We're going to look at the enzyme kinetics for noncompetitive inhibition, so this is
where we have an enzyme reacting with a substrate to make an enzyme complex, and then this enzyme
complex can react to recreate the enzyme and form our product, and we're going to assume
this step is irreversible, or the other way of looking at it is that we're assuming we're
at low concentrations of our product.
We're going to derive our rate expression, and then what we're going to do is look at
the case, so noncompetitive inhibition has also been referred to as mixed inhibition.
The idea here is that we have an inhibitor present that can also react with the enzyme
to make a complex, and this complex now is inactive, meaning it can't react further to
product.
It can of course dissociate back to the enzyme plus the inhibitor.
So an additional step is that the inhibitor could react with the complex, instead of just
reacting with the enzyme.
It's also possible to form another complex which has both the enzyme and the complex,
and then the final step is that we can form the enzyme-substrate inhibitor complex by
having the enzyme inhibitor react with the substrate to form the same complex.
So we're going to use the idea of rate-determining step, and we're going to say this formation
of product, so the rate that product forms, is the rate constant for that second step
times the concentration of this enzyme-substrate complex.
The other steps are all in equilibrium.
Now we're going to write the equilibrium expression for the reverse step because of the form that
this takes that's the commonly used form, so we're going to write a dissociation constant,
which is the enzyme-substrate complex dissociating.
At equilibrium we would have a concentration of enzyme, concentration of substrate divided
by the concentration of the complex.
Now what we're going to assume is that for the substrate in this step here reacting with
the enzyme inhibitor complex, it has the same dissociation constant as the substrate reacting
with just the enzyme.
In other words, the binding of the inhibitor doesn't affect the equilibrium for the binding
of the substrate, so what that means is then we're going to have the same dissociation
constant, and I'll write down the expression, and then we're going to have another dissociation
constant for the inhibitor reacting with the enzyme, and again it's the same form, concentration
of enzyme, concentration of inhibitor, and then finally we're going to assume the same
dissociation constant for this step.
Again the idea is that the inhibitor reacting with the enzyme or the enzyme-substrate complex
has the same dissociation constant.
The fact that the substrate is bound to the enzyme does not affect the equilibrium for
the inhibitor to bind.
So I'll write this down, so the final piece we need to obtain the rate expression is that
the total enzyme concentration does not change.
In other words, the enzyme is a catalyst, if you'd like, so it's either the free enzyme,
or it forms a complex with the substrate, or it forms a complex with the inhibitor,
or it forms a complex with both of them.
Add all this up, we should have the same concentration of total enzyme as a function of time.
So what we're going to do now is substitute into this equation using these equilibrium
expressions that we just wrote down, so let me pause and do that substitution.
Now, our goal is to obtain an expression for the concentration the enzyme complex we can
substitute here, so we get a rate expression in terms of the measurable variables, namely
the substrate concentration and the inhibitor concentration.
So this equation for the balance here, we could solve for the free enzyme concentration,
I could substitute back into this equation and get the enzyme-substrate concentration.
I could then substitute here.
However, first I have to substitute here.
I'm going to use this equation, so then I can take and solve this equation for the free
enzyme concentration.
So I've made this substitution.
These other three terms are the same, so this means I can now write the total enzyme concentration,
I can factor out just the enzyme alone, its concentration, let me do that.
So here I've factored out, so I can now rearrange this to solve for this enzyme concentration.
So now I'm almost finished.
I'm going to go back over here, and let's divide, so we don't overlap the equations,
the rate that I'm making product, the rate constant for that rate determining step, and
the concentration of the enzyme-substrate complex I can get form here, enzyme concentration,
substrate concentration, divided by KM, and now I can substitute this here and I'll get
a final rate expression, and so let's create a little more room, let's do that down here.
So the rate constant, E_T, substrate concentration, KM, and now this term in parentheses.
So we're essentially finished, but let's put it in a form that's a little more commonly
used, so what we're going to do is take this term, we're going to call that Vmax, this
is the standard notation that you will see, so the rate that we're making product is going
to be Vmax, the substrate concentration, now for the denominator I can rearrange this,
I can put together terms if I first multiply through by KM, I'll do that, and now I'm going
to put together in the denominator, so the rate, let's move up here, the rate of product
formation, so what I want is the two terms that contain KM, 1 plus some initiator concentration
over K_I.
The two terms that contain the substrate concentration, so 1 plus initiator over K_I, and so we can
see that this term and this term are the same, so we could factor those out, and we end up
then with our final expression.
Let me stop and write down this, so now I have a rate expression for enzyme kinetics
with noncompetitive inhibition that depends on the inhibitor and the substrate concentration.
We can compare that to the case where there's no inhibition, so we can see the similarity
in the form and we've added this additional term in the denominator, which says of course
the rate we're making produce is going to be slower when we have inhibition.