# The following rate equation represents _____________ inhibition. $V=\frac{V_{max} [S]}{K_m (1+\frac{I}{K_i})+[S]}$

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The following rate equation represents _____________ inhibition.

$V=\frac{V_{max} [S]}{K_m (1+\frac{I}{K_i})+[S]}$

(a) competitive

(b) mixed

(c) non-competitive

(d) uncompetitive

I got this question in class test.

I'd like to ask this question from Fundamentals in section Enzyme Kinetics Fundamentals of Enzyme Technology

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The correct choice is (a) competitive

Easiest explanation: Competitive inhibition occurs when both substrate and inhibitor compete for binding to the active site of the enzyme. In this kind of inhibition, Ki is much greater than total inhibitor concentration and ESI complex is not formed. This kind of inhibition is mostly seen when the substrate concentration are low. This inhibition can be overcome at sufficiently high substrate concentrations as Vmax remain unaffected. Hence the rate equation is represented by $V=\frac{V_{max} [S]}{K_m (1+\frac{I}{K_i})+[S]}$

Where,

Km = Miachelis Menton constant, Ki = inhibitor constant, I = Inhibitor concentration and S = Substrate concentration.

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