Michaelis–Menten insight: when substrate concentration [S] equals 0.1 * KM, what is the approximate initial velocity v as a fraction of Vmax for a simple enzyme reaction?
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A0.1 * Vmax
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B0.3 * Vmax
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C0.5 * Vmax
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D0.7 * Vmax
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E0.9 * Vmax
Answer
Correct Answer: 0.1 * Vmax
Explanation
Introduction / Context:The Michaelis–Menten equation relates initial velocity v to substrate concentration [S], maximum velocity Vmax, and the Michaelis constant KM. Estimating v/Vmax at particular fractions of KM is a practical skill for interpreting kinetic regimes (substrate-limited vs near-saturation).
Given Data / Assumptions:
- Reaction follows v = (Vmax * [S]) / (KM + [S]).
- Given [S] = 0.1 * KM.
- We seek v as a fraction of Vmax.
Concept / Approach:Substitute [S] = 0.1 KM into the Michaelis–Menten equation and simplify algebraically to get a numeric fraction of Vmax. This illustrates how low-substrate conditions (well below KM) produce velocities much less than Vmax and roughly proportional to [S].
Step-by-Step Solution:
1) Start with v = (Vmax * [S]) / (KM + [S]).2) Substitute [S] = 0.1 KM → v = Vmax * (0.1 KM) / (KM + 0.1 KM).3) Factor KM: v = Vmax * 0.1 / (1 + 0.1) = Vmax * 0.1 / 1.1.4) Compute: 0.1 / 1.1 ≈ 0.0909 → approximately 0.1 Vmax.Verification / Alternative check:At [S] = KM, v = 0.5 Vmax. Since 0.1 KM is ten-fold lower, the velocity should be far below 0.5 Vmax, consistent with ~0.09–0.10 Vmax.
Why Other Options Are Wrong:
- 0.3, 0.5, 0.7, 0.9 Vmax: these correspond to much higher [S] relative to KM; they overestimate v in the low-substrate regime.
Common Pitfalls:Forgetting to divide by (1 + [S]/KM) after substituting; confusing KM with a saturation threshold rather than the [S] at half Vmax.
Final Answer:0.1 * Vmax