Enzymes: Inhibition

the essence of noncompetitive inhibition is that the

inhibitor binds and makes some fraction of enzyme inactive. The remainder of the enzyme is functionally the same in the absence of inhibitor. There is less total functional enzyme.

Inhibitor binding to ES or ES* stops

product formation by making the enzyme catalytically inactive.

PALA inhibits the

pyrimidine biosynthetic enzyme, aspartate transcarbamoylase ("activates" aspartate for ring closure reaction to form the cyclic structure leading to uridines synthesis; uridine may then be utilized for synthesis of cytidine and thmidine).

Key enzyme involved in the replication of DNA

Drug action ultimately retards DNA replication in cancerous cells.

Two rxns may be ongoing:

E+I<=>EI and ES+I<=>ESI

The rxn is:

EI<=>E+I, so we can measure Ki=[E][I]/[EI]

Depending on the complexity of the inhibitor binding, and the resulting kinetic changes seen, the inhibition may be reffered to

alternatively as mixed or pure noncompetitive inhibition.

PALA is

an anticancer medication.

Based on all of the three previous graphs, in what direction would Km and Vmax be changed?

both Km and Vmax are decreased.

the rate plot shows that

both Vmax and Km have changed.

noncompetitive inhibition, Vmax is

decreased (y-intercept shifts).

An example of this type of drug is

fluorouracil. This is an anticancer drug that inactivates thymidylate synthetase, an enzyme required for DNA synthesis.

a plot of rate vs. substrate will show a

lower maximum rate.

The inhibition may be somewhat complex, with

multiple binding sites (e.g. inhibitor binds to both E and ES separately and simultaneously, with different dissociation constants).

In the pure form of noncompetitive inhibition, no matter what complex events are going on in the background, the

net effect is that Km is not changed and so we consider substrate binding affinity to also be unchanged.

with pure noncompetitive inhibition apparent Km is

not changed (x-intercept does not shift)

Since the slope of the L-B line is

not changed, it follows that both apparent Km and Vmax must be different in the presence of an uncompetitive inhibitor.

The structure of the inhibitor need

not resemble the substrate, and the forces and stress of inhibitor binding is believed to distort the active site.

Some inhibitors covalently modify the enzyme and take it

out of action permanently. Examples are the drugs penicillin and aspirin.

Pyrrole 2-carboxylate is also a

planar molecule and resembles the transition state more than either of the proline isomers.

The rxn mechanism involves a

planar transition state in which the alpha-C is a carbanion.

In fact, the induced fit model of enzyme action predicts

pyrrole 2-carboxylate will have greater affinity for the enzyme than either D- or L-proline.

A good example of a transition analog is

pyrrole 2-carboxylate, which inhibits the enzyme proline racemase.

the binding of noncompetitive inhibition is

reversible (just like competitive inhibition binding to the active site is a reversible process).

Ordinary competitive inhibition can be recognized by a

reversible rxn at the substance binding site with no change in Vmax and an apparent increase in Km.

Noncompetitive Inhibition

since the inhibitor-binding site in this case is different than the substrate-binding site, one can add more substrate and still not completely offset the inhibition (i.e. cannot reach Vmax).

The loss of enzyme activity that occurs with an irreversible inhibitor that acts at the active site can be

slowed by the presence of increasing amounts of substrate.

A Lineweaver-Burk plot will yield a line with a

steeper slope in the presence of inhibitor. However, that line will intersect the y-axis at the same point as uninhibited enzyme, which is 1/Vmax.

That is, the enzyme works best when it is more favorable to have

the enzyme bend and twist the substrate(s) into the transition state.

The inhibitor can dissociate from the other site, and

the enzyme is restored to its original state.

noncompetitive

the inhibitors binds to a site on the enzyme that is removed from the active site, but upon binding of inhibitor, the enzyme is non-functional

uncompetitive

the inhibitors binds to the ES complex, but does not bind to free enzyme; thus it may distort the active site and render the enzyme catalytically inactive.

In uncompetitive inhibition, the inhibitor binds to

the same form of the E:S complex, but does not bind to free enzyme.

Restoration of enzymatic activity requires

the synthesis of new protein.

The best competitive inhibitor would resemble

the transition state, more than the substrate or product.

Although we think of enzymes are being optimized to bind their substrates, in fact,

they function best when they stabilize the transition state.

The slope of the line is increased by the factor

(1+[I]/Ki), where Ki is the equilibrium dissociation constant for EI.

If we let K' equal the dissociation constant for the tendency of the inhibition to reverse itself and restore the enzyme to activity

(ESI=>ES+I), then we can write K'=[ES][I]/[ESI]

In chemical rxn form, we are seeing:

E+I+S <=> EI+ES ES<=>E+P

Recall the equation: E+S<=>ES<=>ES*<=>E+P

Here the inhibitor binds to ES or ES*: e.g. ES+I<=>ESI

The enzyme proline racemase interconverts

L- and D-proline.

So if we measure rate (velocity) as a function of [S] in the presence of inhibitor, at high [S] the effects of the inhibitor are eliminated and we recover full velocity, i.e. v= Vmax

The [S] concentration at which we achieve half-maximal velocity is a higher value, thus the apparent Km has increased.

However, the entire catalytic process fails and a stable EI complex is formed.

The enzyme has used its catalytic machinery to inactivate itself, hence the term "suicide".

A substance prepared specifically to resemble the transition state is called a

Transition State Analog.

Lineweaver-Burk plot - a competitive inhibitor has no effect on

Vmax but the apparent Km has increased.

What would a competitive inhibitor look like plotted on an Eadi-Hofstee plot?

Vmax is not changed (so same y-intercept) Km is increased to a higher value (greater slope) Vmax/Km smaller (x-intercept is lower, or in other words x-intercept shifts to the left). So the line would be a steeper in slope, with same Vmax and shifted to the left of the line shown.

Vmax decreased by factor:

Vmax-Inh= Vmax/(1+[I]/Ki)

These inhibitors are characterized by a time dependent loss of activity.

When enzyme, substrate and inhibitor are mixed, the rate of rxn will be the fastest at the beginning, but decrease until there is either no active enzyme left or no inhibitor left.

The E-H plots shows

a flatter slope and falling y-intercept; the x-intercept is unchanged.

A suicide inhibitor binds to the

active site and is acted upon by the enzyme just like a normal substrate.

Action may be at the

active site or at an alternate site

this requires that the inhibitor affect catalytic function, but not

affect substrate binding, so additional substrate does not overcome uncompetitive inhibition.

In a simplistic stepwise mechanism, binding of inhibitor occurs

after substrate has bound to enzyme.

a double reciprocal (L-B) plot in the presence of inhibitor will show an

increased slope and a change in the y-intercept. The x-intercept remains the same.

competitive inhibition can be offset by

increasing [S], and thus overcoming the effects of the inhibitor.

Competitive

inhibitors binds to the active site of the enzyme and "competes" with the substrate for occupation of the site (that type is modeled in the previous slide).

Ki is an

inverse measure of the affinity of the inhibitor for the enzyme, just like Km is an inverse measure of the affinity of the substrate for the enzyme.