Enzyme Kinetics with a Competitive Inhibitor
Students assay an enzyme that converts substrate X to product Y. In trial 1 they use enzyme plus increasing concentrations of X and record reaction rate. In trial 2 they repeat the experiment with a fixed concentration of compound Z added to every tube. Results: without Z, the maximum rate (Vmax) is 40 µmol/min and the substrate concentration giving half-maximal rate (Km) is 5 mM. With Z, Vmax is still 40 µmol/min but Km rises to 20 mM. A third trial uses compound W: Vmax falls to 20 µmol/min while Km stays at 5 mM. The students also note that raising substrate concentration in trial 2 eventually restores the rate to 40 µmol/min, but in trial 3 it never does.
1. Classify compounds Z and W as competitive or non-competitive inhibitors, justifying each with the data.
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Compound Z - competitive inhibitor.
- Vmax unchanged at 40 µmol/min: enough substrate can still saturate all enzyme active sites, so the maximum is reached.
- Km rises from 5 to 20 mM: more substrate is needed to reach half-maximal rate because inhibitor and substrate compete for the same active site.
- Raising substrate restores full rate: consistent with competition being overcome by mass action.
Compound W - non-competitive inhibitor.
- Vmax falls to 20 µmol/min: some enzyme molecules are effectively removed from the reaction, so saturation cannot reach the original maximum.
- Km unchanged at 5 mM: the affinity of the remaining active enzyme for substrate is unaffected, so half-maximal rate still occurs at 5 mM.
- Raising substrate never restores full rate: consistent with inhibition at a site other than the active site (allosteric), which substrate cannot outcompete.
The decisive judgement is the pattern of the two parameters together, not either alone.
2. A student claims compound Z 'reduces enzyme activity'. Evaluate this claim using the concept of enzyme saturation and explain what would happen to reaction rate at a substrate concentration of 50 mM with and without Z.
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The claim is imprecise. Compound Z does not reduce the enzyme's maximum capacity (Vmax is unchanged); it reduces the apparent affinity for substrate, raising Km fourfold. At any given sub-saturating substrate concentration the observed rate is lower, but the enzyme can still reach full speed if substrate is plentiful.
At 50 mM substrate:
- Without Z: 50 mM is ten times Km (5 mM), so the enzyme is essentially saturated and the rate is close to Vmax, about 40 µmol/min.
- With Z: 50 mM is 2.5 times the new Km (20 mM). Using the Michaelis–Menten relationship, rate = Vmax × [S]/(Km + [S]) = 40 × 50/70 ≈ 28.6 µmol/min.
So at 50 mM the inhibitor still reduces rate, but the gap has narrowed compared with low substrate concentrations, and it would close entirely at very high substrate. This shows why 'reduces activity' must be qualified: the effect of a competitive inhibitor is substrate-concentration dependent, whereas a non-competitive inhibitor's effect on Vmax is not.