Example 1
An initial rate is 6 units/min. Active enzyme doubles with substrate still abundant. Predict the rate.
Under the proportional model, 2×6=12 units/min. The abundant-substrate condition is essential.
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Year 12 · SACE Stage 2
Enzyme concentration and inhibitors
Calculate fixed rate changes and classify inhibition from supplied conditions.
Method and worked examples
At abundant substrate and otherwise unchanged conditions, initial reaction rate is approximately proportional to active enzyme concentration. Doubling active enzyme can therefore double the initial rate. This proportional rule fails when substrate or another requirement becomes limiting. Use the stated conditions rather than assuming every enzyme addition gives the same increase.
A competitive inhibitor competes for the substrate’s active-site binding region. In a simple reversible model, sufficiently high substrate can reduce its effect. A pure non-competitive inhibitor reduces active catalytic capacity without changing substrate affinity; adding substrate does not restore the original maximum rate. Percentage inhibition is (uninhibited rate − inhibited rate) divided by uninhibited rate, multiplied by 100. Classify only the mechanism actually specified.
Example 1
Example 2
An inhibitor lowers rate from 20 to 12 units/min. Find percentage inhibition.
The reduction is 8 units/min. Percentage inhibition =208×100=40%.
Watch out: The proportional prediction requires sufficient substrate and otherwise comparable conditions.