Question

An enzyme (E) catalyzes the biochemical reaction \(A \rightarrow B\) with \(k_{\text{cat}}=500~\text{s}^{-1}\). If the initial reaction velocity \((V_0)\) is \(10~\mu\text{M}\cdot\text{s}^{-1}\) at the total enzyme concentration \([E_t]=30~\text{nM}\) and substrate concentration \([A]=40~\mu\text{M}\), the value of \(K_m\) (in \(\mu\text{M}\)) is ________________.

Hint:

Use \(V_{\max}=k_{\text{cat}}[E_t]\) and avoid unit slips by converting nM to \(\mu\)M before substituting.

Answer 1

Correct Answer: 20

Step 1: \(V_{\max}=k_{\text{cat}}[E_t]=500\times 30~\text{nM s}^{-1}=15000~\text{nM s}^{-1}=15~\mu\text{M s}^{-1}\).
Step 2: Michaelis–Menten: \(V_0=\dfrac{V_{\max}[S]}{K_m+[S]}\). With \(V_0=10\), \([S]=40\), \(V_{\max}=15\): \[ 10=\frac{15\times 40}{K_m+40}\Rightarrow 10K_m+400=600\Rightarrow K_m=20~\mu\text{M}. \]