If the concentration of substrate and inhibitor is kept the same in case of competitive inhibition, which molecule will bind to protein?
The substrate S and competitive inhibitor I will bind in proportion to their affinities*. Some of the enzyme will be bound to the inhibitor, some will be bound to the substrate, and some will not be bound to either.
*To be more accurate, it is not the affinity (dissociation constant Kd) of the substrate that is relevant here, it is the Km.
The rate of the enzyme reaction in the absence of inhibitor is
Vo = Vmax[S]/(Km + [S])
The rate of the enzyme reaction in the presence of competitive inhibitor is
Vi = Vmax[S]/{Km(1 + [I]/Ki) + [S]}
[S] is the substrate concentration and [I] is the inhibitor concentration.
The fraction of enzyme that is bound to inhibitor in the presence of substrate is equal to 1 minus the ratio of the rate of the inhibited enzyme Vi to that of the uninhibited enzyme Vo:
1-Vi/Vo= 1 - (Km + [S])/{Km(1 + [I]/Ki) + [S]}
The remaining enzyme is either free or bound to its substrate. The fraction of the remaining enzyme that is bound to substrate is Vo/Vmax = [S]/(Km + [S]).
Here is an example. Suppose [S] = Km and [I] = Ki.
The fraction of enzyme bound to inhibitor is 1/3.
The fraction of the 2/3 remaining enzyme bound to substrate is 1/2, hence 1/3 of the total enzyme is bound to substrate.
Finally, 1/3 of the total enzyme is not bound to either substrate or inhibitor.
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