An enzyme follows Michaelis-Menten kinetics with Km = 2 mM and Vmax = 100 nmol/min. A competitive inhibitor is added at a concentration equaling its Ki. Substrate concentration is maintained at 2 mM. What is the observed reaction velocity?

• A 33 nmol/min ✓
• B 50 nmol/min
• C 25 nmol/min
• D 100 nmol/min

Explanation

With a competitive inhibitor at [I] = Ki, the apparent Km increases to Km(app) = Km(1 + [I]/Ki) = 2 mM × (1 + 1) = 4 mM. Vmax is unchanged. Using the Michaelis-Menten equation: v = Vmax × [S] / (Km(app) + [S]) = 100 × 2/(4+2) = 100 × 2/6 = 33.3 nmol/min ≈ 33 nmol/min. Competitive inhibitors increase apparent Km without changing Vmax, so at [S] = Km, v = Vmax/2 without inhibitor (50 nmol/min), but with inhibitor the effective Km doubles, reducing velocity to approximately one-third of Vmax.

Reference: Harper's Illustrated Biochemistry, 32nd ed.

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