Chalk−1

Biology · Introductory biology · Worked example

Tell competitive from noncompetitive inhibition

An enzyme is assayed at 1 to 32 mM substrate alone and with each of two inhibitors, A and B. Fitting each series gives Vmax = 20 µmol/min and Km = 4 mM without inhibitor, Vmax = 20 and Km = 12 mM with A, and Vmax = 10 and Km = 4 mM with B. How does each inhibitor act?

Compare each fit with the control

Put the parameters side by side. Inhibitor A leaves Vmax at 20 but triples Km; inhibitor B leaves Km at 4 but halves Vmax.

Fitted parameters
SeriesVmax (µmol/min)Km (mM)
No inhibitor204
Inhibitor A2012
Inhibitor B104

Identify inhibitor A

An unchanged Vmax with a larger apparent Km is the signature of competitive inhibition: the inhibitor competes for the active site, so more substrate is needed to reach any given rate. In the simple model the apparent Km is Km(1 + [I]/Kᵢ), so here 1 + [I]/Kᵢ = 3.

4⁢(1+2)=12

Identify inhibitor B

A lower Vmax with Km unchanged is the signature of pure noncompetitive inhibition: the inhibitor binds away from the active site, and the enzyme it holds cannot work at any substrate level. Here Vmax is divided by 1 + [I]/Kᵢ = 2.

201+1=10

Confirm at high substrate

At [S] = 100 mM the competitive inhibitor is nearly outcompeted, but the noncompetitive one still halves the rate.

20⁢(100)4+100≈19.2320⁢(100)12+100≈17.8610⁢(100)4+100≈9.62

Know what the fit cannot show

These teaching data follow the model exactly, so the fits come out exact. Real rates are noisy, and a mechanism is best supported by several inhibitor concentrations and by binding evidence, not by one pair of fitted numbers.

Result

Inhibitor A is competitive: Km rises from 4 to 12 mM while Vmax stays at 20 µmol/min. Inhibitor B is noncompetitive: Vmax halves to 10 µmol/min while Km stays at 4 mM.

Your turn

A competitive inhibitor is present at half its Kᵢ, so [I]/Kᵢ = 0.5. If Km = 4 mM without it, what is the apparent Km, and what happens to Vmax?

Show the answer and explanation

The apparent Km is 6 mM; Vmax is unchanged.

Apparent Km = Km(1 + [I]/Kᵢ) = 4(1.5) = 6 mM. A competitive inhibitor does not change Vmax.

4⁢(1+0.5)=6

Keep exploring

The enzyme kinetics tool opens with all three series and compares each inhibitor’s fitted Vmax and Km with the control’s: 1 and 3 times for A, 0.5 and 1 times for B.

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