There's two types of graphs you should definitely know, with respect to inhibition. But first you must understand the BASIC enzyme model and the corresponding graph of how it operates.
Basically, the most important thing is Vmax, which is the maximum rate at which a particular enzyme functions. Vmax for a particular enzyme, say alcohol dehydrogenase, depends on ALL of the alcohol dehydrogenases in that organism/test tube/ whatever. Essentially, Vmax is a quantitative way to measure the catalytic ability of a particular enzyme, and Vmax depends on the enzyme concentration, usually denoted a [E]. If we increase [E] it makes sense that Vmax would increase because there are more of that enzyme in solution and that enzyme can, in-turn, catalyze whatever reaction that it is specific for.
Km is the Michaelis constant its the substrate concentration at which ONE-HALF of Vmax is achieved. Km is inversely proportional to the affinity of that enzyme for its substrate. Think about that for a second. This means that the smaller the Km value, the less substrate will be needed to achieve the Vmax, which means that the enzyme has a higher affinity for that substrate if less of that substrate it needed reach the Vmax value.
For competitive inhibitors, Vmax DOES NOT change because a competitive inhibitor simply competes w/ the substrate for the active site. Nothing happens to the overall [E] concentration, since all the enzymes are still intact, which is why Vmax does not change. The only thing that is affected is Km, which INCREASES, which means that the enzyme has a lower affinity for its substrate. This should make sense to you, because the competitive inhibitor is COMPETING with the regular substrate for that enzymes active site, and some of the enzymes in solution will inevitably be occupied by the inhibitor
For noncompetitive inhibitors, the inhibitor binds to the allosteric site and causes a conformational change in the enzyme. This alters the active site, so its as if the enzyme is being inactivate. This is like reducing the [E], since the inhibitor is, after all, physically altering the shape of the enzyme and inactivating it. This is why Vmax decreases when a non-competitive inhibitor is present in solution. The Km is not affected in the presence of a non-competitive inhibitor.
The concept is really not that hard. Not trying to sound condescending, but just go back an re-read everything carefully and study the graph. Don't worry about calculating Vmax or Km, as those are beyond the scope of the MCAT. good luck😀
