LECTURE 4: Principles of Enzyme CatalysisReading: Berg, Tymoczko & Stryer: Chapter 8

An ENZYMEENZYME is a biomolecular catalyst that accelerates the rate of a specific reaction

Enzymes DO NOT make a chemical reaction more energetically favorable;They only ACCELERATE the rate of the reaction towards its energetic equilibrium

Enzymes work by stabilizing chemical transition states, the high-energy intermediatesthat normally act as a barrier to spontaneous reaction

Most enzymes are FOLDED PROTEINS: proteins have the ability to fold into scaffoldswith binding surfaces for substrates that position the substrates for chemical reaction

A few enzymes are RNA molecules! RNAs also have ability to adopt tertiary structures.Some RNAs (called RIBOZYMES) act as enzymes catalyzing their own site-specific cleavage

or that of other RNA molecules

As catalysts, enzymes are NOT CONSUMED during reactions.

S + E -----> ES -------> P + E

In some cases, enyzmes are chemically modified during catalysis, but return to theiroriginal form after reaction cycle to allow further catalysis of substrate

S + E -----> E*S -------> P + E

Proteases Are Examples of Enzymes That Catalyze An Energetically Favored Process

Fig 8.0new (peptide hydryolys)

Peptide hydrolysis is an energetically favorable process, but normally occurs very slowly.PROTEASES are enzymes that catalyze peptide hydrolysis.Some proteases are rather NONSELECTIVE (e.g., papain)

Other proteases are VERY SELECTIVE (e.g., trypsin, thrombin, fibrin)

Fig 8.1new

Some Enzymes Employ Cofactors

Some enzymes use cofactors as part of the active site in enzymatic catalysis

APOENZYME + COFACTOR --------> HOLOENZYME

Many cofactors cannot be synthesized by humans, and must be obtainedthrough diet as vitamins and minerals

Tab8.2new

Free Energy of Biochemical Reactions

For reaction A + B C + D

G is the differential in free energy between the products vs. reactants

If G < 0, reaction is energetically favorable I.e., reactants will convert to products as system moves to equilibrium

If G = 0, reaction is already at equilibrium I.e., there will be no NET conversion of reactants to products

If G > 0, reaction is disfavored I.e., products will convert to reactants as system moves to equilibrium; the reverse reaction is favored

Free energy G is usually expressed in units kcal/mol or kcal x mol-1

Standard Free Energy is Related to Equilibrium Constant

For reaction A + B C + D

G = G o + RT ln[C] [D]

[A] [B]

G is free energy change when reactant and product concentrations are [A],[B],[C],[D]

Go is free energy change when reactant and product concentrations are each 1M

WHAT DOES THIS MEAN?

Go is measure of whether reactants or products are favored if all components are at same concentration

The actual concentration of reactants and products impacts on G;

even if Go is unfavorable, high ratio of reactants to products can give favorable G

Go can be related to the equilibrium constant, Keq, of the reaction

Keq =[C]eq[D]eq

[A]eq[B]eq

Since G o = - RT lnKeq = - 2.3RT log10Keq

G o = - RT lnKeq = - 2.3RT log10Keq

Tab8.3new

Enzymes Accelerate Rate Constant Without Altering Equilibrium Constant

S PWithout enzyme

With enzyme S P

Keq = kf / kr

Keq = kf / kr

=

Fig8.2new

Enzymes Stabilize Reaction Transition State(s)

Fig8.3new

Properties of Enzyme Active Sites

Active site consists of atoms on residue side chains that are brought together by the fold

Most of the enzyme structure is a scaffold to precisely position active site residues

Active site uses range of noncovalent bonding mechanisms to bind substrate

By binding multiple substrates in a favorable interspatial relationship and/or

by altering charge distribution (resonance) within substrates, Gtransition ismuch smaller than would be spontaneously

Fig8.7new Fig8.8new

The Michaelis-Menton Model of Enzyme Function

E + S ES E + Pk2

k-1

k1

k-2

At time=0, if [P]=0, then E + S ES E + Pk-1

k1

Vo = kcat [ES] [ES] determined by [E], [S], and rate constants

THESE EQUATIONS CAN BE SOLVED TO EXPRESS THE REACTION RATE AS A FUNCTION

OF THE SUBSTRATE CONCENTRATION [S] AND TWO INHERENT PROPERTIESOF THE ENZYME: KM AND kcat

Vo = VMAX

[S]

[S] + KM

Michaelis-Menton Equation

k2 kcat=

kcat

where

VMAX = kcat [E]

KM = k-1 /k1 = [E][S]/[ES]and

The Michaelis-Menton Equation: Meaning Behind The Terms

Fig8.12new

Vo = VMAX

[S]

[S] + KM

No matter how large the substrate concentration,

reaction rate can never exceed VMAX

VMAX reflects the TURNOVER RATE of substrate

molecules through the enzyme (kcat)

and the enzyme concentration

KM is the substrate concentration at which

reaction rate is HALF MAXIMAL

KM reflects the BINDING AFFINITY of the enzyme

for the substrate;

The higher the affinity, the smaller is Km

By performing experiments to calculate Voat different substrate concentrations,

properties VMAX and Km can be calculated

If the enzyme concentration is known, VMAX can

be used to calculate kcat

Lineweaver-Burk Plot Facilitates Calculation of KM and VMAX

Vo = VMAX

[S]

[S] + KM

By inverting equation, get:

Vo

1 1

VMAX

= + KM

VMAX( )1

[S]

Fig8.13new

Tab8.5new

Many Enzymatic Reactions Proceed Through Fixed Sequential Steps

PyrLac.p223new

TransAm.p224new

Reaction Intermediate May Utilize Covalently Modified Enzyme or Cofactor

Enzyme Inhibition

Many small molecules can bind to enzymes and inhibit them.

Inhibitors can be described as REVERSIBLE or IRREVERSIBLE.

Inhibitors may be naturally occuring within the homologous organism orin a heterologous organism

Other inhibitors are synthetic and have been developed as pharmaceuticalsfor research and clinical applications

COMPETITIVE INHIBITORS act by occupying the enzyme active sitein place of the substrate

NONCOMPETITIVE INHIBITORS bind away from the active site, but their bindingexerts allosteric effects that prevents bound substrate conversion to product

Fig8.15AnewFig8.15Bnew Fig8.15Dnew

Kinetics of Competitive Inhibition

Fig8.17new

Vo = VMAX

[S]

[S] + KM ( Ki

[ I ]1 + )

Ki reflects the AFFINITY OF INHIBITOR

for the enzyme

Inhibitor in effect raises the apparent KM term

The potency of inhibitor determined by Ki

Therefore, the amount of substrateneeded for half-maximum rate

is increased

THERE IS NO EFFECT ON VMAX

I.e., a competitive inhibitor can beovercome by sufficiently high

substrate concentration

Kinetics of Noncompetitive Inhibition

Fig8.19new

( Ki

[ I ]1 + )

Vo =[S]

[S] + KM

VMAX xAs before,

Ki reflects the AFFINITY OF INHIBITOR

for the enzyme

Inhibitor reduces VMAX of the enzyme

The potency of inhibitor determined by Ki

The inhibitor does not affect KM , i.e.,

the binding of substrate to enzyme