A man consumes a large quantity of alcohol. Hepatic alcohol dehydrogenase has a very low Km for ethanol compared with the blood concentrations achieved after drinking. The rate of ethanol elimination under these conditions is best described as:
- A First-order, directly proportional to blood ethanol concentration
- B Zero-order, constant regardless of blood ethanol concentration ✓
- C Exponentially declining over time
- D Second-order, dependent on both ethanol and NAD+ concentrations equally
Correct answer: B. Zero-order, constant regardless of blood ethanol concentration
Explanation
When substrate concentration greatly exceeds Km, the enzyme is saturated and works at Vmax, so velocity becomes independent of substrate concentration: this is zero-order kinetics. Alcohol dehydrogenase is saturated even at low drinking levels, so ethanol is eliminated at a fixed rate (roughly 7 to 10 g per hour), which is why blood levels fall linearly rather than exponentially. This contrasts with most drugs, which follow first-order elimination at therapeutic concentrations.
Reference: Lippincott Illustrated Reviews: Biochemistry, 8th ed.
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