GATE 2026 BT Q.21: solution and answer

Official answer: CSource: Official answer key, GATE 2026 BT (IIT Guwahati)

1. Michaelis-Menten rate law: $v=\frac{V_{\max }[S]}{K_M+[S]}$, with $V_{\max }=k_{\text{cat}}[E{]}_0$. Here $v$ is the initial reaction rate, $[S]$ the substrate concentration, $k_{\text{cat}}$ the turnover number (units ${\text{s}}^{-1}$), $K_M$ the Michaelis constant (a concentration), $V_{\max }$ the maximum rate and $[E{]}_0$ the total enzyme concentration. (The paper prints $k_{\text{cat}}$ as $k_{\text{Cat}}$; it is the same quantity.)

2. At low substrate, $[S]\ll K_M$, the denominator is approximately $K_M$, so $v\approx \frac{k_{\text{cat}}}{K_M}[E{]}_0[S]$. This is a second-order rate law: first order in enzyme and first order in substrate.

3. The coefficient $k_{\text{cat}}/K_M$ (units ${\text{M}}^{-1}{\text{s}}^{-1}$) is the apparent second-order rate constant for free enzyme meeting free substrate and converting it to product (at $[S]\ll K_M$ almost all the enzyme is free, so $[E]\approx [E{]}_0$). It combines how fast the enzyme turns substrate over ($k_{\text{cat}}$) with how well it binds substrate ($1/K_M$). This ratio is called the catalytic efficiency, or specificity constant. Its upper limit is set by diffusion, about $10^8$ to $10^9{\text{M}}^{-1}{\text{s}}^{-1}$.

4. Check the other options. (A) $k_{\text{cat}}$ alone is the turnover number; it ignores substrate binding. (B) $V_{\max }/k_{\text{cat}}=[E{]}_0$, which is just the enzyme concentration. (D) $k_{\text{cat}}/V_{\max }=1/[E{]}_0$. Neither (B) nor (D) is a property of the enzyme's efficiency.

5. So catalytic efficiency $=k_{\text{cat}}/K_M$, which is option (C).