The Phosphate Analogue That Was Not Non-Competitive
A final-year student purifies alkaline phosphatase from calf intestine and measures initial rates at 37 °C and pH 9.8 using p-nitrophenyl phosphate as substrate. Rates are nmol product formed per minute per µg protein. The substrate series is run without inhibitor and again with 2 mM of a phosphate analogue. The student concludes the analogue is non-competitive because 'the inhibited rate is lower at every substrate concentration'. A supervisor disputes this. The student has one bench session left and can test substrate up to 50 mM, but the analogue precipitates above 3 mM. A Lineweaver–Burk template is available.
| Substrate (mM) | Rate without inhibitor | Rate with 2 mM analogue |
|---|---|---|
| 0.5 | 10 | 5 |
| 1.0 | 17 | 10 |
| 2.0 | 25 | 17 |
| 4.0 | 33 | 25 |
| 8.0 | 40 | 33 |
1. Evaluate the student's conclusion that the analogue is a non-competitive inhibitor. Use the data to estimate Vmax and Km for both series and state the most likely inhibition type, with justification.
Read the worked answer
Lower rates at every [S] occur with competitive, non-competitive and mixed inhibition, so the student's observation does not identify the type. A Lineweaver–Burk plot (1/v against 1/[S]) gives, without inhibitor, points (2.0, 0.100), (1.0, 0.059), (0.50, 0.040), (0.25, 0.030), (0.125, 0.025); with analogue, (2.0, 0.200), (1.0, 0.100), (0.50, 0.059), (0.25, 0.040), (0.125, 0.030). Both lines extrapolate to a similar y-intercept of about 0.02, so Vmax ≈ 50 nmol min⁻¹ µg⁻¹ is unchanged. The x-intercepts differ: about −0.5 mM⁻¹ (Km ≈ 2 mM) without inhibitor and about −0.25 mM⁻¹ (apparent Km ≈ 4 mM) with it. A non-competitive inhibitor would lower Vmax and leave Km unchanged. The pattern here is competitive: Vmax unchanged, apparent Km doubled. The student should conclude competitive inhibition, not non-competitive.
2. Explain at the molecular level why the apparent Km changes but Vmax does not, and what this implies for inhibition in a cell where substrate concentration is low.
Read the worked answer
The analogue and p-nitrophenyl phosphate compete reversibly for the same active-site pocket. When the analogue occupies the site, substrate cannot bind, so at a given [S] fewer enzyme molecules are active. Raising [S] increases the chance that a substrate molecule, rather than the analogue, occupies the site; at very high [S] the inhibitor is outcompeted and the enzyme can still be fully saturated, so Vmax is unchanged. Km is the substrate concentration giving half Vmax: because more substrate is needed to half-saturate the enzyme in the presence of inhibitor, the apparent Km rises, here from about 2 mM to about 4 mM. In a cell where substrate concentration is near or below Km, a competitive inhibitor produces a large fractional inhibition; the effect is greatest at low [S] and can be relieved by raising [S] or removing the inhibitor. This matters for phosphate analogues and for drugs designed as competitive inhibitors.
3. The student can test one more substrate concentration, 16 mM, with and without 2 mM inhibitor. Predict the results and explain how they would confirm your conclusion.
Read the worked answer
At 16 mM substrate the uninhibited rate should be about 44–45 (approaching Vmax 50) and the inhibited rate about 40. The absolute gap remains, but the ratio inhibited/uninhibited rises from 33/40 = 0.83 at 8 mM to about 0.89 at 16 mM and approaches 1 as [S] rises further. A competitive inhibitor gives exactly this convergence. A non-competitive inhibitor with the same Vmax reduction would keep the ratio roughly constant, about 0.7, at all [S]. Testing 16 mM is feasible because the analogue is used at 2 mM, below its 3 mM solubility limit; testing 50 mM substrate is also possible, but the analogue concentration cannot be increased. The prediction is therefore a narrowing gap, confirming competitive inhibition. If the gap stayed constant, the conclusion would be non-competitive or mixed and the model would need revision.