For a saturable enzyme with Vmax = 1200 mg/h and Km = 6 mg/L, the intrinsic clearance at very low substrate is equal to which value?

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Multiple Choice

For a saturable enzyme with Vmax = 1200 mg/h and Km = 6 mg/L, the intrinsic clearance at very low substrate is equal to which value?

Explanation:
Intrinsic clearance at very low substrate concentration corresponds to Vmax divided by Km. In Michaelis–Menten terms, v = (Vmax × S) / (Km + S). When S is much smaller than Km, v ≈ (Vmax/Km) × S, so CLint = v/S ≈ Vmax/Km. With Vmax = 1200 mg/h and Km = 6 mg/L, CLint ≈ 1200 / 6 = 200 L/h.

Intrinsic clearance at very low substrate concentration corresponds to Vmax divided by Km. In Michaelis–Menten terms, v = (Vmax × S) / (Km + S). When S is much smaller than Km, v ≈ (Vmax/Km) × S, so CLint = v/S ≈ Vmax/Km. With Vmax = 1200 mg/h and Km = 6 mg/L, CLint ≈ 1200 / 6 = 200 L/h.

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