Two mutant forms of an enzyme are studied. Mutant X has kcat = 100 s⁻¹ and Km = 1 mM. Mutant Y has kcat = 500 s⁻¹ and Km = 50 mM. At a substrate concentration of 0.05 mM, which statement is correct?

• A Mutant Y is faster because it has the higher kcat
• B Mutant X is faster because kcat/Km, not kcat alone, governs efficiency at low substrate concentration ✓
• C Both mutants operate at identical velocity because their kcat/Km ratios are equal
• D Neither mutant functions measurably at this substrate concentration

Explanation

When [S] is far below Km, the Michaelis-Menten equation reduces to v = (kcat/Km)[E][S]. Mutant X has kcat/Km = 100,000 M⁻¹s⁻¹ while mutant Y has 10,000 M⁻¹s⁻¹, so X works tenfold faster despite the lower turnover number. Mutant Y only overtakes X once [S] approaches its much larger Km. This comparison tests whether students can apply catalytic efficiency rather than memorising kcat alone.

Reference: Lehninger Principles of Biochemistry, 7th ed.

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