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MCAT · Chemical and Physical Foundations · Biochemistry
Enzyme kinetics is part of Biochemistry in the MCAT Chemical and Physical Foundations section. The questions below are written in the style of the real test, each with the answer and a worked explanation.
Question types you will see here include reading a Michaelis-Menten curve, lineweaver-Burk intercepts and slope, competitive inhibition, noncompetitive and uncompetitive and irreversible inhibition.
Common traps: reading Km as the substrate at Vmax; letting Vmax depend on substrate; treating rate as proportional to substrate at saturation.
Question 1
An enzyme obeying Michaelis–Menten kinetics has Vmax = 80 µM/min and reaches 40 µM/min at [S] = 4 mM. What is the rate at [S] = 12 mM?
60 µM/min
80 µM/min
120 µM/min
240 µM/min
Answer: A. 60 µM/min
Half of Vmax at 4 mM means Km = 4 mM, so v = 80 × 12/(4 + 12) = 60 µM/min (A). B assumes the enzyme is already saturated, while C and D extrapolate linearly (3 × 40 and 3 × 80), which a hyperbolic curve that never exceeds Vmax rules out.
Question 2
An enzyme has Km = 5 mM. A competitive inhibitor with Ki = 5 mM is added at 10 mM. What is the apparent Km?
5 mM
10 mM
15 mM
30 mM
Answer: C. 15 mM
A competitive inhibitor raises the apparent Km by the factor (1 + [I]/Ki) = 1 + 10/5 = 3, so Km,app = 5 × 3 = 15 mM. Vmax is unchanged, because enough substrate still outcompetes the inhibitor.
Question 3
An enzyme has Km = 1 mM. A competitive inhibitor with Ki = 1 mM is added at 3 mM. What is the apparent Km?
0.25 mM
1 mM
3 mM
4 mM
Answer: D. 4 mM
A competitive inhibitor raises the apparent Km by the factor (1 + [I]/Ki) = 1 + 3/1 = 4, so Km,app = 1 × 4 = 4 mM. Vmax is unchanged, because enough substrate still outcompetes the inhibitor.
Question 4
Enzyme P has kcat = 100 s⁻¹ and Km = 10 µM for a substrate. Enzyme Q has kcat = 1000 s⁻¹ and Km = 1 mM for the same substrate. Which statement is correct?
P has 10 times the catalytic efficiency of Q.
Q has 10 times the catalytic efficiency of P.
Q has 100 times the catalytic efficiency of P.
P and Q have equal catalytic efficiency.
Answer: A. P has 10 times the catalytic efficiency of Q.
Catalytic efficiency is kcat/Km: for P, 100 s⁻¹ / (1 × 10⁻⁵ M) = 1 × 10⁷ M⁻¹s⁻¹; for Q, 1000 s⁻¹ / (1 × 10⁻³ M) = 1 × 10⁶ M⁻¹s⁻¹, so P is 10 times more efficient. Choosing Q looks only at kcat and ignores Q's 100-fold weaker substrate affinity.
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