MCAT practice · Biological and Biochemical Foundations · 1A Proteins and enzymes
A new inhibitor of a bacterial phosphatase
An original practice passage written for this site, not an AAMC item. Exam pace is about 8 minutes for a passage like this.
Some Gram-negative pathogens rely on periplasmic phosphatases to scavenge phosphate from their hosts, which makes these enzymes candidate drug targets. In a screen of a small-molecule library, researchers found that compound X slowed a purified periplasmic phosphatase, here called enzyme P. To learn how X acts, they characterized it from initial-rate data.
Enzyme P (2 nM) was assayed at 25 °C and pH 8.0 with the colorless substrate p-nitrophenyl phosphate (pNPP). Hydrolysis releases p-nitrophenolate, which absorbs light at 405 nm, so product formation was followed continuously in a spectrophotometer. Each initial velocity (v₀) was taken from the linear first 30 s of a reaction, during which less than 5% of the substrate was consumed. In Experiment 1, v₀ was measured at five pNPP concentrations from 0.25 to 4.0 mM, with 0, 5 or 10 µM compound X present. Compound X slowed hydrolysis at every substrate concentration tested. The results are shown as a double-reciprocal plot in Figure 1.
Experiment 2 asked whether inhibition persists after X is removed. Enzyme P (200 nM) was preincubated for 30 min with buffer or with 50 µM X. Each sample was then diluted 100-fold, either into plain assay buffer or into assay buffer that also contained 50 µM X, and its activity was measured at 2.0 mM pNPP. Every assay therefore contained 2 nM enzyme. A final reaction contained 50 µM X and 2.0 mM pNPP but no enzyme. The results are summarized in Table 1.
The researchers are now preparing to crystallize enzyme P with the compound bound.
- No compound X
- 5 µM X
- 10 µM X
| Preincubation | Diluted into | X in assay (µM) | Activity (% of control) |
|---|---|---|---|
| Enzyme + buffer | Buffer | 0 | 100 ± 4 |
| Enzyme + 50 µM X | Buffer with 50 µM X | 50 | 33 ± 3 |
| Enzyme + 50 µM X | Buffer | 0.5 | 97 ± 5 |
| No enzyme; 50 µM X | Buffer with 50 µM X | 50 | < 1 |
Data are illustrative, constructed for practice.
1. Based on Figure 1, what are the apparent Km and Vmax of enzyme P in the presence of 10 µM compound X?
- A. Wrong: this is the pattern for noncompetitive inhibition (Km unchanged, Vmax lowered). In Figure 1 the 10 µM line meets the y-axis at the same 1/v₀ as the control, so Vmax is unchanged, and its slope is three times the control's, so Km is not 0.50 mM.
- B. Wrong: 0.67 is the magnitude of the x-intercept, which is 1/Km rather than Km. Taking its reciprocal gives an apparent Km of 1.5 mM.
- C. Wrong: Km is right, but 27 µM/min comes from reading the slope (0.0375) as 1/Vmax. The slope is Km/Vmax; Vmax comes from the y-intercept, 0.025 min/µM, which gives 40 µM/min.
- D. Right: the y-intercept is 1/Vmax = 0.025 min/µM, so Vmax = 40 µM/min, the same as without X. The x-intercept is −1/Km = −0.667 mM⁻¹, so the apparent Km is 1.5 mM (equivalently, slope/intercept = 0.0375/0.025 = 1.5 mM).
2. The results of Experiment 1 are most consistent with compound X binding to:
- A. Wrong: an inhibitor that binds free enzyme and ES equally (noncompetitive) lowers Vmax with Km unchanged, so the lines would share an x-intercept. Here the x-intercepts move toward zero as [X] rises while the y-intercept stays fixed.
- B. Wrong: removing active enzyme permanently would lower Vmax, but Vmax is unchanged in Figure 1, and Table 1 shows that activity returns to 97% of control when X is diluted away, so binding is reversible.
- C. Right: the lines share a y-intercept (Vmax unchanged) and the apparent Km rises with [X]. That is competitive inhibition: X and substrate compete for the free enzyme's active site, so enough substrate outcompetes X.
- D. Wrong: an inhibitor that binds only the ES complex lowers both Km and Vmax and gives parallel double-reciprocal lines. In Figure 1 the slopes differ and the lines share a y-intercept.
3. The reaction in Table 1 that contained compound X and pNPP but no enzyme was included mainly to rule out which possibility?
- A. Wrong: whether X changes Vmax or Km is answered by Experiment 1, which varies substrate concentration. A reaction lacking enzyme has no Vmax or Km to measure.
- B. Right: without enzyme, any change at 405 nm would have to come from X itself or from pNPP hydrolyzing on its own in the presence of X. The rate below 1% of control shows the signal in the other assays comes from enzyme P.
- C. Wrong: reversibility is tested by the third row of Table 1, where enzyme preincubated with X regains 97% of control activity after dilution. A reaction with no enzyme cannot address it.
- D. Wrong: saturation is a property of the enzyme and is read from how v₀ changes with [S] in Experiment 1. A reaction with no enzyme says nothing about it.
4. Suppose the sample in the second row of Table 1, with 50 µM X in the assay, were measured at 20 mM pNPP instead of 2.0 mM and compared with a buffer control also measured at 20 mM. Its activity would most likely be:
- A. Wrong: substrate overcomes a competitive inhibitor fully only as [S] becomes far larger than the apparent Km. With an apparent Km near 5.5 mM at 50 µM X, 20 mM pNPP is not high enough.
- B. Right: X is competitive, so raising [S] recovers activity. Figure 1 shows the apparent Km rising 0.5 mM per 5 µM X, so at 50 µM it is about 5.5 mM. At 20 mM pNPP the sample runs at about 20/25.5 = 78% of Vmax versus 98% for the control, about 80% of control.
- C. Wrong: a fixed fraction of activity regardless of [S] is the pattern of noncompetitive inhibition, which lowers Vmax. X leaves Vmax unchanged, so its effect depends on [S].
- D. Wrong: this reverses the effect of substrate. For a competitive inhibitor, more substrate displaces X from the active site, so the fraction of activity that remains rises.
5. Compared with the uninhibited reaction, how does adding 10 µM compound X change the hydrolysis of pNPP by enzyme P?
- A. Wrong: this is internally inconsistent. ΔG°′ fixes Keq, so if ΔG°′ is unchanged the equilibrium yield of product is unchanged too.
- B. Wrong: Keq is set by the free energies of substrate and products. An inhibitor, like the enzyme itself, changes only how fast equilibrium is reached.
- C. Right: X reduces the rate of catalysis, but neither enzyme nor inhibitor changes ΔG°′ or Keq. The same equilibrium is reached, only more slowly.
- D. Wrong: ΔG°′ is a property of the reaction, not of the catalyst. Inhibiting the enzyme slows the reaction without making it thermodynamically less favorable.
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