Question

The $K_M$ and $v_{\max}$ of an enzyme are 4 mM and 0.1 nM h\(^{-1}\) respectively. In the presence of 1.5 mM inhibitor, the $K'_M$ and $v'_{\max}$ become 6 mM and 0.1 nM h\(^{-1}\). The inhibition constant $K_i$ (correct to 1 decimal place) is ............ mM.

Hint:

Competitive inhibition increases $K_M$ but leaves $v_{\max}$ unchanged.

Answer 1

Correct Answer: 2.9

Step 1: Identify the type of inhibition.
Because \(v_{\max}\) remains unchanged while \(K_M\) increases, this is *competitive inhibition*. For competitive inhibition: \[ K'_M = K_M \left(1 + \frac{[I]}{K_i}\right) \] Step 2: Substitute the values.
\[ 6 = 4\left(1 + \frac{1.5}{K_i}\right) \] \[ \frac{6}{4} = 1 + \frac{1.5}{K_i} \] \[ 1.5 = 1 + \frac{1.5}{K_i} \] \[ 0.5 = \frac{1.5}{K_i} \] Step 3: Solve for \(K_i\).
\[ K_i = \frac{1.5}{0.5} = 3.0\ \text{mM} \] Step 4: Conclusion.
The inhibition constant is 3.0 mM.