In this screencast, I'm going to derive the enzyme kinetics, the rate expression, for
enzyme reactions where we have competitive inhibition.
Let's first write down the reactions.
What we're looking at is an enzyme reacts with a substrate to make a complex, well it
turns out for competitive inhibition, it means there's some inhibitor species that can also,
and this is reversible, can also react with the enzyme to make a complex and the important
thing is that this complex is inactive for making our product.
The final step is that the complex can react to recreate the enzyme, so therefore the enzyme
is a catalyst, and to make the product, and this has a rate constant k, and we're going
to do this at low P concentration, and what that means is then we don't worry about the
reverse reaction.
So we're going to apply the idea of rate determining step to this system, and what that means is
the rate of reaction is the rate of this last step.
So the rate that we're making product is the rate constant times the concentration of this
complex.
Now of course, we want the rate expression in terms of concentrations of things we can
measure, so applying the rate-determining step says we're going to assume this first
step is in equilibrium, it's quasi-equilibrium.
I'm going to write an equilibrium constant for the reverse step just because of the normal
terminology, so we're going to say there's a dissociation constant, and concentration
of E, concentration of S, over the complex, so normal equilibrium expression but for the
reverse reaction.
We're going to do the same thing for the second step.
So then the other piece of information we need to write the rate expression in terms
of measurable variables is that the total concentration of enzyme doesn't change.
In other words, the enzyme is not bound up in the complex, or it's complexed with the
substrate, or with the inhibitor.
And what we're going to do is use the expressions here and substitute both of these into this
expression.
So let me make that substitution.
So now I have an expression that's just in terms of concentration of the enzymes.
Let me factor that out.
So now I can solve for the concentration of E by just moving the term on the right to
the denominator on the left.
Now of course what I want is the concentration of ES, so I can use this equation to substitute
in to get the concentration of ES, because the concentration of ES is going to be concentration
of E, concentration of substrate divided by KM, and I have a concentration from E here,
so I can make this substitution, so now I have an expression for ES I can substitute
back here to get the rate expression as we make this substitution, and then I'm just
going to write this in the form that it's more commonly used, and so we can compare
to the enzyme kinetics, Michaelis-Menten kinetics without inhibition, so this term we refer
to as Vmax, concentration of substrate, and we're just going to rearrange the term on
the bottom slightly.
So I'm going to keep these two terms in the parentheses but multiply so that this KM and
this KM cancel, and so this last term then is just the substrate concentration.
So this is our final answer.
Rate expression, enzyme kinetics for competitive inhibition, both the enzyme and the inhibitor
compete for the substrate, and if we compare this to the normal Michaelis-Menten expression,
so you can see they look similar and what we've done with inhibition, we've added an
additional term here.
We can look at this now as a new effective M, so this number, this term in the denominator
is bigger than KM and the rate is going to be slower when we have inhibition present.