GATE 2025 BT Q.32: solution and answer

Official answer: 0.94 to 0.96Source: Official answer key, GATE 2025 BT (IIT Roorkee)

1. Symbols: $v_0$ is the initial reaction velocity, $V_{\max }$ is the maximum velocity, reached when the enzyme is fully saturated with substrate, $K_m$ is the Michaelis constant (the substrate concentration at which $v_0=V_{\max }/2$), and $[S]$ is the substrate concentration.

2. Michaelis-Menten equation: $v_0=\frac{V_{\max }[S]}{K_m+[S]}$. Dividing both sides by $V_{\max }$ gives $\frac{v_0}{V_{\max }}=\frac{[S]}{K_m+[S]}$.

3. Substitute $[S]=20K_m$: $$\frac{v_0}{V_{\max }}=\frac{20K_m}{K_m+20K_m}=\frac{20K_m}{21K_m}=\frac{20}{21}$$

4. $K_m$ cancels, so the ratio depends only on how many multiples of $K_m$ the substrate concentration is: $\frac{20}{21}=0.9524$.

5. Rounded to two decimal places: $v_0/V_{\max }=0.95$. At twenty times $K_m$ the enzyme runs at about 95% of its maximum rate: close to saturation, but not fully saturated.