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Biology LibreTexts

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.

05 noncompinhib.gif

 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

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


\[K_{is}=K_{ii}\] 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!