# 4.11: Enzyme Inhibition

## No Effect On $$V_{MAX}$$

How do we study competitive inhibition. It is typically done as follows. First one performs a set of V vs. [S] reactions without inhibitor (20 or so tubes, with buffer and constant amounts of enzyme, varying amounts of substrate, equal reaction times). V vs. [S] is plotted, as well as 1/V vs. 1/[S], if desired. Next, a second set of reactions is performed in the same manner as before, except that a fixed amount of the methotrexate inhibitor is added to each tube. At low concentrations of substrate, the inhibitor competes for the enzyme effectively, but at high concentrations of substrate, the inhibitor will have a much reduced effect, since the substrate outcompetes it, due to its higher concentration (remember that the inhibitor is at fixed concentration). Graphically, the results of these experiments are shown above. Notice that at high substrate concentrations, the competitive inhibitor has essentially no effect, causing the Vmax for the enzyme to remain unchanged. To reiterate, this is due to the fact that at high substrate concentrations, the inhibitor doesn’t compete well. However, at lower substrate concentrations it does.

## Increased KM

Note that the apparent KM of the enzyme for the substrate increases (-1/KM gets closer to zero - red line above) when the inhibitor is present, thus illustrating the better competition of the inhibitor at lower substrate concentrations. It may not be obvious why we call the changed KM the apparent KM of the enzyme. The reason is that the inhibitor doesn’t actually change the enzyme’s affinity for the folate substrate. It only appears to do so. This is because of the way that competitive inhibition works. When the competitive inhibitor binds the enzyme, it is effectively ‘taken out of action.’ Inactive enzymes have NO affinity for substrate and no activity either. We can’t measure KM for an inactive enzyme.

The enzyme molecules that are not bound by methotrexate can, in fact, bind folate and are active. Methotrexate has no effect on them and their KM values are unchanged. Why then, does KM appear higher in the presence of a competitive inhibitor. The reason is that the competitive inhibitor is reducing the amount of active enzyme at lower concentrations of substrate. When the amount of enzyme is reduced, one must have more substrate to supply the reduced amount of enzyme sufficiently to get to Vmax/2.

It is worth noting that in competitive inhibition, the percentage of
inactive enzyme changes drastically over the range of [S] values
used. To start, at low [S] values, the greatest percentage of the
enzyme is inhibited. At high [S], no significant percentage of
enzyme is inhibited. This is not always the case, as we shall see
in non-competitive inhibition.