Another type of inhibition that is important for
us to understand is that of non-competitive inhibition.
It's fundamentally different from a competitive
inhibition and we can see it depicted on the screen here.
On the left again we have the enzyme with its
normal substrate which catalyzes a reaction.
However, the enzyme has a site on it that if
it's properly targeted by an inhibitor
the inhibitor can bind to
it and keep the enzyme from
functioning properly with the substrate in the active
site. That is shown in the image on the right.
Now when this happens, the
has a fundamentally different way of interacting
with the enzyme then what we saw before.
They effected by binding at a different location and
by binding at a different location they do not compete
okay? Now this changes the parameters of the
velocity that we have been studying considerably.
And because the inhibitor doesn't compete with
its substrate and the substrate can't out
way it by doing a reaction with
not a lot more substrate.
It means that in every reaction that we do
what happens is that we are inhibiting
a fixed amount of an enzyme.
It doesn't matter how much enzyme that we add, there
is always the same amount of enzyme inhibited.
In the first reactions, the competitively inhibited reactions,
we saw that as we added more substrate
the substrate out-competed the inhibitor and
it was as if the inhibitor disappeared.
So the quantity of enzyme being an inhibited
was changing, the more substrate we added
the more normal enzyme we had.
With a non-competitive inhibitor, we don't have that.
It doesn't matter how much substrate we have;
because, they are not competing for the same site. The
non-competitive inhibitors always going to
knock-out the same amount of enzyme
in every tube irrespective of how
much substrate is added to it.
That means that we have
changed the amount of enzyme.
And if we change the amount of enzyme, we
have already talked about the limitations
of an enzyme in studying of the Vmax.
Remember the factory analogy?
In the factory analogy I said, that if we added an
extra factory would double the amount of product.
What if the factory only worked half a day?
If the factory only worked half a day, it
would make half the amount of product.
We have changed the numbers of workers.
So, what if we use enough inhibitor that we
only have half the amount of enzyme?
Well we will change Vmax accordingly.
So when we have a non-competitive inhibitor
we are changing the amount of enzyme
and in changing the amount of
enzyme, we changed the value of Vmax.
So Vmax decreases for a non-competitive inhibitor.
That wasn't the case for a
competitive inhibitor, right?
Now we can only measure Km for an active enzyme.
And not surprisingly if we
change the amount of enzyme
Km, the affinity of the enzyme for
the substrate doesn't change; because,
the enzyme is still the enzyme when it's active
and we are only studying an active enzyme. So the
Km value doesn't change for non-competitive inhibition.
In a Lineweaver Burk plot we see something
different than we saw with the competitive inhibition,
but consist in it what I just told you.
In green again, we see the linear
the linear plot showing, of course, the uninhibited reaction.
In blue we see the non-competitively inhibited
reaction and we notice that
the two lines cross at -1/Km.
Well this is consistent with what we learned
in the last plot which is that the Km value
doesn't change. They should cross at that point.
However we see the blue line has a
higher slope than does the green line,
meaning that the crossing of the
y-axis is at a higher point.
Now they may seem kinda intuitive
that if we decrease the Vmax
we actually are raising the value of that line.
But remember we are doing a reciprocal.
So by decreasing Vmax, 1/Vmax actually increases.