Exam 2

Question 1

2007

1. The hormones epinephrine and norepinephrine are derived biosynthetically from:
   
   1. histidine.
   2. isoleucine.
   3. tryptophan.
   4. tyrosine
   5. arginine

Answer 1

tyrosine

Question 2

2007

1. Which of the following statements correctly describes the pathway by which purine nucleotides are synthesized?
   
   1. The nitrogen in the purine base that is bonded to ribose in the nucleotide is derived originally from glycine.
   2. Purine deoxynucleotides are made by the same path as ribonucleotides, followed by reduction of the ribose moiety.
   3. The pathway occurs only in plants and bacteria, not in animals.
   4. The purine rings are first synthesized, then condensed with ribose phosphate.
   5. The first enzyme in the path is aspartate transcarbamoylase (ATCase).

Answer 2

Purine deoxynucleotides are made by the same path as ribonucleotides, followed by reduction of the ribose moiety.

Question 3

2007

1. Which of the following is not true of the reaction catalyzed by ribonucleotide reductase?
   
   1. Its mechanism involves formation of a free radical.
   2. There is a separate enzyme for each nucleotide (ADP, CDP, GDP, UDP).
   3. Thioredoxin acts as an essential electron carrier.
   4. Glutathione is part of the path of electron transfer.
   5. It acts on nucleoside diphosphates.

Answer 3

There is a separate enzyme for each nucleotide (ADP, CDP, GDP, UDP).

Question 4

2007

1. The most direct precursors of the nitrogens of UMP are:
   
   1. glutamate and carbamoyl phosphate.
   2. aspartate and bicarbonate.
   3. glutamate and aspartate.
   4. glutamine and aspartate.
   5. glutamine and carbamoyl phosphate.

Answer 4

glutamine and aspartate

Question 5

2007

1. Which one of the following statements is true?
   
   1. Muscle cannot use fatty acids as an energy source.
   2. The brain prefers glucose as an energy source, but can use ketone bodies.
   3. In a well-fed human, about equal amounts of energy are stored as glycogen and as triacylglycerol.
   4. Fatty acids cannot be used as an energy source in humans because humans lack the enzymes of the glyoxylate cycle.
   5. Amino acids are a preferable energy source over fatty acids.

Answer 5

The brain prefers glucose as an energy source, but can use ketone bodies.

Question 6

2007

1. A cell that is unable to synthesize or obtain tetrahydrofolic acid (H₄ folate) would probably be deficient in the biosynthesis of:
2. CMP.
3. UMP.
4. orotate.
5. thymidylate (TMP)
6. all of the above

Answer 6

• thymidylate (TMP)

Question 7

2007

1. Elevated epinephrine levels do not normally stimulate:
2. glycogen breakdown in muscle.
3. fatty acid mobilization in adipose tissue.
4. glycogen synthesis in liver.
5. glycolysis in muscle.
6. gluconeogenesis in liver.

Answer 7

glycogen synthesis in liver.

Question 8

2007

1. One amino acid directly involved in the purine biosynthetic pathway is:
2. alanine.
3. leucine.
4. aspartate.
5. tryptophan
6. glutamate.

Answer 8

aspartate.

Question 9

2007

1. Epinephrine triggers an increased rate of glycolysis in muscle by:
2. the Pasteur effect.
3. inhibition of the Corr Cycle
4. conversion of glycogen phosphorylase a to glycogen phosphorylase b.
5. activation of phosphofructokinase-1.
6. activation of hexokinase.

Answer 9

activation of phosphofructokinase-1.

Question 10

2007

1. Glucose labeled with ¹⁴C in C-3 and C-4 is completely converted to acetyl-CoA via glycolysis and the pyruvate dehydrogenase complex. What percentage of the acetyl-CoA molecules formed will be labeled with ¹⁴C, and in which position of the acetyl moiety will the ¹⁴C label be found?
2. 100% of the acetyl-CoA will be labeled at C-2.
3. 50°% of the acetyl-CoA will be labeled, all at C-2 (methyl).
4. No label will be found in the acetyl-CoA molecules.
5. 100% of the acetyl-CoA will be labeled at C-1 (carboxyl).
6. Not enough information is given to answer this question.

Answer 10

No label will be found in the acetyl-CoA molecules.

Question 11

2007

1. Which one of the following reactions, cycles, or pathways is not found in plant systems?
   
   1. The rubisco reaction
   2. The urea cycle
   3. The Calvin cycle
   4. The gluconeogenesis pathway
   5. The glyoxalate cycle

Answer 11

The urea cycle

Question 12

2007

1. Which of the following is not true of the citric acid cycle?
   
   1. In the presence of malonate, one would expect succinate to accumulate.
   2. Oxaloacetate is used as a substrate but is not consumed in the cycle.
   3. Succinate dehydrogenase channels electrons directly into the electron transfer chain.
   4. The condensing enzyme is subject to allosteric regulation by ATP and NADH.
   5. All enzymes of the cycle are located in the cytoplasm, except succinate dehydrogenase, which is bound to the inner mitochondrial membrane.

Answer 12

All enzymes of the cycle are located in the cytoplasm, except succinate dehydrogenase, which is bound to the inner mitochondrial membrane.

Question 13

2007

1. The coenzyme involved in a transaminase reaction is:
   
   1. pyridoxal phosphate (PLP).
   2. thiamine pyrophosphate (TPP).
   3. nicotinarnide adenine dinucleotide phosphate (NADP+).
   4. biotin phosphate.
   5. lipoic acid.

Answer 13

pyridoxal phosphate (PLP).

Question 14

2007

1. Acetyl-CoA labeled with ¹⁴C in both of its acetate carbon atoms is incubated with unlabeled oxaloacetate and a crude tissue preparation capable of carrying out the reactions of the citric acid cycle. After one turn of the cycle, oxaloacetate would have ¹⁴C in:
2. the keto carbon and one of the carboxyl carbons.
3. none of its carbon atoms.
4. no pattern that is predictable from the information provided.
5. the two carboxyl carbons.
6. all four carbon atoms.

Answer 14

all four carbon atoms.

Question 15

2007

1. Which of the following is not true of the reaction catalyzed by glutamate dehydrogenase?
   
   1. The enzyme can use either NAD+ or NADP+ as a cofactor.
   2. NH4+ is produced.
   3. The enzyme is specific for glutamate, but the reaction is also involved in the oxidation of the other 19 amino acids.
   4. The reaction is similar to transamination in that it involves the coenzyme pyridoxal phosphate (PLP)
   5. α-Ketoglutarate is produced from an amino acid.

Answer 15

The reaction is similar to transamination in that it involves the coenzyme pyridoxal phosphate (PLP)

Question 16

2007

1. The two moles of CO₂ produced in the first turn of the citric acid cycle have their origin in the:
   
   1. carboxyl group of acetate and the keto group of oxaloacetate.
   2. carboxyl and methylene carbons of oxaloacetate
   3. carboxyl group of acetate and a carboxyl group of oxaloacetate.
   4. two carboxyl groups derived from oxaloacetate.
   5. two carbon atoms of acetate.

Answer 16

two carboxyl groups derived from oxaloacetate.

Question 17

2007

1. Which of these directly donates a nitrogen atom for the formation of urea during the urea cycle?
2. adenine
3. creatine
4. glutamate
5. ornithine
6. aspartate

Answer 17

aspartate

Question 18

2007

1. Which of the following answers completes the sentence correctly? The rate of flow of electrons through the electron transport chain is most directly regulated by
   
   1. the rate of oxidative phosphorylation. D. the ATP:ADP ratio.
   2. the catalytic rate of cytochrome oxidase. E. the concentration of acetyl CoA.
   3. feedback inhibition by H2O.
   4. the ATP:ADP ratio.
   5. the concentration of acetyl CoA.

Answer 18

the ATP:ADP ratio.

Question 19

2007

1. The human genetic disease phenylketonuria (PKU) can result from:
2. inability to catabolize ketone bodies.
3. inability to convert phenylalanine to tyrosine.
4. production of enzymes containing no phenylalanine.
5. deficiency of protein in the diet.
6. inability to synthesize phenylalanine.

Answer 19

inability to convert phenylalanine to tyrosine.

Question 20

2007

1. The inner mitochondrial membrane contains translocases—that is, specific transport proteins—for which pairs of substances?
2. MP and ADP
3. Citrate and pyruvate
4. Oxaloacetate and aspartate
5. NAD+ and NADH
6. Glycerol 3-phosphate and dihydroxyacetone phosphate

Answer 20

Glycerol 3-phosphate and dihydroxyacetone phosphate

Question 21

2007

1. In photophosphorylation, absorption of light energy in chloroplast “light reactions” leads to:
2. absorption of CO₂ and release of O₂.
3. use of iron-sulfur proteins.
4. synthesis of ATP and oxidation of NADPH.
5. hydrolysis of ATP and reduction of NADP+.
6. absorption of O₂ and release of CO₂

Answer 21

absorption of CO₂ and release of O₂

Question 22

2007

1. Which of the following experimental observations does not provide evidence supporting the chemiosmotic model of oxidative phosphorylation?
2. A system of bacteriorhodopsin and ATPase can produce ATP in synthetic vesicles when light causes proton pumping.
3. A proton gradient is generated across the inner membrane of mitochondria during electron transport.
4. Submitochondrial particles show a symmetric distribution of the respiratory chain components and ​ATP synthase across the inner mitochondrial membrane.
5. ATP is synthesized when a proton gradient is imposed on mitochondria.
6. A closed membrane or vesicle compartment is required for oxidative phosphorylation.

Answer 22

Submitochondrial particles show a symmetric distribution of the respiratory chain components and ​ATP synthase across the inner mitochondrial membrane.

Question 23

2007

1. In the photolytic cleavage of water by the oxygen-evolving complex [2H₂O → 4 H⁺ + 4e^- + O₂], how many photons of light are required?

A. 4

B. 8

C. 1

D. 2

E. 6

Answer 23

B. 8

Question 24

2007

1. If electron transfer in tightly coupled mitochondria is blocked (with antimycin A) between cytochrome b and cytochrome c₁, then:
   
   1. energy diverted from the cytochromes will be used to make ATP, and the P/O ratio will rise.
   2. electron transfer from succinate to 0 2 will continue unabated.
   3. electron transfer from NADH will cease, but Oz uptake will continue.
   4. all ATP synthesis will stop.
   5. ATP synthesis will continue, but the P/O ratio will drop to one.

Answer 24

all ATP synthesis will stop.

Question 25

2007

1. Cyclic electron flow in chloroplasts produces:
2. NADPH, but not ATP or Hz
3. O₂, but not ATP or NADPH.
4. ATP and O₂, but not NADPH.
5. NADPH, and ATP, but not O₂
6. ATP, but not NADPH or O₂

Answer 25

ATP, but not NADPH or O₂

Question 26

2007

1. 2,4-Dinitrophenol and oligomycin inhibit mitochondrial oxidative phosphorylation. 2,4-Dinitrophenol is an uncoupling agent; oligomycin blocks the ATP synthesis reaction itself. Therefore, 2,4-dinitropheno1will:
   
   1. allow electron transfer in the presence of oligomycin.
   2. do none of the above.
   3. diminish Oi consumption in the presence of oligomycin
   4. allow oxidative phosphorylation in the presence of oligomycin.
   5. block electron transfer in the presence of oligomycin.

Answer 26

allow electron transfer in the presence of oligomycin.

Question 27

2007

1. In the carbon assimilation (“dark”) reactions of photosynthesis, the biosynthesis of l mol of hexose from 6 mol of carbon dioxide requires:
2. no NADPH and 12 mol of ATP.
3. 18 mol of NADPH and 18 mol of ATP.
4. 18 mol of NADPH and 12 mol of ATP.
5. 12 mol of NADPH and 18 mol of ATP.
6. 12 mol of NADPH and 12 mol of ATP.

Answer 27

12 mol of NADPH and 18 mol of ATP.

Question 28

2007

1. In what order do the following five steps occur in the photochemical reaction centers?
2. Excitation of the chlorophyll a molecule at the reaction center
3. Replacement of the electron in the reaction center chlorophyll
4. Light excitation of antenna chlorophyll molecule
5. Passage of excited electron to electron-transfer chain
6. Exiton transfer to neighboring chlorophyll

A. 3-5-1-4-2

B. 4-2-3-5-1

C. 1-2-3-4-5

D. 3-2-5-4-1

E. 5-4-3-2-1

Answer 28

A. 3-5-1-4-2

Question 29

2007

1. Which of these chloroplast enzymes is not regulated by light?
2. Sedoheptulose 1,7-bisphosphatase
3. Fructose 1,6-bisphosphatase
4. Glyceraldehyde-phosphate dehydrogenase
5. Ribulose 5-phosphate kinase
6. All of the above are regulated by light.

Answer 29

All of the above are regulated by light.

Question 30

2010

Neutral triglycerides are biosynthesized starting with the reaction of

• A. glycerol and fatty acids
• B. glycerol and fatty acyl CoA
• C. dihydroxyacetone phosphate and fatty acyl CoA
• D. L-glycerol-3-phosphate and fatty acyl CoA
• E. glycerol and acetyl CoA

Answer 30

L-glycerol-3-phosphate and fatty acyl CoA

Question 31

2010

Ketone bodies are formed in the liver and transported to the extrahepatic tissues mainly as:

• A. acetoacetyl-CoA.
• B. acetone.
• C. beta-hydroxybutyric acid.
• D. beta-hydroxybutyryl-CoA.
• E. lactic acid.

Answer 31

C. beta-hydroxybutyric acid.

Question 32

2010

A sample of malonyl-CoA synthesized from radioactive (14C-labeled) HCO-3 and unlabeled acetyl-CoA is used in fatty acid synthesis. In which carbon(s) will the final fatty acid be labeled? (Recall that the carboxyl carbon is C-1.)

• A. Every carbon
• B. Every odd-numbered carbon
• C. Every even-numbered carbon
• D. Only the carbon farthest from C-1
• E. No part of the molecule will be labeled.

Answer 32

E. No part of the molecule will be labeled.

Question 33

2010

The synthesis of palmitate requires:

• A. 8 acetyl-CoA
• B. 14 NADH
• C. 7 ATP
• D. A and C
• E. A, B, and C

Answer 33

D. A and C

Question 34

2010

How are fatty oxidation and synthesis controlled so that futile cycling does not occur?

• A. They occur in different cellular compartments.
• B. They employ different electron carriers.
• C. The product of the first oxidation reaction inhibits the rate-limiting step of biosynthesis.
• D. A and B
• E. A, B, and C

Answer 34

D. A and B

Question 35

2010

Which of the following correctly describes the citric acid cycle?

• A. Oxygen is required to regenerate electron acceptors.
• B. Citrate is dehydrated and hydrated by the same enzyme, aconitase.
• C. Ten high-energy phosphate bonds are eventually formed as a result of one round of the cycle.
• D. Succinate dehydrogenase links the citric acid cycle to oxidative phosphorylation.
• E. All of the above are true.

Answer 35

E. All of the above are true.

Question 36

You have discovered a compound that inhibits fumarase. How many moles of ATP would you expect to be generated from one mole of acetyl-CoA in the presence of this inhibitor? (P/O for NADH 2.5, for FADH2 1.5)

• A. 5
• B. 6
• C. 6.5
• D. 7.5
• E. No ATP would form under these conditions

Answer 36

D. 7.5

Question 37

2010

Which of the following metabolites when converted to its normal product does not lead to oxidative phosphorylation?

• A. fumaric acid
• B. isocitric acid
• C. Į-ketoglutaric acid
• D. succinic acid
• E. none of the above

Answer 37

A. fumaric acid

Question 38

2010

The proton-motive force generated by the electron transfer chain:

• A. includes a pH-gradient component.
• B. includes an electrical-potential-gradient component.
• C. is used for active transport processes.
• D. is used to synthesize ATP.
• E. has all of the above characteristics.

Answer 38

E. has all of the above characteristics.

Question 39

2010

If electron transfer in tightly coupled mitochondria is blocked (with antimycin A) between cytochrome b and cytochrome c1, then:

• A. all ATP synthesis will stop.
• B. ATP synthesis will continue, but the P/O ratio will drop to one.
• C. electron transfer from NADH will cease, but O2 uptake will continue.
• D. electron transfer from succinate to O2 will continue unabated.
• E. energy diverted from the cytochromes will be used to make ATP, and the P/O ratio will rise.

Answer 39

A. all ATP synthesis will stop.

Question 40

2010

In normal mitochondria, the rate of NADH consumption (oxidation) will:

• A. be increased in active muscle, decreased in inactive muscle.
• B. be very low if the ATP synthase is inhibited, but increase when an uncoupler is added.
• C. decrease if mitochondrial ADP is depleted.
• D. decrease when cyanide is used to prevent electron transfer through the cytochrome a + a3 complex.
• E. All of the above are true.

Answer 40

E. All of the above are true.

Question 41

2010

The relative concentrations of ATP and ADP control the cellular rates of:

• A. glycolysis.
• B. oxidative phosphorylation.
• C. pyruvate oxidation.
• D. the citric acid cycle.
• E. all of the above.

Answer 41

E. all of the above.

Question 42

2010

When the ‘G’° of the ATP synthesis reaction is measured on the surface of the ATP synthase enzyme, it is found to be close to zero. This is thought to be due to:

• A. a very low energy of activation.
• B. enzyme-induced oxygen exchange.
• C. stabilization of ADP relative to ATP by enzyme binding.
• D. stabilization of ATP relative to ADP by enzyme binding.
• E. none of the above.

Answer 42

D. stabilization of ATP relative to ADP by enzyme binding.

Question 43

2010

Of all the components of the “Z scheme,” which has the lowest E’° (i.e., is the best reducing agent)?

• A. P700
• B. P700*
• C. O2
• D. H2O
• E. NADPH

Answer 43

B. P700*

Question 44

2010

Which of the following statements about cyclic photophosphorylation and noncyclic photophosphorylation is correct?

• A. Cyclic photophosphorylation involves only photosystem II and produces only ATP; noncyclic photophos phorylation involves photosystems I and II and produces only ATP.
• B. Both pathways liberate oxygen.
• C. Both pathways involve photosystems I and II.
• D. Cyclic photophosphorylation reduces NADP+ and liberates oxygen; noncyclic photophosphorylation reduces NADP+ but does not liberate oxygen.
• E. Noncyclic photophosphorylation reduces NADP+ and liberates oxygen; cyclic photophosphorylation produces ATP but does not liberate oxygen.

Answer 44

E. Noncyclic photophosphorylation reduces NADP+ and liberates oxygen; cyclic photophosphorylation produces ATP but does not liberate oxygen.

Question 45

2010

Which equation correctly summarizes the carbon fixation reactions of photosynthesis?

• A. 6H2O + 6CO2 → 6O2 + C6H12O6
• B. 6CO2 + 12NADPH + 18ATP + 12H+ → 12NADP+ + 18ADP + 18Pi + C6H12O6
• C. 6H2O + 6ADP + 6Pi + 6NADP → 6O2 + 6ATP + NADPH + 6H+
• D. 2H2O → O2 + 4H+ + 4e-
• E. 2C3O3H5 + CO2 → 2C3O3H3 + H2O + CH2O

Answer 45

B. 6CO2 + 12NADPH + 18ATP + 12H+ → 12NADP+ + 18ADP + 18Pi + C6H12O6

Question 46

2010

18. Ribulose 1,5 -bisphosphate carboxylase/oxygenase is arguably the most important enzyme on earth because nearly all life is dependent, ultimately, on its action. The reactions catalyzed by this enzyme are influenced by:

• A. pH.
• B. substrate concentration.
• C. Mg2+ concentration.
• D. temperature.
• E. all of the above

Answer 46

E. all of the above

Question 47

2010

Under hot, dry conditions, temperate grass species (C3 grasses) are outcompeted by other grasses, notably crab-grass (C4 plants). Why?

• A. C4 plants do not use as much ATP to fix CO2 to hexose as do C3 plants.
• B. C4 plants use CO2 more efficiently under hot conditions because they increase [CO2] in bundle-sheath cells.
• C. The Km of the oxygenase reaction of rubisco is higher in C3 plants.
• D. C4 plants are able to bypass the Calvin cycle and so save energy.
• E. All of the above are true.

Answer 47

B. C4 plants use CO2 more efficiently under hot conditions because they increase [CO2] in bundle-sheath cells

Question 48

2010

Which of the following a characteristic of many aminotransferase reactions?

• A. They have a large, negative ΔG’°
• B. The amino group is transferred to an α-keto acid (such as α-ketoglutarate) to form the corresponding amino acid.
• C. The amino group is transferred from an ammonia molecule.
• D. They are catalyzed by the same enzyme.
• E. They require the cofactor S-adenosylmethionine.

Answer 48

B. The amino group is transferred to an Į-keto acid (such as Į-ketoglutarate) to form the corresponding amino acid.

Question 49

2010

The citric acid cycle and the urea cycle overlap to form what has sometimes been called the “Krebs bicycle.” Which of the following statements is relevant to the interactions between these two metabolic cycles?

• A. Oxaloacetate is converted to aspartate.
• B. Aspartate combines with citrulline to produce argininosuccinate in the cytosol.
• C. Argininosuccinate is cleaved to fumarate and arginine.
• D. Fumarate is a citric acid cycle intermediate.
• E. All of the above are true.

Answer 49

E. All of the above are true.

Question 50

2010

Early in its life, a tadpole lives in an aqueous environment and excretes much of its excess nitrogen as ammonia. Once it matures into an adult, the frog spends more time on dry land and becomes ureotelic. Which of the following enzyme activities would be most likely to increase drastically in the adult frog?

• A. Carbamoyl phosphate synthetase I
• B. Glutamine synthetase
• C. Glutaminase
• D. α-Ketoglutarate dehydrogenase
• E. Carboxypeptidase

Answer 50

A. Carbamoyl phosphate synthetase I

Question 51

2010

Which cofactor involved in amino acid degradation is correctly matched to the one-carbon group it transfers?

• A. Biotin: CHO
• B. Tetrahydrofolate: CHOH
• C. S-adenosylmethionine: CH2OH
• D. Pyridoxal phosphate: CH3
• E. Pepsinogen: CO2

Answer 51

B. Tetrahydrofolate: CHOH

Question 52

2010

Which cofactor is essential to all transamination reactions?

• A. PRPP
• B. CoA
• C. ATP
• D. PLP
• E. adoMet (S-adenosylmethionine)

Answer 52

D. PLP

Question 53

2010

What does the term “essential” mean in the terms of amino acids in the human diet?

• A. Necessary for all protein synthesis
• B. Only available in animal protein
• C. Cannot be synthesized by humans
• D. Cannot be coded for by DNA
• E. Cannot be degraded in the liver

Answer 53

C. Cannot be synthesized by humans

Question 54

2010

In nucleotide metabolism, all of the following are true except:

• A. The committed step in purine biosynthesis is the transfer of an amino group to PRPP.
• B. Purine and pyrimidine biosynthesis are regulated by end-product inhibition.
• C. Nucleotides can be synthesized in a single reaction via salvage pathways.
• D. De novo pyrimidine synthesis begins with a molecule of PRPP.
• E. Orotidylate is the common precursor in the biosynthesis of pyrimidines, and inosinate is the common precursor in the biosynthesis of the purines ATP and GTP.

Answer 54

D. De novo pyrimidine synthesis begins with a molecule of PRPP.

Question 55

2010

Which of the following describes the activity and regulation of ribonucleotide reductase and/or its importance to the cell?

• A. Both its activity and its substrate specificity are regulated by the binding of effector molecules.
• B. ATP increases the overall activity of the enzyme.
• C. Control of the enzyme’s activity ensures a balanced pool of precursors for DNA synthesis.
• D. Ba1anced pools of deoxyribonucleotides are necessary in DNA synthesis, given the complementary basepairing of nucleotides in double-stranded DNA.
• E. All of the above are true.

Answer 55

E. All of the above are true.

Question 56

2010

De novo purine biosynthesis is distinguished from de novo pyrimidine biosynthesis by:

• A. condensation of the completed purine ring with ribose phosphate
• B. incorporation of CO2.
• C. inhibition by azaserine (a glutamine analog).
• D. participation of aspartate.
• E. participation of PRPP (phosphoribosyl pyrophosphate).

Answer 56

B. incorporation of CO2.

Question 57

2010

A cell that is unable to synthesize or obtain tetrahydrofolic acid (H4 folate) would probably be deficient in the biosynthesis of:

• A. CMP
• B. GMP
• C. orotate
• D. thymidylate (TMP)
• E. UMP.

Answer 57

D. thymidylate (TMP)

Question 58

2010

An intermediate of purine degradation in humans is:

• A. glutamate.
• B. NH4 + .
• C. succinate.
• D. urea.
• E. uric acid

Answer 58

E. uric acid

Question 59

2012

How many tritium atoms (3 H) are incorporated into palmitate (C16:0) when fatty acid synthesis is carried out in vitro with the following labeled substrate?

• OOCC³H2C-S-CoA O
• A. 1
• B. 2
• C. 7
• D. 14
• E. 10

Answer 59

C. 7

Question 60

2012

Which of the following is not specifically required in the synthesis of fatty acids?

• A. biotin
• B. acetyl-CoA
• C. malonyl-CoA
• D. NADH
• E. HCO3 - (CO2)

Answer 60

D. NADH

Question 61

2012

The rate-limiting step in fatty acid synthesis is:

• A. formation of acetyl-CoA from acetate.
• B. the reaction catalyzed by acetyl-CoA carboxylase.
• C. condensation of acetyl-CoA and malonyl-CoA.
• D. the reduction of the acetoacetyl group to a β-hydroxybutyryl group.
• E. formation of malonyl-CoA from malonate and coenzyme A

Answer 61

B. the reaction catalyzed by acetyl-CoA carboxylase.

Question 62

2012

The effect of insulin-stimulated uptake of glucose into adipocytes will be:

• A. Inhibition of fatty acid synthesis.
• B. Increase in fatty acid synthesis.
• C. Increase in triglyceride synthesis.
• D. Decrease in triglyceride synthesis.
• E. B and C

Answer 62

E. B and C

Question 63

2012

Oxaloacetate uniformly labeled with ¹⁴C (i.e., with equal amounts of ¹⁴C in each of its carbon atoms) is condensed with unlabeled acetyl-CoA. After a single pass through the citric acid cycle back to oxaloacetate, what fraction of the original radioactivity will be found in the oxaloacetate?

• A. all
• B. 1/2
• C. 1/3
• D. 1/4
• E. 3/4

Answer 63

B. 1/2

Question 64

2012

Malonate is a competitive inhibitor of succinate dehydrogenase. If malonate is added to a mitochondrial preparation that is oxidizing pyruvate as a substrate, which of the following compounds would you expect to decrease in concentration?

• A. Citrate
• B. Fumarate
• C. Isocitrate
• D. Pyruvate
• E. Succinate

Answer 64

B. Fumarate

Question 65

2012

The conversion of 1 mol of pyruvate to 3 mol of CO2 via pyruvate dehydrogenase and the citric acid cycle also yields ________ mol of NADH, _______ mol of FADH2, and _____ mol of ATP (or GTP).

• A. 2; 2; 2
• B. 3; 1; 1
• C. 3; 2; 0
• D. 4; 1; 1
• E. 4; 2; 1

Answer 65

D. 4; 1; 1

Question 66

2012

A function of the glyoxylate cycle, in conjunction with the citric acid cycle, is to accomplish the:

• A. complete oxidation of acetyl-CoA to CO2 plus reduced coenzymes.
• B. net conversion of lipid to carbohydrate.
• C. net synthesis of four-carbon dicarboxylic acids from acetyl-CoA.
• D. net synthesis of long-chain fatty acids from citric acid cycle intermediates.
• E. both B and C are correct

Answer 66

E. both B and C are correct

Question 67

2012

If pyruvate (CH3COCOO- ) is introduced into the citric acid cycle, after 1 turn of the cycle, the loss of CO2 would come from:

• A. C-3
• B. C-2
• C. C-1
• D. C-1, C-2 and C-3
• E. C-1 and C-2

Answer 67

C. C-1

Question 68

2012

If acetyl CoA (C^yH3C^xOSCoA) entered the citric acid cycle, after 1 turn of the cycle its carbon atoms would appear in oxaloacetate at which position?

• A. none
• B. x = 1 y = 4
• C. x = 2 y = 3
• D. x = 1, 4 y = 2, 3
• E. x = 1, 3 y = 2, 4

Answer 68

D. x = 1, 4 y = 2, 3

Question 69

2012

Cyanide, oligomycin, and 2,4-dinitrophenol (DNP) are inhibitors of mitochondrial aerobic phosphorylation. Which of the following statements correctly describes the mode of action of the three inhibitors?

• A. Cyanide and 2,4-dinitrophenol inhibit the respiratory chain, and oligomycin inhibits the synthesis of ATP.
• B. Cyanide inhibits the respiratory chain, whereas oligomycin and 2,4-dinitrophenol inhibit the synthesis of ATP.
• C. Cyanide, oligomycin, and 2,4-dinitrophenol compete with O2 for cytochrome oxidase (Complex IV).
• D. Oligomycin and cyanide inhibit synthesis of ATP; 2,4-dinitrophenol inhibits the respiratory chain.
• E. Oligomycin inhibits the respiratory chain, whereas cyanide and 2,4-dinitrophenol prevent the synthesis of ATP.

Answer 69

B. Cyanide inhibits the respiratory chain, whereas oligomycin and 2,4-dinitrophenol inhibit the synthesis of ATP.

Question 70

2012

Upon the addition of 2,4.-dinitrophenol (DNP) to a suspension of mitochondria carrying out oxidative phosphorylation linked to the oxidation of malate, all of the following occur except:

• A. oxygen consumption decreases.
• B. oxygen consumption increases.
• C. the P/O ratio drops from a value of approximately 2.5 to 0.
• D. the proton gradient dissipates.
• E. the rate of transport of electrons from NADH to O2 becomes maximal

Answer 70

A. oxygen consumption decreases

Question 71

2012

Which of the following statements about the light reactions in photosynthetic plants is false?

• A. A membrane-bound ATPase couples ATP synthesis to electron transfer.
• B. No CO2 is fixed in the light reactions.
• C. The ultimate electron acceptor is O2.
• D. The ultimate source of electrons for the process is H2O.
• E. There are two distinct photosystems, linked together by an electron transfer chain.

Answer 71

C. The ultimate electron acceptor is O2.

Question 72

2012

Oxidative phosphorylation and photophosphorylation share all of the following except:

• A. chlorophyll.
• B. involvement of cytochromes.
• C. participation of quinones.
• D. proton pumping across a membrane to create electrochemical potential.
• E. use of iron-sulfur proteins.

Answer 72

A. chlorophyll.

Question 73

2012

Suppose ribulose -5- phosphate labeled with 14C in carbon 1 is used as a substrate in the dark reactions of photosynthesis. In which carbon of 3-phosphoglycerate (3PG) will the label appear?

• A. C-1
• B. C-2
• C. C-3
• D. A and C
• E. None

Answer 73

C. C-3

Question 74

If you were to determine the P/O ration for the oxidation of α ketoglutarate in the presence of some malonate. The P/O ration that you observe would be: A. 0 B. 1.5 C. 2.5 D. 3.5 E. 4.5

Question 76

2012

If you were to determine the P/O ration for the oxidation of α ketoglutarate in the presence of some malonate. The P/O ration that you observe would be:

• A. 0
• B. 1.5
• C. 2.5
• D. 3.5
• E. 4.5

Answer 76

D. 3.5

Question 77

2012

If oxidation of acetyl CoA yields 10 ATPs per mole through the citric acid cycle, how many ATP’s will be formed from the oxidation of 1 mole of alanine to CO2, H2O and urea?

• A. 10
• B. 12.5
• C. 10.5
• D. 13
• E. 15

Answer 77

E. 15

Question 78

2012

Transamination from alanine to α-ketoglutarate requires the coenzyme:

• A. thiamine pyrophosphate (TPP).
• B. NADH.
• C. biotin.
• D. pyridoxal phosphate (PLP).
• E. No coenzyme is involved.

Answer 78

D. pyridoxal phosphate (PLP).

Question 79

2012

Which of the following conversions requires more than one step?

1. alanine → pyruvate
2. glutamate → α-ketoglutarate
3. aspartate → oxaloacetate
4. serine → pyruvate
5. phenylalanine → succinate

• A. l, 2 and 4
• B. l,3 and 5
• C. 2,4 and 5
• D. l and 4
• E. 4 and 5

Answer 79

E. 4 and 5

Question 80

2012

Which of the following enzymes is not involved in the assimilation of inorganic nitrogen into an organic molecule?

• A. Arginase
• B. Glutamate dehydrogenase
• C. Glutamate synthase
• D. Glutamine synthetase
• E. Nitrogenase

Answer 80

A. Arginase

Question 81

2012

Bile pigments are:

• A. formed in the degradation of heme.
• B. generated by oxidation of sterols.
• C. responsible for light reception in the vertebrate eye.
• D. secreted from the pancreas
• E. the products of purine degradation.

Answer 81

A. formed in the degradation of heme.

Question 82

2012

The hormones epinephrine and norepinephrine are derived biosynthetically from:

• A. arginine.
• B. histidine.
• C. isoleucine.
• D. tryptophan.
• E. tyrosine.

Answer 82

E. tyrosine.

Question 83

2012

In amino acid catabolism, the first reaction for many amino acids is a(n):

• A. decarboxylation requiring thiamine pyrophosphate (TPP).
• B. hydroxylation requiring NADPH and O2.
• C. oxidative deamination requiring NAD+ .
• D. reduction requiring pyridoxal phosphate (PLP).
• E. transamination requiring pyridoxal phosphate (PLP).

Answer 83

E. transamination requiring pyridoxal phosphate (PLP).

Question 84

2012

Urea synthesis in mammals takes place primarily in tissues of the:

• A. brain.
• B. kidney.
• C. liver
• D. skeletal muscle.
• E. small intestine.

Answer 84

C. liver

Question 85

2012

If a person’s urine contains unusually high concentrations of urea, which one of the following diets has he or she probably been eating recently?

• A. High carbohydrate, very low protein
• B. Very high carbohydrate, no protein, no fat
• C. Very very high fat, high carbohydrate, no protein
• D. Very high fat, very low protein
• E. Very low carbohydrate, very high protein

Answer 85

E. Very low carbohydrate, very high protein

Question 86

2012

The ribosyl phosphate moiety needed for the synthesis of orotidylate, inosinate, and guanylate is provided most directly by:

• A. 5-phosphoribosyl 1-pyrophosphate.
• B. adenosine 5’-phosphate.
• C. guanosine 5’-phosphate.
• D. ribose 5-phosphate.
• E. ribulose 5-phosphate.

Answer 86

A. 5-phosphoribosyl 1-pyrophosphate.

Question 87

2012

CMP, UMP, and TMP all have ________________ as a common precursor.

• A. Adenosine
• B. Aspartate
• C. glutamine
• D. inosine
• E. S-adenosyl methionine

Answer 87

B. Aspartate

Question 88

2012

Which of the following molecules yields uric acid when degraded?

• 1. DNA
• 2. FAD
• 3. CTP
• 4. PRPP
• 5. β-alanine
• 6. urea
• 7. NAD+
• A. 1 and 3
• B. 2
• C. 4 and 6
• D. 7
• E. A, B and D

Answer 88

E. A, B and D

Question 89

Suppose you have decided to eat this exam, as opposed to take this exam. You lack an appropriate enzyme activity required to digest the cellulose fibrils in the paper into sugar monomers that can be used to obtain energy. This enzyme activity is:

• A. α-D-glucosidase
• B. β-D-glucosidase E. sucrase (invertase)
• C. cellulose debranching enzyme
• D. α-galactosidase (BeanoTM)

Answer 89

B. β-D-glucosidase E. sucrase (invertase)

Question 90

Using the definition presented in your lecture notes, which of the following is a non-classical enzyme:

• A. The ribozyme RNAse P
• B. The glucose transporter GLUT2
• C. asprtate transcarbamoyalase
• D. aldolase
• E. triose phosphate isomerase (TIM)

Answer 90

B. The glucose transporter GLUT2

Question 91

The product of the fifth step of a metabolic pathway inhibits the enzyme activity of the first step. This phenomena is called:

• A. substrate level inhibition
• B. a futile cycle
• C. feedback inhibition
• D. a substrate cycle
• E. catalytic perfection

Answer 91

C. feedback inhibition

Question 92

An intein is

• A. a protease that catalyzes that removal of peptides from mature proteins
• B. a non-classical enzyme
• C. a self splicing molecule of RNA
• D. a reducing agent used in catabolic reactions
• E. a self-splicing protein

Answer 92

E. a self-splicing protein

Question 93

Which of the following would not rotate the direction of plane-polarized light?

• A. D-allose
• B. 2-amino-D-mannose
• C. The aldaric acid of D-galactose
• D. β-D-glucose – 6 phosphate
• E. None of the above; since all of these molecules have chiral centers they will all rotate the direction of planepolarized light

Answer 93

C. The aldaric acid of D-galactose

Question 94

Substrate channeling:

• A. is an irreversible covalent modification
• B. is an irreversible non-covalent modification
• C. excludes the product of an enzyme catalyzed reaction from solvent water
• D. requires ATP
• E. is an example of oxidation-reduction

Answer 94

C. excludes the product of an enzyme catalyzed reaction from solvent water

Question 95

Which of the following is a ketose?

Answer 95

Question 96

Which of the following would be the β- anomer of D-galactopyranose

Answer 96

Question 97

An N-linked glycan would be attached to ______ while an O-linked glycan would be attached to:

• A. Asn; Ser
• B. Lys; Tyr
• C. Gln; Thr
• D. Arg; Ser
• E. Lys; Ser

Answer 97

A. Asn; Ser

Question 98

Oseltamivir, also known as Tamiflu, is effective as an antiviral against the influenza virus due to its ability to block the action of which protein associated with the surface of the viral particle?:

• A. Hemagglutinin
• B. Neuraminidase
• C. Lysozyme
• D. Proteinase O
• E. RNA polymerase

Answer 98

B. Neuraminidase

Question 99

Anomers can be interconverted ________.

• A. by isomerases such as triose phosphate isomerase
• B. via a linear intermediate
• C. only in non-reducing sugars
• D. by an enediol intermediate
• E. None of the above. Anomers cannot be interconverted

Answer 99

B. via a linear intermediate

Question 100

Another name for β-2-D-deoxyribose is:

• A. DNA
• B. RNA
• C. β-2-L-deoxyribulose
• D. β-2-D-deoxyarabinose
• E. Trick question, there is only a single name for monosaccharides and their derivatives.

Answer 100

D. β-2-D-deoxyarabinose

Question 101

Which of the following would NOT produce a sugar silver mirror?

• A. β-D-galactopyranosyl (1→4) – β -D-glucopyranose
• B. β-D-glucopyranosyl (1→1) – β -D-fructofuranoside
• C. β-D-glucopyranosyl (1→4) – β -D-glucopyranose
• D. α-D-glucopyranosyl (1→4) – β -D-glucopyranose
• E. α-D-glucopyranosyl (1→1) – α -D-glucopyranose

Answer 101

E. α-D-glucopyranosyl (1→1) – α -D-glucopyranose

Question 102

Facilitated diffusion through a biological membrane is:

• A. Not important for biological systems consuming energy
• B. driven by a difference of solute concentration.
• C. generally irreversible.
• D. driven by ATP hydrolysis.
• E. not specific with respect to the substrate

Answer 102

B. driven by a difference of solute concentration.

Question 103

The specificity of an enzyme catalyzed reaction:

• A. is determined only by its intrinsic K_m for a specific substrate
• B. is determined by only by k_cat for a specific substrate
• C. is measured in units of M^-1 S^-1
• D. is only relevant at high substrate concentration when substrates must compete for enzyme
• E. is a property of only of allosterically regulated enzymes

Answer 103

C. is measured in units of M^-1 S^-1

Question 104

Substrate level phosphorylation:

• A. describes the conversion of ADP into ATP with the addition of inorganic phosphate every place throughout the cell.
• B. describes the formation of ADP by phosphoryl group transfer from 1,3 bisphosphoglycerate.
• C. is the formation of ATP by phosphoryl group transfer from a higher energy compound.
• D. requires an enzyme that is catalyically perfect to function efficiently
• E. is always reversible under the conditions operating within the cell

Answer 104

C. is the formation of ATP by phosphoryl group transfer from a higher energy compound.

Question 105

Mutases are:

• A. enzymes that regulate glycolysis by changing the coding sequence of DNA
• B. proteins that regulate glycolysis by binding to glycolytic enzymes and mutating their function
• C. isomerases that catalyze the transfer of phosphoryl groups from one part of a substrate molecule to another.
• D. Isomerases that catalyze the transfer of phosphoryl groups from one molecule to another, “mutating” it.
• E. It will fold into a beta-pleated sheet.

Answer 105

C. isomerases that catalyze the transfer of phosphoryl groups from one part of a substrate molecule to another.

Question 106

The biosynthesis of lactose (β-D-galactopyranosyl (1→4) – α -D-glucopyranose) couples which two monomers?

• A. UDP-Glucose and Galactose
• B. UDP-Glucose and UDP-Galactose
• C. UDP-Galactose and Glucose
• D. Glucose and Galactose
• E. N-acetyl galactosamine and Glucose

Answer 106

C. UDP-Galactose and Glucose

Question 107

The Entner-Doudoroff pathway evolved earlier in evolution than the EMP pathway. One difference between the two pathways is:

• A. The ED pathway does not generate NADH reducing equivalents.
• B. The ED pathway results in a net of 3 ATP molecules per molecule of glucose while the EMP pathway results in a net of 4 ATP per molecule of glucose
• C. The ED pathway is used by organisms that perform gluconeogenesis but lack phosphofructokinase-1
• D. The ED pathway is the exact reverse of gluconeogenesis but the EMP pathway is regulated with bypass reactions.
• E. The reducing equivalents of the ED pathway are transferred to FAD while the reducing equivalents of the EMP pathway are transferred to NAD+ .

Answer 107

C. The ED pathway is used by organisms that perform gluconeogenesis but lack phosphofructokinase-1

Question 108

The hydrolysis of ATP has a large negative ∆G’o, nevertheless the molecule is stable in solution. This stability is due to:

• A. resonance stabilization
• B. entropy stabilization
• C. the hydrolysis reaction having a large activation energy
• D. ionization of the phosphates
• E. the hydrolysis reaction being endergonic.

Answer 108

C. the hydrolysis reaction having a large activation energy

Question 109

A protein that non-covalently binds to a carbohydrate is called a

• A. glycation
• B. lectin
• C. NS domain
• D. glycoprotein
• E. glycan

Answer 109

B. lectin

Question 110

In many biochemical reactions, the phosphate group acts as a:

• A. radical
• B. electrophile
• C. oxidizing agent
• D. reducing agent
• E. chromaphore

Answer 110

B. electrophile

Question 111

Fructose-2,6-bisphosphate:

• A. is an allosteric inhibitor of Phosphofructokinase-1
• B. is an allosteric enhancer (activator) of Phosphofructokinase -1 and an allosteric inhibitor of fructose-1,6- bisphosphatase-1
• C. is an allosteric enhancer (activator) of fructose-1,6-bisphosphatase and an allosteric inhibitor of phosphofructokinase -1
• D. is an allosteric inhibitor of Phosphofructokinase-2
• E. is an allosteric enhancer (activator) of Phosphofructokinase-2.

Answer 111

B. is an allosteric enhancer (activator) of Phosphofructokinase -1 and an allosteric inhibitor of fructose-1,6- bisphosphatase-1

Question 112

Phosphofructokinase-2 and fructose-2,6-bisphosphatase

• A. are both positively regulated by insulin and negatively regulated by glucagon
• B. are both positively regulated by glucagon and negatively regulated by insulin
• C. are two distinct enzyme activities present on the same protein chain
• D. form a futile cycle
• E. are both directly regulated by pyruvate kinase

Answer 112

C. are two distinct enzyme activities present on the same protein chain

Question 113

The hydrolysis of phosphoenolpyruvate (PEP) proceeds with a ∆G’o of about – 62 kJ/mol. The greatest contributing factors to this reaction are the destabilization of the reactants by electrostatic repulsion and stabilization of the product pyruvate by:

• A. ionization
• B. electrostatic attraction
• C. resonance
• D. tautomerization
• E. circularization

Answer 113

D. tautomerization

Question 114

Amylose and amylopectin are both storage carbohydrate polymers found in plants. A difference between the two polymers is:

• A. the monomer of amylose is glucose and the monomer of amylopectin is galactose
• B. amylose has an α(1→4) connection while amylopectin has a β(1→4) glycosidic bond.
• C. amylose is branched but amylopectin is helical
• D. amylose forms a linear polymer but amylopectin is branched
• E. amylose is composed entirely of carbohydrate while amylopectin has a protein at its core

Answer 114

D. amylose forms a linear polymer but amylopectin is branched

Question 115

Calculate (approximately) the free energy of formation (synthesis) of ATP from ADP and Pi. Assume the following ∆G’o (hydrolysis) = -30 kJ/mol, [ATP] = 2.25 mM, [ADP] = 0.25 mM, [AMP] = 0.02 mM, [Pi] = 1.65 mM. Assume that T = 310 K and that R = 8.315 J/mol.

• A. -52 kJ/mol
• B. 8.32 kJ/mol
• C. 52 kJ/mol
• D. -8.32 kJ/mol
• E. Not enough information is given to solve this problem

Answer 115

C. 52 kJ/mol

Question 116

When pyruvate is the starting substrate for gluconeogenesis, reducing equivalents are transported out of the mitochondria in the form of:

• A. NADH
• B. oxaloacetate
• C. ATP
• D. FADH
• E. malate

Answer 116

E. malate

Question 117

In glycolysis, fructose 1.6-bisphosphate is converted to two products with a standard free energy change (∆G’o) of +23.8 kJ/mol. Under what conditions encountered in a normal (living) cell will the reaction proceed spontaneously (towards the right)?

• A. Under standard conditions, enough energy is released to drive the reaction to the right.
• B. The reaction will NEVER go to the right because ∆G’o is large and positive.
• C. When there is a high concentration of products with respect to fructose 1,6-bisphosphate, the reaction will proceed spontaneously.
• D. When there is a high concentration of fructose 1,6-bisphosphate the reaction will proceed spontaneously.
• E. ∆G’o isn’t relevant because the reaction is catalyzed inside of a living cell and the enzyme shifts the equilibrium to the right, irrespective of ∆G’o

Answer 117

D. When there is a high concentration of fructose 1,6-bisphosphate the reaction will proceed spontaneously

Question 118

Flavin adenine dinucleotide (FAD) differs from Nicotinic adenine dinucleotide (NAD+ ) in that (only one answer is correct):

• A. FAD is neutral and NAD+ is positively charged.
• B. FAD is a reducing agent and NAD+ is an oxidizing agent.
• C. FAD is capable of undergoing one or two electron reduction while NAD++ can only undergo a two-electron reduction.
• D. FAD is used exclusively for catabolism and NAD+ is used exclusively for anabolism.
• E. NAD+ generally remains tightly bound to enzymes while FAD can diffuse freely between different enzymes

Answer 118

C. FAD is capable of undergoing one or two electron reduction while NAD++ can only undergo a two-electron reduction.

Question 119

The mitochondrial membrane contains a transporter for

• A. NADH
• B. Acetyl CoA
• C. GTP
• D. ATP
• E. NADPH

Answer 119

D. ATP

Question 120

Mitchell’s chemiosmotic hypothesis involves all the following except

• A. a membrane impermeable to protons.
• B. Electron transport by the respiratory chain pumps protons out of the mitochondrion.
• C. Proton flow into the mitochondrion depends on the presence of ADP and Pi .
• D. Activity of the ATPase is reversible.
• E. Only proton transport is strictly regulated; other positively charged ions can diffuse freely across the mitochondrial membrane

Answer 120

E. Only proton transport is strictly regulated; other positively charged ions can diffuse freely across the mitochondrial membrane

Question 121

Approximately how many ATP molecules are formed for each mitochondrial NADH that is oxidized to NAD+ by O2 via the electron transport chain?

• A. 1.0
• B. 1.5
• C. 2.0
• D. 2.5
• E. 3.0

Answer 121

D. 2.5

Question 122

Almost all the O2 one consumes in breathing is converted to

• A. acetyl CoA
• B. CO2
• C. CO and then CO2
• D. H2O.
• E. none of the above

Answer 122

B. CO2

Question 123

Uncoupling of mitochondrial oxidative phosphorylation

• A. allows continued mitochondrial ATP formation but halts O2 consumption.
• B. halts all mitochondrial metabolism.
• C. halts mitochondrial ATP formation but allows continued O2 consumption.
• D. slows down the citric acid cycle.
• E. slows the conversion of glucose to pyruvate.

Answer 123

C. halts mitochondrial ATP formation but allows continued O2 consumption.

Question 124

Which of the following statements about cyclic photophodphorylation is correct?

• A. It involves NADPH formation.
• B. It uses electrons supplied by photosystem II.
• C. It is activated when NADP+ is limiting phosphorylation.
• D. It does not generate O2.
• E. It involves a substrate-level

Answer 124

D. It does not generate O2.

Question 125

The observation that the incubation of photosynthetic algae with 14CO2 in the light for a very brief time (5 seconds) led to the formation of 14C labeled 3-phosphoglycerate suggested that 14CO2 was condensing with a two-carbon acceptor. The acceptor was

• A. acetyl CoA
• B. acetyl phosphate
• C. glycol phosphate
• D. glycolic acid phosphate
• E. none of the above

Answer 125

E. none of the above

Question 126

The ratio of ATP to ADP controls the rate of

• A. glycolysis
• B. oxidative phosphorylation
• C. pyruvate oxidation
• D. the citric acid cycle
• E. all of the above

Answer 126

E. all of the above

Question 127

If malonyl CoA is synthesized from radioactive (14C labeled) CO2 and unlabeled acetyl CoA, and the labeled malonate is then used for fatty acid biosynthesis, the final product will have 14C in

• A. every carbon.
• B. every odd-numbered carbon.
• C. every even-numbered carbon.
• D. only the farthest carbon from C1.
• E. no part of the fatty acid molecule.

Answer 127

E. no part of the fatty acid molecule.

Question 128

The rate limiting step in fatty acid synthesis is

• A. formation of acetyl CoA from acetate.
• B. the reaction catalyzed by acetyl CoA carboxylase.
• C. condensation of acetyl CoA and malonyl CoA.
• D. the reduction of the acetoacetyl group to a -hydroxybutyryl group.
• E. the formation of malonyl CoA from malonate and coenzyme A

Answer 128

B. the reaction catalyzed by acetyl CoA carboxylase.

Question 129

Which of the following is synthesized by plants and not by humans?

• A. palmitate (C_16:0)
• B. stearate (C_18:0)
• C. linoleate (C_18:2 ^∆9,12)
• D. pyruvate
• E. phosphatidyl choline

Answer 129

C. linoleate (C_18:2 ^∆9,12)

Question 130

Which of the following conversions requires more than one step?

• 1. alanine to pyruvate
• 2. glutamate to α-ketoglutarate
• 3. aspartate to oxaloacetate
• 4. proline to glutamate
• 5. phenylalanine to succinate
• A. 1, 2 & 4
• B. 1, 3 &5
• C. 2, 4 & 5
• D. 4 & 5
• E. 1 & 4

Answer 130

D. 4 & 5

Question 131

Which of the following compounds is not involved in the production of urea from NH4 + via the urea cycle?

• A. aspartate
• B. ATP
• C. ornithine
• D. malate
• E. carbamoyl phosphate

Answer 131

D. malate

Question 132

If a person’s urine contains an unusually high concentration of urea, which of the following diets has been consumed?

• A. very high carbohydrate, very low protein
• B. very low carbohydrate, very high protein
• C. very high fat, very low protein
• D. very high fat, high carbohydrate, no protein
• E. protein only

Answer 132

E. protein only

Question 133

Which of the following enzymes is not involved in the assimilation of inorganic nitrogen into an organic molecule?

• A. glutamine synthetase
• B. nitrogenase
• C. arginase
• D. glutamate dehydrogenase
• E. glutamate synthase

Answer 133

C. arginase

Question 134

The enzyme ubiquinone:cytochrome C oxidoreductase catalyzes the reactions shown below in which ·Q - represents the free radical form of coenzyme Q. Which of the following is the principle role in electron transport played by these reactions?

• A. They circumvent the possible poisoning of electron transport by reactive oxygen species.
• B. Because coenzyme Q is a small molecule and diffuses freely within the mitochondrial membrane, it speeds electron transport.
• C. They convert electron pairs to single electron transfers because the cytochromes and other downstream carriers can only handle one electron at a time.
• D. They prevent leakage of reducing power from the mitochondrion.
• E. None of the abone is a significant role.

Answer 134

C. They convert electron pairs to single electron transfers because the cytochromes and other downstream carriers can only handle one electron at a time

Question 135

The oxidation of NADH in mitochondria is given by the reaction NADH + H+ + ½ O2 → NAD + + H2O. The standard free energy for this reaction is ∆G°’ = -220 kJ/mol NADH oxidized. The ratio of NAD+ to NADH is normally kept very low by the cell. Imagine a mutation which raises the ratio. What effect would this have on the actual free energy of the reaction?

• A. The actual free energy would become more negative.
• B. The actual free energy would become less negative.
• C. The reaction would run from right to left.
• D. The actual free energy would remain unchanged
• E. Cannot tell from the data given.

Answer 135

B. The actual free energy would become less negative

Question 136

Which of the following statements is not true of the pathway by which purine nucleotides are synthesized?

• A. The amino acid glycine is one of the precursors.
• B. Deoxyribonucleotides are formed from 5-phosphodeoxyribosyl 1-pyrophosphate.
• C. Inosinate is the purine nucleotide that is the precursor of both adenylate and guanylate.
• D. CO2 is required for one of the steps in this pathway.
• E. AMP and GMP are inhibitors

Answer 136

B. Deoxyribonucleotides are formed from 5-phosphodeoxyribosyl 1-pyrophosphate.

Question 137

Deoxyribonucleotides

• A. cannot be synthesized so they must be supplied preformed in the diet.
• B. are synthesized de novo using dPRPP.
• C. are synthesized from ribonucleotides by an enzyme system involving, the reducing agent, thioredoxin.
• D. are synthesized from ribonucleotides by nucleotide kinases.
• E. can be formed only by salvaging free bases.

Answer 137

C. are synthesized from ribonucleotides by an enzyme system involving, the reducing agent, thioredoxin.

Question 138

Precursors for the biosynthesis of the pyrimidine ring system include:

• A. glutamate, NH3, and CO2.
• B. glycine, glutamine, CO2, and aspartate.
• C. glycine and succinyl-CoA.
• D. carbamoyl phosphate and aspartate.

Answer 138

D. carbamoyl phosphate and aspartate.

Question 139

The principal committed step in the biosynthesis of purine nucleotides is

• A. conversion of 5-phosphoribosyl-1-pyrophosphate to 5-phospho--D-ribosylamine.
• B. conversion of 5-phospho--D-ribosylamine to glycinamide ribonucleotide.
• C. the inosine monophosphate synthase reaction.
• D. the IMP dehydrogenase reaction
• E. none of the above

Answer 139

A. conversion of 5-phosphoribosyl-1-pyrophosphate to 5-phospho--D-ribosylamine.

Question 140

E. coli cells were grown for many generations in a medium containing only heavy nitrogen, 15N, so that all the DNA formed a single band when centrifuged in a CsCl (cesium chloride) density gradient. After transfer to a medium containing only light nitrogen, 14N, isolation of DNA after one generation will yield

• A. DNA molecules both of whose strands are uniformly labeled with 14N.
• B. DNA molecules whose strands are each randomly labeled with both isotopes.
• C. DNA molecules both of whose strands are uniformly labeled with 15N.
• D. DNA molecules one of whose strands is labeled with 15N and the other with 14N.
• E. none of the above

Answer 140

D. DNA molecules one of whose strands is labeled with 15N and the other with 14N.

Question 141

Which of the following contributes to the accuracy of DNA replication?

• A. catalysis of nucleotide addition at the 3’-hydroxyl
• B. Watson-Crick base pairing
• C. the shape of the polymerase active site
• D. the 3’ to 5’ exonuclease active site
• E. more than one of the above

Answer 141

E. more than one of the above

Question 142

What is the function of dam methylase?

• A. to act as a blockage of methylation, which would otherwise flood the DNA with methyls
• B. to methylate GATC sequences at their 5’ ends
• C. it is what an experimenter says when her methylase doesn’t work
• D. to methylate the sliding clamp
• E. none of the above

Answer 142

B. to methylate GATC sequences at their 5’ ends

2016: None of the above

methylates DNA strands of some bacteria during DNA mismatch repair (an enzyme that adds a methyl group to the adenine of the sequence 5-GATC-3’ in newly synthesized DNA

Question 143

Which of the following does not provide a direct carbon skeleton for the synthesis of amino acids?

• A. Succinate
• B. α-ketoglutarate
• C. pyruvate
• D. oxaloacetate
• E. ribose 5-phosphate

Answer 143

A. Succinate

Question 144

Which of the following has the (immediate) effect of increasing the rate of glycogen breakdown

• A. Increased concentration of cAMP
• B. Increase in the [AMP]/[ATP] ratio
• C. Increased secretion of glucagon
• D. A and C
• E. A, B and C

Answer 144

C. Increased secretion of glucagon

Question 145

Shown below are the structures of three metabolic intermediates that can be used in amino acid synthesis. Which is correctly paired with its amino acid end product(s)?

• A. #1; serine, glycine, cysteine
• B. #2; alanine, valine, leucine
• C. #1; glutamate, glutamine, proline
• D. #1; histidine
• E. #3; methionine, threonine, lysine

Answer 145

C. #1; glutamate, glutamine, proline

Question 146

Which of the following is not a physiological role of nucleotides?

• A. Allosteric regulators
• B. Intermediates for biosynthetic processes
• C. Components of many proteins
• D. Components of the coenzymes NAD, FAD, and CoA
• E. Intracellular signaling molecules

Answer 146

C. Components of many proteins

Question 147

In the synthesis of lagging strand DNA, polymerization of the Okazaki fragment near the replication fork stops when the RNA primer of the previous strand is reached. The RNA is then removed by which of the following processes?

• A. kinetic proofreading
• B. base excision repair
• C. nick translation
• D. ligation
• E. none of the above

Answer 147

C. nick translation

Question 148

Given that plants can produce ATP and NADPH from photophosphorylation, for what purpose(s) do they reduce CO2 to glucose?

• A. At night, plants need the ATP produced by glycolysis and the citric acid cycle in the dark.
• B. Plants need glucose to produce starch and cellulose.
• C. Plants need glucose as a precursor of components of nucleic acids, lipids, and proteins.
• D. B and C
• E. A, B and C

Answer 148

E. A, B and C

Question 149

Which of the following statements about enzymes which interact with DNA is true?

• A. E. coli polymerase I is unusual in that it possesses only a 5’ to 3’ exonuclease activity.
• B. Endonucleases degrade circular but not linear DNA molecules.
• C. Exonucleases degrade DNA at a free end.
• D. Many DNA polymerases have a proofreading 5’ to 3’ exonuclease activity.
• E. Primases synthesize a short stretch of DNA to prime further synthesis.

Answer 149

C. Exonucleases degrade DNA at a free end.

Question 150

In what order do the following enzymes act in the synthesis of Okazaki fragments?

• A. helicase, primase, polymerase III, polymerase I, ligase
• B. ligase, polymerase I, polymerase III, primase, helicase
• C. primase, polymerase I, helicase, polymerase III, ligase
• D. polymerase I, polymerase III, helicase, ligase, primase
• E. none of the above.

Answer 150

A. helicase, primase, polymerase III, polymerase I, ligase

Question 151

Why is the origin of replication in E. coli (ori) relatively poor in GC base pairs?

• A. The replicative apparatus has trouble fitting GC base pairs into its active site until after it has got started.
• B. GC base pairs occur relatively less often than AT base pairs, so a lower occurrence at ori is not surprising.
• C. Deamination of C makes GC base pairs less stable
• D. Being poor in GC means being relatively enriched in AT base pairs, which are easier for the replicative apparatus to open up in order to start replication.
• E. more than one of the above

Answer 151

D. Being poor in GC means being relatively enriched in AT base pairs, which are easier for the replicative apparatus to open up in order to start replication.

Question 152

Sugar cane is a C4 plant. Which of the following is true for these (often) tropical plants?

• A. The affinity of rubisco for oxygen decreases with increasing temperature.
• B. “C4” refers to the four-carbon compounds by which CO2 is shuttled from mesophyll cells to bundle-sheath cells.
• C. The net result of the C4 strategy is to lower the Km of rubisco for CO2.
• D. C4 plants can outcompete C3 plants in hot environments because they have a different enzyme for carbon fixation.
• E. All of the above are true.

Answer 152

B. “C4” refers to the four-carbon compounds by which CO2 is shuttled from mesophyll cells to bundle-sheath cells.

Question 153

Which of the following amino acid combinations are involved in holding essential magnesium ions in place in DNA polymerase I?

• A. glu and gly
• B. lys and asp
• C. arg and his
• D. asp and asp
• E. none of these

Answer 153

D. asp and asp

Question 154

Which compound can serve as a direct acceptor of an additional amino group derived from amino acid catabolism?

• A. Glutamine
• B. Asparagine
• C. α-Ketoglutarate
• D. Fumarate
• E. Glycerol

Answer 154

C. α-Ketoglutarate

Question 155

Which of the following was a clue that DNA polymerase I is involved primarily in DNA repair?

• A. An E. coli strain lacking Pol I was viable but sensitive to DNA damage.
• B. It falls of the DNA too often (low processivity) to be the primary replicator.
• C. Its rate of catalysis is too slow to be the primary replicator.
• D. more than one of these
• E. none of these

Answer 155

A. An E. coli strain lacking Pol I was viable but sensitive to DNA damage.

Question 156

Not taking into account the NADH generated in the malate dehydrogenase reaction, how many high-energy phosphate bonds are used to form a molecule of urea, starting from ammonia and HCO3 - ?

• A. 1
• B. 2
• C. 3
• D. 4
• E.0

Answer 156

D. 4

Question 157

The type of enzyme known as a phosphoribosyltransferase is involved in all of the following except

• A. salvage of pyrimidine bases
• B. the de novo synthesis of pyrimidine nucleotides.
• C. the de novo synthesis of purine nucleotides
• D. salvage of purine bases.

Answer 157

C. the de novo synthesis of purine nucleotides

Question 158

Deoxyribonucleotides

• A. cannot be synthesized so they must be supplied preformed in the diet.
• B. are synthesized de novo using PRPP.
• C. are synthesized from ribonucleotides by an enzyme system involving thioredoxin.
• D. are synthesized from ribonucleotides by nucleotide kinases.
• E. can be formed only by salvaging free bases.

Answer 158

C. are synthesized from ribonucleotides by an enzyme system involving thioredoxin.

Question 159

Hereditary orotic aciduria is characterized by severe anemia, growth retardation, and high levels of orotic acid excretion. The effect may be in orotate phosphoribosyl transferase, orotidine decarboxyase or both. The preferred treatment for this disease is dietary uridine, which reverses the anemia and decreases the formation of orotic acid. Dihydro orotate dehydrogenase is a mitochondrial enzyme. In the de novo synthesis of pyrimidine nucleotiedes

• A. reactions take place exclusively in the cytosol
• B. a free base is formed as an intermediate
• C. PRPP is required in the rate-limiting step.
• D. UMP and CMP are formed from a common intermediate
• E. UMP inhibition of OMP-decarboxylase is the major control of the process

Answer 159

B. a free base is formed as an intermediate

Question 160

The best estimate of the turnover of DNA comes from a measurement in urine of

• A. uric acid
• B. NH4 + and CO2
• C. β-alanine
• D. β-aminoisobutyrate
• E. cytidine

Answer 160

D. β-aminoisobutyrate

Question 161

Aminotransferases

• A. usually require α-ketoglutarate or glutamine as one of the reacting pair.
• B. catalyze reactions that result in a net use or production of amino acids.
• C. catalyze irreversible reactions.
• D. require pyridoxal phosphate as an essential cofactor for the reaction.
• E. are not able to catalyze transamination reactions with essential amino acids.

Answer 161

D. require pyridoxal phosphate as an essential cofactor for the reaction

Question 162

S-Adenosylmethionine

• A. contains a charged sulfur that carries the methyl group to be transferred.
• B. yields α-ketoglutarate in the reaction in which the methyl is transferred
• C. donates a methyl group in a freely reversible reaction.
• D. generates H2S by transsulfuration.
• E. provides the carbons for the formation of cysteine.

Answer 162

A. contains a charged sulfur that carries the methyl group to be transfer

Question 163

Defects in the metabolism of the branched-chain amino acids are rare but serious. The most common one is called maple syrup urine disease (named from the smell of the urine), which is a deficiency of the branchedchain keto acid dehydrogenase complex. The disease is characterized by mental retardation and a short life span. All of the following are true about the branched-chain amino acids except that they

• A. are essential in the diet.
• B. differ in that one is glucogenic, one is ketogenic and one is classified as both.
• C. are catabolized in a manner that bears a resemblance to β-oxidation of fatty acids.
• D. are oxidized by a dehydrogenase complex to branched-chain acyl CoAs that are a carbon shorter than the parent compound.
• E. are metabolized initially in the liver.

Answer 163

E. are metabolized initially in the liver.

Question 164

In the formation of urea from ammonia, all of the following are correct except

• A. aspartate supplies one of the nitrogens found in urea.
• B. this is an energy-expensive process, utilizing several ATPs.
• C. the rate of the cycle fluctuates with the diet.
• D. fumarate is produced.
• E. ornithine transcarbomoylase catalyze the rate-limiting step

Answer 164

E. ornithine transcarbomoylase catalyze the rate-limiting step

Question 165

Nonessential amino acids

• A. are amino acids other than those required for protein synthesis.
• B. are not utilized in mammalian proteins.
• C. are synthesized by plants and bacteria, but not by humans.
• D. can be synthesized in humans.
• E. may be substituted with other amino acids in proteins.

Answer 165

D. can be synthesized in humans

Question 166

The amino acids serine, alanine, and cysteine can be catabolized to yield:

• A. fumarate
• B. pyruvate
• C. succinate
• D. α-ketoglutarate
• E. none of the above

Answer 166

B. pyruvate

Question 167

What happens at a Holliday junction?

• A. Site specific recombination takes place.
• B. Strand exchange takes place.
• C. Vacationers meet.
• D. Gene inversion can occur.
• E. A, B and D

Answer 167

E. A, B and D

Question 168

What is the function of the clamp in DNA replication?

• A. It keeps DNA polymerase III from falling off the DNA too frequently.
• B. It recruits single stranded DNA binding proteins to stabilize the separated strands.
• C. It keeps the RNA primer in place, as the primer is so small it tends to dissociate.
• D. It opens up the double stranded DNA.
• E. None of these.

Answer 168

A. It keeps DNA polymerase III from falling off the DNA too frequently.

Question 169

What is the role of ATP in the final step of ligation of Okazaki fragments?

• A. It drives replacement of the RNA primers.
• B. It inactivates potentially damaging nucleases.
• C. It activates the 5’-phosphate of the fragment’s first deoxynucleotide.
• D. It activates the 3’hydroxyl of the fragment’s first nucleotide.
• E. ATP has no direct role in ligation

Answer 169

C. It activates the 5’-phosphate of the fragment’s first deoxynucleotide.

Question 170

In methyl-directed mismatch repair of DNA

• A. All bases from GATC to a point just beyond the mismatch are removed by an exonuclease.
• B. Polymerase III replaces the bases, correcting the error.
• C. Ligase seals the nick
• D. The repair apparatus must wait for methylation to be completed.
• E. more than one of the above.

Answer 170

E. more than one of the above.

Question 171

What is the effect of methylating O6 of a guanine base on DNA replication?

• A. It replaces a GC base pair with a CG pair.
• B. The third H-bond of a normal GC pair cannot form, so the G will pair with T instead, leading to replacement of GC with AT.
• C. The strand with the unpaired C will be replicated normally.
• D. There will be no effect at all.
• E. B and C

Answer 171

E. B and C

Question 172

23. The coding strand of the gene encoding the E. coli enzyme β-galactosidase begins with the sequence 5’-ATGACCATGATTACG. What is the sequence of the RNA transcript specified by this part of the gene?

• A. 5’-AUGACCAUGAUUACG
• B. 5’-GCAUUAGUACCAGUA
• C. 5’-UACUGGUACUAAUGC
• D. 5’-CGUAAUCAUGGUCAU these
• E. none of

Answer 172

D. 5’-CGUAAUCAUGGUCAU these

Question 173

Splicing mRNA in eukaryotes

• A. Removes the introns
• B. Removes the exons
• C. Always involves folding of the mRNA
• D. Always involves proteins
• E. none of the above

Answer 173

A. Removes the introns

Question 174

In the presence of σ (sigma) factor

• A. RNA polymerase binds strongly to promoters but not to other regions of DNA.
• B. RNA polymerase does not bind to DNA at all.
• C. RNA polymerase stalls at the initiation site and cannot proceed to make m-RNA
• D. RNA polymerase binds weakly to both promoters and non-promoter DNA.
• E. RNA polymerase binds strongly to both promoters and non-promoter DNA.

Answer 174

A. RNA polymerase binds strongly to promoters but not to other regions of DNA.

Question 175

What is the role of the TATA box in RNA biosynthesis?

• A. It serves as a termination sequence to end transcription.
• B. It causes RNA polymerase to shift from the open to the closed complex.
• C. It ends initiation leading to the elongation phase of transcription.
• D. It is the site of assembly of the pre-initiation complex.
• E. more than one of the above

Answer 175

D. It is the site of assembly of the pre-initiation complex.

Question 176

What is the role of guanosine monophosphate in type I intron splicing?

• A. It attacks the 5’phosphate of the adenine at a U-A junction between an exon and intron.
• B. It forms a U to G phosphodiester at the end of the exon in preparation for joining to the exon beyond the intron.
• C. Its phosphate provides the energy to drive the reaction.
• D. It assists in ligation at the splice site.
• E. It has no role in intron splicing.

Answer 176

A. It attacks the 5’phosphate of the adenine at a U-A junction between an exon and intron.

Question 177

Which of the following involves an RNA lariat?

• A. removal of a group I intron
• B. removal of a group II intron
• C. intron removal by a splicosome
• D. capture of a wild horse….
• E. more than one of the above

Answer 177

E. more than one of the above

Question 178

Which of the following contributes to the high rate of virus mutation?

• A. The virus alters the cell’s DNA replication when it takes control.
• B. Reverse transcriptase has no proofreading exonuclease activity.
• C. The cDNA back-transcribed from the viral DNA is unstable.
• D. The virus inactivates the host cell’s DNA repair mechanisms.
• E. None of the above.

Answer 178

B. Reverse transcriptase has no proofreading exonuclease activity.

Question 179

What caused the death of Henry in the Pete Seeger song, Green and Yeller?

• A. His RNA polymerases stopped working
• B. His DNA polymerases stopped working.
• C. He ate snakes.
• D. He got lost in the woods?
• E. none of the above; he didn’t die

Answer 179

C. He ate snakes

Question 180

The known mechanisms of activation of rubisco or of other enzymes of the Calvin cycle during illumination include all of the following except

• A. increased stromal pH.
• B. relief of the inhibition by the “nocturnal inhibitor” 2-carboxyarabinitol 1- phosphate.
• C. light-driven entry of Mg2+ into the stroma.
• D. phosphorylation by cAMP-dependent protein kinase.
• E. reduction of a disulfide bridge by thioredoxin

Answer 180

D. phosphorylation by cAMP-dependent protein kinase.

Question 181

The spectrum of visible light that reaches the earth is shown below. Which of the pigments would be most efficient at harvesting light whose wavelength is approximately 450 nm?

• A. Chlorophyll a
• B. Chlorophyll b
• C. Phycoerythrin
• D. Lutein
• E. all these would be equally efficient

Answer 181

B. Chlorophyll b

Question 182

The carbon assimilation (“dark”) reactions of photosynthetic plants

• A. cannot occur in the light.
• B. are driven ultimately by the energy of sunlight.
• C. are important to plants, but ultimately of little significance for bacteria and animals.
• D. yield (reduced) NADH.
• E. yield ATP, which is required for light reactions.

Answer 182

B. are driven ultimately by the energy of sunlight.

Question 183

Transketolase requires the coenzyme

• A. pyridoxal phosphate.
• B. thiamine pyrophosphate.
• C. cobalamin (vitamin B12).
• D. tetrahydrofolic acid.
• E. none of these

Answer 183

B. thiamine pyrophosphate

Question 184

The conversion of glutamate to an α-ketoacid and NH4 +

• A. is catalyzed by glutamate dehydrogenase
• B. is a reductive deamination.
• C. is accompanied by ATP hydrolysis catalyzed by the same enzyme.
• D. requires ATP.
• E. does not require any cofactors.

Answer 184

A. is catalyzed by glutamate dehydrogenase

Question 185

Which of these amino acids are both ketogenic and glucogenic?

1. isoleucine
2. valine
3. histidine
4. arginine
5. tyrosine

• A. 1 and 5
• B. 2 and 4
• C. 2,3, and 4
• D. 1, 3, and 5
• E. 2, 4, and 5

Answer 185

A. 1 and 5

Question 186

Why do macromolecular solutes in cesium chloride density gradient centrifugation reach a place in the centrifuge tube and cease to move further toward the tube’s bottom?

• A. The statement is incorrect, as the solutes continue to move as long as the centrifuge continues to run.
• B. The centrifugal force produced by the spinning centrifuge balances exactly the buoyant force of the CsCl solution density.
• C. The macromolecule – DNA or protein – interacts directly with the CsCl.
• D. The CsCl forms a solid mass in the tube which blocks further movement of macromolecular solutes.
• E. none of the above

Answer 186

B. The centrifugal force produced by the spinning centrifuge balances exactly the buoyant force of the CsCl solution density.

Question 187

Glutamine is an N donor in the synthesis of

• A. inosinic acid (IMP)
• B. orotate
• C. UMP
• D. CTP
• E. all of the above

Answer 187

A. inosinic acid (IMP)

Question 188

What is/are the role(s) of the catalytic site magnesium ions in the catalytic activity of DNA polymerase?

• A. One aids in forming AT base pairs, while the other aids in forming GC base pairs.
• B. They stabilize the interactions of aspartic acid side chains with the growing DNA.
• C. One facilitates the attack of the 3’-hydroxyl of the growing DNA on the α–phosphate of the incoming NTP.
• D. One facilitates departure of pyrophosphate from the incoming NTP.
• E. more than one of the above

Answer 188

E. more than one of the above

Question 189

In the initiation of DNA replication in E. coli eight DnaA proteins, each with an ATP bound to it, bind to the DNA at the origin of replication. The effect of this is

• A. prevent a third replication fork from opening.
• B. strain the DNA, forcing its strands to separate at an AT rich region at the beginning of the origin.
• C. conserve ATP for use later in the process.
• D. prevent DnaB from falling off.
• E. none of the above

Answer 189

B. strain the DNA, forcing its strands to separate at an AT rich region at the beginning of the origin.

Question 190

What is/are the role(s) of topoisomerase in replication?

• A. It unwinds the double helix as the replication fork progresses.
• B. It facilitates helicase binding.
• C. It restores double helical structure after the fork passes by.
• D. It relieves supercoiling caused by the unwinding of the DNA double helix.
• E. more than one of the above

2016 Final options

• A. to unwind the double helix as the replication fork moves
• B. to relieve supercoiling stress as the fork moves along the DNA
• C. to stabilize the unwound DNA
• D. to stabilize the Okazaki fragments before they are ligated
• E. none of the above

Answer 190

D. It relieves supercoiling caused by the unwinding of the DNA double helix.

2016 Final option

B. to relieve supercoiling stress as the fork moves along the DNA

Question 191

Which of the following is true?

• A. When each new base is added to the leading strand, it is added at the 5’ end.
• B. New bases are added to the lagging strand until the new DNA overlaps the primer of the previous Okazaki fragment, displacing the RNA primer.
• C. When each base is added to the lagging strand, it is added at the 5’ end.
• D. One β-clamp is loaded onto the DNA for each complete strand.
• E. none of the above is true.

Answer 191

E. none of the above is true.

Question 192

How many high energy phosphate bonds does it take to replace one RNA primer nucleotide with the corresponding deoxyribonucleotide?

• A. none; it is a simple exchange
• B. one
• C. two
• D. three
• E. more than three

Answer 192

C. two

Question 193

How does the Ames test help identify potentially carcinogenic compounds?

• A. The compounds prevent utilization of histidine by the bacteria on the petri dish.
• B. The compounds prevent back mutation that would enable the bacteria on the plate to utilize histidine.
• C. The compounds promote back mutation that would enable the bacteria on the plate to utilize histidine.
• D. The compounds promote back mutation that would enable the bacteria on the plate to grow in the absence of histidine.
• E. none of the above

Answer 193

D. The compounds promote back mutation that would enable the bacteria on the plate to grow in the absence of histidine.

Question 194

Why, after replication, can the mismatch repair apparatus distinguish the base that needs to be replaced from the base opposite it on the other strand?

• A. The template strand is methylated at the A of a specific GATC sequence near the mismatch site, whereas the new strand is not methylated.
• B. The new strand is methylated at the A of a specific GATC sequence near the mismatch site , whereas the template strand is not methylated.
• C. The mismatch prevents the Dam methylase from methylating the mismatched base.
• D. Methylation of the new strand at the A of the GATC sequence activates the mismatch apparatus.
• . none of the above

Answer 194

A. The template strand is methylated at the A of a specific GATC sequence near the mismatch site, whereas the new strand is not methylated.

Question 195

In base excision repair

• A. only the mismatched base is removed and replaced.
• B. two endonuclease cuts remove, with the help of a helicase, a sequence of bases on either side of the mismatch. The sequence is then replaced and the ends ligated.
• C. The first step involves a glycosylase to cut a glycosidic bond.
• D. an endonuclease makes a single cut in the strand that has the mismatch.
• E. C and D

Answer 195

E. C and D

Question 196

Accidental methylation of a guanine base on the hydroxyl at position 6 on the ring, if uncorrected, would lead to

• A. replacement of a GC base pair with a CG base pair.
• B. replacement of a GC base pair by an AT base pair.
• C. loss of double helix stability due to loss of an H-bond between the strands.
• D. transfer of the methyl to the 1-position of an adenine.
• E. none of the above

Answer 196

B. replacement of a GC base pair by an AT base pair.

Question 197

Nucleotide sequences from five species listed in the first column are shown below.

• EC - T L T A A I T T V
• SS - T L V G R L L M D R G F I D E K T V K E A
• SC - T T T G H L I Y K C G G I D K R T I E K F
• HH - T M V G R L L Y E T G S V P E H V I E Q H
• HS - T T T G H L I Y K C G G I D R K T I E K F
• BS - T M V G R I T T V

Which of the five species is most closely related to HH?

• A. EC
• B. SS
• C. SC
• D. HS
• E. cannot tell from data given

In the question above, if one had to insert a gap to improve the alignment, what would be the first and last residues in the sequence of species HH that occurs instead of the gap region?

• A. L, V
• B. L, T
• C. R, I
• D. R, T
• E. cannot tell from data give

Answer 197

B. SS

A. L, V

Question 199

In the UPGMA method evolutionary distance between species can be estimated from

• A. the number of sites in a given aligned sequence that favor one tree over another.
• B. the number of nucleotide differences for a given aligned sequence between the species.
• C. the number of gaps seen after alignment in a given protein between the two species.
• D. the number of GC base pairs in a given aligned sequence between the species.
• E. none of the above

Answer 199

B. the number of nucleotide differences for a given aligned sequence between the species.

Question 200

Yeast can grow anaerobically, as it does in fermentation to make alcohol, or it can grow aerobically, as it does when it makes bread dough rise. When yeast are switched from aerobic conditions to anaerobic

• A. ATP production drops
• B. glucose utilization drops
• C. ATP production rises
• D. glucose consumption rises
• E. more than one of the above

Answer 200

D. glucose consumption rises

Question 201

For any particular enzyme, which of the following will have little or no effect on the number of enzyme molecules present at any one time?

• A. rate of degradation of the messenger RNA that directs synthesis of the protein
• B. rate of transcription of the gene for the enzyme into a messenger RNA
• C. binding of an allosteric effector to the enzyme
• D. metabolic degradation of an inhibitor of the enzyme
• E. All these govern the number of enzyme molecules present in the cell.

Answer 201

C. binding of an allosteric effector to the enzyme

Question 202

Which of the following would probably have a large effect on the flux of a metabolite through a pathway?

• A. An increase in [S] when [S] >> Km of an enzyme in the pathway.
• B. A decrease in [S] when [S] >> Km of an enzyme in the pathway.
• C. An increase in [S] when [S] << Km of an enzyme in the pathway.
• D. The forward and reverse rates of an enzyme in the pathway are nearly the same with forward slightly greater than reverse.
• E. none of the above

Answer 202

C. An increase in [S] when [S] << Km of an enzyme in the pathway.

Question 203

Under what conditions would you expect adenylate kinase activity to be high?

• A. when the ATP to ADP ratio is high
• B. when the concentration of AMP is low
• C. when the concentration of ADP is much less than that of AMP
• D. when the concentration of ADP is much greater than that of ATP
• E. none of these

Answer 203

D. when the concentration of ADP is much greater than that of ATP

Question 204

which of the following statements is true?

• A. Regulatory mechanisms maintain homeostasis.
• B. Control mechanisms regulate the flux through pathways while regulatory mechanisms generally govern homeostasis.
• C. Flux through a metabolic pathway increases when a product of the pathway falls below a certain level.
• D. Flux through a pathway depends on the quantities of the enzymes in the path as well as on the concentrations of allosteric effectors.
• E. All the above are true

Answer 204

E. All the above are true

Question 205

Which enzyme removes supercoiling in replicating DNA ahead of the replication fork?

• A. Gyrase
• B. DNA polymerase
• C. Helicase
• D. Topoisomerase
• E. A and D

Answer 205

E. A and D

Question 206

Which of the following reactions is required for proofreading during DNA replication by DNA polymerase III?

• A. 3’ - 5’ exonuclease activity
• B. 3’ - 5’ endonuclease activity
• C. 3’ - 5’ pyrophosphorolysis activity
• D. 5’ - 3’ exonuclease activity
• E. 5’ - 3’ endonuclease activity

Answer 206

A. 3’ - 5’ exonuclease activity

Question 207

A sequence of the Okazaki fragment undergoing replication on the lagging strand template reads 3’- CGCATGTAGCGA-5’. What is the code of a DNA that is in the template for synthesis of the complementary leading strand of this segment?

• A. 5’-TCGCTACATGCG-3’
• B. 3’-CGCATGTAGGGA-5’
• C. 3’-GCGTACATCGCT-5’
• D. 3’-AGCGATGTACGC-5’
• E. 5’-AGCGATGTACGC-3’

Answer 207

E. 5’-AGCGATGTACGC-3’

Question 208

Which association between complementary bases would require the most energy to break?

• A. A:T
• B. A:U
• C. G:C
• D. G:G
• E. C:C

Answer 208

C. G:C

Question 209

Which of the following are found in both DNA and RNA?

• A. Ribose, phosphate groups, and adenine
• B. Deoxyribose, phosphate groups, and guanine
• C. Phosphate groups, guanine, and cytosine
• D. Phosphate groups, guanine, and thymine
• E. Uracil, guanine, and thymine

Answer 209

C. Phosphate groups, guanine, and cytosine

Question 210

The genome of a typical bacterium contains ~5 x 106 base pairs and can be replicated in about 30 minutes. The human genome is ~600 times larger and if its replication would progress at the same rate as in the bacteria, the entire process would require ~12.5 days, yet the entire human genome can be replicated within several hours. How is this possible?

• A. Eukaryotic DNA is simpler to replicate than prokaryotic DNA.
• B. Human DNA polymerases work much faster than those of prokaryotes.
• C. The nucleosomes of eukaryotic DNA allow for faster DNA replication.
• D. Human DNA contains more origins of replication than prokaryotic DNA.
• E. Human DNA polymerases do not have exonuclease activity.

Answer 210

D. Human DNA contains more origins of replication than prokaryotic DNA.

Question 211

An altered DNA polymerase from bacteriophage T7 is suitable for Sanger sequencing, because:

• A. It strongly discriminates against ddNTPs and has a 3’ - 5’ exonuclease activity.
• B. Phenylalanine located in the active site of the enzyme forms a hydrogen bond with ddNTP allowing for incorporation of ddNTP into DNA.
• C. Tyrosine located in the active site of the enzyme forms a hydrogen bond with the incoming ddNTP allowing for incorporation of ddNTP into DNA, and the exonuclease domain is not active.
• D. It incorporates ddNTPs more readily than dNTPs.
• E. Incorporation of ddNTP.

Answer 211

C. Tyrosine located in the active site of the enzyme forms a hydrogen bond with the incoming ddNTP allowing for incorporation of ddNTP into DNA, and the exonuclease domain is not active.

Question 212

Imagine a life form in which the orientations of strands in the DNA double helix were parallel rather than antiparallel. This life form uses DNA polymerases with similar properties to those displayed by DNA polymerases from human animals on Earth, thus we would expect the following:

• A. DNA replication would be much slower than in eukaryotes from Earth.
• B. DNA replication would occur in the 3’ to 5’ direction on one strand and in a 5’ 3’ direction on the other.
• C. DNA replication would only occur toward one end of a replication bubble.
• D. The DNA polymerase would be unable to carry out proofreading.
• E. Okazaki fragments would have a different size on the lagging-strand.

Answer 212

C. DNA replication would only occur toward one end of a replication bubble.

Question 213

Glutamate is metabolically converted to α-ketoglutarate and NH4 + by a process described as:

• A. deamination.
• B. hydrolysis.
• C. oxidative deamination.
• D. reductive deamination.
• E. transanimation.

Answer 213

C. oxidative deamination.

Question 214

Serine or cysteine may enter the citric acid cycle as acetyl-CoA after conversion to:

• A. oxaloacetate
• B. propionate
• C. pyruvate
• D. succinate
• E. succinyl-CoA

Answer 214

C. pyruvate

Question 215

The enzymatic machinery to fix atmospheric N2 into NH4 + is:

• A. a means of producing ATP when excess N2 is available.
• B. composed of two key proteins, each containing iron.
• C. relatively stable when exposed to O2.
• D. specific to plant cells.
• E. unaffected by the supply of electrons

Answer 215

B. composed of two key proteins, each containing iron

Question 216

If glucose labeled with 14C at C-1 were the starting material for amino acid biosynthesis, the product(s) that would be readily formed is (are):

• A. serine labeled at the carboxyl carbon.
• B. serine labeled at the alpha carbon.
• C. serine labeled at the R-group carbon.
• D. all of the above.
• E. none of the above.

Answer 216

D. all of the above.

Question 217

If a cell were unable to synthesize or obtain tetrahydrofolic acid (H4 folate), it would probably be deficient in the biosynthesis of:

• A. isoleucine
• B. leucine
• C. lysine
• D. methionine
• E. serine

Answer 217

D. methionine

Question 218

Which one of the following statements is true of the biosynthetic pathway for purine nucleotides?

• A. CO2 does not participate in any of the steps in this pathway.
• B. deoxyribonucleotides are formed from 5-phosphodeoxyribosyl 1-pyrophosphate.
• C. inosinate is the purine nucleotide that is the precursor of both adenylate and guanylate.
• D. orotic acid is an essential precursor for purine nucleotides.
• E. The amino acid valine is one of the precursors contributing to purine nucleotides.

Answer 218

C. inosinate is the purine nucleotide that is the precursor of both adenylate and guanylate.

Question 219

Protein involved in DNA metabolism often form covalent intermediates with DNA substrates. Which of the following proteins DOES NOT form a covalent link with the DNA substrate?

• A. Cre recombinase
• B. UDG
• C. Ligase III
• D. A and B
• E. Methylotransferase

Answer 219

B. UDG

Question 220

In nucleotide excision repair, which of the following proteins recognizes thymine dimers in the genomic DNA?

• A. MutH
• B. UvrA and B
• C. MulL
• D. UvrC and D
• E. AlkB

Exam 2020

• A. RecA and B
• B. PARP1 and 2
• C. RuvB and C
• D. UvrC and D
• E. UvrA and B

Answer 220

B. UvrA and B

Question 221

A liver extract is often used in the Ames test:

• A. To inactivate a potential mutagen.
• B. Because some compounds are normally mutagenic to humans only after being metabolized in the body.
• C. To dilute the concentration of mutagen.
• D. Because a liver extract is a nutritious supplement for bacteria.
• E. Because compounds are normally mutagenic in human liver.

Answer 221

B. Because some compounds are normally mutagenic to humans only after being metabolized in the body.

Question 222

What is the most common type of DNA damage caused by ultraviolet (UV) light?

• A. Purine trimers
• B. Pyrimidine dimers
• C. Alkylated bases
• D. Apurinic sites
• E. Oxidized bases

Answer 222

B. Pyrimidine dimers

Question 223

T4 DNA ligase uses the following cofactor:

• A. NAD
• B. NMN
• C. AMN
• D. ATP
• E. A and D

Answer 223

D. ATP

Question 224

Which of the following proteins in NOT an enzyme:

• A. Lyase
• B. AlkB
• C. Methylotransferase
• D. A and B
• E. A and C

Answer 224

C. Methylotransferase

Question 226

During DNA replication in bacteria, the mismatch repair system differentiates between the daughter and the parent DNA strands by the recognition of:

• A. Methylated adenine in the sequence: 5’-GATC-3’ in the parent strand but not in the daughter strand.
• B. Methylated adenine in the sequence: 5’-GATC-3’ in the daughter strand but not in the parent strand.
• C. DNA polymerase bound to the parent strand but not to the daughter strand.
• D. A stretch of single-stranded DNA in the parent strand but not in the daughter strand.
• E. Methylated adenine in the sequence: 5’-CACT-3’ in the parent strand but not in the daughter strand

Answer 226

A. Methylated adenine in the sequence: 5’-GATC-3’ in the parent strand but not in the daughter strand.

Question 227

Which DNA repair pathway would remove the following lesions from DNA?

• A. NER (repairs 1 and 3), BER (repairs 2)
• B. MMR (1), BER (2), NER (3)
• C. BER (1) NER (2), Direct repair (3)
• D. NER (1,2,3)
• E. A and B

Answer 227

A. NER (repairs 1 and 3), BER (repairs 2)

Question 228

When RecBCD helicase/nuclease reaches the Chi sequence (5’‐GCTGGTGG-3’) in DNA the following happens:

• A. RecD binds tightly to Chi; 3’‐degradation by RecB nuclease slows down while 5’-degradation speeds up.
• B. RecC binds tightly to Chi; 3’‐degradation by RecB nuclease slows down while 5’-degradation speeds up.
• C. RecC binds tightly to Chi; 3’‐degradation by RecC nuclease slows down while 5’-degradation speeds up.
• D. RecC binds tightly to Chi; 3’‐degradation by RecB nuclease speeds up while 5’-degradation slows down.
• E. RecBCD facilitates loading of RecA onto ssDNA.

Answer 228

B. RecC binds tightly to Chi; 3’‐degradation by RecB nuclease slows down while 5’-degradation speeds up.

Question 229

E.coli photolyase:

• A. Uses N5 , N10‐methenyl‐tetrahydrofolylpolyglutamate (MTHFpolyGlu) to capture energy of a red-light photon.
• B. Recognizes the distortion in the double helix.
• C. Uses reduced flavin adenine dinucleotide (*FADH-) to donate an electron to pyrimidine dimers in DNA.
• D. B and C
• E. A, B and C

Answer 229

D. B and C

2020: A, B, C

Question 230

Double-strand breaks in DNA are repaired via the following molecular mechanism:

• A. Direct repair (DR) and error prone translesion synthesis (TLS)
• B. Homologous recombination (HR)
• C. Base excision repair (BER)
• D. Site specific recombination
• E. Nonhomologous end joining (NHEJ) and Mismatch Repair (MMR)

2020 answers

• A. Direct repair (DR) and error prone translesion synthesis (TLS)
• B. Canonical Nonhomologous end joining (NHEJ) and Mismatch Repair (MMR)
• C. Alternative Nonhomologous end joining (NHEJ) and Base Excision Repair (BER)
• D. Homologous recombination (HR)
• E. Nucleotide excision repair (NER)

Answer 230

B. Homologous recombination (HR)

Question 231

When BRCA1- or BRCA2-mutant breast cancers that initially respond to cisplatin therapy develop resistance to this drug, they often do so by back mutating their BRCA1/2 alleles to sequences that once again encode functional versions of these proteins. The above observation suggests that cancer cells use one of the DNA repair pathways for the removal of cisplatin lesions from DNA.

• A. Nonhomologous end joining (NHEJ)
• B. Error prone translesion synthesis (TLS)
• C. Mismatch Repair (MMR)
• D. Homologous recombination (HR)
• E. Base excision repair (BER)
• F. Nucleotide excision repair (NER)

Answer 231

D. Homologous recombination (HR)

Question 232

What is the speed at which E.coli replisome synthesizes duplex DNA?

• A. Aproximately 1.24e-6 mph
• B. Approximately 1.24e-3 mph
• C. 100 km/h
• D. I have no idea. I didn’t do the homework.
• E. It is probably B, but may well be A.

Answer 232

A. Aproximately 1.24e-6 mph

Question 233

Which of the following would probably have a large effect on the flux of a metabolite through a pathway?

• A. An increase in [S] when [S] >> Km of an enzyme in the pathway.
• B. A decrease in [S] when [S] >> Km of an enzyme in the pathway.
• C. An increase in [S] when [S] << Km of an enzyme in the pathway.
• D. The forward and reverse rates of an enzyme in the pathway are nearly the same with forward slightly greater than reverse.
• E. none of the above

Answer 233

C. An increase in [S] when [S] << Km of an enzyme in the pathway.

Question 234

The enzymatic machinery to fix atmospheric N2 into NH4 + is:

• A. a means of producing ATP when excess N2 is available.
• B. composed of two key proteins, each containing iron.
• C. relatively stable when exposed to O2.
• D. specific to plant cells.
• E. unaffected by the supply of electrons.

Answer 234

B. composed of two key proteins, each containing iron.

Question 235

The nitrogenase complex uses eight electrons per nitrogen molecule but reduction of one N2 to two ammonia molecules only requires six electrons. What happens to the other two electrons?

• A. Their energy is dissipated as heat.
• B. They restore the nitrogenase to its original state so it can process another nitrogen molecule.
• C. They reduce two protons to a molecule of hydrogen gas.
• D. They restore a cofactor to its original state for reuse.
• E. More than one of the above

Answer 235

C. They reduce two protons to a molecule of hydrogen gas.

Question 236

What is the role of leghemoglobin in nitrogen fixation?

• A. It prevents molecular oxygen from inactivating nitrogenase.
• B. It carries oxygen to the nitrogenase.
• C. It helps destroy inactive nitrogenase so the components can be recycled.
• D. It aids in the synthesis of the MoFe cofactor.
• E. more than one of the above

Answer 236

A. It prevents molecular oxygen from inactivating nitrogenase

Question 237

Glutamine synthetase is a key control point in nitrogen metabolism. At least eight nitrogenous end products can exert allosteric inhibition. If each of these were to be added to an assay one by one, the curve of activity versus glutamic acid concentration would shift

• A. to the left
• B. to the right
• C. to the left for some and the right for others
• D. The overall effect would be a cancellation with the curve left unchanged.
• E. none of the above

Answer 237

B. to the right

Question 238

Including pyruvate and TCA cycle intermediates, how many amino acids can be made by single, direct reactions?

• A. none
• B. one
• C. two
• D. three
• E. more than three

Answer 238

D. three

Question 239

With glutamate as a starting material, how many high energy phosphate bonds are used in making one molecule of urea?

• A. none
• B. two
• C. four
• D. six
• E. None of these is correct.

Answer 239

C. four

Question 240

What is the principal role of inosinate (IMP) in nitrogen metabolism?

• A. It can replace G in DNA repair.
• B. It is an intermediate in tryptophan biosynthesis.
• C. It is an intermediate in the breakdown of aromatic amino acids.
• D. It is at the branch for the biosynthesis of AMP and GMP.
• E. none of these

Answer 240

D. It is at the branch for the biosynthesis of AMP and GMP.

Question 241

What is the role of the arginino-succinate – aspartate shunt in the metabolism of nitrogenous compounds?

• A. It connects the urea and TCA cycles.
• B. It regulates the activity of homogentisate 1,2-dioxygenase.
• C. Although it drains oxaloacetate from the TCA cycle, it replaces the carbons lost with malate.
• D. It combines aspartate from OAA transamination in the mitochondrial matrix with citrulline from the urea cycle.
• E. all the above with one exception

Answer 241

E. all the above with one exception

Question 242

Porphobilinogen is an intermediate in the biosynthesis of heme. How many molecules of succinyl-CoA are required to make one molecule of porphobilinogen? (We didn’t cover this in class in any detail, but it is in the notes.)

• A. one
• B. two
• C. four
• D. eight
• E. none of these

Answer 242

B. two

Question 243

How many high energy phosphate bonds are required to make one molecule of GMP if the starting material is ribose 5-phosphate and proceeds via 5- phosphoribosyl-1-pyrophosphate? The involvement of histidine biosynthesis in production of AICAR may be ignored.

• A. none
• B. two
• C. seven
• D. nine
• E. none of these

Answer 243

D. nine

Question 244

In the conversion of ribonucleoside diphosphates (NDPs) to deoxyribonucleoside triphosphates (dNTPs) by ribonucleotide reductase, why would dATP stimulate production of dCTP and dTTP directly but not dGTP?

• A. dTTP, once produced, stimulates dGTP production, so dATP is not needed to do this.
• B. dCTP, once produced, stimulates dGTP production to maintain base pairing, so dATP is not needed.
• C. Since dGTP stimulates dATP production, having dATP stimulate dGTP production would create a wasteful loop.
• D. DNA breakdown supplies adequate dGMP, which is then phosphorylated twice to yield dGTP.
• E. A and C

Answer 244

E. A and C

Question 245

Disposal of ammonia from sewage and farm animal waste is expensive. The accidental discovery of anaerobic bacteria, the Planctomycetes, anaerobic bacteria that use nitrite as terminal electron acceptor, lowered the cost by 90%. These bacteria carry out what are now called anammox reactions, which produce a proton gradient to drive ATP production. Hydrazine (N2H4) is an intermediate. It is toxic, so unstable it is used as rocket fuel, and diffuses readily through ordinary phospholipid membranes. Why does hydrazine not escape before it can be used?

• A. It is bound to a carrier protein to prevent diffusion.
• B. The anammoxosome, which contains the biochemical apparatus, has a membrane of stacked unstable cyclobutane rings which is impenetrable to hydrazine.
• C. It does escape, but it is produces in such large quantities that enough remains to drive ATP production.
• D. It is used so rapidly that its escape by diffusion cannot occur.
• E. none of the above

Answer 245

B. The anammoxosome, which contains the biochemical apparatus, has a membrane of stacked unstable cyclobutane rings which is impenetrable to hydrazine.

Question 246

Which of the following compounds is the branch point for the synthesis of the aromatic amino acids?

• A. shikimic acid
• B. erythrose 4-phosphate
• C. anthranilic acid
• D. chorismic acid
• E. none of the above

Answer 246

D. chorismic acid

Question 247

How many amino acids are derived at least in part from glycolysis intermediates?

• A. none
• B. one
• C. five
• D. eight
• E. eleven

Answer 247

E. eleven

Question 248

Which of the following is NOT involved in the glutamine synthetase reaction?

• A. glutamate
• B. ATP
• C. gamma-glutamyl phosphate
• D. ammonium ion (NH4+)
• E. alpha-ketoglutarate

Answer 248

E. alpha-ketoglutarate

Question 249

In the nitrogenase reaction which of the following is NOT involved in the reaction?

• A. iron
• B. molybdenum
• C. manganese ions
• D. magnesium ions
• E. a 4-iron-4-sulfur cluster

Answer 249

C. manganese ions

Question 250

An altered DNA polymerase from bacteriophage T7 is suitable for Sanger sequencing, because:

• A. It strongly discriminates against dNTPs and has a 3’-5’ exonuclease activity.
• B. It incorporates ddNTPs more readily than dNTPs.
• C. Tyrosine located in the active site of the enzyme forms a hydrogen bond with the incoming ddNTP, allowing for incorporation of ddNTP into DNA, and its exonuclease domain is not active.
• D. It incorporates chain terminators into DNA.
• E. C and D

Answer 250

E. C and D

Question 251

Azvudine, favipiravir and remdesivir are examples of NRTIs that are currently being tested for the treatment of COVID-19. These compounds act by:

• A. Binding to the ACE-2 receptor, thus preventing the entry of SARS-CoV 2.
• B. Binding to the hydrophobic pocket located 10 Angstroms away from SARSCoV 2 RT’s active site, thus blocking the synthesis of viral DNA.
• C. Enzymatic degradation of the viral RNA.
• D. Chain termination of the viral reverse transcription.
• E. B and D

Answer 251

E. B and D

Question 252

The genome of E.coli contains ~5 x 106 base pairs and can be replicated in about 30 minutes. The human genome is ~600 times larger and if its replication would progress at the same rate as in the bacteria, the entire process would require ~12.5 days, yet the entire human genome can be replicated within several hours. How is this possible?

• A. Eukaryotic DNA is simpler to replicate than prokaryotic DNA, because it contains fewer mis-incorporated nucleotides
• B. Human DNA polymerases synthesize DNA with the rate approximately 500 times higher than DNA polymerases in prokaryotes.
• C. Human DNA contains more origins of replication than prokaryotic DNA.
• D. Ori C sequences are longer in humans than in E.coli.
• E. C and D

Answer 252

C. Human DNA contains more origins of replication than prokaryotic DNA.