Enzyme Kinetics (4.2)

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Welcome! Enzyme Kinetics (4.2) — 64 questions across 3 tests.

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  • Test 1 (4.2) — Enzyme Kinetics
  • Test 2 (4.2) — Enzyme Kinetics
  • Test 3 (4.2) — Enzyme Kinetics

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4.2 Kinetics β€” Test 1
Q1. Site-directed mutagenesis gives WT (kcat=77 s⁻¹, KM=20 Β΅M), M1 (70, 17), M2 (0.15, 20), M3 (0.001, 18). Which has the highest catalytic efficiency?βœ“ Mutant 1
Q2. WT enzyme: kcat=0.37 s⁻¹, KM=3.4 mM. Mutant: kcat=1.1 s⁻¹, KM=0.7 mM. Approximate fold increase in catalytic efficiency?βœ“ 14
Q3. Enzyme multiplicity regulates amino acid biosynthesis. A correct example is:βœ“ Chorismic acid mutase and Aldolase
Q4. Which enzyme form is more active?βœ“ Deadenylated glutamine synthetase
Q5. Glutamine synthetase is regulated reversibly by:βœ“ Adenylylation and deadenylylation
Q6. Most metabolic enzymes operate in vivo:βœ“ Below saturation and below Km
Q7. To estimate substrate concentration in an assay, the enzyme should preferably have a:βœ“ Km above the substrate concentration
Q8. In a Lineweaver–Burk plot, the Y-axis data are expressed as:βœ“ 1/V
Q9. For Michaelis–Menten kinetics, if KM = 5 mM and vβ‚€ = 12.5 Β΅mol/(mLΒ·s) at [S] = 5 mM, what is Vmax?βœ“ 25
Q10. A lipase has kcat = 25.0 s⁻¹ and KM = 0.0048 M. At what [S] does it show one-fifth of its maximum rate?βœ“ 1.2 Γ— 10⁻³ M
Q11. Vmax for an enzyme following simple Michaelis–Menten kinetics, if vβ‚€ = 2 Β΅mol min⁻¹ at [S] = 10 KM?βœ“ 2.2 Β΅mol min⁻¹
Q12. On a Michaelis–Menten plot, the curve plateaus and does not rise further on adding substrate because:βœ“ The active site is saturated with substrate
Q13. Which one is a tandem enzyme?βœ“ Phosphofructokinase 2
Q14. Efficiency of an enzyme is best measured in terms of:βœ“ Kcat/Km
Q15. Which is a unit of catalytic efficiency (kcat/KM) of an enzyme?βœ“ min⁻¹ M⁻¹
Q16. Which correctly shows the Eadie–Hofstee plot?βœ“ V vs V/S
Q17. In Eadie–Hofstee plots, the X-intercept and Y-intercept are respectively:βœ“ Vmax and Vmax/Km
Q18. Two strains: Bsyrup (KM 100 mM, Vmax 1000) and Bsoil (KM 10 mM, Vmax 100), assayed at 10/100/1000 mM glucose. Which statement is correct?βœ“ Bsoil higher Vo at 10 mM than Bsyrup
Q19. Which is true for enzyme-catalysed reactions?βœ“ Vmax can never be achieved
Q20. At substrate concentrations much lower than KM, reaction velocity is:βœ“ Directly proportional to [S]
Q21. Which is obtained by multiplying the Lineweaver–Burk equation by [Sβ‚€]?βœ“ Hanes plot equation
4.2 Kinetics β€” Test 2
Q22. For the Hanes–Woolf plot ([S] on X, [S]/v on Y), which statement is correct?βœ“ The X-intercept is βˆ’Km
Q23. Cooperative ligand binding is described quantitatively by the:βœ“ Hill equation
Q24. If the product is formed in an exponential pattern, the initial velocity can be determined by applying:βœ“ Logarithmic equation
Q25. The SI unit for enzyme activity is the katal. One katal equals:βœ“ 6 Γ— 10⁷ IU
Q26. One katal is (repeat):βœ“ 6 Γ— 10⁷ IU
Q27. In enzyme kinetics, the rate constant kcat refers to:βœ“ Turnover number
Q28. The unit of kcat of an enzyme cannot be:βœ“ s⁻¹
Q29. KM is characteristic of an enzyme–substrate pair and is independent of enzyme amount, but Vmax varies with the:βœ“ Amount of enzyme used
Q30. Glucokinase (kcat 600 s⁻¹) with Et = 20 Β΅mol and [S] = 40 Β΅mol gives vβ‚€ = 9.6 Β΅mol/s. The KM is:βœ“ 10 Β΅moles
Q31. A protein kinase hydrolyses substrate (0.03 mmol/L) at vβ‚€ = 1.5Γ—10⁻³ mmol/L/min with Vmax = 4.5Γ—10⁻³ mmol/L/min. KM is:βœ“ 0.06
Q32. A Lineweaver–Burk plot has slope 0.095 s and Y-intercept 0.0126 sΒ·Β΅M⁻¹. The KM (in mM) is:βœ“ 7.54Γ—10⁻³
Q33. For KM, which statements hold? (a) High KM = high affinity; (b) KM indicates enzyme–substrate affinity; (c) KM defines the [S] for effective catalysis.βœ“ b and c
Q34. The Michaelis–Menten constant KM is:βœ“ Numerically equal to the [S] giving half-maximal velocity
Q35. The Michaelis constant KM is:βœ“ Numerically equal to the [S] giving half-maximal velocity
Q36. A low KM value indicates:βœ“ High substrate affinity
Q37. Identify the true statement about KM:βœ“ Km is independent of enzyme concentration
Q38. The relationship between KM and Kd in an enzyme-catalysed reaction is:βœ“ Km is usually more than Kd
Q39. The substrate KM in an enzyme-catalysed reaction:βœ“ Is never less than Kd
Q40. If a 1/v vs 1/[S] plot is linear, this indicates:βœ“ A pre-equilibration
Q41. For enzymes obeying Michaelis–Menten, a 1/vβ‚€ versus 1/[S] plot is the:βœ“ Lineweaver–Burk plot
Q42. When [S] = 0.5Γ—KM, the velocity vβ‚€ is about:βœ“ 0.3Γ—Vmax
4.2 Kinetics β€” Test 3
Q43. For an enzyme following Michaelis–Menten, when [S] = KM, v equals:βœ“ 0.5 Vmax
Q44. For Michaelis–Menten when [S] β‰ͺ KM, which holds?βœ“ vβ‚€ = (Vmax/Km)[S]
Q45. For an enzyme's Michaelis–Menten equation when [S] is very low compared with KM:βœ“ vβ‚€ = (Vmax/Km)[S]
Q46. At [S] = 2KM, the percentage of Vmax (using v = Vmax/(1+KM/[S])) is:βœ“ 66.70%
Q47. An enzyme has KM = 1 mM and Vmax = 5 nM/s. The reaction velocity at [S] = 0.25 mM is:βœ“ 1 nM/s
Q48. A lipase has kcat = 25.0 s⁻¹ and KM = 0.0048 M. At what [S] does it show one-fifth of its maximum rate?βœ“ 1.2 Γ— 10⁻³ M
Q49. Vo–[S] relationship for an enzyme following Michaelis–Menten has the shape of a:βœ“ Rectangular hyperbola
Q50. An enzyme shows v = 10 Β΅mol/min when [S] = KM. Its Vmax is:βœ“ 20 Β΅mol/min
Q51. When [S] = 0.5Γ—KM, the velocity vβ‚€ is about:βœ“ 0.3Γ—Vmax
Q52. When velocity is plotted against substrate concentration, it shows a:βœ“ Hyperbolic curve
Q53. The rate-determining step of the Michaelis–Menten mechanism is:βœ“ The product formation step
Q54. A pure 47 kDa monomeric kinase has specific activity 200 U/mg. 50 Β΅g is found in 100 mL buffer, assayed at saturating substrate. The reaction rate (1 U = 1 Β΅mol/min) is:βœ“ 1 Γ— 10⁻⁴ M/min
Q55. Which property of an enzyme is responsible for saturation behaviour (a maximum rate insensitive to rising [S])?βœ“ The enzyme has a fixed number of active sites
Q56. 'Specific activity' differs from 'activity' in that specific activity:βœ“ Is the activity (units) per milligram of protein
Q57. An uninduced culture has 228 units of Ξ²-galactosidase per mL and 43 Β΅g protein per mL. The specific activity is:βœ“ 5302
Q58. Which might be the units for specific activity of an enzyme?βœ“ Β΅mol min⁻¹ mg⁻¹
Q59. The specific activity of an enzyme is reported in:βœ“ Units per milligram
Q60. Which might be the units for specific activity of an enzyme?βœ“ Β΅mol min⁻¹ mg⁻¹
Q61. The Michaelis–Menten equation relates the rate of an enzyme reaction to:βœ“ Substrate concentration
Q62. Turnover numbers: E1 = 150 s⁻¹, E2 = 15 s⁻¹. This means:βœ“ Velocities at saturating substrate could be equal if [E2] = 10Γ—[E1]
Q63. Which parameter changes upon doubling enzyme concentration?βœ“ Vmax
Q64. At substrate concentrations where the rate reaches Vmax, the reaction follows ___ kinetics.βœ“ Zero order