# Question 0e124

Apr 10, 2015

The key to solving this problem is to realize that an uncompetitive inhibitor does not change the slope of the double-reciprocal plot that describes the enzyme-catalyzed reaction when no inhibitor is present. This is important because, for an uncompetitive inhibition, both ${V}_{\text{max}}$ and ${K}_{m}$ decrease by the same factor, which is why the slope remains unchanged.

More specifically, both ${V}_{\text{max}}$ and K_m"# decrease by

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

$\left[I\right]$ - the concentration of the inhibitor;
${K}_{i}^{'}$ - the inhibitory constant.

However, this is actually equal to ${\alpha}^{'}$, the degree of inhibition.

As a result, ${V}_{\text{max}}^{'}$ will be equal to

${V}_{\text{max"^(') = V_"max"/(1 + ([I])/K_I^(')) = V_"max}} / \left({\alpha}^{'}\right)$

Therefore,

${V}_{\text{max" = alpha^(') * V_"max"^(') = 3.00 * "9.00 nmol s"^(-1) = "27.0 nmol s}}^{- 1}$

SIDE NOTE Check out this great site on enzyme inhibition, it allows you to play with Lineweaver-Buk plots for various inhibitors

http://higheredbcs.wiley.com/legacy/college/voet/0470129301/guided_exp/guided_exploration_11/michaelis_menten.html