GATE 2024 BT Q.44: solution and answer

Official answer: B, CSource: Official answer key, GATE 2024 BT (IISc Bengaluru)

1. Uncompetitive inhibition scheme: the inhibitor $I$ has no affinity for the free enzyme $E$. It binds only to the enzyme-substrate complex, $ES+I\rightleftharpoons ESI$, with dissociation constant $K_I^{\prime }=[ES][I]/[ESI]$. The $ESI$ complex does not form product.

2. Rate law: let $V_{max}$ be the maximum rate without inhibitor, $K_m$ the Michaelis constant, $[S]$ the substrate concentration, $[I]$ the inhibitor concentration and ${\alpha }^{\prime }=1+[I]/K_I^{\prime }$. Then $$v=\frac{V_{max}[S]}{K_m+{\alpha }^{\prime }[S]}=\frac{(V_{max}/{\alpha }^{\prime })[S]}{(K_m/{\alpha }^{\prime })+[S]}$$

3. Apparent constants: $V_{max}^{app}=V_{max}/{\alpha }^{\prime }$ and $K_m^{app}=K_m/{\alpha }^{\prime }$. Because ${\alpha }^{\prime }>1$ whenever $[I]>0$, both fall by the same factor, so the ratio $V_{max}/K_m$ does not change. On a Lineweaver-Burk plot this shows as parallel lines.

4. (A) False: an inhibitor that binds only the substrate-binding site of free enzyme is a competitive inhibitor. That raises $K_m^{app}$ and leaves $V_{max}$ unchanged. An uncompetitive inhibitor cannot bind there, because in $ES$ that site is already filled by the substrate.

5. (B) True: by definition an uncompetitive inhibitor binds only to the $ES$ complex, usually at a site that the bound substrate creates or exposes.

6. (C) True: even at saturating $[S]$, part of the enzyme is held as inactive $ESI$, so $V_{max}^{app}=V_{max}/{\alpha }^{\prime }<V_{max}$.

7. (D) False: binding to both $E$ and $ES$ is noncompetitive (or mixed) inhibition, not uncompetitive inhibition.

8. Answer: the correct statements are (B) and (C).