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10/28/2010 Enzyme inihibition, mechanism s Enzymes: inhibition and mechanisms Andy Howard Introductory Biochemistry 28 October 2010
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Page 1: 10/28/2010Enzyme inihibition, mechanisms Enzymes: inhibition and mechanisms Andy Howard Introductory Biochemistry 28 October 2010.

10/28/2010Enzyme inihibition, mechanisms

Enzymes: inhibition and

mechanismsAndy Howard

Introductory Biochemistry28 October 2010

Page 2: 10/28/2010Enzyme inihibition, mechanisms Enzymes: inhibition and mechanisms Andy Howard Introductory Biochemistry 28 October 2010.

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Inhibition and mechanisms Inhibition is important in its own right and as a tool for understanding kinetics and mechanisms

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Inhibition & Mechanism Topics Inhibitors Types of inhibitors

Kinetics of inhibition

Pharmaceuticals

What makes an inhibitor a useful drug?

Mechanisms Terminology Activation barrier

E-X* complexes Enthalpy & entropy

Protein Motion

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Distinctions we can make

Inhibitors can be reversible or irreversible

Where do they bind? At the enzyme’s active site At a site distant from the active site.

To what do they bind? To the unliganded enzyme E To the enzyme-intermediate complex or the enzyme-substrate complex (ES)

To both (E or ES)

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Types of inhibitors Irreversible Inhibitor binds without possibility of release

Usually covalent Each inhibition event effectively removes a molecule of enzyme from availability

Reversible Usually noncovalent (ionic or van der Waals)

Several kinds Classifications somewhat superseded by detailed structure-based knowledge of mechanisms, but not entirely

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Types of reversible inhibition

Competitive Inhibitor binds at active site of unliganded enzyme

Prevents binding of substrate Noncompetitive

Inhibitor binds distant from active site (E or ES)

Interferes with turnover Uncompetitive (rare?)

Inhibitor binds only to ES complex Removes ES, interferes with turnover

Mixed(usually Competitive + Noncompetitive)

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How to tell them apart Reversible vs irreversible

dialyze an enzyme-inhibitor complex against a buffer free of inhibitor

if turnover or binding still suffers, it’s irreversible

Competitive vs. other reversible: Structural studies if feasible Kinetics

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Competitive inhibition Put in a lot of

substrate:ability of the inhibitor to getin the way of the binding is hindered:out-competed by sheer #s of substrate molecules.

This kind of inhibition manifests itself as interference with binding, i.e. with an increase of Km

SIc

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Competitive inhibitors don’t affect turnover If the substrates manages to bind even though there is inhibitor present, then it can be turned over just as quickly as if the inhibitor is absent; so the inhibitor influences binding but not turnover.

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Kinetics of competition Competitive inhibitor hinders binding

of substrate but not reaction velocity:

Affects the Km of the enzyme, not Vmax.

Which way does it affect it? Km = amount of substrate that needs to be present to run the reaction velocity up to half its saturation velocity.

Competitive inhibitor requires us to shove more substrate into the reaction in order to achieve that half-maximal velocity.

So: competitive inhibitor increases Km

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L-B: competitive inhibitor

Km goes up so -1/ Km moves toward origin

Vmax unchanged so Y intercept unchanged

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Competitive inhibitor:Quantitation of Ki Define inhibition constant Ki to be the concentration of inhibitor that increases Km by a factor of two.

Km,obs = Km(1+[Ic]/Ki)

So [Ic] that moves Km halfway to the origin is Ki.

If Ki = 100 nM and [Ic] = 1 µM, then we’ll increase Km,obs elevenfold!

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Think about that equation!

Remember that it says Km,obs = Km(1+[Ic]/Ki)

It does NOT say Km,obs = Km[(1+[Ic])/Ki] … which would be nonsensical because [Ic] has dimensions and 1 doesn’t

In fact, Ic and Ki have the same dimensions,so they cancel like they should!

But every year several students get that wrong. Don’t be among them!

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Don’t get lazy!

A competitive inhibitor doesn’t automatically double Km

The amount by which the inhibitor increases Km is dependent on [I]c

If it happens that [I]c = KI, then Km will double, as the equation shows

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Noncompetitive inhibition Inhibitor binds distant from

active site, so it binds to theenzyme whether the substrateis present or absent.

Noncompetitive inhibitor has no influence on how available the binding site for substrate is, so it does not affect Km at all

However, it has a profound inhibitory influence on the speed of the reaction, i.e. turnover. So it reduces Vmax and has no influence on Km.

SI

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L-B for non-competitives

Decrease in Vmax 1/Vmax is larger X-intercept unaffected

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Ki for noncompetitives

Ki defined as concentration of inhibitor that cuts Vmax

in half Vmax,obs =Vmax/(1 + [In]/Ki)

In previous figure the “high” concentration of inhibitor is Ki

If Ki = Ki’, this is pure noncompetitive inhibition

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Same warning as before . . . Correct: Vmax,obs = Vmax/(1 + [In]/Ki)

Incorrect: Vmax,obs = Vmax/[(1 + [In])/Ki]

As in the previous instance, the incorrect formula makes no sense because [In] has dimensions and 1 doesn’t.

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Uncompetitive inhibition

Inhibitor binds only if ES has already formed

It creates a ternary ESI complex This removes ES, so by LeChatlier’s Principle it actually drives the original reaction (E + S ES) to the right; so it decreases Km

But it interferes with turnover so Vmax goes down

If Km and Vmax decrease at the same rate, then it’s classical uncompetitive inhibition.

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L-B for uncompetitives -1/Km moves away from origin 1/Vmax moves away from the origin Slope ( Km/Vmax) is unchanged

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Ki for uncompetitives

Defined as inhibitor concentration that cuts Vmax or Km in half

Easiest to read from Vmax value Vmax,obs = Vmax/(1+[I]u/KI) Iu labeled “high” is Ki in this plot

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iClicker quiz, question 1

1. Treatment of enzyme E with compound Y doubles Km and leaves Vmax unchanged. Compound Y is: (a) an accelerator of the reaction

(b) a competitive inhibitor (c) a non-competitive inhibitor (d) an uncompetitive inhibitor

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iClicker quiz, question 2 2. Treatment of enzyme E with compound X doubles Vmax and leaves Km unchanged. Compound X is: (a) an accelerator of the reaction

(b) a competitive inhibitor (c) a non-competitive inhibitor (d) an uncompetitive inhibitor

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Mixed inhibition

Usually involves interference with both binding and catalysis

Km goes up, Vmax goes down Easy to imagine the mechanism:

Binding of inhibitor alters the active-site configuration to interfere with binding, but it also alters turnover

Same picture as with pure noncompetitive inhibition, but with Ki ≠ Ki’

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Most pharmaceuticals are enzyme inhibitors Some are inhibitors of enzymes that are necessary for functioning of pathogens

Others are inhibitors of some protein whose inappropriate expression in a human causes a disease.

Others are targeted at enzymes that are produced more energetically by tumors than they are by normal tissues.

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Characteristics of Pharmaceutical Inhibitors

Usually competitive, i.e. they raise Km without affecting Vmax

Some are mixed, i.e. Km up, Vmax down

Iterative design work will decrease Ki

from millimolar down to nanomolar Sometimes design work is purely blind HTS; other times, it’s structure-based

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Amprenavir

Competitive inhibitor of HIV protease,Ki = 0.6 nM for HIV-1

No longer sold: mutual interference with rifabutin, which is an antibiotic used against a common HIV secondary bacterial infection, Mycobacterium avium

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When is a good inhibitor a good drug? It needs to be bioavailable and nontoxic

Beautiful 20nM inhibitor is often neither

Modest sacrifices of Ki in improving bioavailability and non-toxicity are okay if Ki is low enough when you start sacrificing

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How do we lessen toxicity and improve bioavailability?

Increase solubility…that often increases Ki because the van der Waals interactions diminish

Solubility makes it easier to get the compound to travel through the bloodstream

Toxicity is often associated with fat storage, which is more likely with insoluble compounds

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Drug-design timeline 2 years of research, 8 years of trials

log Ki

Time, Yrs 1020

-3

-8

Cost/yr, 106 $

10

100

Improving affinity

Toxicity and

bioavailabili

ty

Research Clinical Trials

Preliminary toxicity testing

Stage I clinical trials

Stage II clinical trials

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Atomic-Level Mechanisms

We want to understand atomic-level events during an enzymatically catalyzed reaction.

Sometimes we want to find a way to inhibit an enzyme

in other cases we're looking for more fundamental knowledge, viz. the ways that biological organisms employ chemistry and how enzymes make that chemistry possible.

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How we study mechanisms There are a variety of experimental tools available for understanding mechanisms, including isotopic labeling of substrates, structural methods, and spectroscopic kinetic techniques.

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Overcoming the barrier Simple system:single high-energy transition state intermediate between reactants, products

Free Energy

Reaction Coordinate

RP

G‡

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Intermediates Often there is a quasi-stable

intermediate state midway between reactants & products; transition states on either side

Free Energy

R P

T1T2

I

Reaction Coordinate

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Activation energy & temperature

It’s intuitively sensible that higher temperatures would make it easier to overcome an activation barrier

Rate k(T) = Q0exp(-G‡/RT) G‡ = activation energy or Arrhenius energy

This provides tool for measuring G‡

Svante Arrhenius

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Determining G‡

Rememberk(T) = Q0exp(-G‡/RT)

ln k = lnQ0 - G‡/RT Measure reaction rate as function of temperature

Plot ln k vs 1/T; slope will be -G‡/R

ln k

1/T, K-1

uncatalyzed

catalyzed

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How enzymes alter G‡

Enzymes reduce G‡ by allowing the binding of the transition state into the active site

Binding of the transition state needs to be tighter than the binding of either the reactants or the products.

In fact, the enzyme must stabilize the transition-state complex EX‡ more than it stabilizes the substrate complex ES (see section 14.2).

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Dissociation constants for ES and EX* Dissociation constant for ES:Ks = [E][S]/[ES]

Dissociation constant for EX‡:KT = [E][X‡]/[EX‡]

Transition state theory says the ratio of reaction rates is related to the ratio of these:

ke/ku = Ks / KT

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What makes EX‡ more stable than ES? Intrinsic (enthalpic) binding energy

of ES makes it a lower-energy species than E+S; but we want EX* to be lower.

ES loses entropy relative to E + S ES is sometimes strained, distorted, or desolvated relative to E+S

So if EX‡ is less strained and has more entropy, we win

See section 14.3

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How tight is the binding?

Section 14.4 gives some examples Transition-state analogs are stable molecules that are geometrically and electrostatically similar to transition states

Sometimes the analogs bind ~ 160 - 40000 times more avidly than substrates

1,6-hydrate of purine nucleoside binds to adenosine deaminase with KI = 3*10-

13M

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G‡ and Entropy

Effect is partly entropic: When a substrate binds,it loses a lot of entropy.

Thus the entropic disadvantage of (say) a bimolecular reaction is soaked up in the process of binding the first of the two substrates into the enzyme's active site.

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Enthalpy and transition states Often an enthalpic component to the reduction in G‡ as well

Ionic or hydrophobic interactions between the enzyme's active site residues and the components of the transition state make that transition state more stable.

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Two ways to change G‡ Reactants bound by enzyme are properly positioned

Get into transition-state geometry more readily

Transition state is stabilized

E AB

E AB

A+B A+BA-B A-B

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The protein moves as well!

Changes to active-site conformation: Help with substrate binding Position the catalytic groups Induce formation of an NAC Help to break or make bonds Facilitate conversion of S to P

Sometimes involve networks of concerted amino acid changes


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