Molecular Biology Principles And Practice 2nd ed by Micheal Cox - Test Bank

Molecular Biology Principles And Practice 2nd ed by Micheal Cox – Test Bank

 

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Molecular Biology- Principles and Practice 2e

Cox et. al

Chapter 5

 

1. Induced fit can be best described as the result of:

 

1. mutations that increase ligand binding.
2. a decrease in K_a.
3. an increase in K_d.
4. the conformational change of the ligand due to binding to protein.
5. the conformational change in protein due to binding to ligand.

 

Answer: E

Section: 5.1

Level: Easy

Blooms: Understanding

 

2. The Kd for a protein-ligand interaction is:

 

1. the rate at which the protein binds ligand.
2. the square root of the equilibrium constant.
3. the fraction of ligand-binding sites occupied at equilibrium.
4. the ligand concentration at which half of the available ligand-binding sites are occupied.
5. None of these choices is correct.

 

Answer: E

Section: 5.1

Level: Medium

Blooms: Understanding

 

3. Avidin binds biotin with a K_d of ~10⁻¹⁵ mol/L. A protein of interest can be covalently linked to biotin and subsequently isolated by incubating with beads coated with avidin. If the concentration of biotinylated protein is 10^-8 M at the beginning of the assay and the beads provided an excess of avidin-binding sites, then one could expect the beads to bind:

 

1. a small fraction of the protein in the sample.
2. approximately half the protein in the sample.
3. almost all of the protein in the sample.
4. 10–15 mol of protein per 1 mL of sample.
5. This cannot be predicted.

 

Answer: C

Section: 5.1

Level: Medium

Blooms: Understanding

 

4. Determine the Ka for the biotin-avidin reaction in Question 3.

 

1. 10⁻¹⁵ M
2. 10¹⁵ M
3. 10⁻¹⁵ M⁻¹
4. 10¹⁵ M⁻¹
5. 10⁸ M

 

Answer: D

Section: 5.1

Level: Easy

Blooms: Applying

 

5. An E. coli strain has been found to constitutively express the lac operon despite the presence of the Lac repressor and absence of allolactose.  It would be appropriate to surmise that a mutation in the Lac repressor has:

 

1. increased the K_d of the repressor.
2. decreased the K_d of the repressor.
3. increased binding of the repressor to DNA.
4. increased binding of allolactose to the repressor.
5. increased the rate of growth of E. coli.

 

Answer: A

Section: 5.1

Level: Hard

Blooms: Applying

 

6. Positive cooperativity would most likely be observed when a protein is:

 

10. a single polypeptide with a K_d = 10^-10.
11. a single polypeptide with three ligand-binding sites of increasing K_d values.
12. comprised of two identical subunits, each of which binds ligand.
13. comprised of two ligand-binding sites where the K_d for the second site decreases after the first ligand binds.
14. comprised of subunits A and B, where ligand binding to A facilitates ligand binding to B.

 

Answer: E

Section: 5.1

Level: Medium

Blooms: Understanding

 

7. The Kd values of antibody-to-antigen binding range from 10^-12 M to 10^-8 M. Antibody X binds to wild-type E. coli ssDNA-binding protein, showing a value of K_d =10^-11 M.   When antibody X binds to a mutant derived ssDNA-binding protein that contains a single amino acid substitution, a value of K_d = 10^-7 M is determined.    We can conclude that:

 

1. the amino acid substitution has increased the affinity of antibody X.
2. the amino acid substitution has decreased the affinity of antibody X.
3. the amino acid substitution has increased the affinity of ssDNA-binding protein to DNA.
4. the amino acid substitution has decreased the affinity of ssDNA-binding protein to DNA.
5. None of these choices is correct.

 

Answer: B

Section: 5.1

Level: Medium

Blooms: Applying

 

8. Which of the following is least important for the binding of a protein to single-stranded DNA?

 

1. hydrophobic interactions between guanine and tyrosine
2. hydrophobic interactions between adenine and phenylalanine
3. attraction between phosphate groups and lysine
4. attraction between phosphate groups and arginine
5. hydrogen bonding between deoxyribose and tyrosine

 

Answer: E

Section: 5.1

Level: Medium

Blooms: Understanding

 

9. E. coli SSB₃₅  differs from SSB₆₅ due to:

 

1. primary sequences at amino termini.
2. number of protomers.
3. number of OB folds.
4. extent of interaction with DNA.
5. regions of contiguous A=T pairs.

 

Answer: D

Section: 5.1

Level: Medium

Blooms: Remembering

 

10. The Lac repressor:

 

1. binds to the Lac promoter.
2. only binds specifically to DNA.
3. functions as a dimer.
4. functions as a tetramer.
5. prevents bending of DNA.

 

Answer: D

Section: 5.1

Level: Hard

Blooms: Remembering

 

11. Mg²⁺ is a ______________­­­­ for DNA polymerase, while nicotinamide adenine dinucleotide (NAD) is a _____________ for salicylate hydroxylase, which is called a(n) _____________if NAD is not bound.

 

1. cofactor / coenzyme / holoenzyme
2. coenzyme / cofactor / holoenzyme
3. cofactor / coenzyme / apoenzyme
4. coenzyme / cofactor / apoenzyme
5. None of these choices is correct.

 

Answer: C

Section: 5.2

Level: Easy

Blooms: Remembering

 

12. One of the consequences of malnutrition is lack of vitamin B₆. Enzymatic reactions requiring ______________­­­­ will be halted.

 

1. transfer of acyl groups
2. transfer of alkyl groups
3. transfer of methyl groups
4. pyridoxal phosphate
5. tetrahydrofolate

 

Answer: D

Section: 5.2

Level: Easy

Blooms: Remembering

 

13. As DNA polymerase synthesizes a new phosphodiester bond, all the following occur except:

 

1. amino acid residue extracts H⁺ from 3′ hydroxyl of deoxyribose.
2. oxygen of 3′ hydroxyl attacks α phosphoryl group of nucleoside triphosphate.
3. phosphate is displaced and protonated by an acidic amino acid residue.
4. hydrogen bonds between DNA strands help position nucleotides.
5. metal ions lower pK_a within the active site to facilitate base catalysis.

 

Answer: C

Section: 5.2

Level: Medium

Blooms: Remembering

 

14. Which of the following statements about activation energy is not true?

 

1. The activation energy of a reaction is lowered by enzymes.
2. The activation energy is the difference between the energy level of the ground state and the transition state.
3. The activation energy is the difference between the free energy of the product of the catalyzed and uncatalyzed reaction.
4. The activation energy represents an energy barrier to the transition from substrate to product.
5. Activation energy is related to the rate of the reaction.

 

Answer: C

Section: 5.2

Level: Hard

Blooms: Understanding

 

 

15. The following will increase the rate of reaction:

 

1. free energy of product lower than that of substrate
2. free energy of product lower than that of transition state
3. interactions between atoms during the transition state
4. enzyme complementary to substrate
5. number of covalent groups combining with substrate during the reaction

 

Answer: C

Section: 5.2

Level: Medium

Blooms: Understanding

 

 

16. Steady state kinetics describes a reaction with the assumption that:

 

1. there is only one ES intermediate.
2. the formation and dissociation of the [ES] complex occurs at equal rates.
3. the rate of the reaction is constant.
4. V_max occurs when the enzyme is saturated.
5. There are no assumptions required.

 

Answer: B

Section: 5.2

Level: Medium

Blooms: Understanding

 

17. When an enzyme shows Michaelis-Menton kinetics, Km is equal to [S] when the reaction rate is:

 

1. 0.5 V_max.
2. 0.5 V₀.
3. 0.1 V_max.
4. 0.1 V₀.
5. V_max [S].

 

Answer: A

Section: 5.2

Level: Medium

Blooms: Understanding

 

18. The K_m for enzyme Q is 1 × 10^-6 M. Q catalyzes the reaction R→S at a velocity of 30 µmole/min when the concentration of substrate is 0.02 M; hence, V_max is approximately 30 µmole/min. Assuming Michaelis-Menten kinetics is followed, what will the reaction velocity be when [S] is 10^-7 M?

 

1. 30 µmole/min

B   3.0 × 10^-7 µmole/min

2. 2.7 µmole/min
3. 2.7 × 10^-12 µmole/min
4. 0.27 µmole/min

 

Answer: C

Section: 5.2

Level: Hard

Blooms: Applying

 

19. While characterizing a new tyrosine phosphatase, researchers found that with the total concentration [E_t] of phosphatase at 50 nM, V_max = 20 µM s^-1. What is the turnover number (k_cat) for this phosphatase?

 

1. 40 M s^-1
2. 400 M s^-1
3. 4 s^-1
4. 40 s-1
5. 400 s-1

 

Answer: E

Section: 5.2

Level: Hard

Blooms: Applying

 

20. The researchers continued to characterize the new phosphatase. When the total concentration of phosphatase [E_t] is at 2 nM, and the concentration of substrate, phosphotyrosine, is at 30 µM, then V₀ = 600 nM s^-1. What is the K_m for this phosphatase?

 

1. 1 µm
2. 10 µm
3. 100 µm
4. 1 µm s^-1
5. 1 µm s-1

 

Answer: B

Section: 5.2

Level: Hard

Blooms: Applying

 

21. In the DNA ligase reaction, the lysine residue:

 

1. serves as the nucleophile displacing the phosphate group 3′ to the nicked site.
2. serves as the electrophile displacing the phosphate group 3′ to the nicked site.
3. remains phosphorylated after the ligase reaction.
4. serves as a nucleophile, which displaces the pyrophosphate from ATP.
5. remains bound to AMP, ready to repeat the reaction.

 

Answer: D

Section: 5.2

Level: Medium

Blooms: Understanding

 

22. The adenylation of DNA ligase is:

 

1. an unstable initial step in the reaction.
2. occurs at a lysine residue in the DNA binding domain.
3. dependent on the presence of DNA.
4. independent of the presence of DNA.
5. more efficient at lower pH.

 

Answer: D

Section: 5.2

Level: Hard

Blooms: Remembering

 

23. Which of the following is mismatched regarding helicase activity?

 

1. ATP coupling stoichiometry: ATP consumed per base pair traversed
2. step size: average number of base pairs covered per ATP
3. processivity: numbers of base pairs replaced
4. directionality: defined direction of movement
5. oligomeric state: helicase functioning in multiples

 

Answer: C

Section: 5.3

Level: Easy

Blooms: Understanding

 

24. Which of the following helicase superfamilies are not known to contain a RecA-like domain?

 

1. superfamily 1
2. superfamily 3
3. superfamily 6
4. superfamilies 1 and 3
5. superfamilies 3 and 6

 

Answer: E

Section: 5.3

Level: Easy

Blooms: Remembering

 

25. A mutation in the ruvB gene mostly likely will:

 

1. reduce resistance to UV light.
2. increase binding of RuvA to DNA.
3. increase binding of RuvB to DNA.
4. stabilize the Holliday intermediate.
5. increase recombination of DNA.

 

Answer: A

Section: 5.3

Level: Easy

Blooms: Remembering

 

26. The hexameric model of DNA unwinding proposes that:

 

1. DNA strands are separated prior to helicase binding.
2. one dimer is responsible for ATP hydrolysis.
3. two dimers rotate the responsibility of ATP hydrolysis.
4. three dimers rotate the responsibility of ATP hydrolysis.
5. dimer conformational changes drive ATP hydrolysis.

 

Answer: D

Section: 5.3

Level: Medium

Blooms: Understanding

 

27. The “pin” structure of a helicase:

 

1. serves as a spool for the DNA.
2. holds the helicase dimers in the correct position.
3. allows rewinding of the DNA after transcription.
4. separates the complementary strands of DNA.
5. prevents a translocation mechanism.

 

Answer: D

Section: 5.3

Level: Medium

Blooms: Understanding

 

28. Figure 5-17 shows the approximate Km for the hydrolysis of ATP by helicase PcrA to be:
29. 3 µM.
30. 5 µM.
31. 10 µM.
32. 15 µM.
33. 20 µM.

 

Answer: A

Section: 5.3

Level: Hard

Blooms: Applying

 

29. Helicase PcrA follows classic Michaelis-Menten kinetics. In this reaction,   ATP serves as a substrate for hydrolysis, while the DNA is most appropriately called the _____________.

 

1. product.
2. substrate.
3. intermediate.
4. coenzyme.
5. ligand.

 

Answer: E

Section: 5.3

Level: Medium

Blooms: Applying

 

30. Which of the following statements about allosteric control of enzymatic activity is true?

 

1. Allosteric proteins are generally composed of a single protein.
2. Heterotropic allosteric effectors compete with substrate for binding sites.
3. Homotropic allosteric effectors compete with product for binding sites.
4. Allosteric effectors enhance the hyperbolic nature of V₀ versus [S] kinetic plots.
5. Allosteric effectors give rise to the sigmoidal nature of V₀ versus [S] kinetic plots.

 

Answer: E

Section: 5.4

Level: Medium

Blooms: Understanding

 

31. A student prepares lysate A from cells with high levels of cAMP and lysate B from cells with very low levels of cAMP. If these lysates were subjected to DNase, it could be predicted that cAMP receptor protein binding sites would be:

 

1. not present in either lysate.
2. protected in lysate A.
3. protected in lysate B.
4. equally protected in both lysates.
5. not occupied in either lysate.

 

Answer: B

Section: 5.4

Level: Hard

Blooms: Applying

 

32. An allosteric enzyme most clearly displays positive cooperativity when:

 

1. K_m= K_0.5.
2. an increase in substrate increases the rate of reaction.
3. a decrease in substrate concentration is observed soon after initiating the reaction.
4. a minimal increase in modulator is associated with a large increase in V₀.
5. a minimal decrease in modulator is associated with a large increase in V₀.

 

Answer: D

Section: 5.4

Level: Hard

Blooms: Understanding

 

33. Autoinhibition can reduce enzymatic activity by:

 

1. reducing the quantity of active enzyme in the cell.
2. preventing self-assembly of oligomer.
3. requiring a ligand for activity.
4. requiring an auxiliary protein for activity.
5. All of the choices given are true.

 

Answer: E

Section: 5.4

Level: Easy

Blooms: Remembering

 

34. A mutation creates a deletion in the C-terminal region of the bacterial RecA protein of E. coli rendering it Rec^–. It would be reasonable to surmise this mutation might:

 

1. increase autoinhibition of RecA.
2. increase RecA binding to DNA.
3. decrease autoinhibition of RecA.
4. be both A and B.
5. be both B and C.

 

Answer: E

Section: 5.4

Level: Hard

Blooms: Understanding

 

35. The amino acid most commonly modified by phosphorylation is:

 

1. cysteine.
2. tyrosine.
3. tryptophan.
4. alanine.
5. leucine.

 

Answer: B

Section: 5.4

Level: Easy

Blooms: Remembering

 

36. Referring to Figure 5-21, which of the following modifications requires ATP?
37. methylation
38. ubiquitination
39. myristoylation
40. acetylation
41. adenylylation

 

Answer: E

Section: 5.4

Level: Easy

Blooms: Remembering

 

37. Poliovirus is an RNA virus that uses host cell ribosomes to synthesize a polyprotein immediately after entering the cell. In order for functional viral proteins to be produced, one would expect to observe the mechanism of:

 

1. phosphorylation.
2. protein methylation.
3. autoinhibition.
4. ADP-ribosylation.
5. proteolytic cleavage.

 

Answer: E

Section: 5.4

Level: Medium

Blooms: Understanding

 

38. Which of the following type of reversible enzyme inhibitor always lowers the measured Km of the enzyme it binds?

 

1. competitive
2. uncompetitive
3. mixed
4. uncompetitive and mixed
5. competitive, uncompetitive, and mixed

 

Answer: B

Section: Highlight 5-1

Level: Medium

Blooms: Understanding

 

39. Which of the following type of reversible enzyme inhibitor binds to a site distinct from the substrate active site?

 

1. competitive
2. uncompetitive
3. mixed
4. uncompetitive and mixed
5. competitive, uncompetitive, and mixed

 

Answer: D

Section: Highlight 5-1

Level: Easy

Blooms: Understanding

 

40. Sulfanilamide is an antibacterial drug that inhibits dihydropteroate synthetase, an enzyme that converts para-aminobenzoic acid (PABA) to folic acid (which is required for the synthesis of bacterial nucleic acids). Sulfanilamide is a structural analog to PABA so most likely is a(n)______________ inhibitor of dihydropteroate synthetase.

 

1. analogic inhibitor
2. uncompetitive inhibitor
3. mixed inhibitor
4. competitive inhibitor
5. allosteric inhibitor

 

Answer: D

Section: Highlight 5-1

Level: Easy

Blooms: Understanding

 

41. The apparent V_max of dihydropteroate synthetase will _____________ when sulfanilamide is present while the Km of this enzyme will ______________.

 

1. decrease / increase
2. not change / increase
3. decrease / decrease
4. not change / decrease
5. not change / not change

 

Answer: B

Section: Highlight 5-1

Level: Hard

Blooms: Applying

 

42. Researchers would like to create HIV protease inhibitors that will not become ineffective due to viral strain resistance. One approach is to design a:

 

1. protease with no aspartate residues.
2. protease inhibitor and treat all patients in a community simultaneously.

C   protease inhibitor that will not be mutated in the aspartate region.

1. protease inhibitor that binds to highly conserved protease sequences.
2. random screening program that tests hundreds of drug candidates.

 

Answer: D

Section: Highlight 5-2

Level: Hard

Blooms: Applying

 

43. Monod’s group found that that appearance of β-galactosidase correlated with that of galactoside permease. This ultimately led his group to the conclusion that:

 

1. constitutive expression involved two genes.
2. β-galactosidase production is contingent on the presence of galactosidase.
3. the expression of β-galactosidase and permease is under the control of another gene.
4. a mutation in the β-galactosidase gene would decrease expression of a permease gene.
5. a mutation in the regulator gene would decrease expression of galactosidase and permease.

 

Answer: C

Section: How We Know

Level: Hard

Blooms: Understanding

 

44. The DNA-binding site for the lactose repressor was identified by:

 

1. causing cleavage and subsequent methylation at purine residues.
2. methylating and hence protecting the DNA.
3. comparing DNA sequences protected by wild-type and mutant repressors.

D   noting the DNA sequences that were protected from methylation by the repressor.

1. reading the DNA sequence that was methylated by DMS after repressor binding.

 

Answer: D

Section: How We Know

Level: Medium

Blooms: Understanding