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
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Competitive inhibition increases the slope of the Lineweaver–Burk plot but keeps the y-intercept unchanged.
Competitive inhibition follows the classic Michaelis–Menten mechanism where the inhibitor competes with the substrate for the active site of the enzyme. This changes the apparent Michaelis constant $K_m$ but does not change $V_{\max}$.
Step 1: Write the Lineweaver–Burk form.
For competitive inhibition, the rate equation becomes:
\[
\frac{1}{r_s} = \frac{K_m\left(1+\frac{[I]}{K_I}\right)}{V_{\max}} \cdot \frac{1}{[S]} + \frac{1}{V_{\max}}
\]
Important features:
– The slope increases with increasing inhibitor concentration (because slope $\propto 1 + \frac{[I]}{K_I}$).
– The y-intercept stays constant because $V_{\max}$ does not change.
Step 2: Interpret the graph.
In competitive inhibition, Lineweaver–Burk plots show lines that:
– intersect at the same y-intercept, and
– have increasing slopes as inhibitor concentration [I] increases.
Step 3: Match with the options.
Among the given plots, option (A) correctly shows:
– a common y-intercept (same $1/V_{\max}$),
– higher slopes for increasing inhibitor concentration.
Therefore, option (A) matches the expected behavior of competitive inhibition.
