# In competitive inhibition, what happens to Vmax and Km if [I] = Ki?

In competitive inhibition, what happens to Vmax and Km if [I] = Ki?

(a) Lowers to 0.5 Vmax and 0.5 Km

(b) Vmax is unchanged and Km increases 2Km

(c) Lowers to 0.5 Vmax and Km remains unchanged

(d) Lowers to 0.67 Vmax and Km increases to 2Km

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Question is taken from Fundamentals in division Enzyme Kinetics Fundamentals of Enzyme Technology

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The correct option is (b) Vmax is unchanged and Km increases 2Km

Easiest explanation: Competitive inhibition is one wherein the inhibitor and the substrate compete for the active site. Inhibitor and substrate are said to be structurally similar. Thus, the rate equation for competitive inhibition is given by $V=\frac{V_{max} [S]}{K_m (1+\frac{I}{K_i})+[S]}$. According to this equation, Vmax remains unchanged and Km increases 2Km.

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