Inhibitor Choice When Substrate Concentration Is Fixed by Substrate Inhibition
A thermophilic amidase (Vmax 120 µmol min⁻¹ mg⁻¹, Km 0.40 mM for its acyl substrate) is being characterised for an industrial deacylation step. The enzyme is strongly substrate-inhibited above 0.15 mM, so all kinetic work must be done at a fixed free-substrate concentration of 0.10 mM. A screen of a small-molecule library returned two hits, and each was profiled at 0.50 mM in the same buffer (pH 8.0, 55 °C): compound P, a substrate analogue, raised the apparent Km to 1.20 mM while Vmax stayed at 120 µmol min⁻¹ mg⁻¹; compound Q, a chelator, left Km at 0.40 mM but lowered Vmax to 30 µmol min⁻¹ mg⁻¹. Both compounds are soluble and stable under assay conditions, and neither absorbs at the wavelength used to follow product formation. The team must decide which compound to take forward as a mechanistic probe and whether either can serve as a reliable reaction quench.
1. Calculate the uninhibited rate at 0.10 mM substrate and the rates in the presence of P and Q. Express each effect as a percentage inhibition and assign an inhibition type to each compound.
Read the worked answer
Use the Michaelis–Menten equation v = Vmax[S]/(Km + [S]) with [S] = 0.10 mM.
- Uninhibited: v = 120 × 0.10/(0.40 + 0.10) = 12/0.50 = 24 µmol min⁻¹ mg⁻¹.
- With P (Km 1.20, Vmax 120): v = 12/(1.20 + 0.10) = 12/1.30 = 9.2 µmol min⁻¹ mg⁻¹ → inhibition = (24 − 9.2)/24 = 62%.
- With Q (Km 0.40, Vmax 30): v = 3.0/0.50 = 6.0 µmol min⁻¹ mg⁻¹ → inhibition = (24 − 6.0)/24 = 75%.
Mechanistic assignment: P raises Km with Vmax unchanged, the signature of competitive inhibition (P binds the free enzyme and competes with substrate). Q lowers Vmax with Km unchanged, the signature of non-competitive (or irreversible/inactivation-type) loss of active enzyme; a chelator stripping a catalytic metal is consistent with this, and the effect would not be reversed by adding more substrate.
2. Estimate Ki for P from the measurement at 0.50 mM, and predict the inhibition at 5 mM substrate. Explain why the [S] constraint changes the practical meaning of that Ki.
Read the worked answer
For competitive inhibition, Km(app) = Km(1 + [I]/Ki).
1.20 = 0.40(1 + 0.50/Ki) → 3.0 = 1 + 0.50/Ki → 0.50/Ki = 2.0 → Ki = 0.25 mM.
Prediction at 5 mM (hypothetical): uninhibited v = 120 × 5/5.40 = 111.1; with P, Km(app) = 1.20, v = 600/6.20 = 96.8 → inhibition only 13%. This illustrates the defining feature of competitive inhibition: at [S] ≫ Km the inhibitor is outcompeted, so its potency is substrate-dependent.
The practical point is that this escape route is closed here. The enzyme cannot be run above 0.15 mM because of substrate inhibition, so P is effectively locked at ~62% inhibition, and its Ki of 0.25 mM overstates how useful it is under process conditions. Potency must be judged at the working [S], not from Ki alone.
3. Recommend (a) which compound to use as a reversible mechanistic probe and (b) whether Q can be used to quench reactions. Justify each choice from the mechanism implied by the constants.
Read the worked answer
(a) P is the probe. It is a classical reversible competitive inhibitor: Vmax is untouched, so the enzyme is not destroyed, and the effect can be titrated and reversed by raising [S] or by dialysis. It therefore supports mechanistic work (mapping the substrate pocket, varying [S] to confirm competitiveness). The caveat is that any IC50 reported for P must state the [S] used, because the value shifts with substrate concentration.
(b) Q is unsuitable as a probe but plausible as a quench. A fall in Vmax with unchanged Km indicates that a fraction of the enzyme has been inactivated, consistent with irreversible removal of a catalytic metal. That makes the inhibition poorly reversible and concentration/time-dependent, so it cannot be used to define Ki. As a quench it works because it stops turnover rapidly and completely, but two controls are needed: show that the quench is immediate (stopped-time assay) and confirm that Q does not interfere with the detection chemistry. A dialysis or dilution experiment (activity not restored after removing Q) would confirm irreversibility.