03-05-2006, 12:07 PM
BIOCHEMISTRY TEST
You have 546 questions in this exam.
1. Increasing the concentration of dGTP directly causes ribonucleotide reductase to
A. increase the rate of production of dGDP.
B. increase the rate of production of dADP.
C. decrease the rate of production of rCDP.
D. decrease the rate of production of dADP.
E. increase the rate of production of dTTP.
Show answer
Correct Answer: B
Feedback A: Has no direct effect (Increased dGTP increases dATP production, which eventually would decrease RNR activity overall, but dGTP has no immediate effect on the enzyme except to increase dADP formation)
Feedback C: Has no direct effect (Increased dGTP increases dATP production, which eventually would decrease RNR activity overall, but dGTP has no immediate effect on the enzyme except to increase dADP formation)
Feedback D: This rate increases
Feedback E: Has no direct effect (Increased dGTP increases dATP production, which eventually would decrease RNR activity overall, but dGTP has no immediate effect on the enzyme except to increase dADP formation)
2. Glycogen phosphorylase is activated most directly by
A. epinephrine.
B. phosphorylase kinase.
C. phosphorylase phosphatase.
D. cAMP.
E. glucagon.
Show answer
Correct Answer: B
Feedback A: See B
Feedback B: glucagon or epinephrine ---> increased cAMP ---> active PKA ---> active phosphorylase kinase ---> active glycogen phosporylase
Feedback C: See B
Feedback D: See B
Feedback E: See B
3. Which of the following cofactors is most directly concerned with a CO2 fixation reaction?
A. biotin
B. vitamin D
C. coenzyme A
D. thiamine pyrophosphate
E. lipoic acid
Show answer
Correct Answer: A
Feedback A: Biotin is a carrier of activated CO2 and is a component of acetyl carboxylase, propionyl CoA carboxylase, and pyruvate carboxylase.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
4. Which statement about reactions catalyzed by transaminases (aminotransferases) is false?
A. The equilibrium constant is close to 1.
B. The reaction involves a Schiff base intermediate.
C. Pyridoxal phosphate is covalently bound to the enzyme protein through the epsilon amino group of a lysine residue.
D. The reactions are important for the biosynthesis of non-essential amino acids.
E. Ammonia is liberated.
Show answer
Correct Answer: E
Feedback A: Aminotransferases can participate in degradation or biosynthesis depending on substrate concentrations.
Feedback B:
Feedback C:
Feedback D: Aminotransferases most commonly transfer amino groups to alpha-ketoglutarate forming glutamate. Glutamate can then be the amino group donor for amino acid biosynthesis.
Feedback E: False, an amino group is transferred to an alpha-keto acid.
5. The enzyme in muscle which is most directly stimulated by epinephrine is
A. glycogen synthase.
B. phosphofructokinase.
C. isocitrate dehydrogenase
D. glucose-6-phosphatase.
E. adenylate cyclase.
Show answer
Correct Answer: E
Feedback A: inactivated due to epinephrine.
Feedback B: decreased stimulation of PFK-1
Feedback C: very indirect modification
Feedback D: Changes in its activity are determined by synthesis and degradation.
Feedback E: increased intracellular cAMP.
6. Acetyl-CoA regulates gluconeogenesis by activation of
A. phosphoenolpyruvate carboxykinase.
B. pyruvate kinase.
C. pyruvate carboxylase.
D. lactate dehydrogenase.
E. fructose 1,6-bisphosphatase.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Which catalyzes: pyruvate + CO2 ---> oxaloacetate.
Feedback D:
Feedback E:
7. Which compound does NOT cross the inner mitochondrial membrane due to lack of a specific transport protein?
A. malate
B. glutamine
C. NADH
D. citrate
E. ATP
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: NADH reducing potential must be shuttled across the membrane by another molecule.
Feedback D:
Feedback E:
8. All of the following statements are correct with regard to chromatin EXCEPT
A. Core nucleosomes consist of 8 histone protein molecules and about 146 base pairs of DNA.
B. Histones have been highly conserved during evolution.
C. Almost all of the DNA in a eukaryotic cell is assembled into nucleosomes.
D. Assembly of DNA into nucleosomes compacts it about 100-fold in length.
E. Nucleosome cores are separated by variable amounts of linker DNA.
Show answer
Correct Answer: D
Feedback A: see D
Feedback B: see D
Feedback C: see D
Feedback D: Nucleosomes effect about a 5-7 fold compaction of the length of the DNA
Feedback E: see D
9. Which is LEAST likely to be characteristic of a regulated enzyme in a metabolic pathway?
A. Catalyzes the slowest step of the pathway
B. Has a short half-life in vivo
C. Concentration is hormonally regulated
D. Activity under allosteric control
E. Catalyzes a physiologically reversible reaction
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Irreversible reactions are usually regulated, especially those that involve ATP.
10. Which of the following are examples of tertiary (3°) structure? 1. the sequence of amino acids in a protein 2. the alpha-helix of hemoglobin 3. the folding of alpha-helix onto beta-sheet of triose phosphate isomerase 4. hydrogen bonding of one beta -sheet to another beta-sheet 5. juxtaposition of amino acids that are separated by a "long distance" in primary (1°) structure 6. the un-ordered portions of carboxy peptidase A 7. juxtaposition of hemoglobin beta-7 Glu with alpha-68 Asn
A. all the even-numbered items
B. all the odd-numbered items
C. 3, 4, and 5
D. 3, 4, 5, and 7
E. 3, 4, 5, 6, and 7
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: 1-primary, 2-secondary, 6-hard to classify--primary or secondary, 7-quaternary
Feedback D:
Feedback E:
11. The H1 histone
A. is one of the four histones in the nucleosome core.
B. helps form the octamer nucleosome complex.
C. is rich in acidic amino acid residues.
D. is present at a 1:1 weight ratio with DNA.
E. helps fold or stabilize the "beads-on-a-string" structure into higher order structures.
Show answer
Correct Answer: E
Feedback A: No, the core histones are H2A, H2B, H3, and H4
Feedback B: No, the octamer contains only H2A, H2B, H3, and H4
Feedback C: It is rich in basic amino acids
Feedback D: Overall, all histones (H1 and H2A, H2B, H3, and H4) are about 1:1 with DNA
Feedback E:
12. Which mitochondrial enzyme requires acetyl-CoA as a substrate?
A. Citrate synthase
B. Succinyl-CoA synthase
C. Succinic dehydrogenase
D. Pyruvate dehydrogenase
E. Isocitrate dehydrogenase
Show answer
Correct Answer: A
Feedback A: Catalyzes: acetyl CoA + oxaloacetate ---> citrate.
Feedback B: CoA is required coenzyme, but not acetyl CoA.
Feedback C:
Feedback D: Produces acetyl CoA from pyruvate.
Feedback E:
13. Transfer of the methyl group from 5-methyl tetrahydrofolate to homocysteine to form methionine requires
A. NADPH
B. vitamin B12
C. FAD
D. pyridoxal phosphate
E. none of the above
Show answer
Correct Answer: B
Feedback A:
Feedback B: Methycobalamin, a derivative of B12, is the coenzyme that mediates the transfer of methyl groups in methionine biosynthesis.
Feedback C:
Feedback D: Aminotransferases require this coenzyme.
Feedback E:
14. All of the following are true of bacterial plasmids or episomes EXCEPT
A. Episomes like the F factor can be transferred from one cell to another.
B. Plasmids or episomes can confer resistance to many antibiotics simultaneously.
C. Plasmids or episomes can be transferred from one species of bacteria to another.
D. Since plasmids must be replicated, they confer a disadvantage to their hosts and can therefore be expected to disappear from human pathogens soon.
E. Genes can be transferred from bacterial genomes to plasmids at a low frequency.
Show answer
Correct Answer: D
Feedback A: This is true, they can pass through a pilus into another cell.
Feedback B: This is true.
Feedback C: This is true
Feedback D: The disadvantages caused by œcarrying plasmids are outweighed by the advantages they provide; this advantage increases as we use antibiotics.
Feedback E: Strange but true.
15. If one mole of glucose is metabolized to carbon dioxide and water via glycolysis and the tricarboxylic acid cycle, and the glycerol phosphate shuttle is operative, the net conversion of ADP (or equivalent) to ATP (or equivalent) would theoretically be
A. 12 moles.
B. 24 moles.
C. 32 moles.
D. 36 moles.
E. 38 moles.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: The 2NADH produced by glycolysis are converted to 2FADH2 in the mitochondria by the glycerol phosphate shuttle, so they are worth 2ATP each.
Feedback E: With the malate-aspartate shuttle the 2NADH from glycolysis are converted to 2NADH in the mitochondria, and are worth 3ATP each.
16. The cofactor for transamination is
A. nicotinamide.
B. biotin.
C. thiamine pyrophosphate.
D. pyridoxal phosphate.
E. vitamin B12.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: A vitamin B6 derrivative that is a prosthetic group for each transaminase.
Feedback E:
17. Which enzyme catalyzes a reaction in which carbon dioxide (or bicarbonate) is neither a substrate nor a product?
A. isocitrate dehydrogenase
B. pyruvate carboxylase
C. carboxypeptidase
D. phosphoenolpyruvate carboxykinase
E. alpha-ketoglutarate dehydrogenase
Show answer
Correct Answer: C
Feedback A: CO2 produced.
Feedback B: CO2 is a substrate.
Feedback C: Cleaves peptide bonds, no CO2 involved.
Feedback D: CO2 produced.
Feedback E: CO2 produced.
18. 5-phosphoribosyl-1-pyrophosphate (PRPP) is an intermediate in
A. the de novo synthesis of purine nucleotides.
B. the de novo synthesis of pyrimidine nucleotides.
C. the salvage pathway for the synthesis of purine nucleotides.
D. A and C.
E. A, B and C.
Show answer
Correct Answer: E
Feedback A: True (it is activated as the committed step for purine synthesis), but so are B and C
Feedback B: True (orotidylate is formed by reacting the free base with PRPP), but so are A and C
Feedback C: True (PRPP reacts with free bases like guanine and hypoxanthine through HGPRTase), but so are A and B.
Feedback D: True, but B is also correct
Feedback E:
19. Which statements are FALSE? In mitochondria, tricarboxylic acid cycle reactions generally proceed more
(1) rapidly as the ADP concentration rises.
(2) slowly as the ADP concentration rises.
(3) slowly as the NADH concentration rises.
(4) rapidly as the NADH concentration rises.
(5) rapidly as the oxaloacetate concentration increases.
A. 1 and 3
B. 2 and 4
C. 3 and 5
D. 1 and 4
E. 2 and 5
Show answer
Correct Answer: B
Feedback A:
Feedback B: High levels of ATP and NADH signal that the cell™s energy level is high, high ADP and NAD+ signal that more energy is needed.
Feedback C:
Feedback D:
Feedback E:
20. In Lesch-Nyhan syndrome, lack of HGPRTase activity should result in higher than normal tissue concentrations of all of the following EXCEPT
A. adenine.
B. guanine.
C. uric acid.
D. hypoxanthine.
E. PRPP.
Show answer
Correct Answer: A
Feedback A: Adenine levels are not directly affected by loss of this enzyme.
Feedback B: This substrate for HGPRTase would accumulate abnormally in the absence of the enzyme.
Feedback C: Increased hypoxanthine leads to increased xanthine and therefore increased uric acid.
Feedback D: This substrate for HGPRTase would accumulate abnormally in the absence of the enzyme.
Feedback E: This substrate for HGPRTase would accumulate abnormally in the absence of the enzyme.
21. Which of the following represents the correct number of moles of products formed as one mole of acetate (in the form of acetyl-CoA) moves through one turn of the citric acid cycle? (respectively)
GTP
CO2
NADH
FADH2
A. 1 1 3 1
B. 2 2 3 1
C. 1 2 3 1
D. 1 2 2 1
E. 1 1 2 2
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Memorize the cycle.
Feedback D:
Feedback E:
22. How many net moles of ATP are produced per glucose equivalent when glycogen is the substrate for anaerobic glycolysis?
A. 0
B. 1
C. 2
D. 3
E. -2
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: glycogen ---> glucose-1-phosphate ---> glucose-6-phosphate ---> glycolysis. Because glycose-6-phosphate enters glycolysis the reaction catalyzed by glucokinase or hexokinase is not carried out and one less ATP is used.
Feedback E:
23. One of the strands of a double-stranded, 10 kbp DNA duplex has the following numbers of base residues: adenine(a) - 3800, thymine(t) - 2600. The base composition of the whole double-stranded molecule will be
A. a - 5200, t - 5200, g - 4800, c - 4800
B. a - 7600, t - 7600, g - 2400, c - 2400
C. a - 6400, t - 6400, g - 3600, c - 3600
D. a - 6400, t - 3600, g - 6400, c - 3600
E. There is not enough information to determine the overall base composition
Show answer
Correct Answer: C
Feedback A: If one strand has 3800 A and 2600 T, the other strand has 3800 T and 2600 A, so the duplex has 6400 A and 6400 T, with the remainder being G and C (3600 each).
Feedback B: If one strand has 3800 A and 2600 T, the other strand has 3800 T and 2600 A, so the duplex has 6400 A and 6400 T, with the remainder being G and C (3600 each).
Feedback C:
Feedback D: If one strand has 3800 A and 2600 T, the other strand has 3800 T and 2600 A, so the duplex has 6400 A and 6400 T, with the remainder being G and C (3600 each).
Feedback E: If one strand has 3800 A and 2600 T, the other strand has 3800 T and 2600 A, so the duplex has 6400 A and 6400 T, with the remainder being G and C (3600 each).
24. A major control point of the citric acid cycle is exerted at the level of
A. succinate dehydrogenase.
B. citrate synthase.
C. succinyl CoA synthase.
D. malate dehydrogenase.
E. aconitase.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Citrate synthase, isocitrate dehydrogenase, and alpha-ketoglutarate dehydrogenase are the regulated enzymes of TCA cycle.
Feedback C:
Feedback D:
Feedback E:
25. Secondary structure of proteins
A. is maintained by hydrogen bonds.
B. refers to proteins consisting of two (or more) subunits held together by noncovalent forces.
C. encompasses any hydrogen-bonded interaction found in proteins.
D. refers to proteins consisting of one or more polypeptide chains plus a nonprotein moiety.
E. is found only in fibrous proteins.
Show answer
Correct Answer: A
Feedback A: alpha-helix and beta-sheet.
Feedback B: This is quaternary.
Feedback C: No, H-bond interactions between residues far apart linearly would be tertiary interactions.
Feedback D: This is quaternary.
Feedback E: Any protein.
26. Each of the following statements concerning DNA is true EXCEPT which one?
A. C does not have to equal G in single stranded DNA.
B. DNA can be found in either circular or linear forms.
C. The two DNA strands of the double helix are antiparallel.
D. Hydrogen bonds alone hold the DNA helix together.
E. Base pairs lie in a plane perpendicular to the long axis of the helix.
Show answer
Correct Answer: D
Feedback A: This is true. For example, a single-stranded DNA could be all C and have no G.
Feedback B: This is true; most bacterial plasmids are circles while most mammalian chromosomes are linear.
Feedback C: Please don™t tell me you picked this one, I just don™t want to know.
Feedback D: No, they contribute but they are only one part of the picture.
Feedback E: This is true.
27. The reduction potential for a reaction
A. is an oxidation potential.
B. is unrelated to the free energy of the reaction.
C. is negative for a spontaneous process.
D. can be used to decide if one substance will reduce another.
E. is not dependent on reactant and/or product concentration.
Show answer
Correct Answer: D
Feedback A: Has the opposite sign of the reduction potential.
Feedback B:
Feedback C: Positive for a spontaneous redox reaction.
Feedback D: More negative (smaller) reduction potentials donate electrons to more positive (larger) reduction potentials.
Feedback E:
28. Glutamine
A. is not incorporated into proteins because there is no codon or tRNA for glutamine.
B. is the principal donor of amino groups in transamination reactions.
C. donates its alpha-amino group in the de novo biosynthesis of both purine and pyrimidine nucleotides.
D. is synthesized from glutamate primarily in the liver.
E. represents an important non-toxic transport form of ammonia.
Show answer
Correct Answer: E
Feedback A: There are codons and tRNA™s for both glutamine and glutamate.
Feedback B: Glutamine is the ultimate product of tissue transamination. Glutamine is transported to the liver where nitrogen metabolism to urea takes place.
Feedback C: Donates its gamma-amide in both syntheses.
Feedback D: Synthesis occurs primarily in the tissues and then glutamine is broken down into glutamate and ammonia in the liver.
Feedback E: See B.
29. As a cell initiates DNA replication, it contains sufficient pools of dNTPs to complete
A. about 1% or less of DNA replication without making new dNTPs.
B. about 10% of DNA replication without making new dNTPs.
C. 100% of DNA replication without making new dNTPs.
D. about 10 times more dNTPs than it needs to replicate the genome once.
E. no replication at all.
Show answer
Correct Answer: A
Feedback A: About 0.1% in mammals, 1% in bacteria
Feedback B: About 0.1% in mammals, 1% in bacteria
Feedback C: About 0.1% in mammals, 1% in bacteria
Feedback D: About 0.1% in mammals, 1% in bacteria
Feedback E: About 0.1% in mammals, 1% in bacteria
30. The uncoupling of oxidative phosphorylation in the human might be of physiological importance because it
A. allows storage of nutrients.
B. produces water.
C. increases carbon dioxide level in the blood.
D. produces heat.
E. raises the oxygen level in the blood.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: The energy produced by electron transport is released as heat rather than being used to synthesize ATP.
Feedback E:
31. The condition of LACTOSE INTOLERANCE can be caused by defective or deficient
A. lactose
B. UDP-galactose epimerase
C. galactose-1-phosphate uridyl transferase
D. lactase
E. galactase
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
32. During the second day of a fast, after liver glycogen is depleted, blood glucose
A. is derived mainly from fatty acids of adipose tissue.
B. is derived mainly from muscle glycogen.
C. is derived mainly from the amino acids of liver proteins.
D. falls to low levels (e.g., 20 mg/100 ml) until eating is resumed.
E. is derived mainly from the amino acids of muscle proteins.
Show answer
Correct Answer: E
Feedback A: After approximately 3 weeks of starvation muscle almost exclusively oxidizes fatty acids for energy.
Feedback B:
Feedback C:
Feedback D:
Feedback E: Rapid breakdown of muslce protein takes place the first few days of starvation to provide amino acids for gluconeogenesis.
33. A complete lack of adenosine deaminase causes SCID: severe combined immuno-deficiency. Which of the following is LEAST true?
A. Loss of the enzyme causes increased levels of dATP because there is less turnover of adenosine nucleosides in general.
B. Increased dATP decreases the concentration of all rNTPs, blocking RNA synthesis.
C. Increased dATP inhibits ribonucleotide reductase, such that de novo production of all dNDPs is inhibited.
D. Adenosine deaminase loss causes SCID because T cells are particularly sensitive to DNA replication inhibition.
E. It is not clear why adenosine deaminase loss affects T cells so specifically since other cells have the same kind of nucleotide synthesis regulation and might be expected to be equally affected.
Show answer
Correct Answer: B
Feedback A: This is true, failure to degrade adenosine forms causes their concentrations to rise.
Feedback B: Lack of dNTPs blocks DNA synthesis, not RNA synthesis
Feedback C: This is true, dATP levels rise (see A) which shuts down all RNR activities
Feedback D: This is true; the reason is not fully known but it is observed to be the case.
Feedback E: This is true, see D.
34. In the oxidation of NADH by the electron transport system, energy released is coupled to the formation of ATP from ADP plus inorganic phosphate. For every mole of NADH oxidized, how many moles of ATP formed are formed?
A. 1
B. 5
C. 3
D. 6
E. None of the above
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: This is for NADH in the mitochondria. NADH outside of the mitochondria produce different numbers of ATP depending on the shuttle system used to transfer the reducing potential from the cytosol to the mitochondria.
Feedback D:
Feedback E:
35. In analysis of isoenzymes of serum CPK for diagnosis of an AMI
A. CPK above six times normal total CPK indicates an AMI.
B. CPK-MM above four times normal total CPK indicates an AMI.
C. CPK-BB above 1% of total normal, and total CPK of six times normal CPK indicates an AMI.
D. CPK-MB above 3% of total normal CPK indicates an AMI.
E. Total CPK above six times normal, with CPK BB in the normal range indicates an AMI.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: CPK-MB is only 1-3% in other muscle, but 15-20% in heart muscle.
Feedback E:
36. Which of the following is true about the structure and properties of ATP?
A. The structure contains 3 phosphoanhydride bonds and one glycosidic bond.
B. At physiological pH, the molecule has a net charge of approximately -6.
C. The transfer of a phosphoryl group (phosphate residue) is the only biological reaction of ATP that cleaves a high-energy bond.
D. The high free energy of hydrolysis results, in part, from electrical repulsion between phosphate residues.
E. The standard free energy of hydrolysis of ATP is highter than that of other all phosphate derivatives in the glycolytic pathway.
Show answer
Correct Answer: D
Feedback A: 2 phosphoanhydride bonds and one glycosidic bond.
Feedback B:
Feedback C:
Feedback D:
Feedback E: No, hydrolysis of PEP gives more energy than hydrolysis of ATP.
37. Which of the following species of denatured DNA will renature most rapidly in solution, under appropriate conditions of ionic strength, pH, and temperature?
A. Human liver nuclear DNA
B. Vaccinia (viral) DNA
C. E. coli DNA
D. Yeast nuclear DNA
E. Mouse neuronal DNA
Show answer
Correct Answer: B
Feedback A: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
Feedback B: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
Feedback C: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
Feedback D: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
Feedback E: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
38. When oxidative phosphorylation in mitochondria is uncoupled
A. AMP is formed and O2 consumption stops.
B. the oxidation of acetyl CoA in the tricarboxylic acid cycle stops.
C. mitochondrial metabolism stops.
D. ATP synthesis stops but O2 consumption continues.
E. the cytochromes lose their heme groups into the cytoplasm.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Electron transport continues to reduce O2, but the proton gradient is not maintained so ATP is not synthesized.
Feedback E:
39. Oxaloacetate moves through the mitochondrial membrane
A. after conversion to glycerol phosphate.
B. by passive diffusion.
C. after reaction with carnitine.
D. after oxidation to pyruvate.
E. after reduction to malate.
Show answer
Correct Answer: E
Feedback A: DHAP + NADH ---> glycerol-3-phosphate + NAD in the glycerolphosphate shuttle.
Feedback B:
Feedback C: Long-chain fatty acids are transported into the mitochondria for oxidation by carnitine.
Feedback D:
Feedback E: The Malate-Aspartate shuttle for NADH.
40. The absence of which enzyme results in PKU (phenylketonuria)?
A. Tyrosine hydroxylase
B. Amino acid oxidase
C. Phenylalanine hydroxylase
D. Tyrosine transaminase
E. Phenylalanine transaminase
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Phenylalanine is not hydroxylated to tyrosine, and builds up to toxic levels.
Feedback D:
Feedback E:
41. All of the following statements are consistent with the chemiosmotic hypothesis of oxidative phosphorylation EXCEPT which one?
A. Agents which uncouple oxidative phosphorylation from electron transport disrupt permeability barriers such that mitochondrial membranes become permeable to protons.
B. The term electrochemical gradient or proton gradient operationally describes the free energy generated by electron transport which drives phosphorylation of ADP.
C. Protons are transferred into the mitochondria during electron transport establishing a pH gradient.
D. Intact mitochondrial membranes are necessary for oxidative phosphorylation to occur.
E. Electron carriers of the electron transport chain are vectorially oriented.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Protons are transferred from the matrix to the intermembrane space of the mitochondria.
Feedback D: The proton gradient must be set up across the inner membrane.
Feedback E:
42. Which statement about the pyruvate dehydrogenase complex is true?
A. Lipoamide functions in the decarboxylation of pyruvate.
B. Coenzyme A is covalently bound to a lysine residue in one of the proteins of the complex.
C. Phosphorylation catalyzed by a kinase results in decreased activity.
D. NADH is an allosteric activator of pyruvate oxidation.
E. The function of FAD is to oxidize NADH.
Show answer
Correct Answer: C
Feedback A: TPP functions in decarboxylation; lipoamide transfers the acetyl group to CoA.
Feedback B: Lipoamide is bound to lysine.
Feedback C: Increased acetyl CoA/CoA or NADH/NAD ratios activate the kinase which phosphorylates and deactivates PDH.
Feedback D: Increased NADH causes indirect inhibition of PDH.
Feedback E: FADH2 reduces NAD to NADH.
43. Which of the following is the single most important factor accounting for the high rates of enzyme catalysis?
A. Enzymes bring the substrate [S] into close proximity of the coenzyme.
B. Certain enzyme amino acid side chains interact with the substrate and stabilizing reactive intermediate(s).
C. The "free energy" of activation is increased during enzyme catalysis.
D. Enzymes contain prosthetic groups which promote catalysis.
E. Enzymes are globular proteins which contain hydrophobic and hydrophilic amino acids.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Enzymes stablaize the transition state.
Feedback C: Acitvation energy is lowered by enzymes.
Feedback D: True, but not the best explanation.
Feedback E: Does not explain catalysis.
44. Oxidative phosphorylation in mitochondria is carried out by enzymes
A. on the outer membrane.
B. on the inner membrane.
C. in the matrix.
D. in the intermembrane space.
E. all of the above are correct..
Show answer
Correct Answer: B
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
45. What change from normal is likely to result from the ingestion of ethanol?
A. increased reduction of pyruvate
B. increased conversion of beta-hydroxybutyrate to acetoacetate
C. decreased ketogenesis due to lack of acetyl-CoA
D. increased production of oxaloacetate from malate
E. increased production of dihydroxyacetone phosphate from glycerol phosphate
Show answer
Correct Answer: A
Feedback A: Ethanol metabolism requires reduction of NAD to NADH , and as in anaerobic glycolysis, NAD is regenerated by reduction of pyruvate to lactate.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
46. Glucagon
A. is released from the pancreas in response to lowered blood glucose.
B. inactivates liver phosphorylase.
C. increases glucose release from muscle glycogen.
D. utilizes cAMP as a second messenger in muscle cells.
E. increases glucose uptake into hepatocytes.
Show answer
Correct Answer: A
Feedback A:
Feedback B: Activates the phosphorylase for glycogen degradation.
Feedback C: Glucagon does not affect muscle.
Feedback D: See C.
Feedback E: Insulin increased glucose uptake when blood glucose is high.
47. Muscle can carry out all of the following processes EXCEPT
A. transamination of branched chain amino acids.
B. synthesis of glycogen.
C. synthesis of glucose from alanine.
D. oxidation of ketone bodies.
E. phosphorylation of creatine.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Muscle cannot carry out gluconeogenesis because pyruvate caboxylase is found only in liver and kidney cells.
Feedback D:
Feedback E:
48. All of the following are likely to occur following the ingestion of a very large amount of carbohydrate EXCEPT
A. the synthesis of triacylglycerol in the liver.
B. the synthesis of glycogen in the liver.
C. the oxidation of glucose by the pentose phosphate pathway in liver.
D. lipolysis in adipose tissue.
E. the synthesis and release of VLDL from the liver.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: High glucose levels lead to high insulin which supports synthesis of storage molecules, not breakdown.
Feedback E:
49. Catalysis by chymotrypsin depends on the concerted participation of three amino acid side chains (catalytic triad) which are contributed by which of the following amino acids?
A. Arginine, histidine, and serine
B. Histidine, serine, and glutamate
C. Serine, glutamate, and aspartate
D. Aspartate, serine, and histidine
E. Serine, histidine, and arginine
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Memorize it. Know the part each plays in the mechanism.
Feedback E:
50. Which of the following is the most accurate description of phosphofructokinase-1?
A. This enzyme uses fructose-6-phosphate as a substrate and converts it to fructose-2,6- diphosphate.
B. This enzyme is inhibited by ATP, citrate, and fructose-2,6-biphosphate.
C. This enzyme catalyzes a fully reversible reaction under physiological conditions.
D. This activity of this enzyme is indirectly increased by cyclic AMP.
E. No statement above is accurate.
Show answer
Correct Answer: E
Feedback A: PFK-2 makes fructose-2,6-bisphosphate, PFK-1 produces fructose-1,6-bisphosphate.
Feedback B: Inhibited by ATP and citrate, stimulated by fructose-2,6-bisphosphate and AMP.
Feedback C: The committed step of glycolysis, irreversible.
Feedback D: Indirectly decreased because increased cAMP causes PFK-2/FBP-2 complex to act as a phosphatase.
Feedback E:
51. All of the following amino acids are required as precursors for collagen (procollagen) synthesis EXCEPT
A. lysine.
B. glycine.
C. proline.
D. hydroxyproline.
E. phenylalanine.
Show answer
Correct Answer: D
Feedback A: Commonly hydroxylated to hydroxylysine in collagen.
Feedback B: Found every third residue.
Feedback C: Commonly hydroxylated to hydroxyproline.
Feedback D: Proline is the precursor which undergoes posttranslational modification to hydroxyproline.
Feedback E:
52. Which of the following statements is an accurate comparison between chymotrypsinogen and chymotrypsin?
A. Chymotrypsinogen has a single amino terminus and is inactive.
B. Chymotrypsinogen is a storage form of chymotrypsin and is active.
C. Chymotrysinogen and chymotrypsin are readily interconvertable forms of this hydrolase.
D. More than one of the above is an accurate statement.
E. None of the above is an accurate statement.
Show answer
Correct Answer: A
Feedback A: Chymotrypsinogen is a single polypeptide chain that is cleaved to give chymotrypsin (3 chains that are linked by 2 disulfide bonds),
Feedback B: Chymotrypsinogen is inactive.
Feedback C: No, cleavage of chymotrypsinogen to chymotrypsin is irreversible.
Feedback D:
Feedback E:
53. Reactions occurring during anaerobic GLYCOLYSIS in liver which require the input (use of) of ATP include
A. pyruvate kinase and glucokinase
B. pyruvate kinase and phosphofructokinase
C. glyceraldehyde-3-phosphate dehydrogenase and pyruvate kinase
D. glucokinase and phosphofructokinase
E. glucokinase and glyceraldehyde-3-phosphodehydrogenase
Show answer
Correct Answer: D
Feedback A: See C.
Feedback B: See C.
Feedback C: Glyceraldehyde-3-phosphate dehydrogenase produces NADH, pyruvate kinase produces ATP.
Feedback D:
Feedback E: See C.
54. The conversion of the two nitrogen atoms of glutamine to the two nitrogen atoms of urea requires the participation of urea cycle enzymes and
A. glutaminase only.
B. glutamate dehydrogenase only.
C. glutaminase and aspartate aminotransferase (transaminase).
D. glutaminase and glutamate dehydrogenase.
E. glutamate dehydrogenase and aspartate aminotransferase.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: One urea nitrogen comes from glutamine and enters the urea cycle as carbamoyl phosphate, the second nitrogen enters the cycle as aspartate.
Feedback D:
Feedback E:
55. Which molecule can donate a one-carbon unit most directly to tetrahydrofolate?
A. serine
B. methionine
C. glutamate
D. histidine
E. choline
Show answer
Correct Answer: A
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
56. "Induced fit" is observed during catalysis by
A. adenylate cyclase.
B. glycogen phosphorylase "b".
C. chymotrypsin.
D. carboxypeptidase A.
E. cyclic AMP dependent protein kinase.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Binding of substrate alters the conformation of the enzyme.
Feedback E:
57. Overall totally anaerobic glycolysis provides less ATP than aerobic glycolysis because
A. less or no NADH is recovered during anaerobic glycolysis.
B. less NAD is available under anaerobic conditions, so glycolysis cannot continue.
C. lactate produced anaerobically inhibits phosphofructokinase-1.
D. lactate is not a substrate for pyruvate dehydrogenase.
E. more than one of the above is correct.
Show answer
Correct Answer: A
Feedback A: The NADH is oxidized to NAD+ by the conversion of pyruvate to lactate, so that the NAD+ can be used for another round of glycolysis. The NADH is thus not available for oxidative phosphorylation.
Feedback B: See A.
Feedback C: Citrate and ATP inhibit PFK-1.
Feedback D: True, but it is the NADH that is important because it and FADH2 transfer reducing potential to electron transport. The TCA cycle only makes one GTP without electron transport.
Feedback E:
58. The protective mechanism preventing accumulation of ammonia in most tissues resides in which enzyme?
A. aspartate aminotransferase (GOT)
B. glutaminase
C. gamma-glutamyl transpeptidase
D. glutamine synthetase
E. argininosuccinate synthetase
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Glutamine removes ammonia from tissues and transports it to the liver for further metabolism to urea.
Feedback E:
59. Changes in enzyme activities IN SITU often involve "allosteric effects." This regulation most commonly is seen through changes in
A. Vmax.
B. Km.
C. enzyme concentration.
D. phosphorylation of reactive serines.
E. Vmax and Km.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Allosteric modifiers most often change an enzyme™s affinity for substrate; high Km indicates low affinity.
Feedback C:
Feedback D:
Feedback E:
60. Phosphofructokinase-2 was described as a "tandem" or bifunctional enzyme. This means that
A. this enzyme can catalyze more than one reaction.
B. this enzyme hydrolyzes fructose-1,6-diphosphate and may also catalyze the formation of this same compound.
C. this enzyme has "group" or "class" specificity.
D. this enzyme catalyzes the formation of fructose-1,6-diphosphate.
E. this enzyme catalyzed the hydrolysis of fructose-2,6-diphosphate to give fructose-2-phosphate.
Show answer
Correct Answer: A
Feedback A: It has two parts: the kinase that catalyzes phosphorylation and the phosphatase that catalyzes dephosphorylation.
Feedback B: Hydrolysis and phosphorylation that converts between fructose-2,6-bisphosphate and fructose-6-phosphate.
Feedback C:
Feedback D: See B.
Feedback E: See B.
61. An important source of reducing power for biosynthesis is the reaction catalyzed by
A. malate dehydrogenase.
B. pyruvate dehydrogenase.
C. acyl-CoA dehydrogenase.
D. glycerol-3-phosphate dehydrogenase.
E. glucose-6-phosphate dehydrogenase.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Part of the HMP pathway that produces NADPH which is needed for biosynthesis of fatty acids, steroids, etc.
62. Argininosuccinate is an intermediate in the de novo biosynthesis of
A. heme.
B. adenosine monophosphate.
C. uridine monophosphate.
D. creatine.
E. urea.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: arininosuccinate ---> fumarate + arginine ---> urea + ornithine
63. What is the net yield of NADH when glucose 6-phosphate is converted to lactate by anaerobic glycolysis?
A. 0
B. 1
C. 2
D. 3
E. 4
Show answer
Correct Answer: A
Feedback A: Glyceraldehyde-3-phosphate dehydrogenase produces NADH and lactate dehydrogenase oxidizes NADH to NAD+.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
64. An International Unit (IU) is that quantity of enzyme which can convert substrate to product at a rate of
A. 1 micromole/min.
B. 1 micromole/sec.
C. 1 millimole/min.
D. 1 millimole/sec.
E. 1 nanomole/sec.
Show answer
Correct Answer: A
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
65. You have just finished a meal containing 100 grams of carbohydrate, 50 grams of fat and 25 grams of protein. How many kilocalories did you consume?
A. 350
B. 650
C. 950
D. 1250
E. 1550
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: 100g carb (4 Cal/g) + 50g fat (9 Cal/g) + 25g protein (4 Cal/g) = 950 Cal or kcal.
Feedback D:
Feedback E:
66. All of the following can increase the activity of muscle glycogen phosphorylase "b" (dependent) except for
A. epinephrine.
B. cyclic AMP.
C. glycogen phosphorylase kinase.
D. 5'-adenosine monophosphate (AMP).
E. glucagon.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Glucagon does not effect muscle.
67. The major rate-limiting step of glycolysis in liver cells is
A. the conversion of glucose to glucose 6-phosphate.
B. the conversion of glucose 6-phosphate to fructose 6-phosphate.
C. the conversion of fructose 6-phosphate to fructose 1,6-bisphosphate.
D. the aldolase reaction.
E. none of the above.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: The committed step catalyzed by PFK-1.
Feedback D:
Feedback E:
68. A new enzyme has been isolated and shown to have a native molecular weight of 240,00. In the presence of 8M urea the molecular weight of the enzyme was estimated to be 120,000. When beta- mercaptoethanol was added to the urea system, the molecular weight was found to be 60,000. Which of the following best describes the subunit or polypeptide structure of the enzyme?
A. The enzyme consists of two subunits with no interchain disulfide bridges.
B. The enzyme consists of four polypeptide chains with no interchain disulfide bridges.
C. The enzyme consists of two polypeptide chains with disulfide bridges.
D. The enzyme consists of four polypeptide chains interconnected by disulfide bridges.
E. The enzyme consists of two pairs of polypeptide chains with each pair interconnected by a disulfide bridge.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Urea disrupts the noncovalent interactions between the 2 subunits and beta-mercaptoethanol disrupts the disulfide bonds in each subunit.
69. Assume that the STANDARD FREE ENERGY change for the reaction:
ATP ---> ADP + Pi
is -7.0 Cal/mole. What is the best estimate of the "actual" free energy change when the following concentrations exist:
[ATP = 1x10-4], [ADP = 1x10-5], and [Pi = 1x10-4]?
Use R = 2.0 cal/mole K°, and T = 300 K°.
A. -7.3 Cal
B. -10.0 Cal
C. -10.3 Cal
D. -12.3 Cal
E. -13.9 Cal
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: G = Go + 2.3RT log K G = -7000 cal + 2.3(2)(300) log(10-4 X 10-5/10-4) G = -13900 cal or -13.9 Cal
70. Which citric acid cycle intermediate helps to regulate the rate of glycolysis by directly influencing the activity of phosphofructokinase?
A. Citrate
B. Succinate
C. Alpha-ketoglutarate
D. Oxaloacetate
E. Malate
Show answer
Correct Answer: A
Feedback A: Inhibits PFK-1.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
71. Which primers will amplify my favorite gene if it is in the context below? 5'-GGTACAGTC-MY FAVORITE GENE-CTAGATCAT-3'
A. 5'-GACTGTACC and 5'-CTAGATCAT
B. 5'-GGTACAGTC and 5'-ATGATCTAG
C. 5'-GGTACAGTC and 5'-CTAGATCAT
D. 5'-GACTGTACC and 5'-ATGATCTAG
E. 5'-GGTACAGTC and 5'-GACTGTACC
Show answer
Correct Answer: B
Feedback A: Both primers point away from the target.
Feedback B: Yes, these anneal to the sequences flanking the target and are oriented with their 3™ ends towards the target.
Feedback C: The first primer is correct, but the second one points away from the target.
Feedback D: Both primers point the same way, with one pointing away from the target.
Feedback E: The first primer is correct, but the second one is nonsense, it is complementary to the second flank but parallel.
72. The carbon atom in urea synthesized in the liver arises most directly from
A. aspartate.
B. CO2.
C. fumarate.
D. ornithine.
E. glutamate.
Show answer
Correct Answer: B
Feedback B: The CO2 comes into the urea cycle through the synthesis of carbamoyl phosphate.
Feedback C:
Feedback D:
Feedback E:
73. Phosphocreatine is important in metabolism because it
A. is a compound formed by oxidative phosphorylation which can transfer a phosphate group to ADP.
B. serves in synthetic reactions as the source of the phosphate groups of CDP-choline.
C. is formed by substrate-level phosphorylation and irreversibly transfers a phosphate group to ADP and forms creatinine.
D. reacts with ADP to give ATP and creatine.
E. is the source of a nitrogen and the carbon atoms of urea.
Show answer
Correct Answer: D
Feedback A: Phosphocreatine is substrate level phosporylation not oxidative phosphorylation.
Feedback B:
Feedback C: The reaction is reversible.
Feedback D: Phosphocreatine is a temporary storage form of ATP in muscle.
Feedback E:
74. DNA sequencing depends on generating a set of molecules that terminate at every possible position. This is done by stopping the synthesis
A. using modified nucleotides lacking the 3' hydroxyl group.
B. using a DNA polymerase purified from a thermophilic organism.
C. using modified nucleotides phosphorylated at the 5' hydroxyl.
D. using modified nucleotides with deaminated bases.
E. using a non-processive DNA polymerase.
Show answer
Correct Answer: A
Feedback A: Yes, the 3™ hydroxyl is needed for elongation so removing it causes termination
Feedback B: This can be done, but does not contribute to the mechanism of chain termination
Feedback C: No, normal nucleotides have 5™ phosphates
Feedback D: No, this would just make damaged DNA, it would not terminate the elongation.
Feedback E: No, this would just make short products, it would not allow specific termination.
75. Each of the following may be used to estimate the molecular weight of a protein EXCEPT
A. sucrose density gradient centrifugation.
B. ion exchange chromatography.
C. SDS (sodium dodecyl sulfate) gel electrophoresis.
D. gel filtration chromatography.
E. sedimentation and diffusion measurements.
Show answer
Correct Answer: B
Feedback B: Separates molecules based on charge differences.
Feedback C:
Feedback D:
Feedback E:
76. AB arises from a 770 amino acid membrane protein called APP (amyloid precursor protein). APP is mostly processed by alpha and gamma secretases to yield, among other fragments, a small peptide p3 that causes little harm. Enhanced processing by beta-secretase and gamma-secretase generates the 40 or 42 residue AB peptide that results in Alzheimer amyloid deposits. APP spans the lipid bilayer between residues 671 and 770, and the last few residues of AB are in the bilayer. From what you know about membrane proteins which one of the following sequences identifies the last 5 amino acids in the 42 residue AB peptide.
A. Asp-Ala-Glu-Phe-Arg
B. Gly-Val-Val-Ile-Ala
C. His-Asp-Ser-Gly-Tyr
D. Lys-Val-His-His-Gln
E. Glu-Asp-Val-Gly-Ser
Show answer
Correct Answer: B
Feedback B: Because the last few residues of the AB peptide are in the membrane they need to by hydrophobic (non-polar).
Feedback C:
Feedback D:
Feedback E:
77. The glycerol phosphate shuttle
A. results in production of dihydroxyacetone phosphate in the matrix of mitochondria.
B. transfers electrons from cytosolic NADH to the respiratory chain of mitochondria.
C. inhibits glycolysis.
D. supports phosphorylation of three moles of ADP per mole of cytosolic NADH oxidized.
E. uses 3-phosphoglyceric acid as a substrate.
Show answer
Correct Answer: B
Feedback A: DHAP is transferred out of the mitochondria.
Feedback B:
Feedback C:
Feedback D: Transfers NADH reducing potential into the mitochondria as FADH2, which supports the phosphorylation of 2ATP.
Feedback E:
78. When modifying the Southern blot procedure to do a Northern blot, you would want to avoid the step in which
A. the nucleic acid molecules are separated by electrophoresis in agarose.
B. the gel is soaked in sodium hydroxide to denature the nucleic acids.
C. the nucleic acids are transferred to a solid matrix by capillary action.
D. a denatured probe is added to the solid matrix and incubated at 65° C.
E. probe that has hybridized to the immobilized nucleic acids is detected.
Show answer
Correct Answer: B
Feedback A: No, this is necessary to gather size information
Feedback B: Yes, RNA is sensitive to hydrolysis in solutions with high pH values, so this would tend to destroy the RNA that is being tested in a Northern.
Feedback C: No, this is the same in Southerns and Northerns.
Feedback D: No, this is the same in Southerns and Northerns.
Feedback E: No, this is the same in Southerns and Northerns.
79. Urea is the product of an enzyme in liver which catalyzes the hydrolysis of
A. carbamoyl phosphate.
B. citrulline.
C. ornithine.
D. argininosuccinate.
E. arginine.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: carbamoyl phosphate + ornithine ---> citrulline + aspartate ---> argininosuccinate ---> arginine ---> urea
80. From what you know about protein chemistry, which peptide migrates fastest to the cathode (or negative charged electrode) at pH 6.0.
A. Asp-Ala-Glu-Phe-Arg
B. Gly-Val-Val-Ile-Ala
C. His-Asp-Ser-Gly-Tyr
D. Lys-Val-His-His-Gln
E. Glu-Asp-Val-Gly-Ser
Show answer
Correct Answer: D
Feedback A: -1, 0, -1, 0, +1 = -1
Feedback B: 0, 0, 0, 0, 0
Feedback C: +1, -1, 0, 0, 0 = 0
Feedback D: +1, 0, +1, +1, 0 = +3
Feedback E: -1, -1, 0, 0, 0 = -2 This peptide would migrate fastest toward the anode.
81. Which of the following is a substrate for RNA synthesis, but not DNA synthesis?
A. adenosine triphosphate (ATP)
B. deoxyuridine triphosphate (dUTP)
C. guanosine-5', 5'-guanosine triphosphate (GpppG)
D. cyclic adenosine monophosphate (cAMP)
E. pseudouridine triphosphate (psiTP)
Show answer
Correct Answer: A
Feedback A: Yes, this a ribonucleoside triphosphate suitable for synthesis of RNA.
Feedback B: No, dNTPs are not found in RNA
Feedback C: No, this is not a substrate for an RNA polymerase.
Feedback D: No, this has no free hydroxyl to participate in the polymerization reaction.
Feedback E: No, this is not a standard NTP used by RNA polymerase.
82. Ammonium ion in the urine arises primarily from
A. the action of glutamate dehydrogenase in the kidney.
B. the action of glutaminase in the kidney on the amide nitrogen of glutamine.
C. the action of glutamine aminotransferase in the kidney.
D. the action of glutaminase in the kidney on the alpha-amino group of glutamine.
E. the action of asparaginase in the kidney.
Show answer
Correct Answer: B
Feedback A:
Feedback B: In the proximal tubules glutamine is deaminated to glutamate and ammonia.
Feedback C:
Feedback D:
Feedback E:
83. The compound which accepts a two-carbon fragment from acetyl-CoA to initiate the citric acid cycle is
A. oxalosuccinate.
B. oxalate.
C. oxaloacetate.
D. oxytocin.
E. oxybutyrate.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: acetyl CoA + oxaloacetate ---> citrate
Feedback D:
Feedback E:
84. Ketosis may develop as a result of which of the circumstances below?
A. glucose synthesis from fatty acids during glucose deprivation
B. acute attack of gout
C. carnitine transferase I deficiency
D. hyperglycemia
E. fat mobilization to support gluconeogenesis
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Ketone bodies are produced because more acetyl CoA is derived from fatty acid oxidation than can be used in TCA cycle.
85. The alpha-amino groups on most mammalian proteins are acetylated. That is they have the structure: CH3-C(=O)-NH-C-peptide However, the alpha-amino groups on alpha- and beta- hemoglobin subunits are not acetylated. Their acetylation would likely:
A. prevent binding of protons to Hb
B. prevent BPG binding to Hb
C. Reduce CO2 elimination by the lungs
D. A + C
E. B + C
Show answer
Correct Answer: E
Feedback A: Protons bind to alpha-amino groups of the alpha chains and to carboxyl terminal His residues of the beta chains. The beta chains would still bind protons.
Feedback B: Yes, the terminal amino groups of the beta chains aid in bonding BPG.
Feedback C: Alpha-amino groups react with CO2: -NH2 + CO2 ---> NHCOO- + H+
Feedback D:
Feedback E:
86. A person with Xeroderma pigmentosum
A. is missing the enzyme that reverses thymine dimers
B. is sensitive to ultraviolet radiation (UV)
C. is sensitive primarily to ionizing radiation (x-rays)
D. is prone to cancers and should be treated prophylactically with radiation therapy
E. is missing one of the 7 genes that encode type II topoisomerases
Show answer
Correct Answer: B
Feedback A: No, that is called photolyase, and has not been found in humans.
Feedback C: No, they are sensitive to UV
Feedback D: No, due to the defect in damage repair, radiation therapy is not indicated.
Feedback E: There are 7 classes of XP, but none are known to have a defect in a type II topoisomerase.
87. A person regularly eating a high protein diet (diet 1) changes to diet 2 which contains adequate but substantially less protein. After a period of several days on diet 2, what will occur?
A. Nitrogen balance will be positive.
B. Nitrogen balance will be negative.
C. Daily output of urea will increase over that while on diet 1.
D. Daily output of urea will decrease below that while on diet 1.
E. Two of the above are true.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Consumed amino acids are used for protein replacement and excess amino acids are catabolized for energy. There is no real storage form of amino acids. Because diet 2 has less protein intake, there is less excess protein to catabolize.
Feedback E:
88. The three stranded collagen superhelix is characterized by the following properties EXCEPT :
A. Glycine is present every third residue
B. H-bonds are parallel to the helix axis
C. Some lysines and prolines are hydroxylated
D. Sugars can be attached to hydroxylated lysine residues
E. No exceptions
Show answer
Correct Answer: B
Feedback A:
Feedback B: The H-bonds occur btween the three strands and are more perpendicular to the axis.
Feedback C:
Feedback D:
Feedback E:
89. The DNA that is copied by the two replication forks traveling in opposite directions from a common site of initiation is called:
A. a duplex
B. an origin
C. a battery
D. a replicon
E. a transposon
Show answer
Correct Answer: D
Feedback A: No, a duplex is the name for the two stranded form of DNA
Feedback B: No, the origin is the site of initiation itself.
Feedback C: No, a battery is a group of adjacent origins that fire simultaneously
Feedback D:
Feedback E: No, this is a piece of DNA capable of moving to other sites in the genome.
90. Lysosomal enzymes are often maximally active at pH 5.0 or less. In many cases, they are inactive at pH 7.0 which makes some biological sense in that lysosomal rupture would be lethal if the released proteases and nucleases were significantly active at cytosolic pH (~ 7.0). The amino acid R group most affected by a change in pH from 5 to 7 is:
A. arginine
B. aspartate
C. cysteine
D. histidine
E. lysine
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Histidine™s imidazole ring readily accepts or donates protons. pKa = 6.5.
Feedback E:
91. The processivity of DNA polymerases
A. increases in the presence of ligases
B. decreases in the presence of helicases
C. increases in the presence of single strand binding proteins
D. increases in the presence of Okazaki factors
E. decreases in the presence of clamp proteins
Show answer
Correct Answer: C
Feedback A: No, ligases seal nicks in DNA, they do not influence the tendency to move forward or fall off.
Feedback B: Some helicases affect polymerase processivity, but they increase this property.
Feedback C: Yes, SSBs straighten the template and increase both the rate and processivity of DNA polymerases
Feedback D: No, there is no such thing as an Okazaki factor, and the size of Okazaki fragments are the result, rather than the cause, of processivity changes.
Feedback E: Clamp proteins increase the processivity of DNA polymerases.
92. McArdle's disease is due to a muscle
A. debrancher enzyme defect.
B. glucosidase defect.
C. phosphofructokinase defect.
D. phosphorylase absence.
E. phosphohexose isomerase deficiency.
Show answer
Correct Answer: D
Feedback A: Cori™s Disease of Andersen™s Disease
Feedback B: Pompe™s Disease
Feedback C: Glycogen storage disease VII
Feedback D: Cannot degrade glycogen; limited ability to perform strenuous exercise.
Feedback E:
93. The following secondary structures can be classified according to the length of, say ten residues, in the specific conformation. From longest to shortest they are:
A. beta strand, collagen strand, alpha-helix, pi-helix
B. Collagen strand, beta strand, alpha-helix, pi-helix
C. alpha-helix, beta strand, pi-helix, collagen strand
D. alpha-helix, pi-helix, beta strand, collagen strand
E. Collagen strand, alpha-helix, beta strand, pi-helix
Show answer
Correct Answer: A
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
94. A bacterial culture grown for many generations in a "heavy" (15N)medium was transferred to a "light" (15N) medium. After the DNA had been copied 3 times in light medium, what would be the relative distribution of DNA containing two light strands (LL), one heavy and one light strand (HL), and two heavy strands (HH)?
A. 2 HH:2 LL
B. 1 HH:1HL:6 LL
C. 0 HH:1 HL: 7LL
D. 0 HH:1 HL: 3LL
E. 1 HH:2 HL: 1LL
Show answer
Correct Answer: D
Feedback A: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL. There is no HH after the first round.
Feedback B: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL. There is no HH after the first round.
Feedback C: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL. This ratio would result after 4 rounds
Feedback D: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL.
Feedback E: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL. There is no HH after the first round.
95. Mutations can alter protein structure and lead to devastating diseases. Consider the following amino acid changes: 1. Tyr to Phe 2. Asp to Asn 3. Ile to Arg 4. Arg to Lys 5. His to Asn 6. Trp to Gly 7. Ala to Pro Which change is most likely to alter the absorbance of a protein at 280 nm?
A. 2
B. 3
C. 4
D. 5
E. 6
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Glycine is unique in being optically inactive, and tryptophan is aromatic.
96. In methylation-directed mismatch repair in bacteria
A. MutL binds to mismatches in duplex DNA
B. The methylated strand is judged to be the older strand
C. MutL binds to methylated DNA strands
D. One mismatched base is excised leaving a single nucleotide gap to repair
E. RecA makes an incision only at hemimethylated sites
Show answer
Correct Answer: B
Feedback A: No, MutS has this activity, the activity is MutL remains poorly understood.
Feedback B: Yes, since newly synthesized DNA does not have this post-synthetic modification, the methylated strand is older.
Feedback C: No, MutH seems to have this property, the activity is MutL remains poorly understood.
Feedback D: No, this would be base-excision repair.
Feedback E: An incision is made at hemi-methylated sites, but by MutH, not RecA
97. Mutations can alter protein structure and lead to devastating diseases. Consider the following amino acid changes: 1. Tyr to Phe 2. Asp to Asn 3. Ile to Arg 4. Arg to Lys 5. His to Asn 6. Trp to Gly 7. Ala to Pro Which change is most likely to disrupt the folding of a globular protein?
A. 1
B. 2
C. 3
D. 4
E. 5
Show answer
Correct Answer: C
Feedback A: Both are aromatic.
Feedback B: Only one group is different and both are polar.
Feedback C: Ile is non-polar, Arg is very polar and its R group is approximately twice as long as Ile.
Feedback D: Both have long R groups and both are bases.
Feedback E: Both are polar.
98. All of the following statements concerning eukaryotic DNA replication are true EXCEPT
A. single-strand binding proteins coat some DNA intermediates in replication
B. replication of half of the DNA chains occurs in short discontinuous pieces
C. helicases unwind the parental strands for the polymerases
D. a single type of DNA polymerase acts on both leading and lagging strands
E. ring-shaped clamp proteins promote high processivity of DNA polymerase
Show answer
Correct Answer: D
Feedback A: SSBs are needed for all phases of DNA metabolism, including replication.
Feedback B: This is true, describing the synthesis of Okazaki fragments on the lagging strand.
Feedback C: While the precise identity of the eukaryotic replicative helicase is not known, this is very likely to be true given the rate of replication fork movement.
Feedback D: This is unlikely to be true since both Pol alpha and Pol delta seem to be needed for eukaryotic replication, although their exact roles remain unknown.
Feedback E: This is true, the clamp is called PCNA.
99. Substrate Km #1 2.5 #2 12.0 #3 32.0 Comparison of the respective Km values for the hydrolysis of the peptide bond of substrates 1-3 by chymotrypsin shows that
A. The Vmax of substrate #1 must be greater than that for #2 or #3.
B. The Vmax for substrate #3 must be greater than that for #1 or #2.
C. A lower concentration of #3 will be needed to attain Vmax.
D. A lower concentration of #1 will be needed to attain Vmax.
E. Substrate #3 has the highest affinity for the active stie of chymotrypsin.
Show answer
Correct Answer: D
Feedback A: The Km tells us the concentration of substrate needed to reach 1/2 Vmax for a particular enzyme, but provides no comparison for the actual velocities.
Feedback B: See A.
Feedback C: Since Km, which tells us the concentration needed to reach 1/2 Vmax, is highest for #3, #3 must need the highest concentration to reach full Vmax.
Feedback D: See C.
Feedback E: Higher Km indicates less affinity of the enzyme for the substrate.
100. Ionizing radiation is capable of producing genetic mutations by
A. Altering bases so that they form different base pair interactions
B. Causing deletions of DNA sequences
C. Causing inversions of DNA sequences
D. Causing translocations of one part of a chromosome onto another
E. A-D are all correct.
Show answer
Correct Answer: E
Feedback A: True, but so are all of the choices, so the answer is E
Feedback B: True, but so are all of the choices, so the answer is E
Feedback C: True, but so are all of the choices, so the answer is E
Feedback D: True, but so are all of the choices, so the answer is E
Feedback E:
101. Which of the following cases represents a metabolically important control?
A. Substrate oxidation in mitochondria is enhanced by elevated ATP levels.
B. The activity of liver hexokinase is stimulated by high levels of its product, glucose 6-phosphate.
C. Acetyl-CoA inhibits pyruvate carboxylase in the liver.
D. Phosphofructokinase (PFK) is inhibited by high levels of its substrate, MgATP.
E. Glucagon stimulates hepatic glycogen synthase.
Show answer
Correct Answer: D
Feedback A: Increased ATP shuts down enzymes of glycolysis and TCA cycle, so oxidation is decreased.
Feedback B: Hexokinase is inhibited by glucose-6-phosphate.
Feedback C: The activity of pyruvate carboxylase requires acetyl CoA.
Feedback D: ATP and citrate inhib
You have 546 questions in this exam.
1. Increasing the concentration of dGTP directly causes ribonucleotide reductase to
A. increase the rate of production of dGDP.
B. increase the rate of production of dADP.
C. decrease the rate of production of rCDP.
D. decrease the rate of production of dADP.
E. increase the rate of production of dTTP.
Show answer
Correct Answer: B
Feedback A: Has no direct effect (Increased dGTP increases dATP production, which eventually would decrease RNR activity overall, but dGTP has no immediate effect on the enzyme except to increase dADP formation)
Feedback C: Has no direct effect (Increased dGTP increases dATP production, which eventually would decrease RNR activity overall, but dGTP has no immediate effect on the enzyme except to increase dADP formation)
Feedback D: This rate increases
Feedback E: Has no direct effect (Increased dGTP increases dATP production, which eventually would decrease RNR activity overall, but dGTP has no immediate effect on the enzyme except to increase dADP formation)
2. Glycogen phosphorylase is activated most directly by
A. epinephrine.
B. phosphorylase kinase.
C. phosphorylase phosphatase.
D. cAMP.
E. glucagon.
Show answer
Correct Answer: B
Feedback A: See B
Feedback B: glucagon or epinephrine ---> increased cAMP ---> active PKA ---> active phosphorylase kinase ---> active glycogen phosporylase
Feedback C: See B
Feedback D: See B
Feedback E: See B
3. Which of the following cofactors is most directly concerned with a CO2 fixation reaction?
A. biotin
B. vitamin D
C. coenzyme A
D. thiamine pyrophosphate
E. lipoic acid
Show answer
Correct Answer: A
Feedback A: Biotin is a carrier of activated CO2 and is a component of acetyl carboxylase, propionyl CoA carboxylase, and pyruvate carboxylase.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
4. Which statement about reactions catalyzed by transaminases (aminotransferases) is false?
A. The equilibrium constant is close to 1.
B. The reaction involves a Schiff base intermediate.
C. Pyridoxal phosphate is covalently bound to the enzyme protein through the epsilon amino group of a lysine residue.
D. The reactions are important for the biosynthesis of non-essential amino acids.
E. Ammonia is liberated.
Show answer
Correct Answer: E
Feedback A: Aminotransferases can participate in degradation or biosynthesis depending on substrate concentrations.
Feedback B:
Feedback C:
Feedback D: Aminotransferases most commonly transfer amino groups to alpha-ketoglutarate forming glutamate. Glutamate can then be the amino group donor for amino acid biosynthesis.
Feedback E: False, an amino group is transferred to an alpha-keto acid.
5. The enzyme in muscle which is most directly stimulated by epinephrine is
A. glycogen synthase.
B. phosphofructokinase.
C. isocitrate dehydrogenase
D. glucose-6-phosphatase.
E. adenylate cyclase.
Show answer
Correct Answer: E
Feedback A: inactivated due to epinephrine.
Feedback B: decreased stimulation of PFK-1
Feedback C: very indirect modification
Feedback D: Changes in its activity are determined by synthesis and degradation.
Feedback E: increased intracellular cAMP.
6. Acetyl-CoA regulates gluconeogenesis by activation of
A. phosphoenolpyruvate carboxykinase.
B. pyruvate kinase.
C. pyruvate carboxylase.
D. lactate dehydrogenase.
E. fructose 1,6-bisphosphatase.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Which catalyzes: pyruvate + CO2 ---> oxaloacetate.
Feedback D:
Feedback E:
7. Which compound does NOT cross the inner mitochondrial membrane due to lack of a specific transport protein?
A. malate
B. glutamine
C. NADH
D. citrate
E. ATP
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: NADH reducing potential must be shuttled across the membrane by another molecule.
Feedback D:
Feedback E:
8. All of the following statements are correct with regard to chromatin EXCEPT
A. Core nucleosomes consist of 8 histone protein molecules and about 146 base pairs of DNA.
B. Histones have been highly conserved during evolution.
C. Almost all of the DNA in a eukaryotic cell is assembled into nucleosomes.
D. Assembly of DNA into nucleosomes compacts it about 100-fold in length.
E. Nucleosome cores are separated by variable amounts of linker DNA.
Show answer
Correct Answer: D
Feedback A: see D
Feedback B: see D
Feedback C: see D
Feedback D: Nucleosomes effect about a 5-7 fold compaction of the length of the DNA
Feedback E: see D
9. Which is LEAST likely to be characteristic of a regulated enzyme in a metabolic pathway?
A. Catalyzes the slowest step of the pathway
B. Has a short half-life in vivo
C. Concentration is hormonally regulated
D. Activity under allosteric control
E. Catalyzes a physiologically reversible reaction
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Irreversible reactions are usually regulated, especially those that involve ATP.
10. Which of the following are examples of tertiary (3°) structure? 1. the sequence of amino acids in a protein 2. the alpha-helix of hemoglobin 3. the folding of alpha-helix onto beta-sheet of triose phosphate isomerase 4. hydrogen bonding of one beta -sheet to another beta-sheet 5. juxtaposition of amino acids that are separated by a "long distance" in primary (1°) structure 6. the un-ordered portions of carboxy peptidase A 7. juxtaposition of hemoglobin beta-7 Glu with alpha-68 Asn
A. all the even-numbered items
B. all the odd-numbered items
C. 3, 4, and 5
D. 3, 4, 5, and 7
E. 3, 4, 5, 6, and 7
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: 1-primary, 2-secondary, 6-hard to classify--primary or secondary, 7-quaternary
Feedback D:
Feedback E:
11. The H1 histone
A. is one of the four histones in the nucleosome core.
B. helps form the octamer nucleosome complex.
C. is rich in acidic amino acid residues.
D. is present at a 1:1 weight ratio with DNA.
E. helps fold or stabilize the "beads-on-a-string" structure into higher order structures.
Show answer
Correct Answer: E
Feedback A: No, the core histones are H2A, H2B, H3, and H4
Feedback B: No, the octamer contains only H2A, H2B, H3, and H4
Feedback C: It is rich in basic amino acids
Feedback D: Overall, all histones (H1 and H2A, H2B, H3, and H4) are about 1:1 with DNA
Feedback E:
12. Which mitochondrial enzyme requires acetyl-CoA as a substrate?
A. Citrate synthase
B. Succinyl-CoA synthase
C. Succinic dehydrogenase
D. Pyruvate dehydrogenase
E. Isocitrate dehydrogenase
Show answer
Correct Answer: A
Feedback A: Catalyzes: acetyl CoA + oxaloacetate ---> citrate.
Feedback B: CoA is required coenzyme, but not acetyl CoA.
Feedback C:
Feedback D: Produces acetyl CoA from pyruvate.
Feedback E:
13. Transfer of the methyl group from 5-methyl tetrahydrofolate to homocysteine to form methionine requires
A. NADPH
B. vitamin B12
C. FAD
D. pyridoxal phosphate
E. none of the above
Show answer
Correct Answer: B
Feedback A:
Feedback B: Methycobalamin, a derivative of B12, is the coenzyme that mediates the transfer of methyl groups in methionine biosynthesis.
Feedback C:
Feedback D: Aminotransferases require this coenzyme.
Feedback E:
14. All of the following are true of bacterial plasmids or episomes EXCEPT
A. Episomes like the F factor can be transferred from one cell to another.
B. Plasmids or episomes can confer resistance to many antibiotics simultaneously.
C. Plasmids or episomes can be transferred from one species of bacteria to another.
D. Since plasmids must be replicated, they confer a disadvantage to their hosts and can therefore be expected to disappear from human pathogens soon.
E. Genes can be transferred from bacterial genomes to plasmids at a low frequency.
Show answer
Correct Answer: D
Feedback A: This is true, they can pass through a pilus into another cell.
Feedback B: This is true.
Feedback C: This is true
Feedback D: The disadvantages caused by œcarrying plasmids are outweighed by the advantages they provide; this advantage increases as we use antibiotics.
Feedback E: Strange but true.
15. If one mole of glucose is metabolized to carbon dioxide and water via glycolysis and the tricarboxylic acid cycle, and the glycerol phosphate shuttle is operative, the net conversion of ADP (or equivalent) to ATP (or equivalent) would theoretically be
A. 12 moles.
B. 24 moles.
C. 32 moles.
D. 36 moles.
E. 38 moles.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: The 2NADH produced by glycolysis are converted to 2FADH2 in the mitochondria by the glycerol phosphate shuttle, so they are worth 2ATP each.
Feedback E: With the malate-aspartate shuttle the 2NADH from glycolysis are converted to 2NADH in the mitochondria, and are worth 3ATP each.
16. The cofactor for transamination is
A. nicotinamide.
B. biotin.
C. thiamine pyrophosphate.
D. pyridoxal phosphate.
E. vitamin B12.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: A vitamin B6 derrivative that is a prosthetic group for each transaminase.
Feedback E:
17. Which enzyme catalyzes a reaction in which carbon dioxide (or bicarbonate) is neither a substrate nor a product?
A. isocitrate dehydrogenase
B. pyruvate carboxylase
C. carboxypeptidase
D. phosphoenolpyruvate carboxykinase
E. alpha-ketoglutarate dehydrogenase
Show answer
Correct Answer: C
Feedback A: CO2 produced.
Feedback B: CO2 is a substrate.
Feedback C: Cleaves peptide bonds, no CO2 involved.
Feedback D: CO2 produced.
Feedback E: CO2 produced.
18. 5-phosphoribosyl-1-pyrophosphate (PRPP) is an intermediate in
A. the de novo synthesis of purine nucleotides.
B. the de novo synthesis of pyrimidine nucleotides.
C. the salvage pathway for the synthesis of purine nucleotides.
D. A and C.
E. A, B and C.
Show answer
Correct Answer: E
Feedback A: True (it is activated as the committed step for purine synthesis), but so are B and C
Feedback B: True (orotidylate is formed by reacting the free base with PRPP), but so are A and C
Feedback C: True (PRPP reacts with free bases like guanine and hypoxanthine through HGPRTase), but so are A and B.
Feedback D: True, but B is also correct
Feedback E:
19. Which statements are FALSE? In mitochondria, tricarboxylic acid cycle reactions generally proceed more
(1) rapidly as the ADP concentration rises.
(2) slowly as the ADP concentration rises.
(3) slowly as the NADH concentration rises.
(4) rapidly as the NADH concentration rises.
(5) rapidly as the oxaloacetate concentration increases.
A. 1 and 3
B. 2 and 4
C. 3 and 5
D. 1 and 4
E. 2 and 5
Show answer
Correct Answer: B
Feedback A:
Feedback B: High levels of ATP and NADH signal that the cell™s energy level is high, high ADP and NAD+ signal that more energy is needed.
Feedback C:
Feedback D:
Feedback E:
20. In Lesch-Nyhan syndrome, lack of HGPRTase activity should result in higher than normal tissue concentrations of all of the following EXCEPT
A. adenine.
B. guanine.
C. uric acid.
D. hypoxanthine.
E. PRPP.
Show answer
Correct Answer: A
Feedback A: Adenine levels are not directly affected by loss of this enzyme.
Feedback B: This substrate for HGPRTase would accumulate abnormally in the absence of the enzyme.
Feedback C: Increased hypoxanthine leads to increased xanthine and therefore increased uric acid.
Feedback D: This substrate for HGPRTase would accumulate abnormally in the absence of the enzyme.
Feedback E: This substrate for HGPRTase would accumulate abnormally in the absence of the enzyme.
21. Which of the following represents the correct number of moles of products formed as one mole of acetate (in the form of acetyl-CoA) moves through one turn of the citric acid cycle? (respectively)
GTP
CO2
NADH
FADH2
A. 1 1 3 1
B. 2 2 3 1
C. 1 2 3 1
D. 1 2 2 1
E. 1 1 2 2
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Memorize the cycle.
Feedback D:
Feedback E:
22. How many net moles of ATP are produced per glucose equivalent when glycogen is the substrate for anaerobic glycolysis?
A. 0
B. 1
C. 2
D. 3
E. -2
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: glycogen ---> glucose-1-phosphate ---> glucose-6-phosphate ---> glycolysis. Because glycose-6-phosphate enters glycolysis the reaction catalyzed by glucokinase or hexokinase is not carried out and one less ATP is used.
Feedback E:
23. One of the strands of a double-stranded, 10 kbp DNA duplex has the following numbers of base residues: adenine(a) - 3800, thymine(t) - 2600. The base composition of the whole double-stranded molecule will be
A. a - 5200, t - 5200, g - 4800, c - 4800
B. a - 7600, t - 7600, g - 2400, c - 2400
C. a - 6400, t - 6400, g - 3600, c - 3600
D. a - 6400, t - 3600, g - 6400, c - 3600
E. There is not enough information to determine the overall base composition
Show answer
Correct Answer: C
Feedback A: If one strand has 3800 A and 2600 T, the other strand has 3800 T and 2600 A, so the duplex has 6400 A and 6400 T, with the remainder being G and C (3600 each).
Feedback B: If one strand has 3800 A and 2600 T, the other strand has 3800 T and 2600 A, so the duplex has 6400 A and 6400 T, with the remainder being G and C (3600 each).
Feedback C:
Feedback D: If one strand has 3800 A and 2600 T, the other strand has 3800 T and 2600 A, so the duplex has 6400 A and 6400 T, with the remainder being G and C (3600 each).
Feedback E: If one strand has 3800 A and 2600 T, the other strand has 3800 T and 2600 A, so the duplex has 6400 A and 6400 T, with the remainder being G and C (3600 each).
24. A major control point of the citric acid cycle is exerted at the level of
A. succinate dehydrogenase.
B. citrate synthase.
C. succinyl CoA synthase.
D. malate dehydrogenase.
E. aconitase.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Citrate synthase, isocitrate dehydrogenase, and alpha-ketoglutarate dehydrogenase are the regulated enzymes of TCA cycle.
Feedback C:
Feedback D:
Feedback E:
25. Secondary structure of proteins
A. is maintained by hydrogen bonds.
B. refers to proteins consisting of two (or more) subunits held together by noncovalent forces.
C. encompasses any hydrogen-bonded interaction found in proteins.
D. refers to proteins consisting of one or more polypeptide chains plus a nonprotein moiety.
E. is found only in fibrous proteins.
Show answer
Correct Answer: A
Feedback A: alpha-helix and beta-sheet.
Feedback B: This is quaternary.
Feedback C: No, H-bond interactions between residues far apart linearly would be tertiary interactions.
Feedback D: This is quaternary.
Feedback E: Any protein.
26. Each of the following statements concerning DNA is true EXCEPT which one?
A. C does not have to equal G in single stranded DNA.
B. DNA can be found in either circular or linear forms.
C. The two DNA strands of the double helix are antiparallel.
D. Hydrogen bonds alone hold the DNA helix together.
E. Base pairs lie in a plane perpendicular to the long axis of the helix.
Show answer
Correct Answer: D
Feedback A: This is true. For example, a single-stranded DNA could be all C and have no G.
Feedback B: This is true; most bacterial plasmids are circles while most mammalian chromosomes are linear.
Feedback C: Please don™t tell me you picked this one, I just don™t want to know.
Feedback D: No, they contribute but they are only one part of the picture.
Feedback E: This is true.
27. The reduction potential for a reaction
A. is an oxidation potential.
B. is unrelated to the free energy of the reaction.
C. is negative for a spontaneous process.
D. can be used to decide if one substance will reduce another.
E. is not dependent on reactant and/or product concentration.
Show answer
Correct Answer: D
Feedback A: Has the opposite sign of the reduction potential.
Feedback B:
Feedback C: Positive for a spontaneous redox reaction.
Feedback D: More negative (smaller) reduction potentials donate electrons to more positive (larger) reduction potentials.
Feedback E:
28. Glutamine
A. is not incorporated into proteins because there is no codon or tRNA for glutamine.
B. is the principal donor of amino groups in transamination reactions.
C. donates its alpha-amino group in the de novo biosynthesis of both purine and pyrimidine nucleotides.
D. is synthesized from glutamate primarily in the liver.
E. represents an important non-toxic transport form of ammonia.
Show answer
Correct Answer: E
Feedback A: There are codons and tRNA™s for both glutamine and glutamate.
Feedback B: Glutamine is the ultimate product of tissue transamination. Glutamine is transported to the liver where nitrogen metabolism to urea takes place.
Feedback C: Donates its gamma-amide in both syntheses.
Feedback D: Synthesis occurs primarily in the tissues and then glutamine is broken down into glutamate and ammonia in the liver.
Feedback E: See B.
29. As a cell initiates DNA replication, it contains sufficient pools of dNTPs to complete
A. about 1% or less of DNA replication without making new dNTPs.
B. about 10% of DNA replication without making new dNTPs.
C. 100% of DNA replication without making new dNTPs.
D. about 10 times more dNTPs than it needs to replicate the genome once.
E. no replication at all.
Show answer
Correct Answer: A
Feedback A: About 0.1% in mammals, 1% in bacteria
Feedback B: About 0.1% in mammals, 1% in bacteria
Feedback C: About 0.1% in mammals, 1% in bacteria
Feedback D: About 0.1% in mammals, 1% in bacteria
Feedback E: About 0.1% in mammals, 1% in bacteria
30. The uncoupling of oxidative phosphorylation in the human might be of physiological importance because it
A. allows storage of nutrients.
B. produces water.
C. increases carbon dioxide level in the blood.
D. produces heat.
E. raises the oxygen level in the blood.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: The energy produced by electron transport is released as heat rather than being used to synthesize ATP.
Feedback E:
31. The condition of LACTOSE INTOLERANCE can be caused by defective or deficient
A. lactose
B. UDP-galactose epimerase
C. galactose-1-phosphate uridyl transferase
D. lactase
E. galactase
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
32. During the second day of a fast, after liver glycogen is depleted, blood glucose
A. is derived mainly from fatty acids of adipose tissue.
B. is derived mainly from muscle glycogen.
C. is derived mainly from the amino acids of liver proteins.
D. falls to low levels (e.g., 20 mg/100 ml) until eating is resumed.
E. is derived mainly from the amino acids of muscle proteins.
Show answer
Correct Answer: E
Feedback A: After approximately 3 weeks of starvation muscle almost exclusively oxidizes fatty acids for energy.
Feedback B:
Feedback C:
Feedback D:
Feedback E: Rapid breakdown of muslce protein takes place the first few days of starvation to provide amino acids for gluconeogenesis.
33. A complete lack of adenosine deaminase causes SCID: severe combined immuno-deficiency. Which of the following is LEAST true?
A. Loss of the enzyme causes increased levels of dATP because there is less turnover of adenosine nucleosides in general.
B. Increased dATP decreases the concentration of all rNTPs, blocking RNA synthesis.
C. Increased dATP inhibits ribonucleotide reductase, such that de novo production of all dNDPs is inhibited.
D. Adenosine deaminase loss causes SCID because T cells are particularly sensitive to DNA replication inhibition.
E. It is not clear why adenosine deaminase loss affects T cells so specifically since other cells have the same kind of nucleotide synthesis regulation and might be expected to be equally affected.
Show answer
Correct Answer: B
Feedback A: This is true, failure to degrade adenosine forms causes their concentrations to rise.
Feedback B: Lack of dNTPs blocks DNA synthesis, not RNA synthesis
Feedback C: This is true, dATP levels rise (see A) which shuts down all RNR activities
Feedback D: This is true; the reason is not fully known but it is observed to be the case.
Feedback E: This is true, see D.
34. In the oxidation of NADH by the electron transport system, energy released is coupled to the formation of ATP from ADP plus inorganic phosphate. For every mole of NADH oxidized, how many moles of ATP formed are formed?
A. 1
B. 5
C. 3
D. 6
E. None of the above
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: This is for NADH in the mitochondria. NADH outside of the mitochondria produce different numbers of ATP depending on the shuttle system used to transfer the reducing potential from the cytosol to the mitochondria.
Feedback D:
Feedback E:
35. In analysis of isoenzymes of serum CPK for diagnosis of an AMI
A. CPK above six times normal total CPK indicates an AMI.
B. CPK-MM above four times normal total CPK indicates an AMI.
C. CPK-BB above 1% of total normal, and total CPK of six times normal CPK indicates an AMI.
D. CPK-MB above 3% of total normal CPK indicates an AMI.
E. Total CPK above six times normal, with CPK BB in the normal range indicates an AMI.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: CPK-MB is only 1-3% in other muscle, but 15-20% in heart muscle.
Feedback E:
36. Which of the following is true about the structure and properties of ATP?
A. The structure contains 3 phosphoanhydride bonds and one glycosidic bond.
B. At physiological pH, the molecule has a net charge of approximately -6.
C. The transfer of a phosphoryl group (phosphate residue) is the only biological reaction of ATP that cleaves a high-energy bond.
D. The high free energy of hydrolysis results, in part, from electrical repulsion between phosphate residues.
E. The standard free energy of hydrolysis of ATP is highter than that of other all phosphate derivatives in the glycolytic pathway.
Show answer
Correct Answer: D
Feedback A: 2 phosphoanhydride bonds and one glycosidic bond.
Feedback B:
Feedback C:
Feedback D:
Feedback E: No, hydrolysis of PEP gives more energy than hydrolysis of ATP.
37. Which of the following species of denatured DNA will renature most rapidly in solution, under appropriate conditions of ionic strength, pH, and temperature?
A. Human liver nuclear DNA
B. Vaccinia (viral) DNA
C. E. coli DNA
D. Yeast nuclear DNA
E. Mouse neuronal DNA
Show answer
Correct Answer: B
Feedback A: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
Feedback B: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
Feedback C: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
Feedback D: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
Feedback E: The RATE of reannealing depends on the concentration of complementary strands. Given the same overall DNA concentration (a fact that should have been stressed in this question), the more complex the genome the LOWER the concentration of any given sequence (and its complement) so the slower the renaturation. Therefore, the least complex genome will anneal the fastest, in this case the viral DNA (B).
38. When oxidative phosphorylation in mitochondria is uncoupled
A. AMP is formed and O2 consumption stops.
B. the oxidation of acetyl CoA in the tricarboxylic acid cycle stops.
C. mitochondrial metabolism stops.
D. ATP synthesis stops but O2 consumption continues.
E. the cytochromes lose their heme groups into the cytoplasm.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Electron transport continues to reduce O2, but the proton gradient is not maintained so ATP is not synthesized.
Feedback E:
39. Oxaloacetate moves through the mitochondrial membrane
A. after conversion to glycerol phosphate.
B. by passive diffusion.
C. after reaction with carnitine.
D. after oxidation to pyruvate.
E. after reduction to malate.
Show answer
Correct Answer: E
Feedback A: DHAP + NADH ---> glycerol-3-phosphate + NAD in the glycerolphosphate shuttle.
Feedback B:
Feedback C: Long-chain fatty acids are transported into the mitochondria for oxidation by carnitine.
Feedback D:
Feedback E: The Malate-Aspartate shuttle for NADH.
40. The absence of which enzyme results in PKU (phenylketonuria)?
A. Tyrosine hydroxylase
B. Amino acid oxidase
C. Phenylalanine hydroxylase
D. Tyrosine transaminase
E. Phenylalanine transaminase
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Phenylalanine is not hydroxylated to tyrosine, and builds up to toxic levels.
Feedback D:
Feedback E:
41. All of the following statements are consistent with the chemiosmotic hypothesis of oxidative phosphorylation EXCEPT which one?
A. Agents which uncouple oxidative phosphorylation from electron transport disrupt permeability barriers such that mitochondrial membranes become permeable to protons.
B. The term electrochemical gradient or proton gradient operationally describes the free energy generated by electron transport which drives phosphorylation of ADP.
C. Protons are transferred into the mitochondria during electron transport establishing a pH gradient.
D. Intact mitochondrial membranes are necessary for oxidative phosphorylation to occur.
E. Electron carriers of the electron transport chain are vectorially oriented.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Protons are transferred from the matrix to the intermembrane space of the mitochondria.
Feedback D: The proton gradient must be set up across the inner membrane.
Feedback E:
42. Which statement about the pyruvate dehydrogenase complex is true?
A. Lipoamide functions in the decarboxylation of pyruvate.
B. Coenzyme A is covalently bound to a lysine residue in one of the proteins of the complex.
C. Phosphorylation catalyzed by a kinase results in decreased activity.
D. NADH is an allosteric activator of pyruvate oxidation.
E. The function of FAD is to oxidize NADH.
Show answer
Correct Answer: C
Feedback A: TPP functions in decarboxylation; lipoamide transfers the acetyl group to CoA.
Feedback B: Lipoamide is bound to lysine.
Feedback C: Increased acetyl CoA/CoA or NADH/NAD ratios activate the kinase which phosphorylates and deactivates PDH.
Feedback D: Increased NADH causes indirect inhibition of PDH.
Feedback E: FADH2 reduces NAD to NADH.
43. Which of the following is the single most important factor accounting for the high rates of enzyme catalysis?
A. Enzymes bring the substrate [S] into close proximity of the coenzyme.
B. Certain enzyme amino acid side chains interact with the substrate and stabilizing reactive intermediate(s).
C. The "free energy" of activation is increased during enzyme catalysis.
D. Enzymes contain prosthetic groups which promote catalysis.
E. Enzymes are globular proteins which contain hydrophobic and hydrophilic amino acids.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Enzymes stablaize the transition state.
Feedback C: Acitvation energy is lowered by enzymes.
Feedback D: True, but not the best explanation.
Feedback E: Does not explain catalysis.
44. Oxidative phosphorylation in mitochondria is carried out by enzymes
A. on the outer membrane.
B. on the inner membrane.
C. in the matrix.
D. in the intermembrane space.
E. all of the above are correct..
Show answer
Correct Answer: B
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
45. What change from normal is likely to result from the ingestion of ethanol?
A. increased reduction of pyruvate
B. increased conversion of beta-hydroxybutyrate to acetoacetate
C. decreased ketogenesis due to lack of acetyl-CoA
D. increased production of oxaloacetate from malate
E. increased production of dihydroxyacetone phosphate from glycerol phosphate
Show answer
Correct Answer: A
Feedback A: Ethanol metabolism requires reduction of NAD to NADH , and as in anaerobic glycolysis, NAD is regenerated by reduction of pyruvate to lactate.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
46. Glucagon
A. is released from the pancreas in response to lowered blood glucose.
B. inactivates liver phosphorylase.
C. increases glucose release from muscle glycogen.
D. utilizes cAMP as a second messenger in muscle cells.
E. increases glucose uptake into hepatocytes.
Show answer
Correct Answer: A
Feedback A:
Feedback B: Activates the phosphorylase for glycogen degradation.
Feedback C: Glucagon does not affect muscle.
Feedback D: See C.
Feedback E: Insulin increased glucose uptake when blood glucose is high.
47. Muscle can carry out all of the following processes EXCEPT
A. transamination of branched chain amino acids.
B. synthesis of glycogen.
C. synthesis of glucose from alanine.
D. oxidation of ketone bodies.
E. phosphorylation of creatine.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: Muscle cannot carry out gluconeogenesis because pyruvate caboxylase is found only in liver and kidney cells.
Feedback D:
Feedback E:
48. All of the following are likely to occur following the ingestion of a very large amount of carbohydrate EXCEPT
A. the synthesis of triacylglycerol in the liver.
B. the synthesis of glycogen in the liver.
C. the oxidation of glucose by the pentose phosphate pathway in liver.
D. lipolysis in adipose tissue.
E. the synthesis and release of VLDL from the liver.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: High glucose levels lead to high insulin which supports synthesis of storage molecules, not breakdown.
Feedback E:
49. Catalysis by chymotrypsin depends on the concerted participation of three amino acid side chains (catalytic triad) which are contributed by which of the following amino acids?
A. Arginine, histidine, and serine
B. Histidine, serine, and glutamate
C. Serine, glutamate, and aspartate
D. Aspartate, serine, and histidine
E. Serine, histidine, and arginine
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Memorize it. Know the part each plays in the mechanism.
Feedback E:
50. Which of the following is the most accurate description of phosphofructokinase-1?
A. This enzyme uses fructose-6-phosphate as a substrate and converts it to fructose-2,6- diphosphate.
B. This enzyme is inhibited by ATP, citrate, and fructose-2,6-biphosphate.
C. This enzyme catalyzes a fully reversible reaction under physiological conditions.
D. This activity of this enzyme is indirectly increased by cyclic AMP.
E. No statement above is accurate.
Show answer
Correct Answer: E
Feedback A: PFK-2 makes fructose-2,6-bisphosphate, PFK-1 produces fructose-1,6-bisphosphate.
Feedback B: Inhibited by ATP and citrate, stimulated by fructose-2,6-bisphosphate and AMP.
Feedback C: The committed step of glycolysis, irreversible.
Feedback D: Indirectly decreased because increased cAMP causes PFK-2/FBP-2 complex to act as a phosphatase.
Feedback E:
51. All of the following amino acids are required as precursors for collagen (procollagen) synthesis EXCEPT
A. lysine.
B. glycine.
C. proline.
D. hydroxyproline.
E. phenylalanine.
Show answer
Correct Answer: D
Feedback A: Commonly hydroxylated to hydroxylysine in collagen.
Feedback B: Found every third residue.
Feedback C: Commonly hydroxylated to hydroxyproline.
Feedback D: Proline is the precursor which undergoes posttranslational modification to hydroxyproline.
Feedback E:
52. Which of the following statements is an accurate comparison between chymotrypsinogen and chymotrypsin?
A. Chymotrypsinogen has a single amino terminus and is inactive.
B. Chymotrypsinogen is a storage form of chymotrypsin and is active.
C. Chymotrysinogen and chymotrypsin are readily interconvertable forms of this hydrolase.
D. More than one of the above is an accurate statement.
E. None of the above is an accurate statement.
Show answer
Correct Answer: A
Feedback A: Chymotrypsinogen is a single polypeptide chain that is cleaved to give chymotrypsin (3 chains that are linked by 2 disulfide bonds),
Feedback B: Chymotrypsinogen is inactive.
Feedback C: No, cleavage of chymotrypsinogen to chymotrypsin is irreversible.
Feedback D:
Feedback E:
53. Reactions occurring during anaerobic GLYCOLYSIS in liver which require the input (use of) of ATP include
A. pyruvate kinase and glucokinase
B. pyruvate kinase and phosphofructokinase
C. glyceraldehyde-3-phosphate dehydrogenase and pyruvate kinase
D. glucokinase and phosphofructokinase
E. glucokinase and glyceraldehyde-3-phosphodehydrogenase
Show answer
Correct Answer: D
Feedback A: See C.
Feedback B: See C.
Feedback C: Glyceraldehyde-3-phosphate dehydrogenase produces NADH, pyruvate kinase produces ATP.
Feedback D:
Feedback E: See C.
54. The conversion of the two nitrogen atoms of glutamine to the two nitrogen atoms of urea requires the participation of urea cycle enzymes and
A. glutaminase only.
B. glutamate dehydrogenase only.
C. glutaminase and aspartate aminotransferase (transaminase).
D. glutaminase and glutamate dehydrogenase.
E. glutamate dehydrogenase and aspartate aminotransferase.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: One urea nitrogen comes from glutamine and enters the urea cycle as carbamoyl phosphate, the second nitrogen enters the cycle as aspartate.
Feedback D:
Feedback E:
55. Which molecule can donate a one-carbon unit most directly to tetrahydrofolate?
A. serine
B. methionine
C. glutamate
D. histidine
E. choline
Show answer
Correct Answer: A
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
56. "Induced fit" is observed during catalysis by
A. adenylate cyclase.
B. glycogen phosphorylase "b".
C. chymotrypsin.
D. carboxypeptidase A.
E. cyclic AMP dependent protein kinase.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Binding of substrate alters the conformation of the enzyme.
Feedback E:
57. Overall totally anaerobic glycolysis provides less ATP than aerobic glycolysis because
A. less or no NADH is recovered during anaerobic glycolysis.
B. less NAD is available under anaerobic conditions, so glycolysis cannot continue.
C. lactate produced anaerobically inhibits phosphofructokinase-1.
D. lactate is not a substrate for pyruvate dehydrogenase.
E. more than one of the above is correct.
Show answer
Correct Answer: A
Feedback A: The NADH is oxidized to NAD+ by the conversion of pyruvate to lactate, so that the NAD+ can be used for another round of glycolysis. The NADH is thus not available for oxidative phosphorylation.
Feedback B: See A.
Feedback C: Citrate and ATP inhibit PFK-1.
Feedback D: True, but it is the NADH that is important because it and FADH2 transfer reducing potential to electron transport. The TCA cycle only makes one GTP without electron transport.
Feedback E:
58. The protective mechanism preventing accumulation of ammonia in most tissues resides in which enzyme?
A. aspartate aminotransferase (GOT)
B. glutaminase
C. gamma-glutamyl transpeptidase
D. glutamine synthetase
E. argininosuccinate synthetase
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Glutamine removes ammonia from tissues and transports it to the liver for further metabolism to urea.
Feedback E:
59. Changes in enzyme activities IN SITU often involve "allosteric effects." This regulation most commonly is seen through changes in
A. Vmax.
B. Km.
C. enzyme concentration.
D. phosphorylation of reactive serines.
E. Vmax and Km.
Show answer
Correct Answer: B
Feedback A:
Feedback B: Allosteric modifiers most often change an enzyme™s affinity for substrate; high Km indicates low affinity.
Feedback C:
Feedback D:
Feedback E:
60. Phosphofructokinase-2 was described as a "tandem" or bifunctional enzyme. This means that
A. this enzyme can catalyze more than one reaction.
B. this enzyme hydrolyzes fructose-1,6-diphosphate and may also catalyze the formation of this same compound.
C. this enzyme has "group" or "class" specificity.
D. this enzyme catalyzes the formation of fructose-1,6-diphosphate.
E. this enzyme catalyzed the hydrolysis of fructose-2,6-diphosphate to give fructose-2-phosphate.
Show answer
Correct Answer: A
Feedback A: It has two parts: the kinase that catalyzes phosphorylation and the phosphatase that catalyzes dephosphorylation.
Feedback B: Hydrolysis and phosphorylation that converts between fructose-2,6-bisphosphate and fructose-6-phosphate.
Feedback C:
Feedback D: See B.
Feedback E: See B.
61. An important source of reducing power for biosynthesis is the reaction catalyzed by
A. malate dehydrogenase.
B. pyruvate dehydrogenase.
C. acyl-CoA dehydrogenase.
D. glycerol-3-phosphate dehydrogenase.
E. glucose-6-phosphate dehydrogenase.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Part of the HMP pathway that produces NADPH which is needed for biosynthesis of fatty acids, steroids, etc.
62. Argininosuccinate is an intermediate in the de novo biosynthesis of
A. heme.
B. adenosine monophosphate.
C. uridine monophosphate.
D. creatine.
E. urea.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: arininosuccinate ---> fumarate + arginine ---> urea + ornithine
63. What is the net yield of NADH when glucose 6-phosphate is converted to lactate by anaerobic glycolysis?
A. 0
B. 1
C. 2
D. 3
E. 4
Show answer
Correct Answer: A
Feedback A: Glyceraldehyde-3-phosphate dehydrogenase produces NADH and lactate dehydrogenase oxidizes NADH to NAD+.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
64. An International Unit (IU) is that quantity of enzyme which can convert substrate to product at a rate of
A. 1 micromole/min.
B. 1 micromole/sec.
C. 1 millimole/min.
D. 1 millimole/sec.
E. 1 nanomole/sec.
Show answer
Correct Answer: A
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
65. You have just finished a meal containing 100 grams of carbohydrate, 50 grams of fat and 25 grams of protein. How many kilocalories did you consume?
A. 350
B. 650
C. 950
D. 1250
E. 1550
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: 100g carb (4 Cal/g) + 50g fat (9 Cal/g) + 25g protein (4 Cal/g) = 950 Cal or kcal.
Feedback D:
Feedback E:
66. All of the following can increase the activity of muscle glycogen phosphorylase "b" (dependent) except for
A. epinephrine.
B. cyclic AMP.
C. glycogen phosphorylase kinase.
D. 5'-adenosine monophosphate (AMP).
E. glucagon.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Glucagon does not effect muscle.
67. The major rate-limiting step of glycolysis in liver cells is
A. the conversion of glucose to glucose 6-phosphate.
B. the conversion of glucose 6-phosphate to fructose 6-phosphate.
C. the conversion of fructose 6-phosphate to fructose 1,6-bisphosphate.
D. the aldolase reaction.
E. none of the above.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: The committed step catalyzed by PFK-1.
Feedback D:
Feedback E:
68. A new enzyme has been isolated and shown to have a native molecular weight of 240,00. In the presence of 8M urea the molecular weight of the enzyme was estimated to be 120,000. When beta- mercaptoethanol was added to the urea system, the molecular weight was found to be 60,000. Which of the following best describes the subunit or polypeptide structure of the enzyme?
A. The enzyme consists of two subunits with no interchain disulfide bridges.
B. The enzyme consists of four polypeptide chains with no interchain disulfide bridges.
C. The enzyme consists of two polypeptide chains with disulfide bridges.
D. The enzyme consists of four polypeptide chains interconnected by disulfide bridges.
E. The enzyme consists of two pairs of polypeptide chains with each pair interconnected by a disulfide bridge.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Urea disrupts the noncovalent interactions between the 2 subunits and beta-mercaptoethanol disrupts the disulfide bonds in each subunit.
69. Assume that the STANDARD FREE ENERGY change for the reaction:
ATP ---> ADP + Pi
is -7.0 Cal/mole. What is the best estimate of the "actual" free energy change when the following concentrations exist:
[ATP = 1x10-4], [ADP = 1x10-5], and [Pi = 1x10-4]?
Use R = 2.0 cal/mole K°, and T = 300 K°.
A. -7.3 Cal
B. -10.0 Cal
C. -10.3 Cal
D. -12.3 Cal
E. -13.9 Cal
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: G = Go + 2.3RT log K G = -7000 cal + 2.3(2)(300) log(10-4 X 10-5/10-4) G = -13900 cal or -13.9 Cal
70. Which citric acid cycle intermediate helps to regulate the rate of glycolysis by directly influencing the activity of phosphofructokinase?
A. Citrate
B. Succinate
C. Alpha-ketoglutarate
D. Oxaloacetate
E. Malate
Show answer
Correct Answer: A
Feedback A: Inhibits PFK-1.
Feedback B:
Feedback C:
Feedback D:
Feedback E:
71. Which primers will amplify my favorite gene if it is in the context below? 5'-GGTACAGTC-MY FAVORITE GENE-CTAGATCAT-3'
A. 5'-GACTGTACC and 5'-CTAGATCAT
B. 5'-GGTACAGTC and 5'-ATGATCTAG
C. 5'-GGTACAGTC and 5'-CTAGATCAT
D. 5'-GACTGTACC and 5'-ATGATCTAG
E. 5'-GGTACAGTC and 5'-GACTGTACC
Show answer
Correct Answer: B
Feedback A: Both primers point away from the target.
Feedback B: Yes, these anneal to the sequences flanking the target and are oriented with their 3™ ends towards the target.
Feedback C: The first primer is correct, but the second one points away from the target.
Feedback D: Both primers point the same way, with one pointing away from the target.
Feedback E: The first primer is correct, but the second one is nonsense, it is complementary to the second flank but parallel.
72. The carbon atom in urea synthesized in the liver arises most directly from
A. aspartate.
B. CO2.
C. fumarate.
D. ornithine.
E. glutamate.
Show answer
Correct Answer: B
Feedback B: The CO2 comes into the urea cycle through the synthesis of carbamoyl phosphate.
Feedback C:
Feedback D:
Feedback E:
73. Phosphocreatine is important in metabolism because it
A. is a compound formed by oxidative phosphorylation which can transfer a phosphate group to ADP.
B. serves in synthetic reactions as the source of the phosphate groups of CDP-choline.
C. is formed by substrate-level phosphorylation and irreversibly transfers a phosphate group to ADP and forms creatinine.
D. reacts with ADP to give ATP and creatine.
E. is the source of a nitrogen and the carbon atoms of urea.
Show answer
Correct Answer: D
Feedback A: Phosphocreatine is substrate level phosporylation not oxidative phosphorylation.
Feedback B:
Feedback C: The reaction is reversible.
Feedback D: Phosphocreatine is a temporary storage form of ATP in muscle.
Feedback E:
74. DNA sequencing depends on generating a set of molecules that terminate at every possible position. This is done by stopping the synthesis
A. using modified nucleotides lacking the 3' hydroxyl group.
B. using a DNA polymerase purified from a thermophilic organism.
C. using modified nucleotides phosphorylated at the 5' hydroxyl.
D. using modified nucleotides with deaminated bases.
E. using a non-processive DNA polymerase.
Show answer
Correct Answer: A
Feedback A: Yes, the 3™ hydroxyl is needed for elongation so removing it causes termination
Feedback B: This can be done, but does not contribute to the mechanism of chain termination
Feedback C: No, normal nucleotides have 5™ phosphates
Feedback D: No, this would just make damaged DNA, it would not terminate the elongation.
Feedback E: No, this would just make short products, it would not allow specific termination.
75. Each of the following may be used to estimate the molecular weight of a protein EXCEPT
A. sucrose density gradient centrifugation.
B. ion exchange chromatography.
C. SDS (sodium dodecyl sulfate) gel electrophoresis.
D. gel filtration chromatography.
E. sedimentation and diffusion measurements.
Show answer
Correct Answer: B
Feedback B: Separates molecules based on charge differences.
Feedback C:
Feedback D:
Feedback E:
76. AB arises from a 770 amino acid membrane protein called APP (amyloid precursor protein). APP is mostly processed by alpha and gamma secretases to yield, among other fragments, a small peptide p3 that causes little harm. Enhanced processing by beta-secretase and gamma-secretase generates the 40 or 42 residue AB peptide that results in Alzheimer amyloid deposits. APP spans the lipid bilayer between residues 671 and 770, and the last few residues of AB are in the bilayer. From what you know about membrane proteins which one of the following sequences identifies the last 5 amino acids in the 42 residue AB peptide.
A. Asp-Ala-Glu-Phe-Arg
B. Gly-Val-Val-Ile-Ala
C. His-Asp-Ser-Gly-Tyr
D. Lys-Val-His-His-Gln
E. Glu-Asp-Val-Gly-Ser
Show answer
Correct Answer: B
Feedback B: Because the last few residues of the AB peptide are in the membrane they need to by hydrophobic (non-polar).
Feedback C:
Feedback D:
Feedback E:
77. The glycerol phosphate shuttle
A. results in production of dihydroxyacetone phosphate in the matrix of mitochondria.
B. transfers electrons from cytosolic NADH to the respiratory chain of mitochondria.
C. inhibits glycolysis.
D. supports phosphorylation of three moles of ADP per mole of cytosolic NADH oxidized.
E. uses 3-phosphoglyceric acid as a substrate.
Show answer
Correct Answer: B
Feedback A: DHAP is transferred out of the mitochondria.
Feedback B:
Feedback C:
Feedback D: Transfers NADH reducing potential into the mitochondria as FADH2, which supports the phosphorylation of 2ATP.
Feedback E:
78. When modifying the Southern blot procedure to do a Northern blot, you would want to avoid the step in which
A. the nucleic acid molecules are separated by electrophoresis in agarose.
B. the gel is soaked in sodium hydroxide to denature the nucleic acids.
C. the nucleic acids are transferred to a solid matrix by capillary action.
D. a denatured probe is added to the solid matrix and incubated at 65° C.
E. probe that has hybridized to the immobilized nucleic acids is detected.
Show answer
Correct Answer: B
Feedback A: No, this is necessary to gather size information
Feedback B: Yes, RNA is sensitive to hydrolysis in solutions with high pH values, so this would tend to destroy the RNA that is being tested in a Northern.
Feedback C: No, this is the same in Southerns and Northerns.
Feedback D: No, this is the same in Southerns and Northerns.
Feedback E: No, this is the same in Southerns and Northerns.
79. Urea is the product of an enzyme in liver which catalyzes the hydrolysis of
A. carbamoyl phosphate.
B. citrulline.
C. ornithine.
D. argininosuccinate.
E. arginine.
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: carbamoyl phosphate + ornithine ---> citrulline + aspartate ---> argininosuccinate ---> arginine ---> urea
80. From what you know about protein chemistry, which peptide migrates fastest to the cathode (or negative charged electrode) at pH 6.0.
A. Asp-Ala-Glu-Phe-Arg
B. Gly-Val-Val-Ile-Ala
C. His-Asp-Ser-Gly-Tyr
D. Lys-Val-His-His-Gln
E. Glu-Asp-Val-Gly-Ser
Show answer
Correct Answer: D
Feedback A: -1, 0, -1, 0, +1 = -1
Feedback B: 0, 0, 0, 0, 0
Feedback C: +1, -1, 0, 0, 0 = 0
Feedback D: +1, 0, +1, +1, 0 = +3
Feedback E: -1, -1, 0, 0, 0 = -2 This peptide would migrate fastest toward the anode.
81. Which of the following is a substrate for RNA synthesis, but not DNA synthesis?
A. adenosine triphosphate (ATP)
B. deoxyuridine triphosphate (dUTP)
C. guanosine-5', 5'-guanosine triphosphate (GpppG)
D. cyclic adenosine monophosphate (cAMP)
E. pseudouridine triphosphate (psiTP)
Show answer
Correct Answer: A
Feedback A: Yes, this a ribonucleoside triphosphate suitable for synthesis of RNA.
Feedback B: No, dNTPs are not found in RNA
Feedback C: No, this is not a substrate for an RNA polymerase.
Feedback D: No, this has no free hydroxyl to participate in the polymerization reaction.
Feedback E: No, this is not a standard NTP used by RNA polymerase.
82. Ammonium ion in the urine arises primarily from
A. the action of glutamate dehydrogenase in the kidney.
B. the action of glutaminase in the kidney on the amide nitrogen of glutamine.
C. the action of glutamine aminotransferase in the kidney.
D. the action of glutaminase in the kidney on the alpha-amino group of glutamine.
E. the action of asparaginase in the kidney.
Show answer
Correct Answer: B
Feedback A:
Feedback B: In the proximal tubules glutamine is deaminated to glutamate and ammonia.
Feedback C:
Feedback D:
Feedback E:
83. The compound which accepts a two-carbon fragment from acetyl-CoA to initiate the citric acid cycle is
A. oxalosuccinate.
B. oxalate.
C. oxaloacetate.
D. oxytocin.
E. oxybutyrate.
Show answer
Correct Answer: C
Feedback A:
Feedback B:
Feedback C: acetyl CoA + oxaloacetate ---> citrate
Feedback D:
Feedback E:
84. Ketosis may develop as a result of which of the circumstances below?
A. glucose synthesis from fatty acids during glucose deprivation
B. acute attack of gout
C. carnitine transferase I deficiency
D. hyperglycemia
E. fat mobilization to support gluconeogenesis
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Ketone bodies are produced because more acetyl CoA is derived from fatty acid oxidation than can be used in TCA cycle.
85. The alpha-amino groups on most mammalian proteins are acetylated. That is they have the structure: CH3-C(=O)-NH-C-peptide However, the alpha-amino groups on alpha- and beta- hemoglobin subunits are not acetylated. Their acetylation would likely:
A. prevent binding of protons to Hb
B. prevent BPG binding to Hb
C. Reduce CO2 elimination by the lungs
D. A + C
E. B + C
Show answer
Correct Answer: E
Feedback A: Protons bind to alpha-amino groups of the alpha chains and to carboxyl terminal His residues of the beta chains. The beta chains would still bind protons.
Feedback B: Yes, the terminal amino groups of the beta chains aid in bonding BPG.
Feedback C: Alpha-amino groups react with CO2: -NH2 + CO2 ---> NHCOO- + H+
Feedback D:
Feedback E:
86. A person with Xeroderma pigmentosum
A. is missing the enzyme that reverses thymine dimers
B. is sensitive to ultraviolet radiation (UV)
C. is sensitive primarily to ionizing radiation (x-rays)
D. is prone to cancers and should be treated prophylactically with radiation therapy
E. is missing one of the 7 genes that encode type II topoisomerases
Show answer
Correct Answer: B
Feedback A: No, that is called photolyase, and has not been found in humans.
Feedback C: No, they are sensitive to UV
Feedback D: No, due to the defect in damage repair, radiation therapy is not indicated.
Feedback E: There are 7 classes of XP, but none are known to have a defect in a type II topoisomerase.
87. A person regularly eating a high protein diet (diet 1) changes to diet 2 which contains adequate but substantially less protein. After a period of several days on diet 2, what will occur?
A. Nitrogen balance will be positive.
B. Nitrogen balance will be negative.
C. Daily output of urea will increase over that while on diet 1.
D. Daily output of urea will decrease below that while on diet 1.
E. Two of the above are true.
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Consumed amino acids are used for protein replacement and excess amino acids are catabolized for energy. There is no real storage form of amino acids. Because diet 2 has less protein intake, there is less excess protein to catabolize.
Feedback E:
88. The three stranded collagen superhelix is characterized by the following properties EXCEPT :
A. Glycine is present every third residue
B. H-bonds are parallel to the helix axis
C. Some lysines and prolines are hydroxylated
D. Sugars can be attached to hydroxylated lysine residues
E. No exceptions
Show answer
Correct Answer: B
Feedback A:
Feedback B: The H-bonds occur btween the three strands and are more perpendicular to the axis.
Feedback C:
Feedback D:
Feedback E:
89. The DNA that is copied by the two replication forks traveling in opposite directions from a common site of initiation is called:
A. a duplex
B. an origin
C. a battery
D. a replicon
E. a transposon
Show answer
Correct Answer: D
Feedback A: No, a duplex is the name for the two stranded form of DNA
Feedback B: No, the origin is the site of initiation itself.
Feedback C: No, a battery is a group of adjacent origins that fire simultaneously
Feedback D:
Feedback E: No, this is a piece of DNA capable of moving to other sites in the genome.
90. Lysosomal enzymes are often maximally active at pH 5.0 or less. In many cases, they are inactive at pH 7.0 which makes some biological sense in that lysosomal rupture would be lethal if the released proteases and nucleases were significantly active at cytosolic pH (~ 7.0). The amino acid R group most affected by a change in pH from 5 to 7 is:
A. arginine
B. aspartate
C. cysteine
D. histidine
E. lysine
Show answer
Correct Answer: D
Feedback A:
Feedback B:
Feedback C:
Feedback D: Histidine™s imidazole ring readily accepts or donates protons. pKa = 6.5.
Feedback E:
91. The processivity of DNA polymerases
A. increases in the presence of ligases
B. decreases in the presence of helicases
C. increases in the presence of single strand binding proteins
D. increases in the presence of Okazaki factors
E. decreases in the presence of clamp proteins
Show answer
Correct Answer: C
Feedback A: No, ligases seal nicks in DNA, they do not influence the tendency to move forward or fall off.
Feedback B: Some helicases affect polymerase processivity, but they increase this property.
Feedback C: Yes, SSBs straighten the template and increase both the rate and processivity of DNA polymerases
Feedback D: No, there is no such thing as an Okazaki factor, and the size of Okazaki fragments are the result, rather than the cause, of processivity changes.
Feedback E: Clamp proteins increase the processivity of DNA polymerases.
92. McArdle's disease is due to a muscle
A. debrancher enzyme defect.
B. glucosidase defect.
C. phosphofructokinase defect.
D. phosphorylase absence.
E. phosphohexose isomerase deficiency.
Show answer
Correct Answer: D
Feedback A: Cori™s Disease of Andersen™s Disease
Feedback B: Pompe™s Disease
Feedback C: Glycogen storage disease VII
Feedback D: Cannot degrade glycogen; limited ability to perform strenuous exercise.
Feedback E:
93. The following secondary structures can be classified according to the length of, say ten residues, in the specific conformation. From longest to shortest they are:
A. beta strand, collagen strand, alpha-helix, pi-helix
B. Collagen strand, beta strand, alpha-helix, pi-helix
C. alpha-helix, beta strand, pi-helix, collagen strand
D. alpha-helix, pi-helix, beta strand, collagen strand
E. Collagen strand, alpha-helix, beta strand, pi-helix
Show answer
Correct Answer: A
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E:
94. A bacterial culture grown for many generations in a "heavy" (15N)medium was transferred to a "light" (15N) medium. After the DNA had been copied 3 times in light medium, what would be the relative distribution of DNA containing two light strands (LL), one heavy and one light strand (HL), and two heavy strands (HH)?
A. 2 HH:2 LL
B. 1 HH:1HL:6 LL
C. 0 HH:1 HL: 7LL
D. 0 HH:1 HL: 3LL
E. 1 HH:2 HL: 1LL
Show answer
Correct Answer: D
Feedback A: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL. There is no HH after the first round.
Feedback B: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL. There is no HH after the first round.
Feedback C: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL. This ratio would result after 4 rounds
Feedback D: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL.
Feedback E: After 1 round the DNA is all HL, after 2 it is 1 HL:1 LL, and after 3 it is 1 HL:3 LL. There is no HH after the first round.
95. Mutations can alter protein structure and lead to devastating diseases. Consider the following amino acid changes: 1. Tyr to Phe 2. Asp to Asn 3. Ile to Arg 4. Arg to Lys 5. His to Asn 6. Trp to Gly 7. Ala to Pro Which change is most likely to alter the absorbance of a protein at 280 nm?
A. 2
B. 3
C. 4
D. 5
E. 6
Show answer
Correct Answer: E
Feedback A:
Feedback B:
Feedback C:
Feedback D:
Feedback E: Glycine is unique in being optically inactive, and tryptophan is aromatic.
96. In methylation-directed mismatch repair in bacteria
A. MutL binds to mismatches in duplex DNA
B. The methylated strand is judged to be the older strand
C. MutL binds to methylated DNA strands
D. One mismatched base is excised leaving a single nucleotide gap to repair
E. RecA makes an incision only at hemimethylated sites
Show answer
Correct Answer: B
Feedback A: No, MutS has this activity, the activity is MutL remains poorly understood.
Feedback B: Yes, since newly synthesized DNA does not have this post-synthetic modification, the methylated strand is older.
Feedback C: No, MutH seems to have this property, the activity is MutL remains poorly understood.
Feedback D: No, this would be base-excision repair.
Feedback E: An incision is made at hemi-methylated sites, but by MutH, not RecA
97. Mutations can alter protein structure and lead to devastating diseases. Consider the following amino acid changes: 1. Tyr to Phe 2. Asp to Asn 3. Ile to Arg 4. Arg to Lys 5. His to Asn 6. Trp to Gly 7. Ala to Pro Which change is most likely to disrupt the folding of a globular protein?
A. 1
B. 2
C. 3
D. 4
E. 5
Show answer
Correct Answer: C
Feedback A: Both are aromatic.
Feedback B: Only one group is different and both are polar.
Feedback C: Ile is non-polar, Arg is very polar and its R group is approximately twice as long as Ile.
Feedback D: Both have long R groups and both are bases.
Feedback E: Both are polar.
98. All of the following statements concerning eukaryotic DNA replication are true EXCEPT
A. single-strand binding proteins coat some DNA intermediates in replication
B. replication of half of the DNA chains occurs in short discontinuous pieces
C. helicases unwind the parental strands for the polymerases
D. a single type of DNA polymerase acts on both leading and lagging strands
E. ring-shaped clamp proteins promote high processivity of DNA polymerase
Show answer
Correct Answer: D
Feedback A: SSBs are needed for all phases of DNA metabolism, including replication.
Feedback B: This is true, describing the synthesis of Okazaki fragments on the lagging strand.
Feedback C: While the precise identity of the eukaryotic replicative helicase is not known, this is very likely to be true given the rate of replication fork movement.
Feedback D: This is unlikely to be true since both Pol alpha and Pol delta seem to be needed for eukaryotic replication, although their exact roles remain unknown.
Feedback E: This is true, the clamp is called PCNA.
99. Substrate Km #1 2.5 #2 12.0 #3 32.0 Comparison of the respective Km values for the hydrolysis of the peptide bond of substrates 1-3 by chymotrypsin shows that
A. The Vmax of substrate #1 must be greater than that for #2 or #3.
B. The Vmax for substrate #3 must be greater than that for #1 or #2.
C. A lower concentration of #3 will be needed to attain Vmax.
D. A lower concentration of #1 will be needed to attain Vmax.
E. Substrate #3 has the highest affinity for the active stie of chymotrypsin.
Show answer
Correct Answer: D
Feedback A: The Km tells us the concentration of substrate needed to reach 1/2 Vmax for a particular enzyme, but provides no comparison for the actual velocities.
Feedback B: See A.
Feedback C: Since Km, which tells us the concentration needed to reach 1/2 Vmax, is highest for #3, #3 must need the highest concentration to reach full Vmax.
Feedback D: See C.
Feedback E: Higher Km indicates less affinity of the enzyme for the substrate.
100. Ionizing radiation is capable of producing genetic mutations by
A. Altering bases so that they form different base pair interactions
B. Causing deletions of DNA sequences
C. Causing inversions of DNA sequences
D. Causing translocations of one part of a chromosome onto another
E. A-D are all correct.
Show answer
Correct Answer: E
Feedback A: True, but so are all of the choices, so the answer is E
Feedback B: True, but so are all of the choices, so the answer is E
Feedback C: True, but so are all of the choices, so the answer is E
Feedback D: True, but so are all of the choices, so the answer is E
Feedback E:
101. Which of the following cases represents a metabolically important control?
A. Substrate oxidation in mitochondria is enhanced by elevated ATP levels.
B. The activity of liver hexokinase is stimulated by high levels of its product, glucose 6-phosphate.
C. Acetyl-CoA inhibits pyruvate carboxylase in the liver.
D. Phosphofructokinase (PFK) is inhibited by high levels of its substrate, MgATP.
E. Glucagon stimulates hepatic glycogen synthase.
Show answer
Correct Answer: D
Feedback A: Increased ATP shuts down enzymes of glycolysis and TCA cycle, so oxidation is decreased.
Feedback B: Hexokinase is inhibited by glucose-6-phosphate.
Feedback C: The activity of pyruvate carboxylase requires acetyl CoA.
Feedback D: ATP and citrate inhib