During a biochemistry practical, a student measures the rate of an enzyme-catalyzed reaction at several substrate concentrations and obtains a Km of 2 mM and a Vmax of 100 nmol/min. If the substrate concentration in the assay tube is exactly equal to the Km, what fraction of the enzyme's active sites are occupied by substrate at that moment?
- A 25%
- B 75%
- C 50% ✓
- D 100%
Correct answer: C. 50%
Explanation
According to the Michaelis-Menten equation, v = Vmax × [S] / (Km + [S]). When [S] = Km, v = Vmax/2, meaning exactly half of the active sites are occupied (50% saturation). This is in fact the operational definition of Km — the substrate concentration at which reaction velocity equals half the maximum velocity.
Reference: Harper's Illustrated Biochemistry, 32nd ed.
High-yield for: NEET PGINI-CETNExTFMGEUSMLEPLABMRCP
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