Which of the following is true regarding non-competitive inhibition?

Correct Answer: D. Constant km and decrease Vmax

Non-competitive inhibition occurs when an inhibitor binds to the enzyme at a site OTHER than the active site, preventing catalysis regardless of substrate concentration. The critical distinction is that the inhibitor binds with equal affinity to both the free enzyme (E) and the enzyme-substrate complex (ES), forming both EI and ESI complexes. Since the inhibitor does not compete with substrate for the active site, increasing substrate concentration cannot overcome the inhibition—hence the name "non-competitive." Mathematically, non-competitive inhibition affects only the maximum velocity (Vmax), reducing it by a factor of (1 + [I]/Ki), while the Michaelis constant (Km) remains unchanged. This is because Km is a measure of substrate affinity for the active site, which is unaffected when the inhibitor binds elsewhere. In Indian clinical practice, examples include statins inhibiting HMG-CoA reductase allosterically and certain antiretroviral drugs inhibiting reverse transcriptase non-competitively. The Lineweaver-Burk plot shows parallel lines (same x-intercept = −1/Km, different y-intercepts), confirming constant Km and decreased Vmax.

Why the other options are wrong

A. Decreased km and decreased Vmax — This describes a mixed inhibition pattern where the inhibitor affects both substrate binding and catalytic efficiency. Non-competitive inhibition specifically leaves Km unchanged because the inhibitor does not interfere with substrate-active site interaction. Decreased Km would only occur if the inhibitor increased substrate affinity, which is not the mechanism here. This is a common trap for students confusing non-competitive with mixed inhibition. B. Increased km and unchanged Vmax — This is the hallmark of COMPETITIVE inhibition, where the inhibitor competes directly with substrate for the active site. Competitive inhibitors increase Km (apparent decrease in substrate affinity) but Vmax remains unchanged because excess substrate can eventually saturate the enzyme. Students often confuse competitive and non-competitive mechanisms—the key discriminator is whether the inhibitor binds the active site (competitive) or elsewhere (non-competitive). C. Constant km and increased Vmax — This is biochemically impossible under normal inhibition conditions. An inhibitor cannot increase Vmax; it can only decrease or leave it unchanged. This option may trap students who misremember the kinetic equations or confuse inhibition with enzyme activation or allosteric enhancement. No physiological non-competitive inhibitor increases catalytic capacity.

High-Yield Facts

Mnemonics

COMP vs NON-COMP COMPetitive = COMPetes for active site → Km ↑, Vmax ↔. NON-COMP = binds elsewhere → Km ↔, Vmax ↓. Lineweaver-Burk Lines Competitive = lines intersect on y-axis (same Vmax). Non-competitive = parallel lines (same Km). Use this to distinguish kinetic patterns in exam questions.

NBE Trap

NBE pairs "non-competitive" with "unchanged Vmax" (option B, which is actually competitive inhibition) to trap students who conflate the two mechanisms. The correct answer requires recognizing that non-competitive inhibitors reduce Vmax precisely because they bind outside the active site and cannot be outcompeted by substrate.

Clinical Pearl

In Indian clinical practice, recognizing non-competitive inhibition is crucial for understanding why increasing substrate (e.g., increasing dietary cholesterol) cannot overcome statin inhibition of HMG-CoA reductase—the drug binds allosterically, not competitively. This principle guides dosing strategies in lipid-lowering therapy across Indian hospitals.

_Reference: Harper Biochemistry Ch. 8 (Enzyme Kinetics); KD Tripathi Pharmacology Ch. 1 (Drug-Enzyme Interactions)_