An enzyme follows Michaelis-Menten kinetics. When substrate concentration equals the Km, the reaction velocity (v) = Vmax/2. A competitive inhibitor doubles the apparent Km without changing Vmax. At substrate concentration = Km (without inhibitor), what fraction of Vmax is achieved in the presence of the competitive inhibitor (assuming inhibitor concentration equals Ki)?
- A Vmax/3 ✓
- B Vmax/4
- C Vmax/2
- D Vmax/5
Explanation
With a competitive inhibitor where [I] = Ki, the apparent Km becomes Km_app = Km × (1 + [I]/Ki) = Km × 2 = 2Km. Using the Michaelis-Menten equation: v = Vmax × [S] / (Km_app + [S]). Substituting [S] = Km (original Km value, not doubled): v = Vmax × Km / (2Km + Km) = Vmax × Km / 3Km = Vmax/3. This is less than Vmax/2 (which would be obtained at [S] = Km in the absence of inhibitor). This mathematical exercise illustrates that competitive inhibitors are more potently inhibitory at substrate concentrations below Km_app, and that the apparent Km must be exceeded to achieve v = Vmax/2. Physiologically, this is why competitive inhibitors are overcome by high substrate concentrations.
Reference: Harper's Illustrated Biochemistry, 32nd ed.
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