Question

The $K_m$ and $v_{max}$ of lactate dehydrogenase for conversion of pyruvate to lactate are 1 mM and 5 nM s$^{-1}$ respectively. At 0.25 mM pyruvate, the velocity of the reaction catalyzed by lactate dehydrogenase is .............. nM s$^{-1}$.

Hint:

When substrate concentration equals $K_m$, the enzyme runs at half of $v_{max}$.

Answer 1

Correct Answer: 1

Step 1: Use Michaelis–Menten equation.
\[ v = \frac{v_{max}[S]}{K_m + [S]} \] Step 2: Substitute values.
\[ v = \frac{5(0.25)}{1 + 0.25} \] \[ v = \frac{1.25}{1.25} = 1\ \text{nM s}^{-1} \] Step 3: Conclusion.
The reaction velocity is 1 nM s$^{-1}$.