An enzyme follows Michaelis-Menten kinetics. A competitive inhibitor doubles the apparent Km without changing Vmax. If the original Km = 2 mM and the substrate concentration used in the assay is 8 mM, what is the velocity as a fraction of Vmax in the presence of the inhibitor?
- A V = 0.5 Vmax
- B V = 0.67 Vmax ✓
- C V = 0.8 Vmax
- D V = 0.33 Vmax
Correct answer: B. V = 0.67 Vmax
Explanation
With competitive inhibitor, apparent Km doubles: Km(app) = 4 mM. Using Michaelis-Menten: V = Vmax × [S] / (Km(app) + [S]) = Vmax × 8 / (4 + 8) = Vmax × 8/12 = 0.67 Vmax. Without inhibitor: V = Vmax × 8/(2+8) = 0.8 Vmax. This illustrates how competitive inhibitors reduce velocity at a given substrate concentration (by increasing effective Km) but can be overcome by excess substrate (Vmax unchanged). Clinically, competitive inhibitors have direct application in drug design — e.g., statins competitively inhibit HMG-CoA reductase.
Reference: Harper's Illustrated Biochemistry, 32nd ed.
High-yield for: NEET PGINI-CETNExTFMGEUSMLEPLABMRCP
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