The inhibitor that high substrate cannot outrun
A student investigates a digestive protease with a colorimetric assay that measures product formed in 5 minutes. Substrate concentrations of 1, 2, 5, 10 and 20 mmol L⁻¹ give initial rates of 6.7, 11.4, 20.0, 26.7 and 32.0 µmol min⁻¹. The whole experiment is then repeated with 5 µmol L⁻¹ of a compound extracted from a local plant added to every tube, and the rates become 4.2, 7.1, 12.5, 16.7 and 20.0 µmol min⁻¹.
The student writes: "The inhibitor is competitive. At the lowest substrate concentration it cuts the rate a lot (6.7 down to 4.2), and the substrate and inhibitor must be competing for the same site. Raising the substrate concentration should therefore restore the original maximum rate."
A classmate says the data do not support this. The teacher asks for an analysis of both curves, a decision on the type of inhibition, and a prediction of what would be seen at 100 mmol L⁻¹.
1. Estimate Vmax and Km for the uninhibited and inhibited reactions from the data, showing the reasoning rather than quoting a formula, and classify the inhibition.
Read the worked answer
For the uninhibited enzyme, the rate at 5 mmol L⁻¹ is exactly 20.0 µmol min⁻¹. If this is half of Vmax, Vmax = 40, and Km = 5 mmol L⁻¹, because the rate is half maximal when [S] = Km. Check against the other points with v = 40[S]/(5 + [S]): 1 → 6.7; 2 → 11.4; 10 → 26.7; 20 → 32.0. All match, so Vmax = 40 µmol min⁻¹, Km = 5 mmol L⁻¹.
For the inhibited data, the same test: the rate at 5 mmol L⁻¹ is 12.5, and if that is half of a new maximum of 25, then Km is again 5. Check with v = 25[S]/(5 + [S]): 1 → 4.2; 2 → 7.1; 10 → 16.7; 20 → 20.0. All match, so Vmax = 25 µmol min⁻¹, Km = 5 mmol L⁻¹.
The inhibitor lowers Vmax by 37.5 % and leaves Km unchanged: that is non-competitive inhibition. A pure competitive inhibitor would leave Vmax untouched and raise Km, and a pure uncompetitive inhibitor would lower both.
2. Identify the flaw in the student's argument, using the data as evidence, and state the general rule for reading inhibition type from rate data.
Read the worked answer
The student uses a low-substrate observation, which cannot discriminate. All reversible inhibitors reduce the rate at low [S], because the enzyme is mostly free and any binding removes it from the reaction. The diagnostic behaviour appears at high [S], where a competitive inhibitor is outcompeted and Vmax is restored, while a non-competitive one still leaves a fraction of enzyme permanently unable to turn over.
The cleanest evidence here is the ratio of inhibited to uninhibited rate at each concentration: 4.2/6.7 = 0.63, 7.1/11.4 = 0.62, 12.5/20.0 = 0.625, 16.7/26.7 = 0.63, 20.0/32.0 = 0.625. The inhibition is a constant 37.5 % at every substrate concentration, including the highest one available. A competitive inhibitor's effect would shrink as [S] rises; this one does not shrink at all.
The rule: compare Vmax and Km. Vmax falls and Km is unchanged → non-competitive; Km rises with Vmax unchanged → competitive; both fall → uncompetitive.
3. Predict the rate at 100 mmol L⁻¹ with the inhibitor present, and describe one experiment that would establish whether the inhibition is reversible, explaining what result would settle the question.
Read the worked answer
With Vmax = 25 and Km = 5, at 100 mmol L⁻¹ the rate is 25 × 100/(5 + 100) = 23.8 µmol min⁻¹, not 40. The curve has simply levelled off at a lower plateau; no amount of substrate recovers the lost activity, which is why the student's proposed test - keep adding substrate - would fail and would itself be the experiment that refutes the competitive claim.
To test reversibility, separate inhibitor from enzyme and see whether activity returns. Dialyse the inhibited enzyme against a large volume of buffer, or pass it through a gel-filtration column, then re-assay. If Vmax returns to 40 µmol min⁻¹, the inhibitor binds reversibly and can dissociate; if activity stays at 25, the compound has bound irreversibly or denatured the enzyme, and the constant fractional reduction is a chemical modification rather than an equilibrium. A supporting check is to incubate enzyme with inhibitor before adding substrate, varying the pre-incubation time: a progressive loss of activity with time indicates irreversible, time-dependent inactivation.