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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 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 2
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 3
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.
Question 4
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.
Catalase, a heme enzyme found in peroxisomes of liver cells, converts hydrogen peroxide to water and oxygen: 2 H₂O₂ → 2 H₂O + O₂. A student prepares a purified liver catalase solution and measures initial rates of oxygen evolution at 37 °C and pH 7.0 at several peroxide concentrations. Each assay contains the same amount of enzyme. Rates are expressed as µmol of H₂O₂ consumed per minute.
| [H₂O₂] (mM) | Rate (µmol/min) |
|---|---|
| 5 | 13.3 |
| 10 | 22.9 |
| 25 | 40.0 |
| 50 | 53.3 |
| 100 | 64.0 |
| 200 | 71.1 |
A double-reciprocal plot of 1/rate versus 1/[H₂O₂] for these data is a straight line with a y-intercept of 0.0125 min/µmol and an x-intercept of –0.040 mM⁻¹. Catalase is a tetramer of identical subunits, each carrying one heme group at its active site. The student also measures the effect of temperature on a fixed substrate concentration of 50 mM.
| Temperature (°C) | Relative activity (%) |
|---|---|
| 20 | 45 |
| 30 | 75 |
| 37 | 100 |
| 50 | 70 |
| 60 | 15 |
Hydrogen peroxide is formed as a toxic by-product of several oxidase enzymes that are located in the peroxisome.
Question 5
A second catalase preparation gives a double-reciprocal line with the same x-intercept as in Table 1 but a y-intercept twice as large. Which values describe the second preparation?
Vmax = 40 µmol/min, Km = 12.5 mM
Vmax = 160 µmol/min, Km = 25 mM
Vmax = 80 µmol/min, Km = 50 mM
Vmax = 40 µmol/min, Km = 25 mM
Answer: D. Vmax = 40 µmol/min, Km = 25 mM
The y-intercept is 1/Vmax, so doubling it halves Vmax to 40 µmol/min. The x-intercept is –1/Km and is unchanged, so Km stays at 25 mM. A halves Km as well, B reads a larger intercept as a larger Vmax, and C changes Km instead of Vmax.
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