For an enzyme in the absence of inhibitor Vmax is 10 μM.s-1 and its Km is 2 mM. In the presence of a 1 mM inhibitor the apparent Vmax is 4 μM.s-1 and the apparent Km is 4 mM.Select the correct statementGroup of answer choicesalpha = 5; alpha' = 2.5alpha’ = 2alpha and alpha’ = 2.5alpha = 2.5 and alpha’ = 5alpha’ = 5alpha = 2.5
Question
For an enzyme in the absence of inhibitor Vmax is 10 μM.s-1 and its Km is 2 mM. In the presence of a 1 mM inhibitor the apparent Vmax is 4 μM.s-1 and the apparent Km is 4 mM.Select the correct statementGroup of answer choicesalpha = 5; alpha' = 2.5alpha’ = 2alpha and alpha’ = 2.5alpha = 2.5 and alpha’ = 5alpha’ = 5alpha = 2.5
Solution
In enzyme kinetics, alpha (α) and alpha prime (α') are used to describe the effects of inhibitors on enzyme activity.
Alpha (α) is the ratio of the apparent Km in the presence of inhibitor to the Km in the absence of inhibitor.
Alpha prime (α') is the ratio of the apparent Vmax in the presence of inhibitor to the Vmax in the absence of inhibitor.
Given the data:
Km (without inhibitor) = 2 mM Km (with inhibitor) = 4 mM Vmax (without inhibitor) = 10 μM.s-1 Vmax (with inhibitor) = 4 μM.s-1
We can calculate α and α' as follows:
α = Km (with inhibitor) / Km (without inhibitor) = 4 mM / 2 mM = 2
α' = Vmax (with inhibitor) / Vmax (without inhibitor) = 4 μM.s-1 / 10 μM.s-1 = 0.4
So, none of the provided answer choices are correct based on these calculations.
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