Non-competitive Inhibition: What is the primary kinetic effect on an enzyme-catalyzed reaction?

Difficulty: Medium

Correct Answer: It decreases Vmax while Km (for the true non-competitive case) remains essentially unchanged.

Explanation:


Introduction:
Non-competitive inhibition reflects inhibitor binding to an allosteric site on E and often also on ES, reducing catalytic turnover independent of substrate binding at the active site. This question probes the hallmark kinetic signature.


Given Data / Assumptions:

  • Classical non-competitive model (special case of mixed inhibition with identical Ki on E and ES).
  • Michaelis–Menten single-substrate behavior.


Concept / Approach:

Because the inhibitor decreases the fraction of active enzyme capable of catalysis, Vmax drops. Since substrate affinity is not primarily altered in the true non-competitive limit, Km remains roughly the same. On Lineweaver–Burk plots, 1/Vmax increases (y-intercept rises), while x-intercept (−1/Km) stays near constant; slope increases accordingly.


Step-by-Step Solution:

1) Inhibitor binds E and ES at a distinct site ⇒ catalytic turnover number kcat is effectively reduced.2) Vmax = kcat * [E]_total ⇒ Vmax decreases as active fraction falls.3) Km remains approximately constant in the true non-competitive case since substrate binding affinity is not selectively impaired.4) Therefore, velocity cannot increase under ordinary conditions as inhibitor concentration rises.


Verification / Alternative check:

Graphically, lines for different inhibitor concentrations intersect on the x-axis but have higher y-intercepts, confirming lower Vmax with similar Km.


Why Other Options Are Wrong:

B: Exclusive ES binding describes uncompetitive inhibition, not non-competitive. C: B is incorrect, so both cannot be true. D and E: Contradict the defining reduction of Vmax under inhibition.


Common Pitfalls:

Confusing non-competitive with uncompetitive (both lower Vmax, but effects on Km differ), or assuming inhibitors can enhance velocity under typical steady-state conditions.


Final Answer:

It decreases Vmax while Km (for the true non-competitive case) remains essentially unchanged.

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