# Question #1604e

Apr 9, 2015

You're dealing with non-competitive inhibition for which the degree of inhibition is defined as

${\alpha}^{'} = 1 + \frac{\left[I\right]}{{K}_{I}^{'}}$, where

$\left[I\right]$ - the concentration of the inhibitor;

It doesn't get much more straightforward than this, so just plug and play

${\alpha}^{'} = 1 + \frac{\left[I\right]}{{K}_{I}^{'}} \implies {K}_{I}^{'} = \frac{\left[I\right]}{{\alpha}^{'} - 1}$

${K}_{I}^{'} = \text{125 nmol/L"/(3.50 -1) = "50.0 nmol/L}$