Classifying Reversible Enzyme Inhibition from Rate Data
An undergraduate assays an enzyme that converts a single substrate to a coloured product. With 0.50 μM enzyme and no inhibitor, the initial rates at substrate concentrations of 2.0, 5.0 and 10.0 mM are 5.7, 10.0 and 13.3 μmol min⁻¹ respectively. The same runs are repeated with 5.0 mM of a reversible inhibitor present, giving 3.3, 6.7 and 10.0 μmol min⁻¹. The student labels the compound 'non-competitive because the rate falls at every substrate concentration', and proposes raising the substrate concentration as a way to restore the uninhibited rate. His supervisor asks him to justify the classification from the pattern of the kinetic constants, to state what happens at 20 mM substrate, and to predict the effect of raising the inhibitor concentration to 15 mM. The assay is confirmed to be linear over the times used, and no substrate inhibition or aggregation is present.
1. Classify the inhibition from the data, with reasoning.
Read the worked answer
Estimate the constants from v = Vmax[S]/(Km + [S]).
No inhibitor: using the 5 and 10 mM points, 10.0 = 5V/(Km + 5) and 13.3 = 10V/(Km + 10) → Vmax = 20 μmol min⁻¹, Km = 5.0 mM (check at 2 mM: 20 × 2/7 = 5.7 ✓).
With inhibitor: at 10 mM, 20 × 10/(10 + 10) = 10.0 ✓, so Vmax is again 20 μmol min⁻¹, but Km has risen to 10 mM.
Unchanged Vmax with increased Km is the signature of competitive inhibition: the inhibitor occupies the substrate site and can be displaced. Non-competitive inhibition would lower Vmax and leave Km unchanged. 'The rate falls' alone distinguishes nothing - it is true of every inhibitor.
2. Predict the rate at 20 mM substrate, with and without inhibitor, and estimate Ki.
Read the worked answer
Uninhibited: v = 20(20)/(5 + 20) = 16.0 μmol min⁻¹. With 5.0 mM inhibitor (Km_app = 10 mM): v = 20(20)/(10 + 20) = 13.3 μmol min⁻¹.
So raising [S] recovers part of the rate but never all of it at any finite concentration - the two curves converge only as [S] → ∞. That is why the student's proposal is qualitatively right but overstated.
From Km_app = Km(1 + [I]/Ki): 10 = 5(1 + 5.0/Ki) → 2 = 1 + 5.0/Ki → Ki = 5.0 mM.
3. What rate is expected at 20 mM substrate if the inhibitor is raised to 15 mM?
Read the worked answer
Km_app = Km(1 + [I]/Ki) = 5(1 + 15/5) = 5 × 4 = 20 mM; Vmax stays 20 μmol min⁻¹.
v = 20(20)/(20 + 20) = 10.0 μmol min⁻¹, i.e. 62 % of the uninhibited 16.0 μmol min⁻¹ at the same substrate concentration.
The prediction is quantitative and testable: if the measured rate at 15 mM inhibitor falls well below 10 μmol min⁻¹, or if Vmax drops, the competitive assignment is wrong and a mixed or non-competitive model must be fitted instead.