Difficulty: Easy
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:
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:
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)
Discussion & Comments