Understanding how enzymes are inhibited
has important implications both for our
understanding of the mechanism of enzymatic
action and with medical considerations.
In this lecture I will talk
about two primary things.
Reversible enzyme inhibitors and
also irreversible enzyme inhibitors.
Cells of course rely on
enzymes to catalyze reactions
and that reliance on enzymes allow us to
be able to control cells if we can control
enzymes. Then it is a consideration
particularly if we have a bacterium,
for example, that we want to
stop from infecting something, or a cancer
cell that we want to stop from spreading.
So inhibiting enzymes is an important
consideration for us for health purposes.
I want to spend some time talking about three
different types of inhibition of enzymes.
And the first of these that I will talk
about is called competitive inhibition.
You can see this is shown schematically
on the screen. The enzyme
with its normal substrate is shown on the left.
The enzyme binds to the substrate and
converts the substrate into product.
On the right we see that same enzyme that
is the target of an inhibitor of it.
And in this case the target inhibitor
looks like the original substrate.
It fits in the active site of the enzyme.
The same way that the normal substrate did.
But there is something about the inhibitor that the
enzyme can't manipulate. It can't do anything with it.
And that causes the enzyme to sort of sit and spin
it's wheels while it's bound to that inhibitor.
That inhibitor is called the competitive inhibitor
and the competitive inhibitor has
the properties I have shown here.
That it looks like the substrate
and binds to the active site.
Now on the screen here
you can see a couple of
The bottom molecule is a molecule that is used
by an enzyme called dihydrofolate reductase.
The enzyme dihydrofolate reductase uses this
molecule and converts it into a product
where the product is used to make nucleotides,
very important for nucleotides.
The molecule above that is called
methotrexate. And methotrexate is
very similar to dihydrofolate.
However, there is an important
difference to it. And the difference
prohibits the enzyme dihydrofolate
reductase from acting.
Well methotrexate is an inhibitor
that enzyme and by inhibiting
an enzyme that makes nucleotides
that's specific for a cell one could imagine
that one could stop that cell from dividing.
And that's exactly what
this inhibitor is used for.
Now let's study the effects of that
competitive inhibitor on an enzyme.
If we take an enzyme and
we compare the V versus S plot
of an uninhibited reaction with an inhibited reaction,
we will get something like what we see on the screen here.
Now I need to explain how this was done.
I have described how we used say 20 tubes
to generate the data that's used to make
the first line that is the enzyme plus varying about of
substrate, each tube has a different amount of substrate,
and a buffer are used and we measure the
velocity by measuring the
the concentration of products produced over time.
If we want to study the inhibitor reaction,
we want to remember that we wanna have one variable.
And the one variable we said we
have is substrate concentration.
That means that we can't
vary the amount of inhibitor.
So when we do the second set of reactions, we have
the same amount of enzyme, we have the same buffer
and we have the same amount
of inhibitor in each tube.
But we have varying amounts of substrate.
What happens when we do
that? Well when we do that
we see that the reaction starts off
and it's at a lower rate.
That's not too surprising; because, there is an inhibitor
their that's inhibiting enzyme. The velocity is lower.
But as we go to increase in amount of
substrate, we see that the inhibitor
keeps rising, and rising and rising and by
the end it's actually rising fast enough
that it is getting in the range of the
velocity of the uninhibited reaction,
okay? We see that the difference
between two curves is decreasing.
Now I will cut to the chase here and cutting
to the chase, I will tell you that
if we go to very very large amounts
of substrate we will discover
that the two enzymes have the same Vmax.
Now why is that the case? Why does a
competitively inhibited reaction has a same Vmax
as no inhibitor what's so ever?
The answer is do it the way
that the experiment was setup.
I said that we had a
fixed amount of inhibitor.
It gigantic concentrations
of substrate what happens
Well the substrate is much more likely the
substrate would be found by the enzyme than
the inhibitor would be found by the enzymes.
At low concentrations they compete pretty well.
But at high concentrations where I might have a million
times as much substrate as I have inhibitor.
The difference between the uninhibited and
the inhibited is difficult for me to see.
In addition to the Vmax not changing
for a competitively inhibited reaction,
something does change in the reaction
and the thing that changes is the Km.
Since the two reactions that is the
unhibited and the inhibited reaction
have the same Vmax, they have the same Vmax/2.
So if we plot on each curve the Km
value which we get from Vmax/2
we discover that the Km value for the
uninhibited reaction is as we would expect.
But the Km for the competitive
inhibited reaction increases.
Now that increase is indicating an apparent
change in the affinity of the enzyme for the substrate.
Now that I say apparent; because, it doesn't
actually change the affinity of the enzyme
for the substrate. That's a deeper topic than
I talk about here. But the apparent Km increases
making it seem that the enzyme is loosing
its affinity for its substrate.
This is shown graphically in another
way using a Lineweaver Burk Plot.
So remember with the Lineweaver Burk, we take
the same data that we had for the V versus S plot
and we invert all the data and then plot
that on an inverted plot, as you see here 1/Vo
versus 1 over the concentration of S.
When we do that, we see that,
not surprisingly the V versus S data
comes to a line as shown in green
with the y-intercept corresponding to 1/Vmax
and an x-intercept corresponding to -1/Km.
When we plot the competitive inhibitor, we see exactly what
we learned in our last plot which was the Vmax is the same
and the two lines cross at the y-axis.
And since the Km value increased
for the competitive inhibition
what we see then is that the -1/Km gets closer to 0.
The Lineweaver Burk plot shows us very
graphically what's happening with that inhibition.