Regulation, Inhibition & Clinical Enzymology (4.4)

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Welcome! Regulation, Inhibition & Clinical Enzymology (4.4) — 71 questions across 3 tests.

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  • Test 1 (4.4) — Regulation, Inhibition & Clinical
  • Test 2 (4.4) — Regulation, Inhibition & Clinical
  • Test 3 (4.4) — Regulation, Inhibition & Clinical

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4.4 Regulation/Clinical โ€” Test 1
Q1. An enzyme reaction was studied with and without an inhibitor. The Michaelis-Menten plot showed Vmax did not change at different inhibitor concentrations. The inhibition is:โœ“ Competitive
Q2. A non-competitive inhibitor:โœ“ Kโ‚˜ unaltered, Vโ‚˜โ‚โ‚“ decreases
Q3. Competitive inhibition is mediated when the inhibitor:โœ“ Binds reversibly at the active site
Q4. Mixed reversible inhibition is due to:โœ“ Inhibitor binds a separate site and may bind E or ES
Q5. In non-competitive inhibition:โœ“ Inhibitor binds a non-active site, may bind E or ES
Q6. When velocity is plotted against substrate concentration for a Michaelis-Menten enzyme, the curve is:โœ“ Hyperbolic
Q7. An inhibitor binds the enzyme whether or not substrate is bound, but prefers one form. This is:โœ“ Mixed inhibition
Q8. How does a Lineweaver-Burk plot change with a non-competitive inhibitor (Ki < Ki')?โœ“ X-intercept unchanged
Q9. Which kind of inhibition can have Kโ‚˜ unaltered?โœ“ Non-competitive
Q10. Uncompetitive inhibition was concluded from a Lineweaver-Burk plot because:โœ“ Vโ‚˜โ‚โ‚“ and Kโ‚˜ decrease, inhibitor binds ES
Q11. Which enzyme is activated by acid hydrolysis of its proenzyme form?โœ“ Pepsin
Q12. Assertion: a very low amount of inhibitor can act as an activator for allosteric enzymes. Reason: allosteric enzymes follow Michaelis-Menten kinetics.โœ“ A true, R false
Q13. Allosteric enzymes show:โœ“ Sigmoidal kinetics
Q14. Plotting initial velocity vs substrate concentration for an allosteric enzyme gives:โœ“ Sigmoid
Q15. All statements about allosteric enzymes are true EXCEPT:โœ“ They display Michaelis-Menten kinetics
Q16. The allosteric inhibitor binds to the:โœ“ Enzyme, causing a conformational change
Q17. An allosteric inhibitor binds to:โœ“ Enzyme, causing conformational change
Q18. An allosteric inhibitor of an enzyme:โœ“ Enacts feedback regulation
Q19. Allosteric enzymes regulate product formation by:โœ“ Feedback inhibition
Q20. Which best describes an allosteric binding site?โœ“ A site separate from the active site that affects activity when occupied
Q21. Arsenate binds the -SH group of an enzyme to inhibit it. This mode of inhibition is:โœ“ Non-competitive
Q22. A competitive inhibitor of an enzyme:โœ“ Increases Kโ‚˜
Q23. An inhibitor increases Kโ‚˜ but does not affect Vโ‚˜โ‚โ‚“. It is:โœ“ Competitive
Q24. Succinate dehydrogenase is inhibited by oxaloacetate. This inhibition is:โœ“ Reversible and competitive
Q25. On a Lineweaver-Burk plot, the slope equals:โœ“ Kโ‚˜/Vโ‚˜โ‚โ‚“
4.4 Regulation/Clinical โ€” Test 2
Q26. The action of which inhibitors can be reversed by adding excess substrate?โœ“ Competitive inhibitors
Q27. Which is NOT a characteristic of competitive inhibition?โœ“ There is no change in Kโ‚˜
Q28. During competitive inhibition, which statements are correct? (1 substrate competes; 2 inhibitor binds free enzyme; 3 excess substrate overcomes; 4 two sites; 5 Vโ‚˜โ‚โ‚“ unchanged, Kโ‚˜ increases)โœ“ 1,2,3,5
Q29. During competitive inhibition, which are correct? (1 competes; 2 binds free enzyme; 3 excess substrate overcomes; 4 active + modification site; 5 Vโ‚˜โ‚โ‚“ unchanged, Kโ‚˜ increases, affinity decreases)โœ“ 1,2,3,5
Q30. Succinate dehydrogenase converts succinate to fumarate. With competitive inhibitor malonate:โœ“ Kโ‚˜ increases, Vโ‚˜โ‚โ‚“ unchanged
Q31. Kinetic effect of competitive inhibition:โœ“ Increase Kโ‚˜ without affecting Vโ‚˜โ‚โ‚“
Q32. Succinate dehydrogenase + malonate (competitive inhibitor):โœ“ Kโ‚˜ increases, Vโ‚˜โ‚โ‚“ unchanged
Q33. Succinate dehydrogenase with reversible competitive inhibitor malonate:โœ“ Kโ‚˜ increases, Vโ‚˜โ‚โ‚“ remains same
Q34. A competitive inhibitor of an enzyme:โœ“ Increases Kโ‚˜ without affecting Vโ‚˜โ‚โ‚“
Q35. On a Lineweaver-Burk plot, the slope in the presence of a competitive inhibitor is:โœ“ Kโ‚˜/Vโ‚˜โ‚โ‚“
Q36. A characteristic feature of an allosteric protein is:โœ“ Cooperativity
Q37. When end-product inhibition by two or more products on one enzyme is strictly additive, it is called:โœ“ Cumulative
Q38. In feedback regulation, the end product binds to the:โœ“ Allosteric site
Q39. Glutamine synthetase is regulated reversibly by:โœ“ Adenylylation and deadenylylation
Q40. Irreversible inhibitors often form covalent bonds with:โœ“ Ser or Cys residues at/near the active site
Q41. Irreversible inhibitors often form covalent bonds with (repeat):โœ“ Ser or Cys at/near active site
Q42. Which type of enzyme regulation is irreversible?โœ“ Zymogen activation
Q43. Kinase activity is inhibited by which cofactor analog?โœ“ ATP-ฮณ-S (ATPฮณP-32 / non-hydrolysable)
Q44. Lineweaver-Burk plot with inhibitor: x-intercept near the same point, y-intercept much higher. The inhibition is:โœ“ Non-competitive
Q45. LB plot: x-intercept very near the no-inhibitor value but y-intercept much higher. Nature of inhibition?โœ“ Non-competitive
Q46. A non-competitive inhibitor causes:โœ“ Decrease of Vโ‚˜โ‚โ‚“
Q47. The degree of inhibition is independent of initial substrate concentration. The inhibition is:โœ“ Non-competitive
Q48. A non-competitive inhibitor leads to:โœ“ A decrease in observed Vโ‚˜โ‚โ‚“
Q49. On a Lineweaver-Burk plot, non-competitive inhibition shows:โœ“ Same Kโ‚˜, different Vโ‚˜โ‚โ‚“ (lines meet on x-axis)
Q50. A sigmoidal plot of [S] versus velocity indicates:โœ“ Cooperative binding
4.4 Regulation/Clinical โ€” Test 3
Q51. A sigmoidal plot of [S] versus reaction velocity indicates:โœ“ Cooperative binding
Q52. Positive cooperativity of allosteric enzymes is best described by:โœ“ T > R; ligand binding raises R, easing binding
Q53. Positive cooperativity of allosteric enzymes (repeat):โœ“ T > R; ligand raises R, easing binding
Q54. The enzyme reaction v = Vโ‚˜โ‚โ‚“[S]/(Kโ‚˜ + [S] + [S]ยฒ/Ks) describes which curve?โœ“ Substrate inhibition
Q55. Which is true for an uncompetitive inhibitor?โœ“ It increases the affinity of substrate for enzyme
Q56. The inhibition that occurs when the inhibitor binds only the enzyme-substrate complex is:โœ“ Uncompetitive
Q57. Enzyme inhibition that reduces both Kโ‚˜ and Vโ‚˜โ‚โ‚“ is:โœ“ Uncompetitive
Q58. If an inhibitor decreases both Kโ‚˜ and Vโ‚˜โ‚โ‚“, it is:โœ“ Uncompetitive
Q59. If addition of an inhibitor decreases both Kโ‚˜ and Vโ‚˜โ‚โ‚“, the mechanism is:โœ“ Uncompetitive
Q60. Feedback inhibition differs from repression because feedback inhibition:โœ“ Stops the action of pre-existing enzymes
Q61. In single-substrate reactions, the competitive-inhibition Lineweaver-Burk equation correctly includes the factor:โœ“ (1+[I]/Ki) on the slope term only
Q62. Adding X raised the allosteric constant L from 100 to 1000; adding Y lowered L from 100 to 10. Which inference is correct?โœ“ X is a negative regulator
Q63. An enzyme has Kโ‚˜ = 6 mM, Vโ‚˜โ‚โ‚“ = 25 ยตmol/min. With inhibitor (10 mM), Kโ‚˜ = 6 mM and Vโ‚˜โ‚โ‚“ = 20 ยตmol/min. The Ki is:โœ“ 40 mM
Q64. From a 1/v vs 1/[S] plot, both slope and y-intercept increase with inhibitor. The inhibitor binds:โœ“ Both E and ES (mixed)
Q65. Which statement about reversible inhibitors is/are correct? P: uncompetitive binds only ES; Q: non-competitive binds a different site; R: competitive binds the substrate site.โœ“ P, Q and R
Q66. Which statement is NOT true?โœ“ A transition-state analogue increases the rate of product formation
Q67. Match the enzyme with the tissue whose damage chiefly raises it (LDH is a broad, non-specific marker):โœ“ i-1, ii-2, iii-3, iv-4, v-5
Q68. Which is a glycosylated form of hemoglobin used to assess long-term blood glucose control?โœ“ Hemoglobin A1c
Q69. Which cardiac marker is considered most specific for myocardial infarction?โœ“ Troponin
Q70. Modern glucometers measure blood glucose by:โœ“ Rate of electron flow from glucose to an electrode
Q71. Serum cholesterol is decreased in:โœ“ Thyrotoxicosis