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:
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:
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.
Discussion & Comments