Enzyme Inhibition Kinetics

Introduction

Many drugs, poisons and natural regulators work by slowing enzymes. How they do so leaves a kinetic signature. By measuring initial rates at several substrate and inhibitor concentrations, you can tell whether an inhibitor competes for the active site, binds only after substrate, or does both. This page extends the Michaelis–Menten model to reversible inhibition and shows how the three classic patterns appear in the apparent Vmax and KM.

Core explanation

Reversible inhibitors bind and release rapidly, so their binding is treated as an equilibrium. Two dissociation constants describe them: Ki for EI ⇌ E + I, and K′i for ESI ⇌ ES + I. Define α = 1 + [I]/Ki and α′ = 1 + [I]/K′i. Adding the inhibited forms to the enzyme conservation equation gives the general result

v = Vmax[S] ÷ (αKM + α′[S])

which can be rearranged to the Michaelis–Menten form with apparent constants Vmax,app = Vmax/α′ and KM,app = αKM/α′.

Competitive inhibition (binds E only; α′ = 1). The inhibitor usually resembles the substrate and occupies the active site. Vmax is unchanged, because enough substrate outcompetes the inhibitor, but KM,app = αKM increases. On a Lineweaver–Burk plot the lines for different [I] meet on the 1/v axis.

Uncompetitive inhibition (binds ES only; α = 1). The inhibitor binds a site that exists only after substrate has bound. Both Vmax and KM are divided by α′, so their ratio, the slope KM/Vmax, is unchanged. Double-reciprocal lines are parallel. Paradoxically, the apparent affinity for substrate seems to increase, because the inhibitor pulls ES into the dead-end ESI complex.

Mixed inhibition (binds both, Ki ≠ K′i). Vmax falls and KM may rise or fall. Lines intersect to the left of the 1/v axis. The special case Ki = K′i is non-competitive inhibition: Vmax falls by α while KM is unchanged, and lines meet on the 1/[S] axis.

Physical interpretation. Only uncompetitive and mixed inhibitors lower Vmax, because only they trap enzyme that is already bound to substrate; no amount of substrate can reverse this. A competitive inhibitor can always be swamped by high substrate.

Irreversible inhibitors differ entirely: they covalently modify the enzyme, effectively reducing [E]₀. They lower Vmax in a time-dependent way and cannot be treated with equilibrium constants.

Formulae

α = 1 + [I]/Ki; α′ = 1 + [I]/K′i.

General: v = Vmax[S] ÷ (αKM + α′[S]).

Competitive: KM,app = αKM, Vmax,app = Vmax. Uncompetitive: KM,app = KM/α′, Vmax,app = Vmax/α′. Non-competitive: KM,app = KM, Vmax,app = Vmax/α.

Step-by-step reasoning

To diagnose an inhibitor:

1. Measure initial rates over a range of [S] at zero and at two or more [I]. 2. Fit each set to find Vmax,app and KM,app. 3. If Vmax is unchanged, the inhibitor is competitive. 4. If Vmax and KM fall by the same factor, it is uncompetitive. 5. Otherwise it is mixed; use the changes to find Ki and K′i.

Visual explanation

On a Lineweaver–Burk plot, draw the uninhibited line and one inhibited line for each type. Competitive: the lines pivot about a common point on the 1/v axis, with steeper slope. Uncompetitive: the inhibited line is shifted straight upward, parallel to the original. Mixed: the lines cross in the second quadrant, left of the vertical axis.

Real-world analogy

A competitive inhibitor is someone sitting in your seat at the cinema: arrive in greater numbers and you eventually get seats. An uncompetitive inhibitor is someone who locks you in once you are seated: more people arriving does not help, because every seated person can be trapped.

Real-world example

Statin drugs such as atorvastatin competitively inhibit HMG-CoA reductase, a key enzyme in cholesterol synthesis, by mimicking its substrate. Methanol poisoning is treated using ethanol or fomepizole, which compete with methanol for alcohol dehydrogenase and slow formation of toxic metabolites.

Why?

Why does an uncompetitive inhibitor lower KM? Binding of inhibitor to ES removes ES from the equilibrium, so by Le Chatelier's principle more E and S combine to replace it. The enzyme appears to bind substrate more tightly, although the extra complex is inactive.

Common misconception

"All inhibitors lower Vmax." A purely competitive inhibitor leaves Vmax unchanged; its effect disappears at sufficiently high substrate concentration. Only inhibitors that bind the ES complex reduce the saturated rate.

Worked example

Question: An enzyme has KM = 20 μmol dm⁻³. With 5.0 μmol dm⁻³ of a competitive inhibitor, the apparent KM is 60 μmol dm⁻³ and Vmax is unchanged. Find Ki.

Reasoning: α = KM,app/KM = 60/20 = 3.0. Then 1 + [I]/Ki = 3.0, so [I]/Ki = 2.0 and Ki = 5.0/2.0.

Answer: Ki = 2.5 μmol dm⁻³.

Quick check

1. On a Lineweaver–Burk plot, which inhibition type gives parallel lines for different inhibitor concentrations? Answer: Uncompetitive inhibition, because the slope KM/Vmax is unchanged while the intercept increases.

Exam focus

Learn the table of apparent Vmax and KM for each type and the matching Lineweaver–Burk patterns. Be able to calculate Ki from α. Explain physically why competitive inhibition can be overcome by high substrate while uncompetitive cannot.

Advanced insight

In drug discovery, potency is often reported as IC₅₀, the inhibitor concentration giving 50% inhibition under set assay conditions. For a competitive inhibitor, the Cheng–Prusoff relation IC₅₀ = Ki(1 + [S]/KM) shows that IC₅₀ depends on substrate concentration, so IC₅₀ values from different assays cannot be compared directly without converting to Ki.

Summary

Reversible inhibitors are classified by which enzyme forms they bind. Competitive inhibitors raise KM without changing Vmax; uncompetitive inhibitors lower Vmax and KM equally; mixed inhibitors lower Vmax and shift KM. Double-reciprocal plots and fitted apparent constants reveal the type and give Ki.

Practice questions

1. A drug raises the apparent KM of an enzyme but not its Vmax. Which inhibition type is this? Answer: Competitive inhibition. 2. With an uncompetitive inhibitor at [I] = K′i, by what factors do Vmax and KM change? Answer: α′ = 2, so both are halved. 3. A non-competitive inhibitor with Ki = 4.0 μmol dm⁻³ is present at 12 μmol dm⁻³. What fraction of the original Vmax remains? Answer: α = 1 + 12/4.0 = 4.0, so Vmax,app is one quarter, 25%, of the original. 4. Why do irreversible inhibitors not fit these equilibrium models? Answer: They form covalent bonds with the enzyme, so they remove active enzyme permanently and their effect grows with time rather than reaching a binding equilibrium.