In this screencast, we're going to derive the enzyme kinetics expression where we have
what's referred to as uncompetitive inhibition.
The idea is that we have an enzyme, it reacts with a substrate to make a complex.
This complex then can react with an inhibitor to make an enzyme-substrate inhibitor complex,
and the important thing here is this now is not active, so it's inactive for making our
product, only the enzyme-substrate can react to make the enzyme again and a product.
This last step is irreversible, these first two are reversible.
What we're going to do is derive a rate expression using the idea of rate determining step, so
what that means is we're going to assume the first step is in equilibrium.
We're going to write the equilibrium expression for the reverse reaction for convenience,
so we're going to write what's referred to as a dissociation constant, which is dissociation
of this enzyme-substrate complex, concentration of enzyme, concentration of substrate over
concentration of the complex.
We're going to do the same thing for the second step; we're going to assume this is also in
equilibrium.
I'm going to write a similar type of expression.
Finally, the rate determining step, if this is the rate constant, the rate that we make
product would be the rate constant times the concentration of complex, so we need to write
this in terms of measurable variables such as substrate concentration, inhibitor concentration,
and enzyme concentration.
Keep in mind, we're assuming this third step is irreversible because low concentration
of our product P. So one other piece of information we need to now derive the rate expression
is that the total enzyme concentration does not change, and that's the enzyme that's not
in the complex, plus the enzyme that's in the complex with the substrate, plus the enzyme-substrate-inhibitor
complex.
Now what we want to do is make a substitution.
We want to solve for the concentration, what we're going to do is end up with an expression
for the enzyme concentration, because we can substitute for this, and for this from these
two equations.
So let me make that substitution.
So I've made the substitution, but I can't yet solve for enzyme concentration until I
make a second substitution, and that's substituting for E*S again using this expression here.
So I have the enzyme concentration, let's go back and look at our rate expression, and
the rate that we're making product is k, and I'm going to substitute from this equation
so that I have concentration of enzyme, concentration of substrate, divided by K_M, so the rate
of product formation is k, I'm going to substitute in for the enzyme concentration, here's the
substrate concentration, denominator then K_M times this denominator over here, let
me copy it, so essentially we're finished, but let's do a couple things.
We're going to take this term, and we'll call it Vmax, so the rate of product formation
Vmax, substrate concentration, and in the denominator we're just going to multiply through
by K_M, first term K_M, and then we're going to factor out the substrate concentration,
this becomes 1 plus inhibitor concentration over K_I.
So this now is our final answer, it's the rate expression for enzyme kinetics where
we have uncompetitive inhibition, and if we compare it to the Michaelis-Menten expression,
it appears similar, so this is the Michaelis-Menten expression, but we have this additional term
in the denominator which is just telling us the rate is going to be slower, of course
because of inhibition, but it's similar in form to the Michaelis-Menten expression.