# Noncompetitive Inhibition

Noncompetitive inhibition occurs when $$I$$ binds to both $$E$$ and $$ES$$. We will look at only the special case in which the dissociation constants of $$I$$ for $$E$$ and $$ES$$ are the same. In the more general case, the $$K_d$$ values differ, and the inhibition is called mixed. Since inhibition occurs, we will hypothesize that $$ESI$$ cannot form the product ($$P$$). It is a deadend complex which has only one fate, to return to $$ES$$ or $$EI$$. This is illustrated in the mechanism at the top of the figure below, and in the molecular cartoon beneath it.

Let us assume for ease of equation derivation that $$I$$ binds reversibly to $$E$$ with a dissociation constant of $$K_{is}$$ (as we denoted for competitive inhibition) and to $$ES$$ with a dissociation constant $$K_{ii}$$ (as we noted for uncompetitive inhibition). Assume for noncompetitive inhibition that $$K_{is} = K_{ii}$$. A look at the top mechanism shows that in the presence of $$I$$, as $$S$$ increases to infinity, not all of $$E$$ is converted to $$ES$$. That is, there is a finite amount of ESI, even at infinite $$S$$. Now remember that $V_m = k_{cat}E_o$ if and only if all $$E$$ is in the form $$ES$$. Under these conditions, the apparent $$V_m$$, $$V_{mapp}$$ is less than the real $$V_m$$ without inhibitor. In contrast, theapparent $$K_m$$, $$K_{mapp}$$, will not change since $$I$$ binds to both $$E$$ and $$ES$$ with the same affinity, and hence will not perturb that equilibrium, as deduced from Le Châtelier's Principle.

The double reciprocal plot (Lineweaver Burk plot) offers a great way to visualize the inhibition (bottom of figure above). In the presence of $$I$$, just $$V_m$$ will decrease. Therefore, $$-1/K_m$$, the x-intercept will stay the same and $$1/V_m$$ will get more positive. Therefore the plots will consists of a series of lines intersecting on the x axis, which is the hallmark of noncompetitive inhibition. You should be able to figure out how the plots would appear if $$K_{is}$$ is different from $$K_{ii}$$ (mixed inhibition).

An equation, shown in the diagram above can be derived which shows the effect of the uncompetitive inhibitor on the velocity of the reaction. In the denominator, $$K_m$$ is multiplied by $$1+I/K_{is}$$, and $$S$$ by $$1+I/K_{ii}$$. We would like to rearrange this equation to show how $$K_m$$ and $$V_m$$ are affected by the inhibitor, not $$S$$, which obviously is not the case. Rearranging the equation as shown above shows that

$K_{mapp} = \dfrac{K_m(1+I/K_{is})}{1+I/K_{ii}} = K_m$

when

$K_{is}=K_{ii}$ and $V_{mapp} = \dfrac{V_m}{1+I/K_{ii}}.$

This shows that the $$K_m$$ is unchanged and $$V_m$$ decreases as we predicted. The Lineweaver Burk plot shows a series of lines intersecting on the x axis. Both the slope and the y intercept are changed, which are reflected in the names of the two dissociation constants, $$K_{is}$$ and $$K_{ii}$$. Note that if $$I$$ is zero, $$K_{mapp}$$ = $$K_m$$ and $$V_{mapp} = V_m$$.

If you can apply Le Châtelier' Principle, you should be able to draw the Lineweaver-Burk plots for any scenario of inhibition or even the opposite case, enzyme activation!