Enzyme Kinetics Under Inhibitor Treatment
A student studies an enzyme that converts substrate S to product P. Without inhibitor, Vmax = 100 µmol/min and Km = 2 mM. With inhibitor A (10 µM), Vmax falls to 50 µmol/min but Km remains 2 mM. With inhibitor B (10 µM), Vmax stays at 100 µmol/min but Km rises to 8 mM. With inhibitor C (10 µM), both Vmax and Km change: Vmax = 60 µmol/min and Km = 5 mM. The student must decide which inhibitor is competitive, which is noncompetitive, and which is mixed, then predict behavior at a substrate concentration of 2 mM.
1. Classify inhibitors A, B, and C, and justify each assignment using the kinetic parameters.
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Inhibitor A: Vmax decreases, Km unchanged - noncompetitive inhibition. The inhibitor binds equally well to free enzyme and enzyme–substrate complex, so some enzyme is permanently inactivated; apparent Vmax drops while substrate affinity (Km) is unchanged. Inhibitor B: Vmax unchanged, Km increases - competitive inhibition. The inhibitor competes with substrate at the active site; high substrate can outcompete it, so Vmax is reachable, but more substrate is needed to reach half-maximal rate, raising apparent Km. Inhibitor C: both Vmax and Km change - mixed inhibition. The inhibitor binds both free enzyme and ES complex but with different affinities, altering both parameters. The direction of the Km change (increase) indicates the inhibitor prefers free enzyme. These assignments follow directly from the defining kinetic signatures rather than from any memorized drug list.
2. At [S] = 2 mM, calculate the uninhibited rate and the rate with inhibitor B, and explain why the two differ the way they do.
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Use the Michaelis–Menten equation: v = Vmax[S]/(Km + [S]). Uninhibited: v = 100 × 2/(2 + 2) = 200/4 = 50 µmol/min. With inhibitor B (competitive): Vmax = 100, apparent Km = 8, so v = 100 × 2/(8 + 2) = 200/10 = 20 µmol/min. The rate is lower with B because at 2 mM the substrate is far below the raised Km, so a smaller fraction of enzyme is substrate-bound. The key prediction is that raising [S] to very high values (e.g., 100 mM) would restore the rate toward 100 µmol/min with B, whereas with A the rate would plateau at 50 µmol/min no matter how much substrate is added. This distinguishes competitive from noncompetitive inhibition experimentally.