How does a competitive inhibitor affect Km and Vmax ?

Question

BlackTom AI · Accepted Answer

Competitive inhibitors bind reversibly to the enzyme’s active site, directly competing with substrate molecules for that same site.

1. Relevant knowledge points and concepts
- Vmax is the maximal reaction velocity when the enzyme is saturated with substrate.
- Km is the Michaelis constant, the substrate concentration at which reaction velocity is half of Vmax; it reflects substrate affinity (higher Km = lower apparent affinity).
- Competitive inhibition: inhibitor (I) and substrate (S) compete for the active site; binding of I prevents S binding but does not change the catalytic capability of enzyme molecules that are substrate-bound.

2. Step-by-step reasoning
- At low-to-moderate [S], the inhibitor reduces the fraction of enzyme available to bind substrate, so more substrate is required to reach half-maximal velocity — apparent Km increases.
- At very high [S], substrate outcompetes the inhibitor for the active site, so the enzyme can still reach the same Vmax as without inhibitor.

3. Equations and intermediate steps
- Michaelis–Menten with competitive inhibitor: v = Vmax [S] / (Km (1 + [I]/Ki) + [S]).
- Define apparent Km, Km,app = Km (1 + [I]/Ki). This shows Km,app > Km when [I] > 0.
- At saturating [S] (>> Km,app), v → Vmax, so Vmax is unchanged.
- Lineweaver–Burk form: 1/v = (Km(1 + [I]/Ki)/Vmax)(1/[S]) + 1/Vmax. The y-intercept (1/Vmax) is constant, slope increases with [I], indicating increased apparent Km but unchanged Vmax.

4. Why the correct answer is correct
- The algebraic modification and the kinetic interpretation both show Km is multiplied by (1 + [I]/Ki) while Vmax remains the same, so the apparent Km increases and Vmax is unchanged.

5. Why other options are incorrect
- "Vmax decreases, Km unchanged": describes pure noncompetitive inhibition (or some mixed types), not competitive.
- "Apparent Km decreases, Vmax unchanged": inconsistent with competition; Km would not decrease (that occurs in uncompetitive inhibition where both Km and Vmax decrease).
- "Both Km and Vmax decrease": matches uncompetitive inhibition, not competitive.
- "Both Km and Vmax increase": no common inhibition type gives both increasing; competitive increases only Km.

6. Conclusion
The correct answer is: Apparent Km increases, Vmax unchanged, matching the provided answer.

How does a competitive inhibitor affect Km and Vmax ?单项选择题

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How does a competitive inhibitor affect Km and Vmax ?单项选择题

类似问题

Which type of enzyme inhibitor does not affect the maximal velocity of a reaction?

Which of the following sentences best describes what happens during competitive inhibition of an enzyme?

A competitive inhibitor of an enzyme:

A competitive inhibitor of an enzyme:

This type of inhibition is _______________. We know because __________________. In this type of inhibition, ___________ is forbidden.

9. This figure describes inhibition of an enzyme by ____________________.

Which of the following is NOT correct about a competitive inhibitor of an enzyme?

更多留学生实用工具

希望你的学习变得更简单

This type of inhibition is ___. We know because . In this type of inhibition, _ is forbidden.