphate by increasing the phosphoryl-transfer potential. c. The free energy of the thioester intermediate is higher than that of GAP, NAD+, and H2O. d. The thioester intermediate is more stable than 1,3-BPG. e. The free energy of the thioester intermediate is lower than that of 1,3-BPG. .. Powered by Cognero.

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Chapter 16 ANSWER: a 31. What enzyme can catalyze the formation of the glycolytic intermediate with a high phosphoryl transfer potential from 3-phosphoglycerate? a. pyruvate kinase b. phosphofructokinase c. phosphoglycerate mutase d. phosphoglycerate kinase e. aldolase ANSWER: d 32. Which statement about the role of isomerization in glycolysis is FALSE? a. Formation of the monosaccharide that is more stable in the open-chain form is of high importance for the first stage of glycolysis. b. Isomerization of hexoketose to hexoaldose is the base for the regulatory interplay between phosphofructokinase and glucokinase. c. The isomerization of glucose 6-phosphate to fructose 6-phosphate is a conversion of an aldose into a ketose. d. The intramolecular shift of the phosphate group increases energy yield from the milk sugar. e. The phosphoryl shift near the end of glycolysis creates the base for the subsequent internal oxidation–reduction that will markedly elevate the transfer potential of that phosphoryl group.

ANSWER: b 33. How many ATP molecules are required for fructose to enter glycolysis in the liver? a. 0 b. 1 c. 2 d. 3 e. 4 ANSWER: c 34. The energy of which reaction yields ATP in glycolysis? a. oxidation of carbon b. acyl-phosphate formation c. attachment of Pi d. oxidation of NADH e. reduction of NAD+ ANSWER: b 35. Choose the molecule that donates the phosphate group in glycolysis. a. glucose 6-phosphate .. Powered by Cognero.

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Chapter 16 b. fructose 6-phosphate c. 3-phosphoglycerate d. 1,3-bisphosphoglycerate e. fructose 1,6-bisphosphate ANSWER: d 36. Calculate the net yield of ATP from glycolysis after consumption of 5 molecules of lactose. Consider that lactose is FULLY processed via glycolysis. a. 5 b. 10 c. 15 d. 20 e. 25 ANSWER: c 37. What two 3-carbon molecules are generated by the cleavage of fructose 1,6-bisphosphate? a. glyceraldehyde 3-phosphate and 3-phosphoglycerate b. glyceraldehyde 3-phosphate and dihydroxyacetone phosphate c. pyruvate and phosphoenolpyruvate d. enolase and 2-phosphoglycerate e. glyceraldehyde 3-phosphate and pyruvate ANSWER: b 38. What reaction is catalyzed by aldolase? a. isomerization of DHAP to GAP b. ligation of GAP and DHAP c. reversible cleavage of F-1,6-BP to DHAP and GAP d. cleavage of DHAP to GAP e. irreversible aldol condensation of DHAP and GAP ANSWER: c 39. What is the function of a thioester intermediate such as the one formed from GAP? a. It speeds up the actual reaction so that a larger amount of product can be made. b. The thioester shifts the equilibrium of the first stage of the reaction. c. The thioester allows the two-step reaction to be coupled so the second reaction, the energetically unfavorable phosphorylation, can proceed. d. The thioester intermediate induces a conformational change that alters enzyme specificity. e. The thioester prevents the formation of metabolically unfavorable side products.

ANSWER: c 40. What is substrate-level phosphorylation? .. Powered by Cognero.

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Chapter 16 a. phosphorylation of AMP by ATP with 3-phosphoglycerate as a substrate b. ATP synthesis when the phosphate donor is a substrate with a high phosphoryl-transfer potential c. phosphorylation of glycolytic intermediates that have a high phosphoryl-transfer potential d. phosphorylation of ATP coupled to an ion gradient e. ATP and AMP synthesis from two molecules of ADP ANSWER: b 41. Fructose can enter glycolysis at two distinct points, depending on the tissue. How is fructose metabolized in adipose tissue? a. Fructose is cleaved to two molecules of GAP. b. Fructose is converted to fructose 1-phosphate. c. Fructose is converted to fructose 6-phosphate. d. Fructose is cleaved to GAP and DHAP. e. Fructose is converted to glucose, which enters the pathway.

ANSWER: c 42. How and why are pyruvate kinase isozymes regulated differently? a. In muscle, phosphorylation of pyruvate kinase diminishes its activity in response to low bloodglucose levels. b. In muscle, dephosphorylation of pyruvate kinase is activated in response to high levels of fructose 1,6-bisphosphate. c. In the liver, phosphorylation of pyruvate kinase diminishes its activity in response to low bloodglucose levels. d. In the liver, dephosphorylation of pyruvate kinase is activated in response to high levels of fructose 1,6-bisphosphate. e. In the liver, activation of the GLUT2 receptors increases ATP synthesis leading to a decrease in pyruvate kinase activity.

ANSWER: c 43. What is a common mechanistic feature of kinases? a. Phosphoryl groups are transferred from AMP to an acceptor. b. Binding of a substrate induces cleft closing. c. They convert aldoses to ketoses. d. They require a divalent metal ion such as Mg2+ for activity. e. The entire bound molecule of glucose becomes surrounded by the enzyme.

ANSWER: d 44. What is the function of glyceraldehyde 3-phosphate dehydrogenase? a. oxidation by NAD+ and formation of acyl-phosphate b. oxidation of the alcohol to an aldehyde c. dehydration and dephosphorylation of GAP .. Powered by Cognero.

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Chapter 16 d. hydrolysis of GAP e. catalyzes the transfer of a phosphoryl group from ATP to an acceptor ANSWER: a 45. What type of enzyme catalyzes the intramolecular shift of a chemical group? a. hydrolase b. kinase c. dehydrogenase d. mutase e. isomerase ANSWER: d 46. What is the primary metabolic fate of pyruvate? a. ethanol b. lactate c. acetyl CoA d. oxaloacetate e. malate ANSWER: d 47. Lactose intolerance is caused by a deficiency of: a. lactase. b. elastase. c. lactose. d. sucrose. e. fructose.

ANSWER: a 48. What is the purpose of phosphorylating glucose in cytosol? a. to trap glucose in the cell b. to destabilize glucose and facilitate the next series of metabolic steps c. to convert it to a more soluble form d. to receive more ATP from its metabolism e. to trap glucose in the cell, destabilize it and facilitate the next series of metabolic steps ANSWER: e 49. Glycolysis is the name given to a metabolic pathway occurring in many different cell types. It consists of 10 enzymatic steps that convert glucose to pyruvate. Glycolysis is an example of: a. aerobic metabolism. b. anabolic metabolism. c. a net reductive process. .. Powered by Cognero.

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Chapter 16 d. catabolic metabolism. e. oxidative phosphorylation.

ANSWER: d 50. What enzyme helps regenerate NAD+ during lactic acid fermentation? a. lactate dehydrogenase b. hexokinase c. glucokinase d. phosphofructokinase e. aldolase ANSWER: a 51. Which two isomerization reactions occur in glycolysis? a. glucose 6-phosphate →fructose 6-phosphate AND dihydroxyacetone phosphate →glyceraldehyde 3phosphate b. glucose 6-phosphate →glucose AND dihydroxyacetone phosphate →pyruvate c. glucose 6-phosphate →fructose AND dihydroxyacetone phosphate →glycolic acid d. glucose 6-phosphate →fructose 6-phosphate AND dihydroxyacetone phosphate →acetyl-CoA e. glucose 6-phosphate →fructose 1,6-bisphosphate AND dihydroxyacetone phosphate →glucose-6phosphate ANSWER: a 52. Glycolysis is an anaerobic process. What does this mean, and why did it evolve this way? a. An anaerobic process does not utilize carbohydrates. Glycolysis evolved this way due to an absence of carbohydrates in muscle tissues. b. An anaerobic process can only occur in muscle cells. Glycolysis evolved this way to meet the demands of rigorous exercise. c. An anaerobic process requires oxygen to proceed. Glycolysis evolved this way because it originated in the ocean which has a high dissolved oxygen concentration. d. An anaerobic process is more efficient at producing ATP than an aerobic process. Glycolysis evolved this way because there was a need for higher stores of ATP. e. An anaerobic process does not require oxygen. Glycolysis evolved before substantial amounts of oxygen accumulated in the atmosphere.

ANSWER: e 53. Why is phosphofructokinase, rather than hexokinase, the key control point of glycolysis? a. Phosphofructokinase is more abundant in cells than hexokinase, making it a better target for regulation. b. Because glucose-6-phosphate is not only an intermediate in glycolysis, it is also involved in glycogen synthesis and the pentose phosphate pathway. Phosphofructokinase catalyzes the first unique and irreversible reaction in glycolysis. c. Hexokinase is inhibited by high levels of fructose-6-phosphate, whereas phosphofructokinase is not .. Powered by Cognero.

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Chapter 16 affected by feedback inhibition. d. Phosphofructokinase is primarily active in the liver, whereas hexokinase is predominantly active in muscle cells. e. Hexokinase activity is regulated by hormonal signals, while phosphofructokinase activity is not influenced by hormonal changes.

ANSWER: b 54. Which of the following enzymes catalyzes the first step of glycolysis? a. hexokinase b. pyruvate kinase c. glucokinase d. phosphofructokinase e. mutase ANSWER: a 55. Cleavage of fructose 1,6-bisphosphate yields what products? a. two aldoses b. two ketoses c. an aldose and a ketose d. only a ketose e. none of the above ANSWER: c 56. What enzyme rapidly and reversibly converts dihydroxyacetone phosphate to glyceraldehyde 3-phosphate? a. lactase b. hexokinase c. pyruvate kinase d. triose phosphate isomerase e. glucokinase ANSWER: d 57. What product is formed in the first substrate level phosphorylation in glycolysis? a. pyruvate b. 3-phosphoglycerate c. 1,3-bisphosphoglycerate d. 2-phosphoglycerate e. phosphoenolpyruvate ANSWER: b 58. For each molecule of glucose converted to pyruvate in the glycolytic pathway _____ molecules of ATP are used initially (Stage 1) and _____ molecules of ATP are produced (Stage 2) for an overall yield of _____ .. Powered by Cognero.

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Chapter 16 molecules of ATP/glucose. The overall “ATP equation” is what? a. -2 + 4 = 2 b. -1 + 4 = 3 c. -2 + 5 = 3 d. -1 + 2 = 1 e. 2 + 2 = 4 ANSWER: a 59. Phosphofructokinase is allosterically inhibited by _____ and activated by _____? a. AMP, Pi b. ADP, AMP c. citrate, ATP d. ATP, PEP e. ATP, AMP ANSWER: e 60. The regulation of the glycolytic pathway involves which of the following? a. feedback inhibition by ATP b. allosteric inhibition by ATP c. allosteric stimulation by AMP d. all of the above e. none of the above ANSWER: d 61. Muscle cannot release its stored glycogen into circulation to maintain blood glucose levels because it lacks which enzyme? a. glucose-6-phosphatase b. phosphofructokinase c. glycogen phosphorylase d. phosphoglucomutase e. hexokinase ANSWER: a 62. The electrons removed from glucose via NADH formation can be used to _____. a. form 3-phosphoglycerate b. form ethanol in the liver c. make ATP in mitochondria under anaerobic conditions d. reduce pyruvate to lactate, which regenerates NAD+ e. release glucose-6-phosphate ANSWER: d .. Powered by Cognero.

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Chapter 16 63. Which of the following statements 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]/[AMP] ratio.

ANSWER: b 64. Which of these cofactors 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 65. In comparison with the resting state, actively contracting human muscle tissue has _____. a. a higher concentration of ATP b. a higher rate of lactate formation c. lower glucose consumption d. a lower rate of oxygen consumption e. a lower ratio of NADH to NAD+ ANSWER: b 66. Inorganic fluoride inhibits enolase. In an anaerobic system that is metabolizing glucose as a substrate, which of the following compounds would you expect to increase in concentration following the addition of fluoride? a. 2-phosphoglycerate b. glucose c. glyoxylate d. phosphoenolpyruvate e. pyruvate ANSWER: a 67. The ultimate electron acceptor in the fermentation of glucose to ethanol is _____. a. acetaldehyde b. acetate c. ethanol d. NAD+ e. pyruvate ANSWER: a .. Powered by Cognero.

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Chapter 16 68. What is the event that ends the second stage of glycolysis? a. synthesis of ATP b. synthesis of glucose 6-phosphate c. synthesis of triose phosphates d. synthesis of pyruvate e. synthesis of NADH ANSWER: d

.. Powered by Cognero.

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Chapter 17 1. What is the main fuel for the brain and red blood cells? a. galactose b. glucose c. cellulose d. fructose e. lactose ANSWER: b 2. Taking into account the daily glucose requirement in a typical adult human being, calculate the weekly requirement for the whole body. a. 1120 g b. 840 g c. 1330 g d. 140 g e. 1960 g ANSWER: a 3. What are the glycolytic enzymes? a. fructose 1,6-bisphosphatase and pyruvate carboxylase b. pyruvate carboxylase and phosphofructokinase c. pyruvate kinase and phosphofructokinase d. glucose 6-phosphatase and fructose 1,6-bisphosphatase e. phosphoenolpyruvate carboxykinase and phosphofructokinase ANSWER: c 4. What is the reason active skeletal muscle forms lactate through lactic acid fermentation? a. Skeletal muscles don't have enough time to relax. b. Lactic acid is an unstable acid. c. The level of glucose is too high. d. The rate of glycolysis exceeds the rate at which muscle can process pyruvate aerobically. e. There is not enough hemoglobin in blood during the work of active skeletal muscle.

ANSWER: d 5. Why is gluconeogenesis not a reversal of glycolysis? a. All glycolysis reactions are completely different from those in gluconeogenesis. b. Gluconeogenesis is the main fuel during fasting. c. Glycolysis is more difficult and complicated than gluconeogenesis. d. The free Gibbs energy of several reactions is positive under typical cellular conditions. e. Glycolysis comprises irreversible steps which are bypassed in gluconeogenesis by gluconeogenic enzymes.

ANSWER: e .. Powered by Cognero.

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Chapter 17 6. Which of the following statements is TRUE? a. Pyruvate carboxylase is a trimer. b. Pyruvate carboxylase requires biotin during carboxylation of pyruvate. c. Pyruvate carboxylation takes place in two stages. d. Acetyl CoA has an essential effect on the second partial reaction of pyruvate carboxylation. e. The first step of gluconeogenesis occurs in the cytoplasm.

ANSWER: b 7. What are the consequences of biotin deficiency? a. lethargy and depression b. anorexia and insomnia c. headache and nausea d. blood-clotting disorder e. hepatitis ANSWER: a 8. How many bypassed reactions does gluconeogenesis have? a. 1 b. 2 c. 3 d. 6 e. 10 ANSWER: c 9. What phosphoryl donor is used during oxaloacetate decarboxylation and phosphorylation? a. ADP b. ATP c. GDP d. GTP e. NTP ANSWER: d 10. Where does the final step in the generation of free glucose primarily take place? a. cytoplasm of any cells b. mitochondria of hepatocytes c. mitochondria of skeleton muscle cells d. endoplasmic reticulum of hepatocytes e. endoplasmic reticulum of cardiac muscle ANSWER: d .. Powered by Cognero.

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Chapter 17 11. The formation of phosphoenolpyruvate from pyruvate in the gluconeogenic pathway has a ΔG° of: a. +7.8 kJ mol–1. b. –0.8 kJ mol–1. c. +0.8 kJ mol–1. d. +31 kJ mol–1. e. –31 kJ mol–1.

ANSWER: c 12. What reactions of glycolysis are bypassed by the four steps of gluconeogenesis? Select all that apply. a. reaction catalyzed by the hexokinase reaction b. reaction catalyzed by the pyruvate carboxylase reaction c. reaction catalyzed by the phosphofructokinase reaction d. reaction catalyzed by the phosphoenolpyruvate carboxykinase reaction e. reaction catalyzed by the pyruvate kinase reaction ANSWER: a, c, e 13. Gluconeogenesis often ends with the formation of glucose 6-phosphate. Why does this happen? a. Most tissues lack glucose 6-phosphatase as a membrane transporter. b. Glucose 6-phosphate cannot convert into glucose because the reaction is energetically unfavorable. c. Glucose 6-phosphate is not hydrolyzed in the endoplasmic reticulum lumen. d. Glucose 6-phosphate is a stable compound. e. The formation of glucose 6-phosphate is the final step of gluconeogenesis.

ANSWER: a 14. What is the main role of the extra cost of gluconeogenesis rather than glycolysis in terms of nucleoside triphosphate molecules? a. Nucleoside triphosphates provide more energy for active muscle activity. b. Additional nucleoside triphosphates are needed to turn an energetically unfavorable process into a favorable one. c. Energy is used for lipid breakdown. d. Nucleoside triphosphate molecules are used for nucleic acid metabolism. e. Energy is saved up as reserves.

ANSWER: b 15. Which biological molecules are responsible for controlling the rate of gluconeogenesis? a. fructose 6-phosphate amino acids b. amino acids and nucleic acids c. glucose 6-phosphate and glucose d. pyruvate and fructose 1,6-bisphosphate .. Powered by Cognero.

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Chapter 17 e. lactate and glycerol ANSWER: e 16. What enzyme catalyzes formation of fructose 2,6-bisphosphate from fructose 6-phosphate? a. phosphofructokinase 2 b. pyruvate carboxylase c. glucose 6-phosphatase d. phosphoenolpyruvate carboxykinase e. fructose bisphosphatase 2 ANSWER: a 17. What is the atomic mass of the polypeptide chain containing both PFK2 and FBPase2? a. 102 kDa b. 66 kDa c. 40 kDa d. 150 kDa e. 55 kDa ANSWER: e 18. What metabolic circumstance turns off the gene PEPCK in type 2 diabetes? a. insulin resistance b. hypoglycemia c. high level of amino acids in blood d. hypoglycemia e. phosphaturia ANSWER: a 19. What factor may be a contributing factor for the development of type 2 diabetes? a. viral respiratory infection b. emotional disorder c. obesity d. type 1 diabetes e. headache ANSWER: c 20. What is ΔG of gluconeogenesis? a. ΔG = 2 b. ΔG = 1 c. ΔG > 0 d. ΔG < 0 e. ΔG = 0 .. Powered by Cognero.

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Chapter 17 ANSWER: d 21. What are the precursors of glucose? Select all that apply. a. fatty acids b. amino acids c. lactate d. glycerol e. glycogen ANSWER: b, c, d 22. In the following sequence of consecutive reactions, choose the numbers of CORRECT conversions: pyruvate →1 fructose 1,6-bisphosphate →2 malate →3 oxaloacetate →4 glucose a. 3 b. 1 and 4 c. 2 d. 2 and 4 e. 1 ANSWER: a 23. How many GTP and ATP molecules are required for the synthesis of four molecules of glucose from eight molecules of pyruvate? a. 2 and 4, respectively b. 8 and 16, respectively c. 4 and 4, respectively d. 8 and 4, respectively e. 6 and 12, respectively ANSWER: b 24. Which statement describes appropriate pyruvate kinase activity if there is enough ATP and alanine? If there is enough ATP and alanine, pyruvate kinase: a. is inhibited by citric acid. b. catalyzes the transfer of a phosphate group from fructose 6-phosphate to ADP. c. catalyzes formation of oxaloacetate from pyruvate. d. is inhibited only by alanine. e. is inhibited by these factors.

ANSWER: e 25. Choose reaction that takes place in mitochondria. a. formation of glucose from fructose 6-phosphate b. formation of fructose 6-phosphate from fructose 1,6-bisphosphate c. activation of fructose 1,6-bisphosphatase by citrate .. Powered by Cognero.

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Chapter 17 d. phosphorylation of phosphofructokinase 2/ fructose bisphosphatase 2 by protein kinase A e. formation of oxaloacetate from pyruvate ANSWER: e 26. When blood glucose is: a. scarce, glucagon concentration in the blood falls and insulin concentration rises. b. scarce, glucagon concentration in the blood rises and insulin concentration rises. c. abundant, glucagon concentration in the blood falls and insulin concentration rises. d. abundant, glucagon concentration in the blood rises and insulin concentration rises. e. abundant, glucagon concentration in the blood falls and insulin concentration falls.

ANSWER: c 27. What is the final product of gluconeogenesis in most tissues? a. fructose 6-phosphate b. fructose 1,6-bisphosphate c. glucose d. glucose 6-phosphate e. phosphoenolpyruvate ANSWER: d 28. What is the name of a disorder when blood glucose rises to abnormally high levels? a. hyperglycemia b. hypertonia c. hyperplasia d. hypoglycemia e. polyuria ANSWER: a 29. The typical symptom of malignant hyperthermia is: a. blurred vision caused by inability to use glucose as a fuel. b. excessive thirst due to changing blood osmolarity. c. high body temperature due to an abnormally high level of blood glucose. d. high body temperature due to uncontrolled ATP hydrolysis. e. low body temperature due to uncontrolled ATP formation.

ANSWER: d 30. How many high-energy phosphate bonds are consumed in gluconeogenesis? a. 3 b. 6 c. 2 d. 4 .. Powered by Cognero.

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Chapter 17 e. 1 ANSWER: b 31. What would be the result of a mutation in a gene coding pyruvate carboxylase? a. acidosis and hypoglycemia b. hyperglycemia c. liver cirrhosis d. polyuria e. malignant hyperthermia ANSWER: a 32. What do you call a pair of reactions: phosphorylation of fructose 6-phosphate to fructose 1,6-bisphosphate in the glycolytic pathway and its hydrolysis back to fructose 6-phosphate in the gluconeogenic pathway? a. isomerization b. glyoxylate cycle c. Calvin cycle d. citric acid cycle e. substrate cycle ANSWER: e 33. What is the location of glucose 6-phosphatase? a. mitochondria b. cytoplasm c. membrane bound in the endoplasmic reticulum d. lysosome e. cell nucleus ANSWER: c 34. What technique was used to detect both fructose 6-phosphate and fructose 1,6-bisphosphate in gluconeogenesis? a. lateral flow test b. isotope-labelling method c. electron-microscopic method d. ELISA e. immunoscreening ANSWER: b 35. What substrate is used to synthesize glucose in the absence of oxygen? a. phosphoenolpyruvate b. oxaloacetate c. citrate .. Powered by Cognero.

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Chapter 17 d. lactate e. fructose ANSWER: d 36. Which statement is FALSE? a. Fructose 1,6-bisphosphate is converted to fructose 6-phosphate in the endoplasmic reticulum. b. T1 protein transports glucose 6-phosphate into the lumen of the endoplasmic reticulum. c. T2 protein transports Pi into the cytoplasm. d. T3 protein transports glucose into the cytoplasm. e. Free glucose is not formed in the cytoplasm.

ANSWER: a 37. How many amino acids does the kinase domain of phosphofructokinase 2/ fructose bisphosphatase 2 have? a. 282 b. 220 c. 218 d. 438 e. 470 ANSWER: c 38. Choose the statement that sums up reciprocal regulation of gluconeogenesis and glycolysis. a. When glucose is abundant, gluconeogenesis will predominate. b. When glucose is scarce, glycolysis will predominate. c. When glucose is abundant, glycolysis will predominate. d. When glucose is abundant, glycogenolysis will predominate. e. When glucose is abundant, glycolysis will be stopped.

ANSWER: c 39. Select all that apply. High levels of ATP and citrate: a. indicate a high-energy–well-fed state. b. indicate remote gluconeogenesis. c. inhibit glycolysis. d. activate glycolysis. e. indicate formation high-transfer-potential molecules.

ANSWER: a, b 40. Hormonal activation of cyclic AMP levels will: a. activate protein kinase A phosphorylation of FBPase2. b. phosphorylate PFK2 on a tyrosine residue. c. lead to the activation of PFK. .. Powered by Cognero.

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Chapter 17 d. activate the PKC phosphorylation of PFK2. e. increase the activation of gluconeogenesis.

ANSWER: e 41. The enzymes involved in shuttling carbons in gluconeogenesis from the mitochondria to the cytosol are called: a. malate dehydrogenase. b. citrate synthase. c. oxaloacetate transferase. d. oxaloacetate reductase. e. glucose 1,6-bisphosphatase.

ANSWER: a 42. Phosphofructokinase (PFK) is a highly regulated enzyme. What statement about PFK is CORRECT? a. AMP and ADP both bind to and stabilize the inactive conformation of fructose 6-phosphate. b. ATP can overcome the inhibition by citrate. c. Citrate is an inhibitor of PFK. d. Acidic conditions from anaerobic metabolism activate PFK. e. Fructose 2,6-bisphosphate is a potent activator of phosphofructokinase (PFK).

ANSWER: e 43. The bifunctional enzyme is also known as: a. phosphofructokinase I. b. phosphofructokinase II. c. fructose 1-6 phosphatase. d. protein kinase 2. e. phosphoenolpyruvate carboxykinase.

ANSWER: b 44. What is the essential compound required for carboxylation of pyruvate in humans? a. biotin b. glucose c. calcium d. oxaloacetate e. alanine ANSWER: a 45. What strategy does the liver use to maintain adequate levels of glucose in the blood for use by other tissues? a. Glucose 6-phosphatase has a low KM for glucose 6-phosphate so that glycogen is formed only when glucose is plentiful. b. Transporters named T1, T2, and T3 are responsible for transporting glucose 6-phosphate in and .. Powered by Cognero.

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Chapter 17 glucose and inorganic phosphate out of the mitochondrion for gluconeogenesis. c. Pyruvate is transported out of the mitochondrion via the oxaloacetate shuttle when glucose levels in the blood are low. d. Glucose 6-phosphatase is bound to the lumen side of the ER, where the products of this enzyme reaction are then transported back to the cytoplasm. e. None of the answers is correct.

ANSWER: d 46. What is the process of glucose creation from lactate, pyruvate, and amino acids? a. saccharification b. glycogenolysis c. gluconeogenesis d. glycolysis e. hydrolysis ANSWER: c 47. High blood sugar after a meal _____ the level of insulin released by the pancreas. a. increases b. decreases c. has no effect on d. chronically activates e. chronically inhibits ANSWER: a 48. Glycerol, lactate, and amino acids contribute carbon precursors in the formation of glucose; however, the path that glycerol takes is strikingly different from the other precursors. Explain how it differs. a. Glycerol is decarboxylated to acetyl CoA and enters gluconeogenesis as pyruvate. b. Glycerol enters gluconeogenesis as a breakdown product of triacylglycerols in the form of dihydroxyacetone phosphate. c. Glycerol undergoes a reversible reduction/oxidation reaction to form phosphoenolpyruvate. d. Glycerol is first oxidized in a reaction requiring NAD+ and then phosphorylated in a reversible reaction of glycolysis. e. Glycerol is oxidized and enters gluconeogenesis as glyceraldehyde phosphate.

ANSWER: b 49. Glycerol from fats is modified first by glycerol kinase and then by a second enzyme to enter gluconeogenesis at the _______ intermediate.

ANSWER: dihydroxyacetone phosphate 50. In which of the following tissues is the major site for gluconeogenesis located? The major site for gluconeogenesis is located in: a. the brain. .. Powered by Cognero.

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Chapter 17 b. the liver. c. striated muscle. d. adipose tissue. e. red blood cells.

ANSWER: b 51. The major non-carbohydrate precursors of glucose – lactate, amino acids, and glycerol – are usually converted into what initial substrate of gluconeogenesis? a. glucose b. 2-phosphoglycerate c. pyruvate d. glycerol e. fructose 6-phosphate ANSWER: c 52. Which enzyme catalyzes the initial step in gluconeogenesis? a. pyruvate carboxylase b. phosphoenolpyruvate carboxykinase c. glycerol kinase d. lactate dehydrogenase e. enolase ANSWER: a 53. What role does biotin play in the pyruvate carboxylation reaction? a. Biotin aids in the dephosphorylation of ATP to power the carboxylation reaction. b. Biotin helps stabilize the pyruvate molecule in the active site. c. Biotin serves as the carrier of activated CO2 into the active site. d. Biotin binds acetyl CoA in the active site. e. Biotin promotes an increased rate of carboxylation when pyruvate levels are high.

ANSWER: c 54. How many stages does the carboxylation of pyruvate take? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: c 55. Which enzyme oxidizes malate back into oxaloacetate in the cytosol? a. malate dehydrogenase .. Powered by Cognero.

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Chapter 17 b. NAD+-linked malate dehydrogenase c. phosphoenolpyruvate carboxykinase d. oxaloacetate transferase e. oxaloacetate reductase ANSWER: b 56. What molecule does the oxidation of malate to regenerate oxaloacetate provide for later steps in gluconeogenesis? a. ATP b. GTP c. NADH d. FADH2 e. ADP ANSWER: c 57. A total of ___ nucleoside triphosphates must be hydrolyzed during gluconeogenesis in comparison to the ___ molecules of ATP hydrolyzed during glycolysis. a. six, two b. four, two c. two, six d. two, four e. four, six ANSWER: a 58. What single amino acid residue is phosphorylated to regulate the activities of PFK2 and FBPase2? a. threonine b. tyrosine c. serine d. histidine e. phenylalanine ANSWER: c 59. What advantage is there in cardiac and slow-twitch (type 1) skeletal muscles using lactate as a primary source of pyruvate instead of glucose? a. More glucose is freed up in the bloodstream to be used by active skeletal muscles. b. This is the only means higher organisms have of removing lactate, a dead-end substrate. c. The use of lactate allows skeletal muscles to generate ATP under high oxygen conditions. d. The additional fate of lactate prevents the liver from being overwhelmed during periods of high muscle activity. e. Cardiac and slow-twitch (type 1) skeletal muscles do not use lactate as a primary source of pyruvate.

ANSWER: a .. Powered by Cognero.

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Chapter 17 60. What organs participate in the Cori cycle? Select all that apply. a. liver b. muscles c. stomach d. brain e. heart ANSWER: a, b 61. Which irreversible step acts as the primary control point for gluconeogenesis? a. The conversion of pyruvate into oxaloacetate by pyruvate carboxylase. b. The conversion of oxaloacetate into malate by malate dehydrogenase. c. The conversion of fructose 6-phosphate into glucose 6-phosphate by phosphoglucose isomerase. d. The conversion of fructose 1,6-biphosphate into fructose 6-phosphate by FBPase. e. The conversion of 2-phosphoglycerate into 3-phosphoglycerate by phosphoglycerate mutase.

ANSWER: d 62. Which of the following statements is FALSE? a. Most precursors are converted into pyruvate prior to entering the gluconeogenic cycle. b. Glycerol can enter both the glycolytic and gluconeogenic pathways at dihydroxyacetone phosphate. c. Amino acids enter the gluconeogenic process as either pyruvate or oxaloacetate. d. Glycerol is a precursor of glucose, but animals cannot process fatty acids into glucose. e. Active skeletal muscle forms lactate through lactic acid fermentation when the rate of glycolysis exceeds the rate at which muscle can process pyruvate anaerobically.

ANSWER: e 63. Biotin cannot be carboxylated without the presence of which molecule binding pyruvate carboxylase? a. acetyl CoA b. vitamin C c. ATP d. HCO3 e. Mg2+ ANSWER: a 64. What are the two major precursors to glucose in the liver? a. lactate and alanine b. lactate and glycerol c. alanine and pyruvate d. glycerol and pyruvate e. alanine and glycerol ANSWER: a .. Powered by Cognero.

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Chapter 17 65. What two processes are simultaneously performed by PEPCK to convert oxaloacetate into phosphoenolpyruvate? a. carboxylation and hydrolysis b. phosphorylation and decarboxylation c. phosphorylation and hydrolysis d. oxidation and carboxylation e. oxidation and hydrolysis ANSWER: b 66. Pyruvate carboxylase consists of how many subunits? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: d 67. Why are substrate cycles biologically important? a. They allow for the amplification of metabolic signals based on changes in the net flux between reciprocal pathways. b. They allow for the activation of reciprocal pathways at the same time. c. They allow for a large degree of exchange between two otherwise irreversible pathways. d. They ensure perfect metabolic control between reciprocal pathways. e. They prevent the complete activation of reciprocal pathways at the same time.

ANSWER: a 68. In glycolysis, pyruvate is formed from phosphoenolpyruvate in one step. However, if the reverse reaction in gluconeogenesis was catalyzed by pyruvate kinase, the phosphorylation of pyruvate to phosphoenolpyruvate would require a ΔG°’ of +31 kJ/mol-1. How is this energy barrier overcome in the gluconeogenic pathway? a. The decarboxylation reaction catalyzed by PEPCK drives the otherwise highly endergonic phosphorylation reaction. b. The decarboxylation reaction catalyzed by PEPCK drives the otherwise highly exergonic phosphorylation reaction. c. The carboxylation reaction catalyzed by pyruvate carboxylase drives the otherwise highly endergonic phosphorylation reaction. d. The carboxylation reaction catalyzed by pyruvate carboxylase drives the otherwise highly exergonic phosphorylation reaction. e. Pyruvate carboxylase is not involved in the formation of phosphoenolpyruvate.

ANSWER: a 69. The conversion of glycerol into dihydroxyacetone phosphate requires ___ ATP and ___ NAD+. .. Powered by Cognero.

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Chapter 17 a. one, one b. two, one c. one, two d. one, zero e. zero, one ANSWER: a 70. What substrate is generated when the carbon skeleton of some amino acids is used as fuel in muscles? a. serine b. pyruvate c. alanine d. lactate e. tyrosine ANSWER: c

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Chapter 18 1. What prosthetic group does the E1 component of the pyruvate dehydrogenase complex require for proper activity? a. coenzyme A b. FAD c. NADH d. TPP e. NAD+ ANSWER: d 2. The E2 component of the pyruvate dehydrogenase complex contains a lipoamide cofactor that is covalently attached to: a. the histidine residue of the enzyme. b. the lysine residue of the enzyme. c. the serine residue of the enzyme. d. the arginine residue of the enzyme. e. FAD. ANSWER: b 3. What is the critical functional group of lipoamide responsible for carrying acetyl groups and maintaining the free energy of CoA? a. thioester group b. amide group c. thioether group d. carbamoyl group e. sulfhydryl group ANSWER: e 4. What subunit of the pyruvate dehydrogenase regenerates the oxidized form of lipoamide? a. E3 b. dihydrolipoyl transacetylase c. E2 d. E1 e. pyruvate dehydrogenase component ANSWER: a 5. Beriberi is caused by a deficiency of: a. flavine. b. adenine. c. thiamine. d. pantothenate. .. Powered by Cognero.

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Chapter 18 e. niacin.

ANSWER: c 6. What is acetyl CoA used for if NADH levels are high? a. oxidation to CO2 b. synthesis of fatty acids c. synthesis of glucose d. generation of energy e. synthesis of pyruvate ANSWER: b 7. What molecule can act as an activated carrier of acyl groups? a. flavin adenine dinucleotide b. pyruvate c. pantothenic acid d. coenzyme A e. thiamine pyrophosphate ANSWER: d 8. What statement concerning acetyl CoA is FALSE? a. Acetyl CoA contains a pyrimidine ring. b. Acetyl CoA is the fuel for the citric acid cycle. c. The pyruvate dehydrogenase complex is not the only source of acetyl CoA. d. Acetyl CoA is an esterified thiol. e. In resting human muscle, pyruvate is first converted into acetyl CoA. ANSWER: a 9. What vitamin is essential for thiamine pyrophosphate synthesis in humans? a. B1 b. B5 c. B2 d. B6 e. PP ANSWER: a 10. How many molecules of CO2 and NADH will be produced by conversion of pyruvate from one glucose molecule into acetyl CoA? a. 1 molecule of CO2; 1 molecule of NADH b. 4 molecules of CO2; 4 molecules of NADH .. Powered by Cognero.

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Chapter 18 c. 2 molecules of CO2; 2 molecules of NADH d. 2 molecules of CO2; 1 molecule of NADH e. 4 molecules of CO2; 2 molecules of NADH ANSWER: c 11. What conditions are essential for conversion of pyruvate into acetyl coenzyme A rather than into lactic acid or ethanol? a. anaerobic conditions b. aerobic conditions c. acidic conditions d. alkaline medium e. high redox potential ANSWER: b 12. What is the site for oxidative decarboxylation of pyruvate? a. liver b. cytoplasm c. endoplasmic reticulum d. mitochondrial intermembrane space e. mitochondrial matrix ANSWER: e 13. What process links glycolysis that takes place in the cytoplasm of the cell with the citric acid cycle carried out in mitochondria? a. pyruvate transport into mitochondria b. acetyl CoA transport into mitochondria c. lactate transport into mitochondria d. lipoamide transport into mitochondria e. PDH transport into mitochondria ANSWER: a 14. What statement CORRECTLY describes the structure of the transacetylase core monomer in E. coli? a. It is composed of about 45 copies of the E1 and about 10 copies of the E3 enzymes. b. It contains E3-binding protein (E3-BP). c. It consists of a small C-terminal lipoamide domain, a small domain that interacts with E3 within the PDH complex and a larger enzymatic domain at the amino terminus. d. It is composed of a small C-terminal lipoamide domain and a larger enzymatic domain at the amino terminus. e. It contains a small N-terminal lipoamide domain and a small domain that interacts with E3 within the PDH complex. .. Powered by Cognero.

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Chapter 18 ANSWER: c 15. What part of the PDH complex is a flavoprotein? a. dihydrolipoyl transacetylase b. pyruvate dehydrogenase kinase c. dihydrolipoyl dehydrogenase d. pyruvate dehydrogenase phosphatase e. pyruvate dehydrogenase component ANSWER: c 16. What statement CORRECTLY describes the structure of the transacetylase core monomer in mammals? a. It is composed of about 45 copies of the E1 and about 10 copies of the E3 enzymes. b. It contains E3-binding protein (E3-BP). c. It consists of a small C-terminal lipoamide domain, a small domain that interacts with E3 within the PDH complex and a larger enzymatic domain at the amino terminus. d. It is composed of C-terminal enzymatic domain and a larger lipoamide domain at the amino terminus. e. It contains a small N-terminal lipoamide domain and a small domain that interacts with E3 within the PDH complex.

ANSWER: b 17. What serves as a stoichiometric coenzyme in the PDH complex reaction? Select all that apply. a. lipoic acid b. NAD+ c. thiamine pyrophosphate d. FAD e. CoA ANSWER: b, e 18. What serves as a catalytic coenzyme in the PDH complex reaction? Select all that apply. a. lipoic acid b. NAD+ c. TPP d. FAD e. CoA ANSWER: a, c, d 19. What is the rate-limiting step in the synthesis of acetyl CoA? a. decarboxylation b. oxidation .. Powered by Cognero.

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Chapter 18 c. transfer to CoA d. transfer to lipoamide e. regeneration of lipoamide ANSWER: a 20. Choose the intermediate that is formed when pyruvate binds to PDH. a. carboxyethyl-TPP b. carboxymethyl-TPP c. oxyethyl-TPP d. hydroxymethyl-TPP e. hydroxyethyl-TPP ANSWER: e 21. What molecule or structure inserts into the 20-Å-long hydrophobic channel in E1, forcing its active site to take the acetyl group? a. lysine side chain b. CoA c. dihydrolipoamide d. lipoamide arm e. transacetylase domain ANSWER: d 22. Where is the active site of E2 located? a. in the long hydrophobic channel inside the enzyme b. on the long flexible arm c. closely to the E1 active site d. on the surface of the hollow cube e. at the subunit interface inside the cube ANSWER: e 23. PDH is inactive when it is: a. dephosphorylated. b. phosphorylated. c. acylated. d. oxidized. e. anchored in the membrane.

ANSWER: b 24. What coenzyme accepts electrons to reoxidize dihydrolipoamide in PDH? a. TPP .. Powered by Cognero.

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Chapter 18 b. CoA c. FAD d. NAD+ e. NADP ANSWER: c 25. What is necessary to reactivate PDH to start the next catalytic cycle? a. protonation of the TPP carbanion b. oxidation of dihydrolipoamide c. phosphorylation d. dephosphorylation e. substantial ratio of ATP/ADP ANSWER: b 26. How are the three active sites of PDH connected to each other in E. coli? a. The subunits of PDH are bound together covalently. b. The subunits of PDH are linked by hydrophobic interactions. c. The subunits of PDH are linked together with the aid of another protein, E3-BP. d. The subunits of PDH are linked together with the aid of NAD+ and CoA. e. The long flexible arm of E2 carries the substrate from one active site to another.

ANSWER: e 27. The Warburg effect is the conversion of glucose to _____ of oxygen. a. lactate in the presence b. lactate in the absence c. fatty acids in the absence d. fatty acids in the presence e. acetyl CoA in the presence ANSWER: a 28. What factor promotes aerobic glycolysis? a. high insulin level b. pyruvate dehydrogenase kinase deficiency c. enhanced activity of pyruvate dehydrogenase kinase d. enhanced activity of pyruvate dehydrogenase phosphatase e. epinephrine rush ANSWER: c 29. What condition does NOT cause lactic acidosis? a. Warburg effect .. Powered by Cognero.

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Chapter 18 b. pyruvate dehydrogenase phosphatase deficiency c. diabetes mellitus d. increase in mitochondrial Ca2+ e. synthesis of hypoxia inducible factor-1 ANSWER: d 30. What covalent modification of the PDH complex is used to regulate its activity? a. phosphorylation b. glycosylation c. methylation d. adenylation e. ubiquitination ANSWER: a 31. If a lot of acetyl CoA and NADH were produced from fatty acid degradation, the PDH would be: a. subjected to degradation. b. dephosphorylated. c. activated. d. inhibited. e. dissociated.

ANSWER: d 32. What conditions activate PDH? a. NAD+/NADH ratio is low b. acetyl CoA/CoA ratio is high c. the mitochondrial Ca2+ level is high d. ADP/ATP ratio is low e. the energy charge is high ANSWER: c 33. What conditions inhibit PDH? a. NADH/NAD+ ratio is low b. acetyl CoA/CoA ratio is high c. the mitochondrial Ca2+ level is high d. ATP/ADP ratio is low e. the energy charge is low ANSWER: b 34. What is the treatment for mercury poisoning? a. a ketogenic diet (high-fat, low-carbohydrate) .. Powered by Cognero.

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Chapter 18 b. hemodialysis c. the administration of painkillers d. the administration of sulfhydryl reagents with adjacent sulfhydryl groups e. pharmacological inhibition of pyruvate dehydrogenase kinase ANSWER: d 35. What is the CORRECT order of steps of conversion of pyruvate into acetyl CoA? a. decarboxylation, oxidation, transfer to CoA b. decarboxylation, transfer to CoA, transfer to lipoamide, oxidation c. oxidation, decarboxylation, transfer to CoA d. oxidation, transfer to lipoamide, decarboxylation e. oxidation, decarboxylation, transfer to CoA, recovery of lipoamide ANSWER: a 36. What subunit of the PDH decarboxylates pyruvate? a. E1 b. E2 c. E3 d. E4 e. E5 ANSWER: a 37. What coenzyme of the PDH complex functions as a "flexible swinging arm" when it transfers the reaction intermediate from one active site to the other? a. FAD b. NAD+ c. lipoamide d. thiamine pyrophosphate e. coenzyme A ANSWER: c 38. Pyruvate dehydrogenase is _____ when ATP/ADP ratios are high. Select all that apply. a. activated b. inhibited c. phosphorylated d. acetylated e. methylated ANSWER: b, c 39. What disease has symptoms similar to those of beriberi? .. Powered by Cognero.

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Chapter 18 a. arsenite poisoning b. lactic acidosis c. type II diabetes d. scurvy e. cancer ANSWER: a 40. What conditions will activate pyruvate dehydrogenase kinase, which catalyzes phosphorylation and inactivation of E1 in the pyruvate dehydrogenase complex? Select all that apply. a. elevated concentrations of NADH and ATP b. elevated concentrations of NAD+ and ADP c. elevated concentration of Ca2+ d. elevated concentration of insulin e. elevated concentration of acetyl CoA ANSWER: a, e 41. Why does acetyl CoA, a product of fatty acid catabolism, influence the pyruvate dehydrogenase complex (PDH), a control point in carbohydrate catabolism? a. When fatty acid breakdown is inhibited, ADP is low, causing a decrease in the activity of PDH. b. When fatty acid breakdown is inhibited, PDH is inhibited by acetyl CoA due to the phosphatase activity. c. When fatty acid breakdown is high, ADP is low and PDH is inhibited by the phosphatase activity. d. When fatty acid breakdown is high, PDH is inhibited by acetyl CoA due to the PDH kinase activity. e. When fatty acid breakdown is high, NADH levels drop, causing an increase in the E2 activity.

ANSWER: d 42. What statement illustrates the role of Ca2+ in regulation of the PDH complex? a. Ca2+ activates phosphatase activity, deactivating PDH activity. b. When Ca2+ is elevated in the cell, muscle contractions initiate and PDH activity is stimulated. c. Insulin causes a decrease in Ca2+, activating pyruvate dehydrogenase kinase activity. d. In liver, Ca2+ stimulates pyruvate dehydrogenase kinase, increasing conversion of pyruvate to acetyl CoA. e. Insulin causes an increase in intercellular Ca2+, inhibiting pyruvate dehydrogenase phosphatase activity.

ANSWER: b 43. The ΔG°' for the sum of reactions in the pyruvate dehydrogenase complex is –33.4 kJ/mol. What is the primary contribution to this large and negative free-energy change? a. In E2, two electrons are transferred from FADH2 to NADH, a thermodynamically favored transfer. .. Powered by Cognero.

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Chapter 18 b. The product of the reaction, acetyl CoA, is shuttled into the mitochondria, keeping the cytosolic concentration low. c. Free energy from the decarboxylation step drives the formation of NADH. d. In E3, two electrons are transferred from FADH2 to NAD+, a thermodynamically favored transfer. e. The oxidation of dyhydrolipoamide by E3 gives the largest contribution to ΔG°' of the whole process.

ANSWER: c 44. What condition is a result of PDH phosphatase deficiency? a. overstimulated PDH b. low blood glucose c. elevated levels of lactic acid in the blood d. high levels of acetyl CoA e. increased synthesis of fatty acids ANSWER: c 45. What enzyme is responsible for the following reaction? pyruvate + coenzyme A + NAD+ → acetyl CoA + NADH + H+ + CO2 a. acetyl CoA synthetase b. pyruvate decarboxylase c. pyruvate dehydrogenase complex d. dihydrolipoyl dehydrogenase e. transacetylase ANSWER: c 46. In addition to pyruvate dehydrogenase, what other enzyme in the citric acid cycle has a key thiamine pyrophosphate coenzyme? a. isocitrate dehydrogenase b. α-ketoglutarate dehydrogenase c. citrate synthase (in bacteria) d. phosphatase e. fumarase ANSWER: b 47. Select all that apply. Milling and polishing rice results in: a. a loss of the husk of rice. b. white rice. c. a potential for beriberi. d. a loss of thiamine pyrophosphate. e. a potential for chronic elevated plasma lactate.

ANSWER: a, b, c, d .. Powered by Cognero.

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Chapter 18 48. A high-energy charge, which results in increasing concentration of NADH, _______ the activity of PDH. ANSWER: inhibits 49. Which enzyme catalyzes the decarboxylation of pyruvate in the pyruvate dehydrogenase complex? a. dihydrolipoyl transacetylase b. dihydrolipoyl dehydrogenase c. pyruvate dehydrogenase d. citrate synthase e. succinate dehydrogenase ANSWER: c 50. Where does glycolysis occur in eukaryotic cells? a. mitochondria b. nucleus c. cytoplasm d. endoplasmic reticulum e. Golgi apparatus ANSWER: c 51. What is the direct product formed from pyruvate by the pyruvate dehydrogenase complex? a. oxaloacetate b. citrate c. succinyl CoA d. acetyl CoA e. fumarate ANSWER: d 52. Which vitamin precursor is required for the activity of the pyruvate dehydrogenase complex? a. riboflavin b. niacin c. vitamin D d. vitamin C e. thiamine ANSWER: e 53. What is the effect of high concentrations of ATP on the pyruvate dehydrogenase complex? a. It activates the complex. b. It inhibits the complex. c. It has no effect. d. It increases the production of pyruvate. e. It stimulates dihydrolipoyl transacetylase directly. .. Powered by Cognero.

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Chapter 18 ANSWER: b 54. Which of the following cofactors is NOT required by the pyruvate dehydrogenase complex? a. thiamine pyrophosphate b. lipoic acid c. coenzyme A d. biotin e. NAD+ ANSWER: d 55. How does arsenite affect the pyruvate dehydrogenase complex? a. It inhibits pyruvate dehydrogenase. b. It binds to thiamine pyrophosphate. c. It inhibits dihydrolipoyl dehydrogenase. d. It stimulates dihydrolipoyl transacetylase. e. It activates pyruvate dehydrogenase kinase.

ANSWER: c 56. Which enzyme activity regenerates the active transacetylase in the pyruvate dehydrogenase complex? a. pyruvate dehydrogenase b. dihydrolipoyl transacetylase c. dihydrolipoyl dehydrogenase d. succinate dehydrogenase e. isocitrate dehydrogenase ANSWER: c 57. What stimulates the phosphatase associated with the pyruvate dehydrogenase complex? a. Ca2+ ions b. low ADP concentration c. high ATP concentration d. high NADH concentration e. acetyl CoA ANSWER: a 58. What is the role of dihydrolipoyl transacetylase in the pyruvate dehydrogenase complex? a. decarboxylation of pyruvate b. formation of acetyl CoA c. regeneration of active transacetylase d. inhibition of pyruvate dehydrogenase e. oxidation of NADH ANSWER: b .. Powered by Cognero.

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Chapter 18 59. Which of the following molecules allosterically inhibits the pyruvate dehydrogenase complex? a. FADH2 b. oxaloacetate c. NADH d. ADP e. pyruvate ANSWER: c 60. Where in eukaryotic cells does the citric acid cycle occur? a. cytoplasm b. mitochondria c. nucleus d. endoplasmic reticulum e. Golgi apparatus ANSWER: b 61. The activity of which enzyme in the pyruvate dehydrogenase complex is inhibited by mercury? a. pyruvate dehydrogenase b. dihydrolipoyl transacetylase c. dihydrolipoyl dehydrogenase d. citrate synthase e. aconitase ANSWER: c 62. What process links glycolysis and the citric acid cycle? a. oxidative phosphorylation b. fermentation c. oxidative decarboxylation of pyruvate d. substrate-level phosphorylation e. beta-oxidation ANSWER: c 63. Which cofactor is essential for the activity of dihydrolipoyl dehydrogenase? a. NADH b. FAD c. ATP d. ADP e. GTP ANSWER: b .. Powered by Cognero.

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Chapter 18 64. How do high NADH concentrations affect the activity of the pyruvate dehydrogenase complex? a. More pyruvate is converted to acetyl CoA by the pyruvate dehydrogenase complex. b. NADH inhibits pyruvate dehydrogenase kinase, increasing pyruvate dehydrogenase complex activity. c. NADH inhibits dihydrolipoyl dehydrogenase, reducing pyruvate dehydrogenase complex activity. d. NADH stimulates pyruvate dehydrogenase kinase, activating the pyruvate dehydrogenase complex. e. NADH stimulates pyruvate dehydrogenase phosphatase, reducing pyruvate dehydrogenase complex activity.

ANSWER: c 65. What is the consequence of pyruvate dehydrogenase complex inhibition by phosphorylation? a. increased production of acetyl CoA b. decreased production of NADH c. increased glycolysis d. decreased entry of pyruvate into the citric acid cycle e. increased activity of dihydrolipoyl transacetylase ANSWER: d 66. Which compound is a product of the pyruvate dehydrogenase complex and an entry molecule for the citric acid cycle? a. oxaloacetate b. citrate c. acetyl CoA d. malate e. succinyl CoA ANSWER: c 67. What role does thiamine pyrophosphate play in the pyruvate dehydrogenase complex? a. acts as an electron carrier b. facilitates the decarboxylation of pyruvate c. forms acetyl CoA d. regenerates the active enzyme e. inhibits the complex under high ATP conditions ANSWER: b 68. Acetyl CoA is formed from pyruvate by a _____ reaction. a. dehydration b. combustion c. reduction d. dephosphorylation e. oxidative decarboxylation .. Powered by Cognero.

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Chapter 18 ANSWER: e

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Chapter 19 1. What conditions are required to obtain energy from acetyl CoA in the citric acid cycle? Select all that apply. a. availability of O2 b. high energy charge c. anaerobic conditions d. low energy charge e. availability of CO2 ANSWER: a, d 2. What is NOT a part of the citric acid cycle? a. capturing of high-energy electrons b. electron-transport chain c. oxidation of acetyl CoA d. synthesis of tricarboxylic acids e. formation of NADH and FADH2 ANSWER: b 3. What is the term for conversion of acetyl CoA into energy in the form of ATP in the presence of oxygen? a. cellular respiration b. citric acid cycle c. oxidative phosphorylation d. electron-transport chain e. proton gradient ANSWER: a 4. What is the major purpose of the citric acid cycle? a. reduction of NAD+ leading to the capture of high-transfer-potential electrons b. synthesis of citric acid c. utilization of pyruvate d. oxidation of carbon atoms leading to the capture of high-transfer-potential electrons e. generation of ATP due to the capture of high-transfer-potential electrons ANSWER: d 5. What functions of the citric acid cycle are CORRECT? Select all that apply. a. It produces NADH and FADH2. b. It provides intermediates for gluconeogenesis. c. It provides intermediates for the synthesis of fatty acids. d. It bypasses oxidative decarboxylation. e. It produces FAD. ANSWER: a, b, c .. Powered by Cognero.

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Chapter 19 6. What is the name for a reaction that replenishes pathway components? a. aerobic b. anaerobic c. anaplerotic d. anabolic e. allosteric ANSWER: c 7. What metabolic motif is common between the citric acid cycle and synthesis of fatty acids? a. oxidation-oxidation-oxidation b. dehydrogenation-dehydrogenation c. reduction-reduction-reduction d. condensation-condensation e. oxidation-hydration-oxidation ANSWER: e 8. What amino acid residue catalyzes the transfer of a phosphoryl group in the course of one of the steps of the citric acid cycle? a. histidine b. glutamate c. glutamine d. serine e. cysteine ANSWER: a 9. What description of the second stage of the TCA cycle is TRUE? a. It starts with a four-carbon molecule and ends with a six-carbon molecule. b. It starts with a four-carbon molecule and ends with a four-carbon molecule. c. It starts with a six-carbon molecule and ends with a five-carbon molecule. d. It starts with a four-carbon molecule and ends with a five-carbon molecule. e. It starts with a five-carbon molecule and ends with a four-carbon molecule.

ANSWER: b 10. What enzyme of the TCA cycle is similar to the pyruvate dehydrogenase complex? a. citrate synthase b. isocitrate lyase c. malate dehydrogenase d. α-ketoglutarate dehydrogenase e. isocitrate dehydrogenase ANSWER: d .. Powered by Cognero.

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Chapter 19 11. What enzyme of the TCA cycle produces L-malate? a. fumarase b. aconitase c. malate synthase d. isocitrate lyase e. succinate dehydrogenase ANSWER: a 12. What intermediate is formed upon isomerization of citrate to isocitrate? a. cis-aconitate b. semi-citrate c. trans-aconitate d. semi-isocitrate e. cis-aconitase ANSWER: a 13. How does the conformation of the citrate synthase change upon catalyzing the reaction? a. open → closed → completely enclosed b. open → closed c. completely enclosed → closed → open d. closed → open e. unfit → fit ANSWER: a 14. What causes the initial change in the conformation of citrate synthase? a. binding of acetyl CoA b. binding of oxaloacetate c. binding of water d. binding of ATP e. binding of NAD+ ANSWER: b 15. Choose the CORRECT order of steps of the reaction catalyzed by citrate synthase. a. binding of two oxaloacetate molecules → binding of two acetyl CoA molecules → cleavage and release of two CoA molecules → release of two citryl CoA molecules b. binding of oxaloacetate → binding of acetyl CoA → condensation to citryl CoA → cleavage and release of CoA → release of citrate c. binding of acetyl CoA → binding of oxaloacetate → condensation to citrate → cleavage and release of CoA → release of citrate d. binding of two acetyl CoA molecules → binding of two oxaloacetate molecules → condensation to two citryl CoA → cleavage and release of two citrate and two CoA molecules .. Powered by Cognero.

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Chapter 19 e. binding of oxaloacetate → binding of acetyl CoA → condensation to citrate and release of CoA ANSWER: b 16. What enzyme catalyzes the first redox reaction of the TCA cycle? a. pyruvate dehydrogenase b. isocitrate lyase c. α-ketoglutarate dehydrogenase d. isocitrate dehydrogenase e. malate dehydrogenase ANSWER: d 17. Select all that apply. Intermediates of the reactions in the TCA cycle are: a. citryl CoA. b. cis-aconitate. c. oxalosuccinate. d. succinyl CoA. e. succinyl phosphate.

ANSWER: a, b, c, e 18. What coenzymes are used by α-ketoglutarate dehydrogenase? Select all that apply. a. CoA b. TPP c. FAD d. NAD+ e. lipoic acid ANSWER: a, b, c, d, e 19. What are anaplerotic reactions of the citric acid cycle? Select all that apply. a. glutamate → α-ketoglutarate b. α-ketoglutarate → glutamate c. pyruvate → oxaloacetate d. malate → oxaloacetate e. isocitrate → α-ketoglutarate ANSWER: a, c 20. What functional group actually yields carbon dioxide? a. acetyl b. carboxyl c. keto d. hydroxyl .. Powered by Cognero.

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Chapter 19 e. sulfhydryl ANSWER: b 21. The rate of the TCA cycle: a. increases if α-ketoglutarate dehydrogenase is stimulated. b. decreases if isocitrate dehydrogenase is stimulated. c. increases if α-ketoglutarate dehydrogenase is inhibited. d. decreases if isocitrate lyase is stimulated. e. increases if citrate synthase is inhibited.

ANSWER: a 22. What statement about the biosynthetic role of the citric acid cycle is FALSE? a. Oxaloacetate can be converted to glucose. b. Amino acids can be synthesized from α-ketoglutarate. c. Succinate is the direct precursor to porphyrins. d. Amino acids can be synthesized from oxaloacetate. e. Citrate can be used for the synthesis of fatty acids.

ANSWER: c 23. Why does FAD, rather than NAD+, act as a hydrogen acceptor in the reaction of succinate dehydrogenase? a. ΔG°' is too high. b. FAD is an enzyme-bound coenzyme. c. ΔG°' is too low. d. Concentration of NAD+ is insufficient. e. Concentration of FAD is sufficient.

ANSWER: c 24. What processes characterize the citric acid cycle as catabolic? Select all that apply. a. carboxylation of pyruvate b. link with the electron-transport chain c. formation of α-ketoglutarate from glutamate d. substrate-level phosphorylation e. oxidative decarboxylation ANSWER: b, d, e 25. What enzyme links the TCA cycle to the electron-transport chain? a. citrate synthase b. α-ketoglutarate dehydrogenase c. succinate dehydrogenase d. fumarase .. Powered by Cognero.

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Chapter 19 e. malate dehydrogenase ANSWER: c 26. What enzyme requires a compound with a high acyl-transfer potential? a. isocitrate dehydrogenase b. succinate dehydrogenase c. succinate thiokinase d. isocitrate lyase e. malate dehydrogenase ANSWER: c 27. Choose the energy source required for synthase- and synthetase-catalyzed reactions of the TCA cycle. a. high phosphoryl-transfer potential b. oxidative decarboxylation c. thioester bond d. oxidative phosphorylation e. high acetyl-transfer potential ANSWER: c 28. What catalytic strategies are employed by citrate synthase? Select all that apply. a. induced fit b. covalent catalysis c. autocatalysis d. general acid–base catalysis e. metal ion catalysis ANSWER: a, c 29. What is the main acceptor of the phosphoryl group in the TCA cycle in the liver? a. ADP b. GDP c. CoA d. succinate e. TPP ANSWER: b 30. Choose the CORRECT order of steps of the reaction catalyzed by succinyl CoA synthetase. a. Phosphorus atom of orthophosphate attacks the carbonyl carbon of succinyl CoA and displaces CoA → succinyl phosphate is formed → in another site, the phosphoryl group is transferred to ADP. b. Oxygen atom of orthophosphate attacks the carbonyl carbon of succinyl CoA and displaces CoA → succinyl phosphate is formed → His residue removes the phosphoryl group → phosphohistidine is formed and succinate is released → in another site, the phosphoryl group is transferred to ADP. .. Powered by Cognero.

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Chapter 19 c. ATP is hydrolyzed and phosphohistidine is formed → phosphohistidine swings in another site → phosphoryl group displaces CoA and succinyl phosphate is formed. d. Oxygen atom of orthophosphate attacks the carboxylic carbon of succinyl CoA and displaces CoA → succinyl phosphate is formed → in another site, the phosphoryl group is transferred from succinyl phosphate to His residue. e. Oxygen atom of orthophosphate attacks the carboxylic carbon of succinyl CoA and displaces CoA → succinyl phosphate is formed → in another site, the phosphoryl group is transferred to ADP. ANSWER: b 31. What reaction of the TCA cycle is, in part, driven by oxidative phosphorylation rather than a difference in free energy? a. citrate synthase b. fumarase c. malate synthase d. succinate dehydrogenase e. malate dehydrogenase ANSWER: e 32. What enzymes of the TCA cycle are allosteric? Select all that apply. a. isocitrate dehydrogenase b. α-ketoglutarate dehydrogenase c. citrate synthase (in bacteria) d. succinate dehydrogenase e. pyruvate dehydrogenase ANSWER: a, b, c 33. What statement about regulation of the TCA cycle is FALSE? a. α-Ketoglutarate dehydrogenase, like pyruvate dehydrogenase, is inhibited when the concentration of ATP is high. b. Both allosterically regulated enzymes of the TCA cycle are stimulated directly by the ADP. c. ATP and NADH are the only common regulators between three key control points of the citric acid cycle. d. The unique committed step of the cycle is hardly distinguishable due to involvement of its components in many metabolic pathways. e. Regulation of enzymes involved in the metabolism of citrate allows accounting for anabolic requirements of the cell.

ANSWER: b 34. How many NADH are produced after the TCA cycle due to aerobic catabolism of three molecules of glucose? a. 3 b. 9 .. Powered by Cognero.

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Chapter 19 c. 18 d. 27 e. 30 ANSWER: e 35. How many electrons are transferred in the redox reactions of the TCA cycle upon catabolism of one molecule of glucose? a. 4 b. 8 c. 14 d. 16 e. 18 ANSWER: d 36. What role does isocitrate lyase play in allowing plants to grow on acetate? a. Isocitrate lyase hydrolyzes isocitrate to form malate and acetate; acetate is used to synthesize fatty acids. b. Isocitrate lyase hydrolyzes isocitrate to form malate and acetate; acetate is used to synthesize glucose. c. Succinate formed by isocitrate lyase is an intermediate in glucose synthesis via gluconeogenesis. d. One of the isocitrate lyase products is a potent inhibitor of pyruvate kinase. e. One of the isocitrate lyase products is an intermediate in steroid synthesis.

ANSWER: c 37. What enzyme catalyzes reactions that theoretically could follow second-order kinetics? a. succinyl CoA synthetase b. succinate dehydrogenase c. isocitrate dehydrogenase d. succinate thiokinase e. α-ketoglutarate dehydrogenase ANSWER: c 38. What enzymes capture high-energy electrons? Select all that apply. a. pyruvate dehydrogenase complex b. citrate synthase c. isocitrate dehydrogenase d. aconitase e. α-ketoglutarate dehydrogenase ANSWER: a, c, e 39. The direct movement of substrates from one enzyme to the next is called: .. Powered by Cognero.

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Chapter 19 a. a protein complex. b. substrate channeling. c. a linker coenzyme. d. a cell with sufficient available water. e. an electron acceptor.

ANSWER: b 40. In muscle, the enzyme that catalyzes a substrate-level phosphorylation is: a. nucleoside triphosphate transferase. b. protein kinase C. c. GTP kinase. d. succinyl CoA synthetase. e. ATP-GTP transferase.

ANSWER: d 41. Isomerization of citrate is catalyzed by: a. citrate synthase. b. aldolase. c. α-ketoglutarate dehydrogenase. d. aconitase. e. citrate isomerase.

ANSWER: d 42. What is the chemical change involved in the conversion of citrate into isocitrate? a. hydration followed by dehydration b. oxidation followed by dehydration c. oxidation followed by reduction d. dehydration followed by hydration e. hydration followed by oxidation ANSWER: d 43. The citric acid cycle (CAC) is activated in the presence of oxygen, but what is the link between the CAC and oxygen? a. Oxygen is an allosteric activator for citrate synthase. b. The presence of O2 in the mitochondrial matrix releases CO2 into the cytosol. c. The one substrate-level phosphorylation in the CAC can occur in the absence of oxygen. d. A primary product of the CAC is NADH, the principal electron donor to O2. e. The iron–sulfur center requires oxygen to be in the appropriate oxidation state.

ANSWER: d .. Powered by Cognero.

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Chapter 19 44. In what step of the citric acid cycle is FADH2 formed? a. conversion of succinate to malate b. conversion of succinate to oxaloacetate c. conversion of succinate to fumarate d. conversion of malate to oxaloacetate e. conversion of malate to fumarate ANSWER: c 45. Although we study the citric acid cycle as the final stage in the oxidation of carbon from glucose, an indepth look at the cycle shows intermediates entering and leaving the cycle from a number of metabolic pathways. With all of these demands on the cycle, how does it maintain a minimal level of oxaloacetate (OAA) to allow the cycle to function? a. OAA can be formed by the condensation of two moles of acetyl CoA and occurs when the energy charge of the cell is high. b. The rate of the cycle increases when the cell has high levels of NADH. c. Isocitrate dehydrogenase is allosterically inhibited by ADP, which signifies the need for more energy. d. OAA is formed directly via the deamination of glutamate. e. OAA is synthesized via pyruvate carboxylase in an anaplerotic reaction that occurs when acetyl CoA is present.

ANSWER: e 46. Chemically, why is it necessary for citrate to undergo an isomerization to isocitrate prior to decarboxylation? a. This conversion forms an unstable α-ketoacid, which drives the cycle forward. b. The aconitase reaction is a thermodynamically unfavorable hydration reaction followed by a highly favored dehydration. c. The tertiary alcohol in citrate does not favor the oxidative carboxylation that occurs next; however, the secondary alcohol of isocitrate does. d. Citrate induces a major structural rearrangement in aconitase, leading to the creation of the binding site for the water molecule in the hydration step. e. Only when the alcohol is attached to C-2 can it block the activity of prolyl hydroxylase 2, ensuring synthesis of glycolytic proteins.

ANSWER: c 47. The _______ is an organelle in plants and some bacteria where two-carbon acetyl units are converted into four-carbon units (succinate) for glucose synthesis, energy production, and biosynthesis.

ANSWER: glyoxysome 48. Organisms that can convert fat into sugar use the _____ cycle. a. Q b. fatty acid .. Powered by Cognero.

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Chapter 19 c. lipoxylate d. glyoxylate e. glucuronic acid ANSWER: d 49. Which enzyme catalyzes the oxidative decarboxylation of isocitrate to α-ketoglutarate? a. dihydrolipoyl transacetylase b. dihydrolipoyl dehydrogenase c. isocitrate dehydrogenase d. citrate synthase e. succinate dehydrogenase ANSWER: c 50. What molecule is produced when succinyl CoA is converted to succinate? a. NADH b. FADH2 c. GTP d. CO2 e. NAD+ ANSWER: c 51. Which of the following is the first molecule formed in the citric acid cycle after acetyl CoA condenses with oxaloacetate? a. malate b. citrate c. succinate d. fumarate e. isocitrate ANSWER: b 52. What is the primary role of NAD+ in the citric acid cycle? a. to act as a substrate for citrate synthase b. to donate electrons to the electron-transport chain c. to accept electrons during oxidation reactions d. to form acetyl CoA e. to release CO2 ANSWER: c 53. Which enzyme is responsible for the oxidative decarboxylation of α-ketoglutarate? a. isocitrate dehydrogenase b. succinyl-CoA synthetase .. Powered by Cognero.

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Chapter 19 c. α-ketoglutarate dehydrogenase d. fumarase e. malate dehydrogenase ANSWER: c 54. During the citric acid cycle, the cleavage of which molecule directly drives the generation of ATP or GTP? a. α-ketoglutarate b. citrate c. isocitrate d. succinyl CoA e. malate ANSWER: d 55. What is regenerated at the end of the citric acid cycle to enable the cycle to continue? a. citrate b. oxaloacetate c. succinyl CoA d. fumarate e. α-ketoglutarate ANSWER: b 56. How many molecules of ATP are ultimately generated per molecule of acetyl CoA that is oxidized through the citric acid cycle? a. 6 b. 8 c. 10 d. 12 e. 14 ANSWER: c 57. Which enzyme's activity is diminished by a high energy charge in the cell? a. citrate synthase b. fumarase c. α-ketoglutarate dehydrogenase d. aconitase e. nucleoside diphosphokinase ANSWER: c 58. Which of the following molecules is NOT directly involved in the citric acid cycle? a. NAD+ b. FAD .. Powered by Cognero.

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Chapter 19 c. GTP d. ATP e. biotin ANSWER: e 59. What is the role of FAD in the citric acid cycle? a. to act as an electron donor b. to act as an electron acceptor c. to form oxaloacetate d. to form ATP e. to release CO2 ANSWER: b 60. In which reaction of the citric acid cycle is CO2 first released? a. the conversion of citrate to isocitrate b. the conversion of isocitrate to α-ketoglutarate c. the conversion of α-ketoglutarate to succinyl CoA d. the conversion of succinate to fumarate e. the conversion of malate to oxaloacetate ANSWER: b 61. Which compound is a product of the citric acid cycle and is also a substrate for the electron-transport chain? a. fumarate b. succinyl CoA c. malate d. NADH e. citrate ANSWER: d 62. What is the final product of the citric acid cycle that re-enters the cycle to react with acetyl CoA? a. citrate b. succinate c. fumarate d. malate e. oxaloacetate ANSWER: e 63. Which enzyme in the citric acid cycle is regulated by ATP levels in the cell? a. aconitase b. α-ketoglutarate dehydrogenase c. fumarase .. Powered by Cognero.

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Chapter 19 d. malate dehydrogenase e. citrate synthase ANSWER: b 64. How many pairs of electrons are transferred to NAD+ during one complete turn of the citric acid cycle? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: c 65. What is the primary function of the citric acid cycle in cellular respiration? a. to generate heat b. to produce oxygen c. to synthesize glucose d. to harvest high-energy electrons e. to store fat ANSWER: d 66. What compound is formed by the hydration of fumarate in the citric acid cycle? a. citrate b. isocitrate c. malate d. succinate e. oxaloacetate ANSWER: c 67. Which molecule serves as a building block for nucleotide bases and proteins, derived from the citric acid cycle intermediates? a. citrate b. malate c. succinyl CoA d. α-ketoglutarate e. fumarate ANSWER: d 68. What process regenerates NAD+ and FAD needed for the citric acid cycle to continue? a. glycolysis b. electron-transport chain c. fermentation .. Powered by Cognero.

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Chapter 19 d. photosynthesis e. beta-oxidation ANSWER: b

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Chapter 20 1. What accepts electrons in the electron-transport chain? a. O2 b. H2O2 c. NADPH d. NADP+ e. FADH2 ANSWER: a 2. Which of the following donate electrons in the electron-transport chain? Select all that apply. a. O2 b. H2O2 c. NADH d. NAD+ e. FADH2 ANSWER: c, e 3. Coenzyme Q is also called: a. NADH. b. oxidoreductase. c. ubiquinone. d. Complex II. e. Rieske center.

ANSWER: c 4. How many complexes of the electron-transport chain use energy released by the electron flow to pump protons? a. 0 b. 1 c. 2 d. 3 e. 4 ANSWER: d 5. Which of the following statements does NOT correctly describe cellular respiration? a. the process of generating high-transfer-potential electrons by the citric acid cycle, their flow through the respiratory chain, and the accompanying synthesis of ATP b. an ATP-generating process in which an inorganic compound serves as the electron donor for the ultimate electron acceptor c. an ATP-generating process in which an organic compound serves as the electron donor for the ultimate electron acceptor .. Powered by Cognero.

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Chapter 20 d. an ATP-generating process in which an inorganic compound serves as the ultimate electron acceptor e. an ATP-generating process in which an organic compound serves as the ultimate electron acceptor ANSWER: e 6. What are cristae and what is their function? a. They are internal ridges of the outer mitochondrial membrane; they increase the surface area of the outer mitochondrial membrane. b. They are internal ridges of the plasma membrane on prokaryotes; they increase the surface area of the plasma membrane. c. They are internal ridges of the inner mitochondrial membrane; they increase the surface area of the inner mitochondrial membrane. d. They are organelles of the mitochondrion; they play a role of the sites for oxidative. phosphorylation. e. They are organelles of the mitochondrion; they play a role as the site for ATP synthesis.

ANSWER: c 7. What makes the outer mitochondrial membrane quite permeable? a. proton pump b. voltage gated active transporter c. ATP-powered pump d. mitochondrial porin e. pores in the membrane ANSWER: d 8. What is the other name of Complex I? a. succinate-Q reductase b. cytochrome c oxidase c. Q-cytochrome c oxidoreductase d. NADH-Q oxidoreductase e. cytochrome c ANSWER: d 9. What is the other name of Complex II? a. succinate-Q reductase b. cytochrome c c. Q-cytochrome c oxidoreductase d. cytochrome c oxidase e. NADH-Q oxidoreductase ANSWER: a 10. What is the other name of Complex III? a. Q-cytochrome c oxidoreductase .. Powered by Cognero.

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Chapter 20 b. succinate-Q reductase c. NADH-Q oxidoreductase d. cytochrome c oxidase e. cytochrome c ANSWER: a 11. What is the other name of Complex IV? a. Q-cytochrome c oxidoreductase b. cytochrome c c. succinate-Q reductase d. NADH-Q oxidoreductase e. cytochrome c oxidase ANSWER: e 12. Why are mitochondria thought to be semiautonomous organelles? Select all that apply. a. They contain many proteins encoded by nuclear DNA. b. They are the only sites for oxidative phosphorylation in the cell. c. They can exist outside of the cell. d. They contain their own DNA. e. They have an outer quite permeable membrane.

ANSWER: a, b, d 13. What proves mitochondrial emergence through an endosymbiotic event? Select all that apply. a. specific transcription machinery b. circular DNA c. oxidative phosphorylation d. double membrane e. specific translation machinery ANSWER: a, b, d, e 14. What is a redox couple? a. two different substances, one of which acts as an oxidant and the other as a reductant b. a substance that can exist in both an oxidized form and a reduced form c. two isomers that act as oxidants d. two isomers that act as reductants e. a substance that can exist only in an oxidized form or only in a reduced form ANSWER: b 15. Calculate ΔG°′ (in kilojoules per mole) for the reaction where ethanol + NAD+ are converted to acetaldehyde + NADH. E0' for the ethanol → acetaldehyde reaction is 0.20 V and E0' for the NAD+ → NADH .. Powered by Cognero.

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Chapter 20 reaction is –0.32 V. (F = 96.48 kJ mol–1 V–1) a. 100.34 b. 5.53 c. 23.98 d. 11.58 e. 23.16 ANSWER: e 16. In the initial reaction of NADH-Q oxidoreductase, what prosthetic group of this complex accepts two highpotential electrons from NADH? a. FAD b. FMN c. flavin d. FMNH2 e. FADH2 ANSWER: b 17. What is the function of the respirasome? Select all that apply. a. directly conducts ATP synthesis b. generates ROS c. facilitates a rapid transfer of the substrate d. prevents the release of reaction intermediates e. facilitates scavenging of ROS ANSWER: c, d 18. What complexes are NOT included into the respirasome? Select all that apply. a. Complex I b. Complex II c. Complex III d. Complex IV e. ATP synthase ANSWER: b, e 19. What proteins of the electron-transport chain contain iron and in what form? Select all that apply. a. iron–sulfur proteins; in the form of Fe–S clusters b. iron–sulfur proteins; in the form of a heme prosthetic group c. nonheme iron proteins; in the form of a nonheme prosthetic group d. cytochromes; in the form of Fe–S clusters e. cytochromes; in the form of a heme prosthetic group ANSWER: a, e .. Powered by Cognero.

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Chapter 20 20. What does NOT participate in, nor is a component of, the electron-transport chain? a. acetyl CoA b. nonheme-iron proteins c. coenzyme Q d. cytochrome c1 e. NADH ANSWER: a 21. What metal is present in ALL complexes of the electron-transport chain? a. copper b. iron c. magnesium d. potassium e. manganese ANSWER: b 22. What component of the electron-transport chain is a quinone derivative with a long isoprenoid tail? a. cytochrome b. coenzyme Q c. catalase d. succinate-Q reductase e. ubiquitin ANSWER: b 23. What properties does a long tail of the isoprenoid give to ubiquinone? a. makes it hydrophilic and allows it to diffuse rapidly within the inner mitochondrial membrane b. makes it hydrophobic and allows it to diffuse rapidly within the mitochondrial membranes c. makes it hydrophobic and allows it to diffuse rapidly within the inner mitochondrial membrane d. anchors it in the membrane so protons and electrons could shuttle through it e. provides several oxidation states ANSWER: c 24. What definition concerning the oxidation states of quinones is TRUE? a. Q is the fully oxidized state with two hydroxyl groups. b. QH-– is a semiquinone radical anion with one hydroxyl group. c. The addition of one electron results in the non-radical semiquinone form. d. QH- is a semiquinone intermediate with one hydroxyl group. e. QH2 is the fully reduced state with two keto groups.

ANSWER: d .. Powered by Cognero.

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Chapter 20 25. What prosthetic group is present in succinate-Q reductase, Q-cytochrome c oxidoreductase, and NADH-Q oxidoreductase? a. semiquinone b. FMN c. coenzyme Q d. heme group e. iron–sulfur cluster ANSWER: e 26. What prosthetic group is present only in succinate-Q reductase? a. semiquinone b. FMN c. coenzyme Q d. heme group e. FAD ANSWER: e 27. What mitochondrial protein is crucial for the synthesis of Fe–S clusters? And what type of mutation is responsible for the Friedreich's ataxia? a. amytal; trinucleotide expansion b. amytal; trinucleotide deletion c. frataxin; trinucleotide expansion d. frataxin; trinucleotide deletion e. frataxin; trinucleotide inversion ANSWER: c 28. What protein catalyzes the reduction of O2? a. cytochrome c oxidase b. cytochrome c c. coenzyme Q d. catalase e. Q-cytochrome c oxidoreductase ANSWER: a 29. What is the largest complex in the electron-transport chain? a. cytochrome c oxidase b. succinate-Q reductase c. Q-cytochrome c oxidoreductase d. NADH-Q oxidoreductase e. cytochrome c reductase ANSWER: d .. Powered by Cognero.

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Chapter 20 30. What proton pump is L-shaped, with a hydrophobic horizontal arm lying in the membrane and a hydrophilic vertical arm projecting into the matrix? a. Q-cytochrome c oxidoreductase b. NADH-Q oxidoreductase c. cytochrome c oxidase d. ATPase e. succinate-Q reductase ANSWER: b 31. What complex has seven Fe–S clusters? a. ATPase b. succinate-Q reductase c. cytochrome c oxidase d. Q-cytochrome c oxidoreductase e. NADH-Q oxiodoreductase ANSWER: e 32. What statement concerning the structural elements required for proton pumping is FALSE? a. The intermembrane space half-channels are joined by a series of β-hairpin-helix-connecting elements βH. b. An enclosed Q chamber exists near the junction of the hydrophilic portion and the membraneembedded portion. c. A hydrophilic funnel connects the Q chamber to a water-lined channel into which the half-channels open. d. The matrix half-channels are connected by a long horizontal helix (HL). e. Four proton half-channels consist of vertical and horizontal helices.

ANSWER: e 33. What is the term for the mechanism for the coupling of electron transfer from Q to cytochrome c to the transmembrane proton transport? a. Q pool b. Q cycle c. cellular respiration d. ubiquitin cycle e. cytochrome c pool ANSWER: b 34. What makes aerobic organisms aerobic? a. They need to breathe. b. They use oxidative phosphorylation. c. Reduction of oxygen to useful water molecules takes place. .. Powered by Cognero.

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Chapter 20 d. The final acceptor in the electron-transport chain is oxygen. e. Oxygen is the prevailing gas in the atmosphere.

ANSWER: d 35. What is the name of protons that participate in the reduction of O2 to H2O? a. chemical protons b. additional protons c. reducing protons d. pumped protons e. redox protons ANSWER: a 36. What is the reaction in which a single reactant is converted into two different products with differing oxidation states? a. dismutation b. isomerization c. racemization d. synthesis e. cleavage ANSWER: a 37. From a biochemical point of view, is the net effect of exercise negative and why? a. No, it is not. The amount of superoxide dismutase decreases in the cell. b. No, it is not. The increase in superoxide dismutase more effectively protects the cell during periods of rest. c. No, it is not. An increase in ROS is always beneficial, as they have been implicated in the control of cell differentiation. d. Yes, it is. The elevated aerobic metabolism during exercise causes more harmful ROS to be generated. e. Yes, it is. An increase in ROS never brings benefits.

ANSWER: b 38. The only lipid-soluble antioxidant synthesized by human beings is: a. superoxide dismutase. b. catalase. c. ubiquinol. d. vitamins E. e. vitamins C. ANSWER: c 39. Choose the CORRECT path taken by a pair of electrons as they travel down the electron-transport chain. .. Powered by Cognero.

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Chapter 20 a. NADH → Complex I → CoQ → Complex III → Cyt c → Complex IV → O2 → Complex I → CoQ → Complex II → Cyt c → Complex IV → O2 b. FADH2 → Complex I → CoQ → Complex III → Cyt c → Complex IV → O2 → Complex II → CoQ → Complex III → CoQ → Complex IV → O2 c. NADH → Complex I → Complex II → Complex III → Cyt c → Complex IV → O2 → Complex I → CoQ → Complex I → Cyt c → Complex I → O2 d. FADH2 → Complex II → CoQ → Complex III → Cyt c → Complex IV → O2 → Complex III → CoQ → Complex III → Cyt c → Complex IV → Complex I → O2 e. NADH → Complex I → CoQ → Complex III → Cyt c → Complex IV → O2 and FADH2 → Complex II → CoQ → Complex III → Cyt c → Complex IV → O2 ANSWER: e 40. Cytochrome c oxidase contains two heme A groups and three _______ ions.

ANSWER: copper 41. An electron flow down the electron-transport chain leads to: a. the transport of protons across the inner mitochondrial membrane from the matrix to the intermembrane space. b. the transport of protons across the inner mitochondrial membrane from the intermembrane space into the matrix. c. the coupled synthesis of GTP. d. a dangerous imbalance of K+ ions across the mitochondrial membrane. e. a dangerous imbalance of Na+ ions across the mitochondrial membrane.

ANSWER: a 42. Protons do NOT pump: a. Complex I. b. Complex II. c. Complex III. d. Complex IV. e. Complex V ANSWER: b 43. What is the rational way to say that electrons flow DOWN the electron-transport chain? a. Electrons flow from oxidized carriers to reduced carriers in discrete steps like a staircase. b. Electrons flow from half-reactions with more positive redox potentials to more negative. c. Electrons flow from reactions that continuously generate negative free-energy values. d. Electrons flow from the outer surface to the inner surface, a top-down perception. e. Because we think of electrons flowing like a liquid, the only direction electrons can flow is down. .. Powered by Cognero.

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Chapter 20 ANSWER: c 44. Electrons are not very soluble in hydrophobic environments such as a bilipid membrane. What is the evidence that electrons move from complex to complex through the lipid membrane? a. Cytochrome c is the electron carrier in the membrane and undergoes a head-over-heel flip to set up the proton gradient. b. Cytochrome c is the electron carrier in the membrane and contains a hydrophobic porphyrin center. c. CoQ is the electron carrier within the membrane and seems to be confined to the respirasome. d. CoQ is the electron carrier within the membrane and contains a hydrophobic porphyrin center. e. FeS clusters are the carriers within the membrane, and the electrons move from one cluster to the next in a chain-like fashion.

ANSWER: c 45. In the Rieske center, the iron–sulfur center is coordinated to the amino acid: a. His. b. Cys. c. Ala. d. Met. e. Gly. ANSWER: a 46. The Q cycle: a. transfers electrons from a two-electron carrier to a one-electron carrier. b. recycles protons. c. acts as an NADH exchanger within the cytosol. d. transfers electrons from a four-electron carrier to a two-electron carrier. e. acts as an ATP exchanger within the extracellular compartment.

ANSWER: a 47. Select all that apply. Cytochrome is a: a. protein that transfers electrons. b. chloroplast protein that accumulates electrons. c. protein that pumps ATP. d. protein that contains a heme prosthetic group. e. protein that contains an iron sulfur prosthetic group.

ANSWER: a, d 48. Carbon monoxide is considered toxic because it acts on Complex IV by binding to its heme group, preventing the transfer of electrons to oxygen. How would the addition of carbon monoxide to actively respiring mitochondria affect the relative oxidation–reduction state of all components of the electron-transport chain? a. All four complexes would remain oxidized because they function as a multisystem complex. b. Complexes I and III would be reduced, but Complexes II and IV would be oxidized. .. Powered by Cognero.

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Chapter 20 c. Complexes I, II, and III would be oxidized, but Complex IV would remain reduced. d. Complexes I, II, and III would be reduced, and Complex IV would be oxidized. e. All four complexes would remain reduced because they function as a multisystem complex.

ANSWER: d 49. What pathologic condition does NOT result from free-radical injury? a. emphysema b. Parkinson disease c. diabetes d. atherogenesis e. asthma ANSWER: e 50. Why does spent ATP need to be recycled back to usable ATP? a. Humans only possess less than 1% of the daily required amount of ATP. b. ATP can only be recycled 1 time per day. c. Humans do not need a large amount of ATP to function. d. Spent ATP contains toxins that can lead to illness over time. e. Humans possess more than enough ATP, so spent ATP is not needed and is recycled.

ANSWER: a 51. Which complexes in the electron transport chain are responsible for pumping protons into the intermembrane space? a. Complex I, Complex II, and Complex III b. Complex II, Complex III, and Complex IV c. Complex I, Complex III, and Complex V d. Complex I, Complex III, and Complex IV e. Complex III, Complex IV, and Complex V ANSWER: d 52. Which of the following components of the electron transport chain does not directly contribute to the proton gradient? a. NADH-Q oxidoreductase b. Q-cytochrome c oxidoreductase c. succinate-Q reductase d. cytochrome c oxidase e. ubiquinone ANSWER: c 53. Which molecule is reduced to form water at the end of the electron transport chain? a. H2O2 .. Powered by Cognero.

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Chapter 20 b. NADP+ c. O2 d. CO2 e. ATP ANSWER: c 54. Which two processes are combined in oxidative phosphorylation? a. oxidation and glycolysis b. reduction and ATP synthesis c. the citric acid cycle and oxidation-reduction d. oxidation and ATP synthesis e. anaerobic respiration and phosphorylation ANSWER: d 55. Which complex is involved in both the citric acid cycle and the electron transport chain? a. Complex I b. Complex II c. Complex III d. Complex IV e. Complex V ANSWER: b 56. What is the role of cytochrome c in the electron transport chain? a. to pump protons across the membrane b. to transfer electrons between Complex III and Complex IV c. to synthesize ATP d. to act as the final electron acceptor e. to reduce NAD+ ANSWER: b 57. What is the primary benefit of having the electron transport chain located in the inner mitochondrial membrane? a. enhanced ATP production b. increased surface area for reactions c. facilitation of oxygen transport d. prevention of free radical formation e. direct regulation by nuclear DNA ANSWER: b 58. How does the inner mitochondrial membrane aid in the formation of a proton gradient? a. by containing electron carriers that pump protons across the membrane .. Powered by Cognero.

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Chapter 20 b. by facilitating the diffusion of protons into the matrix c. by preventing the entry of protons into the intermembrane space d. by transporting ATP out of the mitochondria e. by acting as a barrier to electron flow ANSWER: a 59. Why is the inner mitochondrial membrane highly impermeable to ions? a. to maintain the proton gradient necessary for ATP synthesis b. to protect the mitochondrial matrix from toxins c. to facilitate the diffusion of oxygen d. to support the structure of the mitochondrion e. to prevent the leakage of NADH ANSWER: a 60. What does a negative reduction potential indicate? a. The oxidized form of a substance has a lower affinity for electrons than H2 does. b. The oxidized form of a substance has a higher affinity for electrons than H2 does. c. The oxidized form of a substance has the same affinity for electrons as H2 does. d. The reaction is not proceeding under standard conditions. e. None of the above.

ANSWER: a 61. The membrane potential created by the electron transport chain is primarily used for which purpose? a. ATP synthesis b. fatty acid oxidation c. DNA replication d. protein synthesis e. glucose transport ANSWER: a 62. How are electrons transferred between complexes in the electron transport chain? a. through direct contact between complexes b. by small, mobile carriers like ubiquinone and cytochrome c c. via ATP molecules d. through the matrix of the mitochondrion e. by diffusion through the membrane ANSWER: b 63. Which complex in the electron transport chain directly transfers electrons to oxygen? a. Complex I b. Complex II .. Powered by Cognero.

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Chapter 20 c. Complex III d. Complex IV e. Complex V ANSWER: d 64. Which molecule carries electrons from Complex I to Complex III? a. cytochrome c b. NADH c. FADH2 d. ubiquinone (coenzyme Q) e. ATP ANSWER: d 65. The movement of electrons through the electron transport chain primarily serves to _____. a. generate heat b. produce ATP directly c. create a proton gradient across the inner mitochondrial membrane d. oxidize glucose e. synthesize NADH ANSWER: c 66. What is the immediate result of electron transfer through Complex IV? a. NADH is oxidized. b. Protons are pumped into the intermembrane space. c. ATP is synthesized directly. d. FAD is reduced to FADH2. e. Oxygen is released as a byproduct.

ANSWER: b 67. Which step in the electron transport chain directly produces water? a. electron transfer to Complex I b. reduction of ubiquinone c. oxidation of NADH d. reduction of oxygen at Complex IV e. proton pumping by Complex III ANSWER: d 68. Which molecule donates electrons to Complex II? a. NADH b. FADH2 c. ubiquinone .. Powered by Cognero.

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Chapter 20 d. cytochrome c e. oxygen ANSWER: b 69. How does the movement of electrons through the electron transport chain lead to ATP synthesis? a. by directly phosphorylating ADP b. by generating a proton gradient that drives ATP synthase c. by oxidizing glucose d. by reducing NADP+ e. by transporting ATP out of the mitochondria ANSWER: b

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Chapter 21 1. What is the more preferred name for a molecular assembly in the inner mitochondrial membrane, which carries out synthesis of ATP? a. F1F0 ATPase b. mitochondrial ATPase c. ATP synthase d. NADH e. the proton-motive force ANSWER: c 2. What is the essence of Mitchell's chemiosmotic hypothesis? a. Electron transport and ATP synthesis are coupled by a proton gradient across the inner mitochondrial membrane. b. Electron transport and ATP synthesis are coupled by an electron gradient across the inner mitochondrial membrane. c. The proton concentration becomes higher in the matrix, and an electric field with a negative matrix side is generated. d. The respiratory chain and ATP synthase are biochemically separate systems, linked only by an electron-motive force. e. The transfer of protons leads to the pumping of electrons from the matrix to the cytoplasmic side of the inner mitochondrial membrane.

ANSWER: a 3. What explains a name of ATP synthase such as F1F0 ATP synthase? a. ATP synthase generates free energy (F). b. ATP synthase consists of four α and four β subunits. c. ATP synthase consists of an F0 subunit in the mitochondrial matrix and an F1 subunit in the inner membrane. d. ATP synthase provides a chemical gradient and a charge gradient. e. ATP synthase consists of an F0 subunit in the inner membrane and an F1 subunit in the mitochondrial matrix.

ANSWER: e 4. Which of the following is the proton-motive force composed of? Select all that apply. a. free energy b. a pH gradient c. a concentration of ATPase d. a membrane potential e. transport of electrons ANSWER: b, d 5. What conformation of β subunits contributes to form ATP? .. Powered by Cognero.

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Chapter 21 a. F1 conformation b. open conformation c. loose conformation d. tight conformation e. O conformation ANSWER: d 6. The rotation of the γ subunit leads to the change of: a. the structure and functions of the F1 subunit. b. the structure and functions of β subunits. c. the charge of the inner membrane. d. a method of ATP formation. e. the functions of the F0 subunit.

ANSWER: b 7. What promotes the rotation of the γ subunit? a. hydrolysis of ADP b. hydrolysis of ATP c. formation of ATP d. change in the conformation of the β subunit e. binding of nucleotides ANSWER: b 8. The proton channel of ATP synthase consists of: a. 8–14 pairs of α helices. b. 7 α helices. c. 8–14 α helices. d. 8–14 half-channels. e. 8–14 a subunits.

ANSWER: a 9. What condition is required for a c subunit to pass into the hydrophobic region? a. proton transfer through the proton pump b. proton binding to the glutamate residue c. electron transfer through the electron-transport chain d. electron binding to the aspartate residue e. releasing of protons to the proton-rich environment ANSWER: b 10. How many ATP molecules will be synthesized if a γ subunit rotates three times by 360 degrees? .. Powered by Cognero.

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Chapter 21 a. 4.5 molecules b. 7.5 molecules c. 9 molecules d. 10 molecules e. 3.3 molecules ANSWER: c 11. Choose the CORRECT sequence of oxidative phosphorylation. a. transfer of electrons through the electron-transport chain → movement of protons through the halfchannels → binding of the actin filament to the γ subunit → rotation of the α subunit → rotation of the β subunit → synthesis and releasing of ADP b. electron transport from NADH to a half-channel → movement of protons through the respiratory chain → synthesis and releasing of ADP c. transfer of electrons through the electron-transport chain → movement of protons through the halfchannels → rotation of the a subunit → rotation of the β subunit → synthesis and releasing of ADP d. transfer of electrons through the electron-transport chain → generation of a proton gradient → movement of protons through the half-channels → rotation of the c ring → rotation of the γ subunit → synthesis and releasing of ATP e. electron transport from NADH to a half-channel → movement of protons through the respiratory chain → rotation of the c ring → rotation of the γ subunit → synthesis and releasing of ATP ANSWER: d 12. What is one way NADH transfers across the mitochondrial membrane? a. transferring H+ from mitochondrial NADH to cytoplasmic NAD+ b. introducing NADH by the glycerol 3-phosphate shuttle c. introducing electrons from NADH by the glycerol 3-phosphate shuttle d. transferring electrons from NADH to FADH2 e. introducing H+ from NADH by dihydroxyacetone phosphate ANSWER: c 13. In the heart and liver, electrons from cytoplasmic NADH are brought into mitochondria by: a. ATP–ADP translocase. b. the malate–aspartate shuttle. c. pyruvate carrier. d. the glycerol–aspartate shuttle. e. phosphate carrier.

ANSWER: b 14. Choose the CORRECT statement about transport of ATP. a. Transport of ATP is not coupled with transport of ADP. b. ATP is moved across the mitochondrial membrane by the 30-kDa specific protein. .. Powered by Cognero.

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Chapter 21 c. ATP has the same negative charge as ADP. d. Transport of ATP occurs without energy expenditure. e. ATP passes freely through the inner mitochondrial membrane.

ANSWER: b 15. What components form the ATP synthasome? Select all that apply. a. the malate–aspartate shuttle b. the phosphate carrier c. ATP–ADP translocase d. guanine nucleotide transporter e. ATP synthase ANSWER: b, c, e 16. What are mitochondrial carriers? a. transport molecules that carry only ATP across the outer mitochondrial membrane b. transmembrane glycoproteins that carry specific ions and charged metabolites across the inner mitochondrial membrane c. transmembrane proteins that carry specific ions and charged metabolites across the inner mitochondrial membrane d. transport molecules that carry the exchange of cytoplasmic ATP for matrix ADP and Pi e. surface proteins that carry specific ions and charged metabolites across the inner mitochondrial membrane ANSWER: c 17. Choose the CORRECT sequence of action of ATP translocase. a. binding of ADP from the cytoplasm with the translocase → eversion of the transporter → releasing ADP into the matrix → binding of ATP from the matrix → back eversion to the primary conformation of ATP translocase → releasing ATP into the cytoplasm b. binding of ATP from the cytoplasm with the translocase → eversion of the transporter → binding of ADP from the matrix → back eversion to the primary conformation of ATP translocase → releasing ADP into the cytoplasm c. electrons enter the mitochondrial electron-transport chain → electrons transfer to ATP translocase → binding of ADP from the cytoplasm with the translocase → eversion of the transporter → binding of ATP from the matrix → releasing ATP into the cytoplasm d. binding of ADP from the cytoplasm with ATP synthase → ADP transfer to the transporter → eversion of the transporter → releasing ADP into the matrix → binding of ATP from the matrix → back eversion to the primary conformation of ATP translocase → releasing ATP into the cytoplasm e. electrons enter the mitochondrial electron-transport chain → electrons transfer to ATP translocase → eversion of the transporter → proton transfer to the translocase → binding of ATP → releasing ATP into the cytoplasm ANSWER: a .. Powered by Cognero.

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Chapter 21 18. How many molecules of ATP are formed by the oxidation of glucose? a. 2.5 b. 26 c. 30 d. 2 e. 1.5 ANSWER: c 19. _____ are components of the F1 subunit of ATP synthase. Select all that apply. a. Three α subunits b. Two ε subunits c. γ subunit d. 14 c subunits e. Subunit ANSWER: a, c 20. Transport through the inner mitochondrial membrane is nearly impossible for: a. oxaloacetate. b. malate. c. aspartate. d. α-ketoglutarate. e. glutamate.

ANSWER: a 21. What is the relationship between obesity and UCP-1? a. Obesity leads to a decrease in brown adipose tissue rich in UCP-1. b. Obesity leads to an increase in brown adipose tissue rich in UCP-1. c. Obesity leads to a decrease in white adipose tissue lacking UCP-1. d. An increase in white adipose tissue leads to obesity. e. An increase in brown adipose tissue leads to obesity.

ANSWER: a 22. The organs most vulnerable to mitochondrial diseases: a. are highly dependent on oxidative phosphorylation. b. are highly dependent on glycolysis. c. consume a lot of energy. d. have a high level of brown fat. e. have a high level of adipose tissue.

ANSWER: a 23. On the graph, point 1 describes the: .. Powered by Cognero.

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Chapter 21

a. increase of O2 consumption after adding ATP. b. increase of O2 consumption after adding ADP. c. increase of O2 consumption with time. d. decrease of O2 consumption after adding ADP. e. decrease of O2 consumption after adding ATP. ANSWER: b 24. What controls the rate of the citric acid cycle? a. availability of NAD+ and FAD b. availability of malate c. availability of NAD+ and Pi d. permeability of the inner mitochondrial membrane e. permeability of the outer mitochondrial membrane ANSWER: a 25. Where is glycerol 3-phosphate reoxidized to dihydroxyacetone phosphate? a. the outer surface of the inner mitochondrial membrane b. the inner surface of the outer mitochondrial membrane c. cytoplasm of the cell d. matrix of the mitochondria e. lumen of the endoplasmic reticulum ANSWER: a 26. What molecule connects the citric acid cycle and the electron-transport chain? a. H2O b. O2 c. NADH .. Powered by Cognero.

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Chapter 21 d. H+ e. ADP ANSWER: c 27. What provides energy for the rotation of bacterial flagella and ATP synthesis? a. proton gradient b. pH gradient c. charge gradient of H+ d. electrochemical gradient for OH– e. ATP gradient ANSWER: a 28. ATP synthesis controls: a. reduction of NADH and FADH2. b. oxidation of NADH and FADH2. c. the flow of electrons through UCP-1. d. the proton gradient. e. reduction of glucose.

ANSWER: b 29. When tissue is deprived of O2, the electron-transport chain is unable to generate a proton gradient and: a. ATP synthase will produce ATP. b. ATP synthase will stop its work. c. IF1 will inhibit the hydrolytic activity of ATP synthase. d. ATP synthase will hydrolyze ATP to ADP and Pi. e. IF1 will activate the hydrolytic activity of ATP synthase.

ANSWER: c 30. What is the product of UCP-1 action? a. ATP b. proton-motive force c. heat d. ADP e. O2 ANSWER: c 31. The direct transporter of H+ through UCP-1 is a(n): a. carboxyl group of a fatty acid. b. α-adrenergic hormone. .. Powered by Cognero.

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Chapter 21 c. NAD+ and FAD. d. malate. e. amino group of histidine.

ANSWER: a 32. Leber hereditary optic neuropathy is inherited from: a. both parents with the X chromosome. b. the father with the Y chromosome. c. the mother with the X chromosome. d. the mother with mitochondrial chromosomes. e. the father with mitochondrial chromosomes.

ANSWER: d 33. A low concentration of atractyloside: a. stops oxidative phosphorylation. b. inhibits the hydrolytic activity of ATP synthase. c. activates ATP synthase. d. activates ATP–ADP translocase. e. inhibits the phosphate carrier.

ANSWER: a 34. How many ATP molecules are generated after a pair of transported electrons flows from NADH to O2? a. 2.5 b. 1.5 c. 2.0 d. 1.0 e. 3.0 ANSWER: a 35. Which process provides reduction of FADH2 using the electron-transport system? a. glycerol 3-phosphate shuttle b. malate–aspartate shuttle c. ATP synthesis d. glycolysis e. hydrolysis of ATP ANSWER: a 36. Acceptor control of oxidative phosphorylation means that the rate of respiration depends on the level of: a. O2. b. cAMP. .. Powered by Cognero.

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Chapter 21 c. ATP. d. ADP. e. AMP. ANSWER: d 37. Choose the compound that blocks the flow of electrons from cytochrome c to oxygen. Select all that apply. a. azide b. thermogenin c. bongkrekic acid d. atractyloside e. cyanide ANSWER: a, e 38. In the glycerol phosphate shuttle, cytoplasmic glycerol phosphate dehydrogenase uses cytoplasmic NADH to reduce _______ to glycerol 3-phosphate.

ANSWER: dihydroxyacetone phosphate 39. How does the rotation of the c ring lead to ATP synthesis? a. The c ring is linked tightly to the gamma and epsilon subunits in the stalk of F1. b. The c ring interacts with the beta subunit. c. The gamma subunit rotates with proton gradient formation inducing the binding-change mechanism. d. All of the answers are correct. e. None of the answers are correct.

ANSWER: d 40. The subunit of ATPase embedded in the inner mitochondrial membrane is the: a. anchor subunit. b. membrane–c ring subunit. c. F0 subunit. d. F1 subunit. e. Fm subunit.

ANSWER: c 41. The proton-motive force consists of _____ gradient(s). a. a chemical and a charge b. a proton c. an electron d. only a chemical e. a chemical, a proton, and an electron ANSWER: a .. Powered by Cognero.

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Chapter 21 42. A diet pill that acts to increase oxygen consumption and a high amount of electron transport without ATP production is likely to be what kind of compound? a. uncoupler b. ATP synthase activator c. site I inhibitor d. site II activator e. cyanide ANSWER: a 43. Suppose there is a mutation in the c subunit of ATP synthase, such that glutamate found in the middle of one of the membrane-spanning helices is converted to valine. What is likely to be the effect on ATP synthesis and why? a. No effect. The middle of the helix is in contact with the hydrophobic center of the lipid bilayer, and valine is readily soluble in lipid. b. No effect. The valine side chain is shorter than the glutamate side chain, so it causes no change in the secondary structure of the helix. c. Inhibition of ATP synthesis. Valine cannot bind a proton, so there will be no proton flow through the inner membrane. d. Inhibition of ATP synthesis. Because valine is hydrophobic, the a subunit will move in the reverse direction, causing hydrolysis, not synthesis, of ATP. e. Increase of ATP synthesis. Because the valine side chain is hydrophobic, the a subunit can move easily without regard for oxidative processes.

ANSWER: c 44. In the malate–aspartate shuttle, electrons from NADH are transferred to _____, forming malate. a. oxaloacetate b. aspartate c. acetate d. glutamate e. None of the answers are correct.

ANSWER: a 45. Why is it NOT surprising that substances such as intermediates for the citric acid cycle, protons, inorganic phosphates, nucleotide phosphates, and many others have their transport across the inner mitochondrial membrane regulated? a. Regulated transport allows for more effective substrate cycling. b. Ultimately, all electrons flow into the electron-transport chain, which regulates electron flow from the matrix to the inner membrane space. c. The experiment where bacteriorhodopsin and ATP synthase were inserted into reconstituted vesicles showed that biochemically, membranes control the movement of electrons in the respiratory chain. d. Many of the reactions in the citric acid cycle and ATP synthesis are driven by accessibility of substrates and differential gradients across the inner membrane. .. Powered by Cognero.

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Chapter 21 e. It is necessary to segregate the enzymes capable of substrate-level phosphorylation during anaerobic respiration.

ANSWER: d 46. An electron flow down the electron-transport chain leads to: a. transport of protons across the inner mitochondrial membrane from inside the matrix to the intermembrane space. b. transport of protons across the inner mitochondrial membrane from the intermembrane space into the matrix. c. coupled synthesis of GTP. d. a dangerous imbalance of K+ ions across the mitochondrial membrane. e. None of the answers is correct.

ANSWER: a 47. What is the reaction of ATP synthase? a. AMP3– + 2HPO42– + H+ ↔ ATP4– + H2O b. ADP3– + HPO42– + H+ ↔ ATP4– + H2O c. ADP3– + HPO42– + 2H+ ↔ ATP4– + H2O d. AMP3– + 2HPO42– + 2H+ ↔ ATP4– + H2O e. None of the answers are correct.

ANSWER: b 48. When glucose is totally oxidized to CO2 and H2O, how many ATP molecules are made by oxidative phosphorylation relative to the maximum yield? a. 12 out of 30 b. 26 out of 30 c. 26 out of 32 d. 12 out of 38 e. None of the answers is correct.

ANSWER: b 49. Which component of the electron-transport chain contributes the most to creating the proton gradient across the inner mitochondrial membrane? a. Complex I b. Complex II c. Complex III d. Complex IV e. ATP synthase ANSWER: d .. Powered by Cognero.

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Chapter 21 50. How is the energy from the proton gradient converted into chemical energy in ATP? a. direct transfer of electrons to ADP b. protons flow through ATP synthase, driving the phosphorylation of ADP. c. oxidation of NADH d. reduction of oxygen e. direct binding of protons to ATP ANSWER: b 51. What happens if the mitochondrial membrane becomes permeable to protons? a. ATP production increases b. ATP synthesis becomes more efficient c. The proton gradient dissipates, reducing ATP production d. The electron-transport chain speeds up e. NADH production increases ANSWER: c 52. Which molecule acts as the final electron acceptor in the electron-transport chain? a. NADH b. FADH2 c. water d. oxygen e. ATP ANSWER: d 53. What is the role of ATP synthase in the mitochondria? a. It synthesizes ATP using the energy from the proton gradient. b. It pumps protons into the mitochondrial matrix. c. It transfers electrons between complexes. d. It oxidizes NADH. e. It transports ATP out of the mitochondria.

ANSWER: a 54. Which shuttle systems transfer reducing equivalents from cytoplasmic NADH into the mitochondria? a. malate-aspartate shuttle and citrate shuttle b. glycerol-3-phosphate shuttle and malate-aspartate shuttle c. citrate shuttle and ubiquinone shuttle d. carnitine shuttle and glycerol-3-phosphate shuttle e. ubiquinone shuttle and malate-aspartate shuttle ANSWER: b 55. How does ATP exit the mitochondria to be used in the cytoplasm? .. Powered by Cognero.

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Chapter 21 a. through diffusion b. via the adenine nucleotide translocator c. by binding to oxygen d. through the electron-transport chain e. via the carnitine shuttle ANSWER: b 56. What is the primary role of the glycerol-3-phosphate shuttle in the cell? a. It generates ATP in the cytoplasm. b. It oxidizes NADH to maintain the proton gradient. c. It transfers electrons from NADH to the electron-transport chain via FADH2. d. It pumps protons into the mitochondrial matrix. e. It carries ATP out of the mitochondria.

ANSWER: c 57. Which molecule is exchanged with ATP as it leaves the mitochondria? a. ADP b. NADH c. FADH2 d. phosphate e. oxygen ANSWER: a 58. Which process is responsible for bringing pyruvate into the mitochondria for further oxidation? a. simple diffusion b. active transport c. pyruvate translocation via a transporter d. endocytosis e. glycolysis ANSWER: c 59. What role does oxygen concentration play in the rate of cellular respiration? a. Low oxygen levels slow cellular respiration by limiting electron-transport chain activity. b. High oxygen levels increase the rate of glycolysis. c. Oxygen concentration does not affect cellular respiration. d. Oxygen inhibits the citric acid cycle. e. Low oxygen levels enhance ATP production.

ANSWER: a 60. How does ADP availability influence the rate of oxidative phosphorylation? a. ADP levels do not affect the rate of oxidative phosphorylation. .. Powered by Cognero.

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Chapter 21 b. High ADP levels increase the rate of ATP synthesis. c. Low ADP levels inhibit ATP synthesis. d. ADP is not involved in oxidative phosphorylation. e. ADP reduces oxygen consumption, slowing the rate of oxidative phosphorylation.

ANSWER: b 61. Which of the following enzymes plays a critical role in both the citric acid cycle and oxidative phosphorylation? a. hexokinase b. succinate dehydrogenase c. pyruvate kinase d. citrate synthase e. malate dehydrogenase ANSWER: b 62. How does NADH concentration affect the rate of the electron-transport chain? a. High levels of NADH increase the activity of the electron-transport chain. b. NADH inhibits the electron-transport chain. c. NADH is not involved in the electron-transport chain. d. Low NADH levels enhance electron-transport chain activity. e. NADH directly generates ATP in the cytoplasm.

ANSWER: a 63. Which factor is most likely to slow cellular respiration? a. an increase in oxygen levels b. a rise in ADP concentration c. increased levels of NADH d. a decrease in glucose availability e. enhanced glycolytic activity ANSWER: d 64. What is the effect of high levels of ATP on the rate of cellular respiration? a. High ATP levels inhibit key enzymes, slowing cellular respiration. b. High ATP levels enhance electron-transport chain activity. c. ATP levels have no effect on cellular respiration. d. High ATP levels increase glucose uptake. e. ATP acts as a catalyst, stimulating cellular respiration.

ANSWER: a 65. Which of the following molecules influences the rate of electron-transport chain activity? a. NADH .. Powered by Cognero.

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Chapter 21 b. oxygen c. FADH2 d. ADP e. all of the above ANSWER: e 66. What is the relationship between glucose availability and the rate of cellular respiration? a. Glucose availability has no effect on cellular respiration. b. Low glucose levels reduce the rate of glycolysis and thus cellular respiration. c. High glucose levels inhibit cellular respiration. d. Glucose availability only affects oxidative phosphorylation. e. Glycolysis is an anaerobic process, so glucose decreases oxygen consumption during cellular respiration.

ANSWER: b 67. How does a decrease in oxygen availability affect the rate of cellular respiration? a. It increases the rate of glycolysis. b. It decreases the activity of the electron-transport chain. c. It has no effect on cellular respiration. d. It stimulates the citric acid cycle. e. It enhances ATP production in the cytoplasm.

ANSWER: b 68. Which factor can significantly impact the overall rate of cellular respiration in a cell? a. the availability of oxygen b. the concentration of ADP c. the levels of ATP d. the availability of substrates like glucose e. all of the above ANSWER: e

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Chapter 22 1. What is the key process to both photosynthesis and cellular respiration? a. generation of high-energy electrons b. synthesis of glucose c. generation of CO2 d. generation of low-energy electrons e. oxidation of H2O ANSWER: a 2. In what part of chloroplasts do the coupled redox reactions occur? a. intermembrane space b. stroma c. thylakoid lumen d. stroma membrane e. thylakoid membrane ANSWER: e 3. Choose the CORRECT statement(s) about pigments. Select all that apply. a. The formyl group of Chla allows it to cover light with the wavelength between 420 and 670 nm. b. Each pigment absorbs the light of a specific range of wavelengths. c. Assembly in light-harvesting complexes increases the range of light wavelengths absorbed by chlorophylls. d. Each pigment always absorbs the light of a wide range of wavelengths. e. Pigments of chloroplast grana absorb the light of longer wavelengths than pigments of stroma.

ANSWER: b, c 4. What molecule gains reducing power in photoinduced charge separation? a. special pair b. electron carrier c. electron acceptor d. electron donor e. proton acceptor ANSWER: c 5. What makes chlorophyll an effective photoreceptor? a. double bonds b. resonance c. magnesium ion d. hydrophilic phytol group e. single bonds ANSWER: b .. Powered by Cognero.

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Chapter 22 6. What is the name of the group of photosynthetic organisms that uses sunlight energy as its only vital source? a. heterotrophs b. chemotrophs c. saprotrophs d. autotrophs e. autoheterotrophs ANSWER: d 7. Choose the INCORRECT aspect concerning carotenoids. a. absorbance of light between 400 and 500 nm b. suppression of damaging photochemical reactions c. absorbance of light of yellow and red spectra d. expansion of absorption spectra of chlorophylls e. absorbance of blue and indigo light ANSWER: c 8. Why are green potatoes poisonous? a. They were exposed to light, leading to the synthesis of highly toxic cocaine. b. They were exposed to light, leading to the formation of a toxic alkaloid, which reduces the potato's starch content. c. They were exposed to light, leading to the production of solanine, a toxic compound. d. They were exposed to light, increasing the production of chlorophyll, which is harmful to humans.

ANSWER: c 9. What nucleotide-based cofactor is of high importance for light-sensitive complexes? a. UDP b. CoA c. NAD+ d. NADP+ e. TPP ANSWER: d 10. Which statement explains the relationship between resonance energy transfer and photoinduced charge separation? a. In resonance energy transfer, an excited electron is transferred to a nearby molecule. b. In photoinduced charge separation, an excited electron moves from accessory pigments to the lightharvesting complex. c. Photoinduced charge separation of pigments in the light-harvesting complex drives the resonance energy transfer in a reaction center. d. In resonance energy transfer, the energy of electron excitement is transferred only to an appropriate .. Powered by Cognero.

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Chapter 22 acceptor. e. Resonance energy transfer from pigments of a light-harvesting complex drives the photoinduced charge separation in a reaction center.

ANSWER: e 11. Photosystem I does NOT contain: a. ferredoxin. b. quinone A1. c. PsaB. d. chlorophyll A0. e. PsaA. ANSWER: a 12. Ferredoxin–NADP+ reductase: a. is able to synthesize one molecule of NADPH by accepting two protons from ferredoxin and taking two electrons from quinone A1. b. uses NADP+ as a cofactor to accept electrons from plastoquinol to reduce ferredoxin. c. is able to synthesize two molecules of NADPH by accepting two electrons from ferredoxin and taking two protons from the stroma. d. is able to bind two molecules of ferredoxin to reduce FAD to FADH2 without formation of an intermediate. e. is able to synthesize one molecule of NADPH by accepting two electrons from ferredoxin and taking two protons from the stroma.

ANSWER: e 13. Which is a CORRECT photosynthetic special pair? a. chlorophyll a – chlorophyll a b. plastoquinone QA – plastoquinone QB c. chlorophyll a – chlorophyll b d. chlorophyll A0 – quinone A1 e. chlorophyll a – chlorophyll A0 ANSWER: a 14. What is the name of the chlorophyll with two H+ ions in place of the central Mg2+ ion? a. plastocyanin b. plastoquinone c. lutein d. pheophytin e. phycoerythrin .. Powered by Cognero.

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Chapter 22 ANSWER: d 15. What amino acid residue of the water-oxidizing complex is the MOST important for water photolysis?

Why? a. tyrosine; because it is charged and can undergo acid–base reactions b. tryptophan; because it is aromatic and electrons can get trapped in its double bonds c. tyrosine; because it is polar and can undergo redox reactions d. tryptophan; because it is charged and can undergo acid–base reactions e. tyrosine; because it is aromatic and electrons can get trapped in its double bonds ANSWER: c 16. Select all that apply. The CF1–CF0 complex: a. transfers protons into the stroma. b. participates in the synthesis of ATP due to substrate-level phosphorylation. c. is a primary active transporter. d. transfers protons into the thylakoid lumen. e. participates in the synthesis of ATP due to the proton-motive force.

ANSWER: a, c, e 17. What molecules are able to carry electrons between components of photosystems I and II? Select all that apply. a. plastocyanin b. ferredoxin c. NADP+ d. plastoquinone QB e. plastoquinone QA ANSWER: a, b 18. What is the side product of light reactions of photosynthesis? a. ATP b. O2 c. NADP+ d. NADPH e. proton gradient ANSWER: b 19. What is NOT generated in cyclic photophosphorylation? a. ATP b. NADPH c. Fdred .. Powered by Cognero.

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Chapter 22 d. Fdox e. proton gradient ANSWER: b 20. Which special pair has the highest reduction potential? a. P680 b. P680* c. P700 d. P700* e. WOC ANSWER: a 21. How much ATP is produced in cyclic photophosphorylation upon absorption of six photons? a. 1 b. 2 c. 3 d. 6 e. 12 ANSWER: e 22. What is the other name for noncyclic photophosphorylation? a. light-independent phosphorylation b. cyclic phosphorylation c. Z scheme of photosynthesis d. dark phosphorylation e. proton-motive force ANSWER: c 23. In light reactions of photosynthesis, the proton-motive force (Δp) does NOT contribute to: a. ΔpH. b. proton gradients. c. thylakoid membrane nonpermeability to protons. d. Δψ. e. thylakoid membrane permeability to ions other than H+.

ANSWER: e 24. Which statement about the water-oxidizing complex is TRUE? a. Yzoz extracts electrons from manganese ions, which contribute to the oxidation of H2O in the manganese center. b. Tyrosine residues Yz extract electrons from water molecules to yield O2. .. Powered by Cognero.

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Chapter 22 c. Yzox reduces manganese ions of the manganese center, which contributes to binding of H2O by the center. d. Two molecules of water contained in the manganese center are oxidized to yield O2 upon oxidation of Yz. e. Four molecules of H2O in the manganese center form two molecules of O2 as soon as one Yzox is reduced.

ANSWER: a 25. How much H+ will be pumped by ATP synthase, and how much ATP will be generated if ATP synthase has 12 subunit III components? a. 12 H+; 3 ATP b. 12 H+; 12 ATP c. 3 H+; 12 ATP d. 3 H+; 3 ATP e. 4 H+; 4 ATP ANSWER: a 26. The stimulation of ATP synthesis by the ε subunit of CF1 is NOT contributed to by: a. increasing proton-motive force. b. decreasing Fdred. c. catalyzing the reaction by cytochrome b6f. d. catalyzing reaction by Fd–NADP+ reductase. e. increased Fdred.

ANSWER: b 27. Maximal activity of chloroplastic ATP synthase is achieved when: a. the concentration of an oxidized ferredoxin has increased. b. two specific cysteines of the γ subunit are reduced. c. the concentration of a reduced thioredoxin has decreased. d. the specific disulfide bond of the γ subunit is formed. e. the concentration of a reduced ferredoxin has decreased.

ANSWER: b 28. Where are the products of the light reactions transported for the subsequent dark reactions to proceed? a. stroma b. thylakoid lumen c. granum d. thylakoid membrane .. Powered by Cognero.

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Chapter 22 e. stroma lamellae ANSWER: a 29. Choose the CORRECT order of reactions that are conducted by photosystem I. a. P700 is activated by a photon → pheophytin accepts two high-energy electrons → two bound plastoquinones are reduced → cytochrome b6f oxidizes plastoquinol molecules to reduce FAD → two protons are pumped to the stroma b. P680 is activated by two photons → two Chl A0 accept two high-energy electrons → two molecules of ferredoxin are reduced → ferredoxin–NADP+ reductase oxidizes one ferredoxin at a time to reduce FAD to FADH- and then to FADH2 → the proton-motive force is created c. P700 is activated by a photon → Chl A0 accepts two high-energy electrons → one plastoquinone QA is fully reduced → two electrons are passed on one 4Fe–4S cluster → two molecules of ferredoxin are reduced → ferredoxin–NADP+ reductase takes two protons from the stroma and reduces one FAD d. P680 is activated by a photon → pheophytin accepts one high-energy electron → one mobile plastiquinone is fully reduced → cytochrome b6f oxidizes plastoquinol to reduce FAD → two protons are pumped to the thylakoid lumen e. P700 is activated by two photons → two Chl A0 accept two high-energy electrons → one quinone A1 is fully reduced → two electrons are passed on one 4Fe–4S cluster → two molecules of ferredoxin are reduced → ferredoxin–NADP+ reductase takes two protons from the stroma and reduces one NADP+ ANSWER: e 30. Choose the CORRECT order of components that transfer electrons in light reactions of photosynthesis. a. P700 → A0 → A1 → 4Fe–4S → ferredoxin → P680 → plastocyanin → QA → QB → cytochrome b6f → pheophytin → NADPH → H2O b. H2O → P680 → pheophytin → plastoquinones → cytochrome b6f → plastocyanin → P700 → A0 → A1 → 4Fe–4S → ferredoxin → NADPH c. P680 → pheophytin → plastoquinones → cytochrome b6f → plastocyanin → WOC → P700 → A0 → A1 → 4Fe–4S → ferredoxin → NADPH d. H2O → P680 → plastocyanin → plastoquinones → cytochrome b6f → pheophytin → P700 → A0 → A1 → ferredoxin → 4Fe–4S → NADPH e. WOC → P680 → plastocyanin → QA → QB → Fd-NAD+ reductase → NADH → pheophytin → P700 → A0 → A1 → ferredoxin → 4Fe–4S → H2O ANSWER: b 31. How many photons are required for photolysis of water to proceed? a. 1 b. 2 .. Powered by Cognero.

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Chapter 22 c. 3 d. 4 e. 5 ANSWER: d 32. How many electrons are transferred to plastocyanin upon absorption of six photons? a. 2 b. 4 c. 6 d. 12 e. 18 ANSWER: c 33. How many protons are transferred into the thylakoid lumen by photosystem I after absorption of six photons? a. 2 b. 4 c. 6 d. 12 e. 18 ANSWER: c 34. Choose the INCORRECT statement about nonphotochemical quenching. a. NPQ requires the proton gradient across a thylakoid membrane. b. Nonphotochemical quenching disrupts the proper resonance energy transfer at the light stage of photosynthesis. c. Photosystem II subunit S is involved in NPQ. d. NPQ dissipates the energy from the light-harvesting complex as heat. e. Nonphotochemical quenching disrupts the proper photoinduced charge separation in the reaction center.

ANSWER: e 35. Why do reaction centers have different pigments rather than only chlorophylls? Select all that apply. a. increasing the coverage of the visible spectrum of light b. arranging light-harvesting complexes c. compensating for the high density of chlorophylls d. narrowing the coverage of the visible spectrum of light e. quantitatively increasing the absorption of photons ANSWER: a, b, e 36. Select all that apply. The main outcomes of photosynthetic light reactions are: .. Powered by Cognero.

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Chapter 22 a. absorption of the light energy. b. proton transport across the membrane. c. glucose synthesis. d. water oxidation. e. utilization of reducing power.

ANSWER: a, b, d 37. Determine E 0 for the reaction NADP+ + H2O → NADPH + 1/2O2 + H+ if standard reduction potentials for NADP+ → NAPDH and 1/2O2 + H+ → H2O are –0.32 V and +0.82 V, respectively. Consider that the Faraday constant is 96.48 kJ mol–1 V–1. a. –0.50 V b. +0.50 V c. –1.14 V d. +1.14 V e. –0.26 V ANSWER: c 38. Determine the ΔG° for the reaction NADP+ + H2O → NADPH + 1/2O2 + H+ if standard reduction potentials for NADP+ → NAPDH and 1/2O2 + H+ → H2O are –0.32 V and +0.82 V, respectively. Consider that the Faraday constant is 96.48 kJ mol–1 V–1. a. 96.5 kJ/mol b. –96.5 kJ/mol c. 50.18 kJ/mol d. –220.02 kJ/mol e. 220.02 kJ/mol ANSWER: e 39. P700 is the pigment reaction center for _______.

ANSWER: photosystem I 40. The transfer of excitation energy from one molecule to one nearby through electromagnetic interactions is called _______.

ANSWER: resonance energy transfer 41. In the overall stoichiometry of light reactions, eight photons of light generate: a. 5 NADPH. b. 4 NADH. c. 3 ATP. d. 4 NADH and 3 ATP. e. 4 NADPH, 4 NADH, and 3 ATP. .. Powered by Cognero.

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Chapter 22 ANSWER: c 42. What is the original source of electrons used to neutralize the special pair in photosystem II? a. reduced cytochrome bf b. NADH c. H2O d. membrane lipids e. ferredoxin ANSWER: c 43. How many total protons are transferred into the thylakoid lumen after absorption of four photons? a. 2 b. 12 c. 18 d. 4 e. 8 ANSWER: b 44. What is the implication of the difference in permeability of the mitochondrial membrane versus the thylakoid membrane to Mg2+ and Cl–? a. The thylakoid membrane is less permeable to Mg+, Cl–, and H+; therefore, more energy is needed to pump protons across the membrane. b. The mitochondrial membrane being permeable to Mg2+ and Cl– means that the electron flow occurs only with transport of one Mg2+ and one Cl–. c. The thylakoid membrane being permeable to Mg2+ and Cl– means that the electron flow occurs only with transport of one Mg2+ and one Cl–. d. The mitochondrial membrane being permeable to Mg2+ and Cl– means that although protons are pumped across the membrane, no membrane potential is generated. e. The thylakoid membrane being permeable to Mg2+ and Cl– means that although protons are pumped across the membrane, no membrane potential is generated.

ANSWER: e 45. Select all that apply. Stacked and unstacked regions of the thylakoid are arranged such that: a. photosystem I is primarily located in the unstacked regions. b. photosystem II is primarily located in the stacked region. c. ATP synthesis occurs mainly in unstacked regions. d. photosystem I is located in the stacked regions. e. photosystem II is located in the unstacked regions.

ANSWER: a, b, c .. Powered by Cognero.

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Chapter 22 46. How many chloroplasts does a typical plant cell have? a. 1–100 b. 200–500 c. > 1000 d. 500–800 e. > 2000 ANSWER: a 47. Light absorption induces electron transfer from P680 to: a. pheophytin. b. QH2. c. cytochrome c. d. chlorophyll PP. e. carotenoid.

ANSWER: a 48. Complete the following reaction for photosystem I: Pc(Cu+) + Fdox + light → a. → Pc(Cu+) + (Cu+)Fdred b. → Pc(Cu2+) + Fdred c. → Pc + Fdred(Cu2+) d. → Pc(Cu3+) + Fdred e. → Pc(Cu2+) + Fdox ANSWER: b 49. Choose the CORRECT order of reactions that are conducted by photosystem II. a. P700 is activated by a photon → pheophytin accepts two high-energy electrons → two bound plastoquinones are reduced → cytochrome b6f oxidizes plastoquinol molecules to reduce FAD → two protons are pumped to the stroma b. P680 is activated by two photons → two Chl A0 accept two high-energy electrons → two molecules of ferredoxin are reduced → ferredoxin–NADP+ reductase oxidizes one ferredoxin at a time to reduce FAD to FADH- and then to FADH2 → the proton-motive force is created c. P700 is activated by a photon → Chl A0 accepts two high-energy electrons → one plastoquinone QA is fully reduced → two electrons are passed on to one 4Fe–4S cluster → two molecules of ferredoxin are reduced → ferredoxin–NADP+ reductase takes two protons from the stroma and reduces one FAD d. P680 is activated by a photon → pheophytin accepts one high-energy electron → one mobile plastoquinone is fully reduced → cytochrome b6f oxidizes plastoquinol to reduce FAD → two protons are pumped to the thylakoid lumen e. P700 is activated by two photons → two Chl A0 accept two high-energy electrons → one quinone A1 is fully reduced → two electrons are passed on to one 4Fe–4S cluster → two molecules of ferredoxin .. Powered by Cognero.

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Chapter 22 are reduced → ferredoxin–NADP+ reductase takes two protons from the stroma and reduces one NADP+ ANSWER: d 50. Organisms on earth are capable of converting about _____ of solar energy into chemical energy. a. 1% b. 2% c. 50% d. 60% e. 100% ANSWER: b 51. When grown under low oxygen conditions, Halobacterium salinarum generates additional ATP by using which two membrane proteins? a. ATP synthase b. bacteriorhodopsin c. ferredoxin d. photosystem I e. photosystem II ANSWER: a, b 52. Photosynthesis is composed of what two parts? a. cellular respiration and the light reactions b. cellular respiration and the dark reactions c. the Calvin–Benson cycle and the dark reactions d. the light-independent reactions and the dark reactions e. the light reactions and the dark reactions ANSWER: e 53. For eukaryotes, photosynthesis takes place in _____.

ANSWER: chloroplasts 54. Which is a depiction of electron transfer?

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Chapter 22 a.

b.

c.

d.

ANSWER: c 55. Consider the figure to answer the following question. Which chlorophyll molecule would absorb light that emits at 670 nm?

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Chapter 22

a. chlorophyll a b. chlorophyll b ANSWER: a 56. Consider the figure to answer the following question. Which pigment is best suited to absorb 525 nm wavelength sunlight?

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Chapter 22

a. β-carotene b. chlorophyll a c. chlorophyll b d. lutein e. phycocyanin f. phycoerythrin ANSWER: f 57. Green potatoes may produce solanine. This molecule is toxic to humans because it inhibits ______, an enzyme crucial for controlling nerve impulse transmission. a. acetylcholinesterase b. alkaloid c. ATP synthase d. morphine e. NADP+ reductase ANSWER: a 58. Which photosystem will respond to light emitting at 650 nm? a. PS I b. PS II c. both d. neither .. Powered by Cognero.

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Chapter 22 ANSWER: c 59. Ferredoxin is a strong _____.

ANSWER: reductant 60. When electrons flow from PS II to PS I, they pass through the cytochrome b6f complex, a(n) _____. a. electron carrier b. electron pump c. proton carrier d. proton pump ANSWER: d 61. The oxidation of 4 molecules of water generates how many electrons? a. 2 b. 4 c. 6 d. 8 ANSWER: d 62. Cooperation between photosystem I and photosystem II creates an electron flow pathway from H2O to NADP+ called the _________.

ANSWER: Z scheme of photosynthesis 63. The activity of ATP synthase is sensitive to the _____ in the chloroplast. a. light conditions b. proton flow c. redox conditions d. reduction conditions ANSWER: c 64. In cyclic photophosphorylation, ATP is generated _____ the formation of NADPH. a. dependent on b. independently of ANSWER: b 65. In the Z scheme of photosynthesis, how many photons are required to produce one ATP molecule? a. 2 b. 2.7 c. 3 d. 8 ANSWER: b .. Powered by Cognero.

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Chapter 22 66. The _____ of most plants are differentiated into stacked and unstacked regions.

ANSWER: thylakoids 67. The currency of readily available biosynthetic reducing power in cells is _____.

ANSWER: NADPH 68. Cellular respiration is the oxidation of glucose and the reduction of O2 to generate ATP. _____ is the reverse process.

ANSWER: photosynthesis

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Chapter 23 1. In what part of the chloroplasts does the Calvin cycle take place? a. matrix b. stroma c. thylakoid lumen d. cristae e. intermembrane space ANSWER: b 2. During what stage of the Calvin cycle are hexose sugars formed? a. stage 1 b. stage 4 c. stage 2 d. stage 3 e. stage 5 ANSWER: c 3. What is the name of the enzyme that catalyzes the first stage in the Calvin cycle? a. ribose 3,5-diphosphate oxygenase b. ribulose 1,5-diphosphate carboxylase/oxygenase c. ribulose 5-phosphate carboxylase d. ribose 4,5-bisphosphate carboxylase/oxygenase e. ribulose 1,5-bisphosphate carboxylase/oxygenase ANSWER: e 4. Choose the CORRECT thermodynamic characteristic of the first reaction in the Calvin cycle. a. ΔG°' > 0; exergonic reaction b. ΔG°' < 0; endergonic reaction c. ΔG°' < 0; exergonic reaction d. ΔG°' = 0; endergonic reaction e. ΔG°' > 0; endergonic reaction ANSWER: c 5. Which reaction is the rate-limiting step in hexose synthesis? a. reduction of 3-phosphoglycerate to form fructose 6-phosphate b. regeneration of ribulose 1,5-bisphosphate c. formation of glyceraldehyde 3-phosphate d. fixation of CO2 e. formation of ribulose 5-phosphate from ribose 5-phosphate and xylulose 5-phosphate ANSWER: d 6. What are the steps of carbon dioxide fixation? Select all that apply. .. Powered by Cognero.

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Chapter 23 a. Hydrolysis of an unstable six-carbon compound to two molecules of 3-phosphoglycerate. b. Conversion of an unstable six-carbon compound to a stable five-carbon compound. c. Conversion of ribulose 1,5-bisphosphate into an enediolate intermediate. d. Hydrolysis of an enediolate intermediate to 3-phosphoglycerate and two molecules of CO2. e. Condensation of the CO2 molecule with the enediolate intermediate.

ANSWER: a, c, e 7. What statement concerning rubisco is FALSE? a. Small subunits enhance catalytic activity of large subunits. b. Only a small subunit has a regulatory site. c. Rubisco is very abundant in chloroplasts. d. Active sites lie only in the large subunits. e. Rubisco consists of eight large and eight small subunits.

ANSWER: b 8. The full name (not the abbreviation) of the adaptation that temporally separates CO2 accumulation from CO2 utilization by absorbing CO2 at night is _______.

ANSWER: Crassulacean acid metabolism 9. Rubisco requires a bound CO2 for catalytic activity. What happens with rubisco when CO2 is bound? Select all that apply. a. CO2 forms a carbamate with ribulose 1,5-bisphosphate. b. CO2 forms a carbamate with lysine 201 of the large subunit. c. CO2 binds to a Mg2+ ion. d. Ribulose 1,5-bisphosphate forms a carbamate with alanine 150 of the large subunit. e. The carbamate binds to a Mg2+ ion.

ANSWER: b, e 10. What happens to rubisco in the absence of CO2? a. Rubisco activase uses ATP to allow the release of a Mg2+ ion. b. Rubisco degenerates. c. Rubisco inactivates by tight binding to ribulose 1,5-bisphosphate. d. Rubisco inactivates by losing two small and two large subunits. e. Rubisco activase binds to ribulose 1,5-bisphosphate and activates rubisco.

ANSWER: c 11. Select all that apply. The isomeric forms of the hexose phosphate product of the Calvin cycle are: a. glucose 1-phosphate. b. glucose 6-phosphate. .. Powered by Cognero.

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Chapter 23 c. fructose 1,6-bisphosphate. d. fructose 6-phosphate. e. xylulose 5-phosphate.

ANSWER: a, b, d 12. What are the isomeric forms of the hexose phosphate products of the Calvin cycle collectively called? a. glucose monophosphate pool b. fructose monophosphate pool c. hexose monophosphate pool d. monophosphates pool e. phosphates pool ANSWER: c 13. Select all that apply. Enzymes that have major roles in the rearrangement of carbon atoms in the final stage of the Calvin cycle are: a. phosphoribulose kinase. b. aldolase. c. rubisco activase. d. phosphopentose isomerase. e. transketolase.

ANSWER: b, e 14. Choose the CORRECT statement. a. Starch can't be in the form of amylose. b. Starch is a polymer of fructose residues. c. Starch is never stored in the cytoplasm. d. Starch can't be in the form of amylopectin. e. Starch is synthesized in mitochondria.

ANSWER: c 15. Sucrose 6-phosphate is formed by the reaction between: a. fructose UDP-glucose and UDP-glucose. b. glucose 6-phosphate and UDP-fructose. c. fructose UDP-glucose and UDP-fructose. d. glucose and UDP-fructose. e. fructose 6-phosphate and UDP-glucose.

ANSWER: e 16. Light reactions do NOT lead to an increase in: a. stromal concentrations of Mg2+. .. Powered by Cognero.

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Chapter 23 b. stromal concentration of H+. c. stromal concentration of NADPH. d. stromal concentration of reduced ferredoxin. e. pH in the thylakoid lumen.

ANSWER: b 17. Which is a function of thioredoxin? a. The oxidized form of thioredoxin inactivates enzymes of the Calvin cycle. b. Thioredoxin reduces ferredoxin by ferredoxin–thioredoxin reductase. c. Reduced thioredoxin can cleave disulfide bonds in enzymes. d. The oxidized form of thioredoxin can destroy ferredoxin. e. Reduced thioredoxin inhibits rubisco.

ANSWER: c 18. The wasteful side reaction of rubisco is binding of _____ instead of CO2. a. O3 b. O2 c. N2 d. CO e. NO2 ANSWER: b 19. How many carbon atoms does phosphoglycolate have? a. 2 b. 3 c. 6 d. 5 e. 4 ANSWER: a 20. Why is photorespiration said to be wasteful? a. CO2 is consumed and O2 is released. b. Only five of the six carbon atoms of two molecules of glycolate are recycled. c. Two molecules of glycolate are not recycled, and the carbon skeleton is not recovered. d. The organic carbon is converted into CO2 without the production of an energy-rich metabolite. e. One carbon atom of glycolate is lost as CO. ANSWER: d 21. In the C4 pathway, oxaloacetate carries: .. Powered by Cognero.

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Chapter 23 a. O2 from bundle-sheath cells to mesophyll cells. b. CO2 from mesophyll cells to a stoma. c. O2 from bundle-sheath cells to vascular tissue. d. CO2 from a stoma to vascular tissue. e. CO2 from mesophyll cells to bundle-sheath cells.

ANSWER: e 22. What enzyme catalyzes the reaction of oxaloacetate formation? a. phosphopentose epimerase b. phosphoenolpyruvate carboxylase c. pyruvate–Pi dikinase d. aldolase e. rubisco ANSWER: b 23. What are stomata? a. ABC transporters b. voltage-dependent ion channels c. microscopic openings in the leaves d. the other name for vascular tissue e. stroma of bundle-sheath cells and mesophyll cells ANSWER: c 24. What is the difference between the number of ATP molecules per molecule of hexose in C4 and C3 plants? a. 20 b. 12 c. 30 d. 10 e. 18 ANSWER: b 25. What level of photorespiration is characteristic of tropical plants with C4? a. very high b. high c. moderate d. zero e. low ANSWER: e .. Powered by Cognero.

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Chapter 23 26. C3 plants consume _____ molecules of ATP per hexose molecule formed in the absence of photorespiration. a. 12 b. 30 c. 18 d. 10 e. 48 ANSWER: c 27. CAM plants grow in: a. steppes. b. wet tropics. c. deserts. d. swamps. e. mountains.

ANSWER: c 28. In the CAM plants, carbon dioxide is fixed by the C4 pathway into _____, which is stored in ______. a. pyruvate; stroma b. oxaloacetate; stroma c. malate; vacuoles d. malate; stroma e. oxaloacetate; vacuoles ANSWER: c 29. What statement concerning CAM plants is FALSE? a. They decarboxylate oxaloacetate during the day to deliver CO2 to the Calvin cycle. b. CAM stands for crassulacean acid metabolism. c. They separate CO2 accumulation from CO2 utilization in time. d. They cannot generate CO2 as rapidly as C3 and C4 plants. e. They grow more slowly than C3 and C4 plants.

ANSWER: a 30. What pathway of photosynthesis is activated by an increased concentration of carbon dioxide? a. CAM b. C4 c. C3 d. C2 e. Calvin cycle ANSWER: e .. Powered by Cognero.

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Chapter 23 31. Which statement about differences between C4 and C3 plants is FALSE? a. C3 and C4 plants have anatomical differences of the leaf. b. C3 plants use rubisco to directly fix CO2, while C4 uses a second pathway (PEP carboxylase) to fix CO2. c. C3 is called C3 because carbon dioxide is fixed into a three-carbon sugar, while C4 pathways fix CO2 into a four-carbon sugar. d. C3 plants comprise the majority of plants on Earth, and C4 plants are found in drier climates. e. C3 plants consume 30 molecules of ATP per hexose molecule formed, while C4 plants consume 12.

ANSWER: e 32. Name the pathway that requires a sufficient amount of CO2 for dark reactions to proceed to minimize wasteful photorespiration. a. C3 b. C4 c. CAM d. C2 e. glycolysis ANSWER: b 33. Which enzyme is involved in both the Calvin cycle and the pentose phosphate pathway? a. pyruvate-Pi dikinase b. transketolase c. phosphoenolpyruvate carboxylase d. ferredoxin–thioredoxin reductase e. sucrose 6-phosphate synthase ANSWER: b 34. What is the main transporting carbohydrate in plants? a. lactose b. maltose c. glucose d. sucrose e. xylulose ANSWER: d 35. What is the final enzyme in the C4 pathway? a. pyruvate–Pi dikinase .. Powered by Cognero.

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Chapter 23 b. transketolase c. aldolase d. rubisco activase e. phosphopentose isomerase ANSWER: a 36. Ribulose 1,5-bisphosphate needs to be regenerated because it is: a. used as a chemical fuel at the expense of NADPH and ATP. b. used for sucrose synthesis. c. the acceptor of CO2 in the first stage of the Calvin cycle. d. used in the gluconeogenic pathway. e. unstable.

ANSWER: c 37. How many molecules of NADPH are used to bring carbon dioxide to form hexoses in the Calvin cycle? a. 2 b. 3 c. 4 d. 5 e. 1 ANSWER: a 38. What enzyme catalyzes the transformation of erythrose 4-phosphate and dihydroxyacetone phosphate to sedoheptulose 1,7-bisphosphate? a. phosphopentose isomerase b. phosphopentose epimerase c. transketolase d. aldolase e. sedoheptulose 1,7-bisphosphate phosphatase ANSWER: d 39. The more leaves receive light, the more _____ will be active, and the more CO2 will be converted into sugar. a. pyruvate–Pi dikinase b. aldolase c. rubisco d. rubisco activase e. transketolase ANSWER: c .. Powered by Cognero.

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Chapter 23 40. What is the source of carbons for the Calvin cycle? a. glucose b. carbon dioxide c. glycogen d. glyoxylate e. sucrose ANSWER: b 41. Which coenzyme is required by glyceraldehyde 3-phosphate dehydrogenase in chloroplasts to convert 3phosphoglycerate into glyceraldehyde 3-phosphate? a. NADH b. NADPH c. NAD+ d. NADP+ e. thiamine pyrophosphate ANSWER: b 42. The C4 pathway is necessary for tropical plants because: a. at high temperatures, the plants cannot maintain sufficient water levels within the cells. b. they do not have proper day/light cycles to maintain the balance of CO2 necessary for carbohydrate storage. c. at high temperatures, the oxygenase activity of rubisco is high. d. at high temperatures, transpiration stops. e. they always keep stomata of their leaves open.

ANSWER: c 43. What is the energy cost of the C4 pathway? a. Thirty molecules of ATP are used per hexose molecule made. b. Eighteen molecules of ATP are used per hexose molecule made. c. Thirty molecules of NADPH are used per hexose molecule made. d. Eighteen molecules of ATP and 12 of NADPH are used per hexose molecule made. e. Fifteen molecules of NADPH are used per hexose molecule made.

ANSWER: a 44. _____ round(s) of the Calvin cycle is/are required for the production of hexose. a. One b. Three c. Six d. Twelve e. Nine .. Powered by Cognero.

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Chapter 23 ANSWER: c 45. On a molecular level, how might global warming affect carbon dioxide fixation? a. Carbon dioxide, being a denser gas than oxygen, will cause more plants to sequester CO2 in mesophyll cells. b. An increase in temperatures leads to an increase in water loss in leafy plants compared to succulents; thus, light reactions are inhibited. c. Rubisco's oxygenase activity decreases with an increase in temperature, requiring more plants to use the C3 cycle. d. Ribusco's oxygenase activity increases with an increase in temperature, requiring more plants to use the C4 pathway. e. Ribusco's carboxylase activity increases with an increase in temperature, requiring more plants to use the C3 pathway.

ANSWER: d 46. Knowing what you do about the distribution of the light reaction enzymes, where would you expect to find the Calvin cycle enzymes and why? a. the luminal side of the thylakoid membrane, as this is the location of photolysis of water b. the luminal side of the thylakoid membrane, as this is the location of NADPH and ATP synthesis c. thylakoid integral membrane proteins that derive energy from proton pumping d. the stromal side of the thylakoid membrane, as this is the location of NADPH and ATP formation e. the stromal side of the thylakoid membrane, as this is the location of photolysis of water ANSWER: d 47. Plants store glucose as _____ and _____. a. starch; sucrose b. fructose; sucrose c. starch; fructose d. glucose; starch e. galactose; fructose ANSWER: a 48. What is the MOST abundant protein on Earth? a. ribulose 1.5-bisphosphate b. aldolase c. rubisco d. phosphopentose epimerase e. transketolase ANSWER: c 49. Which form of thioredoxin activates certain Calvin cycle enzymes? .. Powered by Cognero.

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Chapter 23 a. reduced b. oxidized c. dimeric d. trimeric e. phosphorylated ANSWER: a 50. The energetic equivalent of _____ ATP molecule(s) is consumed in transporting CO2 to the chloroplasts of the bundle-sheath cells. a. one b. two c. three d. four e. six ANSWER: b 51. How do light reactions modify the activity of Rubisco? a. Light reactions lead to the generation of ATP and NADPH. b. Reduced ferredoxin and NADPH formed by light-driven reactions regulate enzymes of the CalvinBenson cycle. c. Carbamate formation is favored by high concentrations of Mg2+ and high stromal pH, which are formed by light-driven proton pumps. d. Light reactions regulate the exchange of oxygen and carbon dioxide. e. Rubisco is a key enzyme in light reactions.

ANSWER: c 52. How many stages does the Calvin-Benson cycle consist of? a. one b. two c. three d. four e. five ANSWER: c 53. Which of the following statements about the Calvin-Benson cycle is TRUE. a. Stage one of the cycle involves the fixation of oxygen to ribulose 1,5-biphosphate. b. Stage four involves the reduction of 3-phosphoglycerate into hexose sugars. c. Stage two involves the reduction of 3-phosphoglycerate into pentose sugars. d. Stage three involves the regeneration of ribulose 1,5-biphosphate e. Stage three involves the regeneration of 3 phosphoglycerate.

ANSWER: d .. Powered by Cognero.

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Chapter 23 54. How many total subunits does Rubisco contain? a. 16 b. 8 c. 4 d. 2 e. 12 ANSWER: a 55. Which statement about Rubisco’s subunits is TRUE? a. The S chains contain a regulatory site that controls Rubisco activity. b. The L chains are approximately 13-kDa in size. c. The S chains are approximately 55-kDa in size. d. The S chains enhance the catalytic activity of the L chains. e. The L chains consist only of a catalytic site.

ANSWER: d 56. Which Calvin-Benson cycle enzyme also participates in glycolysis? a. aldolase b. transketolase c. Rubisco d. glyceraldehyde 3-phosphate dehydrogenase e. phosphopentose isomerase ANSWER: a 57. Which Calvin-Benson cycle enzyme is responsible for the formation of xylulose 5-phosphate? a. aldolase b. transketolase c. Rubisco d. glyceraldehyde 3-phosphate dehydrogenase e. phosphopentose isomerase ANSWER: b 58. Transketolase and aldolase generate the five carbon sugars ribose 5-phosphate and _____. a. ribose 1,5-biphosphate b. 2-deoxyribose c. xylulose 5-phosphate d. ribulose 5-phosphate e. arabinose ANSWER: c .. Powered by Cognero.

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Chapter 23 59. What is amylose? a. A form of starch consisting of glucose molecules joined by α-1,4 linkages only. b. A form of starch consisting of glucose molecules joined by α-1,6 linkages only. c. A form of starch consisting of glucose molecules joined by α-1,4 linkages and α-1,6 linkages. d. A disaccharide synthesized in the cytoplasm. e. A monosaccharide synthesized in the cytoplasm.

ANSWER: a 60. Starch comes in the form of ____ and ____. a. amylose; sucrose b. amylopectin; sucrose c. hexose; sucrose d. hexose; amylose e. amylose; amylopectin ANSWER: e 61. The CETCH cycle catalyzes the conversion of two CO2 molecules into one ____ molecule. a. glyoxylate b. sucrose c. glutamate d. starch e. glyceraldehyde 3-phosphate ANSWER: a 62. Starch is synthesized in ____, while sucrose is synthesized in _____ a. chloroplasts; cytoplasm b. cytoplasm; chloroplasts c. leaves; roots d. roots; leaves e. xylem; phloem ANSWER: a 63. Which of the following statements about PEP carboxylase is FALSE? a. PEP carboxylase has a higher affinity for CO2 than Rubisco. b. PEP carboxylase incorporates the carbon isotope 13CO2 more readily than Rubisco. c. PEP carboxylase is used in the C3 pathway to concentrate CO2 around Rubisco. d. PEP carboxylase converts CO2 and phosphoenolpyruvate into oxaloacetate. e. PEP carboxylase fixes CO2 in the mesophyll cells.

ANSWER: c 64. By what process is CO2 released from malate in the C4 and CAM pathways? .. Powered by Cognero.

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Chapter 23 a. carboxylation b. decarboxylation c. phosphorylation d. reduction e. oxidation ANSWER: b 65. What is another name for the C4 pathway? a. Calvin-Benson cycle b. crassulacean acid metabolism c. Hatch-Slack pathway d. photorespiration e. Kreb’s cycle ANSWER: c 66. Photorespiration is the salvage pathway used when Rubisco incorporates ____ instead of CO2. a. O2 b. ATP c. NADPH d. Mg2+ e. CO ANSWER: a 67. Which of the following enzymes are inhibited by reduced thioredoxin? a. Rubisco b. phosphoribulokinase c. glyceraldehyde 3-phoshpate dehydrogenase d. glucose 6-phosphate dehydrogenase e. NADP+-malate dehydrogenase ANSWER: d 68. What are the products of Rubisco oxygenase activity? Select all that apply. a. 3-phosphoglycerate b. fructose 1,6-biphosphate c. phosphoglycolate d. ribulose 1,5-biphosphate e. sucrose 6-phosphate ANSWER: a, c 69. Harnessing the CETCH cycle to create an artificial photosynthetic system is an example of _____. a. bioinformatics .. Powered by Cognero.

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Chapter 23 b. biochemistry c. synthetic biology d. genetic modification e. organic chemistry ANSWER: c 70. From where does the C4 pathway derive its name? a. The synthesis of four-carbon compounds to transport CO2 from mesophyll to bundle-sheath cells. b. Rubisco fixation of CO2 to ribulose 1,5-biphosphate produces a four-carbon energetic molecule as opposed to the three-carbon molecule generated in the C3 pathway. c. C4 plants fix CO2 at a higher rate, resulting in four molecules of CO2 fixed for every one molecule fixed in C3 plants. d. The transportation of carbon across four different cell types: mesophyll, bundle-sheath, xylem, and phloem. e. C4 is an alternate pathway to the Calvin-Benson cycle requiring only four enzymes to complete carbon fixation.

ANSWER: a

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Chapter 24 1. What is the intermediate in the interconversion of glucose 1-phosphate and glucose 6-phosphate? a. glucose 1,6-bisphosphate b. α-1,6-glucoside c. free glucose d. a block of three glucosyl residues e. glucose 2-phosphate ANSWER: a 2. What enzyme splits the glycosidic linkage between C-1 and C-4 carbon atoms? a. hexokinase b. glycogen phosphorylase c. phosphoglucomutase d. α-1,6-glucosidase e. phosphodiesterase ANSWER: b 3. What process results in the removal of the glucosyl residue from the nonreducing end of the glycogen molecule by the addition of orthophosphate? a. glycolysis b. dephosphorylation c. phosphorolysis d. hydrolysis e. phosphorylation ANSWER: c 4. How many glucose molecules can one glycogen phosphorylase get from the fragment of glycogen that consists of two branches, each containing 11 α-1,4 linkages and the branching comes from the third glucose unit from the reducing end? a. 22 b. 11 c. 12 d. 13 e. 24 ANSWER: c 5. What type of bond is glycogen phosphorylase able to cleave? a. α-1,6-glycosidic bond at a glycogen branch b. α-1,6-glycosidic bond at a glycogen core chain c. β-1,6,-glycosidic bond at a glycogen core chain d. α-1,4-glycosidic bond at a glycogen branch e. β-1,4-glycosidic bond at a glycogen branch .. Powered by Cognero.

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Chapter 24 ANSWER: d 6. What amino acid residue of phosphoglucomutase is able to transfer the phosphoryl group to glucose 1phosphate? a. phosphorylated serine residue b. acetylated lysine residue c. phosphorylated lysine residue d. acetylated serine residue e. phosphorylated tyrosine residue ANSWER: a 7. Choose the CORRECT sequence of steps catalyzed by phosphoglucomutase. a. phosphoryl group is transferred from the serine residue to the C-6 hydroxyl group of glucose 1phosphate → phosphoryl group is transferred from C-1 of glucose 1-phosphate to the serine residue b. phosphoryl group is transferred from the serine residue to the C-1 hydroxyl group of glucose 6phosphate → phosphoryl group is transferred from C-6 of glucose 6-phosphate to the serine residue c. a free orthophosphate is formed → phosphoryl group is transferred from the serine residue to the C-6 hydroxyl group of glucose 1-phosphate d. phosphoryl group is transferred from the tyrosine residue to the C-6 hydroxyl group of glucose 1phosphate → a free orthophosphate is formed e. phosphoryl group is transferred from the lysine residue to the C-6 hydroxyl group of glucose 1phosphate → phosphoryl group is transferred from C-1 of glucose 1-phosphate to the serine residue ANSWER: a 8. The concentration of glucose in the blood to meet the needs of the organism as a whole is maintained by: a. the spleen. b. skeletal muscle. c. smooth muscle. d. the brain. e. the liver.

ANSWER: e 9. What enzymes are present in a single polypeptide chain? Select all that apply. a. glycogen phosphorylase b. transferase c. α-1,6-glucosidase d. α-1,4-glucosidase e. hexokinase ANSWER: b, c 10. What is the result of the activity of transferase? a. a single glucosyl residue on the outer branch .. Powered by Cognero.

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Chapter 24 b. a single glucosyl residue on the core chain c. a short chain of three glucosyl residues on the outer branch d. a short chain of four glucosyl residues on the outer branch e. free glucose and phosphate ANSWER: a 11. Which end of the glycogen molecule has the anomeric carbon that can freely turn to an open or closed conformation? a. reducing b. nonreducing c. free d. blunt e. phosphorylated ANSWER: a 12. Both phosphorylases a and b exist in an equilibrium between the R and T states. Which phosphorylase state is the MOST active? a. phosphorylase a in the R state b. phosphorylase a in the T state c. phosphorylase b in the R state d. phosphorylase b in the T state e. both phosphorylase a and phosphorylase b in the R state ANSWER: a 13. Which enzyme initiates phosphorylation of glycogen phosphorylase? a. protein phosphatase 1 b. phosphorylase a c. protein kinase A d. phosphorylase kinase e. phosphorylase b ANSWER: d 14. Which hormone signifies the starved state? a. glucagon b. insulin c. norepinephrine d. calmodulin e. epinephrine ANSWER: a 15. What hormones are signals for reversible phosphorylation? Select all that apply. .. Powered by Cognero.

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Chapter 24 a. insulin b. epinephrine c. glucagon d. calmodulin e. norepinephrine ANSWER: b, c 16. What is the main trigger for secretion of epinephrine and glucagon? a. low blood-glucose concentration b. high blood-glucose concentration c. low cAMP concentration d. high cAMP concentration e. low Ca2+ concentration ANSWER: a 17. _____ is not stimulated by cAMP. a. Brain phosphorylase b. Phosphofructokinase c. Muscle phosphorylase d. Epinephrine e. Protein kinase A ANSWER: d 18. Choose the CORRECT pair of the enzyme and its allosteric effector. a. liver phosphorylase a – Ca2+ b. muscle phosphorylase b – glucose c. liver phosphorylase a – redox switch d. phosphorylase kinase – AMP e. muscle phosphorylase b – AMP ANSWER: e 19. The target of protein kinase A is the: a. active site of calmodulin. b. γ subunit of phosphorylase kinase. c. δ subunit of phosphorylase kinase. d. β subunit of phosphorylase kinase. e. active site of phosphorylase a.

ANSWER: d 20. Choose the CORRECT statement. .. Powered by Cognero.

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Chapter 24 a. Phosphorylase a has a single phosphorylated serine residue in each subunit. Phosphorylase b has a loop moved out from the active site. b. Phosphorylase a is contained in skeletal muscle cells. Phosphorylase b is contained in liver cells. c. Liver phosphorylase a favors the T state. Muscle phosphorylase b favors the R state. d. Phosphorylase b is activated by glucose molecules. Phosphorylase a is activated by AMP. e. Both phosphorylases have about a 40% difference in the amino acid sequence.

ANSWER: a 21. Select all that apply. The first step of the adenylate cyclase cascade is binding of epinephrine or glucagon to the: a. 7TM receptor on the plasma membrane and activation of Gαs protein. b. 7TM receptor on the plasma membrane and activation of Gαq protein. c. Gαs protein of 7TM receptor on the lumenal side of the endoplasmic reticulum. d. Gαq protein of 7TM receptor on the lumenal side of the endoplasmic reticulum. e. Gαs which activates cAMP formation through activation of the adenylate cyclase.

ANSWER: a, b 22. What is the main fuel for slow-twitch fibers? a. glycogen b. fatty acids c. glucose 6-phosphate d. free glucose e. glucose 1,6-bisphosphate ANSWER: b 23. In McArdle's patients during vigorous exercise, the pH level rises because skeletal muscles synthesize ATP: a. from creatine phosphate and ADP. Creatine formed in this reaction has a guanidium group that is a strong base. b. from creatine phosphate and ADP. Creatine formed in this reaction has a guanidium group that is a strong acid. c. from creatine phosphate and ADP. Creatine formed in this reaction has a calmodulin group that is a strong base. d. by fatty acid oxidation. The product of fatty acid oxidation, oxaloacetate, is a strong acid. e. by fatty acid oxidation. The product of fatty acid oxidation, creatine, is a strong base.

ANSWER: a 24. If phosphodiesterase is inhibited, then concentration of _____ becomes _____ and glycogen phosphorylase is _____. a. cAMP; high; activated b. cAMP; low; activated .. Powered by Cognero.

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Chapter 24 c. cAMP; high; inactivated d. AMP; high; activated e. AMP; low; inactivated ANSWER: a 25. Select all that apply. The targets of epinephrine are: a. β-adrenergic receptors of the liver cells. b. epinephrine receptors of the liver cells. c. β-adrenergic receptors of the skeletal muscle cells. d. δ-adrenergic receptors of the skeletal muscle cells. e. α-adrenergic receptors of the liver cells.

ANSWER: c, e 26. Muscle phosphorylase b is activated by: a. low blood-glucose concentration. b. high blood-glucose concentration. c. high AMP concentration. d. low AMP concentration. e. high ATP concentration.

ANSWER: c 27. The main role of glycogen degradation in the liver is: a. formation of glucose for export to other tissues. b. formation of glucose for itself. c. formation of glucose 1-phosphate for export to other tissues. d. formation of glucose 1,6-bisphosphate for itself. e. production of cAMP for regulation of glycogen degradation.

ANSWER: a 28. What enzyme is present in the endoplasmic reticulum of liver cells? a. glucose 6-phosphatase b. phosphorylase b c. protein phosphatase 1 d. protein kinase A e. phosphoglucomutase ANSWER: a 29. Which of the following enzymes inactivates glycogen phosphorylase? a. protein phosphatase 1 b. phosphoglucomutase c. protein kinase A .. Powered by Cognero.

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Chapter 24 d. hexokinase e. transferase ANSWER: a 30. McArdle's disease results from a lack of: a. glycogen phosphorylase a. b. glycogen phosphorylase b. c. protein phosphatase A. d. phosphofructokinase. e. phosphorylase kinase.

ANSWER: b 31. Choose the CORRECT statement concerning glycogen. a. Glycogen is a polymer of glucose with α-1,6-linkages and with α-1,4-linked branches at about every 12th residue. b. Muscle cells are not able to maintain the blood-glucose level because they lack glucose 6phosphatase. c. Liver cells are able to maintain the blood-glucose level because they are rich in mitochondria. d. Glycogen is a polymer of both amylose and branched amylopectin with α-1,4-linkages. e. Upon glycogen breakdown, glucose residues are sequentially removed from the reducing ends.

ANSWER: b 32. Epinephrine secretion: a. activates protein phosphatase 1 in liver and muscle cells. b. activates adenylate cyclase in liver and muscle cells. c. activates glycogen phosphorylase a in liver cells. d. inhibits glycogen phosphorylase in brain cells. e. inhibits glycogen phosphorylase b in muscle cells.

ANSWER: b 33. Choose the CORRECT sequence of activation steps of phosphorylase kinase. a. low blood-glucose concentration → epinephrine secretion → protein kinase A phosphorylates the α subunit → protein kinase A phosphorylates the β subunit → Ca2+ binds to the δ subunit b. high blood-glucose concentration → epinephrine secretion → Ca2+ binds to the δ subunit → protein kinase A phosphorylates the β subunit → protein kinase A phosphorylates the α subunit c. low blood-glucose concentration → epinephrine secretion → Ca2+ binds to the δ subunit → protein phosphatase 1 dephosphorylates the β subunit → protein phosphatase 1 dephosphorylates the α subunit d. high blood-glucose concentration → epinephrine secretion → protein kinase A phosphorylates the α subunit → protein kinase A phosphorylates the β subunit → Ca2+ binds to the δ subunit .. Powered by Cognero.

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Chapter 24 e. low blood-glucose concentration → epinephrine secretion → Ca2+ binds to the δ subunit → protein kinase A phosphorylates the β subunit → protein kinase A phosphorylates the α subunit ANSWER: e 34. The debranching enzyme releases a free glucose molecule by: a. hydrolysis of the α-1,4-glycosidic bond. b. hydrolysis of the α-1,6-glycosidic bond. c. phosphorylation of the glucose residue. d. dephosphorylation of the glucose residue. e. shifting a phosphoryl group from C-1 to C-6 in the glucose residue.

ANSWER: b 35. A low blood-glucose level: a. shifts the equilibrium of phosphorylase a from the R state to the T state. b. shifts the equilibrium from phosphorylase a to phosphorylase b. c. activates glucagon secretion. d. activates insulin secretion. e. activates protein phosphatase 1.

ANSWER: c 36. Which process slows down in muscle cells after binding epinephrine to β-adrenergic receptors? a. glycogen breakdown b. gluconeogenesis c. glycolysis d. the citric acid cycle e. oxidative phosphorylation ANSWER: b 37. Glucagon is able to bind to the 7TM receptors of: a. liver cells. b. muscle cells. c. pancreas cells. d. neurons. e. epithelial cells.

ANSWER: a 38. What is the first step in glycogen degradation? a. isomerization of glucose 1-phosphate to glucose 6-phosphate b. remodeling of the glycogen substrate c. formation of glucose 6-phosphate from glucose 1-phosphate d. release of glucose 1-phosphate from glycogen .. Powered by Cognero.

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Chapter 24 e. activation of the T state of phosphorylase b ANSWER: d 39. What is the main function of the cAMP cascade? a. highly amplifies the effects of epinephrine and glucagon b. highly amplifies the effects of epinephrine and phosphorylase a c. inhibits phosphorylase kinase d. inhibits phosphodiesterase e. activates phosphofructokinase ANSWER: a 40. What liver enzyme cleaves the phosphate from glucose 6-phosphate? a. α-1,6-glycosidic b. phosphorylase kinase c. α-1,4-glycoside d. phosphorylase a e. glucose 6-phosphatase ANSWER: e 41. The transferase enzyme shifts a block of _____ glycosyl units. a. 3 b. 4 c. 6 d. 12 e. 2 ANSWER: a 42. What molecule must be excluded from the active site of glycogen phosphorylase? a. glucose b. glucose 1-phosphate c. water d. orthophosphate e. glucose 6-phosphate ANSWER: c 43. Why does it make sense that muscle cells are not responsive to glucagon, but both types of cells are responsive to epinephrine? a. Glucagon in the liver signals the need for glycogen synthesis. b. Phosphorylase b in muscle transitions to the T state in the presence of epinephrine so that control is independent of liver phosphorylase b. c. Epinephrine stimulates the transferase enzyme so that glycogen is more readily available in highly .. Powered by Cognero.

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Chapter 24 active muscles. d. Lactate formed in muscle cells is released into the blood where it is taken up by the liver for gluconeogenesis. e. The function of epinephrine is to maintain a nearly constant concentration of glucose in the blood during periods of relaxation.

ANSWER: d 44. You are taking your new puppy for a walk and a vicious-looking dog comes rushing out from an unfenced yard. You pick up your puppy and take off down the road. What is the MOST likely form of the phosphorylase in your muscle? a. phosphorylase a, R state, not phosphorylated b. phosphorylase a, R state, phosphorylated c. phosphorylase a, T state, phosphorylated d. phosphorylase b, R state, phosphorylated e. phosphorylase b, T state, phosphorylated ANSWER: b 45. You are on a long car trip and stop to fill up on gas and sugary snacks. Once back on the road, what is the MOST likely form of phosphorylase in your liver? a. phosphorylase a, R state, not phosphorylated b. phosphorylase a, R state, phosphorylated c. phosphorylase a, T state, phosphorylated d. phosphorylase b, R state, phosphorylated e. phosphorylase b, T state, phosphorylated ANSWER: c 46. Why is the T state of glycogen phosphorylase less active? a. The adjacent amino acids are not phosphorylated and thus catalysis cannot be carried out. b. The active site is partially blocked. c. ATP cannot be bound by the T state. d. The adjacent amino acids are phosphorylated and thus the active state is blocked. e. Phosphorylase in the T state has a ridge structure.

ANSWER: b 47. Which liver enzyme is deficient in Hers disease? a. phosphorylase b. transferase c. glucosidase d. phosphoglucomutase e. phosphofructokinase ANSWER: a .. Powered by Cognero.

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Chapter 24 48. Select all that apply. The major sites of glycogen storage are: a. adipose tissue. b. the liver. c. skeletal muscle. d. the blood. e. the brain.

ANSWER: b, c 49. Inhibition of muscle glycogen phosphorylase by glucose 6-phosphate is an example of what kind of inhibition? a. allosteric heterotropic activation b. allosteric homotropic inhibition c. feedback inhibition d. feedforward inhibition e. mixed inhibition ANSWER: a 50. One mechanism for turning off glycogen degradation involves the removal of a phosphate group from glycogen phosphorylase by the enzyme _______.

ANSWER: protein phosphatase 1 51. Why is glycogen used as the storage form of glucose in animals? a. Stored glucose is osmotically unfavorable to cells compared to glycogen. b. Glycogen is the intermediate between sucrose and glucose. c. Glycogen is the precursor in the biosynthesis of many compounds. d. Glucose polymerizes automatically into glycogen once transported to the cytoplasm. e. Glycogen’s branching structure allows it to be stored in granular shape within skeletal muscle and liver tissue.

ANSWER: a 52. Approximately how many glucose residues are between branches of a glycogen molecule? a. 3 b. 7 c. 12 d. 11 e. 9 ANSWER: c 53. Why is the phosphorolytic cleavage of glycogen more favorable than hydrolytic cleavage? a. Phosphorolytic cleavage releases glucose at a faster rate than hydrolytic cleavage. b. Phosphorolytic cleavage is not readily reversible. .. Powered by Cognero.

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Chapter 24 c. Hydrolytic cleavage is confined only to the cell. d. The product of phosphorolytic cleavage is already phosphorylated. e. Hydrolytic cleavage requires multiple enzymes to complete, while phosphorolytic cleavage is performed solely by phosphorylase.

ANSWER: d 54. Phosphorylase stops cleaving linkages when it reaches a residue _____ residue(s) away from a branch point. a. one b. two c. three d. four e. five ANSWER: d 55. The debranching enzyme combines which two enzymatic activities? Select all that apply. a. transferase b. phosphorylase c. α-1,6-glucosidase d. ligase e. kinase ANSWER: a, c 56. Which enzyme catalyzes the conversion of glucose 1-phosphate into glucose 6-phosphate? a. transferase b. phosphorylase c. phosphoglucomutase d. glucose 6-phosphatase e. protein phosphatase A ANSWER: c 57. Which enzyme activates phosphorylase b? a. phosphorylase kinase b. hexokinase c. phosphoglucomutase d. glucose 6-phosphatase e. protein phosphatase A ANSWER: a 58. Which amino acid of each phosphorylase a subunit is phosphorylated? a. tyrosine 14 .. Powered by Cognero.

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Chapter 24 b. threonine 14 c. glutamic acid 14 d. histidine 14 e. serine 14 ANSWER: e 59. Which are the preferred equilibrium states for phosphorylase a and b? Select all that apply. a. phosphorylase a - R state b. phosphorylase b - R state c. phosphorylase a - T state d. phosphorylase b - T state e. phosphorylase b (phosphorylated) - R state ANSWER: a, d, e 60. Which subunit of phosphorylase kinase contains the active site? a. α b. ꞵ c. ℽ d. δ e. ε ANSWER: c 61. Phosphorylase kinase is activated when ____ binds to the δ subunit? a. Ca2+ b. ATP c. cAMP d. Pi e. insulin ANSWER: a 62. Glycogen breakdown requires how many enzymes? a. one b. two c. three d. four e. five ANSWER: d 63. The process of cleaving a covalent bond by the addition of orthophosphate is called _____. a. phosphorolysis b. hydrolysis .. Powered by Cognero.

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Chapter 24 c. phosphorylation d. glycolysis e. reduction ANSWER: a 64. Why is the ΔG°’for phosphorolysis so small? a. The glycosidic linkage is replaced by a phosphoryl ester linkage that has a nearly equal transfer potential. b. The ratio of orthophosphate to glucose 1-phosphate heavily disfavors phosphorolysis. c. When catalyzed by phosphorylase, the reaction is readily reversible. d. The cell’s pH is high enough to promote phosphorolysis. e. The dephosphorylation of ATP provides additional energy to compensate for the breaking of the glycosidic bond.

ANSWER: a 65. Choose the correct steps for glycogen breakdown. a. α-1,6-glucosidase hydrolyzes the connecting branches→ phosphorylase cleaves α-1,4-glycosidic bonds, leaving four residues on each branch → transferase shifts a block of three glucosyl residues from one branch to another → phosphorylase cleaves α-1,4-glycosidic bonds of the remaining chain b. transferase shifts a block of three glucosyl residues from one outer branch to another → phosphorylase cleaves α-1,4-glycosidic bonds, leaving four residues on each branch → α-1,6glucosidase hydrolyzes the remaining glucosyl residue from the branch→ phosphorylase cleaves α1,4-glycosidic bonds of the remaining chain c. phosphorylase cleaves α-1,6-glycosidic bonds, leaving four residues on each branch → transferase shifts a block of three glucosyl residues from one outer branch to another → α-1,4-glucosidase hydrolyzes the remaining glucosyl residue from the branch→ phosphorylase cleaves α-1,6-glycosidic bonds of the remaining chain d. phosphorylase cleaves α-1,4-glycosidic bonds, leaving four residues on each branch → transferase shifts a block of three glucosyl residues from one outer branch to another → α-1,6-glucosidase hydrolyzes the remaining glucosyl residue from the branch→ phosphorylase cleaves α-1,4-glycosidic bonds of the remaining chain e. phosphorylase cleaves α-1,4-glycosidic bonds, leaving four residues on each branch → transferase shifts a block of three glucosyl residues from one outer branch to another → phosphorylase cleaves α-1,4-glycosidic bonds of the remaining chain → α-1,6-glucosidase hydrolyzes the remaining glucosyl residue from the branch ANSWER: d 66. The debranching enzyme is an example of what kind of enzyme? a. bifunctional b. inhibitor c. monofunctional d. promoter e. trifunctional .. Powered by Cognero.

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Chapter 24 ANSWER: a 67. How does the cell promote glycogen breakdown despite the phosphorolysis reaction having an equilibrium ratio of 3.6? a. During phosphorolysis, the glycosidic linkage is replaced by a phosphoryl ester linkage that has a nearly equal transfer potential. b. The substrate to product (orthophosphate to glucose 1-phosphate) ratio for phosphorolysis is usually greater than 100, substantially favoring the direction of phosphorolysis. c. Phosphoglucomutase is catalytically faster than phosphorylase and converts glucose 1-phosphate to glucose 6-phosphate almost immediately. d. The cell’s in vivo pH is high enough to promote phosphorolysis. e. The dephosphorylation of ATP provides additional phosphate to push the reaction towards phosphorolysis.

ANSWER: b 68. What is another name for the δ subunit of phosphorylase kinase? a. glucomutase b. calmodulin c. glucagon d. phosphorylase e. epinephrine ANSWER: b 69. What protein resides at the core of a glycogen molecule? a. glucomutase b. glycogenin c. glucagon d. α-1,6-glucosidase e. α-1,4-glucosidase ANSWER: b 70. Glycogen is the primary storage molecule for which organisms? Select all that apply. a. plants b. animals c. archaea d. bacteria e. protists ANSWER: b, c, d, e

.. Powered by Cognero.

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Chapter 25 1. How many glucose residues does the glycogen branching enzyme move to form a branch point? a. 2 b. 4 c. 7 d. 16 e. 21 ANSWER: c 2. What acts as a "high-energy handle" for glycogen synthesis? a. uridine triphosphate b. pyrophosphate c. uridine diphosphate glucose d. uridine diphosphate e. phosphoryl group ANSWER: d 3. Which enzyme transfers UDP-glucose to the hydroxyl of an existing glycogen core? a. glycogen synthase b. glycogen synthetase c. glycogen ligase d. glycogen diphosphokinase e. nucleoside diphosphokinase ANSWER: a 4. Which protein can perform autoglycosylation with up to 10–20 glucose units? a. nucleoside diphosphokinase b. the branching enzyme c. protein kinase A d. glycogen synthase e. glycogenin ANSWER: e 5. What is the result of glycogen synthase phosphorylation? a. stimulation of glycogen synthesis b. stimulation of glycogen degradation c. inhibition of glycogen synthesis d. inhibition of glycogen degradation e. acceleration of glycogen synthesis ANSWER: c 6. Select all that apply. Glycogen synthase is converted into the active form by the action of: .. Powered by Cognero.

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Chapter 25 a. protein phosphatase 1 b. glucose 1-phosphate c. glycogen synthase kinase d. glucose 6-phosphate e. glycogen diphosphokinase ANSWER: a, d 7. Insulin stimulates glycogen synthase by inactivating: a. protein phosphatase 1. b. protein kinase A. c. glycogen synthase kinase. d. glycogen phosphorylase. e. nucleoside diphosphokinase.

ANSWER: c 8. The activated glucose donor of glycogen synthesis is: a. glucose 1-phosphate. b. glucose 6-phosphate. c. uridine triphosphate glucose. d. glucose 1,6-bisphosphate. e. uridine diphosphate glucose.

ANSWER: e 9. Choose the CORRECT statement. Glycogenin: a. is a homotetramer. b. is autoglycosylated by β-1,4-glucose polymers. c. is the primer for glycose synthase. d. binds glucose units to a tyrosine residue. e. catalyzes the formation of α-1,6-glucose polymers.

ANSWER: d 10. An α-1,6-glycosidic link is created by the: a. branching enzyme. b. glucose transferase. c. glycogen isomerase. d. glycogen synthase. e. glycogenin.

ANSWER: a 11. What results from a loss or inactivation of insulin receptors? a. hypoglycemia .. Powered by Cognero.

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Chapter 25 b. hypoinsulinemia c. type I diabetes d. type II diabetes e. protein phosphatase 1 dysregulation ANSWER: d 12. Select all that apply. After a carbohydrate-rich meal: a. the blood-glucose level increases, leading to glycogen storage in the liver. b. the blood-glucose level increases, leading to glycogen storage in muscle. c. glycogen storage in the muscle becomes exhausted. d. the protein kinase A cascade is activated in muscle. e. the insulin-receptor cascade is activated in the liver.

ANSWER: a, b, e 13. An excessive amount of glucagon relative to that of insulin in patients with type I diabetes leads to _____ in the liver. a. increased glucose transport b. increased glycogen synthesis c. a decrease in fructose 2,6-phosphate d. a decrease in gluconeogenesis e. an increase in glycolysis ANSWER: c 14. What protein is activated in the cytoplasm due to activation of the insulin receptor? a. protein phosphatase 1 b. protein kinase A c. glycogen phosphorylase d. insulin-receptor substrate e. glycogen synthase kinase ANSWER: d 15. _____ is NOT required for synthesis of glycogen. a. Primer b. Phosphorylase kinase c. Glycogenin d. UTP e. Pyrophosphorylase ANSWER: b 16. What is the donor of glycosyl units in autoglycosylation of glycogenin? a. glucose .. Powered by Cognero.

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Chapter 25 b. glucose 1-phosphate c. UDP-glucose d. the branching enzyme e. glycogen ANSWER: c 17. How many molecules of ATP are hydrolyzed to incorporate one dietary glucose molecule into glycogen? a. 1 b. 2 c. 4 d. 6 e. 8 ANSWER: b 18. How many molecules of ATP are used to create one branch on the glycogen chain by breaking an α-1,4 link and forming an α-1,6 link? a. 1 b. 2 c. 3 d. 4 e. 0 ANSWER: e 19. What number of glucosyl units in the main glycogen chain is required for the branching enzyme to cleave the new branch? a. 7 b. 11 c. 4 d. 20 e. 50 ANSWER: b 20. Hydrolysis of _____ drives the formation of the activated precursor for glycogen synthesis from glucose 1phosphate. a. ATP b. UTP c. pyrophosphate d. glucose 1-phosphate e. UDP ANSWER: c .. Powered by Cognero.

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Chapter 25 21. What is the key regulatory enzyme in glycogen synthesis? a. the branching enzyme b. glycogen synthase c. nucleoside diphosphokinase d. glycogen phosphorylase e. glycogenin ANSWER: b 22. Which form of glycogen synthase is active? a. phosphorylated b. unbound to glucose 6-phosphate c. a form d. T state e. b form ANSWER: c 23. An allosteric regulator of glycogen synthase is: a. glucose 6-phosphate. b. glucose 1-phosphate. c. UDP-glucose. d. cAMP. e. ATP. ANSWER: a 24. What is the effect of the glucagon-triggered cAMP cascade on glycogen synthesis in the liver? a. initiation b. activation c. shutting off d. acceleration e. irreversible inhibition ANSWER: c 25. Select all that apply. Activation of protein kinase A leads to: a. activation of glycogen synthesis. b. inhibition of glycogen synthesis. c. inhibition of glycogen phosphorolysis. d. activation of glycogen breakdown. e. inhibition of glycogen breakdown.

ANSWER: b, d 26. What enzymes regulating liver-glycogen metabolism predominate in the fasting state? Select all that apply. .. Powered by Cognero.

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Chapter 25 a. glycogen phosphorylase a b. glycogen synthase a c. glycogen phosphorylase b d. glycogen synthase b e. protein phosphatase 1 ANSWER: a, d 27. What second messenger switches off glycogen synthesis in the epinephrine-triggered cascade? a. insulin b. GTP c. cGMP d. ATP e. cAMP ANSWER: e 28. What converts glycogen synthase into the a form after exercise? a. previous emission of epinephrine b. phosphorylase c. glycogen synthase kinase d. protein kinase A e. protein phosphatase 1 ANSWER: e 29. The regulatory subunit of PP1 does NOT interact with: a. phosphorylase kinase. b. glycogen phosphorylase. c. glycogen synthase. d. adenylate cyclase. e. glycogen.

ANSWER: d 30. What is important for the regulation of liver-glycogen metabolism upon an increase in the blood-glucose concentration? Select all that apply. a. binding of glucose to phosphorylase a b. binding of PP1 to phosphorylase b c. binding of glucose to phosphorylase b d. binding of PP1 to phosphorylase a e. rendering of phosphoserine 14 on phosphorylase a ANSWER: a, d, e 31. What prevents the phosphatase activity of PP1 from always inhibiting glycogen degradation? Select all that .. Powered by Cognero.

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Chapter 25 apply. a. epinephrine cascade b. insulin-receptor cascade c. glucagon cascade d. cAMP cascade e. protein kinase A ANSWER: a, c, d, e 32. Type I diabetes is also known as _______-dependent diabetes mellitus.

ANSWER: insulin 33. What is the characteristic feature of type I diabetes? a. Glucagon-to-insulin ratio is low. b. Glucagon-to-insulin ratio is high. c. Insulin receptors are less sensitive to insulin. d. Insulin receptors are hypersensitive to insulin. e. Glucose is overproduced by the liver and is underused by other organs.

ANSWER: b 34. Very long outer branches of glycogen stored in normal amounts are the result of a mutation in: a. the insulin receptor. b. α-1,4-glucosidase. c. glycogen phosphorylase. d. α-1,6-glucosidase. e. the branching enzyme.

ANSWER: e 35. Very short outer branches of glycogen stored in normal amounts are the result of the mutation in: a. the insulin receptor. b. α-1,4-glucosidase. c. glycogen phosphorylase. d. α-1,6-glucosidase. e. the branching enzyme.

ANSWER: d 36. Glucose is a(n) _____ for glycogen phosphorylase a. a. product b. allosteric activator c. allosteric inhibitor d. cofactor e. suicide inhibitor .. Powered by Cognero.

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Chapter 25 ANSWER: c 37. A complete oxidation of a glucose residue derived from glycogen yields _____ ATP. a. 2 b. 4 c. 7 d. 11 e. 31 ANSWER: e 38. What enzymatic activity does the insulin receptor possess? a. tyrosine kinase b. serine kinase c. protein phosphatase d. threonine kinase e. protein phosphorylase ANSWER: a 39. Insulin triggers a pathway that activates: a. protein kinase A. b. protein kinases. c. glycogen synthase kinase. d. adenylate cyclase. e. glycogen phosphorylase.

ANSWER: b 40. What is the core of the glycogen molecule? a. glucose b. glycogenin c. UDP-glucose d. glycogen synthase e. glycogen phosphorylase ANSWER: b 41. What enzyme is required to synthesize α-1,4-glycosidic bonds in glycogen? a. glycogen synthase b. UDP-glucose phosphorylase c. branching enzyme d. glycogen phosphorylase e. phosphoglucomutase ANSWER: a .. Powered by Cognero.

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Chapter 25 42. An incorrect, tight binding of PP1 to glycogen synthase would result in: a. a hyperactive glycogen synthase. b. hyperglycogen levels. c. an increase in insulin action. d. high blood glucose. e. an activated glycogen synthase.

ANSWER: d 43. ATP is known as the energy currency of the cell; however, ATP is not used directly in any of the enzymatic reactions in glycogen synthesis. How, then, does ATP provide energy currency for glycogen synthesis? a. ATP is used to regenerate GTP in the cGMP cascade. b. ATP is used by diphosphokinase to regenerate UTP. c. Dietary glucose is phosphorylated to glucose 1-phosphate by hexokinase. d. The branching enzyme requires the hydrolysis of ATP. e. An activated glycosyl unit of ADP-glucose is transferred to the chain of glycogen.

ANSWER: b 44. Predict the major consequence of the loss of the gene that encodes the glycogen-targeting subunit of protein phosphatase 1. a. Protein kinase A would not be inhibited by phosphorylating PP1. b. Glycogen synthase a will be maintained in the T form. c. Phosphorylase kinase will be activated only in the presence of cAMP. d. Glucose 6-phosphatase will display reduced activity. e. A loss of this gene would prevent association of the catalytic subunit of PP1 with its substrate.

ANSWER: e 45. Why would you expect the amount of glycogen in type I glycogen-storage disease (von Gierke disease) to be increased? a. Glycogen phosphatase is activated by an increase in glucose 6-phosphate. b. Glucose 6-phosphate, the first metabolite in glycogen synthesis, remains high. c. Epinephrine inhibition of glycogen synthesis is defective. d. cAMP inhibition of glycogen synthesis is defective. e. The R form of glycogen synthase is stabilized permanently.

ANSWER: b 46. Which enzyme catalyzes the following reaction?

Glucose 1-phosphate + UTP → UDP-glucose + PPi a. UDP-glucose pyrophosphorylase b. branching enzyme .. Powered by Cognero.

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Chapter 25 c. glycogen synthase b d. phosphofructokinase e. UDP-glucose phosphorylase ANSWER: a 47. The enzyme that begins the kinase cascade activating glycogen degradation is: a. glycogen synthase. b. phosphorylase kinase. c. PKC. d. PKA. e. tyrosine kinase.

ANSWER: d 48. Which is the true glucose level sensor in the cell? a. insulin b. phosphorylase a c. glucagon d. glycogen synthase e. protein phosphatase I ANSWER: b 49. After exercise, muscle cell glycogen metabolism is regulated by: a. insulin. b. phosphorylase a. c. glucagon. d. glycogen synthase. e. protein phosphatase 1.

ANSWER: e 50. Glycogen metabolism is regulated, up to a point, by a cyclic AMP cascade. At what point do glycogen synthesis pathways diverge from glycogen degradation pathways? a. Phosphorylase kinase converts phosphorylase b to phosphorylase a and glycogen synthase a to glycogen synthase b. b. Protein kinase A leads to the activation of glycogen degradation, and also to the inhibition of glycogen synthase by conversion from a to b. c. Phosphorylase kinase converts phosphorylase a to phosphorylase b and glycogen synthase a to glycogen synthase b. d. Protein kinase A leads to the activation of glycogen degradation, and also to the inhibition of glycogen synthase by conversion from b to a. e. Cyclic AMP converts inactive PKA to active PKA, causing the inactivation of phosphorylase kinase.

ANSWER: b .. Powered by Cognero.

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Chapter 25 51. What is the activated precursor used in glycogen synthesis? a. glucose 1-phosphate b. glucosyl c. UDP-glucose d. glucose e. glycogen ANSWER: c 52. How does inorganic pyrophosphatase help drive the synthesis of UDP-glucose? a. It catalyzes the bond between glucose 1-phosphate and UTP. b. It hydrolyzes pyrophosphate to orthophosphate, keeping the pyrophosphate concentration low. c. It catalyzes the formation of UTP, a substrate required to form UDP-glucose. d. It transfers phosphate from the C-6 carbon of glucose to the C-1 carbon. e. It phosphorylates UDP-glucose pyrophosphorylase, activating the enzyme.

ANSWER: b 53. During synthesis, new glucosyl units are added to the ____ terminal residues of glycogen.

ANSWER: nonreducing 54. Which carbon of the glycogen terminal residue becomes part of the α-1,4-glycosidic bond during glycogen synthesis? a. C-1 b. C-2 c. C-3 d. C-4 e. C-5 ANSWER: d 55. Which carbon of the new glucosyl unit becomes part of the α 1,4-glycosidic bond during glycogen synthesis? a. C-1 b. C-2 c. C-3 d. C-4 e. C-5 ANSWER: a 56. Why does glycogen synthesis require a primer? a. Glycogenin requires a polysaccharide with at least four glucosyl residues to extend the glycogen molecule. b. The primer permits the formation of the initial α-1,6-glycosidic bond, allowing for glycogen’s branching structure. .. Powered by Cognero.

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Chapter 25 c. To extend a glycogen molecule, glycogen synthase requires a polysaccharide with at least four linked glucosyl residues. d. The glucose units are added stepwise, but glycogen synthase cannot bind the glucose donor, UDPglucose, without a primer. e. The core protein of glycogen requires a polysaccharide to bind to and initiate synthesis.

ANSWER: c 57. Why is glycogen branching important? Select all that apply. a. Branching increases the solubility of glycogen. b. Branching allows for more terminal residues to be present, increasing the rate of glycogen synthesis and degradation. c. Branching allows glycogen to be easily stacked, creating structural support for the cell. d. Branching mechanically regulates glycogen degradation by limiting enzyme access to terminal residues. e. Branching reduces the formation of granules.

ANSWER: a, b 58. Phosphorylation has the ____ effect on the enzymatic activity of glycogen synthase and glycogen phosphorylase.

ANSWER: opposite 59. Approximately what percentage of glucose residues in a glycogen molecule are cleaved via phosphorolysis into glucose 1-phosphate? a. 12% b. 94% c. 88% d. 50% e. 30% ANSWER: c 60. Approximately what percentage of glucose residues in a glycogen molecule are cleaved via phosphorolysis into glucose 1-phosphate? a. 12% b. 94% c. 88% d. 50% e. 30% ANSWER: c 61. Which enzyme inactivates phosphorylase a and phosphorylase kinase by dephosphorylating them? a. protein phosphatase 1 b. pyrophosphatase .. Powered by Cognero.

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Chapter 25 c. glycogen synthase d. glycogenin e. α-amylase ANSWER: a 62. The regulatory subunit Gm of PP1 is most prevalent in which tissues? Select all that apply. a. liver b. skeletal muscle c. stomach d. heart e. brain ANSWER: b, d 63. The regulatory subunit GL of PP1 is most prevalent in which tissue? a. liver b. skeletal muscle c. stomach d. heart e. brain ANSWER: a 64. Which of the following statements about glycogen metabolism in the liver is TRUE? a. Phosphorylase b is the glucose sensor of the liver. b. PP1 binds tightly to phosphorylase a only when it is in the T state. c. PP1 only becomes active when it binds to phosphorylase b. d. Glucose binding shifts the allosteric equilibrium of phosphorylase a from the active R state to the inactive T state. e. Activated PP1 converts phosphorylase from the b form to the a form.

ANSWER: d 65. Which of the following statements about glycogen metabolism in the liver is TRUE? a. Phosphorylase b is the glucose sensor of the liver. b. PP1 binds tightly to phosphorylase a only when it is in the T state. c. PP1 only becomes active when it binds to phosphorylase b. d. Glucose binding shifts the allosteric equilibrium of phosphorylase a from the active R state to the inactive T state. e. Activated PP1 converts phosphorylase from the b form to the a form.

ANSWER: d 66. Branching _____ the rate of glycogen synthesis and degradation.

ANSWER: increases .. Powered by Cognero.

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Chapter 25 67. Individuals with Cori disease lack which of the following enzymes? a. glycogen synthase b. glycogenin c. debranching enzyme d. protein kinase A e. glycogen synthase kinase ANSWER: c 68. Overexpression of the GL subunit of PP1 in ethanol-fed mice produced which of the following results? a. No change in the glycogen levels of the genetically modified mice compared to mice fed the same amount of ethanol. b. A drastic decrease in glycogen levels in the genetically modified mice compared to mice not fed on ethanol. c. A drastic decrease in glycogen levels in the genetically modified mice compared to WT mice fed the same amount of ethanol. d. A moderate increase in the glycogen levels of the genetically modified mice compared to WT mice fed the same amount of ethanol. e. A drastic increase in the glycogen levels of the genetically modified mice compared to WT mice fed the same amount of ethanol.

ANSWER: d 69. Approximately what percentage of glucose residues in a glycogen molecule are cleaved via hydrolysis into glucose 1-phosphate? a. 12% b. 94% c. 88% d. 50% e. 30% ANSWER: a 70. What is the percent efficiency of glycogen storage? a. 12% b. 94% c. 88% d. 50% e. 30% ANSWER: b

.. Powered by Cognero.

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Chapter 26 1. Phosphopentose epimerase converts ribulose 5-phosphate to: a. xylulose 5-phosphate. b. ribose 5-phosphate. c. fructose 6-phosphate. d. erythrose 4-phosphate. e. ribulose 3-phosphate.

ANSWER: a 2. The intermediate in the oxidation of glucose 6-phosphate to ribulose 5-phosphate is: a. ribose 5-phosphate. b. xylulose 5-phosphate. c. glucose 1-phosphate. d. 6-phosphogluconate. e. fructose 6-phosphate.

ANSWER: d 3. Which metabolic pathway requires NADPH? a. neurotransmitter biosynthesis b. light reaction of photosynthesis c. ATP synthesis d. pentose phosphate pathway e. glycolysis ANSWER: a 4. What enzyme catalyzes the committed step of the pentose phosphate pathway? a. lactonase b. phosphofructokinase c. phosphopentose isomerase d. glucose 6-phosphate dehydrogenase e. 6-phosphogluconate dehydrogenase ANSWER: d 5. Which statement about the conversion of ribulose 5-phosphate to fructose 6-phosphate is TRUE? a. Fructose 6-phosphate is formed from glyceraldehyde 3-phosphate and sedoheptulose 7-phosphate. b. Glyceraldehyde 3-phosphate is formed after isomerization of C7 carbohydrates. c. Xylulose 5-phosphate is always required to produce ribose 5-phosphate. d. Erythrose 4-phosphate is directly formed from xylulose 5-phosphate. e. Fructose 6-phosphate is formed from erythrose 4-phosphate.

ANSWER: a 6. Choose the enzymes that catalyze reactions with concomitant formation of NADPH. .. Powered by Cognero.

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Chapter 26 a. glucose 6-phosphate dehydrogenase and lactonase b. transketolase and transaldolase c. phosphopentose epimerase and glucose 6-phosphate dehydrogenase d. glucose 6-phosphate dehydrogenase and 6-phosphogluconate dehydrogenase e. phosphofructokinase and 6-phosphogluconate dehydrogenase ANSWER: d 7. Choose the CORRECT pair of epimers. a. glucose 6-phosphate and glucose 1-phosphate b. glucose 6-phosphate and fructose 6-phosphate c. xylulose 5-phosphate and ribulose 5-phosphate d. ribulose 5-phosphate and ribose 5-phosphate e. D-xylulose 5-phosphate and L-xylulose 5-phosphate ANSWER: c 8. Lactonase opens the lactone ring by hydrolyzing the: a. phosphoester link. b. phosphodiester link. c. ester link between C-1 and C-5. d. ester link between C-1 and C-6. e. pyrophosphate link.

ANSWER: c 9. The pentose phosphate pathway: a. is activated by NADH. b. operates in mitochondria. c. is an alternative pathway to gluconeogenesis. d. involves glutathione reductase. e. is primarily anabolic.

ANSWER: e 10. Which carbohydrate is NOT produced in the pentose phosphate pathway? a. fructose 1,6-bisphosphate b. glyceraldehyde 3-phosphate c. fructose 6-phosphate d. glucose 6-phosphate e. erythrose 4-phosphate ANSWER: a 11. Which compound is the reactant in the transaldolase reaction that leads to the formation of fructose 6phosphate? .. Powered by Cognero.

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Chapter 26 a. ribose 5-phosphate b. sedoheptulose 7-phosphate c. glucose 6-phosphate d. fructose 1,6-bisphosphate e. dihydroxyacetone phosphate ANSWER: b 12. Which reaction in the pentose phosphate pathway releases carbon dioxide? a. oxidative decarboxylation of 6-phosphogluconate b. hydrolysis of 6-phosphoglucono-δ-lactone c. oxidation of glucose 6-phosphate d. oxidative decarboxylation of pyruvate e. oxidation of 6-phosphoglucono-δ-lactone ANSWER: a 13. Which reaction is irreversible? a. sedoheptulose 7-phosphate + glyceraldehyde 6-phosphate → fructose 6-phosphate + erythrose 4phosphate b. ribulose 5-phosphate → ribose 5-phosphate c. ribulose 5-phosphate → xylulose 5-phosphate d. glucose 6-phosphate + NADP+ → 6-phosphoglucono-δ-lactone + NADPH + H+ e. oxidized glutathione + NADPH + H+ → reduced glutathione + NADP+ ANSWER: d 14. What are the functions of the pentose phosphate pathway? Select all that apply. a. formation of NADPH b. formation of three- and six-carbon intermediates of glycolysis c. synthesis of pentose sugars for DNA and RNA synthesis d. ATP synthesis e. synthesis of pentose sugars for coenzyme A synthesis ANSWER: a, b, c, e 15. In what tissues is the pentose phosphate pathway active? Select all that apply. a. liver b. mammary gland c. red blood cells d. muscle e. pancreas ANSWER: a, b, c .. Powered by Cognero.

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Chapter 26 16. In rapidly dividing cells: a. the needs for NADPH and ribose 5-phosphate are balanced. b. more NADPH than ribose 5-phosphate is required. c. much more ribose 5-phosphate than NADPH is required. d. the needs for NADPH and ATP are balanced. e. NADPH is needed in excess.

ANSWER: c 17. What is the concomitant product of the reaction catalyzed by pyruvate kinase? a. NADPH b. NAD+ c. ATP d. ADP e. pyruvate ANSWER: c 18. How many NADPH molecules can be produced in mode 3 of the pentose phosphate pathway, when the need for NADPH is greater than the need for ribose 5-phosphate? a. 2 b. 12 c. 106 d. 6 e. 8 ANSWER: b 19. Name the enzyme that catalyzes the reaction shown in the figure.

a. glutathione peroxidase b. glutathione reductase c. pyruvate kinase isozyme d. glucose 6-phosphate dehydrogenase e. cytochrome c peroxidase ANSWER: b 20. What is the respiratory quotient for carbohydrates? a. 1.0 .. Powered by Cognero.

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Chapter 26 b. 0.7 c. 0.8 d. 0.5 e. 1.5 ANSWER: a 21. Hemolytic anemia can be caused by a deficiency of which enzyme? a. glutathione peroxidase b. glutathione reductase c. glucose 6-phosphate dehydrogenase d. 6-phosphogluconate dehydrogenase e. lactonase ANSWER: c 22. What happens when NADPH concentration is high? a. The pentose phosphate pathway halts and glucose 6-phosphate enters glycolysis. b. The nonoxidative phase of the pentose phosphate pathway halts and ribose 5-phosphate enters DNA synthesis. c. The oxidative phase of the pentose phosphate pathway halts and glucose 6-phosphate converts to pentose phosphates. d. The nonoxidative phase of the pentose phosphate pathway halts and ribose 5-phosphate enters gluconeogenesis. e. The pentose phosphate pathway activates and glucose 6-phosphate converts to pentose phosphates.

ANSWER: a 23. The nonoxidative phase of the pentose phosphate pathway is controlled by the: a. availability of ribulose 5-phosphate. b. availability of glucose 6-phosphate. c. concentration of NADPH. d. concentration of NADP+. e. availability of ATP. ANSWER: a 24. Choose the CORRECT sequence of intermediate products in the pentose transfer between the pentose phosphate pathway and gluconeogenesis. a. ribose 5-phosphate → glyceraldehyde 3-phosphate → fructose 1,6-bisphosphate b. ribose 5-phosphate → glyceraldehyde 3-phosphate → fructose 6-phosphate c. ribulose 5-phosphate → fructose 6-phosphate d. fructose 6-phosphate → fructose 1,6-bisphosphate → ribose 5-phosphate e. ribose 5-phosphate → glyceraldehyde 3-phosphate → sedoheptulose 7-phosphate ANSWER: a .. Powered by Cognero.

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Chapter 26 25. What happens when the needs for NADPH and ribose 5-phosphate are balanced? a. Glucose 6-phosphate primarily enters glycolysis, then the product enters the nonoxidative phase of the pentose phosphate pathway. b. Glucose 6-phosphate enters the oxidative phase of the pentose phosphate pathway, then the product enters the nonoxidative phase of the pentose phosphate pathway. c. Glucose 6-phosphate enters the oxidative phase of the pentose phosphate pathway, then the product enters glycolysis. d. Glucose 6-phosphate enters the oxidative phase of the pentose phosphate pathway, then the product enters gluconeogenesis. e. Glucose 6-phosphate primarily enters glycolysis, then the product enters the oxidative phase of the pentose phosphate pathway.

ANSWER: b 26. The actual respiratory quotient is greater than 1 for a hummingbird. What is the source of the extra carbon dioxide? a. the nonoxidative phase of the pentose phosphate pathway b. the oxidative phase of the pentose phosphate pathway c. fatty acid synthesis d. the combination of glycolysis and the pentose phosphate pathway e. the combination of glycolysis and nucleic acids synthesis ANSWER: b 27. What metabolic pathways do anaerobic bacteria use to metabolize pentose sugars? a. the nonoxidative phase of the pentose phosphate pathway and glycolysis b. the pentose phosphate pathway and gluconeogenesis c. the citric acid cycle and the pentose phosphate pathway d. the nonoxidative phase of the pentose phosphate pathway and fermentation e. only fermentation ANSWER: d 28. What are the concomitant products of the conversion of glucose 6-phosphate to pyruvate in mode 4 of the pentose phosphate pathway? Select all that apply. a. NADH b. NADPH c. ADP d. CO2 e. ATP ANSWER: a, b, d, e 29. Under what conditions does the net reaction shown take place? glucose 6-phosphate + 12 NADP+ + 7 H2O → 6 CO2 + 12 NADPH + 12 H+ + Pi .. Powered by Cognero.

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Chapter 26 a. The need for ribose 5-phosphate is greater than the need for NADPH. b. The needs for NADPH and ribose 5-phosphate are balanced. c. Both NADPH and ATP are required. d. The need for NADPH is greater than the need for ribose 5-phosphate. e. Both NADPH and pyrophosphate are required.

ANSWER: d 30. Glucose 6-phosphate dehydrogenase is highly specific for: a. NADPH. b. NADP+. c. NAD+. d. NADH. e. ADP. ANSWER: b 31. Where do the reactions of the pentose phosphate pathway take place? a. mitochondrial matrix b. inner mitochondrial membrane c. cytoplasm d. lumen of the rough endoplasmic reticulum e. stroma of a chloroplast ANSWER: c 32. Choose the CORRECT enzyme–substrate pair from the pentose phosphate pathway. a. glucose 6-phosphate dehydrogenase – 6-phosphoglucono-δ-lactone b. 6-phosphogluconate dehydrogenase – 6-phosphoglucono-δ-lactone c. transaldolase – xylulose 5-phosphate d. transketolase – xylulose 5-phosphate e. lactonase – ribulose 5-phosphate ANSWER: d 33. Choose the CORRECT sequence of intermediate products in the oxidative phase of the pentose phosphate pathway. a. glucose 6-phosphate → 6-phosphoglucono-δ-lactone → 6-phosphogluconate → ribulose 5-phosphate b. glucose 6-phosphate → 6-phosphoglucono-δ-lactone → 6-phosphogluconate → ribose 5-phosphate c. glucose 6-phosphate → 6-phosphogluconate → 6-phosphoglucono-δ-lactone → ribose 5-phosphate d. glucose 6-phosphate → 6-phosphoglucono-δ-lactone → 6-phosphogluconate → ribulose 5-phosphate → fructose 6-phosphate e. glucose 6-phosphate → 6-phosphoglucono-δ-lactone → 6-phosphogluconate → ribulose 5-phosphate → ribose 5-phosphate → pyruvate .. Powered by Cognero.

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Chapter 26 ANSWER: a 34. Since phosphoenolpyruvate inhibits triose phosphate isomerase, the: a. phosphorylated glycolytic intermediates enter the nonoxidative phase of the pentose phosphate pathway. b. phosphorylated glycolytic intermediates enter the oxidative phase of the pentose phosphate pathway. c. conversion of phosphoenolpyruvate to pyruvate is inhibited. d. conversion of pyruvate to phosphoenolpyruvate is activated. e. phosphorylated glycolytic intermediates enter glycolysis.

ANSWER: a 35. Which molecule can reduce reactive oxygen species (ROS)? a. reduced glutathione b. oxidized glutathione c. NADPH d. NADP+ e. glucose 6-phosphate dehydrogenase ANSWER: a 36. What is the function of the pentose phosphate pathway in red blood cells? a. synthesis of fatty acids b. synthesis of steroids c. maintenance of reduced glutathione d. synthesis of proteins e. synthesis of ATP ANSWER: c 37. When both NADPH and ATP are required, the intermediates from the pentose phosphate pathway: a. enter glycolysis. b. enter gluconeogenesis. c. restore glucose 6-phosphate by entering the nonoxidative phase of the pentose phosphate pathway. d. restore glucose 6-phosphate by entering the oxidative phase of the pentose phosphate pathway. e. enter the citric acid cycle.

ANSWER: a 38. What intermediates of the pentose phosphate pathway can enter glycolysis in mode 4? Select all that apply. a. glyceraldehyde 3-phosphate b. fructose 6-phosphate c. ribose 5-phosphate d. glucose 6-phosphate e. fructose 1,6-bisphosphate .. Powered by Cognero.

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Chapter 26 ANSWER: a, b 39. Which compound is a substrate for the pentose phosphate pathway? a. glucose 6-phosphate b. fructose 6-phosphate c. glucose d. glucose 1-phosphate e. ribose 5-phosphate ANSWER: a 40. The use of primaquine: a. will completely conquer malaria in the future. b. causes hemolysis in individuals with a deficiency in glucose 6-phosphate dehydrogenase. c. causes hemolysis in individuals with a deficiency in glutathione peroxidase. d. inhibits the formation of Heinz bodies in the red blood cells. e. inhibits glutathione reductase in the red blood cells.

ANSWER: b 41. ATP is called the energy currency. The currency of biosynthetic reducing power is: a. NADPH. b. NADH. c. AMP. d. ADP. e. FADH2.

ANSWER: a 42. Which reaction is NOT involved in the pentose phosphate pathway? a. C5 + C5 C3 + C7 b. C4 + C5

C3 + C6

c. C3 + C7

C6 + C4

d. C5 + C6

C2 + C9

e. 3C5

2C6 + C3

ANSWER: d 43. In the nonoxidative stage of the pentose phosphate pathway, intermediates of _____ are produced. a. the citric acid cycle b. glycolysis c. glycogen degradation d. fatty acid biosynthesis .. Powered by Cognero.

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Chapter 26 e. lipid degradation ANSWER: b 44. When the cell requires ribose for biosynthesis but does not require NADPH, the formation of ribose 5phosphate is summarized by 5 glucose 6-phosphate + ATP → 6 ribose 5-phosphate + ADP Identify the enzyme that catalyzes the reaction that requires ATP. a. lactonase b. glutathione reductase c. aldolase d. phosphofructokinase e. 6-phosphogluconate dehydrogenase ANSWER: d 45. If C-1 of ribose 5-phosphate is labeled with radioactive 14C, where would this label be after the reaction xylulose 5-phosphate + ribose 5-phosphate glyceraldehyde 3-phosphate + sedoheptulose 7-phosphate a. C-1 of glyceraldehyde 3-phosphate b. C-3 of glyceraldehyde 3-phosphate c. C-1 of sedoheptulose 7-phosphate d. C-2 of sedoheptulose 7-phosphate e. C-3 of sedoheptulose 7-phosphate ANSWER: e 46. How does reduced glutathione maintain the red blood cell structure? Select all that apply. a. protects against falciparum malaria b. maintains the hemoglobin structure c. prevents Heinz body formation d. causes Heinz body formation e. maintains the glutathione peroxidase structure ANSWER: b, c 47. Glucose 6-phosphate dehydrogenase is inhibited by low levels of a. NADPH. b. NADP+. c. ribose 5-phosphate. d. ribulose 5-phosphate. e. NADH. ANSWER: b 48. The link between the pentose phosphate pathway and glycolysis is well documented. How does this link facilitate the growth of rapidly dividing cancer cells? .. Powered by Cognero.

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Chapter 26 a. Rapidly dividing cells meet their NADP+ needs by switching to anaerobic glycolysis. b. Glycolytic intermediates are rapidly depleted due to the increased demand for ATP to synthesize cellular structures, leading to a decreased production of ribose 5-phosphate. c. Accumulated glycolytic intermediates are used by the pentose phosphate pathway to synthesize ribose 5-phosphate. d. Lactonase is allosterically activated by an increase in glycolytic intermediates. e. 6-Phosphogluconate dehydrogenase speeds up ribose 5-phosphate synthesis by switching to NADH formation.

ANSWER: c 49. Several conditions occur due to a deficiency in glucose 6-phosphate dehydrogenase. Why do all of these conditions include anemia as a symptom? a. The oxidized form of glutathione, which normally eliminates peroxides, is depleted in the absence of NADPH. b. Peroxides damage red blood cells due to the absence of NADPH production. c. Red blood cells with diminished glucose 6-phosphate dehydrogenase activity have an overactive response to oxidative stress. d. Reduced glutathione enhances reactive oxygen species formation, leading to cell membrane disruption. e. Red blood cells accumulate NADPH due to the loss of mitochondria.

ANSWER: b 50. The enzyme transketolase transfers a(n) _______-carbon fragment from a ketose to an aldose. Write the number out as a word.

ANSWER: two 51. How many phases does the pentose phosphate pathway consist of? a. one b. two c. three d. four e. five ANSWER: b 52. The _____ phase generates NADPH, the_____ phase interconverts phosphorylated sugars.

ANSWER: oxidative, nonoxidative 53. Choose the CORRECT tissue and function pairing for the pentose phosphate pathway: a. adrenal gland - steroid synthesis b. red blood cells - fatty acid synthesis c. liver - maintenance of reduced glutathione d. ovaries - fatty acid synthesis .. Powered by Cognero.

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Chapter 26 e. adipose tissue - steroid synthesis ANSWER: a 54. Ribose 5-phosphate is a precursor for which molecules? Select all that apply. a. DNA b. RNA c. ribulose 5-phosphate d. ATP e. coenzyme A ANSWER: a, b, d, e 55. Which enzyme(s) participate in the oxidative phase of the pentose phosphate pathway? Select all that apply. a. glucose 6-phosphate dehydrogenase b. phosphopentose isomerase c. lactonase d. transketolase e. 6-phosphogluconate dehydrogenase ANSWER: a, c, e 56. The largest sugar formed by the pentose phosphate pathway has how many carbons? a. six b. nine c. seven d. eight e. five ANSWER: c 57. Which enzymes catalyze the transition into the nonoxidative phase of the pentose phosphate pathway?

Select all that apply. a. glucose 6-phosphate dehydrogenase b. phosphopentose isomerase c. transketolase d. phosphopentose epimerase e. lactonase ANSWER: b, d 58. How many different enzymes participate in the nonoxidative phase of the pentose phosphate pathway? a. three b. eight c. two d. four .. Powered by Cognero.

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Chapter 26 e. five ANSWER: d 59. Which of the following molecules serves as a coenzyme for transketolase? a. coenzyme A b. NADH c. thiamine pyrophosphate d. glyceraldehyde 3-phosphate e. ATP ANSWER: c 60. Which of the following reactions are catalyzed by transketolase? Select all that apply. a. C5 + C5 ⇋ C3 + C7 b. C3 + C7 ⇋ C6 + C4 c. C5 + C5 ⇋ C6 + C4 d. C4 + C5 ⇋ C6 + C3 e. C3+ C7 ⇋ C4 + C5 ANSWER: a, d 61. Which of the following reactions are catalyzed by transaldolase? Select all that apply. a. C5 + C5 ⇋ C3 + C7 b. C3 + C7 ⇋ C6 + C4 c. C5 + C5 ⇋ C6 + C4 d. C4 + C5 ⇋ C6 + C3 e. C3+ C7 ⇋ C4 + C5 ANSWER: b 62. What is the net reaction for transketolase and transaldolase activity in the pentose phosphate pathway? a. 3 C4 ⇋ 2 C5 + C3 b. 3 C3 ⇋ C5 + C4 c. 3 C5 ⇋ 2 C6 + C3 d. 3 C5 ⇋ 2 C7 + C3 e. 3 C6 ⇋ 2 C4 + C7 ANSWER: c 63. What are epimers? a. isomers with a single asymmetric center b. isomers that differ at multiple asymmetric centers c. isomers with multiple asymmetric centers that only differ at one d. isomers in which only one functional group of the molecule differs .. Powered by Cognero.

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Chapter 26 e. molecules with a completely different chemical formula but a transferable functional group ANSWER: c 64. Which substrate competes with NADP+ to bind to glucose 6-phosphate dehydrogenase? a. NADPH b. ATP c. glucose 6-phosphate d. 6-phosphoglucono-δ-lactone e. ribulose 5-phosphate ANSWER: a 65. Under what conditions would ribulose 5-phosphate formed during the oxidative phase of the pentose phosphate pathway be converted to pyruvate? a. when both NADPH and ribose 5-phosphate are required b. when the need for ribose 5-phosphate exceeds the need for NADPH c. when both NADPH and ATP are required d. when both ribose 5-phosphate and ATP are required e. when the need for NADPH exceeds the need for ribose 5-phosphate ANSWER: c 66. In mode 3 of the pentose phosphate pathway, ribulose 5-phosphate is recycled to glucose 6-phosphate by a combination of the nonoxidative phase and enzymes from which pathway? a. glycogenesis b. glycogenolysis c. glycolysis d. citric acid cycle e. gluconeogenesis ANSWER: e 67. Under what conditions would mode 3 of the pentose phosphate pathway be activated? a. when both NADPH and ribose 5-phosphate are required b. when the need for ribose 5-phosphate exceeds the need for NADPH c. when both NADPH and ATP are required d. when both ribose 5-phosphate and ATP are required e. when the need for NADPH exceeds the need for ribose 5-phosphate ANSWER: e 68. Under what conditions would mode 2 of the pentose phosphate pathway be activated? a. when both NADPH and ribose 5-phosphate are required b. when the need for ribose 5-phosphate exceeds the need for NADPH c. when both NADPH and ATP are required .. Powered by Cognero.

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Chapter 26 d. when both ribose 5-phosphate and ATP are required e. when the need for NADPH exceeds the need for ribose 5-phosphate ANSWER: a 69. Under what conditions would mode 1 of the pentose phosphate pathway be activated? a. when both NADPH and ribose 5-phosphate are required b. when the need for ribose 5-phosphate exceeds the need for NADPH c. when both NADPH and ATP are required d. when both ribose 5-phosphate and ATP are required e. when the need for NADPH exceeds the need for ribose 5-phosphate ANSWER: b 70. The pentose phosphate pathway can be considered the functional mirror to which pathway? a. citric acid cycle b. glycolysis c. Calvin-Benson cycle d. fatty acid synthesis e. gluconeogenesis ANSWER: c

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Chapter 27 1. What protein triggers mobilization of triacylglycerols? a. hormone-sensitive lipase b. adipose triglyceride lipase c. perilipin d. phosphorylase kinase e. cyclic AMP ANSWER: c 2. Which process makes the following reaction irreversible?

RCOO– + CoA RCO-CoA +AMP +PPi a. hydrolysis of inorganic pyrophosphate b. hydrolysis ATP c. heat release d. AMP release e. catalysis by acyl CoA synthetase ANSWER: a 3. Which enzyme is bound to the outer mitochondrial membrane? Select all that apply. a. acyl CoA synthetase b. carnitine acyltransferase I c. translocase d. carnitine acyltransferase II e. acyl CoA dehydrogenase ANSWER: a, b 4. The first step of β oxidation is: a. oxidation by FAD. b. oxidation by NAD+. c. hydration by enoyl CoA hydrase. d. thiolysis by CoA. e. activation by CoA. ANSWER: a 5. What stereoisomer is a regular substrate in the β oxidation pathway? a. trans-Δ2-enoyl CoA b. cis-Δ12-enoyl CoA c. trans-Δ3-enoyl CoA d. cis-Δ3-enoyl CoA e. cis-Δ9-enoyl CoA .. Powered by Cognero.

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Chapter 27 ANSWER: a 6. Choose the CORRECT reaction sequence for the utilization of propionyl CoA. a. carboxylation to D-methylmalonyl CoA → isomerization to L-methylmalonyl CoA → conversion to succinyl CoA, which enters the citric acid cycle b. carboxylation to D-methylmalonyl CoA → isomerization to L-methylmalonyl CoA → conversion to acetyl CoA, which enters the citric acid cycle c. carboxylation to L-methylmalonyl CoA → isomerization to D-methylmalonyl CoA → conversion to acetyl CoA, which enters the citric acid cycle d. isomerization to trans-Δ2-enoyl CoA → hydration to L-3-hydroxyacyl CoA → oxidation to 3ketoacyl CoA → thiolysis to acetyl CoA, which enters the citric acid cycle e. isomerization to trans-Δ2-enoyl CoA → hydration to L-3-hydroxyacyl CoA → oxidation to 3ketoacyl CoA → conversion to acetyl CoA, which enters the citric acid cycle ANSWER: a 7. D-glyceraldehyde 3-phosphate can be converted: a. into pyruvate or glucose in the liver. b. only into pyruvate in the liver. c. only into glucose in muscles. d. only into acetyl CoA in muscles. e. only into acetyl CoA in the liver.

ANSWER: a 8. What is the function of cis-Δ3-enoyl CoA isomerase? a. conversion of cis-Δ3 double bond into trans-Δ3 single bond b. conversion of trans-Δ2 double bond to cis-Δ3 double bond c. shift the position of the double bond for the proper work of acyl CoA dehydrogenase d. reduction of 2,4-dienoyl to cis-Δ3-enoyl CoA e. reduction of 2,4-dienoyl to trans-Δ2-enoyl CoA ANSWER: c 9. Phosphorylation of perilipin: a. triggers the release of a coactivator for adipose triglyceride lipase. b. stimulates phosphorylation of protein kinase A. c. activates adenylate cyclase. d. initiates the release of a fatty acid from triacylglycerol. e. activates hormone-sensitive lipase.

ANSWER: a 10. How does activated fatty acid move to the mitochondrial matrix? .. Powered by Cognero.

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Chapter 27 a. The activated fatty acid moves across the outer mitochondrial membrane through a porin channel.

Acetyl carnitine shuttles across the inner mitochondrial membrane by a translocase. b. The activated fatty acid moves across the outer mitochondrial membrane by passive diffusion. Acetyl carnitine moves across the inner mitochondrial membrane by a translocase. c. The activated fatty acid moves across the outer mitochondrial membrane through a porin channel, and shuttles across the outer mitochondrial membrane by a translocase. d. Acetyl carnitine moves across the outer mitochondrial membrane through a porin channel, and shuttles across the outer mitochondrial membrane by a translocase. e. D-glyceraldehyde 3-phosphate moves across the outer mitochondrial membrane through a porin channel, and shuttles across the outer mitochondrial membrane by a translocase.

ANSWER: a 11. Hormone-sensitive lipase acts in: a. fatty acid β oxidation. b. transport of a fatty acid through blood. c. thiolysis in the mitochondrial matrix. d. mobilization of triacylglycerols. e. hydrolysis of triacylglycerols in the mitochondrial matrix.

ANSWER: d 12. Which compound is able to move across the inner mitochondrial membrane? a. glycerol b. free fatty acid c. acyl CoA d. D-glyceraldehyde 3-phosphate e. acyl carnitine ANSWER: e 13. Where is adipose tissue located in a human body? a. under the skin (subcutaneous fat) and around the inner organs (visceral fat) b. under the skin (visceral fat) and around the inner organs (subcutaneous fat) c. under the skin (subcutaneous fat), around the inner organs (visceral fat), and in the liver d. under the skin (subcutaneous fat) and around the inner organs (blubber) e. inside muscle tissue ANSWER: a 14. The products of lipolysis are glycerol and free fatty acids. What is the role of these compounds? a. Glycerol can be phosphorylated to D-glyceraldehyde 3-phosphate, which can enter both glycolytic and gluconeogenic pathways. Free fatty acids can be oxidized to succinyl CoA, which enters the citric acid cycle. b. Free fatty acids can be used in both glycolytic and gluconeogenic pathways in the liver cells.

Glycerol can be used as a fuel inside cells. .. Powered by Cognero.

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Chapter 27 c. Glycerol can be used only in gluconeogenesis in the liver cells. Free fatty acids can be used as a fuel inside cells. d. Glycerol can be used only in glycolysis in the liver cells. Free fatty acids can be used as a fuel inside cells. e. Free fatty acids can be oxidized to acyl CoA, which enters the citric acid cycle. Glycerol can be phosphorylated to D-glyceraldehyde 3-phosphate, which can enter both glycolytic and gluconeogenic pathways.

ANSWER: a 15. Methylmalonyl CoA mutase contains vitamin B12 as its coenzyme. What is the substrate for this enzyme? a. propionyl CoA b. D-glyceraldehyde 3-phosphate c. succinyl CoA d. trans-Δ2-enoyl CoA e. trans-Δ3-enoyl CoA ANSWER: a 16. Identify the compound shown.

a. an intermediate in ketogenesis b. an intermediate in lipolysis c. an intermediate in fatty acid oxidation d. an intermediate in fatty acid activation e. a product of fatty acid activation ANSWER: d 17. Which compound does β oxidation begin from? a. acetyl CoA b. trans-Δ2-enoyl CoA c. acyl CoA d. CoA .. Powered by Cognero.

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Chapter 27 e. acyl carnitine ANSWER: c 18. High levels of acetoacetate in blood lead to a decrease in the rate of lipolysis in _______ tissue.

ANSWER: adipose 19. Which statement about ketone bodies is TRUE? a. Acetone and D-3-hydroxybutyrate can be obtained from acetoacetate. b. Ketone bodies bind to the blood protein albumin, which delivers them to tissues. c. Ketone bodies increase the blood pH. d. Ketone bodies are the main fuel during short-term starvation. e. During starvation, acetoacetate can be captured to synthesize glucose.

ANSWER: a 20. The main site of ketogenesis is: a. the liver cells. b. the muscles cells. c. neurons. d. blood plasma. e. adipocytes.

ANSWER: a 21. How many molecules of acetyl CoA are involved in a single round of ketogenesis? a. 2 b. 3 c. 4 d. 1 e. 0 ANSWER: b 22. Which of the following compounds is toxic? a. acetoacetate b. cobalamin c. hypoglycin d. methylene cyclopropylacetate e. trans-Δ3-enoyl CoA ANSWER: d 23. Acetoacetate is formed from acetyl CoA and can spontaneously turn to: a. acetone. b. acyl CoA. .. Powered by Cognero.

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Chapter 27 c. acetyl CoA. d. glucose. e. pyruvate.

ANSWER: a 24. Untreated, chronic ketone body production will: a. increase blood pH levels. b. decrease blood pH levels. c. increase blood glucose levels. d. decrease hydrogen ion concentration in blood. e. increase bicarbonate concentration in blood.

ANSWER: b 25. The availability of _____ determines whether acetyl CoA, made from fatty acids, can enter the citric acid cycle. a. pyrophosphate b. pyruvate dehydrogenase c. oxaloacetate d. cobalamin e. β hydroxybutyrate ANSWER: c 26. The compound _____ is considered one of the ketone bodies. a. pyrophosphate b. glucagon c. cobalamin d. β hydroxybutyrate e. oxaloacetate ANSWER: d 27. In the absence of insulin, most acetyl CoA: a. enters the citric acid cycle. b. enters ketogenesis. c. enters gluconeogenesis. d. is released into blood. e. is stored as lipid droplets.

ANSWER: b 28. Choose the CORRECT sequence of utilization of D-3-hydroxybutyrate. a. D-3-hydroxybutyrate is oxidized to acetoacetate → acetoacetate converts into acetoacetyl CoA → acetoacetyl CoA converts into 2 molecules of acetyl CoA, which enter the citric acid cycle .. Powered by Cognero.

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Chapter 27 b. D-3-hydroxybutyrate is oxidized to acetoacetate → acetoacetate converts into acetoacetyl CoA → acetoacetyl CoA converts into CO2 and acetyl CoA, which enter the citric acid cycle c. D-3-hydroxybutyrate converts into acetone by D-3-hydroxybutyrate dehydrogenase d. D-3-hydroxybutyrate converts into acetoacetate by D-3-hydroxybutyrate dehydrogenase → acetoacetate spontaneously converts into acetone, which enters gluconeogenesis e. D-3-hydroxybutyrate is oxidized to acetoacetate → acetoacetate converts into acetoacetyl CoA → acetoacetyl CoA converts into 2 molecules of pyruvate, which enter the citric acid cycle ANSWER: a 29. During starvation, the brain begins to consume an appreciable amount of ketone bodies ____ days after the beginning of the starvation state. a. 3 b. 7 c. 14 d. 1 e. 30 ANSWER: a 30. The second priority of metabolism during starvation is: a. providing sufficient glucose to the muscles. b. preserving muscle proteins. c. providing the process of gluconeogenesis. d. preserving liver proteins. e. providing the citric acid cycle.

ANSWER: b 31. During triacylglycerol hydrolysis, hormone-sensitive lipase catalyzes the release of: a. monoacylglycerol and two free fatty acids. b. diacylglycerol and a single free fatty acid. c. glycerol and three free fatty acids. d. monoacylglycerol and a single free fatty acid. e. glycerol and a single free fatty acid.

ANSWER: d 32. Which statement about starvation is TRUE? a. Plasma levels of fatty acids and ketone bodies decrease, whereas the plasma level of glucose increases. b. Carbohydrate reserves are exhausted in a single day. c. Triacylglycerol reserves are exhausted in a single month. d. Plasma levels of fatty acids and glucose bodies decrease, whereas the plasma level of ketone bodies increases. .. Powered by Cognero.

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Chapter 27 e. During starvation, the liver cells convert any available acetyl CoA, lactate, and alanine to glucose for the brain's needs.

ANSWER: b 33. High cAMP concentration: a. inhibits lipolysis in the liver cells. b. activates β oxidation of fatty acids in the mitochondrial matrix. c. decreases the level of free fatty acids in blood. d. increases the level of free fatty acids in blood. e. inhibits β oxidation of fatty acids in the mitochondrial matrix.

ANSWER: d 34. The ketogenic diet: a. kills the intestinal flora. b. decreases blood pH. c. is the only way to treat epilepsy. d. increases seizures in children with epilepsy. e. requires high-fat, adequate-protein, and low-carbohydrate consumption.

ANSWER: e 35. Which enzyme catalyzes the conversion of acetoacetate into D-3-hydroxybutyrate? a. D-3-hydroxybutyrate dehydrogenase b. hydroxymethylglutaryl CoA cleavage enzyme c. 3-ketothiolase d. hydroxymethylglutaryl CoA synthase e. CoA transferase ANSWER: a 36. What can promote ketone body formation? a. accumulation of acetyl CoA b. lack of acetyl CoA c. high insulin level d. low glucagon level e. excessive protein intake ANSWER: a 37. Which compound is an intermediate in ketogenesis? a. 3-hydroxy-3-methyl-glutaryl CoA b. D-3-hydroxybutyrate c. acetone d. D-methylmalonyl CoA .. Powered by Cognero.

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Chapter 27 e. L-methylmalonyl CoA ANSWER: a 38. Which enzyme–substrate pair is CORRECT? a. methylmalonyl CoA mutase – 3-hydroxy-3-methyl-glutaryl CoA b. hydroxymethylglutaryl CoA synthase – L-methylmalonyl CoA c. cis-Δ3-enoyl CoA isomerase – trans-Δ3-enoyl CoA d. thiolase – 3-ketoacyl CoA e. acyl CoA synthetase – acetyl CoA ANSWER: d 39. Select all that apply. Epidemiological evidence suggests that consumption of a large amount of saturated fatty acids and trans fat promotes: a. obesity. b. type 2 diabetes. c. type 1 diabetes. d. atherosclerosis. e. thrombosis.

ANSWER: a, b, d, e 40. The increase in acetyl CoA during starvation inhibits the enzyme: a. acyl CoA dehydrogenase. b. cobalamin. c. glucagon. d. hormone-sensitive lipase. e. pyruvate dehydrogenase.

ANSWER: e 41. β oxidation of myristyl-CoA (14:0) yields: a. 7 acetyl CoA + 7 FADH2 + 7 NADH + 7 H+. b. 6 acetyl CoA + 7 FADH2 + 7 NADH + 7 H+. c. 7 acetyl CoA + 6 FADH2 + 6 NADH + 6 H+. d. 7 acetyl CoA + 7 FAD + 7 NAD+. e. 14 acetyl CoA + 12 FADH2 + 12 NADH + 12 H+.

ANSWER: c 42. How many rounds of β oxidation would be required for a 16-carbon fatty acyl chain to be degraded to acetyl CoA? a. 16 b. 8 .. Powered by Cognero.

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Chapter 27 c. 7 d. 15 e. 10 ANSWER: c 43. The activation of fatty acids for degradation takes place in two steps. What is the intermediate formed and why is activation necessary for β oxidation to occur? a. Acyl adenylate; cAMP formed by this reaction activates protein kinase A. b. Acyl adenylate; only the activated form of the fatty acid can react with coenzyme A. c. Propionate; only the activated form of the fatty acid can react with coenzyme A. d. Acyl carnitine; cAMP formed by this reaction activates protein kinase A. e. Acyl carnitine; this intermediate is needed to fuel the carnitine cycle.

ANSWER: b 44. Exercising first thing in the morning without eating mobilizes lipid stores for fuel. All of the below are involved in this mobilization, EXCEPT: a. glucagon binds 7TM receptors that activate adenylate cyclase. b. perilipin is phosphorylated. c. cAMP stimulates protein kinase A. d. hormone-sensitive lipase completes the mobilization of fatty acids with the production of a free fatty acid and glycerol. e. hormone-sensitive lipase is phosphorylated.

ANSWER: d 45. Which organs or tissues prefer to use ketone bodies such as acetoacetate as a fuel energy source instead of glucose? a. heart muscle b. renal cortex c. brain d. heart muscle and renal cortex e. liver ANSWER: d 46. How does formation of acetyl CoA by β oxidation not only provide energy for cellular respiration but also spare glucose? a. As acetyl CoA levels increase during lipid metabolism, D-3-hydroxybutyrate is reduced to pyruvate for gluconeogenesis. b. Acetyl CoA provides carbon for the formation of oxaloacetate, which in turn can be decarboxylated to form pyruvate. c. Acetyl CoA can be converted to acetoacetate and from there to lactate for anaerobic metabolism. d. Acetyl CoA derived from fatty acids inhibits pyruvate dehydrogenase, the enzyme that converts pyruvate into acetyl CoA. .. Powered by Cognero.

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Chapter 27 e. Acetyl CoA derived from fatty acids activates pyruvate carboxylase, stimulating gluconeogenesis.

ANSWER: d 47. Every biochemistry student knows that most naturally occurring unsaturated fatty acids are cis fatty acids.

Why, then, do so many fat-containing foods come with the notation "Contains no trans fats"? a. Vegetable oils contain polyunsaturated fatty acids prone to oxidation, which then converts them to trans fats. b. Olive oils, normally healthy oils, are converted to trans fat when cooked at high temperatures. Thus, olive oils are used only at low temperatures. c. Vegetable oils are often hydrogenated to form trans unsaturated fatty acids to preserve shelf life. d. Oils that contain an odd number of carbon atoms are particularly susceptible to oxidation to trans fats. Thus, no odd-numbered carbon fatty acids are used in these foods. e. Oils that contain an odd number of carbon atoms are particularly susceptible to reduction to trans fats. Thus, no odd-numbered carbon fatty acids are used in these foods.

ANSWER: c 48. Coenzyme(s) involved in the degradation of saturated fatty acyl CoA include(s): a. FAD. b. NAD+. c. TPP. d. FAD and NAD+. e. All of the answers are correct.

ANSWER: d 49. The hormone _____ induces lipolysis, whereas _____ inhibits the process. a. epinephrine; adrenocorticotropin b. glucagon; insulin c. insulin; norepinephrine d. glucagon; epinephrine e. epinephrine; glucagon ANSWER: b 50. What does muscle weakness and cramping in the skeletal muscle, heart, and kidneys likely indicate? a. excess carnitine in the diet b. loss of hormone-sensitive lipase activity c. malfunctioning or deficient carnitine translocase d. ketone-body production e. ketogenic diet ANSWER: c 51. The ruby-throated hummingbird has one of the highest metabolic rates, nearly doubling its mass before a .. Powered by Cognero.

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Chapter 27 long migratory flight. Which of the following best describes how the hummingbird’s metabolism supports its migration, particularly regarding its energy source? a. The hummingbird stores excess carbohydrates as glycogen, which is then converted into fat to fuel its migration. b. The hummingbird relies exclusively on carbohydrates during migration to fuel its high-energy demands. c. The hummingbird burns only the fat stored in its wings, while the rest of its body uses glucose for energy. d. The hummingbird stores excess protein in its muscles, which is broken down to produce glucose for energy during migration. e. The hummingbird uses fat stored in its body as an energy source, which is broken down into fatty acids via lipolysis, providing energy during non-stop flights.

ANSWER: e 52. Peripheral tissues, such as muscle, access lipid energy reserves stored in adipose tissue through three stages.

Which of the following stages describes how these tissues access stored lipids? Select all that apply. a. The mobilization stage is regulated by insulin, which promotes the storage of fatty acids in adipose tissue. b. Mobilization is the first stage, in which stored triacylglycerols are broken down into free fatty acids and glycerol that are released from the adipose tissue into the bloodstream. c. Activation and transport is the second stage, in which fatty acids are converted to glucose for transport to energy-requiring tissues. d. In the breakdown into acetyl CoA stage, fatty acids undergo β-oxidation in the mitochondria to produce acetyl CoA, which enters the citric acid cycle. e. Activation and transport is the second stage, in which fatty acids must be activated and transported into mitochondria for further degradation.

ANSWER: b, d, e 53. Which of the following statements describes how epinephrine and glucagon induce lipolysis in adipose tissue? a. Epinephrine and glucagon bind to receptors that activate phosphodiesterase, which decreases cyclic AMP (cAMP) levels, promoting lipolysis. b. Epinephrine directly activates lipases in adipose tissue, leading to the hydrolysis of triacylglycerols into fatty acids. c. Glucagon inhibits the activity of adipose triglyceride lipase (ATGL), allowing the breakdown of triacylglycerols into free fatty acids. d. Epinephrine and glucagon directly activate hormone-sensitive lipase without the need for cAMP signaling, triggering lipolysis in adipose tissue. e. Epinephrine and glucagon bind to G-protein-coupled receptors, increasing cAMP levels and activating protein kinase A (PKA), which phosphorylates perilipin and hormone-sensitive lipase, triggering lipolysis.

ANSWER: e .. Powered by Cognero.

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Chapter 27 54. Which of the following statements explains how exercising in a fasted state, after consuming a cup of coffee, can promote fat loss during exercise? Select all that apply. a. The sugar content in the cup of coffee increases glucose levels, which eliminates the body’s dependence on fat for energy during exercise. b. Caffeine inhibits phosphodiesterase, increasing cAMP levels, which activates protein kinase A (PKA) and enhances lipolysis. c. Exercising in a fasted state increases insulin levels, promoting fat storage for energy use during aerobic exercise. d. Fasting decreases the amount of stored glycogen, making the body more likely to use fat as a fuel source during exercise. e. The combination of fasting and the caffeine-induced increase in cAMP levels enhances the activation of enzymes like hormone-sensitive lipase. This promotes fat breakdown during aerobic exercise.

ANSWER: b, d, e 55. Why are the symptoms of carnitine acyltransferase II (CAT II) deficiency and long-chain 3-hydroxyacyl CoA dehydrogenase (LCHAD) deficiency more severe during fasting or prolonged exercise? a. During fasting or prolonged exercise, the body uses stored fatty acids for energy, but the defective enzymes prevent the breakdown of fatty acids. This leads to energy deficits and muscle weakness. b. During fasting or prolonged exercise, the body increases glucose production from fat, which is hindered in these patients, forcing them to rely on less efficient energy sources, resulting in muscle weakness. c. Fasting causes a decrease in blood sugar levels, and the body shifts to using only protein for energy, leading to muscle breakdown and further weakness in these patients. d. During fasting or prolonged exercise, the body compensates by increasing ketone body production, which fails to provide energy in these patients due to impaired fatty acid oxidation, leading to fatigue and weakness. e. Prolonged fasting and exercise result in an excess of fatty acids in the bloodstream, leading to an overload of glucose, which impairs muscle function.

ANSWER: a 56. A medical resident on their first internship is assigned to a patient diagnosed with Chanarin-Dorfman syndrome. The patient has a defective coactivator required by adipose triglyceride lipase (ATGL), leading to an inability to break down fat. What is the most likely consequence of this defect? a. The patient experiences enhanced fatty acid oxidation, leading to a reduction in fat stores throughout the body. b. Fat release from adipose tissue is too efficient, leading to a decrease in fat accumulation throughout the body. c. The liver compensates for ATGL deficiency by increasing fatty acid synthesis, leading to increased levels of fat accumulation. d. The defective ATGL coactivator generates an increase in triglyceride breakdown in muscle tissue, which exacerbates fat storage in other organs. e. Fat accumulates in the body because ATGL is unable to convert triacylglycerols into fatty acids, disrupting normal fat metabolism. .. Powered by Cognero.

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Chapter 27 ANSWER: e 57. During fasting or diabetes, the body shifts to using ketone bodies as alternative fuel due to the depletion of oxaloacetate for gluconeogenesis. Which of the following statements accurately describes this metabolic shift?

Select all that apply. a. In conditions of fasting or diabetes, acetyl CoA is diverted from the citric acid cycle to form ketone bodies like acetoacetate and D-3-hydroxybutyrate due to oxaloacetate depletion. b. During fasting or diabetes, oxaloacetate, which is normally involved in the citric acid cycle, is consumed for gluconeogenesis, leading to the diversion of acetyl CoA towards ketogenesis. c. Muscle cells primarily use acetoacetate and D-3-hydroxybutyrate for energy, but these ketone bodies cannot be utilized by the brain during periods of fasting. d. During fasting or diabetes, elevated levels of ketone bodies in the blood can contribute to acidosis because they are acidic in nature. e. Ketone bodies, such as acetoacetate and D-3-hydroxybutyrate, are produced when there is an excess of glucose, allowing acetyl CoA to bypass the citric acid cycle.

ANSWER: a, b, d 58. The plot below shows the plasma concentration of ketone bodies, glucose, and fatty acids in millimolar (mM) during 8 days of starvation. Choose the answer that correctly labels the data on the plot.

ANSWER: Red line=Ketone bodies, blue line= Glucose, Brown line=Fatty acids 59. During starvation, the body shifts its metabolic priorities to ensure survival. Which of the following statements describes the first and second priorities of the body during starvation? Select all that apply. a. During starvation, the body’s primary priority is to maintain blood-glucose concentration. b. During starvation, the body’s primary priority is to increase gluconeogenesis to replenish glycogen stores and maintain normal blood-glucose levels. c. During starvation, the body’s secondary priority is to maintain blood-glucose concentration. d. During starvation, the body’s secondary priority is switching the cell energy source to protein to maintain the blood-glucose level and prevent muscle breakdown. .. Powered by Cognero.

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Chapter 27 e. During starvation, the body’s secondary priority is to preserve muscle protein by shifting the fuel source from glucose to fatty acids and ketone bodies, particularly in organs that typically rely on glucose.

ANSWER: a, e 60. Protein wasting is a common metabolic response during prolonged starvation or certain diseases. Which of the following statements best explains protein wasting in the body? a. Protein wasting occurs after one day of starvation to provide immediate energy for vital organs. b. Protein wasting occurs after fat stores are depleted, when the body must use muscle proteins to produce glucose through gluconeogenesis. c. Protein wasting occurs after fatty acids are converted into glucose to maintain blood-glucose concentrations. d. Protein wasting is prevented during prolonged starvation because the body always prefers to break down fats over proteins for energy. e. Protein wasting is prevented in prolonged starvation because the body uses ketone bodies as the primary energy source, preserving muscle mass.

ANSWER: b 61. Which of the following statements describes the body’s energy reserves during fasting? a. Glucose, stored in glycogen, is the most abundant and efficient energy molecule in the body and is capable of supporting prolonged survival during fasting. b. The glucose stored in glycogen can provide energy for the body for several weeks, while triacylglycerols provide energy for only a few days. c. Triacylglycerols are the primary energy source, capable of sustaining the body for weeks, while glycogen stores are depleted within days. d. Glycogen can store more energy than fat, so the body would be more efficient in using glycogen instead of triacylglycerols for long-term survival. e. Protein is the most efficient energy source in the body during fasting, providing the longest-lasting energy.

ANSWER: c 62. Which of the following statements correctly describes the phosphorylation of perilipin and hormonesensitive (HS) lipase in the context of lipolysis? Select all that apply. a. Protein kinase A (PKA) phosphorylates perilipin, causing a rearrangement of the lipid droplet and the release of the coactivator for ATGL. b. Phosphorylation of perilipin by PKA results in the activation of HS lipase, which then stimulates lipogenesis. c. Perilipin directly activates HS lipase to access lipid droplets and initiate lipolysis. d. Phosphorylation of HS lipase activates its enzymatic activity; HS lipase then releases a fatty acid from diacylglycerol, resulting in the formation of monoacylglycerol. e. The phosphorylation of both perilipin and hormone-sensitive (HS) lipase is triggered by the binding of epinephrine and glucagon to their respective receptors, which increases cAMP levels.

ANSWER: a, d, e .. Powered by Cognero.

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Chapter 27 63. Which of the following statements describes the four sequential steps involved in the β-oxidation of fatty acids? Select all that apply. a. Oxidation of the fatty acid chain results in the formation of a trans double bond between C-2 and C3. b. Hydration occurs when water is added to the double bond formed during oxidation, resulting in the addition of a hydroxyl group on C-3. c. Reduction occurs when the carbonyl group is reduced to a hydroxyl group, making the fatty acid chain ready for further oxidation. d. A second oxidation step occurs when the hydroxyl group is oxidized to a keto group, generating 3ketoacyl CoA. e. Thiolysis occurs when 3-ketoacyl CoA is cleaved by coenzyme A (CoA), releasing acetyl CoA and a shortened fatty acid chain.

ANSWER: a, b, d, e 64. A doctor specializing in rare diseases evaluates a patient who presents with vomiting, hypoglycemia, and muscle weakness, which resemble symptoms of hypoglycin toxicity. After further testing, the patient is diagnosed with medium-chain acyl CoA dehydrogenase (MCAD) deficiency. Which of the following statements describes the underlying metabolic issue in this patient? a. MCAD deficiency leads to an overproduction of ketone bodies during fasting, which causes metabolic acidosis and muscle weakness. b. The absence of MCAD enzymatic activity prevents the breakdown of glycogen, leading to hypoglycemia. c. MCAD deficiency causes a toxic buildup of acetoacetate, leading to nausea, vomiting, and hypoglycemia, similar to hypoglycin toxicity. d. Deficiency of MCAD enzymatic activity triggers excess β-oxidation of fatty acids, which reduces glucose utilization and results in hypoglycemia. e. The lack of MCAD enzymatic activity leads to an inability to metabolize medium-chain fatty acids, impairing energy production from fat stores during fasting and inhibiting gluconeogenesis. Similar to hypoglycin toxicity, the patient develops hypoglycemia.

ANSWER: e 65. During starvation, the body exhausts its glycogen stores and breaks down fat to produce ketone bodies as an alternative energy source. Which of the following statements is true regarding the utilization of ketone bodies during starvation? a. Under starvation conditions, some acetone may be captured to synthesize small amounts of glucose, but most acetone is exhaled. b. Under starvation conditions, ketone bodies, including acetone, are stored in the liver for future use. c. During starvation conditions, only the liver, but no other tissues, uses ketone bodies for energy. d. Under starvation conditions, acetone is not typically exhaled or excreted in the urine. e. Under starvation conditions, ketone bodies, including acetone, accumulate in the bloodstream but are quickly removed to avoid abnormally high levels.

ANSWER: a .. Powered by Cognero.

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Chapter 27 66. A striking feature of diabetes is the shift in fuel utilization from carbohydrates to fats, with glucose, more abundant than ever, left unused in the bloodstream. Which of the following statements explains this shift in fuel utilization in diabetes? a. Excess glucose in the bloodstream is stored as glycogen, leading to a decrease in fat metabolism. b. In diabetes, the cells in the body become more sensitive to glucose, leading to enhanced glucose reuptake and reduced fat usage. c. Increased insulin production in diabetes forces the body to use glucose, inhibiting fat breakdown. d. Glucagon levels decrease, inhibiting fat breakdown and forcing the body to use glucose. e. Insulin resistance or deficiency prevents glucose from entering cells, leading to the use of fat as the primary energy source.

ANSWER: e 67. Historically, diabetics with diabetic ketosis were sometimes diagnosed as being drunk. Which of the following statements explains why diabetic ketosis can mimic the symptoms of intoxication? Select all that apply. a. Lack of glucose in the brain can lead to odd behaviors in diabetic ketosis patients. b. A high level of insulin in diabetic ketosis decreases the production of acetone, resulting in alcohol intoxication behavior. c. Increased insulin production in diabetes forces the body to use glucose, inhibiting fat breakdown. d. High levels of glucose in the brain lead to odd behaviors in diabetic ketosis patients. e. The body shifts to using ketones instead of glucose for energy, and acetoacetate spontaneously decarboxylates to acetone, creating a breath smell similar to alcohol.

ANSWER: a, e 68. A diabetic person is in a _____________________ state despite having a high blood glucose level. Because insulin is deficient, the entry of glucose into adipose and muscle cells is impaired.

ANSWER: biochemical fasting 69. The absence of insulin in the body has two major biochemical consequences regarding fat metabolism.

Which of the following two statements correctly describes these consequences? a. Adipose cells keep releasing fatty acids into the bloodstream. The liver converts the fatty acids into acetyl CoA through β-oxidation. b. Adipose cells do not release fatty acids into the bloodstream, and therefore, β-oxidation is greatly reduced. c. Increased insulin production in diabetes forces the body to use glucose, inhibiting fat breakdown. d. The liver absorbs glucose and provides oxaloacetate to process fatty acid-derived acetyl CoA. e. The liver cannot absorb glucose and consequently cannot provide oxaloacetate to process fatty acidderived acetyl CoA. ANSWER: a, e 70. Many animals rely on fat stores as their primary energy source during hibernation. Which of the following statements explains the metabolic changes required to prepare for hibernation? a. Before entering hibernation, the production of triacylglycerols is enhanced. .. Powered by Cognero.

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Chapter 27 b. Before entering hibernation, animals increase glucose metabolism to ensure energy availability during hibernation. c. Before entering hibernation, fat breakdown is prioritized over protein breakdown to maintain growth. d. Before entering hibernation, the liver stores large amounts of glucose to provide quick energy during hibernation. e. Before entering hibernation, the breakdown of triacylglycerols is enhanced to store fat. The liver produces ketone bodies to provide an alternative energy source for the brain.

ANSWER: a 71. Many animals rely on fat stores as their primary energy source during hibernation. Which of the following statements explains the metabolic changes during hibernation? a. During hibernation, the body uses stored fatty acids and ketone bodies as energy, while the breakdown of triacylglycerols is minimal to conserve energy. b. During hibernation, animals increase glucose metabolism to ensure energy availability. c. During hibernation, protein breakdown is prioritized over fat breakdown to ensure the brain has an adequate glucose supply. d. During hibernation, the liver stores large amounts of glucose to provide quick energy. e. During hibernation, the body breaks down triacylglycerols to produce fatty acids and ketone bodies.

ANSWER: e 72. Which of the following statements explains how activated fatty acids can cross the outer mitochondrial membrane of mitochondria? a. Activated fatty acids can cross the outer mitochondrial membrane by simple diffusion. b. Activated fatty acids can cross the outer mitochondrial membrane using an ATPase. c. Activated fatty acids never cross the outer mitochondrial membrane. d. Activated fatty acids do not require a protein-based transport system to cross the outer mitochondrial membrane. e. Activated fatty acids can cross the outer mitochondrial membrane through a porin channel.

ANSWER: e

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Chapter 28 1. Fatty acid synthesis primarily takes place in the: a. cytoplasm. b. inner membrane of mitochondria. c. matrix of mitochondria. d. Golgi apparatus. e. nucleoli.

ANSWER: a 2. What is the product of the committed step in fatty acid synthesis? a. C16-acyl ACP b. acetyl CoA c. malonyl CoA d. acetoacetyl ACP e. β-hydroxybutyryl ACP ANSWER: c 3. The committed step in fatty acid synthesis is inactivated by: a. acetyl CoA carboxylase 2. b. citrate. c. protein phosphatase 2A. d. AMP-activated protein kinase. e. malonyl CoA. ANSWER: d 4. Which coenzyme is used in the committed step of fatty acid synthesis? a. biotin b. lipoic acid c. FADH2 d. NADH e. NADPH ANSWER: a 5. Activation of acetyl CoA results in the production of: a. malonyl CoA. b. citrate. c. acetoacetyl CoA. d. acetyl ACP. e. citryl CoA. ANSWER: a .. Powered by Cognero.

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Chapter 28 6. What is the driving force of condensation of carbon units in fatty acid synthesis? a. carboxylation of malonyl CoA b. decarboxylation of acetoacetyl ACP c. NADPH oxidation d. ATP hydrolysis at the previous stage e. entropy ANSWER: d 7. What compound is used as an oxidant in the conversion of stearoyl CoA into oleoyl CoA? a. NADP+ b. O2 c. H2O d. FAD e. cytochrome c ANSWER: b 8. In which compartment are double bonds introduced into saturated fatty acids? a. cytosol b. inner membrane of mitochondria c. matrix of mitochondria d. Golgi apparatus e. endoplasmic reticulum ANSWER: e 9. Animals that have fasted and are then fed high-carbohydrate or low-fat diets show a marked increase in activities of acetyl CoA carboxylase 1 and fatty acid synthase within a few days. What is the cause of this phenomenon? a. Glucose and insulin act at the level of gene transcription. b. Glucose and insulin act at the level of zymogen activation. c. The number of insulin receptors during fasting significantly decreases. d. The increase in epinephrine level during fasting causes activation of acetyl CoA carboxylase 1 and fatty acid synthase. e. The decrease in NADH level during fasting causes activation of acetyl CoA carboxylase 1 and fatty acid synthase.

ANSWER: a 10. What are the products at the stage of fatty acid synthesis when a two-carbon molecule reacts with a threecarbon molecule to give a four-carbon molecule and a one-carbon molecule? Select all that apply. a. acetyl CoA b. acetoacetyl CoA c. malonyl CoA .. Powered by Cognero.

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Chapter 28 d. methyl CoA e. CO2 ANSWER: b, e 11. Select all that apply. Fatty acid synthesis is necessary during: a. fasting. b. embryonic development. c. after a meal. d. lactation. e. sleep.

ANSWER: b, c, d 12. Which coenzymes are required for fatty acid synthesis? Select all that apply. a. biotin b. lipoic acid c. FADH2 d. NADH e. NADPH ANSWER: a, e 13. What are the results of a biotin-deficient diet? Select all that apply. a. increase in cytosolic crotonyl CoA b. decrease in butyryl ACP c. decrease in cytosolic acetyl CoA d. accumulation of cytosolic acetyl CoA e. decrease in fatty acid de novo synthesis ANSWER: b, d, e 14. What are the sources of NADPH for fatty acid synthesis? Select all that apply. a. pentose phosphate pathway b. citric acid cycle c. malic enzyme reaction d. oxidative decarboxylation of malate by malic enzyme in cytosol e. action of cytoplasmic malate dehydrogenase ANSWER: a, c, d 15. Polyunsaturated fatty acids are essential in mammals because they lack the enzymes to: a. introduce double bonds at carbon atoms beyond C-9. b. synthesize unsaturated fatty acids with cycles in their acyl chain. c. introduce double bonds at carbon atoms beyond C-2. .. Powered by Cognero.

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Chapter 28 d. introduce double bonds at carbon atoms beyond C-7. e. synthesize polyunsaturated fatty acids.

ANSWER: a 16. What is the role of citrate in the reaction of acetyl CoA carboxylase 1? a. product inhibitor b. prosthetic group c. mechanism-based inhibitor d. allosteric activator e. allosteric inhibitor ANSWER: d 17. Which state of acetyl CoA carboxylase 1 is active? a. monomer b. trimer c. tetramer d. R state e. dimer ANSWER: d 18. In fatty acid synthesis, what is the direct product of the following metabolic conversion?

a. crotonyl ACP b. D-3-hydroxybutyryl ACP c. butyryl ACP d. D-3-oxybutyryl CoA e. D-3-hydroxyacetyl ACP ANSWER: b 19. How many acetyl CoA molecules should be transported from the mitochondrial matrix to the cytosol to provide all of the carbon atoms for oleate synthesis? a. 6 b. 16 c. 8 d. 9 e. 18 ANSWER: d 20. How many carbon atoms in oleic acid come from malonyl CoA? .. Powered by Cognero.

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Chapter 28 a. 2 b. 12 c. 16 d. 18 e. 9 ANSWER: c 21. Using _____ CoA in the condensation of carbon units is favorable because the condensation is coupled to carboxylation that contributes to a substantial _______ in free energy. a. acetyl; decrease b. malonyl; decrease c. malonyl; increase d. citryl; increase e. acetyl; increase ANSWER: b 22. The role of biotin in the reaction of acetyl CoA carboxylase 1 is: a. acting as a nucleophile and attacking the carbonyl group of acetyl CoA. b. binding of acetyl CoA. c. binding of ATP to the enzyme. d. binding of CO2 that results in production of carboxybiotin. e. covalent attaching to the epsilon-amino group of specific lysine residues in the enzyme.

ANSWER: d 23. How many NADPH molecules are formed when a sufficient quantity of acetyl CoA is transferred to the cytoplasm via the citrate shuttle for synthesis of the stearate? a. 1 b. 8 c. 9 d. 16 e. 18 ANSWER: c 24. Predict the short-time result of triclosan addition to the growth medium for yeasts. a. The amount of acetyl CoA in cells would decrease. b. The amount of crotonyl ACP in cells would increase. c. NADPH concentration in cells would significantly increase. d. Citrate concentration in cells would significantly increase. e. The amount of butyryl ACP in cells would increase.

ANSWER: b .. Powered by Cognero.

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Chapter 28 25. In what organisms are the component enzymes of fatty acid synthesis linked in a single large polypeptide chain? a. mammals b. plants c. archaea d. Gram-positive bacteria e. E. coli ANSWER: a 26. What reaction is catalyzed by the body and the legs of mammalian fatty acid synthase? a. dehydration b. reduction c. modification d. condensation e. CO2 binding ANSWER: d 27. What is the role of ψ domains in mammalian fatty acid synthase? a. ketoreductase catalytic domains b. methyltransferase catalytic domains c. maintenance of the multifunctional enzyme structure d. transfer of intermediates from one catalytic site to another e. linker between functional domains ANSWER: c 28. What is the acetyl group carrier that transports acetyl groups from the mitochondrial matrix to the cytoplasm? a. acetyl CoA b. malonyl CoA c. citrate d. malate e. oxaloacetate ANSWER: c 29. Which sequence of reactions leading to the reduction of the C-3 keto group to the methylene group upon fatty acid synthesis is TRUE? a. reduction – dehydration – reduction b. oxidation – hydration – reduction c. reduction – hydrogenation – dehydration d. reduction – hydrogenation – condensation e. condensation – dehydration – hydrogenation .. Powered by Cognero.

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Chapter 28 ANSWER: a 30. In eukaryotes, saturated fatty acids larger than 18 carbons are formed by: a. elongation reactions catalyzed by enzymes on the endoplasmic reticulum membrane. b. one-carbon additions at the inner mitochondrial membrane. c. elongation reactions catalyzed by enzymes in the peroxisomes. d. two-carbon additions at the inner mitochondrial membrane. e. elongation reactions catalyzed by enzymes in the lumen of the endoplasmic reticulum.

ANSWER: a 31. The function of the acyl CoA synthetase is: a. ATP-dependent lipolysis to release free glyceraldehyde. b. ATP-dependent reduction of acetyl-CoA prior to activation. c. ATP-dependent activation of fatty acids using CoA. d. ATP-dependent activation of acetyl-CoA. e. synthesis of acetyl CoA from three-carbon units.

ANSWER: c 32. Which compound inhibits the entry of fatty acyl CoA into mitochondria when the cell is in a high-energy state? a. mitochondrial malonyl CoA b. carnitine acyltransferase I c. acetyl CoA carboxylase 2 d. citrate e. cytosolic malonyl CoA ANSWER: a 33. Acetylation of serine on prostaglandin synthase inhibits its activity. What eicosanoid hormones will NOT be produced? Select all that apply. a. prostacyclin b. thromboxane c. PGH2-derived prostaglandins d. leukotrienes e. prostaglandin H2 ANSWER: a, b, c, e 34. Select all that apply. Essential fatty acids are: a. palmitate (16:0). b. palmitoleate (16:1). c. linoleate (18:2). d. oleate (18:1). .. Powered by Cognero.

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Chapter 28 e. linolenate (18:3).

ANSWER: c, e 35. What is the difference between the reduction of acetoacetyl ACP to β-hydroxybutyryl ACP in fatty acid synthesis and the corresponding β oxidation in fatty acid degradation? Select all that apply. a. In reduction, the L rather than the D isomer is formed. b. In reduction, the D rather than the L isomer is formed. c. NADPH/NADP+ is a redox coenzyme pair used in reduction, whereas NADH/NAD+ is a redox coenzyme pair used in oxidation. d. NADH/NAD+ is a redox coenzyme pair used in reduction, whereas NADPH/NADP+ is a redox coenzyme pair used in oxidation. e. In reduction, NADPH is the reductant, whereas FAD is the oxidant in the corresponding reaction in β oxidation.

ANSWER: b, c 36. What are ways to regulate the activity of acetyl CoA carboxylase? Select all that apply. a. allosteric inhibitors b. phosphorylation and dephosphorylation c. zymogen activation d. direct binding of cAMP e. allosteric activators ANSWER: b, e 37. Which of the following statements about the regulation of fatty acid synthesis in a cell under low energy charge conditions is FALSE? a. Acetyl CoA carboxylase 1 is switched off by phosphorylation. b. AMP activates AMP-activated protein kinase. c. Acetyl CoA carboxylase 1 exists as inactive dimers unless citrate and ATP levels are high. d. Polymerization of active acetyl CoA carboxylase 1 is facilitated by the protein MIG12. e. Malonyl CoA is an especially effective inhibitor of acetyl CoA carboxylase 2 in the heart and muscle tissues.

ANSWER: e 38. Increasing the activity of adenylate kinase results in the inactivation of: a. acetyl CoA carboxylase. b. protein phosphatase 2A. c. protein kinase C. d. AMP-activated protein kinase. e. citrate synthase.

ANSWER: d .. Powered by Cognero.

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Chapter 28 39. Chronic ethanol ingestion leads to fatty liver due to an increase in: a. NADH. b. NADPH. c. ATP. d. AMP. e. lactate.

ANSWER: a 40. The enzyme that catalyzes the addition of an activated two-carbon compound to an activated three-carbon compound is: a. malonyl transacylase. b. β-ketoacyl synthase. c. β-ketoacyl ACP reductase. d. acetyl transacylase. e. enoyl ACP reductase.

ANSWER: b 41. Insulin leads to: a. activation of fatty acid synthesis. b. inhibition of fatty acid synthesis. c. inhibition of ATP-citrate lyase. d. decrease of protein phosphatase hydrolysis of fatty acid synthesis enzymes. e. decrease in pyruvate transport into mitochondria.

ANSWER: a 42. How are the consumption of corn oil and prostaglandin synthesis related? a. Prostaglandins are synthesized from membrane proteins originally derived from corn oil. b. ω-6 fatty acids found in corn oil inhibit thromboxane synthase, shifting synthesis to prostaglandins. c. Corn oil contains linoleate, which is required for arachidonate synthesis. d. ω-3 fatty acids found in corn oil inhibit thromboxane synthase, shifting synthesis to prostaglandins. e. Prostaglandin synthesis and corn oil are not related.

ANSWER: c 43. Which statement about hormones regulating fatty acid synthesis after exercise or a meal is TRUE? a. Insulin stimulates fatty acid synthesis by activating acetyl CoA carboxylase 1. b. Insulin stimulates the mobilization of fatty acids for β oxidation. c. Epinephrine stimulates the mobilization of fatty acids and stimulates their accumulation as triacylglycerols. d. Glucagon activates carboxylase by enhancing phosphorylation of AMPK. e. Epinephrine stimulates AMPK, preventing phosphorylation of acetyl CoA carboxylase 1.

ANSWER: a .. Powered by Cognero.

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Chapter 28 44. What are the conditions that lead to a "beer gut" due to an excess consumption of alcohol? a. NADH produced from the metabolism of ethanol stimulates the citric acid cycle for glucose-derived acetyl CoA. b. Excess ethanol metabolism leads to an accumulation of NADH that inhibits fatty acid metabolism. c. NADH stimulates citric acid cycle enzymes that stimulate glucose-derived acetyl CoA metabolism. d. NADH inhibits ketone body formation, stimulating glucose rather than fatty acid metabolism. e. The processing of acetate in the liver becomes inefficient, leading to a pH imbalance in liver cells, reducing enzyme efficiency in general.

ANSWER: b 45. The carrier of an acyl chain through the synthetic protein complex is: a. ACP. b. a serum albumin. c. acetyl coenzyme A. d. a biotin-activated complex. e. ψ domain.

ANSWER: a 46. What enzyme catalyzes the committed step in fatty acid synthesis? a. enoyl reductase b. acetyl CoA carboxylase c. transacylase d. 3-hydroxyacyl dehydratase e. β-ketoacyl reductase ANSWER: b 47. _______ serves as a signal molecule for a high-energy state as it stimulates acetyl CoA carboxylase.

ANSWER: Citrate 48. Acetyl CoA carboxylase 1 (ACC1) is switched off by _____ and activated by _____.

ANSWER: phosphorylation, dephosphorylation 49. The synthesis of palmitate requires _____ molecules of NADPH as well as the expenditure of _____.

ANSWER: 14, ATP 50. Which of the following statements accurately describes the transport of acetyl CoA between mitochondria and the cytoplasm? a. Acetyl CoA is converted to citrate in the cytoplasm, and citrate is transported into the mitochondria, where it is converted back to acetyl CoA for fatty acid synthesis. b. Acetyl CoA is transported from mitochondria into the cytoplasm in the form of citrate, where it is cleaved back into acetyl CoA and oxaloacetate. Oxaloacetate is then directly transported back into the mitochondrial matrix. .. Powered by Cognero.

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Chapter 28 c. Acetyl CoA is directly transported from mitochondria into the cytoplasm via an acetyl CoA transporter. d. Acetyl CoA is already present in the cytoplasm, and thus, transport from the mitochondria is not necessary. e. Acetyl CoA is combined with oxaloacetate to form citrate in the mitochondria, and citrate is transported into the cytoplasm, where it is converted back to acetyl CoA for fatty acid synthesis.

ANSWER: e 51. What role does thioesterase play in the synthesis of fatty acids? a. Thioesterase acts as a “ruler” to determine the final length of the fatty acid chain by releasing it from the carrier protein when the chain reaches a specific length. b. Thioesterase determines the rate of fatty acid synthesis. c. Thioesterase facilitates the addition of two-carbon units to the fatty acid chain, controlling the chain elongation process. d. Thioesterase binds to malonyl CoA to help transfer it onto the growing fatty acid chain. e. Thioesterase removes acetyl CoA molecules from the growing fatty acid chain, determining the final chain length.

ANSWER: a 52. Which of the following statements describes the difference between NADPH and NADH in metabolic pathways? a. NADPH is consumed in biosynthetic reactions, whereas NADH is generated in energy-yielding reactions. b. NADPH is used primarily in catabolic reactions to break down molecules for energy, while NADH is used in anabolic reactions to build molecules. c. NADPH is only used in glycolysis, while NADH is involved in the citric acid cycle. d. Both NADPH and NADH are consumed in energy-yielding reactions, with no significant differences in their role. e. NADPH is generated in biosynthetic reactions, whereas NADH is consumed in energy-yielding reactions.

ANSWER: a 53. Which of the following statements are TRUE regarding the synthesis of palmitate? Select all that apply. a. The synthesis of palmitate requires 8 molecules of acetyl CoA. b. The synthesis of palmitate requires 14 molecules of NADH. c. The synthesis of palmitate requires 10 molecules of acetyl CoA. d. The synthesis of palmitate consumes 7 molecules of ATP. e. The synthesis of palmitate produces 8 molecules of NADPH. f. The synthesis of palmitate requires 14 molecules of NADPH. ANSWER: a, d, f 54. Which of the following statements describes the role of omega-enriched foods in the diet? .. Powered by Cognero.

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Chapter 28 a. Omega-enriched foods provide non-essential fatty acids, such as linoleate and linolenate, which can also be synthesized in the body. b. The fatty acids in omega-enriched foods are primarily used to synthesize saturated fatty acids and increase cholesterol levels. c. Omega-enriched foods are rich in omega-3 fatty acids, which are exclusively used for energy storage and have no role in hormone synthesis. d. Omega fatty acids, such as those derived from linoleate, are only used to produce triglycerides, not hormones. e. The essential fatty acids linoleate and linolenate are present in omega-enriched foods and are important precursors for synthesizing other unsaturated fatty acids and certain hormones.

ANSWER: e 55. Which of the following statements describes the regulation of acetyl CoA carboxylase 1 (ACC1)? a. AMP-activated protein kinase (AMPK) inactivates ACC1 by dephosphorylating it, while protein phosphatase 2A (PP2A) has no role in regulating ACC1. b. AMP-activated protein kinase (AMPK) activates ACC1 by dephosphorylating it, while protein phosphatase 2A (PP2A) inactivates ACC1. c. AMP-activated protein kinase (AMPK) inactivates ACC1 by dephosphorylating it, while protein phosphatase 2A (PP2A) activates ACC1 by phosphorylating it. d. AMP-activated protein kinase (AMPK) and protein phosphatase 2A (PP2A) are not involved in the regulation of acetyl CoA carboxylase 1 (ACC1). e. AMP-activated protein kinase (AMPK) inactivates ACC1 by phosphorylating it, while protein phosphatase 2A (PP2A) activates ACC1 by dephosphorylating it.

ANSWER: e 56. Which of the following statements describes the effect of ethanol on gene transcription and liver metabolism? a. Ethanol stimulates the synthesis of the transcription factor SREBP, which increases the expression of genes promoting fatty acid synthesis. Acetaldehyde, on the other hand, inactivates the transcription factor PPAR α, reducing fatty acid oxidation. b. Ethanol stimulates the synthesis of PPAR α, which promotes fatty acid oxidation, while acetaldehyde inhibits SREBP, reducing fatty acid synthesis. c. Ethanol decreases SREBP levels, reducing fatty acid synthesis, while acetaldehyde activates PPAR α, promoting fatty acid oxidation. d. Ethanol inhibits SREBP synthesis and activates PPAR α, leading to a balanced regulation of fatty acid synthesis and oxidation. e. Ethanol has no effect on SREB and PPAR α, and therefore, fatty acid synthesis and oxidation are unaltered.

ANSWER: a 57. Insulin _____ fatty acid synthesis by _____ ACC1, whereas glucagon and epinephrine have the _____ effect.

ANSWER: stimulates, activating, reverse .. Powered by Cognero.

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Chapter 28 58. In how many stages does fatty acid synthesis take place? a. 4 b. 2 c. 3 d. 6 e. 1 ANSWER: c 59. Which of the following statements is FALSE regarding the metabolism of excess ethanol and its impact on cellular function? a. Excess acetyl CoA from ethanol metabolism is primarily used for synthesizing fatty acids. b. Acetaldehyde, a byproduct of ethanol metabolism, is a non-reactive compound that enhances protein function. c. Acetaldehyde modifies proteins, impairing their function and contributing to cellular damage. d. Acetaldehyde inactivates the transcription factor PPAR α, which normally stimulates genes involved in fatty acid oxidation. e. Ethanol is metabolized to acetyl CoA, which can contribute to ketosis.

ANSWER: b 60. Which of the following statements is FALSE regarding the effect of ethanol metabolism on cellular metabolism? a. The citric acid cycle is impaired due to increased NADH and decreased NAD+ availability, which inhibits critical citric acid cycle enzymes. b. Gluconeogenesis is inhibited due to a shortage of pyruvate caused by the shift in lactate dehydrogenase activity induced by a high NADH concentration. c. Glycolysis is slowed due to inhibited GAPDH activity, which requires NAD+ to function. d. Ethanol affects fatty acid metabolism by increasing lipogenesis and reducing fatty acid oxidation. e. Fatty acid metabolism is unaffected by ethanol consumption, with normal rates of lipogenesis and fatty acid oxidation.

ANSWER: e 61. A patient arrives at the emergency room with hypoxia, lactic acidosis, and hypoglycemia, likely due to inhibition of gluconeogenesis. What potential diagnosis does the medical intern consider? a. type 1 diabetes b. pancreatitis c. hyperthyroidism d. kidney disease e. chronic alcohol consumption ANSWER: e 62. Chronic alcohol consumption leads to an _____ in the NADH/NAD+ ratio, which promotes the conversion of pyruvate to lactate, resulting in a buildup of lactic acid. .. Powered by Cognero.

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Chapter 28 ANSWER: increase 63. Which of the following statements explains the metabolic consequences of chronic alcohol consumption on the NADH/NAD+ ratio and its relation to hypoglycemia and lactic acidosis? a. Chronic alcohol consumption increases the NADH/NAD+ ratio, promoting gluconeogenesis and reducing lactic acid production. b. Chronic alcohol consumption increases the NADH/NAD+ ratio, which inhibits pyruvate dehydrogenase, causing pyruvate to be converted to lactate and leading to lactic acidosis. It also inhibits gluconeogenesis, contributing to hypoglycemia. c. Chronic alcohol consumption decreases the NADH/NAD+ ratio, which activates pyruvate dehydrogenase, causing pyruvate to be converted to lactate and leading to lactic acidosis. It also activates gluconeogenesis, contributing to hypoglycemia. d. Chronic alcohol consumption leads to a decrease in NADH, causing an increase in gluconeogenesis, lactate clearance, and prevention of hypoglycemia. e. Hypoxia resulting from the increase in oxidative phosphorylation required to process the excess NADH generated after alcohol consumption improves lactate clearance and prevents the buildup of lactic acid, reducing the risk of hypoglycemia.

ANSWER: b 64. A biochemist grows cell cultures in the presence of 14C-palmitate at different concentrations of Firsocostat (also known as ND-630). After 6 hours of exposure to Firsocostat, the biochemist measures the amount of 14CO2 produced and obtains the following plot. What is the biochemist likely trying to verify with this experiment?

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Chapter 28

a. the effect of Firsocostat on cell growth and proliferation b. the effect of Firsocostat on fatty acid synthesis, shown by measuring CO2 production c. the role of Firsocostat in protein synthesis d. the effect of Firsocostat on glucose metabolism e. the effect of Firsocostat on fatty acid oxidation, shown by measuring CO2 production ANSWER: e 65. Which statement differentiates γ-hydroxybutyric acid (GHB) and β-hydroxybutyric acid (BHB)? a. GHB and BHB are both ketone bodies produced during the breakdown of fat, without major differences in function. b. BHB acts as a neurotransmitter in the brain, while GHB is primarily used as an energy source during ketogenesis. c. BHB is a psychoactive compound that affects the central nervous system, while GHB is a ketone body involved in energy metabolism. d. GHB is synthesized in the liver, while BHB is generated in the kidneys. e. BHB is a ketone body involved in energy metabolism, while GHB is a psychoactive compound that affects the central nervous system.

ANSWER: e 66. _____ and mammalian-induced gene 12 (MIG12) work together to regulate acetyl CoA carboxylase 1 .. Powered by Cognero.

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Chapter 28 (ACC1) and thus contribute to fatty acid metabolism regulation. This coordination ensures that fatty acid synthesis is upregulated when energy (in the form of acetyl CoA and citrate) is abundant.

ANSWER: Citrate 67. Citrate and MIG12 promote the polymerization of ACC1 dimers into active filaments for fatty acid synthesis. _____ does so by acting as an allosteric activator, while _____ acts as a cofactor, directly enhancing ACC1 polymerization and activity.

ANSWER: Citrate, MIG12

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Chapter 29 1. The storage form of fatty acids is: a. cholesterol. b. phospholipid. c. triacylglycerol. d. sphingolipid. e. diacylglycerol.

ANSWER: c 2. Where are triacylglycerols synthesized? a. ER of the liver cells b. bloodstream from low-density lipoproteins c. cells of adipose tissue d. bloodstream from lipoprotein particles e. cytoplasm of the liver cells ANSWER: a 3. What is the last common precursor for the synthesis of triacylglycerols and phospholipids? a. dihydroxyacetone phosphate b. diacylglycerol c. cytidine monophosphate d. ethanolamine e. phosphatidate ANSWER: e 4. Which is NOT a phospholipid? a. cardiolipin b. phosphatidylinositol c. phosphatidylglycerol d. cerebroside e. phosphatidylethanolamine ANSWER: d 5. CDP-diacylglycerol is NOT a precursor of: a. cardiolipin. b. phosphatidylinositol. c. triacylglycerol. d. CMP. e. diphosphatidylglycerol.

ANSWER: c 6. Which lipids can be synthesized from cerebroside by the addition of activated sugars? .. Powered by Cognero.

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Chapter 29 a. gangliosides b. sphingomyelin c. ceramide d. cardiolipin e. phosphatidate ANSWER: a 7. The committed step in cholesterol formation is the synthesis of: a. HMG CoA. b. 3-isopentenyl pyrophosphate. c. squalene. d. isoprene. e. mevalonate.

ANSWER: e 8. Select all that apply. Hydroxylation of cholesterol by cytochrome P450 requires: a. NADPH. b. activated O2. c. PLP. d. NADH. e. HMG CoA. ANSWER: a, b 9. What CANNOT repress the expression of genes of the cholesterol biosynthetic pathway? a. binding of SREBP by SCAP b. high cholesterol level c. inhibited transport of vesicles from the endoplasmic reticulum to the Golgi complex d. binding of SRE by SCAP e. transport of SCAP-SREBP to the Golgi apparatus ANSWER: e 10. When energy charge is: a. high, the synthesis of cholesterol is stimulated by phosphorylation. b. high, the synthesis of cholesterol is inhibited. c. low, the synthesis of cholesterol is inhibited. d. low, the synthesis of cholesterol is stimulated by phosphorylation. e. low, dephosphorylation of 3-hydroxy-3-methylglutaryl CoA reductase is stimulated.

ANSWER: c 11. A prominent means to reduce the risk of atherosclerosis is stimulation of: .. Powered by Cognero.

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Chapter 29 a. reverse transport of cholesterol. b. formation of cholesterol-bound HDL. c. direct transport of cholesterol. d. receptor-mediated endocytosis. e. formation of cholesterol-bound LDL. ANSWER: a 12. What serves as an intermediate for the formation of phosphatidate? a. CDP-diacylglycerol b. lysophosphatidate c. diacylglycerol d. glycerol 3-phosphate e. acyl CoA ANSWER: b 13. What lipoproteins are used to transport lipids from the liver to the blood? a. chylomicrons b. very low-density lipoproteins c. intermediate-density lipoproteins d. low-density lipoproteins e. high-density lipoproteins ANSWER: b 14. Which statement about phospholipid synthesis is TRUE? a. It usually takes place in microsomes of the liver cells. b. Phosphatidylethanolamine is synthesized from phosphatidate and ethanolamine. c. Hydrolysis of pyrophosphate can drive the activation of both diacylglycerol and alcohol. d. Phosphatidylglycerol can react with inositol to yield cardiolipin. e. Only the formation of CDP-diacylglycerol is driven by the hydrolysis of pyrophosphate.

ANSWER: c 15. What lipoproteins are used to transport lipids from blood or tissues to the liver? Select all that apply. a. chylomicrons b. very low-density lipoproteins c. intermediate-density lipoproteins d. low-density lipoproteins e. high-density lipoproteins ANSWER: c, d, e 16. What is a TRUE cellular sensor of cholesterol? a. SREBP .. Powered by Cognero.

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Chapter 29 b. SCAP c. SRE d. LDL e. HDL ANSWER: b 17. A disrupted formation of phosphorylcholine would impair the synthesis of: a. sphingolipids. b. sphingomyelin. c. phosphatidylcholine. d. cerebroside. e. ganglioside.

ANSWER: b 18. Select all that apply. The deleterious effects of excess ethanol consumption are due to: a. overconsumption of NADH. b. overproduction of NADH. c. generation of free radicals. d. overconsumption of NADPH. e. overproduction of NADPH. ANSWER: b, c, d 19. Which statement about bile is FALSE? a. Polar and nonpolar regions of cholesterol make bile a good detergent. b. Glycocholate is the major bile salt. c. Excess cholesterol in the bile leads to the formation of gallbladder stones. d. The addition of hydroxyl groups and glycine residues to cholesterol leads to the formation of glycocholate. e. Bile salts solubilize dietary lipids upon their release into the small intestine.

ANSWER: a 20. _____ is NOT formed from cholesterol. a. Xanthoma b. Pregnenolone c. Vitamin D d. Limonene e. Glucocorticoids ANSWER: d 21. What condition is caused by vitamin D deficiency? a. atherosclerosis .. Powered by Cognero.

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Chapter 29 b. osteomalacia c. cholelithiasis d. Niemann–Pick disease e. Tangier disease ANSWER: b 22. What functional group differs between phosphatidate and diacylglycerol? a. phosphate b. pyrophosphate c. CDP d. CMP e. acyl ANSWER: a 23. The increase in the concentration of DAG is due to: a. inhibited activity of phosphatidic acid phosphatase. b. stimulated activity of diacylglycerol kinase. c. stimulated activity of phosphatidic acid phosphatase. d. hydrolysis of pyrophosphate. e. hydrolysis of CTP. ANSWER: c 24. The increase in the concentration of phosphatidylinositol is due to: a. inhibited activity of phosphatidic acid phosphatase. b. inhibited activity of diacylglycerol kinase. c. stimulated activity of phosphatidic acid phosphatase. d. hydrolysis of pyrophosphate. e. hydrolysis of CTP. ANSWER: a 25. In lipid metabolism, acetyl CoA is able to convert to a fatty acid and: a. all phospholipids. b. phosphatidylcholine. c. cholesterol. d. ceramide. e. all sphingolipids.

ANSWER: c 26. What reaction activates squalene prior to its cyclization to lanosterol? a. protonation of carbocation b. protonation of epoxide to form a carbocation .. Powered by Cognero.

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Chapter 29 c. oxidation of the C-2 and C-3 atoms d. esterification e. oxidation of the C-30 atom ANSWER: c 27. Which statement about steroids is TRUE? a. Cytochrome P450 monooxygenase introduces oxygen atoms into calcitriol. b. Cholesterol is excreted from the organism only in the form of bile salts. c. All classes of steroid hormones have a different number of carbon atoms. d. Cytochrome c oxidase requires cholesterol for full activity. e. All steroids are usually hydrophilic, so they can be transported without the help of lipoprotein particles.

ANSWER: a 28. What inhibits the translation of HMG-CoA reductase mRNA? Select all that apply. a. binding of SREPB to the sterol regulatory element b. dietary cholesterol c. phosphorylation of SREPB d. metabolites of mevalonate e. endogenous cholesterol ANSWER: b, d 29. The similarity of phospholipids and sphingolipids is the: a. requirement of activated diacylglycerol for their synthesis. b. synthesis from triacylglycerols. c. requirement of phosphatidate for their synthesis. d. availability of two fatty acid residues in their structure. e. availability of diacylglycerol in their structure.

ANSWER: d 30. Which statement about sphingolipids is TRUE? a. Condensation of palmitoyl CoA and serine followed by oxidation, acylation, and reduction yields ceramide. b. All three classes of sphingolipids contain carbohydrate substituents. c. Two hydroxyl groups of ceramide serve as sites for different substituents. d. Condensation of palmitoyl CoA and serine followed by reduction, acylation, and oxidation yields ceramide. e. Only sphingomyelin and cerebroside contain carbohydrate substituents.

ANSWER: d 31. How many isoprene units does squalene contain? .. Powered by Cognero.

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Chapter 29 a. 1 b. 3 c. 5 d. 6 e. 30 ANSWER: d 32. How does the number of carbon atoms change upon synthesis of cholesterol starting from mevalonate? a. C6 → C5 → C10 → C15 → C30 b. C5 → C10 → C15 → C30 c. C5 → C6 → C10 → C15 → C30 d. C5 → C10 → C15 → C30 → C27 e. C6 → C5 → C10 → C15 → C30 → C27 ANSWER: e 33. Hydrolysis of ATP is required for the: a. formation of phosphatidate from GAP. b. activation of diacylglycerol. c. activation of ethanolamine. d. activation of inositol. e. formation of ceramide.

ANSWER: c 34. Choose the CORRECT statement concerning receptor-mediated endocytosis in cholesterol metabolism. a. Apoprotein A binds a specific receptor localized in coated pits on the cell membranes of the peripheral tissues. b. The acidic environment of the endosome induces the specific receptor to relinquish the lipoprotein particle. c. The released unesterified cholesterol is either used for membrane biosynthesis or stored in the cell. d. Unesterified cholesterol is directed to the Golgi complex to be used for membrane biosynthesis. e. Apoprotein B-100 binds an HDL receptor localized in coated pits on the cell membranes of the liver.

ANSWER: d 35. What molecules enhance the activity of phosphatidic acid phosphatase (PAP)? Select all that apply. a. CDP-diacylglycerol b. sphingosine c. phosphatidylinositol d. dihydrosphingosine e. cardiolipin ANSWER: a, c, e .. Powered by Cognero.

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Chapter 29 36. Select all that apply. The synthesis of triacylglycerol will increase in the case of: a. increased concentration of dihydroxyacetone phosphate. b. lipin 1 stimulation by cardiolipin. c. proteolytic release of the SREBP. d. hydrolysis of the phosphoester bond of phosphatidate. e. phosphorylation of diacylglycerol.

ANSWER: a, b, d 37. The result of the knockdown of the enzyme catalyzing the committed step in cholesterol formation is: a. condensation of acetyl CoA with mevalonate. b. dephosphorylation of phosphatidate. c. phosphorylation of mevalonate. d. reduction of 3-ketosphinganine. e. hydroxylation of mevalonate.

ANSWER: c 38. In previous chapters, biosynthetic pathways were commonly seen to be regulated by phosphorylation– dephosphorylation mechanisms. Which mechanism for steroid biosynthesis uses an alternative method? a. The sterol regulatory element binding protein (SREBP) migrates to the nucleus and binds the sterol regulatory element (SRE), which stimulates transcription. b. SREBP can be inactivated by the cleavage from the endoplasmic reticulum membrane; thus, it can no longer block steroid synthesis. c. CTP reacts with acetoacetate to provide an activated substrate to start cholesterol synthesis. d. CTP reacts with acetate to provide an activated substrate to start cholesterol synthesis. e. CTP is required to provide the phosphate for the conversion of mevalonate to 5-phosphomevalonate, unlike other pathways that use ATP. ANSWER: a 39. Select all that apply. The role of lipoprotein particles is to: a. solubilize hydrophobic lipids. b. aid in clot formation. c. contain cell-targeting signals. d. reduce the amount of cholesterol in the blood. e. control phosphorous metabolism.

ANSWER: a, c 40. Feedback regulation of cholesterol synthesis is mainly controlled at the step catalyzed by the enzyme: a. 3-hydroxy-3-methylglutaryl (HMG) CoA reductase. b. geranyl transferase. c. HMG transferase. .. Powered by Cognero.

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Chapter 29 d. farnesyl reductase. e. HMG-mutase.

ANSWER: a 41. Tay–Sachs disease is caused by an inability to degrade: a. sphingosine. b. gangliosides. c. dipalmitoylphosphatidylcholine. d. ceramide. e. acetylcholine.

ANSWER: b 42. HMG CoA is synthesized from acetyl CoA and: a. oxaloacetate. b. acetoacetyl CoA. c. acetyl CoA. d. farnesyl pyrophosphate. e. mevalonate.

ANSWER: b 43. A(n) _______ is composed of a fatty acid linked to sphingosine via an amide bond.

ANSWER: ceramide 44. Respiratory distress syndrome is caused by a failure in the biosynthetic pathway of: a. sphingosine. b. gangliosides. c. dipalmitoylphosphatidylcholine. d. cholesterol. e. bile salts.

ANSWER: c 45. The major carrier(s) of dietary fat from the intestine is/are: a. VLDL. b. chylomicrons. c. HDL. d. LDL. e. IDL. ANSWER: b 46. The LDL receptors on the plasma membrane are localized in coated pits that contain the receptors of which of the following proteins? .. Powered by Cognero.

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Chapter 29 a. clathrin b. lipoprotein A c. adrenodoxin d. HMG-CoA reductase e. sterol regulatory element binding protein (SREBP) ANSWER: a 47. If you were considering designing a drug to regulate blood pressure, which of the following families of steroids would you target? a. androgens b. estrogens c. glucocorticoids d. mineralocorticoids e. progestogens ANSWER: d 48. Your grandfather has been told he has high cholesterol and has been given one of the statins to bring it down. He asks if you can think of any reason that he should not double the dose so that it will reduce his blood cholesterol faster. What do you tell him? a. Yes, statins block the reabsorption of bile, effectively blocking the absorption of cholesterol. b. Yes, statins decrease the number of LDL receptors on the surface of intestinal mucosal cells, thereby decreasing cholesterol uptake into the blood. c. Yes, statins increase the number of LDL receptors on the surface of blood vessels, increasing the internalization of cholesterol into cells and clearing blood. d. No, at high levels, statins have a stimulatory effect, rather than inhibitory effect, on HMG-CoA reductase. e. No, some cholesterol synthesis is essential. Cholesterol is crucial for the proper functioning of cell membranes and serves as precursors for bile salts and steroid hormones.

ANSWER: e 49. A friend, who is struggling with weight gain, reads about a new miracle supplement that blocks the formation of phosphatidate, the precursor for triacylglycerols. Would taking this supplement work as a weight reduction scheme? a. Yes, many weight loss drugs target enzyme-catalyzed reactions that occur early in this biosynthetic pathway. b. Yes, a decrease in the amount of phosphatidate leads to activation of key β-oxidation enzymes, which drive lipid metabolism over glucose metabolism. c. Yes, a decrease in the amount of phosphatidate leads to an increase in cellular glycerol 3-phosphate, which stimulates glycolytic enzymes. d. No, because this would also block formation of phospholipids, and cell membrane integrity would be impacted. e. No, triacylglycerol synthesis takes place in the mitochondria, and it is unlikely that there is a .. Powered by Cognero.

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Chapter 29 transporter for the supplement in the inner membrane.

ANSWER: d 50. You are prone to sunburn, but your doctor tells you not to apply your sunscreen until you have been outside for five minutes. Why? a. Sunscreen blocks UV radiation necessary for cholesterol synthesis in the skin, where it is converted to vitamin D. b. Sunlight is necessary for the conversion of cholesterol to vitamin D, a hormone necessary for bone health. c. Skin pigmentation, which occurs when you get a light tan, blocks UV radiation and reduces the incidence of skin cancers. d. Vitamin D in the skin blocks UV radiation, but only for a short period of time; thus, it is a natural sun block. e. Sunlight promotes HDL mobilization and clearing of cholesterol in blood.

ANSWER: b 51. How does cholera toxin gain access to cells and contribute to the development of cholera? a. Cholera toxin binds phospholipids in the plasma membrane, causing cell lysis and diarrhea. b. Cholera toxin directly activates the immune system, triggering an inflammatory response and constipation. c. Cholera toxin binds carbohydrates in the plasma membrane, blocking water absorption in the intestine and leading to diarrhea. d. Cholera toxin diffuses through the plasma membrane, allowing the development of cholera. e. Cholera toxin binds to gangliosides in the plasma membrane, leading to toxin internalization and subsequent disruption of signaling pathways.

ANSWER: e 52. Which of the following statements about β-N-acetylhexosaminidase and its role in Tay-Sachs disease are FALSE? Select all that apply. a. β-N-Acetylhexosaminidase deficiency causes Tay-Sachs disease and leads to the accumulation of GM2 gangliosides in the brain. b. An activating mutation in β-N-acetylhexosaminidase leads to the accumulation of GM2 gangliosides in the brain, resulting in Tay-Sachs disease. c. β-N-Acetylhexosaminidase is responsible for breaking down GM2 gangliosides in neurons. d. β-N-Acetylhexosaminidase is an enzyme involved in the degradation of glycolipids and glycoproteins but does not affect ganglioside metabolism. e. β-N-Acetylhexosaminidase is required for the breakdown of specific glycosphingolipids, with deficiencies leading to Tay-Sachs disease.

ANSWER: b, d 53. Tay-Sachs and Sandhoff diseases are rare genetic disorders that affect the breakdown of lipids in the brain, leading to severe neurological symptoms. While the two diseases share similar clinical features, they differ in the enzymes involved in the process. Which of the following statements correctly compares Sandhoff and Tay.. Powered by Cognero.

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Chapter 29 Sachs diseases? a. Tay-Sachs disease causes GM2 ganglioside accumulation in the brain, while Sandhoff disease only affects the breakdown of other glycosphingolipids, not GM2 gangliosides. b. Both diseases are caused by a deficiency in hexosaminidase A, leading to the accumulation of GM2 gangliosides in the brain. c. Each disease has distinct clinical features due to different enzyme deficiencies, making them easily distinguishable from one another. d. Both diseases are caused by a deficiency in hexosaminidase B, leading to a deficiency of GM2 gangliosides in the brain. e. Tay-Sachs disease is caused by a deficiency in hexosaminidase A, while Sandhoff disease results from a deficiency in both hexosaminidase A and B. ANSWER: e 54. A traveler presents to the emergency room with symptoms of diarrhea, abdominal cramps, and nausea. The patient recently returned from a trip abroad. Laboratory testing revealed that her symptoms are caused by enterotoxigenic Escherichia coli, which produces a toxin that binds gangliosides on the surface of intestinal cells. Which of the following describes the mechanism by which these bacteria can cause diarrhea? a. The toxin binds to enzymes located at the intestinal brush border, disrupting carbohydrate absorption and causing osmotic diarrhea. b. The toxin stimulates the release of prostaglandins, increasing intestinal motility and fluid secretion. c. The toxin inhibits protein synthesis in intestinal cells, leading to cell death and leakage of cell contents into the stool. d. The toxin stimulates insulin release, which increases fluid secretion. e. The toxin increases cyclic AMP (cAMP) levels, leading to the secretion of chloride and water into the lumen of the intestine.

ANSWER: e 55. Which of the following statements about phosphatidylcholine synthesis is TRUE? a. Phosphatidylethanolamine can be demethylated to form phosphatidylcholine. b. In mammals, phosphatidylcholine is only synthesized through the methylation of phosphatidylethanolamine using S-adenosylmethionine as the methyl donor. c. In mammals, phosphatidylcholine is synthesized exclusively from dietary choline. d. In mammals, phosphatidylcholine cannot be synthesized through a pathway that utilizes dietary choline. e. In mammals, phosphatidylcholine can be synthesized both by methylation of phosphatidylethanolamine using S-adenosylmethionine and through a pathway that utilizes dietary choline.

ANSWER: e 56. Which statement about sphingomyelin is FALSE? a. The addition of phosphatidylcholine to ceramide results in the formation of sphingomyelin and diacylglycerol. b. Sphingomyelin is a component of the myelin sheath covering many nerve fibers. .. Powered by Cognero.

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Chapter 29 c. Sphingomyelin is synthesized by the addition of a glucose molecule to ceramide. d. Sphingomyelin plays an important role in maintaining the structural integrity of cell membranes. e. Sphingomyelin is a common type of sphingolipid.

ANSWER: c 57. The biosynthesis of ceramide involves the following steps. Select all that apply. a. condensation b. oxidation c. acylation d. reduction e. denaturation ANSWER: a, b, c, d 58. Which of the following steps are involved in the synthesis of ceramide from palmitoyl CoA? Select all that apply. a. Palmitoyl CoA is condensed with serine to form 3- ketosphinganine. b. 3-Ketosphinganine is oxidated to form ceramide. c. Dihydrosphingosine is acylated to form dihydroceramide. d. 3-Ketosphinganine is reduced to dihydrosphingosine. e. Dihydroceramide is desaturated to form ceramide.

ANSWER: a, c, d, e 59. Which of the following statements about the conversion of cholesterol to bile salts, such as glycocholate, is TRUE? a. Hydroxyl groups are added to cholesterol, making it more hydrophobic. b. Glycocholate is formed by the addition of hydroxyl groups to cholesterol and then conjugating it with glucose. c. Glycocholate is formed when cholesterol is conjugated with glycine but does not undergo any hydroxylation. d. Glycocholate is formed when cholesterol is conjugated with phosphatidylcholine. e. The addition of hydroxyl groups to cholesterol increases its water solubility, which is important for its role in lipid emulsification.

ANSWER: e 60. The _____ form of isoprene can be polymerized to form natural rubber.

ANSWER: cis 61. How many isoprene units are present in the fragrant compounds myrcene in bay leaves and limonene in lemon oil? a. 0 b. 1 c. 2 .. Powered by Cognero.

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Chapter 29 d. 3 e. 4 ANSWER: c 62. Geraniol (from geraniums), menthol (from peppermint oil), and citronellal (from citronella oil) are known for their aromatic properties. Which of the following statements correctly compares these compounds based on their functional groups? a. Geraniol and menthol are aldehydes, while citronellal is an alcohol. b. Geraniol is an alcohol, menthol is an aldehyde, and citronellal is a ketone. c. Geraniol and menthol are alcohols, while citronellal is a ketone. d. Geraniol is an aldehyde, menthol is an alcohol, and citronellal is a ketone. e. Geraniol and menthol are alcohols, while citronellal is an aldehyde.

ANSWER: e 63. β-Carotene is an important pigment in many fruits and vegetables. Which of the following statements describe the properties and function of β-carotene? Select all that apply. a. β-Carotene contains extended networks of conjugated single and double bonds, which allow it to absorb light. b. β-Carotene is a protein that aids in the absorption of light in plants. c. β-Carotene is primarily used by plants to absorb light, particularly green and red wavelengths, for photosynthesis. d. β-Carotene serves as a light-harvesting molecule in photosynthetic assemblies. e. The conjugated polyene structure in β-carotene contributes to its bright orange color.

ANSWER: a, d, e 64. A ___________ is a compound that has a chain of conjugated single and double bonds, which allows for unique interactions with light and provides other chemical properties.

ANSWER: polyene 65. _______ is an _______ with a variety of practical applications that include its use for sealants and adhesives, solvents and lubricants, and flavors and fragrances.

ANSWER: Farnesene, isoprenoid 66. Coenzyme Q10 is a naturally occurring antioxidant and a crucial component of the energy-producing processes in mitochondria. How many isoprene units are present in this compound? a. 0 b. 1 c. 2 d. 5 e. 10 ANSWER: e .. Powered by Cognero.

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Chapter 29 67. Which of the following statements regarding farnesyl and its role in protein targeting are FALSE? Select all that apply. a. Farnesyl contains three isoprene units in its structure. b. Proteins are targeted to membranes by the covalent attachment of a modified farnesyl unit to their carboxyl-terminal cysteine residue. c. Farnesylation involves the attachment of a farnesyl group to the amino terminus of proteins, promoting their interaction with cellular membranes. d. Farnesyl is a monoterpene that consists of two isoprene units. e. The attachment of a farnesyl group to proteins aids in their membrane anchoring and functional localization.

ANSWER: c, d 68. Which of the following statements regarding cholesterol biosynthesis from lanosterol are FALSE? Select all that apply. a. Lanosterol (C30) undergoes modifications, including the removal of three carbon atoms, to form cholesterol (C27). b. Lanosterol is an intermediate molecule in the cholesterol biosynthesis pathway and is formed from squalene. c. Lanosterol is produced from cholesterol as a precursor and then undergoes further modifications to produce squalene. d. The conversion of lanosterol to cholesterol is part of the sterol biosynthetic pathway, which takes place primarily in the cytosol and endoplasmic reticulum of cells. e. Cholesterol is synthesized in mitochondria from lanosterol through a complex series of reactions.

ANSWER: c, e 69. Which of the following is NOT a major class of steroid hormones? a. thyroid hormones b. progestagens c. glucocorticoids d. mineralocorticoids e. androgens f. estrogens ANSWER: a 70. Which statement about steroid hormones is FALSE? a. Progesterone prepares the lining of the uterus for the implantation of an ovum and is essential for the maintenance of pregnancy. b. Estradiol, along with progesterone, participates in the ovarian cycle. c. Glucocorticoids (e.g., cortisol) promote gluconeogenesis, promote glycogen formation, enhance fat and protein breakdown, and inhibit inflammation. d. Mineralocorticoids (e.g., aldosterone) act on the kidney to regulate salt balance, volume, and blood pressure. .. Powered by Cognero.

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Chapter 29 e. Testosterone is responsible for the development of male secondary sex characteristics, maintaining the testes, and promoting muscle mass. f. Thyroid hormones act as steroid hormones by directly regulating metabolism, growth, development, and overall body function.

ANSWER: f 71. Which statement about the use of steroid hormones in breast and ovarian cancers is FALSE? a. Aromatase inhibitors are commonly used in adjuvant therapy to reduce the risk of recurrence. b. Breast and ovarian cancers depend on estrogen for growth. c. Anastrozole (Arimidex) is a competitive inhibitor of aromatase, decreasing estrogen production. d. Exemestane (Aromasin) covalently modifies and inactivates aromatase, an enzyme responsible for estrogen synthesis. e. Estrogen receptor antagonists, like tamoxifen, are commonly used in the treatment of breast cancer to block estrogen signaling. f. Thyroid hormone antagonists are commonly used in the treatment of breast and ovarian cancers to regulate metabolism, growth, development, and overall body function.

ANSWER: f

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Chapter 30 1. What is the major site of amino acid degradation in mammals? a. kidneys b. stomach c. liver d. spleen e. intestine ANSWER: c 2. What compound is oxidatively deaminated to yield an ammonium ion? a. amino acid b. glutamate c. aspartate d. α-ketoglutarate e. oxaloacetate ANSWER: b 3. What enzymes catalyze the transfer of an α-amino group from an α-amino acid to an α-ketoacid? a. aminotransferases b. dehydrogenases c. synthetases d. transcarbamoylases e. hydroxylases ANSWER: a 4. Which product is regenerated in oxidative deamination of glutamate? a. pyruvate b. α-ketoglutarate c. oxaloacetate d. aspartate e. alanine ANSWER: b 5. What complex is formed on the enzyme when a product is replaced by a substrate before the reaction is complete? a. failed complex b. ineffective complex c. unproductive complex d. worthless complex e. abortive complex ANSWER: e .. Powered by Cognero.

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Chapter 30 6. What can be used as a reducing power by glutamate dehydrogenase? a. NAD+ b. NADPH c. NADH d. either NADH or NADPH e. acetyl CoA ANSWER: d 7. The presence of alanine and aspartate aminotransferase in the blood is an indication of: a. heart disease. b. spleen damage. c. liver damage. d. kidney failure. e. intestinal obstruction.

ANSWER: c 8. What is a normal blood value for aspartate aminotransferase activity? a. 5–30 units/L b. 40–70 units/L c. 40–125 units/L d. 90–120 units/L e. 110–150 units/L ANSWER: c 9. The α-amino groups of which amino acids can be directly deaminated into an ammonium ion? a. alanine and asparagine b. isoleucine and leucine c. arginine and serine d. serine and threonine e. valine and tyrosine ANSWER: d 10. In what principal transport forms are nitrogen transported from muscle to the liver? a. alanine and glutamine b. alanine and leucine c. arginine and threonine d. serine and threonine e. valine and leucine ANSWER: a 11. What is the final product formed after alanine is taken up in the liver and converted into pyruvate? .. Powered by Cognero.

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Chapter 30 a. urea b. glucose c. lactate d. glutamate e. glycogen ANSWER: b 12. What form is an ammonium ion converted into in most terrestrial vertebrates? a. urea b. glucose c. serine d. aspartate e. glycogen ANSWER: a 13. Where does the urea cycle begin? a. endoplasmic reticulum b. cytoplasm c. lysosome d. mitochondria e. cell membrane ANSWER: d 14. What enzyme catalyzes the committed reaction of the urea cycle? a. carbamoyl phosphate synthetase I b. carbamoyl phosphate synthetase II c. carbamoyl phosphate dehydrogenase I d. carboxyphosphate synthetase I e. carboxyphosphate synthetase II ANSWER: a 15. How many molecules of ATP does the synthesis of carbamoyl phosphate require? a. 0 b. 1 c. 2 d. 3 e. 5 ANSWER: c 16. What allosteric regulator is required for carbamoyl phosphate synthetase activity? a. acetyl CoA .. Powered by Cognero.

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Chapter 30 b. glutamate c. acetylglutamate d. N-acetylglutamate e. carboxyphosphate ANSWER: d 17. N-Acetylglutamate synthase is activated by: a. alanine. b. arginine. c. asparagine. d. glutamate. e. serine.

ANSWER: b 18. What amino acid is formed in a reaction catalyzed by ornithine transcarbamoylase? a. glutamine b. threonine c. arginine d. citrulline e. ornithine ANSWER: d 19. In which cellular compartment does citrulline condense with aspartate, the donor of the second amino group of urea? a. endoplasmic reticulum b. cytoplasm c. vacuole d. mitochondria e. cell membrane ANSWER: b 20. Which amino acid is hydrolyzed to generate urea and ornithine in the urea cycle? a. arginine b. alanine c. valine d. leucine e. isoleucine ANSWER: a 21. How much urea is excreted by a human being per year? a. 10 kg .. Powered by Cognero.

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Chapter 30 b. 20 kg c. 30 kg d. 40 kg e. 50 kg ANSWER: a 22. Which urea cycle product is a precursor for glucose synthesis? a. glutamate b. aspartate c. fumarate d. oxaloacetate e. malate ANSWER: c 23. How often do urea cycle disorders occur? a. 10 in 15,000 births b. 1 in 15,000 births c. 1 in 100,000 births d. 1 in 150,000 births e. 100 in 150,000 births ANSWER: b 24. All defects in the urea cycle lead to an elevated level of ammonium ion in blood, which is called: a. hypoammonemia. b. ammonemia. c. hyperammonemia. d. anemia. e. toxaemia.

ANSWER: c 25. What is the third stage of liver damage from excessive ethanol consumption? a. alcoholic hepatitis b. fatty liver c. coma d. death e. cirrhosis ANSWER: e 26. Choose the molecule into which the carbon skeletons of 20 fundamental amino acids are not funneled. a. pyruvate b. fumarate .. Powered by Cognero.

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Chapter 30 c. acetyl CoA d. oxaloacetate e. malate ANSWER: e 27. What are amino acids that are degraded to acetyl CoA or acetoacetyl CoA called? a. acetylated amino acids b. ketogenic amino acids c. glucogenic amino acids d. glycogenic amino acids e. ketone amino acids ANSWER: b 28. What amino acids are solely ketogenic? a. alanine and valine b. valine and leucine c. leucine and lysine d. tryptophan and tyrosine e. tyrosine and leucine ANSWER: c 29. How many amino acids are classified as solely glucogenic? a. 10 b. 11 c. 12 d. 13 e. 14 ANSWER: e 30. What amino acid can be converted into serine by the enzymatic addition of a hydroxymethyl group? a. glycine b. valine c. alanine d. arginine e. leucine ANSWER: a 31. What molecules can aspartate and asparagine be converted into upon their degradation? Select all that apply. a. glutamate b. acetate c. fumarate .. Powered by Cognero.

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Chapter 30 d. oxaloacetate e. malate ANSWER: c, d 32. Deficiency of what vitamin is indicated by methylmalonic acid accumulation? a. B1 b. B2 c. B5 d. B6 e. B12 ANSWER: e 33. What enzyme initiates degradation of threonine? a. ornithine catalase b. ornithine deaminase c. threonine deaminase d. threonine amylase e. threonine catalase ANSWER: c 34. What molecule is the direct product of the adenylation of methionine? a. S-adenosyl methionine b. S-adenosyl homocysteine c. homocysteine d. cysteine e. succinyl CoA ANSWER: a 35. What molecule is dehydrogenated to yield β-methylcrotonyl CoA in the degradation of leucine? a. α-ketoisocaproate b. β-methylglutaconyl c. pyruvate carboxylase d. isovaleryl CoA e. succinyl CoA ANSWER: d 36. What amino acid is formed upon degradation of phenylalanine? a. tyrosine b. valine c. leucine .. Powered by Cognero.

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Chapter 30 d. serine e. threonine ANSWER: a 37. What enzymes catalyze nearly all cleavages of aromatic rings in biological systems? a. catalases b. proteases c. monooxygenases d. dioxygenases e. hydrogenases ANSWER: d 38. How often does phenylketonuria occur? a. 1 in 1000 births b. 1 in 10,000 births c. 1 in 15,000 births d. 1 in 20,000 births e. 1 in 100,000 births ANSWER: b 39. Phenylketonuria is caused by an absence or a deficiency of _______.

ANSWER: phenylalanine hydroxylase 40. Which amino acids are glucogenic? Select all that apply. a. asparagine b. methionine c. valine d. lysine e. serine ANSWER: a, b, c, e 41. Which prosthetic group is used in deamination reactions catalyzed by dehydratases? a. pyridoxal phosphate b. NAD+ c. NADP d. CoA e. FADH2 ANSWER: a 42. Excessive alcohol consumption not only makes you overweight, it can also lead to ammonia poisoning.

How is alcohol metabolism linked to increased levels of blood ammonia? .. Powered by Cognero.

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Chapter 30 a. Excessive alcohol consumption can cause liver damage, the primary site of urea formation. A loss of liver function can lead to increased ammonia in the blood. b. Excessive ethanol can halt the urea cycle at the formation of carbamoyl phosphate. c. Excessive alcohol consumption makes the liver more acidic, driving the equilibrium from ammonium ion to free ammonia. d. Alcohol acts as an allosteric inhibitor of carbamoyl phosphate, reducing the rate of ammonium ion removal. e. Excessive alcohol facilitates the formation of an abortive complex for glutamate.

ANSWER: a 43. In the urea cycle, free NH3 is coupled with carboxyphosphate to form: a. uretic phosphate. b. pyruvate. c. carbamic acid. d. carbamoyl phosphate. e. N-acetylaspartate.

ANSWER: c 44. Which amino acid is a metabolite in the urea cycle but is not used as a building block of proteins? Select all that apply. a. ornithine b. citrulline c. glutamate d. asparagine e. argininosuccinate ANSWER: a, b, e 45. In the urea cycle, the carbon skeleton of aspartate is preserved as: a. succinate. b. fumarate. c. urea. d. malate. e. asparagine.

ANSWER: b 46. Ammoniotelic organisms excrete excess nitrogen as: a. N2H8. b. NH4+. c. urea. d. arginine. .. Powered by Cognero.

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Chapter 30 e. alanine.

ANSWER: b 47. Uricotelic organisms release nitrogen as: a. NH4+. b. arginine. c. uric acid. d. alanine. e. N2H8.

ANSWER: c 48. Which amino acids supply carbons for an eventual entry into metabolism as succinyl CoA? a. methionine, valine, leucine b. methionine, isoleucine, valine c. isoleucine, valine, leucine d. valine, leucine, threonine e. valine, alanine, arginine ANSWER: b 49. Which amino acids can be ketogenic? Select all that apply. a. asparagine b. tyrosine c. valine d. lysine e. threonine ANSWER: b, d, e 50. In the urea cycle, the second nitrogen of urea enters the cycle in the form of: a. alanine. b. glutamine. c. ornithine. d. aspartate. e. arginine.

ANSWER: d 51. Excess amino acids: a. can be excreted in waste. b. can be stored. c. can be used as metabolic fuel. d. can be excreted and stored. .. Powered by Cognero.

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Chapter 30 e. can be stored and used as metabolic fuel.

ANSWER: c 52. The degradation and resynthesis of proteins is called _____.

ANSWER: protein turnover 53. Which statements are TRUE about ubiquitin? Select all that apply. a. It is a large, 486-amino-acid protein. b. It is a small, 76-amino-acid protein. c. It is present in all eukaryotic cells. d. Its primary role is to serve as a tag that marks proteins for destruction. e. Its primary role is to serve as the signal for the start of protein synthesis.

ANSWER: b, c, d 54. How many enzymes participate in the attachment of ubiquitin to a protein?

ANSWER: 3 55. What is the function of ubiquitin-activating enzyme? a. It brings E2 and the target protein together. b. It forms an acyl adenylate at the C-terminal carboxylate of ubiquitin, releasing pyrophosphate, and transfers a sulfhydryl group of a key cysteine residue to ubiquitin. c. It catalyzes the shuttling of activated ubiquitin to a sulfhydryl group of E2. d. It catalyzes the transfer of ubiquitin from E2 to an ε-amino group on the target protein.

ANSWER: b 56. A specific sequence of amino acids that indicates a protein should be degraded is called a _____.

ANSWER: degron 57. When the amino group is removed from an amino acid, what remains is the _____.

ANSWER: carbon skeleton 58. The transport of amino groups and carbon skeletons from muscle to the liver occurs in the _____ cycle.

ANSWER: glucose–alanine 59. How many molecules of ATP are consumed in the formation of 3 molecules of urea? a. 1 b. 3 c. 6 d. 9 ANSWER: d 60. What occurs in stage 1 of liver damage from excessive ethanol consumption? .. Powered by Cognero.

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Chapter 30 a. alcoholic hepatitis develops b. cirrhosis develops c. fatty liver develops d. groups of cells die and inflammation results ANSWER: c 61. In a hibernating bear, _____ is still salvaged to produce urea and passed to the bladder. Urea is then absorbed into the blood and released into the intestine where it is used to synthesize amino acids and _____.

ANSWER: nitrogen, proteins 62. Identify the function of the 26S proteasome. a. It adds ubiquitin to proteins. b. It digests ubiquitinated proteins. c. It makes ubiquitinated proteins. d. It removes ubiquitin from proteins.

ANSWER: b 63. What components are in the 26S proteasome? Select all that apply. a. a 19S catalytic core b. a 19S regulatory unit c. a 20S catalytic core d. a single 26S unit e. a 28S regulatory unit ANSWER: b, c 64. The degradation of leucine yields _____. Select all that apply. a. acetyl coenzyme A b. acetoacetate c. carbon dioxide d. hydrogen monoxide e. succinyl coenzyme A ANSWER: a, b 65. The degradation of isoleucine yields _____. Select all that apply. a. acetyl coenzyme A b. acetoacetate c. carbon dioxide d. hydrogen monoxide e. succinyl coenzyme A ANSWER: a, e .. Powered by Cognero.

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Chapter 30 66. The degradation of valine yields _____. Select all that apply. a. acetyl coenzyme A b. acetoacetate c. carbon dioxide d. hydrogen monoxide e. succinyl coenzyme A ANSWER: c, e 67. The inherited metabolic disorder caused by the absence of homogentisate oxidase is called _____.

ANSWER: alkaptonuria 68. _____ has been used as a marker for disease for centuries.

ANSWER: Urine 69. Pyruvate is the entry point of which amino acids? Select all that apply. a. alanine b. cysteine c. proline d. serine e. threonine ANSWER: a, b, d 70. α-Ketoglutarate is the entry point of which amino acids? Select all that apply. a. arginine b. cysteine c. glutamine d. histidine e. proline ANSWER: a, c, d, e

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Chapter 31 1. Nitrogen fixation is the: a. fixing of gaseous N2 into an unreactive form. b. conversion of gaseous N2 into a biologically useful form of nitrogen (NH3). c. conversion of gaseous NH3 into a biologically useful form of nitrogen (N2). d. reduction of nitrogen to a peptide bond. e. transfer of electrons to ATP. ANSWER: b 2. What is the bond strength in the N2 molecule? a. 225 kJ mol–1 b. 300 kJ mol–1 c. 940 kJ mol–1 d. 240 kJ mol–1 e. 800 kJ mol–1 ANSWER: c 3. Nitrogenase is a protein of the nitrogenase complex that: a. provides electrons with a high reducing power. b. uses electrons to reduce N2 to NH3. c. carries out oxidation of N2 to NH3. d. transfers electrons from ADP to ATP. e. carries out reduction of N2 to NO3.

ANSWER: b 4. Reductase is the Fe protein composed of two: a. polypeptide chains linked by a 4Fe–4S cluster. b. polypeptide chains linked by three disulfide bonds. c. α subunits and two β subunits. d. polypeptide chains linked by a 2Fe–2S cluster.

ANSWER: a 5. What is the site of conversion of N2 to NH3? a. reductase b. bacterial membrane c. glutamine synthetase d. nitrogenase e. ATP receptor ANSWER: d .. Powered by Cognero.

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Chapter 31 6. Why is the biological reduction of N2 to NH3 an eight-electron process, not six? a. The biological reaction generates an additional 1 mol of H2 per each mole of N2 fixed. b. Two electrons are necessary for ATP. c. The biological reaction generates an additional 1 mol of NH4+ per each mole of N2 fixed. d. Two electrons are necessary for the formation of ferredoxin. e. Two electrons are necessary for the reductase.

ANSWER: a 7. What enzyme catalyzes the formation of glutamine? a. glutamate dehydrogenase b. glutamine synthetase c. homocysteine methyltransferase d. 3-phosphoglycerate dehydrogenase e. threonine deaminase ANSWER: b 8. How many biosynthetic families of amino acids are defined? a. 4 b. 3 c. 6 d. 21 e. 2 ANSWER: c 9. Amino acids are divided into biosynthetic families according to the fact that there are: a. six types of important metabolic products b. different families of bacteria which are able to convert N2 gas into NH3. c. six types of important metabolic precursors. d. different ways of converting N2 into NH3. e. three types of important metabolic precursors.

ANSWER: c 10. What is the source of the amino acid glutamine in eukaryotic cells? a. citric acid cycle b. glycolysis c. pentose phosphate pathway d. nitrogen fixation e. glycolysis and pentose phosphate pathway ANSWER: a .. Powered by Cognero.

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Chapter 31 11. Which statements about pyridoxine are TRUE? a. Pyridoxine is a precursor to pyridoxamine phosphate and adenosine monophosphate. b. Pyridoxine is also called vitamin B12. c. A deficiency of pyridoxine results in fatigue, cheilosis, glossitis, and stomatitis. d. The deficiency of pyridoxine is common in the United States. e. Pyridoxine serves as a cofactor in a host of enzymes, including transaminases.

ANSWER: c, e 12. Because their sufficiency is dependent on a particular set of conditions, amino acids such as arginine and tyrosine are sometimes called _______ amino acids.

ANSWER: conditionally essential 13. Select all essential amino acids: a. leucine b. glycine c. alanine d. histidine e. tryptophan ANSWER: a, d, e 14. Nonessential amino acids are those that: a. cannot be synthesized by a human and must be supplied in the diet. b. can be synthesized in the organism if dietary content is insufficient. c. can be synthesized in the organism if glutamine is sufficient. d. are synthesized by bacteria. e. can be biosynthesized in a large number of steps.

ANSWER: b 15. α-Ketoglutarate is converted into: a. alanine by transfer of an NH3 group. b. alanine by a reaction with oxaloacetate. c. glutamate by a reaction with oxaloacetate. d. glutamate that is part of alanine synthesis. e. pyruvate that is part of alanine synthesis.

ANSWER: d 16. What group is bound to both N5 and N10 of tetrahydrofolate in the medium reduced state? a. methyl group b. formyl group .. Powered by Cognero.

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Chapter 31 c. methenyl group d. formimino group e. methylene group ANSWER: e 17. Which statement about folic acid is FALSE? a. A lack of folic acid during prenatal development results in spinal-cord defects. b. Folic acid deficiency can result in scurvy. c. Green vegetables, such as spinach and broccoli, are good sources of folic acid. d. Folic acid is a precursor of tetrahydrofolate. e. Folic acid deficiency can result in anencephaly (lack of a brain).

ANSWER: b 18. Choose the CORRECT sequence of glycine formation. a. oxidation of 3-phosphoglycerate → transamination of 3-phosphohydroxypyruvate → hydrolysis of 3phosphoserine → transfer of the side-chain methylene group of serine to tetrahydrofolate → glycine b. oxidation of 3-phosphoglycerate → oxidation of 3-phosphohydroxypyruvate → hydrolysis of 3phosphoserine → glycine c. transamination of 3-phosphoglycerate → hydrolysis of 3-phosphohydroxypyruvate → hydrolysis of 3-phosphoserine → transfer of the side-chain methylene group of cysteine to tetrahydrofolate → glycine d. transamination of 3-phosphohydroxypyruvate → hydrolysis of 3-phosphoserine → oxidation of 3phosphoglycerate →transfer of the side-chain methylene group of serine to tetrahydrofolate → glycine e. transamination of 3-phosphohydroxypyruvate → oxidation of 3-phosphoglycerate → transfer of the side-chain amino group of serine to tetrahydrofolate → glycine ANSWER: a 19. What prosthetic group do all aminotransferases contain? a. tetrahydrofolate b. cobalamin c. pyridoxamine d. pyridoxal phosphate e. pyridoxine ANSWER: d 20. What coenzyme mediates the transfer of a methyl group upon formation of methionine from homocysteine? a. tetrahydrofolate b. methylcobalamin c. folic acid d. pteridine e. methionine .. Powered by Cognero.

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Chapter 31 ANSWER: b 21. What is the function of cystathionine β-synthase? a. stimulation of homocysteine formation b. transfer of the methyl group c. stimulation of serine formation d. combination of homocysteine and serine to form cystathionine e. combination of glycine and serine to form cystathionine ANSWER: d 22. Select all that apply. Amino acid synthesis is generally regulated by: a. turnover. b. diet. c. feedback inhibition. d. feedback activation. e. enzyme multiplicity.

ANSWER: c, d, e 23. What does enzyme multiplicity mean? a. One enzyme may catalyze the committed steps of different metabolic pathways. b. Multiple enzymes that are either 1) catalytically identical or 2) similar but have different allosteric properties may catalyze the committed step of a metabolic pathway. c. The enzyme that catalyzes one step of the pathway is inhibited by another enzyme of a parallel pathway. d. It refers to the opportunity of enzymes to activate committed steps of different pathways. e. One enzyme activates a metabolic pathway, and the other inhibits the pathway.

ANSWER: b 24. What tripeptide consists of a cysteine residue flanked by a glycine residue and a glutamate residue? a. thyroxine b. serotonin c. glutathione d. histamine e. sphingosine ANSWER: c 25. What enzyme is regulated by cumulative feedback inhibition? a. threonine deaminase b. 3-phosphoglycerate dehydrogenase c. guanine synthase d. glutamine synthetase .. Powered by Cognero.

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Chapter 31 e. glutamine dehydrogenase ANSWER: d 26. What does cumulative feedback inhibition mean? a. Each inhibitor can reduce the activity of an enzyme, even when other inhibitors are exerting their own maximal inhibition. b. Multiple enzymes that are catalytically identical or similar but have different allosteric properties, may catalyze the committed step of a metabolic pathway. c. One enzyme may catalyze the committed step of a different metabolic pathway. d. One enzyme activates the metabolic pathway, and another inhibits the pathway. e. The committed step can be catalyzed by two or more enzymes with different regulatory properties.

ANSWER: a 27. What molecules provide NH3 for incorporation into amino acids? Select all that apply. a. β-ketoglutarate b. serine c. glutamate d. aspartate e. glutamine ANSWER: c, e 28. What enzyme is regulated by feedback inhibition and activation? a. glutamine synthetase b. adenosine monophosphate synthetase c. 3-phosphoglycerate dehydrogenase d. threonine deaminase e. glutamate dehydrogenase ANSWER: d 29. What types of melanin exist? Select all that apply. a. albinism b. black eumelanin c. brown eumelanin d. pheomelanin ANSWER: b, c, d 30. What are large and highly cross-linked polymers of 5,6-dihydroxyindole and 5,6-dihydroxyindole-2carboxylicacid? a. melanin b. pheomelanin c. eumelanin .. Powered by Cognero.

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Chapter 31 d. tyrosine e. glutathione ANSWER: c 31. What inhibits the nitrogenase complex? a. N2 b. O2 c. NH4+ d. ATP e. ADP ANSWER: b 32. What is the minimal number of ATP required for the reduction of 2 molecules of N2? a. 2 b. 4 c. 8 d. 16 e. 32 ANSWER: e 33. An example of a reaction controlled by enzyme multiplicity is phosphorylation of: a. asparagine by aspartokinases. b. aspartate by aspartokinases. c. glutamine by glutamine synthetases. d. ammonia by nitrogenase. e. tyrosine by phenylalanine hydroxylases.

ANSWER: b 34. Select all that apply. ATP is hydrolyzed by: a. MoFe proteins. b. Fe proteins. c. glutamine synthetases. d. methionine synthases. e. 3-phosphoglycerate dehydrogenases.

ANSWER: b, c 35. Which statement about the activated methyl cycle is TRUE? a. Synthesis of SAM relies solely on hydrolysis of pyrophosphate. b. S-adenosylhomocysteine, rather than just homocysteine, is methylated to lower the free energy of methyl transfer. .. Powered by Cognero.

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Chapter 31 c. The most reduced form of the one-carbon unit on tetrahydrofolate is required to methylate homocysteine. d. The high transfer potential of the methyl group from methylcobalamin allows restoration of methionine. e. The ATP hydrolyzed in the synthesis of SAM is restored in the activated methyl cycle in the reaction of methionine synthase.

ANSWER: c 36. In transamination reactions, the amino group is transferred to pyruvate from: a. glutamine. b. pyruvate. c. pyridoxal phosphate. d. pyridoxamine phosphate. e. glutamate.

ANSWER: e 37. What organisms can synthesize essential amino acids? Select all that apply. a. bacteria b. humans c. plants d. mammals e. archaea ANSWER: a, c, e 38. Threonine deaminase is allosterically inhibited by: a. α-ketobutyrate. b. valine. c. isoleucine. d. threonine. e. leucine.

ANSWER: c 39. How much glutamine can be synthesized upon reduction of 1 N2 molecules by the nitrogenase complex, and how much ATP will be required for that? a. 1 glutamine molecule; at least 16 ATP b. 2 glutamine molecules; at least 17 ATP c. 1 glutamine molecule; at least 32 ATP d. 2 glutamine molecules; at least 16 ATP e. 1 glutamine molecule; at least 34 ATP ANSWER: d .. Powered by Cognero.

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Chapter 31 40. This amino acid, in high levels, is correlated with the damage of cells lining blood vessels. a. serine b. cysteine c. S-adenosylmethionine d. citrulline e. homocysteine ANSWER: e 41. The carbon skeletons for amino acids are intermediates found in: a. glycolysis. b. the citric acid cycle. c. the pentose phosphate pathway. d. All of the answers are correct. e. None of the answers are correct.

ANSWER: d 42. Organisms capable of carrying out reduction of atmospheric nitrogen include: a. some bacteria and archaea. b. higher eukaryotic organisms, such as mammals. c. all plants. d. higher eukaryotic organisms, such as birds. e. fungi.

ANSWER: a 43. The activated methyl cycle involves production and use of: a. glutamine. b. glycine. c. heterocysteine. d. homocysteine. e. methanol.

ANSWER: d 44. The making of carbon–carbon bonds requires energy. How does the activated methyl cycle provide energy for methyl group transfer to a wide variety of acceptors? a. The side-chain methylene group of serine is transferred to tetrahydrofolate, a carrier of one-carbon units. b. A fully oxidized one-carbon unit, CO2, is carried by biotin. c. The most reduced form of one-carbon groups carried by tetrahydrofolate is the methyl group. d. Transmethylation reactions are carried out by pyridoxal phosphate–dependent methyltransferases. e. The methyl group of methionine is activated by the transfer of an adenosyl group to the sulfur atom of methionine. .. Powered by Cognero.

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Chapter 31 ANSWER: e 45. S-adenosylmethionine carries: a. methyl groups. b. CO2. c. ammonia. d. carbonyl groups. e. carboxyl groups.

ANSWER: a 46. Insufficient amounts of _____ would not only inhibit protein synthesis but also synthesis of phostidylcholine and phostidylethanolamine. a. glutamine b. histidine c. phenylalanine d. methionine e. alanine ANSWER: d 47. Essential amino acids are synthesized by: a. microorganisms. b. humans. c. plants. d. both microorganisms and humans. e. both microorganisms and plants.

ANSWER: e 48. Through what process might feedback inhibition processes have evolved? a. duplication of genes encoding catalytic domains b. evolution of homologous subunits in the enzyme catalyzing the committed step c. linking of specific regulator domains to catalytic domains d. linking of multiple regulatory domains e. duplication of multiple regulatory domains ANSWER: c 49. Erythrose 4-phosphate is a precursor to the amino acids: a. tryptophan, tyrosine, and phenylalanine. b. tryptophan and phenylalanine. c. tyrosine and phenylalanine. d. tryptophan, tyrosine, phenylalanine, and serine. e. None of the answers are correct. .. Powered by Cognero.

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Chapter 31 ANSWER: a 50. Essential amino acids differ from nonessential amino acids in that: a. nonessential amino acids are synthesized in simple reactions compared to most essential amino acids. b. essential amino acids are generally synthesized directly from the citric acid cycle intermediates, but nonessential amino acids are not. c. microorganisms and animals cannot synthesize essential amino acids, but plants can. d. animals cannot synthesize essential amino acids because they have lost the ability to carry out transamination reactions. e. essential amino acids can be synthesized by the symbiotic bacteria in animals' intestines.

ANSWER: a 51. In transamination reactions, the amino group is transferred to pyridoxal phosphate to form: a. glutamine. b. pyruvate. c. pyridoxal phosphate. d. pyridoxamine phosphate. e. glutamate.

ANSWER: d 52. The nitrogen in amino acids, purines, pyrimidines, and other biomolecules ultimately comes from ________, which makes up about 80% of Earth’s atmosphere.

ANSWER: atmospheric nitrogen N2

53. The amount of N2 fixed by diazotrophic microorganisms has been estimated to be _____ per year. a. 15 kg b. 600 kg c. 940 kg d. 1011 kg e. 2515 kg ANSWER: d 54. In principle, the reduction of N2 to NH3 is a(n) __ electron process. a. 2 b. 4 c. 6 d. 8 e. 12 ANSWER: c 55. To allow ATP synthesis and nitrogenase to function simultaneously, leguminous plants maintain a very low .. Powered by Cognero.

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Chapter 31 concentration of free O2 in their root nodules. This is accomplished by binding O2 to _____________. a. ferredoxin

b. hemoglobin c. leghemoglobin d. nitrogenase e. reductase ANSWER: c 56. What is the role of the reductase component of the nitrogenase complex? a. to transfer electrons from the nitrogenase component to a suitable acceptor b. to transfer electrons from a suitable donor to the nitrogenase component c. to transfer electrons of high oxidation power d. to reduce the inactive N2 gas to NH3 e. to hydrolyze N2 ANSWER: b 57. The nitrogenase component of the nitrogenase complex is also called what? Select all that apply. a. iron protein b. MoFe cofactor c. MoFe protein d. molybdenum–iron protein e. reductase component ANSWER: c, d 58. What is the site of nitrogen fixation? a. ATP receptor b. bacterial membrane c. iron protein d. MoFe cofactor e. reductase protein ANSWER: d 59. Tyrosine requires __ steps to be synthesized from scratch. a. 5 b. 6 c. 8 d. 10 e. 16 ANSWER: d 60. The pathways for the synthesis of essential amino acids require from _____ steps. .. Powered by Cognero.

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Chapter 31 a. 1 to 5 b. 1 to 10 c. 5 to 10 d. 5 to 16 e. 10 to 16 ANSWER: d 61. Which amino acids are formed from 3-phosphoglycerate? Select all that apply. a. alanine b. cysteine c. glutamate d. glycine e. serine ANSWER: b, d, e 62. Tetrahydrofolate is a(n) ____ essential for the synthesis of many amino acids and nucleotides. a. amino acid b. coenzyme c. cofactor d. enzyme e. vitamin ANSWER: b 63. The one-carbon group carried by tetrahydrofolate that is bonded to its nitrogen atom can exist in ____ oxidation states.

a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: c 64. The _____ of synthesis of amino acids depends mainly on the amounts of the biosynthetic enzymes and on their activities.

ANSWER: rate 65. In the regulation of amino acid synthesis, the final product of the pathway often inhibits the enzyme that catalyzes the ________.

ANSWER: committed step 66. In the regulation of the enzyme 3-phosphoglycerate dehydrogenase, the binding of serine to a regulatory site reduces the value of ____ for the enzyme. .. Powered by Cognero.

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Chapter 31 a. kcat b. KM c. ʋ d. Vmax e. catalytic efficiency ANSWER: d 67. Threonine deaminase is allosterically activated by: a. α-ketobutyrate b. isoleucine. c. leucine. d. threonine. e. valine.

ANSWER: e 68. The amide group of glutamine is a source of nitrogen in the biosyntheses of a variety of compounds including: Select all that apply.

a. alanine b. carbamoyl phosphate c. glucosamine 6-phosphate d. glutamic acid e. histidine ANSWER: b, c, e 69. Histamine is derived from _______ by decarboxylation. a. glutamine b. histadine c. serine d. tryptophan e. tyrosine ANSWER: b 70. The neurotransmitter serotonin is synthesized from the amino acid ____________. a. glutamine b. histadine c. serine d. tryptophan e. tyrosine ANSWER: d

.. Powered by Cognero.

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Chapter 32 1. What are the roles of nucleotides? Select all that apply. a. They are activated precursors of the nucleic acid. b. They can serve as the energy currency. c. They contain genes. d. They are signal molecules. e. They can activate biosynthetic metabolites.

ANSWER: a, b, d, e 2. What are the pathway types for nucleotide biosynthesis? Select all that apply. a. anaplerotic b. in vitro c. salvage d. de novo e. allosteric ANSWER: c, d 3. Choose the INCORRECT statement about nucleotides. a. A nucleoside is a phosphate ester of a nucleotide. b. In a nucleoside, as well as in nucleotides, the base is linked to a sugar. c. Deoxyribonucleotides are produced from ribonucleotides. d. Nucleotides contain a base, sugar, and one to three phosphate residues. e. The purine base consists of two heterocyclic rings, whereas the pyrimidine base consists of one such ring.

ANSWER: a 4. What is the complex of enzymes that synthesizes purine nucleotides? a. peroxisome b. endosome c. purinosome d. GFP complex e. purinergic receptor ANSWER: c 5. Choose the CORRECT statement about synthesis of nucleotides. a. Synthesis of the pyrimidine ring begins with attachment to ribose. b. Tetrahydrofolate is required for the synthesis of both purine and pyrimidine nucleotides. c. The purine ring is assembled on the deoxyribose unit. d. Inosinate is important only for the synthesis of purine nucleotides. e. The sugar has to be activated only for the synthesis of the pyrimidine base.

ANSWER: d .. Powered by Cognero.

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Chapter 32 6. What is the compound that can act as a suicide inhibitor? a. thymine b. fluorodeoxyuridylate c. trimethoprim d. aminopterin e. adenylosuccinate ANSWER: b 7. Normally, nucleotide biosynthesis is regulated by: a. suicide inhibition. b. phosphorylation. c. feedback inhibition. d. competitive inhibition. e. purinosomes.

ANSWER: c 8. What metabolite is required for the synthesis of both purine and pyrimidine nucleotides? a. glycine b. tetrahydrofolate c. orotate d. urea e. 5-phosphoribosyl-1-pyrophosphate ANSWER: e 9. What enzyme forms deoxyribonucleotides from ribonucleotides? a. thymidine phosphorylase b. UMP kinase c. nucleoside diphosphate kinase d. ribonucleotide reductase e. AMP kinase ANSWER: d 10. What carriers of electrons are necessary for the transformation of ribonucleotides into deoxyribonucleotides? Select all that apply. a. NADPH b. NADH c. FADH2 d. NAD+ e. thioredoxin ANSWER: a, c, e .. Powered by Cognero.

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Chapter 32 11. Thymidylate is formed from: a. dADP. b. ribonucleoside diphosphate. c. dihydrofolate. d. dUMP. e. dUDP. ANSWER: d 12. What compounds are precursors in the de novo synthesis of the pyrimidine ring? Select all that apply. a. bicarbonate b. aspartic acid c. GTP d. Gly e. ammonia ANSWER: a, b, e 13. What compounds are precursors in the de novo synthesis of the purine ring? Select all that apply. a. bicarbonate b. Gln c. ribose d. Asp e. ammonia ANSWER: b, d, e 14. What is the committed step in purine nucleotide biosynthesis? a. carboxylation of glycinamide ribonucleotide b. formation of phosphoribosylamine c. formation of fumarate d. decarboxylation of orotidylate e. formation of orotate ANSWER: b 15. What is the committed step in pyrimidine nucleotide biosynthesis? a. formation of carbamoylaspartate b. formation of orotate c. phosphorylation of ribose 5-phosphate d. formation of uridylate e. formation of phosphoribosylamine ANSWER: a 16. Choose the enzyme of nucleotide biosynthesis that is unique for DNA synthesis. .. Powered by Cognero.

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Chapter 32 a. thymidylate synthase b. GMP synthetase c. glutamine phosphoribosyl amidotransferase d. UMP kinase e. nucleoside diphosphate kinase ANSWER: a 17. What is the final product of purine catabolism? a. fumarate b. urate c. inosinate d. inosine e. xanthylate ANSWER: b 18. What enzyme is the target for competitive inhibition by methotrexate? a. thymidylate synthase b. thymidylate kinase c. dihydrofolate reductase d. aspartate transcarbamoylase e. nucleoside diphosphate kinase ANSWER: c 19. Choose the reactions of nucleotide biosynthesis pathways that represent reciprocal substrate relations. Select all that apply. a. stimulation of aspartate transcarbamoylase by ATP b. formation of CTP by amination of UTP c. hydrolysis of ATP in the synthesis of GMP d. inhibition of aspartate transcarbamoylase by ATP e. hydrolysis of GTP in the synthesis of ADP ANSWER: a, c 20. What enzymes are sites of feedback inhibition of nucleotide biosynthesis in bacteria? Select all that apply. a. thymidine kinase b. carbamoyl phosphate synthetase c. glutamine phosphoribosyl amidotransferase d. nucleoside diphosphate kinase e. aspartate transcarbamoylase ANSWER: b, c, e 21. Some forms of severe combined immunodeficiency (SCID) result from: .. Powered by Cognero.

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Chapter 32 a. the inability to form inosine by deamination of adenosine. b. bone marrow transplantation. c. reduced production of dATP. d. the inability to remove ribose from ribonucleoside. e. the inability to remove deoxyribose from deoxyribonucleoside.

ANSWER: a 22. What enzyme catalyzes the first step of pyrimidine nucleotides synthesis? a. aspartate transcarbamoylase b. carbamoyl phosphate synthetase II c. dihydroorotase d. carbamoyl phosphate synthetase I e. glutamine phosphoribosyl amidotransferase ANSWER: b 23. What enzymes of nucleotide synthesis are translated as one polypeptide chain? a. CPS II, aspartate transcarbamoylase, and dihydroorotase b. CPS II, glutamine-PRPP amidotransferase, and dihydrofolate reductase c. CPS II, adenine phosphoribosyltransferase, and dihydroorotase d. CPS I, ATCase, and dihydrofolate reductase e. CPS II, adenine phosphoribosyltransferase, and ribonucleotide reductase ANSWER: a 24. What statement about synthesis of deoxyribonucleotide triphosphates is FALSE? a. The first dUMP must be synthesized from UMP to yield thymidine triphosphate (TTP). b. dUDP is synthesized by reduced ribonucleotide reductase. c. Nucleoside diphosphate kinase phosphorylates TDP to yield TTP. d. Specific ADP kinase is required to transform ADP to ATP. e. dCDP can be synthesized if ribonucleotide reductase is reduced by thioredoxin.

ANSWER: d 25. What enzyme of nucleotide biosynthesis is bifunctional? a. PRPP synthetase b. oroate phosphoribosyltransferase c. TMP synthase d. UMP synthase e. orotidylate decarboxylase ANSWER: d 26. The most likely cause of behavioral symptoms of Lesch–Nyhan syndrome is: a. hyperproliferation of dopamine-secreting neurons. .. Powered by Cognero.

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Chapter 32 b. mutation of adenosine deaminase. c. elevated levels of urate in the serum. d. allosteric regulation of ribonucleotide reductase. e. insufficient activation and regulation of G-protein-coupled receptors in neurons.

ANSWER: e 27. Choose the component of the DNA chain. a. thymine b. thymidine c. thymidylate d. thymidine diphosphate e. thymidine triphosphate ANSWER: c 28. What individual reactions are catalyzed by carbamoyl phosphate synthetase II? Select all that apply. a. hydrolysis of carbamoyl phosphate b. cleavage of the amide bond of glutamine c. transfer of the phosphate group from ATP to bicarbonate d. phosphorylation of carbamic acid e. phosphorylation of glutamate ANSWER: b, c, d 29. What statement about salvage pathways of nucleotide synthesis is TRUE? a. Hypoxanthine–guanine phosphoribosyltransferase (HGPRT) recycles guanine and adenine only through the formation of inosinate. b. TMP can be produced from thymidine by degradation. c. HGPRT recognizes only guanine and adenine as substrates, while PRPP is its cosubstrate. d. Phosphorylation of uridine results in formation of uridinylate. e. Thymidylate synthase produces thymidylate from thymine.

ANSWER: d 30. How many allosteric sites are there in ribonucleotide reductase? a. 1 b. 2 c. 4 d. 5 e. 6 ANSWER: c 31. What reaction of nucleotide biosynthesis requires NAD+, rather than NADP, as a cofactor? a. formation of adenylosuccinate .. Powered by Cognero.

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Chapter 32 b. oxidation of hydrated inosinate c. reduction of ribonucleoside diphosphates d. oxidation of xanthine e. reduction of urate ANSWER: b 32. Choose the reaction that is driven by the hydrolysis of pyrophosphate. a. phosphorylation of uridylate b. reduction of ribonucleoside diphosphates c. formation of inosinate from hypoxanthine d. formation of adenylosuccinate e. formation of xanthylate ANSWER: c 33. Suppose you have a cell culture that is capable of de novo synthesis of pyrimidine nucleotides but needs purine nucleotides for survival due to a lack of the required enzymes. What would happen if you supply the culture with excessive amounts of UMP and GMP? a. Excessive GMP would strongly inhibit amidotransferase by feedback inhibition. Excessive UMP would inhibit the formation of ATP. b. Excessive UMP would strongly inhibit ATCase by feedback inhibition. Excessive GMP would inhibit the formation of ATP. c. Excessive GMP would stimulate the synthesis of IMP, but excessive UMP would inhibit its formation. d. Excessive UMP would strongly inhibit ATCase due to feedback inhibition by CTP. Excessive GMP would inhibit the formation of IMP, which could halt cell division. e. Excessive UMP and GMP would completely halt the activity of aspartate transcarbamoylase due to feedback inhibition by GTP. ANSWER: d 34. What statement about synthesis of guanylate is TRUE? a. Two carbonyl groups are present in guanylate: one originates from dehydrogenation by NAD+ and one from bicarbonate. b. Overall, two molecules of glutamine are required: one brings a nitrogen atom within the pyrimidine ring, and one is used to replace carbonyl oxygen by NH2. c. Two molecules of N5, N10-methylenetetrahydrofolate are required to synthesize one guanylate molecule. d. Four molecules of fumarate are formed upon synthesis of two molecules of guanylate. e. Overall, two molecules of aspartate are required: one brings a nitrogen atom within the pyrimidine ring, and one is used to replace carbonyl oxygen with NH2.

ANSWER: b .. Powered by Cognero.

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Chapter 32 35. Choose the CORRECT sequence of metabolites in the synthesis of pyrimidine nucleotides. For the sake of simplicity, some metabolites are omitted. a. carbamoyl phosphate → dihydroorotate → orotate → cytidylate b. carboxyphosphate → carbamoylaspartate → inosinate → uridylate c. carbamoyl phosphate → orotate → dihydroorotate → uridylate d. glutamine → carbamoylaspartate → orotate → uridylate e. carboxyphosphate → orotate → carbamoylaspartate → uridylate ANSWER: a 36. What is the role of glycine in nucleotide biosynthesis? a. It finishes the formation of the pyrimidine ring, bringing two carbon atoms and one nitrogen atom. b. It starts the formation of the imidazole ring by bringing two carbon atoms and one nitrogen atom. c. Its α-amino group allows amination of the carbonyl group of formylglycinamide ribonucleotide. d. It starts the formation of the pyrimidine ring, bringing two nitrogen atoms and one carbon atom. e. It finishes the formation of the imidazole ring by bringing three carbon and three nitrogen atoms.

ANSWER: b 37. Choose the CORRECT statement about regulation of purine nucleotides synthesis. a. The pool of purine nucleotides is balanced by the stimulation of aspartate transcarbamoylase by adenylate. b. Amidotransferase is inhibited by adenylate and guanylate, while PRPP synthase is inhibited by inosinate. c. PRPP synthase is inhibited by adenylate and guanylate, while amidotransferase is inhibited by inosinate. d. The pool of pyrimidine nucleotides is balanced through the reciprocal inhibition of aspartate transcarbamoylase by adenylate. e. The pool of purine nucleotides is balanced by the reciprocal inhibition of pathways branching from inosinate.

ANSWER: e 38. Which statement about the regulation of ribonucleotide reductase is TRUE? a. Binding of dATP to the overall-activity control site enhances formation of deoxyribonucleotides. b. Binding of TTP to the substrate-specificity control site promotes the reduction of GDP and UDP. c. Binding of dATP to the substrate-specificity control site can enhance formation of pyrimidine deoxyribonucleotides. d. Binding of ATP to the overall-activity control site inhibits formation of ribonucleotides. e. Binding of any deoxyribonucleotide to the overall-activity control site inhibits the reduction of ribonucleotides.

ANSWER: c, e 39. How much ATP will be required for the de novo synthesis of one molecule of ATP (starting from PRPP) and for synthesis in the salvage pathway (not accounting for the synthesis of PRPP)? .. Powered by Cognero.

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Chapter 32 a. 5 in the de novo pathway; 1 in the salvage pathway b. 5 in the de novo pathway; 0 in the salvage pathway c. 6 in the de novo pathway; 1 in the salvage pathway d. 6 in the de novo pathway; 0 in the salvage pathway e. 10 in the de novo pathway; 2 in the salvage pathway ANSWER: a 40. Select all that apply. Inosinate can be converted to: a. AMP. b. GMP. c. UMP. d. TMP. e. ADP. ANSWER: a, b 41. Allopurinol is used to treat gout and is an inhibitor of the enzyme a. xanthine oxidase. b. xanthine hydrolase. c. hypoxanthine-guanine phosphoribosyltransferase. d. adenosine deaminase. e. nucleoside phosphorylase.

ANSWER: a 42. Sulfa drugs are analogs of p-aminobenzoate, a compound necessary for the biosynthesis of N10formyltetrahydrofolate. How should these drugs act upon bacterial infections and why are humans not susceptible? a. N10-Formyltetrahydrofolate is a ferredoxin analog and a potent inhibitor of ribonucleotide reductase in bacteria but not humans. b. N10-Formyltetrahydrofolate is a ferredoxin analog and a potent inhibitor of ribonucleotide reductase, but humans consume sufficient quantities of ferredoxin in their diet to saturate reductase. c. N10-Formyltetrahydrofolate is an NADPH analog and a potent inhibitor of ribonucleotide reductase in bacteria but not humans. d. N10-Formyltetrahydrofolate transfers a formyl group twice during de novo synthesis of the purine ring. Bacteria dihydrofolate reductase is blocked by sulfa drugs, whereas the human enzyme is not. e. N10-Formyltetrahydrofolate transfers a formyl group twice during de novo synthesis of the purine ring. Bacteria synthesize their own folate, whereas humans get theirs from eating green plants.

ANSWER: e 43. What position would be labeled in newly synthesized nucleotides if the reaction were done in the presence of [15N]aspartate? .. Powered by Cognero.

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Chapter 32 a. N3 in IMP b. N7 in IMP c. N9 in IMP d. N1 in UTP e. N3 in UTP ANSWER: d 44. Which amino acid side chain in thymidylate synthase activates the ring of dUMP, making C-5 a good nucleophile? a. cysteine b. aspartate c. glutamine d. tyrosine e. serine ANSWER: a 45. Hypoxanthine-guanine phosphoribosyltransferase catalyzes the formation of guanylate and _______.

ANSWER: inosinate 46. How is the purine nucleotide cycle linked to the citric acid cycle? a. The purine nucleotide cycle serves as an anapleurotic reaction for the formation of oxaloacetate. b. The purine nucleotide cycle serves as an anapleurotic reaction for the formation of fumarate. c. The purine nucleotide cycle serves as an anapleurotic reaction for the formation of pyruvate. d. Condensation of IMP with glutamate generates adenylosuccinate, which is hydrolyzed to form succinate. e. Condensation of IMP with glutamine generates adenylosuccinate, which is hydrolyzed to form succinate.

ANSWER: b 47. What intermediate in pyrimidine synthesis will accumulate if a strain of bacteria is lacking in aspartate? a. orotidylate b. orotate c. dihydroorotate d. 5-phosphoribosyl-1-amine e. carbamoyl phosphate ANSWER: e 48. Human beings contain two different carbamoyl phosphate synthetase enzymes. If you wanted to study the enzyme inhibition for the enzyme specific for nucleotide synthesis, which of the following analogs would you choose? a. analog for bicarbonate .. Powered by Cognero.

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Chapter 32 b. analog for carboxyphosphate c. analog for carbamic acid d. analog for NH3 e. analog for glutamine ANSWER: e 49. The framework for a single ring _____ base is assembled first and then attached to ribose sugar. The framework for a bicyclic _____ base is synthesized piece by piece directly onto a ribose-based molecule.

ANSWER: pyrimidine; purine 50. What is TRUE about salvage pathways? a. The nucleotide nitrogenous bases are assembled from simpler compounds. b. Preformed nitrogenous bases are recovered and reconnected to a carbon unit. c. All salvage pathways lead to the synthesis of ribonucleotides. d. The framework for a bicyclic purine base is synthesized piece by piece directly onto a ribose-based molecule. e. Substantial amounts of ATP are saved by skipping most of the energy-requiring steps of de novo nucleotide synthesis.

ANSWER: e 51. What enzymes convert nucleoside monophosphates into triphosphates? a. hydrolases b. kinases c. phosphorylases d. reductases e. transferases ANSWER: b 52. _____ can be recovered from the breakdown products of DNA and RNA by the use of salvage pathways.

Select all that apply. a. Deoxyribonucleotides b. Pyrimidine bases c. Purine bases d. Ribonucleotides e. Ribose sugars ANSWER: b, c 53. Thymine released from degraded DNA is salvaged in two steps. In the first step, thymine is converted into the nucleoside thymidine by _____. a. thymidylate b. thymidine kinase .. Powered by Cognero.

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Chapter 32 c. thymidine nucleotides d. thymidine phosphorylase e. thymidine triphosphate ANSWER: d 54. Thymine released from degraded DNA is salvaged in two steps. In the second step, the nucleoside thymidine is converted into a nucleotide by _____. a. thymidylate b. thymidine kinase c. thymidine nucleotides d. thymidine phosphorylase e. thymidine triphosphate ANSWER: b 55. The first step in purine assembly begins with: a. the oxidation of inosinate to xanthine monophosphate. b. successive steps of activation by displacement. c. successive steps of activation by phosphorylation. d. the attachment to ribose. e. the departure of fumarate.

ANSWER: d 56. Viral thymidine kinase differs from the mammalian enzyme and thus provides a therapeutic target. Herpes simplex infections are treated with acyclovir. Viral thymidine kinase binds acyclovir more than 200 times as tightly as cellular thymidine kinase. Once incorporated into viral DNA, acyclovir triphosphate acts as: a. an activator of bacterial DNA replication. b. a terminator of bacterial DNA replication. c. an activator of viral DNA replication. d. a terminator of viral DNA replication. e. a terminator of mammalian DNA replication.

ANSWER: d 57. Which amino acids are the required precursors to produce the nucleotide inosine monophosphate? Select all that apply. a. aspartate b. cysteine c. glycine d. glutamine e. serine ANSWER: a, c, d .. Powered by Cognero.

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Chapter 32 58. How many ATP equivalents are required for de novo purine biosynthesis? a. 1 b. 2 c. 5 d. 6 e. 8 ANSWER: d 59. Adenylate is synthesized from inosinate by: a. the elimination of an amino group from the C-4. b. the addition of an amino group at C-4. c. the addition of an amino group at C-6. d. the substitution of an amino group for the carbonyl oxygen atom at C-4. e. the substitution of an amino group for the carbonyl oxygen atom at C-6.

ANSWER: e 60. What is the complex formed upon the association of the purine synthesis enzymes? a. aspartate transcarbamoylase b. carbamoyl phosphate synthetase II c. hypoxanthine-guanine phosphoribosyltransferase d. purinosome e. 5-phosphoribosyl-1-pyrophosphate ANSWER: d 61. What is the term for when an enzyme converts a substrate into a reactive inhibitor that halts the enzyme’s catalytic activity? a. competitive inhibition b. feedback inhibition c. noncompetitive inhibition d. reversible inhibition e. suicide inhibition ANSWER: e 62. Which enzymes are sensitive targets of chemotherapy? Select all that apply. a. aspartate transcarbamoylase b. carbamoyl phosphate synthetase II c. dihydrofolate reductase d. hypoxanthine-guanine phosphoribosyltransferase e. thymidylate synthase ANSWER: c, e .. Powered by Cognero.

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Chapter 32 63. Thymidylate is formed by the _____ of deoxyuridylate. a. carbonation b. methylation c. hydration d. decomposition e. precipitation ANSWER: b 64. Stem cells in bone marrow, epithelial cells of the intestinal tract, hair follicles, and other rapidly replicating cells are all susceptible to methotrexate because it is a _____. a. folate agonist b. folate analog c. folate antagonist d. methyl donor e. suicide inhibitor ANSWER: c 65. The phosphorolytic cleavage of nucleosides to free bases and deoxyribose 1-phosphate is catalyzed by what enzymes? a. nucleoside hydrolases b. nucleoside kinases c. nucleoside phosphorylases d. nucleoside reductases e. nucleoside transferases ANSWER: c 66. One hypothesis about why folic acid supplementation is important in pregnancy is that: a. more folate derivatives are needed for the synthesis of DNA precursors for rapidly dividing cells in the neural epithelium. b. more folic acid is needed to bind to and activate G-protein-coupled receptors that regulate the dopamine-secreting neurons. c. folic acid is required for glycogen synthesis and the glycosylation of proteins and lipids. d. the brain has a limited capacity to synthesize folic acid, and needs an external source. e. folate derivatives are needed to transport oxygen to the forming neural tube.

ANSWER: a 67. The _____ has limited capacity for de novo purine synthesis. a. brain b. heart c. kidney d. liver .. Powered by Cognero.

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Chapter 32 e. stomach ANSWER: a 68. The concentration of _____ in the blood of rodents and humans is dependent on whether the organism is fasting or fed. It serves as a signaling molecule for caloric homeostasis. a. adenosine b. cytidine c. guanosine d. thymidine e. uridine ANSWER: e

.. Powered by Cognero.

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Chapter 33 1. What was known about DNA before the 1950s? a. DNA is composed of four nitrogen bases. b. The structure of DNA is a double helix. c. There are certain principles of DNA replication. d. There are different enzymes that take part in DNA transcription. e. The stacked bases interact with van der Waals forces.

ANSWER: a 2. What are the components of a nucleic acid monomer? a. a base, an ester residue, a phosphate b. an amino acid, an ester residue, a base c. a sugar, a base d. a sugar, a phosphate, a base e. an amino acid, a sugar, a phosphate ANSWER: d 3. Why does RNA have a 2′-hydroxyl group in a sugar while DNA does not? a. The hydroxyl group of a DNA sugar is broken rapidly in a water solution. b. The presence of a hydroxyl group is impossible in the negatively charged DNA. c. The absence of a 2′-hydroxyl group in DNA increases its resistance to hydrolysis. d. The hydroxyl group of a DNA sugar is absent in order to avoid possible mutations. e. The oxygen atom of the hydroxyl group is too reactive to be in a hydroxyl form.

ANSWER: c 4. Which statements reflect the Watson–Crick model of DNA? Select all that apply. a. Two helical DNA strands are coiled around a common axis, forming a right-handed double helix. b. The diameter of the helix is 30 Å. c. The purine and pyrimidine bases lie outside of the helix. d. The sugar–phosphate backbone lies outside of the helix. e. The strands of DNA run in the same direction.

ANSWER: a, d 5. Which interactions take part in forming complementary base pairs? a. ionic bonds b. covalent bonds c. hydrogen bonds d. van der Waals forces e. base-stacking interactions ANSWER: c 6. Choose the CORRECT statement about semiconservative replication of DNA. .. Powered by Cognero.

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Chapter 33 a. Daughter double-stranded DNA molecules are identical to the corresponding parental ones. b. One of the strands of each daughter DNA molecule is newly synthesized, whereas the other is passed unchanged from the parent DNA molecule. c. Each daughter DNA molecule is single-stranded and identical to the parent DNA molecule. d. Two daughter DNA molecules are double-stranded and differ from the parent. e. There is one daughter double-stranded DNA molecule, which is identical to the parental one.

ANSWER: b 7. What is the direct consequence of the hydrophobic effect that stabilizes the double helix structure? a. DNA is insoluble. b. Sugars become more sensitive to digestion. c. Phosphodiester bonds are more prone to breaking. d. DNA is a rigid molecule. e. Bases are positioned inside the helix.

ANSWER: e 8. What bonds connect adjacent sugars in a DNA strand? a. ionic bonds b. hydrogen bonds c. peptide bonds d. disulfide bridges e. phosphodiester bonds ANSWER: e 9. Which nitrogen atom of a purine and a pyrimidine bases is bound to a 1′-carbon atom of the sugar? a. N-1 of purine and N-1 of pyrimidine b. N-1 of purine and N-3 of pyrimidine c. N-3 of purine and N-9 of pyrimidine d. N-9 of purine and N-3 of pyrimidine e. N-9 of purine and N-1 of pyrimidine ANSWER: e 10. Choose the CORRECT fact about RNA. a. branched polymer consisting of nucleotides A, G, U, C; the sugar is fructose b. unbranched polymer consisting of nucleotides A, G, C, T; the sugar is deoxyribose c. unbranched polymer consisting of nucleotides A, G, C, U; the sugar is ribose d. unbranched polymer consisting of nucleotides A, G, C, U; the sugar is deoxyribose e. unbranched polymer consisting of nucleotides A, G, C, T; the sugar is glucose ANSWER: c 11. Choose the techniques that Meselson and Stahl used to establish semiconservative replication of DNA. .. Powered by Cognero.

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Chapter 33 a. isotope labeling and density gradient equilibrium sedimentation b. isotope labeling and photoaffinity labeling c. DNA sequencing and photoaffinity labeling d. density gradient equilibrium sedimentation and DNA sequencing e. PCR and stable-isotope probing ANSWER: a 12. How is the melting temperature of a double-stranded DNA defined? a. It's a temperature at which a double-stranded DNA is denatured to a single-stranded DNA. b. It's a temperature at which half of the helical structures are lost. c. It's a temperature at which stacking interactions are lost. d. The temperature is equal to 70ºC. e. It's a temperature of the boiling solution.

ANSWER: b 13. What statement about DNA polarity is TRUE? a. One end of the chain has a free 5′-OH group. The other end of the chain has a free 3′-OH group. b. One end of the chain has a 5′-OH group attached to a phosphoryl group. The other end of the chain has a free 3′-OH group, which is linked to another nucleotide. c. One end of the chain has a free 3′-OH group or 3′-OH group attached to a phosphoryl group. The other end of the chain has a free 5′-OH group, which is linked to another nucleotide. d. One end of the chain has only a free 5′-OH group. The other end of the chain has a free 3′-OH group.

Neither is linked to another nucleotide. e. One end of the chain has a free 5′-OH group or 5′-OH group attached to a phosphoryl group. The other end of the chain has a free 3′-OH group. None is linked to another nucleotide.

ANSWER: e 14. Choose the CORRECT components of a nucleoside. a. a pentose sugar bonded with a hexose sugar b. a base bonded to a sugar and one or more phosphoryl groups c. a base bonded to a sugar d. a base bonded to a sugar and an alcohol residue e. a sugar bonded to a lipid molecule ANSWER: c 15. Choose the temperature that corresponds to DNA annealing. a. <Tm b. >Tm c. =Tm d. 24ºC .. Powered by Cognero.

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Chapter 33 e. 48ºC ANSWER: a 16. Name these forms of DNA in numerical order 1-2-3.

a. A-Z-B b. Z-A-B c. Z-B-A d. A-B-Z e. B-A-Z ANSWER: d 17. How many full turns does a linear B-DNA have if its length is 374 bp? a. 11 b. 110 c. 37 d. 36 e. 35 ANSWER: e 18. Why is negative supercoiling of DNA vital for the cell? a. Negatively supercoiled DNA is less sensitive to mutagens. b. DNA is limited by the cell space and has to fit it. c. Negative supercoiling prepares DNA for the processes requiring separation of DNA strands. d. While DNA is negatively supercoiled, it does not undergo mutations. e. Negative supercoiling makes the DNA molecule more stable.

ANSWER: c .. Powered by Cognero.

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Chapter 33 19. How many chromosomes does the human genome include? a. 22 autosomal chromosomes and 2 sex chromosomes b. 22 pairs of autosomal chromosomes and 2 sex chromosomes c. 46 pairs of autosomal chromosomes and 2 sex chromosomes d. 46 autosomal and 4 sex chromosomes e. 23 pairs of autosomal chromosomes and 1 sex chromosome ANSWER: b 20. What is the reason for the presence of major and minor grooves in the B-DNA form? a. Water solution favors the presence of grooves. b. It is energetically favorable to have such a structure for the DNA molecule. c. Hydrogen bonds between complementary base pairs have a different length. d. Glycosidic bonds of a base pair are not diametrically opposite each other. e. The grooves are essential for the formation of the Z-DNA form.

ANSWER: d 21. What is the name of DNA isomers that have the same nucleotide sequence but a different coiling? a. topoisomers b. stereoisomers c. enantiomers d. cis-trans isomers e. structural isomers ANSWER: a 22. What was shown by the existence of Z-DNA? a. DNA is a long molecule. b. DNA comprises four nitrogen bases. c. DNA has a double-helix structure. d. DNA chains are polar. e. DNA is a flexible molecule.

ANSWER: e 23. What are the most common forms of DNA present under physiological conditions? Select all that apply. a. left-handed double helix b. Z form c. right-handed double helix d. A form e. B form ANSWER: c, e 24. Amino acid residues that compose the tail of each histone in a nucleosome are: .. Powered by Cognero.

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Chapter 33 a. Lys and Ala. b. Ala and Asn. c. Lys and Asp. d. Lys and Arg. e. Arg and Ala. ANSWER: d 25. What structure can be formed from a single-stranded RNA? a. hairpin loop b. double loop c. helix-turn-helix d. helix-loop-helix e. stem-loop ANSWER: e 26. Choose the CORRECT components of a nucleosome. a. a pair of H2A-H2B dimers, (H3)2(H4)2 tetramer, 200-bp DNA, linker DNA b. (H2A)2-(H2B)2 tetramer, a pair of H3-H4 dimers, 200-bp DNA, linker DNA c. a pair of H2A-H2B dimers, (H3)2(H4)2 tetramer, 145-bp DNA, linker DNA d. a pair of H2A-H2B dimers, a pair of H3-H4 dimers, 100 bp DNA, linker DNA e. a pair of H2A-H2B dimers, (H3)2(H4)2 tetramer, 145-bp DNA ANSWER: a 27. What type of DNA has been discovered in the nuclei of eukaryotic cells only recently? a. rDNA b. eccDNA c. ssDNA d. mtDNA e. cpDNA ANSWER: b 28. What is the role of DNA winding around the nucleosome? a. To prevent overwinding during DNA replication. b. Histones in nucleosomes are temporary structures and have to be kept together only during DNA replication. c. It is essential not to let histones disassemble during DNA replication. d. The formed structure defends histones from bacteria. e. This DNA is packed by decreasing its linear extent.

ANSWER: e .. Powered by Cognero.

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Chapter 33 29. What ion assists in stabilization of complexes of nucleic acids? a. Mg2+ b. Ca2+ c. K+ d. Na+ e. Fe2+ ANSWER: a 30. Which nitrogenous bases are distinguished from each other by the presence of a methyl group? a. uracil and cytosine b. adenine and guanine c. uracil and thymine d. pyrimidine and cytosine e. purine and guanine ANSWER: c 31. Cisplatin is a cancer chemotherapy drug that prevents replication and transcription. How does it work? a. The platinic atom of cisplatin forms covalent bonds with carbon atoms in deoxyribose. b. Chloride ligands of cisplatin are displaced by purines on two adjacent bases. c. The molecule of cisplatin degrades phosphodiester bonds. d. Hydrogen bonds between DNA chains are degraded by cisplatin. e. Amino groups of cisplatin are displaced by pyrimidines on two opposite bases.

ANSWER: b 32. Which statement about DNA is FALSE? a. DNA has a double-helix structure. b. DNA is composed of four bases: A, T, G, and C. c. Nucleosides are the monomeric units of DNA. d. The strands of DNA run in opposite directions. e. A-DNA appears when DNA is less hydrated.

ANSWER: c 33. Which statement is true about the presence of DNA grooves in terms of donor–acceptor interactions? a. The O-4 atom of thymine and the O-6 atom of guanine are located in the major groove and can serve as hydrogen acceptors. b. In the major groove, N-7 of guanine or adenine is a potential donor. c. In the minor groove, N-3 of adenine or guanine can serve as a hydrogen donor. d. In the major groove, the amino group attached to C-2 of guanine can be a hydrogen acceptor. e. In the minor groove, N-3 of adenine or guanine and O-2 of thymine or cytosine can serve as hydrogen acceptors. .. Powered by Cognero.

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Chapter 33 ANSWER: e 34. What organisms are exposed to serious devastating influence by oxybenzone? a. human beings b. reptiles c. birds d. corals e. bacteria ANSWER: d 35. Which property of DNA makes the molecule appropriate for storing genetic information? a. length of DNA b. hydrogen bonds between nitrogen bases of different strands c. ability to form three different forms d. presence of A and T nitrogen bases e. the sequence determination of its bases ANSWER: e 36. How does oxybenzone affect DNA structure? a. It modifies thymine forming an oxetane derivative. b. It breaks phosphodiester bonds. c. It unwinds DNA chains. d. It forms stem-loop structures of DNA. e. It modifies deoxyribose forming ribose.

ANSWER: a 37. What would be the effect of the increased length of the double-stranded DNA on Tm? a. Tm would be about 0 ºC. b. Tm would not change. c. Tm would decrease. d. Tm would increase. e. It is difficult to predict.

ANSWER: d 38. Choose the CORRECT order of compaction of DNA in eukaryotes. a. DNA → nucleosome → "beads on a string" → fiber → loops → chromosome b. DNA → fiber → loops → nucleosome → "beads on a string" → chromosome c. DNA → "beads on a string" → fiber → nucleosome → loops → chromosome d. DNA → nucleosome → fiber → loops → "beads on a string" → chromosome e. DNA → nucleosome → loops → fiber → "beads on a string" → chromosome .. Powered by Cognero.

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Chapter 33 ANSWER: a 39. Well-defined and complex structures allow _______ and single-stranded DNA to perform a host of functions that the double-stranded DNA molecule cannot.

ANSWER: RNA 40. The difference in RNA bases compared with DNA bases is that RNA contains: a. A instead of T. b. U instead of G. c. U instead of T. d. A instead of U. e. C instead of G. ANSWER: c 41. What is the actual mode of replication in the cell? a. conservative b. random c. semiconservative d. N-linked e. direct ANSWER: c 42. The shorthand notation pApCpG denotes that: a. a phosphate is attached to the 5′ end of the adenosine nucleotide unit. b. a phosphate is attached to the 3′ end of the guanine nucleotide unit. c. the cytosine nucleotide has a free hydroxyl group. d. a phosphate is attached to the 5′ end of the adenosine nucleotide unit and the cytosine nucleotide has a free hydroxyl group. e. the guanosine nucleotide is attached to the 5′ end of the adenosine nucleotide unit.

ANSWER: a 43. The chemical forces that contribute to the stability of DNA due to base-stacking present in the DNA helix are: a. hydrogen bonds. b. van der Waals interactions. c. disulfide bonds. d. covalent bonds. e. ionic bonds.

ANSWER: b 44. After two generations of replication in the Meselson and Stahl experiment, what was the composition of the two bands? .. Powered by Cognero.

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Chapter 33 a. One band was all 14N and one band was all 15N. b. One band was all 14N and one band was half 14N and half 15N. c. One band was all 15N and one band was half 14N and half 15N. d. One band was all 14N and one band was one quarter 14N and three quarters 15N. e. One band was all 15N and one band was one quarter 14N and three quarters 15N. ANSWER: b 45. Which property of DNA makes it more stable than RNA in the basic environment? a. The negative charge of the phosphate on DNA repels the negative charged base. b. Histones block access to all except a few nucleotides that act as linkers. c. The 2′ –H in DNA is not reactive, whereas the 2′ –OH of RNA is reactive under basic conditions. d. DNA precipitates in basic solutions, making it unreactive. e. RNA forms elaborate structures that are susceptible to alkaline digestion.

ANSWER: c 46. What is a transient copy of genetic information? a. supercoiled RNA b. DNA c. B-DNA d. Z-DNA e. RNA ANSWER: e 47. Which scientist provided data for Watson and Crick's model of DNA? a. Francis Collins b. Barbara McClintock c. Erwin Chargaff d. Rosalind Franklin e. Marcus Rhoades ANSWER: d 48. Which form of DNA is the E. coli genome in? a. supercoiled b. circular c. stem loop d. B-DNA e. A-DNA ANSWER: b 49. Select all that apply. Unwinding prior to ligation results in: .. Powered by Cognero.

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Chapter 33 a. a relaxed DNA. b. a narrowing of a major groove. c. negative supercoiling. d. positive supercoiling. e. restrictase activity.

ANSWER: b, c 50. What is the name of the process of rehybridization of melted DNA? a. denaturation b. Tm c. annealing d. dehybridization e. replication ANSWER: c 51. The sugar in ribonucleic acid is: a. deoxypentose b. deoxyribose c. glucose d. pentose e. ribose ANSWER: e 52. The sugar in deoxyribonucleic acid is: a. deoxypentose b. deoxyribose c. glucose d. pentose e. ribose ANSWER: b 53. Phosphodiester linkages have a __________ which repels nucleophilic species capable of breaking the integrity of both DNA and RNA backbones.

a. electrophile b. negative charge c. neutral charge d. nucleophile e. positive charge ANSWER: b 54. Nucleoside triphosphates are also referred to as ____________, which are the monomers that are linked to form .. Powered by Cognero.

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Chapter 33 RNA and DNA.

a. nucleic acids b. nucleosides c. nucleotides d. nucleosomes e. trinucleotides ANSWER: c 55. The abbreviated notation pGpTpG denotes a trinucleotide of DNA consisting of which building blocks? Select all that apply. a. deoxyadenylate monophosphate

b. deoxycytidylate monophosphate c. deoxyguanylate monophosphate d. deoxythymidine monophosphate e. deoxyuridine monophosphate ANSWER: c, d 56. Which technique did Rosalind Franklin use to examine DNA structures?

ANSWER: X-ray crystallography 57. The stacking of bases one on top of another contributes to the stability of the DNA double helix in which two ways?

a. The distribution of stacked bases allows spacing in the double helix to store immense information. b. Hydrophobic interactions between the stacked bases drive the bases to the interior of the helix and the more polar surfaces to the surrounding water. c. The stacked bases attract one another through hydrogen bonds.

d. The stacked bases attract one another through van der Waals forces. e. Stacked bases allow for complementary base pairs in paired DNA strands.

ANSWER: b, d 58. Which form of DNA is the broadest? a. A-DNA, with a diameter of ∼18 Å b. A-DNA, with a diameter of ∼26 Å c. B-DNA, with a diameter of ∼26 Å d. Z-DNA, with a diameter of ∼18 Å e. Z-DNA, with a diameter of ∼26 Å ANSWER: b 59. How many full turns does A-DNA have if its length is 220 bp? a. 9 b. 12 .. Powered by Cognero.

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Chapter 33 c. 18 d. 20 e. 21 ANSWER: d 60. What is the pitch of Z-DNA if its length is 547 bp? a. 1 Å b. 12 Å c. 46 Å d. 1520 Å e. 2079 Å ANSWER: e 61. Major- and minor-groove sides are created by: a. the distances between glycosidic bonds in a single base pair. b. the distances between hydrogen bonds in a single base pair. c. the hydrophobic interactions between base pairs. d. the interactions between hydrogen bonds in a single base pair. e. the stacking of nitrogenous bases one on top of another.

ANSWER: a 62. Glycosidic bonds are: a. equal distances apart on either side of the major and minor grooves. b. closer together on either side of the major groove and equal distances apart on either side of the minor groove. c. closer together on either side of the major groove and farther apart on either side of the minor groove.

d. closer together on either side of the minor groove and farther apart on either side of the major groove. e. equal distances apart on either side of the minor groove and farther apart on either side of the major groove.

ANSWER: d 63. Which DNA minor groove species can serve as hydrogen donors in donor–acceptor interactions? a. N-3 of adenine or guanine b. O-2 of thymine or cytosine c. amino group attached to C-2 of guanine d. N-7 of guanine or adenine e. amino groups attached to C-6 of adenine ANSWER: c 64. Which DNA minor groove species can serve as hydrogen acceptors in donor–acceptor interactions? Select all that apply.

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Chapter 33 a. N-3 of adenine or guanine b. O-2 of thymine or cytosine c. amino group attached to C-2 of guanine d. N-7 of guanine or adenine e. O-4 of thymine ANSWER: a, b 65. Which DNA major groove species can serve as hydrogen donors in donor–acceptor interactions? Select all that apply.

a. O-6 of guanine b. amino groups attached to C-6 of adenine c. amino group attached to C-4 of cytosine d. N-7 of guanine or adenine e. O-4 of thymine ANSWER: b, c 66. Which DNA major groove species can serve as hydrogen acceptors in donor–acceptor interactions? Select all that apply. a. O-6 of guanine

b. amino groups attached to C-6 of adenine c. amino group attached to C-4 of cytosine d. N-7 of guanine or adenine e. O-4 of thymine ANSWER: a, d, e 67. Which nucleic acid commonly forms a stem-loop motif?

ANSWER: RNA 68. Which nucleic acid commonly forms a circular motif to maintain a compact shape inside of cells?

ANSWER: DNA 69. Human chromosomes in metaphase are compacted by a factor of: a. 7 b. 10 c. 100 d. 104 e. 107 ANSWER: d 70. The smaller complex of the histone octamer and a 145-147-bp DNA fragment is called the _______. a. chromatin .. Powered by Cognero.

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Chapter 33 b. histone c. nucleosome d. nucleosome core particle e. topoisomer ANSWER: d

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Chapter 34 1. Choose the enzyme that primes the synthesis of DNA. a. helicase b. topoisomerase I c. topoisomerase II d. primase e. gyrase ANSWER: d 2. DNA replication in E. coli starts with the binding of DnaA protein to: a. prepriming complex. b. replicon. c. oriC locus. d. primase. e. primosome.

ANSWER: c 3. DNA gyrase is what type of enzyme in bacteria? a. bacterial topoisomerase I b. bacterial topoisomerase II c. helicase d. primase e. DNA polymerase I ANSWER: b 4. What is the function of helicase? a. It removes mismatched nucleotides from the 3′ end of DNA. b. It eliminates supercoils by cleaving DNA. c. It primes synthesis of DNA. d. It utilizes the energy of ATP to add negative supercoils to DNA. e. It utilizes the energy of ATP for separation of strands.

ANSWER: e 5. What are telomeres? a. ends of eukaryotic chromosomes b. ends of prokaryotic chromosomes c. replicon ends d. proofreading mechanisms e. replication fork ANSWER: a 6. In what direction does the DNA strand grow upon replication? .. Powered by Cognero.

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Chapter 34 a. 3′ → 5′ b. 5′ → 3′ c. 3′ → 3′ d. 5′ → 5′ e. both 3′ → 5′ and 5′ → 3′ ANSWER: b 7. The enzyme that removes the RNA primer during DNA replication in E. coli is: a. DNA polymerase I. b. DNA polymerase II. c. DNA polymerase III. d. DNA polymerase IV. e. DNA polymerase V. ANSWER: a 8. Choose the enzyme that does NOT take part in DNA repair. a. DNA polymerase I b. DNA polymerase II c. DNA polymerase III d. DNA polymerase IV e. DNA polymerase V ANSWER: c 9. The frequency of DNA replication error is: a. 1 per 102 nucleotides. b. 1 per 104 nucleotides. c. 1 per 106 nucleotides. d. 1 per 108 nucleotides. e. 1 per 1010 nucleotides.

ANSWER: d 10. Which statement about DNA clamps is TRUE? a. They form ends of eukaryotic chromosomes. b. They consist of β2 subunits. c. They remove mismatched nucleotides. d. They prime the synthesis of DNA. e. They utilize ATP. ANSWER: b .. Powered by Cognero.

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Chapter 34 11. Which unit represents the origin of replication in eukaryotes? a. prepriming complex b. replicon c. oriC locus d. primase e. primosome ANSWER: b 12. DNA pieces that are synthesized on the lagging strand are referred as: a. replication fork. b. Okazaki fragments. c. polymerase. d. Holliday fragments. e. telomere.

ANSWER: b 13. Polymerase III is a dimeric _______, and the lagging strand allows the enzyme to proceed in the same direction with each strand.

ANSWER: holoenzyme 14. RNA-dependent DNA polymerase takes part in the: a. relaxation of supercoiled DNA. b. formation of Okazaki fragments. c. addition of negative supercoils to DNA. d. rejoining of DNA strands. e. replication of telomeres.

ANSWER: e 15. What enzyme can add negative supercoils to DNA? a. helicase b. DNA polymerase I c. DNA polymerase II d. topoisomerase I e. topoisomerase II ANSWER: e 16. What enzyme is involved in relaxation of supercoils? a. helicase b. DNA polymerase I c. DNA polymerase II d. topoisomerase I .. Powered by Cognero.

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Chapter 34 e. topoisomerase II ANSWER: d 17. What is the role of the small RNA primer in DNA replication? a. It is involved in recognition of the replication origin. b. It provides a free 3′ hydroxyl group for DNA polymerase. c. Unlike the leading strand, the lagging strand cannot be replicated without primers. d. It provides a free 5′ phosphate group for DNA polymerase. e. It is used by RNA-dependent DNA polymerase.

ANSWER: b 18. Choose the CORRECT statement about telomeric DNA. a. It repairs DNA. b. It utilizes ATP. c. It forms the origin of replication. d. It contains a lot of tandemly repeated hexanucleotide sequences. e. It synthesizes short stretches of RNA. ANSWER: d 19. Which of the following enzymes is a specific type of RNA-dependent DNA polymerase? a. helicase b. primase c. telomerase d. exonuclease e. gyrase ANSWER: c 20. The formation of a phosphodiester linkage between the 5′-phosphate group at the end of one DNA fragment and the 3′-hydroxyl group at the end of the other fragment is catalyzed by: a. telomerase. b. polymerase. c. gyrase. d. ligase. e. primase.

ANSWER: d 21. What is the function of the exonuclease activity of DNA polymerases? a. to remove incompatible nucleotides from the 3′ end of DNA b. to remove incompatible nucleotides from the 5′ end of DNA c. to start synthesis of complementary strands d. to initiate replication .. Powered by Cognero.

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Chapter 34 e. to catalyze strand separation ANSWER: a 22. The defect in activity of this enzyme causes Werner syndrome. a. endonuclease b. gyrase c. primase d. helicase e. DNA polymerase II ANSWER: d 23. How often does DNA polymerase I remove a correct nucleotide? a. 1 correct nucleotide in 15 b. 1 correct nucleotide in 20 c. 1 correct nucleotide in 25 d. 1 correct nucleotide in 50 e. 1 correct nucleotide in 100 ANSWER: b 24. What are the reasons for having multiple proteins in DNA replication rather than just DNA polymerase?

Select all that apply. a. There are no reasons. The presence of multiple proteins randomly happened in the course of evolution. b. DNA polymerase cannot correct mistakes, so other proteins do that. c. DNA polymerase requires a primer to start synthesis. d. The polymerase cannot by itself separate strands of such a stable molecule as DNA. e. Other proteins transfer nucleotides to the polymerase.

ANSWER: c, d, e 25. The free end of the eukaryotic chromosome is thought to form a unique DNA structure that allows for a complete replication. This DNA structure is a: a. replication fork. b. large duplex loop. c. supercoiled loop. d. displacement loop. e. DNA clamp.

ANSWER: b 26. What prevents two complementary strands unwound by helicase from forming a double helix? a. DnaA proteins b. DnaB proteins .. Powered by Cognero.

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Chapter 34 c. SSB proteins d. oriC locus e. DNA polymerase ANSWER: c 27. Ciprofloxacin is an inhibitor of: a. bacterial topoisomerase I. b. bacterial topoisomerase II. c. bacterial endonuclease. d. helicase. e. primase.

ANSWER: b 28. What is critical during telomerase complex organization? a. prepriming complex b. polymerase activity c. double helix d. oriC locus e. G-rich strand ANSWER: e 29. The step-by-step addition of deoxyribonucleotides to a DNA strand is catalyzed by: a. topoisomerases. b. helicase. c. polymerases. d. gyrase. e. endonuclease.

ANSWER: c 30. Select all that apply. Sliding DNA clamps: a. require clamp loaders. b. form a ring around the front of the fork of DNA. c. allow the polymerase to move along the DNA molecule without falling off. d. increase DNA polymerase processivity. e. are helicases.

ANSWER: a, b, c, d 31. Choose the enzyme that synthesizes a 10 nucleotide-long RNA strand that is complementary to one of the template DNA strands. a. polymerase b. endonuclease .. Powered by Cognero.

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Chapter 34 c. primase d. helicase e. topoisomerase ANSWER: c 32. What is the name of the site of DNA synthesis in prokaryotes? a. origin of replication b. replicon c. prepriming complex d. replication fork e. Okazaki fragment ANSWER: d 33. Choose the molecule that serves to stop replication of DNA. a. polymerase b. DnaA protein c. DnaB protein d. DNA ligase e. Tus protein ANSWER: e 34. Which polymerase initiates replications in eukaryotes? a. polymerase α b. polymerase δ c. polymerase I d. polymerase II e. polymerase III ANSWER: a 35. After polymerase switching in eukaryotes, which polymerase acts to continue replication after synthesis of the first 20 deoxyribonucleotides? a. polymerase α b. polymerase δ c. polymerase I d. polymerase II e. polymerase III ANSWER: b 36. What structure forms when SSB proteins bind to the single-stranded regions of DNA after binding of DnaA and DnaB? a. replication fork .. Powered by Cognero.

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Chapter 34 b. Okazaki fragments c. prepriming complex d. origin of replication e. telomere ANSWER: c 37. What proteins in eukaryotes bind to the origin of replication to form the DNA synthesis initiation complex? a. replicon b. licensing factors c. Okazaki fragments d. DnaA proteins e. DnaB proteins ANSWER: b 38. What does DNA polymerase accuracy depend on? a. polymerase activity b. type of polymerase c. length of DNA d. double helices formation e. Watson–Crick base pairings ANSWER: e 39. What statement is TRUE about the difference between type I and type II topoisomerases? a. Unlike type II, type I topoisomerases catalyze an uphill reaction and do not require energy from hydrolysis of ATP. b. Type I topoisomerases are prokaryotic, while type II are eukaryotic. c. Type II topoisomerases are prokaryotic, while type I are eukaryotic. d. Unlike type I, type II topoisomerases catalyze an uphill reaction and require energy from hydrolysis of ATP. e. Unlike type I, type II topoisomerases eliminate supercoils.

ANSWER: d 40. How can the leading and lagging strands be synthesized in a coordinated fashion? a. Specific enzymes control the size of the DNA strands unwinding. b. Lagging-strand binding proteins inhibit leading-strand replication if the strands become disproportionate in size. c. The lagging strand is looped which allows DNA polymerase III to proceed in the same direction on each strand. d. The leading strand is looped which allows DNA polymerase III to proceed in the same direction on each strand. e. The DNA polymerase III holoenzyme is dimeric and can synthesize two new strands simultaneously. .. Powered by Cognero.

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Chapter 34 ANSWER: c 41. What actually dictates the 5′ → 3′ direction of DNA replication? a. The energy for polymerization is obtained from hydrolysis of pyrophosphate on the 3′ end of a preexisting nucleotide. b. Okazaki fragments are not sterically hindered by 2′-OH found in ribonucleotides, making the 5′-to-3′ synthesis possible on both strands. c. The nucleophilic attack of the 3′-hydroxyl group of the growing strand on the 5′-phosphate group of incoming nucleotide allows energy to be obtained from hydrolysis of pyrophosphate on the 5′ end. d. The triphosphate on 5′-OH undergoes a nucleophilic attack by Mg2+ in DNA polymerase. e. The direction of polymerization is determined by primase, not the nucleotide structure.

ANSWER: c 42. Processivity: a. indicates the error rate for a polymerase. b. is a deletion of one or more bases in the DNA. c. is specifically the speed of replication. d. is the ability to catalyze many consecutive reactions without releasing a substrate. e. indicates the observed rate of the lagging strand replication.

ANSWER: d 43. How is replication specificity dictated? Select all that apply. a. Watson–Crick hydrogen bonding must occur. b. Enzyme interactions with DNA act as a "ruler" to determine whether the properly spaced base pair has been formed. c. The hydrogen bond is broken and reformed to ensure its accurate placement at each base pair. d. Binding of SSB proteins ensures specific action of DNA polymerase. e. The form of the pocket inside DNA polymerase dictates its specificity.

ANSWER: a, b, e 44. Choose CORRECT statements about E. coli DNA polymerases. Select all that apply. a. There are five structural classes. b. All have finger and thumb domains that wrap around the DNA. c. All catalyze the same basic reaction, which requires a metal ion cofactor. d. Only some of them require a primer. e. All of them possess exonuclease activity.

ANSWER: a, b, c 45. Select all that apply. The Klenow fragment: a. has primase activity. b. is an E. coli DNA polymerase fragment. .. Powered by Cognero.

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Chapter 34 c. has exonuclease activity. d. is RNA-dependent DNA polymerase. e. has polymerase activity.

ANSWER: b, c, e 46. Select all that apply. Polymerase I of E. coli has: a. a 3′ → 5′ exonuclease site. b. a 5′ → 3′ exonuclease site. c. an oriC locus binding site. d. primase activity. e. a magnesium binding site.

ANSWER: a, b, e 47. Choose reactions that always require hydrolysis of ATP. Select all that apply. a. formation of the phosphodiester linkage by DNA polymerase I b. unwinding of DNA strands by helicase c. formation of the phosphodiester linkage by DNA ligase d. sliding along template strands e. unwinding of DNA strands by β2 subunits ANSWER: b, c, d, e 48. The mode of DNA replication termed the trombone model describes a mechanism whereby the loop lengthens and shortens like the slide on a trombone. Specifically, what loop is being described? a. The loop refers to the clamp of DNA polymerase. b. The loop is the strand of RNA synthesized by the primase. c. The loop refers to the unwinding of the DNA by topoisomerase I. d. The loop is made by the leading strand as it completes replication. e. The loop is formed by the 3′-to-5′ lagging strand on the DNA template.

ANSWER: e 49. If you were to subject E. coli circular DNA to gel electrophoresis, what would you expect to find in the absence (lane 1) and presence (lane 2) of topoisomerase I? a. Both lanes would look the same because the DNA is still the same size, with only a small portion uncoiled. b. Lane 1 would run farther than lane 2 due to a smaller charge/density ratio of the DNA in lane 1. c. Lane 1 would run farther than lane 2 due to relaxed and therefore slower moving DNA in lane 2. d. Lane 2 would run farther than lane 1 due to a smaller charge/density ratio of the DNA in lane 1. e. Lane 2 would run farther than lane 1 due to relaxed and therefore slower moving DNA in lane 1.

ANSWER: c 50. How would a mutation that reduced uracil formation affect DNA replication? .. Powered by Cognero.

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Chapter 34 a. The effect would be negligible because uracil is found in RNA, not DNA. b. The effect would be minimal because UTP is formed from the emanation of CTP. c. Uracil could be synthesized by a salvage pathway and so the effect would be minimal. d. DNA synthesis depends on dCTP formed by nucleotide reductase of dUTP. e. DNA synthesis depends on the synthesis of a short strand of RNA primer; thus, DNA synthesis would be negatively impacted.

ANSWER: e 51. Choose the enzymes that take part in DNA repair. Select all that apply. a. DNA polymerase I b. DNA polymerase II c. DNA polymerase III d. DNA polymerase IV e. DNA polymerase V ANSWER: a, b, d, e 52. Choose the enzyme that takes part in primer removal. a. DNA polymerase I b. DNA polymerase II c. DNA polymerase III d. DNA polymerase IV e. DNA polymerase V ANSWER: a 53. Choose the enzyme that takes part in replication. a. DNA polymerase I b. DNA polymerase II c. DNA polymerase III d. DNA polymerase IV e. DNA polymerase V ANSWER: c 54. A new DNA strand is assembled directly on a preexisting DNA _____, generating two double-stranded nucleic acid molecules. a. domain b. primer c. strand d. telomere e. template ANSWER: e .. Powered by Cognero.

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Chapter 34 55. The shape of this protein fragment approximates the shape of a right hand with domains that are referred to as the fingers, the thumb, and the palm. a. DNA polymerase II b. Klenow fragment c. Okazaki fragment d. Taco fragment e. Trombone fragment ANSWER: b 56. DNA gets positioned in the _____ domain, or active site, of DNA polymerase I. a. fingers b. fingertips c. palm d. thumb e. wrist ANSWER: c 57. What is the basis of DNA polymerase’s low error rate? a. the binding of SSB proteins b. the binding of a DNA primer c. the induced fit of the correct nucleotide when it binds into DNA polymerase d. the negative supercoiling of DNA molecules e. the use of energy from ATP hydrolysis ANSWER: c 58. Select all that apply. Polymerases II, IV, and V can replicate through regions of damaged DNA. Thus, they are referred to as _____ polymerases. a. error-ignoring b. error-prone c. replication d. translation e. translesion ANSWER: b, e 59. DNA strands must be _____ to expose single-stranded templates for replication. a. completely overwound b. completely unwound c. wound d. locally overwound e. locally unwound ANSWER: e .. Powered by Cognero.

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Chapter 34 60. _____ is an antibiotic that blocks the binding of ATP to gyrase. a. Anthrax b. Ciprofloxacin c. Nalidixic acid d. Novobiocin e. Penicillin ANSWER: d 61. The flopping of mismatched nucleotides outside of the exonuclease active site is the result of: a. Brownian motion. b. induced fit. c. linear motion. d. rotary motion. e. stronger hydrogen bonding.

ANSWER: a 62. Proofreading by DNA polymerase increases the accuracy of replication by a factor of approximately _____. a. 100 b. 101 c. 102 d. 103 e. 104 ANSWER: d 63. Which entity represents the origin of replication in prokaryotes? a. oriC locus b. prepriming complex c. primer d. replicon e. SSB ANSWER: a 64. Which entities comprise the primosome? Select all that apply. a. oriC locus b. prepriming complex c. primer d. primase e. replicon ANSWER: b, d .. Powered by Cognero.

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Chapter 34 65. When DnaA proteins oligomerize, they: a. are binding to one another. b. are pairing with one another. c. are priming one another. d. are repelling one another. e. are unwrapping from each other.

ANSWER: a 66. DNA polymerase III catalyzes the formation of many thousands of phosphodiester linkages before releasing its template, compared with only 20 phosphodiester linkages for _____. a. DNA polymerase I b. DNA polymerase II c. DNA polymerase III d. DNA polymerase IV e. DNA polymerase V ANSWER: a 67. DNA polymerase III begins the synthesis of the leading strand starting from the _____ formed by primase. a. DNA primer b. helicase c. replication fork d. RNA primer e. telomere ANSWER: a 68. In prokaryotes, DNA replication is terminated at: a. helicase sites. b. Okazaki sites. c. telomeres. d. Ter sites. e. Tus sites.

ANSWER: d 69. Select all that apply. DNA synthesis is more complex in eukaryotes than in bacteria because of what differences? a. Bacterial genomes have more origins of replication than eukaryotic ones. b. Bacterial genetic information is contained on one chromosome, whereas eukaryotic genetic information is contained within several chromosomes. c. Bacterial genetic information is contained within several chromosomes, whereas eukaryotic genetic information is contained on one chromosome. d. The circular bacterial genomes require less origins of replication than linear eukaryotic genomes. .. Powered by Cognero.

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Chapter 34 e. The sheer size of eukaryotic genomes is larger.

ANSWER: b, d, e 70. The lagging strand always has an incomplete _____ after the removal of the RNA primer. a. 3’ end b. 5’ end c. replication fork d. telomere e. Ter site ANSWER: b

.. Powered by Cognero.

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Chapter 35 1. Why will two daughter double helices have different sequences in the replication process in case of a point mutation in the maternal helix? a. The daughter double helices will be synthesized while the reparation system is knocked down. b. The daughter double helices will be synthesized with a non-Watson–Crick base-pairing model. c. Mismatched maternal bases lead to different bases for two daughter double helices, according to the Watson–Crick base-pairing model. d. Daughter helices will change themselves spontaneously. e. There will not be two daughter helices, only one of them.

ANSWER: c 2. Which polymerase is responsible for the completion of replication of damaged genomes? a. polymerase I b. polymerase II c. polymerase α d. error-prone polymerase e. telomerase ANSWER: d 3. Why are insertions and deletions of DNA more damaging than a mismatched base pair? a. Replicative polymerases can stall or fall off a damaged template entirely. b. Offspring of the next generation will be sterile c. Synthesized proteins may have low activity. d. The organism will be susceptible to diseases. e. Replicative polymerases will make more mistakes.

ANSWER: a 4. What is the main cause of Huntington disease? a. mismatched base pair in two genes: hMSH2 and hMLH1 b. huntingtin expression in the brain c. presence of glutamine residues within the gene d. from 6 to 31 CAG sequences in huntingtin e. expansion of trinucleotide repeats ANSWER: e 5. How many neurodegenerative diseases are caused by trinucleotide repeats? a. 5 b. about 10 c. about 35 d. more than 40 e. more than 70 ANSWER: d .. Powered by Cognero.

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Chapter 35 6. Select all that apply. Symptoms of Huntington disease are: a. polydipsia. b. uncontrolled movements. c. loss of cognitive facilities. d. enamel hypoplasia. e. personality alterations.

ANSWER: b, c, e 7. Choose the nitrogen base that pairs with a guanine that is mutated by a hydroxyl radical in DNA. a. uracil b. adenine c. cytosine d. thymine e. guanine ANSWER: b 8. What is the name of chemical agents that alter specific bases within DNA after completed replication? a. mutagens b. oncogenes c. acceptors d. replicative inhibitors e. mutants ANSWER: a 9. Which organelle functions are disrupted by trinucleotide repeat expansion in the gene encoding frataxin? a. ribosomes b. mitochondria c. endoplasmic reticulum d. Golgi apparatus e. lysosomes ANSWER: b 10. What is the key role of melanin pigments? a. protection from biological toxins such as muscarine b. absorption of ultraviolet light c. decreasing tumor growth d. participating in oxygenation processes e. breaking the non-Watson–Crick base pair ANSWER: b .. Powered by Cognero.

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Chapter 35 11. _____ act in the formation of 8-oxoguanine. a. Guanosine triphosphate and ozone b. Guanosine and atomic oxygen c. Indole and hydroxyl radical d. Guanine and hydrogen peroxide e. Guanine and hydroxyl radical ANSWER: e 12. Which type of disease does Huntington disease belong to? a. lysosomal storage disease b. protein-aggregation disease c. lipid storage disease d. disorder of carbohydrate metabolism e. sex-linked disorder ANSWER: b 13. The bacteria _____ illustrates the extraordinary power of DNA repair systems. a. Deinococcus radiodurans b. Deinococcus reticulitermes c. Escherichia coli d. Mycobacterium tuberculosis e. Mycobacterium leprae ANSWER: a 14. What is the name of the reaction leading to the formation of hypoxanthine from adenine? a. halogenation b. carboxylation c. deamination d. dehydration e. nitrosation ANSWER: c 15. How does the mismatch repair machinery of E. coli determine which of the bases is incorrect? a. The incorrect base is tagged by a hydroxyl group. b. A newly synthesized strand is vulnerable to exonucleases, so the enzymes work precisely. c. The incorrect base induces protein cascades to activate the repair machinery. d. RAD51 helps to identify the incorrect base. e. The incorrect base is nonmethylated in the newly synthesized daughter strand.

ANSWER: e 16. What are the components of the mismatch repair systems? .. Powered by Cognero.

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Chapter 35 a. protein MutH, endonucleases MutS and MutL b. protein MutL, endonucleases MutS and MutH c. proteins MutL and MutH, endonuclease MutS d. proteins MutS and MutL, endonuclease MutH e. proteins MutS and MutH, endonuclease MutL ANSWER: d 17. Ultraviolet light: a. is able to modify adenine to form hypoxanthine. b. adds a hydrocarbon molecule to nucleotide bases. c. promotes the formation of cross-links between adjacent pyrimidines along one DNA strand. d. converts a G-C base pair into a T-A base pair in a reaction of DNA with aflatoxin B1 epoxide. e. promotes the formation of single- and double-stranded breaks in DNA. ANSWER: c 18. Which enzyme cleaves pyrimidine dimers in the direct repair system? a. DNA photolyase b. DNA polymerase I c. DNA polymerase III d. DNA ligase e. helicase ANSWER: a 19. In nucleotide excision repair of E. coli, AlkA: a. seals the repaired strand. b. binds with damaged DNA and cleaves the glycosidic bonds to release the damaged base. c. recognizes the AP site and nicks the backbone adjacent to the missing base. d. excises the residual deoxyribose phosphate unit. e. inserts an undamaged nucleotide, as dictated by the base on the undamaged complementary strand.

ANSWER: b 20. What is the nitrogen base shown?

a. 3-methyluracil .. Powered by Cognero.

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Chapter 35 b. 5-methylthymine c. thymine d. 3-methylcytosine e. 5-methylcytosine ANSWER: e 21. 3′-to-5′ exonuclease activity of DNA polymerase III of E.coli: a. cleaves a short nucleotide sequence from the recognition site near the 5′ end of DNA. b. cleaves a short nucleotide sequence from the recognition site near the 3′ end of DNA. c. removes mismatched nucleotides from both 3′ and 5′ ends of DNA. d. removes mismatched nucleotides from the 5′ end of DNA. e. replaces mismatched nucleotides from the 3′ end of DNA. ANSWER: b 22. How many nucleotides does the UvrABC enzyme excise from both ends of the damaged side of the DNA strand? a. 6 nucleotides on the 5′ side and 6 nucleotides on the 3′ side b. 5 nucleotides on the 5′ side and 7 nucleotides on the 3′ side c. 7 nucleotides on the 5′ side and 5 nucleotides on the 3′ side d. 4 nucleotides on the 5′ side and 8 nucleotides on the 3′ side e. 8 nucleotides on the 5′ side and 4 nucleotides on the 3′ side ANSWER: e 23. The loss of the _____ occurs in an AP site. a. pyrimidine nucleotide b. nucleotide base c. deoxyribose d. purine nucleotide e. phosphate group ANSWER: b 24. What is a symptom of xeroderma pigmentosum? a. keratoses b. pale skin of the face c. no melanin d. oily skin e. irregular melanin pattern ANSWER: a 25. Choose the CORRECT sequence of reactions that is appropriate for nucleotide excision repair, particularly in the excision of the thymine dimer. .. Powered by Cognero.

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Chapter 35 a. DNA polymerase I enters the gap → DNA ligase joins the DNA chain → UvrABC excinuclease cuts the damaged DNA. b. UvrABC excinuclease cuts the damaged DNA → DNA ligase joins the DNA chain. c. UvrABC excinuclease cuts the damaged DNA → DNA polymerase I enters the gap → DNA ligase joins the DNA chain. d. DNA polymerase I enters the gap → UvrABC excinuclease cuts the damaged DNA → DNA ligase joins the DNA chain. e. DNA polymerase I enters the gap → DNA ligase joins the DNA chain.

ANSWER: c 26. What disease is caused by a mutation in two genes of the DNA mismatch repair system (namely, hMSH2 and hMLH1)? a. xeroderma pigmentosum b. Huntington disease c. Lynch syndrome d. Friedreich ataxia e. skin cancer ANSWER: c 27. Why does DNA have thymine instead of uracil? a. It is energetically unfavorable to make A-U bonds in DNA. b. The repair system can distinguish cytosine and methylated cytosine. c. Uracil retains methylated thymine and the repair system excises it from the chain. d. Deamination of cytosine results in thymine, which is recognized by the repair system and excised from the chain. e. Deamination of cytosine results in uracil, which is recognized by the repair system and excised from the chain.

ANSWER: e 28. A key feature of the Ames test is to add mammalian liver homogenate to: a. compare both bacterial and mammalian activity of the potential carcinogens. b. use mammalian enzymes as a control group. c. test validity. d. provide the enzymes that bacteria lack in order to activate the potential carcinogens mutagens that bacteria cannot convert by their enzymes may develop in a mammalian environment. e. identify mammalian revertants.

ANSWER: d 29. What is the consequence of the replication stall? a. double-strand breaks b. single-strand breaks c. point mutations .. Powered by Cognero.

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Chapter 35 d. incomplete replication of both DNA strands e. recombination between two strands ANSWER: a 30. Excision of an adenine or guanine base while the sugar-phosphate backbone is still intact results in formation of a(n) _______, or AP, site.

ANSWER: apurinic 31. In the anticipation phenomenon: a. siblings of an affected person show the same symptoms of a disease at an older age than did the affected person. b. a disease in first-generation children is determined by the genotype of the mother. c. a disease in children of both the first and the second generations of an affected person is manifested with 100% frequency. d. the first-generation children of an affected person tend to show symptoms of a disease determined by the genotype of the father. e. the first-generation children of an affected parent tend to show symptoms of a disease at an earlier age than did the parent.

ANSWER: e 32. What is the first step of the reparation process? a. proofreading b. DNA synthesis using DNA polymerase c. cleavage of glycosidic bonds d. excision of wrong bases e. ligation of synthesized strands ANSWER: a 33. What is formed by cytochrome P450 upon conversion of aflatoxin B1?

a. hypoxanthine .. Powered by Cognero.

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Chapter 35 b. polycyclic aromatic hydrocarbons c. reactive epoxide d. 8-oxoguanine e. thymine dimer ANSWER: c 34. As the result of gene knockout applied to myogenin, a regulatory protein of mice, the mouse: a. becomes weak. b. dies at birth. c. dies 3 days after birth. d. will have hypertrophied skeletal muscle. e. will be sterile.

ANSWER: b 35. These paired bases show a result of a chemical agent's influence. Give the names to bases 1 and 2, respectively.

a. 8-oxoguanine and thymine b. 8-oxoguanine and adenine c. 5-methylcytosine and adenine d. thymine and hypoxanthine e. cytosine and hypoxanthine ANSWER: e 36. Choose the INCORRECT statement about the recombination of DNA. a. In homologous recombination, parent DNA duplexes align at regions of sequence similarity. b. Recombination is the exchange of segments between two DNA molecules. c. A displacement loop is a three-stranded structure. d. Holiday junction is a common intermediate of DNA recombination, which consists of two strands of DNA. e. Double-strand breaks can be repaired by recombination. .. Powered by Cognero.

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Chapter 35 ANSWER: d 37. What type of enzyme is RAD51 and what is its function? a. isomerase; has to make isomers of the recombination proteins b. transferase; helps to transfer a complex of proteins to damaged DNA c. ATPase; has to bind single-stranded DNA and keep it unwound d. polymerase; has to add dNTPs to synthesize a strand of DNA e. ligase; needs to join newly formed regions of DNA ANSWER: c 38. What is the name of the structure formed by an undamaged double strand (blue) and one damaged strand (red) of DNA during recombination?

a. tight junction b. Holliday junction c. helix-loop-helix d. helix-turn-helix e. D-loop ANSWER: e 39. Which factor(s) directly causes double-strand breaks in DNA? Select all that apply. a. aflatoxin B1 b. infrared radiation c. X-rays d. γ rays e. ionizing radiation ANSWER: c, d, e 40. The three-strand structure involved in strand invasion is: a. the D-loop. b. recombinant DNA. c. an RNA triplet. d. non-Watson–Crick pairing. e. RecA. ANSWER: a 41. Select all that apply. In E. coli, mismatch repair involves: .. Powered by Cognero.

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Chapter 35 a. recognition of a mismatched pair by MutS. b. removal of a mismatched nucleotide by DNA polymerase III. c. cleavage of the backbone by MutH. d. binding of RAD51 to single-stranded DNA. e. synthesis of new DNA by polymerase III. ANSWER: a, c, e 42. Exposure to aflatoxin B1 can lead to: a. the expansion of repeats of three nucleotides. b. DNA damage by alkylation. c. the defective repair of DNA. d. the recombination of DNA. e. double-strand breaks in DNA. ANSWER: b 43. Why does deamination of cytosine to uracil not cause a problem in the next round of DNA replication? a. Although uracil cannot form the third hydrogen bond to guanine, it is recognized by DNA polymerase as a mismatch and the error is corrected in the next round of replication. b. AP endonuclease nicks the backbone, removes uracil, and replaces it with another cytosine nucleotide. c. The repair machinery recognizes uracil in DNA as a mistake and replaces it with a cytosine base. d. Uracil nucleotide undergoes spontaneous depurination and is replaced with cytosine. e. Deamination of cytosine rarely occurs within the coding region and so causes few mutations.

ANSWER: c 44. If you are not genetically predisposed to xeroderma pigmentosum, is it okay to go ahead and get a summer tan? a. Yes, especially for children, because vitamin D formation is critical for bone growth and it requires extensive exposure to UV light. b. Yes, the mutation rate in affected genes is extremely low in the absence of xeroderma pigmentosum so there is nothing to be afraid of. c. No, many defects can arise through UV entering the eyes, which are not protected by the sunscreen. d. No, ultraviolet is still a mutagen and DNA mutations can arise from an inability to repair all photoinduced damage if there were too many. e. No, the defective gene in xeroderma pigmentosum is highly susceptible to UV mutation so the disease can be acquired later in life.

ANSWER: d 45. If you could look at individual strands of DNA and you saw a Holliday junction, what would that tell you? a. A region of DNA that contains a thymidine dimer is undergoing repair. b. DNA repair of a double-strand break is occurring via a recombination mechanism. .. Powered by Cognero.

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Chapter 35 c. DNA with a point mutation is undergoing base-excision repair. d. DNA with a point mutation is undergoing nucleotide-excision repair. e. Mutagenic reversal is occurring, as seen in the Ames test.

ANSWER: b 46. There are 1080 depurination repair events required per cell per hour in the diploid human cell which contains about 6 × 109 base pairs. What is the overall spontaneous rate of depurination per minute? a. 1.08 × 10–8 depurinations per purine per minute b. 1.5 × 10–9 depurinations per purine per minute c. 4.8 × 10–9 depurinations per purine per minute d. 2 × 10–9 depurinations per purine per minute e. 3 × 10–9 depurinations per purine per minute ANSWER: b 47. An assay used to determine carcinogenic potential is the _____ test. a. Huntington disease b. mutagen c. photolyase d. Ames e. direct repair ANSWER: d 48. The process whereby two daughter molecules of DNA are formed by the exchange of genetic material between two parent molecules is known as: a. chemical modification. b. recombination. c. alkylation. d. dimerization. e. replication.

ANSWER: b 49. How many Holliday junctions are formed upon repair of one double-strand break in DNA? a. 1 b. 2 c. 3 d. 4 e. Holliday junctions are formed only when single-strand breaks are repaired.

ANSWER: b 50. The Ames test is a simple and sensitive test for detecting chemical mutagens. What characteristic of the .. Powered by Cognero.

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Chapter 35 genetically modified strain of Salmonella used for this test makes them appropriate for "detecting chemical mutagens"? a. A high proportion of Salmonella reverses the original mutation. b. The addition of the chemical mutagen allows reversal of a mutation in the genes of DNA synthesis. c. The addition of the chemical mutagen allows reversal of a mutation in the genes of histidine synthesis. d. Reversal of the original mutation requires liver homogenates; thus, there are no spontaneous reversals. e. This strain of bacteria is susceptible to very low levels of mutagens and amplifies very small signals.

ANSWER: c 51. Choose the INCORRECT statement. a. Mutagenic mismatches can distort the DNA double helix. b. Mutagenic mismatches cannot be passed on to daughter double helices. c. Mutagenic mismatches can result in daughter double helices with different sequences. d. Mutagenic mismatches can result in permanent changes in DNA sequence. e. Mutagenic mismatches can result in temporary changes in DNA sequence.

ANSWER: b 52. _______ is the creation of two nonidentical daughter cells. a. Meiosis b. Mitosis c. Mosaicism d. Mutagenesis e. Replication ANSWER: c 53. Which are types of errors in DNA replication? Select all that apply. a. deletions b. fragmentations c. insertions d. mismatches e. shifts ANSWER: a, c, d 54. Choose the CORRECT statement a. Translesion polymerases are less prone to error than replicative polymerases. b. Translesion polymerases are more precise than replicative polymerases. c. Translesion polymerases are recruited to DNA regions without errors in DNA sequence in order to create lesions. d. Translesion polymerases can replicate across many lesions in DNA sequences.

.. Powered by Cognero.

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Chapter 35 e. Translesion polymerases never add incorrect nucleotides to growing DNA strands.

ANSWER: d 55. Choose mechanisms for responding to interruptions in DNA replication. Select all that apply. a. alkylation b. base excision repair c. mismatch repair d. recombination e. recruitment of translesion polymerases ANSWER: b, c, d, e 56. In a TG pair, _____ is always treated as the incorrect base and removed by the base excision repair proteins. a. 5-methylcytosine b. adenine c. cytosine d. guanine e. thymine ANSWER: e 57. What three enzymes are essential for the nucleotide excision repair process to remove a pyrimidine dimer in E. coli?

a. DNA ligase b. DNA polymerase I c. DNA polymerase III d. DNA replicase e. UvrABC excinuclease ANSWER: a, b, e 58. An estimated ________ DNA-damaging events take place in a single human cell in a day. a. 100 to 103 b. 102 to 103 c. 102 to 105 d. 103 to 106 e. 104 to 106 ANSWER: e 59. Mismatch repair systems detect mismatches and then recruit a(n) _______ that cleaves the newly synthesized DNA strand close to the lesion. a. endonuclease

b. exonuclease c. ligase .. Powered by Cognero.

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Chapter 35 d. polymerase e. replicase ANSWER: a 60. In a mismatch repair system, a(n) _______ excises incorrect bases so that a polymerase can fill the gap. a. endonuclease b. exonuclease c. ligase d. replicase e. synthase ANSWER: b 61. The endonuclease activation protein of E. coli is called: a. MutA. b. MutH c. MutL. d. MutS. e. MutU ANSWER: c 62. Genes encoding DNA-repair proteins are often ______, meaning that they repress tumor development when at least one copy of the gene is free of a mutation that would eliminate the function of the gene product. a. tumor-defending genes

b. tumor-degrading genes c. tumor-developing genes d. tumor-inciting genes e. tumor-suppressor genes ANSWER: e 63. When both copies of a gene encoding DNA-repair proteins are mutated: a. tumors do not develop at all. b. tumors develop at rates less than those for the population at large. c. tumors develop at rates greater than those for the population at large. d. tumors always develop. e. tumors are present at an organism’s birth.

ANSWER: c 64. Mutations in the hMSH2 and hMLH1 genes account for most cases of hereditary nonpolyposis colorectal cancer, which occurs in as many as ________ people.

a. 1 in 10 b. 1 in 50 .. Powered by Cognero.

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Chapter 35 c. 1 in 100 d. 1 in 200 e. 1 in 1000 ANSWER: d 65. Mutations in the hMSH2 and hMLH1 genes:

Select all that apply. a. lead to 10% of inherited breast cancer cases.

b. lead to extreme sensitivity to ultraviolet light. c. lead to mutations in genes that control cell proliferation, resulting in the onset of cancer. d. likely allow mutations to accumulate throughout an organism’s genome. e. promote the death of cancer cells.

ANSWER: c, d 66. The enzyme poly(ADP-ribose) polymerase (PARP):

Select all that apply. a. is activated by binding to a site of DNA damage.

b. catalyzes the transfer of poly(ADP-ribose) units to itself or proteins required for DNA repair. c. promotes the formation of reactive oxygen species that damage a host of biomolecules. d. serves a critical role in DNA repair. e. is a tumor-suppressor enzyme.

ANSWER: a, b, d 67. The process of a genetic mutation being changed back to the original sequence is called ________. a. correction b. homogenization c. recombination d. repair e. reversion ANSWER: e 68. Gene-knockout and gene-knock-in mice can be generated through DNA __________. a. base excision repair b. direct replication c. mutagenesis d. nucleotide excision repair e. recombination ANSWER: e 69. DNA recombination mechanisms are important in which types of biological processes? Select all that apply. a. base excision repair .. Powered by Cognero.

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Chapter 35 b. generation of antibody diversity c. meiosis d. mutagenesis e. some viral life cycles ANSWER: b, c, e 70. Holliday junctions consist of: a. two strands of DNA that come together to form a cross-like structure. b. two strands of DNA that come together to form two cross-like structures. c. four strands of DNA that come together to form a cross-like structure. d. two strands of DNA that come together to form a looped structure. e. four strands of DNA that come together to form a looped structure.

ANSWER: c

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Chapter 36 1. Which subunit of E. coli RNA polymerase is required only for initiation of RNA synthesis? a. α b. β c. ω d. β' e. σ ANSWER: e 2. Which composition refers to the holoenzyme of E. coli RNA polymerase? a. α2ββ'σω b. α2ββ'σ c. α2ββ'ω d. α2β2σω e. αβ2σω ANSWER: a 3. Which RNA sequence is required for termination? a. Pribnow box b. 35 sequence c. palindromic d. AT-rich sequence e. guanine rich sequence ANSWER: c 4. Which type of RNA molecule does NOT require any modifications? a. transfer RNA b. ribosomal RNA c. deoxyribose RNA d. messenger RNA e. template RNA ANSWER: d 5. Choose the CORRECT statement about ribonuclease P. a. Ribonuclease P contains a catalytically active RNA. b. Ribonuclease P excises 5S, 30S, and 50S rRNA from the primary transcript. c. Ribonuclease P generates the correct 3′ terminus of all tRNA. d. Ribonuclease P cleaves spacer regions from the primary transcript of tRNA. e. Ribonuclease P generates the correct 5′ terminus of all mRNA. ANSWER: a .. Powered by Cognero.

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Chapter 36 6. What is the therapeutic effect of actinomycin D? a. suppression of cell growth b. suppression of cell division c. stimulation of cell division d. suppression of cell growth and division e. stimulation of cell growth ANSWER: a 7. The inhibitory mechanism of RNA synthesis by rifampicin is: a. occupying the channel in the rho protein and preventing RNA release. b. tightly binding to a double helix of DNA and preventing RNA binding. c. occupying the channel for the nascent RNA and preventing elongation of RNA synthesis. d. tightly binding to RNA polymerase and preventing formation of the RNA-DNA hybrid. e. binding to the σ subunit of RNA polymerase to prevent elongation of RNA synthesis.

ANSWER: c 8. In which direction does RNA polymerase read the DNA strand? a. 5′ → 3′ on the coding strand b. 5′ → 3′ on the template strand c. 5′ → 3′ on the antisense strand d. 3′ → 5′ on the coding strand e. 3′ → 5′ on the template strand ANSWER: e 9. Constitutive: a. expression depends on the position of the cell cycle. Regulated expression may depend on the availability of substrates. b. expression ensures continual transcription under particular circumstances. Regulated expression depends only on the availability of substrates. c. expression may depend on the levels of gene products. Regulated expression is activated only by the environmental challenge. d. expression is peculiar to all bacteria. Regulated expression is peculiar only to eukaryotes. e. genes are continually transcribed at a relatively constant rate. Regulated expression may depend on the availability of substrates.

ANSWER: e 10. The possible effect of a mutation in the 3′ end of mRNA leading to an increase in AT-rich regions is: a. breaking of the stem-loop structures that form the terminal signal. It would create indiscriminate stop sites for the mRNA product. b. breaking of the stem-loop structures that form the start signal. It would create indiscriminate start sites for the mRNA product. c. the formation of stem-loop structures. It would create indiscriminate stop sites for the mRNA .. Powered by Cognero.

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Chapter 36 product. d. prevention of mRNA binding to a ribosome. It would block translation of the mRNA product. e. breaking of the structure of mRNA and increasing the number of spacer regions. It would block translation of the mRNA product.

ANSWER: a 11. RNA polymerase forms the open promoter complex by: a. unwinding about 17 base pairs of double-helical DNA. b. unwinding about 35 base pairs of double-helical DNA. c. dissociating the σ subunit. d. binding to the RNA primer. e. hydrolyzing pyrophosphate and unwinding about 17 base pairs of double-helical DNA. ANSWER: a 12. A transcription bubble in E. coli does NOT contain: a. RNA primer. b. RNA polymerase. c. RNA-DNA hybrid helix. d. DNA. e. nascent RNA. ANSWER: a 13. The rho protein acts as a(n): a. kinase. b. isomerase. c. helicase. d. phosphodiesterase. e. nuclease.

ANSWER: c 14. Choose the CORRECT –35 sequence. a. 5′ ----T T G A C A --- start site – 3′ b. 5′ --- T A T A A T --- start site – 3′ c. 3′ --- T T G A C A --- start site – 5′ d. 3′ --- U A U A A U --- start site – 5′ e. 5′ --- U U G A C A – start site – 3′ ANSWER: a 15. Choose the CORRECT pair subunit of E. coli RNA polymerase and its function. a. α subunit – binding to the regulatory RNA primer b. α subunit – binding to DNA .. Powered by Cognero.

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Chapter 36 c. β subunit – take part in hydrolysis of pyrophosphate d. β' subunit – promoter recognition e. ω subunit – binding to the –10 and –35 promoter sequences.

ANSWER: c 16. The single σ subunit of E. coli RNA polymerase: a. tightly binds to the β' subunit of a particular RNA polymerase. b. can assist many polymerases to find the –10 and –35 promoter sequence. c. can assist many polymerases to find the UP element. d. catalyzes hydrolysis of a single pyrophosphate. e. unwinds a 17-bp DNA segment.

ANSWER: b 17. How does the addition of lactose to E. coli induce the expression of galactoside permease? a. Lactose is partially metabolized to allolactose. This carbohydrate binds to the lac repressor and the allolactose–repressor complex leaves the operator site. Transcription of the lac operon increases. b. Lactose is partially metabolized to allolactose. This carbohydrate binds to the lac repressor and the allolactose–repressor complex binds the operator site. Transcription of the lac operon is blocked. c. Lactose binds to the lac repressor and the allolactose–repressor complex leaves the operator DNA. Transcription of the lac operon increases. d. The addition of lactose increases the concentration of cAMP that binds to CAP. The CAP–cAMP complex binds the operator site. Transcription of the lac operon is blocked. e. The addition of lactose increases the concentration of cAMP that binds to CAP. The CAP–cAMP complex binds the repressor of the lac operon. Transcription of the lac operon increases.

ANSWER: a 18. How many mature mRNA are obtained after transcription of the lac operon? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: c 19. If glucose is in full supply, the phosphorylated enzyme IIA transfers a phosphate to: a. form glucose 6-phosphate, leading to a decrease in cAMP and inhibited transcription of the lac operon. b. adenylate cyclase, leading to an increase in cAMP and enhanced transcription of the lac operon. c. adenylate cyclase, leading to an increase in cAMP and inhibited transcription of the lac operon. d. form glucose 6-phosphate, leading to a decrease in cAMP and enhanced transcription of the lac operon. e. CAP to form a CAP–cAMP complex. This complex binds to the regulatory DNA sequence and .. Powered by Cognero.

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Chapter 36 inhibits transcription of the lac operon.

ANSWER: a 20. What is the substrate for the CAP–cAMP complex? a. regulatory sequence between the promoter and the structural genes b. regulatory sequence near position –61 relative to the start site for transcription c. regulatory sequence near position –35 relative to the start site for transcription d. regulatory site inside the operator site e. operator site of the lac operon ANSWER: b 21. What are riboswitches? a. complexes of regulatory proteins with mRNA b. mature mRNA c. regulatory proteins d. segment of mRNA e. complexes of cAMP with mRNA ANSWER: d 22. What is the function of galactoside permease? a. hydrolysis of lactose b. transport of lactase across a cell membrane c. hydrolysis of galactose d. transfer of an acetyl group to galactose e. transport of lactose across a cell membrane ANSWER: e 23. Select all that apply. The lactose operon consists of the: a. promoter site. b. operator site. c. three structural genes. d. terminator site. e. control sites.

ANSWER: a, b, c 24. What is the inducer for the lac operon? a. glucose b. ATP c. galactose d. lactose e. allolactose .. Powered by Cognero.

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Chapter 36 ANSWER: e 25. What is the main stimulus for bioluminescence of Vibrio fischeri? a. high density of the colony b. nutrition availability c. low density of the colony d. luciferase availability e. moonlight ANSWER: a 26. What is the predominant mode of regulation for the matrix biofilm genes? a. negative control b. positive control c. quorum sensing d. catabolite repression e. catabolite sensing ANSWER: c 27. What mechanism provides a different rate of expression of constitutive genes? a. The promoters of these constitutive genes have a different affinity to RNA polymerase. b. The expression of these genes is coordinated by different mechanisms, such as catabolite repression, quorum sensing, positive control, and riboswitches regulation. c. The expression of these genes is coordinated by quorum sensing according to a specific environment change. d. The repressors of these constitutive genes have a different affinity to inducers. e. The expression of these genes is coordinated by riboswitches according to a specific environment change.

ANSWER: a 28. When the concentration of flavin mononucleotide increases: a. it binds to the riboswitch of RNA, and mRNA changes its conformation to prevent the formation of the functional mRNA. b. it binds to the riboswitch of mRNA transcript, and mRNA changes its conformation to prevent the formation of the functional mRNA. c. the riboswitch of mRNA binds to FMN, and mRNA changes its conformation to increase mRNA degradation. d. it binds to the operator site to decrease affinity to RNA polymerase and blocks RNA transcription. e. it binds to the promoter site to decrease affinity to RNA polymerase and blocks RNA transcription.

ANSWER: b 29. When allolactose binds to the lac repressor, the affinity of the: a. repressor for the operator DNA decreases. .. Powered by Cognero.

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Chapter 36 b. repressor for the operator DNA increases. c. repressor for RNA polymerase decreases. d. RNA polymerase for the promoter site increases. e. RNA polymerase for the promoter site decreases.

ANSWER: a 30. The expression of the lac operon is maximal if the concentration of: a. allolactose is high and the concentration of glucose is high. b. galactose is high and the concentration of glucose is low. c. lactose is high and the concentration of glucose is high. d. allolactose is high and the concentration of glucose is low. e. galactose is low and the concentration of glucose is low.

ANSWER: d 31. Choose the CORRECT successive processes on the trp operon when tryptophan concentration is high. a. A ribosome continues translation → terminator is formed → RNA polymerase is released from the DNA → decrease in transcription of the trp operon. b. A ribosome stalls at the attenuator → RNA polymerase continues transcription → increase in transcription of the trp operon. c. Tryptophan binds to the repressor → the repressor binds to RNA polymerase → decrease in transcription of the trp operon. d. Tryptophan binds to the mRNA transcript → terminator is formed → RNA polymerase is released from the DNA → decrease in transcription of the trp operon. e. Tryptophan binds to the regulatory site → decrease in the affinity to RNA polymerase → decrease in transcription of the trp operon.

ANSWER: a 32. An attenuator is a segment of: a. functional mRNA that is able to form two alternative structures. b. mRNA transcript that is able to form two alternative conformations. c. functional mRNA that is able to bind to the inducer. d. functional mRNA that is able to bind to the repressor. e. the transcription regulation.

ANSWER: a 33. As the result of a mutation in the tryptophan attenuator leading to a substitution of tryptophan to arginine, transcription of the trp operon would: a. be blocked. b. be reduced. c. continue with the previous rate. d. be blocked. .. Powered by Cognero.

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Chapter 36 e. be activated.

ANSWER: b 34. What is the major structural regulatory element in mRNA? a. GC-rich segment b. TATTAAT sequence c. repetitive U residues d. stem-and-loop structure e. helix-turn-helix structure ANSWER: d 35. The leader of the trp mRNA encodes: a. two tryptophan residues and an untranslated attenuator with terminator regions. b. five tryptophan residues and an untranslated terminator with attenuator regions. c. two tryptophan residues and an untranslated attenuator with the repressor and the operator sites. d. two tryptophan residues and an untranslated regulatory site that binds to a small regulatory molecule. e. five tryptophan residues and an untranslated attenuator with three stem-loop structures.

ANSWER: a 36. The CAP–cAMP complex binds to the: a. attenuator. b. terminator. c. regulatory site near position –61 relative to the start site. d. regulatory site near position –35 relative to the start site. e. operator site just behind the start site.

ANSWER: c 37. Choose the FALSE statement concerning the regulation of RNA transcription in prokaryotes. a. Attenuation is a mechanism of fine-tuning regulation of trp RNA transcription. b. Riboswitches affect only RNA transcription. c. Biofilm formation is based on quorum sensing. d. The rho protein triggers termination of transcription. e. The binding of CRP to CAP leads to the activation of the lac operon.

ANSWER: b 38. Which nucleotide is the first in RNA transcription in prokaryotes? a. pppA b. ppG c. pppC d. ppU e. pA .. Powered by Cognero.

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Chapter 36 ANSWER: a 39. Which enzyme catalyzes the reaction shown?

a. galactoside permease b. thiogalactoside transacetylase c. β-galactosidase d. α-amylase e. glucoamylase ANSWER: c 40. Select all that apply. The possible forms of transcription control for quorum sensing are: a. negative control. b. positive control. c. catabolic repression. d. phosphorylation. e. riboswitch repression.

ANSWER: a, b, c 41. The Pribnow box is: a. found at +10 base pairs from the start site of an operon. b. found at –10 base pairs from the end site. c. the site where promoters bind. d. a regulator of translation. e. found at +35 base pairs from the start site of an operon.

ANSWER: c 42. In transcription, the 3′-hydroxyl group: a. attacks the α-phosphorous group on an incoming nucleotide. b. binds to Mg2+ in the active site. c. binds to the 5′ ribose. d. attacks the 5′ hydrogen of the incoming nucleotide. e. attacks the γ-phosphorous group on an incoming nucleotide.

ANSWER: a 43. The RNA mutation improves the error rate slightly at the expense of a slightly slow transcription speed. Is it a significant improvement? .. Powered by Cognero.

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Chapter 36 a. No, because transcription takes place at a rate nearly 50 times slower than needed by most cells. b. No, because an error in mRNA affects only one molecule and the error does not become part of the permanent genome. c. Yes, because fewer errors in transcription give bacteria, which turn over rapidly, a survival advantage. d. Yes, because bacteria turn over rapidly, and so protein synthesis runs at maximum capacity. e. Yes, because the transcription bubble is stable only at slow transcription rates.

ANSWER: b 44. If you want to explore a mutation in RNA polymerase that inhibits bacterial growth, which subunit would you target and why? a. σ subunit because it binds to DNA b. α subunit because it takes part in promoter recognition c. α subunit because it binds to DNA d. β subunit because it is required to restore denatured polymerase e. β subunit because it takes part in all stages of catalysis ANSWER: e 45. Is it possible to predict the complete sequence of a tRNA molecule, knowing the DNA sequence that encodes it? a. Yes, tRNA is transcribed in the same process as mRNA. b. Yes, tRNA forms short sequences of Watson–Crick base pairs. c. Yes, primer sequences are regulated in the same way for all types of RNA. d. No, tRNA is modified enzymatically after transcription. e. No, tRNA is not translated, and so sequence data are unavailable.

ANSWER: d 46. What would happen if regions of DNA encoding CAP were deleted? a. Bacterial growth would be inhibited because β-galactosidase would remain low even when glucose levels are low. b. This condition would effectively drive lipid production due to excess glucose metabolism. c. On a molecular level, the lac repressor would stay bound to the operon preventing transcription. d. This would activate the catabolite repression mechanism in the presence of glucose. e. The bacteria would switch to glycogen metabolism.

ANSWER: a 47. What parallels are there between the lac operon and a gene for an allosteric enzyme? a. Both respond to cAMP cascades. b. Both contain regulatory regions and structural genes. c. Transcription of both requires the formation of a hairpin loop. d. Both are sensitive to actinomycin D inhibition. .. Powered by Cognero.

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Chapter 36 e. Transcription of the regulator gene requires binding of the inducer to the promoter.

ANSWER: b 48. An increase in cellular metabolism that inhibits expression of the genes for that metabolism is a process known as: a. quorum sensing. b. riboswitch repression. c. catabolite repression. d. negative control of expression. e. positive control of expression.

ANSWER: c 49. Select all that apply. Functions of RNA polymerase include: a. searching for promoter sites. b. modifying RNA. c. regulating gene expression. d. detecting termination signals. e. unwinding short stretches of DNA. ANSWER: a, d, e 50. The rate of RNA synthesis in E. coli is approximately _______ nucleotides per second.

Write the answer in numeric form.

ANSWER: 50 51. What is a promoter? a. an RNA sequence that encodes for the synthesis of a protein b. a DNA sequence that identifies where the transcription of a specific gene begins c. a DNA sequence that encodes machinery for translation of RNA d. an RNA sequence that promotes transcription e. a DNA sequence that inhibits transcription ANSWER: b 52. Which of the following is the configuration for the core enzyme of RNA polymerase? a. α2ββ'σω b. α2ββ'σ c. α2ββ'ω d. α2β2σω e. αβ2σω ANSWER: c 53. What components are required for RNA polymerase to carry out transcription? Select all that apply. .. Powered by Cognero.

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Chapter 36 a. DNA template b. RNA template c. activated precursors d. transcription factor e. divalent metal ion ANSWER: a, c, e 54. Which ribonucleoside triphosphates serve as building blocks for RNA? Select all that apply. a. ATP b. TTP c. GTP d. CTP e. UTP ANSWER: a, c, d, e 55. The first nucleotide of a transcribed DNA sequence is denoted as _____. a. -35 b. -10 c. +1 d. +2 e. +10 ANSWER: c 56. What are consensus sequences? a. commonly observed sequences that denote the start site of transcription b. commonly observed sequences that provide regions for transcription factors to bind c. commonly observed sequences that denote the site of transcription termination d. commonly observed sequences that create additional binding sites for RNA polymerase e. commonly observed sequences designating genomic features ANSWER: e 57. Which nucleotide location relative to the transcription start site can serve as the start site of an upstream element (UP) sequence? a. +40 b. -40 c. -35 d. -10 e. +10 ANSWER: b 58. Which subunit of RNA polymerase binds to the upstream element? .. Powered by Cognero.

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Chapter 36 a. α b. β c. β’ d. ω e. σ ANSWER: a 59. The σ subunit helps RNA polymerase to locate promoter regions by _____ the enzyme’s affinity for general regions of DNA. ANSWER: decreasing 60. What are the stages of transcription? Select all that apply. a. transcription b. initiation c. modification d. elongation e. termination ANSWER: b, d, e 61. During transcription, what event initiates the transition from initiation to elongation? a. The σ subunit binds to the core enzyme of RNA polymerase. b. The α subunit binds to the upstream element. c. The new RNA chain reaches 9-10 nucleotides in length. d. The transcription bubble is formed. e. The σ subunit is ejected from the transcription complex.

ANSWER: e 62. Why does RNA synthesis allow for lower fidelity than DNA synthesis? a. RNA is not heritable by offspring and a few short-lived, defective transcripts are tolerable. b. RNA polymerase has better nuclease activity than DNA polymerase. c. Multiple codons code for the same amino acid, meaning a few mistakes will not impact future translation. d. RNA synthesis is de novo and does not require a primer. e. The RNA molecules are single-stranded, and thus, there is less opportunity for errors to occur.

ANSWER: a 63. How long is the unwound DNA segment during RNA transcription? a. 10 bp b. 35 bp c. 17 bp d. 16 bp .. Powered by Cognero.

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Chapter 36 e. 36 bp ANSWER: c 64. During transcription, what event initiates the transition from initiation to elongation? a. The σ subunit binds to the core enzyme of RNA polymerase. b. The α subunit binds to the upstream element. c. The new RNA chain reaches 9-10 nucleotides in length. d. The transcription bubble is formed. e. The σ subunit is ejected from the transcription complex.

ANSWER: e 65. An _____ binds to a repressor, causing structural changes that lower the repressor’s affinity for the operator DNA. ANSWER: inducer 66. Which DNA sequence is likely a termination signal for RNA transcription? a. 3’-GCGCATTACGCGUUU-5’ b. 5’-GCGCATTACGCGUUU-3’ c. 3’-GCGCATTACGCGTTT-5’ d. 5’-GCGCATTACGCGTTT-3’ e. 5’-GCGCATTAGCGCUUU-3’ ANSWER: c 67. Rho requires what cofactor to move along an RNA strand? a. Mg2+ b. NADPH c. ATP d. FADH2 e. cAMP ANSWER: c 68. Which enzyme excises the 5S, 16S, and 23S rRNA precursors from the primary transcript? a. rho b. ribonuclease III c. ribonuclease P d. β-galactosidase e. CAP ANSWER: b 69. _____ is a terminal sequence required for the function of all tRNAs. a. UAU .. Powered by Cognero.

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Chapter 36 b. UUU c. GCG d. CCA e. GAG ANSWER: d 70. How does Actinomycin D inhibit transcription? a. Actinomycin D binds tightly to DNA, preventing it from being an effective template. b. Actinomycin D interferes with the formation of the initial phosphodiester linkages, preventing initiation. c. Actinomycin D binds to RNA polymerase, changing the protein’s conformation so the σ subunit cannot bind to the core enzyme. d. Actinomycin D binds tightly to elongating RNA strands, preventing the formation of the hairpin loop for termination. e. Actinomycin D binds to RNA polymerase, causing the protein to disengage from the DNA template.

ANSWER: a

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Chapter 37 1. What do eukaryotes exploit to regulate gene expression? Select all that apply. a. intensive translation b. extensive mRNA processing c. complex transcription d. complex replication e. compartmentalization ANSWER: b, c, e 2. In eukaryotes during transcription: a. all three RNA polymerases recognize all possible promoter sequences. b. each eukaryotic RNA polymerase recognizes one unique promoter. c. both RNA polymerases recognize specific sets of promoters. d. RNA polymerases don't require promoters and operate only through enhancers. e. each RNA polymerase recognizes some set of promoter sequences.

ANSWER: e 3. The genes of 18S, 5.8S, and 28S ribosomal RNA are transcribed: a. by RNA polymerase III. b. separately. c. as a single transcript. d. at the same time as they are translated. e. with 5S rRNA. ANSWER: c 4. The result of association of transcription factors into large complexes which differ in their specificity to genes in different cells is called: a. combinatorial control. b. synergy. c. the hypomethylation effect. d. mediating control. e. a mediator complex.

ANSWER: a 5. What is the distinctive feature of many DNA-binding proteins that are involved in chromatin remodeling? a. helix-turn-helix motif b. acetyllysine binding domain c. zinc-finger motif d. Mg2+-binding domain e. ATP-binding domain ANSWER: b .. Powered by Cognero.

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Chapter 37 6. What is the distinctive feature of RNA polymerase II in comparison with other eukaryotic as well as prokaryotic polymerases? a. regulatory carboxyl-terminal domain (CTD) b. synthesis of microRNA c. synthesis of rRNA d. separation of transcription and translation in space and time e. bromodomain ANSWER: a 7. The transition from initiation to elongation is marked by the: a. dephosphorylation of the CTD domain of RNA polymerase II. b. phosphorylation of the CTD domain of RNA polymerase II. c. binding of a ligand by a steroid hormone receptor. d. binding of the G-protein. e. acetylation of TFIIH. ANSWER: b 8. Select all that apply. When phosphorylated, the carboxyl-terminal domain of RNA polymerase II: a. is insensitive to α-amanitin. b. enhances transcription. c. recruits other factors. d. transcribes ribosomal RNA. e. inhibits transcription.

ANSWER: b, c 9. Eukaryotic promoters: a. are referred to as trans-acting elements. b. have approximately the same composition. c. are referred to as cis-acting elements because they can be located at a distance from the genes they control. d. are referred to as cis-acting elements because they are located on the same molecule of DNA as the genes they control. e. are positioned near the genes they control.

ANSWER: d 10. A common selective estrogen receptor modulator used in breast cancer therapy is: a. tamoxifen. b. α-amanitin. c. acetyllysine binding protein. d. p63. e. CTD. .. Powered by Cognero.

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Chapter 37 ANSWER: a 11. Select all that apply. The promoter for RNA polymerase I can contain: a. ribosomal initiator (rInr). b. downstream promoter element (DPE). c. initiator element (Inr). d. TATA box. e. upstream promoter element (UPE).

ANSWER: a, e 12. Select all that apply. The promoter for RNA polymerase II can contain: a. ribosomal initiator (rInr). b. downstream promoter element (DPE). c. initiator element (Inr). d. TATA box. e. upstream promoter element (UPE).

ANSWER: b, c, d 13. What is the distinctive feature of the promoters for RNA polymerase III? a. location downstream of the start site within the transcribed sequence b. location within the TATA box c. location upstream of the start site within the transcribed sequence d. possession of the TATA box e. very distant location upstream of the start site ANSWER: a 14. Transcription activators can be linked with the preinitiation complex by the: a. enhancer. b. promoter. c. transcription factor. d. CTD. e. mediator.

ANSWER: e 15. Select all that apply. Pluripotent cells: a. are stem cells. b. can develop into any adult cell. c. can develop into any fetal cell. d. can develop into extraembryonic tissues. e. can grow into an entire organism.

ANSWER: a, b, c .. Powered by Cognero.

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Chapter 37 16. Eukaryotic RNA polymerases: a. unlike prokaryotic, require a primer. b. I are located in nucleoli and transcribe genes of rRNA. c. that are insensitive to α-amanitin are RNA polymerases III. d. can synthesize mRNA. e. III are the only polymerases capable of synthesizing rRNA. ANSWER: b 17. Which element is common for promoters of genes with constitutive expression? a. TATA box b. GC box c. DPE d. Inr e. UPE ANSWER: b 18. Which statement about regulation of transcription through eukaryotic promoters is TRUE? a. Deviations from the consensus sequence of the promoter don't decrease its activity in eukaryotes. b. Elements of the eukaryotic promoters need to be present on the template strand to be effectively recognized by RNA polymerase. c. Location of promoter elements on the template strand is required for the promoter to be recognized by transcription factors. d. Promoters are usually present on both strands of DNA, which is important for binding of TBP proteins. e. Elements of eukaryotic promoters need to be present on the coding strand to be effectively recognized by RNA polymerase.

ANSWER: c 19. Which is important for establishing tissue specificity? a. prokaryotic promoter b. TATA box c. eukaryotic promoter d. enhancer e. TFIID ANSWER: d 20. Coactivators mediate expression by: a. blocking RNA polymerase binding. b. loosening the bonding of the histone complex with DNA. c. inhibiting intron splicing. .. Powered by Cognero.

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Chapter 37 d. blocking binding of transcription factors. e. dephosphorylating the CTD domain.

ANSWER: b 21. Upon binding of a ligand: a. estrogen response elements are translocated into the nucleus. b. nuclear hormone receptors are internalized by the cell. c. nuclear receptors bind to the promoters. d. the ligand–nuclear receptor complex can bind specific response elements in the DNA. e. the receptor binds an enhancer.

ANSWER: d 22. Which statement about nuclear hormone receptors is FALSE? a. The DNA-binding domain of nuclear receptors is fully functional regardless of the ligand binding. b. Corepressors act as antagonists of nuclear hormone receptors. c. Ligand binding is required to recruit a coactivator but not a corepressor. d. Binding of the agonist of the estradiol receptor will facilitate transcription by recruiting transcription activators to the estrogen response elements (EREs). e. Binding of receptor agonists doesn't influence the binding of the ligand–receptor complex to response elements.

ANSWER: b 23. How does acetylation of histones influence transcription? Select all that apply. a. decreasing affinity of histones to DNA b. increasing affinity of histones to DNA c. recruiting TBP-associated factors (TAFs) d. initiating dense packaging of chromatin e. recruiting chromatin-remodeling engines ANSWER: a, c, e 24. What is the nature of contact between DNA and the TATA-box-binding protein? a. ionic bonds b. hydrophobic interactions c. protein–protein interaction d. covalent bonds e. complementary binding ANSWER: b 25. What part of nucleosome is crucial to the activation of transcription through remodeling of chromatin? a. histone H2B b. nucleosome core particle .. Powered by Cognero.

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Chapter 37 c. histone tails d. linker DNA e. major DNA groove ANSWER: c 26. The activity of which enzyme that regulates transcription is linked to intermediary metabolism through ATP-citrate lyase? a. histone acetyltransferase (HAT) b. helicase c. RNA polymerase d. histone deacetylase e. methyltransferase ANSWER: a 27. Methylation of DNA: a. attracts corepressors to the DNA. b. usually activates transcription. c. attracts coactivators to the DNA. d. forms methylated islands of nucleotides that are usually near the transcription start site. e. can prevent binding of proteins to DNA. ANSWER: e 28. By changing the expression of transcription factors in fibroblasts, it is possible to obtain a(n): a. pluripotent cell. b. totipotent cell. c. induced pluripotent cell. d. induced totipotent cell. e. organ transplant.

ANSWER: c 29. Which statement about regulation of gene expression is FALSE? a. Hypomethylated regions of DNA can't be hypersensitive to DNase I. b. Most of the mammalian genes can be regulated through the CpG islands. c. Hypersensitivity to DNase I is associated with the action of chromatin-remodeling engines. d. Acetylated histones should be present in the sites hypersensitive to DNase I. e. Hypermethylated regions of DNA are not prone to cleavage by DNase I. ANSWER: a 30. Choose the CORRECT consensus sequence of the given TATA boxes. Account only for 7 bases starting with the first T of the box. The subscripts in the answer options denote the frequency (%) of the base at the corresponding position. .. Powered by Cognero.

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Chapter 37 1 GGGCTATAATACGCCACCTA 2 GAGCTATAATGCTAGCGTCA 3 GAGGTATAGAGCAGGGTTCA 4 CGACTATAAAAGGGGTCTAA 5 TCGGTATAAAGGTGAACAGA a. T100A100T100A100G20A60A40 b. T100A100T100A100A80A60A40 c. T100A100T100A100A90A60G60 d. T100A100T100A100A80A60G60 e. T100A100T100A100A80T50G40 ANSWER: d 31. Choose the CORRECT position at which given TATA boxes actually start. The transcription start site is given in green color.

1 GGGCTATAATACGCCACCTA 2 GAGCTATAATGCTAGCGTCA 3 GAGGTATAGAGCAGGGTTCA 4 CGACTATAAAAGGGGTCTAA 5 TCGGTATAAAGGTGAACAGA a. –10 b. –25 c. –15 d. –13 e. –16 ANSWER: c 32. Which statement about assembly of the preinitiation complex is FALSE? a. First, there is binding of TBP to the TATA box. b. Factors TFIIA and TFIIB bind to TFIID. c. RNA polymerase is recruited by TFIID. d. The factor TFIIE binds to the complex after TFIIF. e. The helicase TFIIH is the last one to bind to the complex.

ANSWER: c 33. What would be the first process to happen to regulate gene expression upon the beginning of hemoglobin synthesis in 35-hour-old chicken embryos? a. binding of a coactivator b. binding of TBP .. Powered by Cognero.

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Chapter 37 c. recruitment of HATs d. recruitment of demethylases e. recruitment of histone deacetylases ANSWER: d 34. What part of the 5′-CpG-3′ sequence is actually methylated? a. hydroxylated carbon of the cytosine base b. carbonyl carbon of the cytosine base c. one of the unsaturated carbons of the cytosine base d. 5′ carbon of deoxyribose e. carbonyl carbon of the guanine base ANSWER: c 35. Which statement about eukaryotic regulatory sequences is FALSE? a. Binding of the mediator requires an enhancer and promoter to be available for transcription activators and factors, respectively. b. The GC box can be located within CpG islands since they can be near the start of transcription and consist of appropriate nucleotides. c. The CAAT and GC boxes also require pairing with the initiator element at the +1 position. d. The regulatory +30 sequence will recruit PIC but without the help of TBP. e. RNA polymerase II isn't able to recognize any of the regulatory sequences by itself.

ANSWER: b 36. Binding of _____ could be the most affected process with a mutation of cysteine residues of the nuclear estradiol receptor. Select all that apply. a. Zn2+ b. DNA c. estradiol d. a corepressor e. coactivators ANSWER: a, b, d 37. Select all that apply. Combinatorial control is exploited by eukaryotes to: a. adjust the high strength of eukaryotic promoters. b. determine cell specificity. c. adjust the probability of transcription initiation. d. contribute to the developmental stage. e. account for the space separation of transcription and translation.

ANSWER: b, c, d 38. Which statement about chromatin remodeling upon expression of genes is TRUE? .. Powered by Cognero.

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Chapter 37 a. First, TBP binds to the promoter and starts assembly of the preinitiation complex. b. HATs are attracted by coactivators to weaken the ionic bonds between the DNA backbone and histones. c. The coactivator binds to acetylated lysine residues of histones. d. The preinitiation complex can be assembled right after histones are acetylated. e. HATs are attracted by the chromatin-remodeling engines to weaken the ionic bonds between DNA bases and unwind the double helix.

ANSWER: b 39. An example of a bifunctional enzyme is: a. TFIIH. b. RNA polymerase III. c. histone acetyltransferase. d. RNA polymerase II. e. TFIID. ANSWER: a 40. Which statement about interactions of transcription factors with DNA is FALSE? a. A minor groove of DNA needs to be bent for TBP to bind the TATA box. b. Changes in the DNA major groove near the gene can inhibit transcription. c. A major groove of DNA is unlikely to contain a TATA box. d. Unwinding of the DNA helix isn't required for the binding of transcription factors. e. Methylation of DNA inhibits transcriptions, but methylation of histones can activate it.

ANSWER: d 41. Triiodothyronine (T3) is a thyroid hormone that has a nuclear receptor but cannot cross the plasma membrane by simple diffusion. Which statement about the regulation of gene expression by T3 is TRUE? a. A specific coactivator binds to the T3–thyroid hormone receptor complex bound with the response elements. b. After binding the thyroid hormone receptor, the formed complex interacts with the specific response elements. c. A corepressor is bound to the specific response elements, preventing binding of the T3–thyroid hormone receptor complex d. A corepressor prevents chromatin remodeling by binding to the T3–thyroid hormone receptor complex. e. A specific coactivator recruits RNA polymerase after binding to the thyroid hormone receptor.

ANSWER: a 42. Which process that activate transcription in eukaryotes is FALSE? The transcription factor binds to the: a. –25 regulatory promoter sequence. .. Powered by Cognero.

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Chapter 37 b. template strand prior to the transcription start. c. +30 regulatory promoter sequence. d. –10 regulatory promoter sequence. e. template strand after the transcription start.

ANSWER: d 43. Which statement about eukaryotic RNA processing is TRUE? a. Splicing of the mRNA precursor is done in the cytoplasm, while modification of the ends is done in the nucleus. b. An unprocessed mRNA precursor is translocated into the cytoplasm to be modified before the translation. c. Transcription yields primary transcript, which can by itself yield a functional protein. d. Splicing of the mRNA precursor is done in the nucleus, while modification of the ends is done in the cytoplasm. e. Functional individual rRNA cannot be produced without splicing.

ANSWER: e 44. Phosphorylation of histones is best known to participate in: a. transcription activation. b. transcription repression. c. DNA reparation. d. DNA hypersensitivity to the DNase I. e. hypomethylation of DNA. ANSWER: c 45. Hypothetical protein X can specifically bind acetylated H3K9 residue. Which protein that it can recruit and their effects on transcription are TRUE? Select all that apply. a. specific methylase, which would activate the transcription by monomethylation of H3K4 b. deacetylase, which would activate the transcription c. other chromatin-remodeling proteins to activate the transcription d. coactivator; to activate the transcription e. corepressor; to repress the transcription ANSWER: a, c, d 46. Which transcription factor is important for transcription elongation as well as for initiation in eukaryotes? a. TFIIA b. TFIIB c. TFIID d. TFIIE e. TFIIF ANSWER: e .. Powered by Cognero.

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Chapter 37 47. RNA-processing enzymes that can act in the course of elongation in eukaryotes are recruited by: a. TFIIB. b. TFIIH. c. phosphorylated CTD of RNA polymerase II. d. activation domain of TFIID. e. dephosphorylated CTD of RNA polymerase II. ANSWER: c 48. Why is the action of nuclear receptors so different from membrane-bound receptors? a. Membrane-bound receptors have no influence on nuclear proteins. b. Nuclear receptors activate transcription factors via methylation reactions, not phosphorylation. c. Membrane-bound receptors activate or inhibit enzymes and do not regulate transcription. d. Nuclear receptors bind steroids, which can cross the cellular membrane; thus, specificity of action resides in the cytoplasm. e. Membrane-bound receptors bind carbohydrate moieties, whereas nuclear receptors bind prostaglandin hormones.

ANSWER: d 49. Why would an inhibitor of RNA polymerase II NOT cause rapid cell death, even if every molecule of RNA polymerase II were inhibited? a. Phosphorylation protects RNA polymerase II from binding of an inhibitor. b. As long as TFIID can bind, RNA polymerase II can stay bound and continue to transcribe proteins. c. There would still be many functional mRNA and viable proteins to continue cellular functions. d. TBP binding to the TATA box protects degradation of RNA polymerase II. e. Downstream core promoter elements initiate RNA polymerase activation.

ANSWER: c 50. When a steroid hormone receptor binds a ligand, it does not change the binding of a receptor to DNA but to the _______.

ANSWER: coactivator 51. In eukaryotic cells, transcription occurs in the ___, while translation occurs in the ____.

ANSWER: nucleus, cytoplasm 52. RNA polymerase I is localized to what specialized structure within eukaryotic cells? a. nucleolus b. nucleoplasm c. cytoplasm d. ribosome e. endoplasmic reticulum ANSWER: a .. Powered by Cognero.

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Chapter 37 53. Trans-acting elements are ________. a. proteins that regulate RNA translation b. DNA sequences that regulate gene expression and are located on the same molecule of DNA c. RNA sequences that regulate RNA synthesis d. proteins that recognize cis-acting elements and regulate RNA synthesis e. DNA sequences that recognize cis-acting elements, regulate RNA synthesis, and are located on the same molecule of DNA ANSWER: d 54. Trans-acting elements are commonly called _________. a. promoters b. enhancers c. chromatin d. transcription factors e. histones ANSWER: d 55. Where is the TATA box usually located in eukaryotic genes? a. -25 b. -10 c. -40 d. +30 e. +25 ANSWER: a 56. Unlike the -35 region in bacteria, the CAAT box and GC box can be effective when present on the ____ strand.

ANSWER: template or antisense 57. The DPE is commonly found in conjunction with an initiator element for transcripts that lack a _______. a. TATA box b. -35 region c. GC box d. promoter e. CAAT box ANSWER: a 58. Which transcription factor binds to the TATA box at the beginning of initiation? a. TFIIA b. TFIIB .. Powered by Cognero.

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Chapter 37 c. TFIIH d. TFIID e. TFIIE ANSWER: d 59. What function does TFIIH perform in the preinitiation complex? Select all that apply. a. TFIIH binds to the TATA box. b. TFIIH unwinds the double helix. c. TFIIH recruits RNA polymerase II. d. TFIIH phosphorylates the carboxyl-terminal domain of RNA polymerase II, marking the switch from initiation to elongation. e. TFIIH stabilizes the pre-initiation complex.

ANSWER: b, d 60. What is an enhancer? a. A sequence that can stimulate transcription but has no promoter activity of its own. b. A sequence that acts as an additional promoter region for RNA polymerase to bind to. c. A sequence that binds to histone complexes. d. A sequence that regulates constitutive expression across all cell types. e. A sequence that inhibits the transcription of certain genes.

ANSWER: a 61. Which of the following statements about enhancers is TRUE? a. Enhancers are effective only when present on the template strand. b. Enhancers can only be located upstream of the transcribed gene. c. Enhancers require a promoter to function. d. A specific enhancer is only effective in certain cell types. e. Enhancers are effective only when present on the coding strand.

ANSWER: d 62. What is the name of a large complex of intermediary proteins that acts as a bridge between transcription factors and RNA polymerase II? a. enhancer b. PIC c. mediator d. promoter e. CTD ANSWER: c 63. How many transcription factors are needed to induce pluripotency in already-differentiated skin cells? a. 3 .. Powered by Cognero.

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Chapter 37 b. 2 c. 4 d. 7 e. 10 ANSWER: c 64. DNA sites bound by nuclear hormone receptors are known as _______. a. response elements b. enhancers c. mediators d. promoter e. transcription factors ANSWER: a 65. What is the consensus sequence for estrogen response elements? a. 5’-TCCAGTNNNACTGGA-3’ b. 5’-AGGTCANNNTGACCT-3’ c. 3’-AGGTCANNNTGACCT-5’ d. 3’-TAGGTCNNNGATCT-5’ e. 5’-TCTAGNNNCTGGAT-3’ ANSWER: b 66. What nucleotide does N stand for in a consensus sequence? a. adenine b. cytosine c. guanine d. thymine e. any of the above ANSWER: e 67. What are the principal domains of nuclear hormone receptors? Select all that apply. a. DNA-binding domain b. hinge domain c. ligand-binding domain d. amino terminal activation domain e. helicase domain ANSWER: a, b, c, d 68. DNA methylation and histone modification are examples of ________. a. activation b. combinatorial control .. Powered by Cognero.

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Chapter 37 c. pluripotency d. coactivators e. epigenetics ANSWER: e 69. Which of the following statements about CpG sequences is FALSE? a. About 70% of CpG sequences are methylated in mammalian genomes. b. The distribution of CpG sequence across mammalian genomes is uniform. c. Carbon 5 of cytosine is often methylated in CpG sequences. d. The absence of 5-methylcytosines near a start site is referred to as hypomethylation. e. Most genes are found in CpG islands.

ANSWER: b 70. Proteins that bind to a receptor after it has bound a steroid are called _________. a. activator proteins b. corepressors c. mediators d. coactivators e. enhancers ANSWER: d

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Chapter 38 1. What is a term for noncoding regions in primary transcripts of eukaryotic genes? a. nonsense b. empty c. intron d. exon e. enhancer ANSWER: c 2. What is a term for coding regions in primary transcripts of eukaryotic genes? a. sense b. consensus c. intron d. exon e. precursor ANSWER: d 3. The small subunit of a eukaryotic ribosome contains: a. 5.8S RNA. b. 5S RNA. c. 18S RNA. d. 28S RNA. e. snRNA. ANSWER: c 4. What snRNP recognizes the 5′ splice site at the first step of splicing in mammalian cells? a. U1 b. U2 c. U3 d. U4 e. U5 ANSWER: a 5. RNA self-splicing demonstrates the role of RNA as a(n): a. intron. b. exon. c. spliceosome. d. nucleophile. e. catalyst.

ANSWER: e 6. The immediate product of RNA polymerase II is often referred to as: .. Powered by Cognero.

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Chapter 38 a. miRNA. b. mRNA. c. pro-mRNA. d. pre-mRNA. e. proto-mRNA. ANSWER: d 7. What is the CORRECT sequence of eukaryotic pre-rRNA processing? a. cleavage → association with snoRNPs → modification of nucleotides → assembly with ribosomal proteins b. association with snoRNPs → modification of nucleotides → assembly with ribosomal proteins → cleavage c. association with snoRNPs → modification of nucleotides → cleavage → assembly with ribosomal proteins d. assembly with ribosomal proteins → modification of nucleotides → association with snoRNPs → cleavage e. cleavage → assembly with ribosomal proteins → association with snoRNPs → modification of nucleotides ANSWER: b 8. The function of small nucleolar ribonucleoproteins (snoRNPs) is to: a. catch small nucleolar RNA and inactivate them. b. direct the base modification in pre-rRNA. c. be part of ribosomes. d. determine the sites of cleavage of pre-rRNA. e. enhance assembly of ribosomes.

ANSWER: b 9. What event would be inhibited by a mutation in a ribozyme that catalyzes a transesterification reaction? a. mRNA termination b. polyadenylation c. mRNA splicing d. 5′ capping e. chain initiation ANSWER: c 10. The 5′ end of mRNA is modified: a. immediately after initiation of RNA synthesis. b. shortly after initiation of RNA synthesis, when the strand is about 100 nucleotides in length. c. shortly after initiation of RNA synthesis, when the strand is about 25 nucleotides in length. d. after the C-terminal domain of RNA polymerase II is dephosphorylated. .. Powered by Cognero.

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Chapter 38 e. after termination of RNA synthesis.

ANSWER: c 11. What structural element of eukaryotic cap 1 is methylated? a. N-7 nitrogen atom of the 3′-terminal guanine and ribose moiety of the third base b. N-5 nitrogen atom of the 5′-terminal guanine and ribose moiety of the first base c. N-7 nitrogen atom and ribose moiety of the 5′-terminal guanine d. N-7 nitrogen atom of the 5′-terminal guanine and ribose moiety of the second base e. N-5 nitrogen atom of the 5′-terminal guanine and ribose moiety of the third base ANSWER: d 12. What sequence in the 3′-end region of pre-mRNA is recognized by the cleavage and polyadenylation specificity factor? a. AAUAAA b. GU-rich region downstream of the cleavage site c. CCA d. AAGUAAGU e. CstF ANSWER: a 13. How many adenosine bases are in a poly(A) tail? a. 5 b. 25 c. 100 d. 250 e. 500 ANSWER: d 14. What event takes place at the second transesterification step of splicing? a. The branch-site adenosine attacks the 5′ splice site. b. The branch-site adenosine attacks the 3′ splice site. c. The 3′ splice-site hydroxyl group attacks the branch-site adenosine. d. The 5′ splice-site hydroxyl group attacks the 3′ splice site. e. The 3′ splice-site hydroxyl group attacks the 5′ splice site.

ANSWER: d 15. What statement about the spliceosome is FALSE? a. The spliceosome may be the most complicated molecular machine in the cell. b. The spliceosome contains more than 300 proteins. c. The spliceosome contains several snoRNAs. d. RNA components of the spliceosome recognize the 5′ and 3′ splice sites and the branch site. .. Powered by Cognero.

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Chapter 38 e. It removes introns and links exons together to form mature mRNA. ANSWER: c 16. What event initiates a spliceosome assembly? a. The complex of U4, U5, and U6 binds to the complex of U1 and U2. b. U2 snRNP binds to the branch point of the pre-mRNA. c. The complex of U4, U5, and U6 snRNPs binds to the pre-mRNA strand. d. U5 snRNP interacts with the 5′ and 3′ splice sites. e. U1 snRNP binds to the 5′ splice site.

ANSWER: e 17. What forms the catalytic center of the spliceosome? a. U5 b. U2 and U6 c. U4–U5–U6 tri-snRNP d. U2, U4, and U6 e. U1 and U5 ANSWER: b 18. Which statement about splicing is FALSE? a. The splicing reaction is completed by the release of excised circular intron from the spliceosome. b. RNA molecules play key roles in directing the alignment of splice sites. c. RNA molecules play key roles in catalysis of the transesterification reactions. d. ATP-powered RNA helicases must unwind RNA helices and allow alternative base-pairing arrangements to form. e. Both transesterification reactions are promoted by a pair of bound magnesium ions.

ANSWER: a 19. What is affected by trans-acting mutations causing defective pre-mRNA splicing? a. pre-mRNA itself b. template strand of DNA c. splicing factors d. noncoding regions of DNA e. RNA polymerase ANSWER: c 20. Which statement about alternative splicing is FALSE? a. Alternative splicing is a widespread mechanism for generating protein diversity. b. Selection of specific splicing sites is determined by the binding of trans-acting splicing factors to cisacting sequences in pre-mRNA. c. Mutations that affect alternative splicing cause diseases such as thalassemia. .. Powered by Cognero.

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Chapter 38 d. Current estimates are that more than 70% of human genes that encode proteins are alternatively spliced. e. Different combinations of exons in the same gene may be spliced into a mature RNA, producing distinct forms of a protein.

ANSWER: c 21. Consider a gene with three positions at which alternative splicing can take place. Assuming that these alternative splicing pathways are regulated independently, what is the total quantity of different mRNA that can be generated. a. 27 b. 9 c. 8 d. 6 e. 3 ANSWER: c 22. What is required for self-splicing in a group I intron? Select all that apply. a. guanosine cofactor b. free 2′-OH group of a specific adenylate of the intron c. free 3′-OH group of a specific adenylate of the intron d. free guanine nucleotide e. complex folding pattern of helices and loops in the intron ANSWER: a, d, e 23. Select all that apply. The polypyrimidine tract: a. is found at the 3′ end of an intron. b. is a consensus sequence. c. contains a stretch of 10 pyrimidines. d. is followed by any base and then by G. e. ends with the invariant AG. ANSWER: a, b, c, e 24. What is required for self-splicing in a group II intron? Select all that apply. a. guanosine cofactor b. free 2′-OH group of a specific adenylate of the intron c. free 3′-OH group of a specific adenylate of the intron d. free guanine nucleotide e. complex folding pattern of helices and loops in the intron ANSWER: b, e 25. What RNA are transcribed by RNA polymerase I? Select all that apply. .. Powered by Cognero.

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Chapter 38 a. 18S rRNA b. 28S rRNA c. tRNA d. 5S RNA e. 5.8S rRNA ANSWER: a, b, e 26. What RNA are transcribed by RNA polymerase III? Select all that apply. a. 18S rRNA b. mRNA c. tRNA d. 5S rRNA e. 5.8S rRNA ANSWER: c, d 27. What does the mammalian pre-rRNA transcript contain? Select all that apply. a. CCA tail b. rRNA of the large ribosomal subunit c. rRNA of the small ribosomal subunit d. multiple methylation sites e. pseudouridine bases ANSWER: b, c, d, e 28. What are the functions of the 5′ cap? Select all that apply. a. The cap contributes to the stability of mRNA by protecting their 5′ ends from phosphatases. b. The cap binds transport proteins to deliver mRNA from the nucleus to the cytoplasm. c. The cap enhances the translation of mRNA. d. The type of the cap (0, 1, or 2) defines the degree of methylation of mRNA bases. e. The cap enhances the lifetime of mRNA by protecting their 5′ ends from nucleases.

ANSWER: a, c, e 29. What enzymes take part in 5′ capping of eukaryotic mRNA? Select all that apply. a. RNA phosphatase b. RNA triphosphatase c. nucleotidyltransferase d. guanylyltransferase e. N-7 guanine methyltransferase ANSWER: b, d, e 30. What components does the eukaryotic polyadenylation complex contain? Select all that apply. a. specific endonuclease .. Powered by Cognero.

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Chapter 38 b. PAS c. CstF d. CPSF e. RNase Z ANSWER: a, c, d 31. Choose the means of generating diverse proteins from one gene. Select all that apply. a. splicing b. alternative splicing c. RNA editing d. RNA methylation e. mutations ANSWER: b, c 32. What proteins are products of RNA editing? Select all that apply. a. K+ channels of octopuses living in tropical and arctic seas b. apolipoproteins apo B-100 and apo B-48 c. antibodies d. blood coagulation factors VII and IX e. globins β and γ ANSWER: a, b 33. What are common structural features of introns in vertebrates? Select all that apply. a. Introns begin with the invariant GU. b. The consensus sequence at the 3′ end includes 10 pyrimidines. c. The consensus sequence at the 5′ end is AGAUAA. d. The polypurine tract at the 3′ end is followed by G. e. Introns end with the invariant AG. ANSWER: a, b, e 34. What interaction allows snRNPs of the spliceosome to bind the pre-mRNA strand? a. electrostatic attraction between phosphoryl groups of the pre-mRNA and two magnesium ions in the catalytic center b. hydrophobic interaction between phosphoryl groups of the pre-mRNA and two magnesium ions in the catalytic center c. van der Waals interaction between pre-mRNA and amino acids of snRNPs d. hydrogen bonding between pre-mRNA and amino acids of snRNPs e. base-pairing between pre-mRNA and complimentary sequences of snRNAs ANSWER: e 35. What is the main function of U5 snRNP in the spliceosome? .. Powered by Cognero.

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Chapter 38 a. It is included in the catalytic center of the spliceosome. b. It holds two magnesium ions bound by phosphate groups. c. An inhibitor masks U6 until the specific splice sites are aligned. d. It holds the 5′ exon and the 3′ exon far away from each other. e. It holds the free 5′ exon and the 3′ exon close together.

ANSWER: e 36. What event changes a codon for glutamine (CAA) to a stop codon (UAA) in the mRNA of apolipoprotein B-100? a. methylation b. amination c. deamination d. hydration e. dehydration ANSWER: c 37. Catalytic RNA is called: a. ribosome. b. ribozyme. c. ribose. d. ribonuclease. e. ribonucleotidase.

ANSWER: b 38. What coordinates the transcription and processing of mRNA by recruiting different proteins? a. CTD of RNA polymerase II b. the nascent pre-mRNA strand c. CCA of RNA polymerase II d. the YSPSTPS domain of RNA polymerase II e. transcription factors, particularly TFIIH ANSWER: a 39. What event in eukaryotic tRNA processing is FALSE? a. modification of base and ribose moieties b. removal of the 3′ trailer by RNase Z c. cleavage of the 5′ leader by RNase P d. CCA addition by tRNA nucleotidyltransferase e. AUU addition to the 3′ end by tRNA nucleotidyltransferase ANSWER: e 40. The ATP-powered _______ unwinds RNA duplex intermediates of splicing. .. Powered by Cognero.

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Chapter 38 ANSWER: helicase 41. What is used to form cap 0 of mRNA? a. S-adenosylmethionine b. cysteine c. biotin d. methanol e. dimethyl-RNA methylase ANSWER: a 42. A loss of 3′ polyadenylation will lead to: a. intensified capping. b. decreased stability of the mRNA. c. intensified splicing of mRNA. d. alternative splicing. e. intensified splicing of tRNA. ANSWER: b 43. Proteins that possess alternative splicing products include: a. antibodies. b. hemoglobin β. c. apolipoprotein. d. RNA polymerase. e. CPSF endonuclease.

ANSWER: a 44. The carboxy-terminal domain (CTD) of RNA polymerase II undergoes reversible phosphorylation by transcription factor TFIIH during transcription. What does this phosphorylation event signal? a. Dephosphorylation signals recruiting of proteins to catalyze polyadenylation. b. Phosphorylation signals recognition of the "stop" codon. c. Dephosphorylation signals recruiting of spliceosome proteins. d. Phosphorylation signals the transition from transcription initiation to elongation. e. Dephosphorylation activates the catalytic function of ribozymes.

ANSWER: d 45. The function of the guanosine cofactor in self-splicing is: a. to provide energy. b. to provide an attacking group. c. a necessary base for RNA editing. d. to provide nucleophilic phosphate group. e. a matrix for splicing. .. Powered by Cognero.

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Chapter 38 ANSWER: b 46. Because RNA polymerase II has no proofreading function, it may provide a mechanism for generating diversity. What is the most likely reason for the lack of proofreading function in RNA polymerase II? a. Small nucleolar ribonucleoproteins direct modification to form pseudouridine bases. b. RNA can be altered after transcription by RNA editing. c. Proofreading would slow down transcription, preventing cells from responding rapidly to changing cellular demands. d. The 5′ leader sequence is removed prior to translation. The synthesis of this extra length of transcript is thermodynamically unfavorable. e. Several 5′ caps are added to mRNA, which shifts the reading frame, depending on the cap.

ANSWER: b 47. You read an article in a popular magazine about a unicellular plant that has DNA with catalytic activity.

Knowing that RNA can have catalytic activity, would you believe this article? a. Yes, DNA could easily be attacked by a guanosine cofactor. b. Yes, if DNA is not associated with proteins that would constrain the formation of DNA-loop structures. c. Yes, because all that is required is a 3′–OH to attack the 3′ splice site. d. No, uracil of RNA is required in the catalytic site. e. No, DNA is constrained by its double helix structure and could not fold into a catalytic structure.

ANSWER: e 48. Why might a single base-pair mutation in eukaryotic mRNA be less serious than one in prokaryotic mRNA? a. If the mutation occurs in the exon, it will not affect the gene product. b. If the mutation occurs in the splice site of a transcript with alternative splicing, only one gene product may be affected. c. If the mutation occurs in the 3′ end of the start site, it will not affect the gene product. d. If the mutation occurs in the intron or not in the splice site of a transcript with alternative splicing, it will not affect the gene product. e. If the mutation occurs in the 5′ end of the start site, it will not affect the gene product.

ANSWER: d 49. Diseases caused by mutations in splicing factors include: a. hemophilia A. b. thalassemia. c. retinitis pigmentosa. d. cystic fibrosis. e. spinal muscle atrophy.

ANSWER: c 50. The percentage of all genetic diseases caused by mutations that affect mRNA splicing is: .. Powered by Cognero.

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Chapter 38 a. 2%. b. 15%. c. 20%. d. 30%. e. 7%.

ANSWER: b 51. What is a primary transcript? a. a processed sequence of RNA consisting only of exons b. a processed sequence of RNA consisting only of introns c. an unprocessed sequence of RNA d. a sequence of DNA coding for a mRNA strand e. a sequence of DNA coding for a tRNA strand ANSWER: c 52. Where in the cell does rRNA processing primarily take place? a. nucleolus b. nucleoplasm c. cytoplasm d. ribosome e. endoplasmic reticulum ANSWER: a 53. Which of the following is not considered a functional RNA? a. mRNA b. rRNA c. tRNA d. miRNA e. snoRNA ANSWER: a 54. Nucleotides on the 5’ end of the tRNA precursor are cleaved by _____. a. poly(A)-polymerase b. RNase Z c. tRNA nucleotidyltransferase d. guanylyltransferase e. RNase P ANSWER: e 55. Nucleotides on the 3’ end of the tRNA precursor are cleaved by _____. a. poly(A)-polymerase .. Powered by Cognero.

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Chapter 38 b. RNase Z c. tRNA nucleotidyltransferase d. guanylyltransferase e. RNase P ANSWER: b 56. How long is the mRNA primary transcript when 5’-capping occurs? a. 50 nucleotides b. 100 nucleotides c. 25 nucleotides d. 75 nucleotides e. 150 nucleotides ANSWER: c 57. What is the PAS sequence for 3’-end processing of pre-mRNA? a. 5’-AAUAAA-3’ b. 5’-AATAAA-3’ c. 5’-AAAUAA-3’ d. 5’-AAATAA-3’ e. 5’-TTATTT-3’ ANSWER: a 58. The GU-rich region is recognized by which component of the polyadenylation complex? a. poly(a) polymerase b. PAS c. CPSF d. CstF e. endonuclease ANSWER: d 59. What is the consensus sequence for the 5’-splice site in vertebrates? a. 3’-AGGUAAGU-5’ b. 5’-AGGUAAGU-3’ c. 3’-AGGTAAGT-5’ d. 5’-AGGTAAGT-3’ e. 5’-UGAAUGGA-3’ ANSWER: b 60. What class of RNA act as essential components of the spliceosome? a. mRNA b. miRNA .. Powered by Cognero.

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Chapter 38 c. snoRNA d. snRNA e. rRNA ANSWER: d 61. What RNA-protein complexes act as key components of the spliceosome? a. CTD b. PAS c. CPSF d. CstF e. snRNPs ANSWER: e 62. Cis-acting mutations can alter which component of mRNA processing? a. spliceosome b. snRNA c. pre-mRNA d. poly(a) tail e. 5’-cap ANSWER: c 63. What is transesterification? a. the reaction of an alcohol and an ester to form a different alcohol and ester b. the reaction of an alcohol and an ester to form a different alcohol c. the reaction of an alcohol and an ester to form a different ester d. the reaction of an alcohol and a carboxylic acid to form an ester e. the reaction of an ester and hydrogen to form an alcohol ANSWER: a 64. What role does U4 play in splicing? a. U4 inhibits U2 until the specific splice sites are aligned. b. U4 holds the two exons in close proximity following the first transesterification event. c. U4 inhibits U6 until the specific splice sites are aligned. d. U4 serves as the catalytic center of the spliceosome, along with U6 e. U4 recognizes and binds the 5’ splice site.

ANSWER: c 65. What shape does the excised intron take when bound to the U2-U5-U6 complex? a. a helix b. a hairpin c. a strand .. Powered by Cognero.

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Chapter 38 d. a lariat e. a spiral ANSWER: d 66. How many Mg2+ ions are bound to the phosphate groups of U6? a. 1 b. 2 c. 5 d. 6 e. 3 ANSWER: b 67. Following cleavage of the 5’ exon during splicing, the intron takes the form of a lariat intermediate via an unusual 5’-2’ _____ bond.

ANSWER: phosphodiester 68. Thalassemia is caused by the following base pair mutation: a. A-to-T b. A-to-C c. A-to-G d. C-to-G e. T-to-A ANSWER: c 69. The carboxyl-terminal domain (CTD) of RNA polymerase II contributes to pre-mRNA processing by recruiting which of the following proteins? Select all that apply. a. capping enzymes b. helicase c. components of the spliceosome d. transcription factors e. specific endonuclease ANSWER: a, c, e 70. What is RNA editing? a. a change in RNA nucleotide sequence prior to transcription b. a change in RNA nucleotide sequence that occurs via splicing after transcription c. a change in RNA nucleotide sequence after transcription via any process other than splicing d. a change in RNA nucleotide sequence after splicing e. a change in RNA nucleotide sequence after translation ANSWER: c .. Powered by Cognero.

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Chapter 39 1. The _______ is the relation between the sequence of bases in DNA (or its RNA transcripts) and the sequence of amino acids in proteins.

ANSWER: genetic code 2. _____ nucleotides encode an amino acid. a. Two b. Three c. Four d. Five e. Six ANSWER: b 3. The group of three bases that encode an amino acid is called: a. codon. b. sequence. c. lexicon. d. DNA. e. RNA. ANSWER: a 4. Codons that specify the same amino acid are called: a. analogues. b. antonyms. c. synonyms. d. equivalents. e. homonyms.

ANSWER: c 5. What amino acids are encoded by only one triplet? Select all that apply. a. leucine b. valine c. arginine d. tryptophan e. methionine ANSWER: d, e 6. What characteristic of the genetic code minimizes deleterious effects of mutations? a. degradation b. universality c. termination d. directionality .. Powered by Cognero.

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Chapter 39 e. degeneracy ANSWER: e 7. What is the name of the phenomenon of a nonrandom use of synonymous codons in genes in different organisms? a. codon dislocation b. nucleotide bias c. sequence bias d. codon bias e. synonym bias ANSWER: d 8. What process may codon bias help to regulate? a. initiation b. transcription c. RNA processing d. translocation e. translation ANSWER: e 9. What molecule serves as an adaptor between the codon and its specified amino acid? a. transfer RNA b. transfer DNA c. transport RNA d. transport DNA e. transmission RNA ANSWER: a 10. In what part of the cell does DNA encode a distinct set of transfer RNA–adaptor molecules that recognize the alternative codons? a. endoplasmic reticulum b. cytoplasm c. lysosome d. mitochondria e. nucleus ANSWER: d 11. How many ribonucleotides does a transfer RNA molecule contain? a. between 33 and 53 b. between 53 and 73 c. between 73 and 93 .. Powered by Cognero.

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Chapter 39 d. between 73 and 83 e. between 93 and 103 ANSWER: c 12. What is the three-dimensional shape of transfer RNA molecules? a. U-shape b. L-shape c. O-shape d. X-shape e. T-shape ANSWER: b 13. What nucleoside in tRNA is formed by deamination of adenosine after synthesis of the primary transcript? a. thymine b. inosine c. adenine d. adenosine e. guanine ANSWER: b 14. In what pattern can all tRNA molecules be arranged when depicted on a two-dimensional surface? a. cloverleaf pattern b. flower pattern c. maple leaf pattern d. petal pattern e. olive leaf pattern ANSWER: a 15. What part of tRNA is phosphorylated? a. 3′ end b. 5′ end c. 3′ CCA d. 5′ CCA e. acceptor stem ANSWER: b 16. What base of an anticodon determines whether a particular tRNA molecule reads one, two, or three kinds of codons? a. first b. second c. third .. Powered by Cognero.

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Chapter 39 d. fourth e. fifth ANSWER: a 17. What tRNA base maximizes the number of codons that this tRNA can recognize? a. inosine b. 5-methylcytosine c. pseudouracil d. dehydrouridine e. guanine ANSWER: a 18. Protein synthesis takes place in: a. lysosomes. b. nuclear pores. c. vacuoles. d. ribosomes. e. mitochondria.

ANSWER: d 19. What chemical structure do the activated intermediates have in protein synthesis? a. amino acid aldehydes b. amines c. amino alcohols d. amino acid esters e. amino acid ketones ANSWER: d 20. What is an amino acid ester of tRNA called? a. aminoacyl-tRNA b. acetyl-tRNA c. aminoacyl adenylate d. succinyl-tRNA e. acetoacetyl-tRNA ANSWER: a 21. To which functional group of the adenylate is the carboxyl group of the amino acid attached in the first step of amino acid activation? a. 5′-hydroxyl group b. 2′-hydroxyl group c. 3′-hydroxyl group .. Powered by Cognero.

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Chapter 39 d. 5′-phosphoryl group e. 3′-phosphoryl group ANSWER: d 22. Which statement about activation of amino acids for protein synthesis is TRUE? a. Aminoacyl-AMP is part of the CCA terminus of tRNA. b. Formation of aminoacyl-tRNA requires hydrolysis of three molecules of ATP. c. The aminoacyl-adenylate intermediate dissociates from the enzyme during activation. d. The pyrophosphate is hydrolyzed in the first step of activation, and it drives the whole activation. e. Aminoacyl-AMP is formed from amino acid and ADP. ANSWER: d 23. The equivalent of how many molecules of ATP is consumed in the synthesis of each aminoacyl-tRNA? a. 2 b. 3 c. 4 d. 5 e. 1 ANSWER: a 24. What prevents the aminoacyl-AMP intermediate from leaving the active site of the aminoacyl-tRNA synthetase? a. covalent bond with catalytic aspartate of the active site b. noncovalent interactions with the site c. noncovalent interactions with the CCA terminus of tRNA d. hydrolysis of the pyrophosphate e. covalent bond with a metal ion ANSWER: b 25. Aminoacyl-tRNA synthetase will incorporate the incorrect amino acid only once in: a. 102 or 103 reactions. b. 104 or 105 reactions. c. 105 or 106 reactions. d. 107 or 108 reactions. e. 109 or 1010 reactions.

ANSWER: b 26. What ion does the binding site of threonyl-tRNA synthetase contain? a. iron b. zinc .. Powered by Cognero.

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Chapter 39 c. magnesium d. calcium e. potassium ANSWER: b 27. What would be the main negative outcome of the frequency if mistakes in protein synthesis were equal to 10–2? a. The protein synthesis would be too slow. b. The protein synthesis would halt. c. All proteins longer than 100 residues would be mutated. d. All proteins shorter than 100 residues would be mutated. e. Absolutely all proteins would be mutated.

ANSWER: c 28. What functional group prevents Thr-tRNAThr from fitting into the editing site of the corresponding aminoacyl-tRNA synthetase? a. butyl group b. hydroxyl group c. acetyl group d. ethyl group e. methyl group ANSWER: e 29. What parts of tRNA are mostly recognized by the synthetase? a. anticodon loop and acceptor stem b. anticodon loop and DHU loop c. acceptor stem and CCA terminus d. CCA terminus and anticodon loop e. Tψc loop and DHU loop ANSWER: a 30. What is the molecular mass of the E. coli ribosome? a. 5.7 Da b. 1.7 Da c. 1.7 MDa d. 2.7 Da e. 2.7 MDa ANSWER: e 31. Which statement about a prokaryotic ribosome is TRUE? a. The large subunit contains more protein and RNA molecules than the small subunit. .. Powered by Cognero.

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Chapter 39 b. The small subunit is based on the 23S rRNA molecule. c. Even though the large subunit has a larger mass, it has a lower sedimentation coefficient. d. Ribosomal RNA has a mostly random folding pattern. e. The coefficient of sedimentation of the whole ribosome equals the sum of coefficients of the subunits.

ANSWER: a 32. What is the molecular basis of genetic code degeneracy? a. pseudouracil bases in the anticodon loop b. wobble base-pairing of the first base of the codon c. low specificity of the aminoacyl-tRNA synthetase d. wobble base-pairing of the third base of the codon e. Watson–Crick base-pairing of the third base of the anticodon ANSWER: d 33. What RNA are critical for prokaryotic ribosomal architecture and function? a. 10S, 12S, 16S b. 5S, 6S, 7S c. 3S, 5S, 16S d. 6S, 16S, 25S e. 5S, 16S, 23S ANSWER: e 34. Using the table of genetic code, which mRNA sequences will contain three glutamic acid residues?

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Chapter 39

a. 5′-UUC AAA GAG GGG CAA GGA UUU UAU-3′ b. 5′-CAA UGU CAG GGG CAA GGA GUU UGU-3′ c. 5′-UUC AAA GAG GAA CAA UUU GAA UAU-3′ d. 5′-CAA UGU CCC GAG CAA GGA CUA UCU-3′ e. 5′-CCC AGU GGU GGG GAA GGA GUU UGU-3′ ANSWER: c 35. Using the genetic code table, choose the DNA template strand that corresponds to the protein sequence with only two positively charged amino acid residues.

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Chapter 39

a. 5′-CGT CAT CTA AGA GCA-3′ b. 3′-GTA TTC CTA AAA GAG-5′ c. 5′-TTT CAT GCA AGA AAA-3′ d. 3′-CGT CAT CTA AGA GCA-5′ e. 3′-GTA TTC CTA AAA GCT-5′ ANSWER: b 36. What is the transcribing and translating complex consisting of RNA polymerase and the 60S ribosome called? a. polysome b. expressome c. multiple ribosome d. autosome e. monosome ANSWER: b 37. Where does the interaction of the polymerase with the ribosome occur? a. carbonyl-terminal domain of the polymerase b. hydroxyl-terminal domain of the polymerase c. carboxyl-terminal domain of the polymerase d. acetyl-terminal domain of the polymerase .. Powered by Cognero.

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Chapter 39 e. acetoacetyl-terminal domain of the polymerase ANSWER: c 38. What processes are closely coupled in space and time in prokaryotes? a. translation and initiation b. transcription and termination c. transcription and translation d. initiation and termination e. transcription and translocation ANSWER: c 39. What is the group of ribosomes bound to an mRNA molecule called? a. polysome b. expressome c. ribosomal complex d. autosome e. monosome ANSWER: a 40. In E. coli, the 70S ribosome is made up of _____ subunits. a. 50S and 20S b. 50S and 30S c. 50S and 10S d. 30S and 40S e. 10S and 60S ANSWER: b 41. What parts of tRNA are the least important for its binding by aminoacyl-tRNA synthetase? Select all that apply. a. acceptor stem b. anticodon loop c. TψC loop d. CCA terminus e. DHU loop ANSWER: c, d, e 42. Using the table of genetic code, choose the CORRECT protein sequence that will be produced from the following sense strand of DNA: 5′-AAA GTA CGG CCT TTG GAT CCC - 3′

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Chapter 39

a. Pro-Stop b. Phe-His-Ala-Gly-Asn-Leu-Val c. Lys-Val-Arg-Pro-Leu-Asp-Pro d. Val-Ile-Gln-Arg-Pro-Tyr-Phe e. Val-Val-Ser-Pro-Leu-Asp-Thr ANSWER: c 43. The initiating mRNA codon in most organisms is: a. UAG. b. AUG. c. AAA. d. UAA. e. UGA. ANSWER: b 44. In what component of the ribosome would a mutation have the GREATEST impact and why? a. RNA, because RNA ensures proper orientation of amino acids. b. RNA, because RNA is thought to provide catalytic activity. c. Protein, because protein is thought to provide catalytic activity. d. Protein, because the recognition sites consist primarily of protein. e. Carbohydrate, because the recognition sites consist primarily of protein. .. Powered by Cognero.

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Chapter 39 ANSWER: b 45. Select all that apply. The ribosome is: a. an example of a catalyst that has survived from the RNA world. b. unique to eukaryotes. c. a ribozyme. d. based mostly on proteins rather than on RNA. e. unique to prokaryotes.

ANSWER: a, c 46. In preparation for attachment to tRNA, amino acids are activated by: a. methylation. b. adenylation. c. demethylation. d. acetylation. e. oxidation.

ANSWER: b 47. The key point for the accuracy of translation is: a. codon degeneracy. b. a wobble at the third base of the codon. c. tRNA specificity in binding the mRNA. d. synthesis of aminoacyl tRNA. e. the structure of mRNA. ANSWER: d 48. Consider the overall reaction of amino acid activation. amino acid + ATP + tRNA ⇌ aminoacyl-tRNA + AMP + PPi If the ATP was labeled with 32P in the α position, where would the label end up? a. aminoacyl-tRNA b. AMP c. AMP and PPi d. PPi e. the next tRNA in the sequence ANSWER: b 49. Abundant amino acids in proteins have the most codons, and the least abundant amino acids have the fewest. Why might this provide a selective advantage? a. Degeneracy decreases the likelihood that a substitution for a base will change the encoded amino acid. b. Degeneracy prevents variation in base composition; therefore, proofreading is not necessary. .. Powered by Cognero.

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Chapter 39 c. Early organisms had overlapping DNA and under those conditions, each amino acid was coded by the same number of codons. d. Thermodynamically, less energy is expended by the cell in making abundant amino acids. e. Redundancy allows for base modifications without affecting the protein structure and function.

ANSWER: a 50. What can occur in bacteria but not eukaryotes? a. RNA can be translated in both 5′ → 3′ and 3′ → 5′ directions. b. Splicing can occur. c. The genetic code is overlapping. d. Translation can start before transcription is completed. e. There is only one tRNA for each amino acid.

ANSWER: d 51. A patient with a rare genetic disorder undergoes whole-genome sequencing to identify the underlying mutation. The sequencing reveals the following sense strand sequence: 5’-TGG ACC CAG AGG TTC-3’, compared to the control sequence: 5’-TGG ACC TAG AGG TTC-3’. This mutation affects the coding sequence of a specific protein. Using the genetic code table, what is the most likely result of this mutation?

a. The mutation does not affect the protein, as the nucleotide change results in the same amino acid being produced. b. The mutation results in the loss of a stop codon, causing the protein to be elongated beyond its .. Powered by Cognero.

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Chapter 39 normal length. c. The mutation does not affect the protein at all since it occurs in a non-coding region. d. The mutation introduces a new amino acid, altering the protein’s structure and function. e. The mutation results in a chain-termination mutation or premature stop codon, leading to a truncated protein that is nonfunctional or partially functional.

ANSWER: d 52. During vertebrate evolution, some aminoacyl-tRNA synthetases acquired additional protein domains that are not required for their core function of aminoacylation or editing. These additional domains provide noncanonical functions. Which of the following are NOT examples of noncanonical functions that these synthetases can perform? Select all that apply. a. regulation of inflammation b. signal transduction c. translation and transcription control d. facilitation of ribosome assembly e. proofreading tRNA for errors ANSWER: d, e 53. Which of the following mutations is likely associated with Charcot-Marie-Tooth disease, a condition characterized by progressive degeneration of motor and sensory neurons? a. a mutation in DNA that causes elongation of the encoded protein beyond its normal length b. a mutation in the mitochondrial DNA polymerase c. a mutation in glutamyl-prolyl tRNA synthetase d. a mutation in ribosomal RNA e. a mutation in glycyl-tRNA synthetase ANSWER: e 54. Which of the following statements about glutamyl-prolyl tRNA synthetase is TRUE? a. Glutamyl-prolyl tRNA synthetase is considered a bifunctional enzyme because it acts as a synthetase and helps regulate the inflammatory response. b. Phosphorylation of glutamyl-prolyl tRNA synthetase results in the translation silencing of mRNA related to inflammation. c. Glutamyl-prolyl tRNA synthetase exclusively acts in protein synthesis and has no role in the inflammatory response. d. Glutamyl-prolyl tRNA synthetase is the only synthetase that does not participate in protein synthesis but has an important role in the regulation of inflammation. e. Glutamyl-prolyl tRNA synthetase enhances chronic inflammation by facilitating the translation of pro-inflammatory genes.

ANSWER: b 55. What would be the consequence if the direction of translation were opposite that of transcription, given that transcription occurs in the 5′ → 3′? .. Powered by Cognero.

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Chapter 39 a. Translation would begin before transcription is completed, leading to incomplete protein synthesis. b. Only fully synthesized mRNA would be able to be translated. c. Translation would no longer require ribosomes. d. The sequence of the protein would be reversed, resulting in no functional proteins. e. Translation would occur in the 3′ → 5’ direction, which would change the genetic code, but no other elements of translation would be altered.

ANSWER: b 56. Which of the following statements about the differences between the genetic codes in ciliated protozoa and human mitochondria is FALSE? Select all that apply. a. Both use UGA as the sole termination signal. b. In ciliated protozoa, UGA is the only termination signal used, while in human mitochondria, AGA and AGG function as stop codons. c. Ciliated protozoa and human mitochondria share the same codon utilization for amino acid codification. d. In human mitochondria, UGA is a stop codon, while in ciliated protozoa, UGA codes for an amino acid. e. In ciliated protozoa, UGA is used for termination, while in human mitochondria, UGA encodes for tryptophan.

ANSWER: a, c, d 57. What amino acids are encoded by six codons? Select all that apply. a. leucine b. tryptophan c. methionine d. serine e. arginine ANSWER: a, d, e 58. The __________ use of synonymous codons in genes across different organisms is a phenomenon known as ______________.

ANSWER: nonrandom, codon bias 59. What is the modified base shown below?

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Chapter 39 a. N6-methyladenosine b. 5-hydroxyuracil c. 8-oxoguanine d. 5-methylcytidine e. inosine ANSWER: e 60. What term describes the flexibility or relaxed pairing rules that occur at the third position of the mRNA codon? a. codon redundancy b. base pairing c. synonym pairing d. anticodon flexibility e. wobble base pairing ANSWER: e 61. What is the modified base shown below?

a. 5-hydroxyuracil b. inosine c. 8-oxoguanine d. N6-methyladenosine e. 5-methylcytidine ANSWER: e 62. How does methylation of tRNA molecules, such as the presence of 5-methylcytidine, affect translation? a. Methylation blocks translation by preventing the binding of tRNA molecules to the ribosome. b. Methylation strengthens the formation of base pairs, stabilizing the interaction of tRNA molecules with the translation machinery. c. Methylation does not affect base pairing, and therefore, base interactions with the translation machinery are unchanged. d. Methylation increases the overall stability of tRNA; without methylation, the tRNA would be rapidly degraded and translation would be impaired. e. Methylation prevents the formation of certain base pairs, thereby making some bases accessible for .. Powered by Cognero.

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Chapter 39 interaction with the translation machinery.

ANSWER: e 63. What is the CCA terminus of tRNA? a. a sequence of three nucleotides (cytosine-cytosine-adenine) located at the 3’ end of the tRNA molecule b. a sequence of three nucleotides (adenine-cytosine-cytosine) located at the 5’ end of the tRNA molecule c. a sequence of three nucleotides (adenine-cytosine-cytosine) located at the 3’ end of the tRNA molecule d. a sequence of three nucleotides (cytosine-adenine-cytosine) located at the 3’ end of the tRNA molecule e. a sequence of three nucleotides (cytosine-adenine-cytosine) located at the 5’ end of the tRNA molecule ANSWER: a 64. Why is the ∆G°’ of the reaction involving the hydrolysis of the ester bond of aminoacyl-tRNA close to zero? a. The reaction uses energy from the hydrolysis of GTP instead of ATP, which shifts the ∆G°’ to a negative value. b. The hydrolysis of the ester bond in aminoacyl-tRNA is highly endergonic, making the reaction very favorable. c. The hydrolysis of the ester bond in aminoacyl-tRNA is highly exergonic, making the reaction very unfavorable. d. The ester bond in the aminoacyl-tRNA is hydrolyzed without any change in free energy. e. The free energy of hydrolysis of the ester bond of aminoacyl-tRNA is similar to the hydrolysis of ATP to AMP and PPi, resulting in a near-zero ∆G°’.

ANSWER: e 65. How is the formation of Ser-tRNAThr prevented, given that serine has a hydroxyl group that can bind to the zinc in the active site of the synthetase? a. The threonyl-tRNA synthetase has an editing site that selectively removes serine from tRNAThr. b. The zinc cofactor is specifically situated in the enzyme active site to prevent any incorrect amino acid from binding to tRNAThr. c. The binding site only recognizes amino acids without hydroxyl groups, preventing serine from attaching to tRNAThr. d. The hydroxyl group on serine binds tightly to the zinc site, preventing binding of tRNAThr to the synthetase. e. The hydroxyl group on serine prevents the attachment of serine to tRNAThr through electrostatic repulsion.

ANSWER: a 66. Most __________________ contain editing sites in addition to acylation sites. .. Powered by Cognero.

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Chapter 39 ANSWER: aminoacyl-tRNA synthetases 67. The active site____________ amino acids that are __________than the correct one because there is __________ room for them.

ANSWER: rejects, larger, insufficient 68. Editing sites ___________ activated species that are __________ than the correct one.

ANSWER: cleave, smaller 69. Proteins are synthesized in the _________________, and the mRNA is read by the ribosome in the _________direction, ____________ the direction of RNA transcription.

ANSWER: amino-to-carboxyl direction, 5′ → 3′, matching 70. The ____________ remains nearly invariant because it is efficient, robust, and crucial for maintaining the functionality of _______ across all life forms, and any changes could disrupt fundamental cellular processes.

ANSWER: genetic code, proteins

.. Powered by Cognero.

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Chapter 40 1. How many stages are involved in proteins synthesis? a. 1 b. 2 c. 3 d. 4 e. 5 ANSWER: c 2. What is the feature of bacterial mRNA? a. Molecules are polycistronic. b. Molecules have many noncoding regions. c. The first translated codon is at the 5′ end. d. Molecules do not have signals that define the beginning of the polypeptide chain. e. Molecules do not have signals that define the end of the polypeptide chain.

ANSWER: a 3. The sequence that interacts with a complementary sequence on the 3′ end of the 16S rRNA is: a. ATP b. ricin c. Shine–Dalgarno d. kanamycin e. N-formylmethionine ANSWER: c 4. What are the components of protein synthesis initiation? Select all that apply. a. fMet-tRNAf b. EF-Tu c. mRNA d. GTPase e. Shine–Dalgarno sequence ANSWER: a, c, d 5. What factors are essential for the initiation of translation? Select all that apply. a. IF1 b. IF2 c. IF3 d. EF-Tu e. EF-Ts ANSWER: a, b, c 6. What components can be linked with aminoacyl-tRNA synthetase? Select all that apply. .. Powered by Cognero.

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Chapter 40 a. EF-Tu b. EF–Tu–GDP complex c. tRNAf d. tRNA e. fMet-tRNAf ANSWER: c, d 7. Choose the initiating codon in bacterial mRNA and an amino acid that is encoded by it. a. threonine; ACG b. serine; UCG c. methionine; UAA d. methionine; AUG e. tryptophan; UGG ANSWER: d 8. What is the main role of tRNAf? a. It is included in the initiation complex to bind with IF-1. b. It formylates the amino group of the methionine molecule. c. It inserts methionine in internal positions during elongation. d. It modifies methionine to start protein synthesis. e. It brings formylmethionine to the ribosome to initiate protein synthesis.

ANSWER: e 9. What is the rate-limiting step in protein synthesis in bacteria? a. formation of the 70S initiation complex b. formation of the 30S initiation complex c. formation of the IF2–GTP–initiator–tRNAf complex d. binding of EF-Tu to aminoacyl-tRNA e. formation of the 50S initiation complex ANSWER: a 10. What is the feature of the mRNA initiation sequence in eukaryotes? a. purine-rich on the 5′ side b. pyrimidine-rich on the 5′ side c. GGQ-sequence d. AUG near the 5′ end e. UAA and UGA sequence on the 5′ side ANSWER: d 11. What catalyzes the formation of the peptide bond in protein synthesis? .. Powered by Cognero.

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Chapter 40 a. RRF b. RF3 c. aminoacyl-tRNA synthetase d. transformylase e. peptidyl transferase center ANSWER: e 12. What complex prevents the 30S subunit from prematurely joining the 50S subunit? a. complex of the 30S subunit with IF2 b. 70S initiation complex c. IF2-tRNAf d. complex of the 30S subunit with IF1 and IF3 e. fMet-tRNAf ANSWER: d 13. What forces tRNA and mRNA to move through the ribosome by a distance, corresponding to one codon? a. pairing of the initiator codon on mRNA with the anticodon of an initiator tRNA molecule b. interactions of EF-Tu with the sarcin–ricin loop c. conformational change in EF-G caused by binding EF-G with the EF-Tu binding site. d. N-formylmethionine e. interactions of the release factors with a loop containing the GGQ sequence ANSWER: c 14. Choose the components of the 30S initiation complex. Select all that apply. a. 30S subunit b. IF2-GTP c. IF1 d. IF3 e. mRNA ANSWER: a, b, e 15. What happens with the A and E sites when fMet-tRNA occupies the P site? a. The stop codon is in the A site; E is empty. b. EF-Tu binds aminoacyl-tRNA in the A site; E is empty. c. The A site is empty; E is occupied by a polypeptide. d. Aminoacyl-tRNA is in the A site; E is empty. e. The A and E sites are empty.

ANSWER: e 16. What is the rate of protein synthesis in a bacterial cell (amino acids per sec)? .. Powered by Cognero.

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Chapter 40 a. 2 b. 10 c. 20 d. 40 e. 80 ANSWER: c 17. What components of translation recognize stop codons? a. release factors b. elongation factors c. termination factors d. anticodons e. tRNA ANSWER: a 18. Choose a pair of homologous RNA of prokaryotes and eukaryotes, respectively. a. 23S and 18S b. 16S and 28S c. 5S and 5.8S d. 3′ end of 23S and 5S e. 5′ end of 23S and 5.8S ANSWER: e 19. What is the function of the 5′ cap in eukaryotes? a. The 5′ cap is an initiating codon. b. It provides binding of the 40S ribosome to Met-tRNA. c. It helps elongation factors to increase the rate of amino acid adding. d. It provides an easily recognizable starting point for translation. e. The function is uncertain.

ANSWER: d 20. Would it be possible for RF3 to catalyze the removal of RF1 and RF2 from the ribosome upon release of the newly synthesized protein? Why? a. Yes, RF1 and RF2 are compact proteins. b. Yes, RF3 is transferase. c. Yes, RF3 is GTPase. d. No, all RFs are elongation factors. e. No, RF1 is GTPase.

ANSWER: c 21. Choose the codons that are recognizable by release factors. Select all that apply. .. Powered by Cognero.

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Chapter 40 a. UAG b. CAG c. UGA d. UAA e. UCA ANSWER: a, c 22. What amino acid participates in eukaryotic initiation? a. Arg b. Lys c. Leu d. Met e. Gln ANSWER: d 23. Why do bacteria, unlike eukaryotes, have multiple start sites? a. Generally, bacterial mRNA has multiple Shine–Dalgarno sequences. b. Bacterial start sites have mutations, so they have one more Shine–Dalgarno sequence. c. Topology of cyclic bacterial mRNA requires conditions to favor the start of translation. d. The preinitiation complex of bacteria recognizes only multiple translation sites. e. Bacteria do not have the 5′ end of mRNA capped.

ANSWER: a 24. What is translocase? a. protein that provides proximity to the amino acid with aminoacyl-tRNA in the P site b. a component of the peptidyl transferase center c. an enzyme that makes the 5′ end capped in eukaryotes d. a GTP-activity factor catalyzing the removal of RF1 or RF2 from the ribosome e. a bacterial elongation factor G ANSWER: e 25. What amino acid sequence with one methylated amino acid does peptidyl transferase center contain? a. CQQ b. GGQ c. DDC d. QCE e. GCG ANSWER: b 26. The _____ of the cell provides organization of the translation machinery in higher eukaryotes. a. cell wall .. Powered by Cognero.

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Chapter 40 b. centriole c. cytoskeleton d. starch granules e. peroxisomes ANSWER: c 27. Choose the CORRECT sequence of the first three acts of the SRP targeting cycle. a. Protein synthesis begins on free ribosomes → after the signal sequence has exited the ribosome, it is bound by SRP; protein synthesis halts → SRP–ribosome complex docks with the ER membrane. b. Protein synthesis begins on free ribosomes → SRP–ribosome complex docks with the ER membrane → the ribosome is released; protein tunnel in the translocon closes. c. After the signal sequence has exited the ribosome, it is bound by SRP → SRP and the SRP receptor hydrolyze bound GTPs → protein is synthesized into ER. d. SRP and the SRP receptor hydrolyze bound GTPs → SRP is free to bind another signal sequence → protein is synthesized into ER. e. The ribosome–nascent polypeptide is transferred to the translocon → the signal peptidase removes the signal sequence as it enters the lumen of ER → the ribosome is released; protein tunnel in the translocon closes.

ANSWER: a 28. Mutations in eukaryotic ____ cause white matter replacement by cerebrospinal fluid, seen in the image in white (shown with arrows).

a. eIF1 b. eIF2 c. eIF-4E d. EF1α e. EF1βγ ANSWER: a 29. What is the function of chloramphenicol? .. Powered by Cognero.

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Chapter 40 a. It inhibits translocation in eukaryotes. b. It inhibits initiation and causes misreading of bacterial mRNA. c. It inhibits the peptidyl transferase activity of the 50S bacterial subunit. d. It binds to the 50S subunit and inhibits translocation in bacteria. e. It binds to the 30S subunit and inhibits the binding of mRNA in bacteria.

ANSWER: c 30. Which statement about protein synthesis in eukaryotes is FALSE? a. Eukaryotic ribosomes consist of a 60S large subunit and a 40S small subunit. b. In eukaryotes, pre-mRNA must be processed more in comparison to prokaryotic pre-mRNA. c. Eukaryotic mRNA has multiple start sites. d. In eukaryotes, an amino acid that participates in initiation is methionine. e. Termination in eukaryotes is carried out by a single release factor named eRF1.

ANSWER: c 31. What is the effect of diphtheria toxin on a cell? a. It prevents capping of the 5′ end of mRNA. b. It blocks the release factor eRF1 by modification of its amino group. c. It catalyzes the covalent modification of elongation factor 2. d. It makes the cell produce bacterial proteins. e. It damages the glycoproteins of the cell membrane.

ANSWER: c 32. Describe the reciprocal relations of ferritin and transferrin-receptor expression levels. a. When iron is high, the amount of transferrin receptor decreases, and little or no new ferritin is synthesized. b. When iron is high, the amount of transferrin receptor increases, and ferritin is synthesized. c. When iron is high, the amount of transferrin receptor increases, and no new ferritin is synthesized. d. When iron is scarce, the amount of transferrin receptor increases, and little or no new ferritin is synthesized. e. When iron is scarce, the amount of transferrin receptor decreases, and ferritin is synthesized.

ANSWER: d 33. RISC is: a. a viral RNA. b. a process that leads to mRNA degradation induced by the presence of dsRNA. c. a dsRNA cleaved into 21-nucleotide fragments. d. one of the single-stranded components of RNase cleavage products. e. a complex of one strand of the former dsRNA with an Argonaute protein.

ANSWER: e .. Powered by Cognero.

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Chapter 40 34. Which is the CORRECT pairing of the length of the signal sequence and the type of amino acids it has? a. 9–14 hydrophobic amino acid residues – seldom positively charged b. 9–12 hydrophobic amino acid residues – sometimes positively charged c. 9–16 negatively charged amino acid residues – sometimes positively charged d. 9–16 positively charged amino acid residues – often hydrophobic e. 15–16 hydrophilic amino acid residues – often hydrophilic ANSWER: b 35. Why is diphtheria toxin lethal? a. The toxin of Corynebacterium diphtheriae inhibits protein synthesis. b. The toxin of Corynebacterium diphtheriae is able to enter the cytoplasm of its target cell. c. The toxin of Corynebacterium diphtheriae is cleaved into two fragments when in the cell. d. People do not get appropriate immunizations. e. Corynebacterium diphtheriae strains are mostly resistant to antibiotics.

ANSWER: a 36. What is the structure that ferritin mRNA includes? a. α helix b. β sheet c. iron-response element d. D-loop e. helix-loop-helix ANSWER: c 37. The _______ is a ribonucleoprotein consisting of a 7S RNA and six different proteins.

ANSWER: signal-recognition particle 38. What toxins inhibit protein-synthesis elongation by covalent modification of rRNA? Select all that apply. a. diphtheria toxin b. tetrodotoxin c. ricin d. α-amanitin e. α-sarcin ANSWER: c, e 39. According to the theory of protein targeting, proteins from the cytoplasm are delivered to different parts of the cell. What are those parts? Select all that apply. a. nucleus b. centrioles c. peroxisomes .. Powered by Cognero.

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Chapter 40 d. lysosomes e. chloroplast ANSWER: a, c, d, e 40. Eukaryotic mRNA is: a. linear. b. circular. c. modified only at the ribose. d. discontinuously translated. e. planar.

ANSWER: b 41. In vanishing white matter disease, nerve cells disappear due to: a. phagocytosis of living cells. b. an autoimmune result of mutations in mRNA. c. mutations in release factors. d. mutations in initiation factor 2. e. mutations in the tRNA charging mechanism.

ANSWER: d 42. Chloramphenicol acts by: a. binding to the peptidase cell-wall enzyme. b. inhibiting peptidyl transferase activity. c. inhibiting protein synthesis. d. binding to fMET-tRNA. e. changing the shape of the mRNA. ANSWER: b 43. What protein of E. coli is responsible for the translocation of mRNA as a result of a conformation change due to hydrolysis of GTP? a. RRF b. EF-Tu c. EF-G d. Shine–Dalgarno e. Rho ANSWER: c 44. An error in the synthesis of an RNA molecule or a protein molecule will not be passed on to subsequent generations. Suggest a reason why there is a proofreading mechanism in protein synthesis but not RNA synthesis. a. Protein proofreading occurs during protein folding events; thus, the rate of protein synthesis is not .. Powered by Cognero.

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Chapter 40 reduced. b. Protein proofreading occurs during posttranslational splicing of proteins (such as insulin); thus, the rate of protein synthesis is not reduced. c. Protein synthesis takes place at a much slower rate than RNA synthesis, thus giving time for improper codon–anticodon binding destabilization. d. The mischarging of some tRNA is corrected by the hydrolytic action of aminoacyl-tRNA synthetase; thus, mistakes are prevented. e. Some tRNAs are proofread in the E site, triggering the hydrolytic action of peptidyl transferase; thus, mistakes are corrected as they are made.

ANSWER: d 45. What would happen if you introduced a substitution of two of the Gs for two Cs in the Shine–Dalgarno sequence? a. Weak base-pairing between mRNA and the 3′ end of the 16S rRNA could terminate initiation. b. Weak base-pairing between the 5′ end of tRNA and the 18S rRNA could cause an insertion of the wrong amino acid. c. Weak base-pairing between the codon and the anticodon sequences could cause an insertion of the wrong amino acid. d. The Shine–Dalgarno sequence allows for the proper pairing of each amino acid with its tRNA, which would lead to an incorrect protein sequence. e. The 16S and 18S ribosomal subunits would not be bound tightly enough, and early chain termination would occur.

ANSWER: a 46. Even if the protein synthesis machinery in the ribosomes is flawless, could there be other pitfalls in the formation of a biologically active protein? Why? a. Yes, a defect in the signal-recognition particle fails to recognize the termination signal. b. Yes, a defect in the signal-recognition particle fails to recognize the signal sequence. c. Yes, the splicing mechanism of the 18S rRNA is defective, so proteins are not spliced properly. d. No, proper protein folding is driven by hydrophobic interactions. e. No, protein tertiary structure is determined by the primary structure.

ANSWER: b 47. Signals that define the beginning and the end of protein synthesis are contained in: a. rRNA. b. tRNA. c. mRNA. d. the ribosome. e. protein.

ANSWER: c 48. What is the order of the tRNA binding sites on the 70S ribosome with respect to the 5′ → 3′ direction of .. Powered by Cognero.

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Chapter 40 mRNA? a. E P A b. P A E c. A E P d. P E A e. E A P ANSWER: a 49. Ricin is made from castor beans and is a strong toxin that: a. removes adenine from 28S rRNA. b. glycosylates tRNA. c. inhibits protein synthesis initiation. d. blocks the A binding site. e. coats the mRNA. ANSWER: a 50. Suppose you add 20 amino acids, each carrying a 14C label, to a protein synthesis system with only one mRNA present. After one minute, you purify the newly synthesized protein and then cleave it through several rounds, first with an aminopeptidase and then a carboxypeptidase. If the majority of the radioactive label is in the carboxypeptidase digest, what does it indicate? a. tRNA molecules are not bound to amino acids until mRNA binds to the 16S ribosomal site. b. The newly synthesized protein is not labeled because it is screened by the 18S rRNA. c. Protein synthesis occurs relative to the 3′-to-5′ direction of mRNA. d. Protein synthesis occurs from the amino-terminal to carboxy-terminal direction. e. Carboxypeptidase has a much higher kcat than aminopeptidase.

ANSWER: d 51. Which of the following proteins is NOT a GTPase? a. initiation factor 2 (IF2) b. elongation factor Tu (EF-Tu) c. release factor 3 (RF3) d. signal recognition particle (SRP) e. elongation factor G (EF-G) f. translocon ANSWER: f 52. Which of the following statements about the translocon are FALSE? Select all that apply. a. The translocon is responsible for folding the protein once it enters the lumen of the endoplasmic reticulum. b. The translocon is a protein-conducting channel that transports the growing polypeptide into the lumen of the endoplasmic reticulum. .. Powered by Cognero.

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Chapter 40 c. The translocon opens when it binds to the ribosome, allowing protein synthesis to resume. d. The translocon facilitates the insertion of proteins throughout the rough endoplasmic reticulum membrane. e. The growing polypeptide chain passes through the translocon channel only after translation has been completed.

ANSWER: a, e 53. Which of the following statements about microRNAs (miRNAs) are FALSE? Select all that apply. a. The dsRNA fragments produced by Dicer are approximately 80 base pairs in length. b. miRNAs are small noncoding RNAs that regulate gene expression posttranscriptionally. c. A mature single-stranded miRNA remains associated with Argonaute, forming the microRISC complex. d. The microRISC complex functions by degrading target mRNA or blocking its translation to inhibit protein synthesis. e. miRNA is generated from larger transcripts produced by RNA polymerase I. ANSWER: a, e 54. Which of the following mutations in tRNA could potentially suppress a frameshift mutation? a. a mutation that prevents the tRNA from binding to any amino acid b. a mutation that causes the tRNA to bind to the mRNA sequence without following the normal anticodon-codon pairing c. a mutation that causes the tRNA to bind only to the first codon of the mRNA sequence d. a mutation that causes the tRNA to stop recognizing codons, halting translation e. a mutation that changes the tRNA anticodon to recognize a codon in the shifted reading frame ANSWER: e 55. In bacteria, the first amino acid incorporated into the nascent protein is typically formylmethionine (fMet).

Which of the following statements regarding the role of fMet in bacterial protein synthesis is TRUE? a. The fMet group is removed in approximately half of the proteins during protein elongation, typically when the nascent chain reaches 10 amino acids in length. b. fMet is required for protein folding and stability in mature proteins. c. fMet is used in both bacterial and eukaryotic translation initiation. d. fMet is never removed, remaining as the first amino acid in all mature bacterial proteins. e. fMet is always removed when the nascent chain reaches 10 amino acids in length; however, it is later reincorporated into the mature protein.

ANSWER: a 56. Which of the following are differences between bacterial and eukaryotic protein synthesis? Select all that apply. a. In eukaryotes, protein synthesis begins with formylmethionine (fMet), while in bacteria, protein synthesis begins with methionine (Met). b. In bacteria, translation and transcription occur simultaneously, while in eukaryotes, translation and .. Powered by Cognero.

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Chapter 40 transcription are spatially separated. c. Eukaryotic ribosomes are smaller and less complex than bacterial ribosomes. d. Eukaryotic protein synthesis involves fewer release factors than bacterial protein synthesis. e. Bacterial ribosomes are 80S, while eukaryotic ribosomes are 70S. ANSWER: b, d 57. How do the differences between eukaryotic and bacterial ribosomes help antibiotics specifically target bacterial infections? a. Antibiotics target the ribosomal RNA in eukaryotes, exploiting structural differences between bacterial and eukaryotic ribosomes. b. Antibiotics only target prokaryotic ribosomes because they are more complex than eukaryotic ribosomes. c. Bacterial ribosomes are structurally distinct and smaller than eukaryotic ribosomes, allowing antibiotics to bind selectively to bacterial ribosomes without affecting eukaryotic ribosomes. d. Antibiotics target eukaryotic ribosomes better because they are more complex than bacterial ribosomes. e. Antibiotics affect both bacterial and eukaryotic ribosomes equally, leading to widespread protein synthesis disruption in both organisms.

ANSWER: c 58. Which of the following statements regarding untranslated regions (UTRs) in mRNA are FALSE? Select all that apply. a. UTRs play a critical role in regulating the stability, localization, and translation of mRNA molecules. b. The 5’-UTR is located at the 3’ end of the mRNA, and the 3’UTR at the 5’ end. c. The 5’-UTR and the 3’-UTR of mRNA are not translated into proteins. d. Both 5’-UTRs and 3’-UTRs can contain binding sites for regulatory proteins and miRNAs that affect the stability and translation of mRNA. e. UTRs are entirely irrelevant to the regulation of gene expression and protein synthesis.

ANSWER: b, e 59. Which of the following factors contribute to ribosome heterogeneity and its potential role in the regulation of gene expression? Select all that apply. a. Nonribosomal proteins can associate with ribosomes, influencing their activity and potentially regulating translation in specific contexts. b. Ribosomes are homogenous in all cell types, meaning they do not differ in their structure or activity based on cell type or physiological state. c. Ribosome composition can be altered in a cell-specific manner, influencing how ribosomes function in different cell types or under different conditions. d. Ribosome concentration is the same in all tissues, and thus, ribosome number does not affect the regulation of gene expression. e. A low ribosome concentration in certain tissues might lead to insufficient production of specific proteins, especially those encoded by mRNA for which translation is not efficiently initiated. .. Powered by Cognero.

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Chapter 40 ANSWER: a, c, e 60. Streptomycin is an antibiotic that prevents the correct initiation of bacterial translation. We have heard that overuse or misuse of antibiotics can lead to the development of resistant bacterial strains. Which of the following mechanisms could potentially contribute to streptomycin resistance in bacteria? Select all that apply. a. Bacteria produce antibiotic-modifying enzymes that chemically alter the antibiotic, making it ineffective. b. Bacteria increase their uptake of streptomycin into the cell, enhancing its ability to inhibit protein synthesis. c. Bacteria increase their efflux of streptomycin out of the cell, thereby decreasing the amount of intracellular antibiotic. d. Streptomycin resistance occurs when the bacteria develop the ability to synthesize the antibiotic, making them immune to its effect. e. Mutations in the formylmethionyl-tRNA recognition site or initiation factors may prevent streptomycin from interfering with translation initiation.

ANSWER: a, c, e 61. Streptomycin is an antibiotic that inhibits protein synthesis in bacteria. Based on the chemical structure of streptomycin shown below, which of the following statements are TRUE? Select all that apply.

a. Streptomycin is a highly basic trisaccharide, consisting of three sugar molecules modified with amino and hydroxyl groups. .. Powered by Cognero.

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Chapter 40 b. Streptomycin contains no sugar molecules in its structure. c. Streptomycin contains four sugar molecules, which make it chemically active against bacteria. d. The amino and hydroxyl groups in streptomycin are not critical for interactions with bacterial ribosomes. e. The amino and hydroxyl groups in streptomycin are critical for interactions with bacterial ribosomes and confer the ability to inhibit protein synthesis.

ANSWER: a, e 62. Puromycin is an antibiotic that resembles the aminoacyl terminus of an aminoacyl-tRNA and disrupts protein synthesis. Based on the chemical structures shown below, which of the following statements are TRUE? Select all that apply.

a. Puromycin mimics the aminoacyl-tRNA structure, allowing it to join the growing polypeptide chain.

This disrupts translation by prematurely terminating protein elongation. b. Puromycin prevents the binding of aminoacyl-tRNA to the ribosome, thereby inhibiting the initiation of translation. c. The adduct formed by puromycin and the growing polypeptide chain is unstable due to the ester linkage present in this antibiotic. d. The amino group in puromycin forms a bond with the carboxyl group of the growing polypeptide chain, producing a stable adduct that rapidly dissociates from the ribosome. e. Puromycin contains an amide group, which allows it to be incorporated into the growing polypeptide chain.

ANSWER: a, d 63. Which of the following statements regarding the mechanism of action of diphtheria toxin are FALSE? Select all that apply. a. Diphtheria toxin causes the posttranslational modification of diphthamide in elongation factor 2, which leads to the activation of protein synthesis. b. Diphthamide is formed by posttranslational modification of a histidine residue in elongation factor 2 (EF2). c. The ADP-ribosylation of diphthamide prevents the translocation activity of elongation factor 2, inhibiting protein translation in eukaryotic cells. .. Powered by Cognero.

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Chapter 40 d. Diphtheria toxin blocks protein synthesis by catalyzing the transfer of an ADP-ribose unit from NAD+ to a modified histidine residue (diphthamide) in elongation factor 2. e. The mechanism of action of diphtheria toxin involves a direct modification of ribosomal RNA, blocking translation.

ANSWER: a, e 64. The endoplasmic reticulum (ER) plays a key role in the synthesis and processing of proteins. Which statements regarding the rough and smooth endoplasmic reticulum are FALSE? a. In eukaryotic cells, a ribosome remains free in the cytoplasm unless directed to the endoplasmic reticulum. b. Ribosomes are bound to the rough ER, giving it a rough appearance, while the smooth ER does not contain ribosomes. c. The smooth ER is responsible for protein synthesis, while the rough ER is involved in lipid metabolism. d. The synthesis of proteins begins with a free ribosome, but if the mRNA contains a specific signal sequence, synthesis is halted until the ribosome docks with the ER membrane. e. The rough ER has a smooth appearance because it lacks ribosomes, while the smooth ER appears rough due to the presence of ribosomes.

ANSWER: c, e 65. Which of the following statements regarding the action of cycloheximide and puromycin on translation in prokaryotic and eukaryotic cells are FALSE? Select all that apply. a. Cycloheximide exclusively inhibits translation in eukaryotic cells, while puromycin is only effective in bacterial cells due to differences in ribosome structure. b. Both cycloheximide and puromycin can inhibit translation by targeting conserved processes in protein synthesis. c. Puromycin and cycloheximide act via different mechanisms to halt protein synthesis. d. Both puromycin, which inhibits protein synthesis by inducing premature release of nascent polypeptides, and cycloheximide, which blocks translocation in eukaryotic ribosomes, are useful antibiotics for blocking protein synthesis in specific cells. e. The fundamental processes of elongation and chain termination during protein synthesis are different in all organisms, making drugs like cycloheximide and puromycin only effective on one cell type.

ANSWER: a, e 66. The SRP receptor (SR) is an integral membrane protein composed of two subunits, SRα and SRβ. Which of the following statements regarding the SRα subunit is TRUE? a. SRα binds to SRβ to form a stable, non-dynamic complex. b. Like SRα, SRβ functions as a GTPase, similar to the signal recognition particle (SRP). c. SRβ and SRα are involved in protein degradation in the cytoplasm. d. SRα is a receptor for mRNA during translation. e. SRα functions as a GTPase, similar to the signal recognition particle (SRP).

ANSWER: e .. Powered by Cognero.

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Chapter 40 67. Considering the role of proteins like transferrin, transferrin receptor, and ferritin in iron homeostasis, how might a defect in ferritin affect cellular function and the synthesis of these proteins? Select all that apply. a. If ferritin cannot properly store iron, cells may experience iron overload, potentially damaging proteins, lipids, and nucleic acids in the cell. b. Ferritin dysfunction would have no impact on the synthesis of these proteins, as ferritin is not directly involved in translation. c. A malfunction in the ability of ferritin to store iron would result in an insufficient iron concentration, causing a decrease in the synthesis of other iron-regulated proteins, such as transferrin receptor. d. A malfunction in ferritin would likely result in an increase in iron levels, which would increase the translation of ferritin mRNA. e. A defect in ferritin would cause an accumulation of iron, resulting in increased degradation of transferrin receptor mRNA. ANSWER: a, d, e 68. Ferritin mRNA includes a stem-loop structure called an _____. This IRE binds the _____, which _____ the translation of ferritin mRNA under low-iron conditions.

ANSWER: iron-response element (IRE), IRE-binding protein, blocks 69. _____ and _____ expression levels are reciprocally related in their responses to changes in iron levels.

ANSWER: Ferritin, transferrin-receptor 70. Approximately 60% of human genes are regulated by one or more _____.

ANSWER: miRNAs

.. Powered by Cognero.

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Chapter 41 1. What is the biological role of restriction enzymes in bacteria? a. repair DNA b. induce DNA crossover c. cleave foreign DNA d. extract genes from the bacteria e. open sticky ends ANSWER: c 2. What is DNA cloning? a. growing bacterial cells b. amplifying DNA c. growing cellular cultures d. finding palindromic sequences e. Identical copies of the gene of interest ANSWER: e 3. What is the direction of reverse transcription? a. from 5′-phosphate to 3′-hydroxyl end b. from 3′-hydroxyl to 5′-phosphate end c. from 5′-hydroxyl to 3′-phosphate end d. from 3′-phosphate to 5′-hydroxyl end e. depends on the primer ANSWER: a 4. Recombinant DNA technologies are possible thanks to: a. restriction endonucleases. b. reverse transcriptases. c. DNA polymerases. d. zinc-finger nucleases. e. ligases.

ANSWER: a 5. DNA degradation by restriction enzyme at a specific site would be prevented by: a. two-fold rotational symmetry. b. DNA sequence larger than 20 kb. c. circular DNA sequence. d. blunt ends of the DNA sequence. e. methylated cytosine base.

ANSWER: e 6. Restriction enzymes are useful in recombinant DNA technologies, as they allow: .. Powered by Cognero.

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Chapter 41 a. gene extraction from the bacteria. b. DNA to be released from the chromatin. c. to obtain smaller and more specific DNA fragments. d. cleavage of methylated DNA. e. obtaining the entire genome from the cell.

ANSWER: c 7. What is the minimal amount of DNA that can be analyzed with gel electrophoresis? a. 50 ng b. 100 ng c. 50 ng/ml d. 5 mg e. 5 µg ANSWER: a 8. Which method allows generating DNA fragments with a determined length? a. screening of the expression vectors b. constructing a plasmid c. gel-electrophoresis after reverse transcription d. Sanger dideoxy method e. DNA cloning ANSWER: d 9. What is a cDNA library? a. collection of vectors made of restriction fragments b. collection of all mRNA of bacterial organisms c. collection of complementary DNA copies of mRNA from tissues or organisms, which are inserted into vectors for further studies d. collection of linker DNA e. collection of sequences made of DNA fragments ANSWER: c 10. What is used as a primer for the synthesis of the first strand of cDNA? a. oligo(A) sequence b. the sequence complementary to the poly(A) tail c. oligo (dC) sequence d. oligo (dG) sequence e. the sequence complementary to the 5′ cap ANSWER: b 11. Reverse transcriptase is normally found in: .. Powered by Cognero.

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Chapter 41 a. plants. b. retroviruses. c. mitochondria. d. the pancreas. e. mammals.

ANSWER: b 12. What is a genomic library? a. collection of vectors made of restriction fragments b. collection of all mRNAs of some organisms c. collection of vectors made of reversely transcribed mRNA d. collection of linker DNA e. collection of sequences made of DNA fragments ANSWER: e 13. Which method can amplify DNA sequences? a. controlled termination of replication b. polymerase chain reaction c. screening of the expression vectors d. constructing a plasmid e. polymerase cloning reaction ANSWER: b 14. Which technique is BEST to study levels of gene expression in multiple genes and tissues? a. quantitative PCR b. gene cloning c. gel electrophoresis d. DNA microarray e. microinjections ANSWER: d 15. What is a southern blotting technique? a. transfer of DNA molecules on the nitrocellulose sheet b. autoradiographic detection of specifically selected molecules on the nitrocellulose sheet c. transfer of specifically selected molecules on the nitrocellulose sheet d. fluorescent detection of specifically selected molecules on the nitrocellulose sheet e. transfer of molecules on agarose gel ANSWER: a 16. Which technique or method can be used to directly identify the presence of a specific RNA sequence? a. southern blotting .. Powered by Cognero.

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Chapter 41 b. western blotting c. northern blotting d. autoradiography e. polymerase chain reaction (PCR) ANSWER: c 17. What order of steps of the polymerase chain reaction cycle is CORRECT? a. annealing → denaturation → elongation b. elongation → annealing → denaturation c. separation → denaturation → annealing d. denaturation → annealing → elongation e. separation → hybridization → annealing ANSWER: d 18. Which statement about cloning with retroviruses is TRUE? a. The double-DNA genome of retroviruses must be produced by reverse transcriptase and incorporated into the host in this form. b. A proviral genome version needs to be injected into the cell rather than the whole virion. c. The striking feature of the retroviral lifecycle is hosting genome editing. d. The insertion of the viral genome into the host cell is done in a highly specific manner. e. Reverse transcription of retroviral guide RNA allows insertion of the viral genome into the host genome.

ANSWER: a 19. Which statement about recombinant DNA technology is FALSE? a. This technology is about constructing a vector and cloning it in the host organism. b. Recombination of the desired DNA fragments is the point of this technology. c. Vector bearing the DNA insert can be cleaved by restriction enzymes to check whether the insertion is correct. d. Southern blotting allows checking for correctness of the desired DNA fragment prior to recombination with the vector. e. To run gel electrophoresis, the constructed vector should be closer to the beginning of the gel than the original fragment.

ANSWER: b 20. What enzyme is required to obtain a single strand of cDNA apart from reverse transcriptase? a. DNA polymerase b. terminal transferase c. oligo(dG) transferase d. annealing enzyme e. DNA ligase .. Powered by Cognero.

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Chapter 41 ANSWER: b 21. Select all that apply. The specificity or stringency of a PCR reaction can be controlled by altering the reaction: a. volume. b. target sequence. c. temperature. d. salt concentration. e. enzyme concentration.

ANSWER: c, d 22. Which statement about CRISPR/Cas9 system is TRUE? a. First, it was discovered in the genome of E. coli. b. The target DNA site is recognized by the 3′ end of the single-guide RNA. c. The protospacer-adjacent motif has to be recognized by sgRNA to activate the NUC domain of Cas9. d. The CRISPR/Cas9 system is more efficient than the TALEN-based system because a single nuclease domain of Cas9 is enough to make a double-stranded nick. e. sgRNA in cells is bound to the Cas9 protein to efficiently couple recognition and cleavage.

ANSWER: a 23. To what must DNA strand(s) from the probe be complementary? a. either of the strands b. both strands together c. only the coding strand d. only the template strand e. mature mRNA strand ANSWER: a 24. What is NOT required for the screening of the cDNA library? a. bacterial colonies b. agarose gel c. nitrocellulose membrane d. X-ray film e. radiolabeled probe ANSWER: d 25. What elements are required to construct an expression vector? Select all that apply. a. promoter b. DNA insert c. ribosome binding site d. poly(A) tail .. Powered by Cognero.

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Chapter 41 e. restriction site(s) ANSWER: a, b, c, e 26. Which step in the screening of expression vectors is NOT required for selecting a cDNA library? a. transformation of bacteria b. transfer onto the nitrocellulose membrane c. autoradiographic visualization d. lysis of bacteria on the replica plate e. addition of the labeled selective marker ANSWER: d 27. During preparation of the genomic library: a. fragments of DNA are separated by gel electrophoresis to isolate fragments longer than 15 kb. b. linkers are attached to the total genomic DNA. c. the total genomic DNA sequence first has to be determined. d. restriction endonucleases are used after the attachment of linkers. e. cloning can be done only with the phage as a vector; a bacterial plasmid cannot be used.

ANSWER: a 28. A usual deoxyribonucleotide differs from its analog used in the controlled termination of the replication method in that it is: a. missing the 3′ hydroxyl group. b. missing the 2′ hydroxyl group. c. missing the 5′ hydroxyl group. d. missing the 5′ phosphate group. e. additional the 2′ hydroxyl group.

ANSWER: a 29. What is a common principle exploited by next-generation sequencing (NGS) methods? a. detection of pyrophosphate b. detection of protons c. detection of nucleotide incorporation d. usage of fluorescence e. analysis of sequence fragments ANSWER: c 30. Is it possible to cleave a plasmid with one restriction enzyme and the sequence of interest with a different restriction enzyme and still obtain a fully functional vector after annealing? a. No, the enzymes have to be the same. b. Yes, if the enzymes produce the same cohesive ends. c. Yes, DNA ligase can fix the noncomplementary bonding between different sticky ends. .. Powered by Cognero.

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Chapter 41 d. No, you first need to account for the properties of the sequence and then decide about the restriction. e. Not unless you have a proper synthetic linker inserted in the plasmid.

ANSWER: b 31. Select all that apply. A polymerase chain reaction can be used in: a. next-generation sequencing. b. Sanger sequencing. c. the screening of DNA libraries. d. gene disruption. e. studies of gene expression.

ANSWER: a, c, e 32. Which statement about PCR is TRUE? a. Knowledge of the full sequence of the target gene is a prerequisite for the reaction. b. High hybridization temperature is the only factor influencing PCR stringency. c. A single target molecule of appropriate length is enough for PCR to proceed. d. To date, the maximal length of the target sequence is limited to 1 kb. e. The exact sequence of the target gene needs to be known to synthesize exactly matching primers.

ANSWER: c 33. Judging by the given depiction of autoradiogram of electrophoresis gel, choose the correct sequence of the template (from 5′ to 3′ end).

a. CTC GTA TCA GCT A b. GAG CAT AGT CGA T c. ATC GAC TAT GCT C d. TAG CTG ATA CGA G e. There is not enough information to conclude.

ANSWER: a 34. What is the reason to clone vectors in eukaryotic cells? Select all that apply. a. ability to create transgenic mice b. proper replication machinery c. more efficient technologies for cloning .. Powered by Cognero.

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Chapter 41 d. proper transcriptional machinery e. proper posttranslational processing ANSWER: d, e 35. Assume you have a template sequence 3′-AAT GCG ATC CGG TAC G-5′. In the strand-termination method, the addition of ddCTP, you can obtain: a. CGTA. b. AATGCGATCCG. c. GCGATCCGGTACG. d. AATGCC. e. TTACGCTAGGC. ANSWER: e 36. To what ends of the initial DNA are the direct and reverse primers bound? a. The direct primer binds to the 3′ end on one strand and reverse to the 5′ end on another strand. b. The reverse primer binds to the 5′ end on one strand and direct to the 5′ end on another strand. c. The direct primer binds to the 3′ end on one strand and reverse to the 5′ end on the same strand. d. There is only one primer, and it binds to the 3′ end of the template strand. e. The direct primer binds to the 3′ end on one strand and reverse to the 3′ end on another strand.

ANSWER: a 37. What groups need to be protected to prevent undesirable side reactions in the synthesis of oligonucleotides by the phosphite triester method? Select all that apply. a. The 5′ hydroxyl group is protected by DMT. b. All reactive groups of the base are protected. c. The 3′ phosphate group is protected by βCE. d. The 3′ phosphate group is protected by phosphoramidites. e. The 5′ phosphate group is protected by βCE. ANSWER: a, c 38. Select all that apply. In the synthesis of oligonucleotides by the solid-phase phosphite triester method: a. The first nucleoside, phosphoramidite, is covalently attached to the solid support material. b. Deprotection of the 5′ end by detritylation always precedes coupling of nucleoside phosphoramidites. c. Deprotection of the 3′ end by oxidation always precedes the coupling of nucleoside phosphoramidites. d. The coupling reaction is directly followed by the oxidation of the 3′ phosphite triester to a naturally occurring phosphotriester. e. The coupling reaction is directly followed by detritylation of the 5′ end to prepare it for the next cycle.

ANSWER: a, b, d .. Powered by Cognero.

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Chapter 41 39. Which of the reactions from the scheme (marked with numbers) has the highest initial number of gene copies and why?

a. the first reaction because it has a lower CT b. the fifth reaction because it has a higher CT c. the first reaction because 2n is the biggest d. the fifth reaction because 2n is the biggest e. the fourth reaction due to optimal CT (about 24) ANSWER: a 40. Which statement about DNA microarray technology is TRUE? a. DNA is isolated from the investigated cells or tissues. b. The samples of different tissues need to be combined prior to loading onto the microarray chip. c. The technology allows various pathological conditions to be revealed. d. Isolated mRNA are labeled with fluorescent tags depending on the tissue type. e. Differently labeled cDNAs are loaded into different wells of the microarray chip.

ANSWER: b 41. Which statement about genome editing machinery is TRUE? a. Proteins with TALE repeats require fusion with restriction enzymes, as they cannot cleave DNA on their own. b. To recognize a six-base DNA sequence, a ZFN needs six zinc-finger domains. c. Six α helices in the TALE repeats can recognize six nucleotides. d. To recognize a six-base DNA sequence, TALENs require 204 amino acids encoded in the TALE repeat array e. TALE repeats are palindromic, which allows them to bind sites in DNA with a similar two-fold rotational symmetry.

ANSWER: d 42. Which genome-editing machinery can insert a required sequence into the genome? a. ZFN fused to the restriction endonuclease b. TALEN fused to the restriction endonuclease .. Powered by Cognero.

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Chapter 41 c. CRISPR/Cas9 system d. two ZFN- or TALEN-fused restriction endonucleases e. None of the available machineries are capable of doing genome-editing.

ANSWER: e 43. What do southern, northern, and western blots detect, respectively? a. DNA, RNA, and protein b. DNA, protein, and RNA c. RNA, DNA, and protein d. protein, DNA, and RNA e. RNA, protein, and DNA ANSWER: a 44. How does the genome differ from the transcriptome? a. The genome is all of the genes in the DNA of a species, whereas the transcriptome is only those that are unique to an individual. b. The genome is all of the genes in the DNA of a cell, whereas the transcriptome is only those that are expressed within a specific tissue. c. The genome is the genes that encode proteins in a cell, whereas the transcriptome is only those proteins that are needed for steady-state levels of a particular cell. d. The transcriptome is all genes that can be transcribed by a cell, whereas the genome is the complete DNA sequence of the cell. e. The genome does not contain untranslated regions of DNA, but the transcriptome includes all transcribed regions of the DNA. ANSWER: d 45. Select all that apply. Plasmids used in recombinant DNA technology typically: a. possess a gene for antibiotic resistance. b. replicate independently of the host genome. c. are circular double-stranded molecules. d. cannot copy independently of the bacterial genome. e. possess a gene for antibiotic resistance and replicate independently of the host genome.

ANSWER: a, b, c, e 46. What would MORE likely yield a more optimal oligonucleotide probe, one with tryptophan and methionine or one with serine and leucine? a. It is impossible to predict the optimal nucleotide sequence, which is why poly-U nucleotides are generally used. b. The amino acids coded by codons, where all three positions can be the same nucleotide, generate the optimal oligonucleotide; therefore, serine and leucine. c. The amino acids coded by codons, where the first two positions are the same, generate the optimal oligonucleotide; therefore, serine and leucine. .. Powered by Cognero.

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Chapter 41 d. The amino acids with the least number of codons would not provide enough variation for all necessary oligonucleotides; therefore, leucine and serine. e. The amino acids with the least number of codons would require the fewest number of nucleotide probes; therefore, tryptophan and methionine.

ANSWER: e 47. Controlling stringency may be the most important parameter in PCR. If you want to check a gene from two distantly related species, for example, humans vs. zooplankton, what kinds of stringency measures would you take? a. Start PCR annealing temperatures high to see if there are similar sequences. b. Lower PCR annealing temperatures slowly if no product is formed. c. Adjust salt concentrations to achieve maximum ionic strength. d. Synthesize primers corresponding to the ends of the gene. e. All of the answers are correct.

ANSWER: e 48. In the controlled termination method of DNA sequencing, reading the gel from _____ gives the sequence in the _____ direction; _____ fragments that were terminated _____ in polymerization move faster down the gel a. top to bottom; 5′ to 3′; shorter; early b. top to bottom; 5′ to 3′; longer; early c. bottom to top; 5′ to 3′; shorter; early d. bottom to top; 5′ to 3′; longer; later e. bottom to top; 3′ to 5′; shorter; early ANSWER: c 49. Which DNA sequence contains a four- to eight-base palindromic site? (Note: Only one strand is shown.) a. CAGTCC b. GCATCC c. CGATTAGC d. GAGAGAGA e. GCATATGC ANSWER: e 50. Bacterial plasmid DNA and bacteriophage DNA are commonly used _______ to introduce foreign DNA into a bacterium.

ANSWER: vectors 51. In Chapter 40, we learned that a virus infecting Corynebacterium diphtheriae plays a role in producing a lethal toxin. Bacteria have endogenous restriction enzymes. Why, then, is the phage DNA that enters Corynebacterium diphtheriae not cleaved by these bacterial restriction enzymes? a. The restriction enzymes of bacteria degrade the phage DNA before it has the chance to infect the bacteria. .. Powered by Cognero.

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Chapter 41 b. The phage DNA probably is not recognized by the bacterial restriction enzymes due to modifications such as methylation or a lack of restriction sites. c. The phage does not enter the bacterial cell, avoiding any interaction with the bacterial restriction enzymes. d. The bacterial restriction enzymes are not capable of cleaving phage DNA due to its complicated tertiary structure. e. The phage DNA and bacterial restriction enzymes are located in different intracellular compartments, preventing any interaction.

ANSWER: b 52. A researcher is performing PCR to amplify a specific DNA sequence. She starts with a sample that contains one molecule of the target DNA. After 10 cycles of PCR, the amount of amplified DNA is measured. Assuming the ideal amplification of DNA with each cycle, how many molecules of the target DNA will the researcher have after 10 cycles? a. 1,024 molecules b. 10,240 molecules c. 2 molecules d. 20 molecules e. 10 molecules ANSWER: a 53. The frequency of a restriction enzyme’s recognition site in a DNA sequence depends on the length of the recognition sequence. For a restriction enzyme that recognizes a 6 base pair (bp) sequence, what is the approximate frequency of finding the recognition site in a random DNA sequence? a. 1 in 256 b. 1 in 1,024 c. 1 in 4,096 d. 1 in 16,384 e. 1 in 65,536 ANSWER: c 54. Which of the following statements about restriction enzymes that produce staggered (sticky) ends and blunt ends are TRUE? Select all that apply. a. Blunt ends are generated when the enzyme cuts both DNA strands at the same positions, resulting in no overhangs. b. Staggered cuts occur when the enzyme cuts both DNA strands at different positions, leaving overhangs. c. Blunt ends are generated when the enzyme cuts both DNA strands at different positions leaving overhangs. d. Blunt ends are harder to ligate together, whereas staggered ends generally have a higher ligation efficiency. e. Neither staggered nor blunt cuts can be used for cloning. .. Powered by Cognero.

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Chapter 41 ANSWER: a, b, d 55. Which of the following is a characteristic of a recognition sequence for a restriction enzyme with two-fold rotational symmetry? a. The recognition sequence is asymmetrical, allowing the enzyme to make a single cut on one strand. b. The enzyme always cuts both DNA strands at the same position within the recognition site, forming blunt ends. c. The enzyme always cuts the DNA strands at different positions, forming sticky ends. d. The sequence reads the same regardless of whether it is read from 5’ to 3’ or 3’ to 5’. e. The sequence reads the same when rotated 180 degrees.

ANSWER: e 56. A patient undergoes whole-genome sequencing, and it is found that retroviruses have been randomly incorporated into their genome. Which of the following can explain this finding? a. The patient’s immune system has developed a resistance to retroviral integration. b. The retroviral genome is converted into a double-helical DNA form by DNA polymerase I and becomes randomly incorporated into the host chromosomal DNA. c. The retroviral DNA genome was introduced directly into the patient’s genome through CRISPRbased editing. d. The retroviral DNA genome was introduced directly into the patient’s genome. e. The retroviral RNA genome is converted into a double-helical DNA form by a reverse transcriptase and then becomes randomly incorporated into the host chromosomal DNA. ANSWER: e 57. Which of the following statements describes the mechanism and application of electroporation? a. The electric field induces the formation of temporary pores in the cell membrane, enabling the introduction of genetic material into the cell. b. Electroporation dismantles the cell wall, allowing large molecules to diffuse through the cell membrane. c. Electroporation is used exclusively for bacterial cells and cannot be applied to eukaryotic cells. d. Electroporation disrupts the cell membrane, allowing proteins, but not genetic material, to enter through physical breaches in the membrane. e. Electroporation is used exclusively for eukaryotic cells and cannot be applied to bacterial cells.

ANSWER: a 58. Which technique is NOT commonly used to introduce genetic material into cells? a. electroporation b. microinjections c. ultracentrifugation d. heat shock transformation e. gene gun ANSWER: c .. Powered by Cognero.

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Chapter 41 59. Which of the following statements about the “shotgun” approach and gene guns is TRUE? Select all that apply. a. The “shotgun” approach involves randomly shearing genomic DNA (gDNA) into fragments, which are then sequenced and assembled to identify genes and other genomic elements. b. Gene guns, like the “shotgun” approach, randomly fragment DNA into small pieces for sequencing. c. The “shotgun” approach can be applied only in animal cells and cannot be used in plant cells. d. Gene guns work by physically inserting DNA-coated tungsten pellets into cells to deliver foreign DNA. e. Gene guns and “shotgun” techniques are similar.

ANSWER: a, d 60. A plasmid and its insert are both digested with a restriction enzyme that produces sticky ends and compatible overhangs. What is the expected proportion of the insert being introduced in the correct orientation compared to the incorrect orientation? a. The insert has a 100% chance of being introduced in the correct orientation. b. The insert is always introduced in the correct orientation since ligation is not a random process. c. The insert has a 75% chance of being introduced in the correct orientation and a 25% chance of being introduced in the incorrect orientation. d. The insert has a 90% chance of being introduced in the correct orientation and a 10% chance of being introduced in the incorrect orientation. e. The insert has a 50% chance of being introduced in the correct orientation and a 50% chance of being introduced in the incorrect orientation.

ANSWER: e 61. Which of the following statements regarding the detection mechanisms used in pyrosequencing and ion semiconductor sequencing is TRUE? a. Pyrosequencing detects the release of pyrophosphate, while ion semiconductor sequencing detects the release of protons during nucleotide incorporation. b. Pyrosequencing detects the release of protons, while ion semiconductor sequencing detects the release of pyrophosphate during nucleotide incorporation. c. In pyrosequencing and ion semiconductor sequencing, radioactive tracers are used to detect nucleotide incorporation. d. Pyrosequencing and ion semiconductor sequencing use pyrophosphate to detect nucleotide incorporation. e. Both pyrosequencing and ion semiconductor sequencing use fluorescent signals to detect the incorporation of nucleotides.

ANSWER: a 62. What is the relation between the following chemical structures:

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Chapter 41

a. A nucleotide contains a hydroxyl group (-OH) on the 3’ carbon of the sugar, whereas a dideoxynucleotide has a hydrogen atom (H) at the same position. b. A nucleotide contains a hydrogen atom (H) on the 3’ carbon of the sugar, whereas a dideoxynucleotide has a hydroxyl group (-OH) at the same position. c. All nucleotides have a hydrogen atom (H) on the 3’ carbon of the sugar. d. All nucleotides have a hydroxyl group (-OH) on the 3’ carbon of the sugar. e. A nucleotide contains a phosphate group, while a dideoxynucleotide lacks a phosphate group.

ANSWER: a 63. Why does primer excess in a PCR reaction prevent the re-annealing of the parent DNA duplex? a. The primer competes with the parent DNA for binding to the polymerase enzyme, preventing duplex formation. b. The primers bind to one DNA strand (5’ to 3’ end), stabilizing one of the strands and preventing the re-annealing of the parent duplex. c. The primers are so long that they form a stable duplex with the template, preventing the original parent DNA strands from coming together. d. The primers initiate DNA replication at multiple sites, completely preventing duplex formation. e. The primers are in large excess and bind to single-stranded, unhybridized DNA, stabilizing the separation and preventing the re-annealing of the parent duplex.

ANSWER: e 64. Which of the following statements regarding homologous recombination is FALSE? Select all that apply. a. The gene to be introduced does not need a flanking sequence. b. Homologous recombination occurs in regions of homology. c. After recombination, the targeted gene is replaced by the foreign gene. d. Homologous recombination is generally limited to specific animal models such as yeast, mice, and flies. e. Introducing point mutations into genes using homologous recombination is relatively simple, whereas knocking out the entire gene can be difficult and time-consuming.

ANSWER: a, e 65. Which of the following statements comparing homologous recombination, the CRISPR/Cas system, and ZFNs/TALENs are FALSE? Select all that apply. a. The CRISPR/Cas system is limited to gene editing in animal models and cannot be used in plants. .. Powered by Cognero.

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Chapter 41 b. Homologous recombination relies on the presence of regions of homology between the target DNA and the donor DNA. c. The CRISPR/Cas system requires a guide RNA to direct the Cas9 enzyme to the target DNA sequence. d. ZFNs and TALENs use sequence-specific DNA-binding proteins to create double-strand breaks at the target site. e. Homologous recombination is a more precise method for creating targeted mutations than CRISPR/Cas or ZFNs/TALENs. ANSWER: a, e 66. Which of the following statements about the restriction enzymes EcoRI and HindIII are FALSE? Select all that apply. a. The sticky ends generated by EcoRI and HindIII can be directly ligated together without any modifications. b. EcoRI and HindIII both recognize the same sequence but cut at different positions within the recognition site. c. If the recognition sites are methylated, EcoRI and HindIII will be unable to cleave the DNA, preventing the formation of sticky ends. d. EcoRI and HindIII were originally isolated from two different bacterial species, E. coli and H. influenzae, respectively. e. EcoRI and HindIII have blunt ends so they are compatible for ligation.

ANSWER: a, b, e 67. Which of the following statements about DNA and RNA microinjection into cells is TRUE? Select all that apply. a. RNA is microinjected into the cell nucleus to alter the genetic code directly. b. DNA is microinjected into the cytoplasm of cells, where it can integrate into the cell’s genome and encode proteins that can later be expressed. c. RNA microinjection into the cytoplasm of cells is used to take advantage of the translational machinery for protein expression. d. RNA microinjection into the cytoplasm of the cells is used to directly alter the cell’s genomic DNA. e. DNA is microinjected into the nucleus of cells, where it may integrate into the genome and potentially alter gene expression.

ANSWER: c, e 68. Which of the following statements comparing retroviruses and CRISPR/Cas9 as genetic manipulation techniques is TRUE? a. Retroviruses are often used to deliver genes as a whole, while CRISPR/Cas9 is a more precise method for genome editing. b. Retroviruses are typically used for precise, targeted gene editing, while CRISPR/Cas9 is used for stable gene delivery. c. With CRISPR/Cas9, a modified gene is randomly inserted into the genome, while retroviruses allow researchers to control where a gene is inserted. .. Powered by Cognero.

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Chapter 41 d. Both retroviruses and CRISPR/Cas9 are used primarily for gene delivery and do not affect the host genome. e. Neither retroviruses nor CRISPR/Cas9 are used for gene delivery.

ANSWER: a 69. A scientist uses polymerase chain reaction (PCR) to amplify a gene of interest from a DNA sample. After several PCR cycles, the expected product is not visible on the gel. The scientist suspects that the primers may not be annealing properly. Which of the following steps should the scientist take to increase the chance of successful amplification? a. The scientist should lower the annealing temperature to increase primer binding efficiency. b. The scientist should increase the concentration of primers to increase the chance of successful amplification. c. The scientist should not change anything and repeat the experiment. d. The scientist should add more Taq polymerase to the reaction to increase amplification. e. The scientist should increase the annealing temperature to ensure more specific primer binding.

ANSWER: e 70. A scientist uses polymerase chain reaction (PCR) to amplify a gene of interest from a DNA sample. After several PCR cycles, the expected product is not visible on the gel. Which of the following steps should the scientist take to increase the chance of successful amplification? a. The scientist should check whether the annealing temperature is too high, preventing efficient primer binding. b. The scientist should ensure the concentration of primers is appropriately in excess. c. The scientist should check whether his Taq polymerase is active and functional. d. The scientist should check whether the annealing temperature is too low, preventing sufficiently specific primer binding. e. All of the above.

ANSWER: e 71. In a typical PCR amplification reaction, how does the number of DNA molecules change over time in the exponential phase? a. The number of DNA molecules decreases exponentially, halving with each cycle. b. The number of DNA molecules increases linearly. c. The number of DNA molecules initially increases but decreases after a few cycles. d. The number of DNA molecules remains constant throughout the cycles. e. The number of DNA molecules increases exponentially, doubling with each cycle.

ANSWER: e

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