GATE 2022 CH – Question 27
In an enzymatic reaction, an inhibitor (I) competes with the substrate (S) to bind with the enzyme (E), thereby reducing the rate of product (P) formation. The competitive inhibition follows the reaction mechanism shown below. Let [S] and [I] be the concentration of S and I, respectively, and $r_s$ be the rate of consumption of S. Assuming pseudo-steady state, the correct plot of $\frac{1}{-r_s}$ vs $\frac{1}{[S]}$ is
![the mechanism E + S ⇌ E·S → E + P (rate constants $k_1$, $k_2$, $k_3$) and E + I ⇌ E·I. The four options A to D each show lines of $\frac{1}{-r_s}$ against $\frac{1}{[S]}$ for no inhibition and for increasing [I]: (A) the lines have a common intercept on the vertical axis and the slope grows with [I], (B) parallel lines with a growing intercept, (C) a common intercept on the vertical axis and the slope grows with [I] (steeper than A), and (D) a different family of lines.](/question-images/GATE_CH_2022_Q27.png)
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Correct answer: (A) Option A in the figure
Explanation
For competitive inhibition, $\frac{1}{-r_s} = \frac{1}{V_{max}} + \frac{K_M}{V_{max}}\left(1 + \frac{[I]}{K_I}\right)\frac{1}{[S]}$. The intercept $\frac{1}{V_{max}}$ does not change with [I], and the slope increases with [I]. So the lines meet at one point on the vertical axis and get steeper with a higher inhibitor concentration.