Under Michaelis–Menten kinetics, the initial velocity v0 approaches V max under which limiting condition on substrate concentration?
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A[S] = Km
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B[S] = 10 * Km
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C1/[S] = 1/Km
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D1/[S] → 0 (i.e., [S] ≫ Km)
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E[S] = 0.1 * Km
Answer
Correct Answer: 1/[S] → 0 (i.e., [S] ≫ Km)
Explanation
Introduction / Context:In saturation kinetics, the rate approaches a maximum as substrate becomes abundant. Recognizing the limiting regime that yields v0 ≈ V max is essential for interpreting enzyme experiments and plotting transformations.
Given Data / Assumptions:
- Michaelis–Menten rate law: v = (Vmax * [S]) / (Km + [S]).
- Vmax reflects catalytic capacity at saturating [S].
- We examine limiting behavior as [S] varies relative to Km.
Concept / Approach:When [S] ≫ Km, the denominator Km + [S] ≈ [S], so v ≈ Vmax * [S]/[S] = Vmax. Equivalently, 1/[S] → 0 on a Lineweaver–Burk plot drives the intercept toward 1/Vmax.
Step-by-Step Solution:
Start with v = (Vmax * [S]) / (Km + [S]).Assume [S] ≫ Km: Km is negligible in the sum.Simplify to v ≈ Vmax * [S]/[S] = Vmax.Thus, the limiting condition is 1/[S] → 0 (very high [S]).Verification / Alternative check:Experimental saturation curves flatten as [S] increases; Eadie–Hofstee plots show approach to Vmax as v stabilizes at high [S].
Why Other Options Are Wrong:
- [S] = Km gives v = 0.5 * Vmax.
- [S] = 10 * Km gives ~0.91 * Vmax, close but not the limiting definition.
- 1/[S] = 1/Km corresponds to [S] = Km (half-maximal).
- [S] = 0.1 * Km is far from saturation.
Common Pitfalls:Equating “large” (e.g., 10 * Km) with the strict mathematical limit; the definition uses the limit [S] ≫ Km.
Final Answer:1/[S] → 0 (i.e., [S] ≫ Km)