Enzyme A has Km = 0.1 mM and Vmax = 100 nmol/min/mg. Enzyme B has Km = 2 mM and Vmax = 100 nmol/min/mg. At a substrate concentration of 0.1 mM, Enzyme A operates at approximately 50% Vmax while Enzyme B operates at approximately 5% Vmax. This difference in Km BEST reflects which property of these enzymes?
- A Enzyme A has lower catalytic efficiency (kcat/Km) than Enzyme B
- B Enzyme A has higher affinity for its substrate than Enzyme B, allowing efficient catalysis at low substrate concentrations ✓
- C Enzyme B is a more potent enzyme in terms of maximum velocity
- D The two enzymes have different allosteric regulatory mechanisms
Explanation
The Michaelis constant Km is inversely related to the affinity of the enzyme for its substrate: a lower Km indicates higher substrate affinity (tighter binding). At any given substrate concentration below saturation, an enzyme with lower Km will operate at a higher fraction of its Vmax. This has physiological significance — hexokinase (low Km, ~0.1 mM for glucose) is saturated at normal blood glucose and serves all tissues, while glucokinase (high Km, ~10 mM) acts as a glucose sensor in hepatocytes and pancreatic beta cells, only becoming significantly active when glucose rises postprandially.
Reference: Harper's Illustrated Biochemistry, 32nd ed.
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