Effect of Changing Kinetic Parameters: For an enzyme with Km = 10 mM and Vmax = 100 mmol/min, at [S] = 100 mM, which statements about velocity changes are correct?
-
AA 10-fold increase in Vmax would increase the reaction velocity approximately 10-fold.
-
BA 10-fold decrease in Km (to 1 mM) would increase the reaction velocity at [S] = 100 mM.
-
CBoth (a) and (b)
-
DA 10-fold increase in Vmax would decrease the reaction velocity 20-fold.
-
ENeither (a) nor (b)
Answer
Correct Answer: Both (a) and (b)
Explanation
Introduction:This problem applies the Michaelis–Menten equation to assess how changes in Vmax and Km affect reaction velocity at a fixed substrate concentration. It tests quantitative intuition at saturating and near-saturating conditions.
Given Data / Assumptions:
- Km = 10 mM.
- Vmax = 100 mmol/min.
- [S] = 100 mM.
- Single-substrate Michaelis–Menten kinetics.
Concept / Approach:
Use v = Vmax * [S] / (Km + [S]). At [S] ≫ Km, velocity approaches Vmax, so changes in Vmax have roughly proportional effects. Decreasing Km at fixed [S] increases the fraction [S]/(Km + [S]), raising v but with diminishing returns as [S] greatly exceeds Km.
Step-by-Step Solution:
1) Baseline: v0 = 100 * 100 / (10 + 100) = 10000 / 110 ≈ 90.91 mmol/min.2) Increase Vmax 10-fold: Vmax' = 1000 ⇒ v' = 1000 * 100 / 110 = 100000 / 110 ≈ 909.09 mmol/min (≈ 10× increase).3) Decrease Km 10-fold: Km' = 1 mM ⇒ v' = 100 * 100 / (1 + 100) = 10000 / 101 ≈ 99.01 mmol/min (increase relative to 90.91).4) Therefore, (a) and (b) are both true; (d) is false.Verification / Alternative check:
Limiting case logic: as [S] → ∞, v → Vmax; thus scaling Vmax scales v. Reducing Km increases saturation fraction toward 1, raising velocity until limited by Vmax.
Why Other Options Are Wrong:
D: Increasing Vmax cannot decrease v; it raises the upper limit. E: Contradicts calculations showing both effects increase v.
Common Pitfalls:
Assuming that once [S] ≫ Km, changing Km has no effect; it still slightly increases v unless saturation is complete.
Final Answer:
Both (a) and (b)