GATE 2026 BT, question 32: Enzyme kinetics — solution and answer

Official key: 20 to 20 · Official answer key, GATE 2026 BT (IIT Guwahati)

1. The turnover number $k_{cat}$ is the number of substrate molecules one enzyme molecule converts to product per unit time when the enzyme is saturated with substrate. It is defined as $k_{cat}=\frac{V_{max}}{[E{]}_T}$, where $V_{max}$ is the maximum velocity and $[E{]}_T$ is the total enzyme concentration.

2. Write both quantities in the same concentration unit. Since $1\mu {\text{mol L}}^{-1}=1\mu \text{M}$, we have $V_{max}=1800\mu {\text{mol L}}^{-1}{\text{ min}}^{-1}=1800\mu {\text{M min}}^{-1}$ and $[E{]}_T=1.5\mu \text{M}$.

3. Divide: $$k_{cat}=\frac{1800\mu {\text{M min}}^{-1}}{1.5\mu \text{M}}=1200{\text{min}}^{-1}$$

4. Convert to per second ($1\text{min}=60\text{s}$): $$k_{cat}=\frac{1200}{60}{\text{s}}^{-1}=20{\text{s}}^{-1}$$