Enzyme Kinetics

Enzymes Are Uniquely Powerful Catalysts

• Enzymes are proteins that can accelerate biochemical reactions by factors of 105 to 1017! This is much higher than chemical catalysts.

• Enzymes can be extremely specific in terms of reaction substrates and products.

• Enzymes catalyze reactions under mild conditions (e.g., pH 7.4, 37ºC).

• The catalytic activities of many enzymes can be regulated by allosteric effectors.

Chemical Kinetics

Irreversible First-Order Reactions

A B

v = d[B]/dt = -d[A]/dt = k[A]

(k = first-order rate constant (s-1))

Change in [A] with time (t):[A]= [A]o e –kt or[A]/[A]o = e –kt

ln([A]/[A]o) = –kt

([A]o = initial concentration)

k

Reversible First-Order Reactions

A B

v = -d[A]/dt = k1[A] - k-1[B]

At equilibrium: k1[A]eq - k-1[B]eq = 0

[B]eq/[A]eq = k1/k-1 = Keq

k1

k-1

Second-Order Reactions

2A P

v = -d[A]/dt = k[A]2

Change in [A] with time:1/[A] = 1/[A]o + kt

A + B P

v = -d[A]/dt = -d[B]/dt = k[A][B]

(k = second-order rate constant (M-1s-1))

k

k

Note: third-order reactions rare, fourth- and higher-order reactions unknown.

Free Energy Diagrams

Keq = e –∆Gº/RT

For A A‡

[A]‡/[A]o = e –∆Gº‡/RT

[A]‡ = [A]o e –∆Gº‡/RT

Keq = equilibrium constant [A]‡ = concentration of molecules having the activation energy[A]o = total concentration of A –∆Gº‡ = standard free energy change of activation (activation energy)

Relationship of Reaction Rate Constant to Activation Energy and

Temperature: The Arrhenius Equationk = A e -Ea/RT

Reaction rate constant (k) determined by activation energy (Ea or ∆Gº‡, applying transition state theory) and temperature (T) and proportional to frequency of forming product(A or Q = kBT/h, where kB = Boltzmann’s constant, h = Planck’s constant):

k = (kBT/h) e -G°‡/RT

k = Q e -G°‡/RT

G = H - T S, so:k = Q e S°‡/R e -H°‡/RT

k = Q e -H°‡/RT

(where Q = Q e S°‡/R)So: ln k = ln Q - H°‡/RT

L-malate fumarate + H20ln k

Relation of Equilibrium Constant to Activation Energy

Keq = k1/k-1

Keq = (Q e -G1°‡/RT)/(Q e -G-1°‡/RT)

Keq = e -(G1°‡ - G-1°‡)/RT

∆G° = G1°‡ - G-1°‡

Keq = e –∆G°/RT

Equilibrium constant Keq says nothing about rate of reaction, only free energy difference between final and initial states. The activation energy barrier opposes reaction in both directions

Effect of a Catalyst on Activation Energy

•Catalysts do not affect GA (initial) or GB (final) and so do not affect overall free energy change (∆G° = GB - GA) or equilibrium constant Keq.•Equilibrium concentrations of A and B still determined solely by overall free energy change. •Catalysts only affect ∆G°‡, lowering the activation energy.•They accelerate both the forward and reverse reaction (increase kinetic rate constants k1 and k-1).

Intermediate States in Multistep Reactions

Enzyme Kinetics

The Effect of Substrate Concentration on Reaction Velocity

Michaelis-Menten Kinetics (1)

v = k2[ES] (Note: k2 also referred to as kcat)

[Enzyme]total = [E]t = [E] + [ES]

How to solve for [ES]?

1. Assume equilibrium, if k-1 >> k2:KS = k-1/k1 = [E][S]/[ES]

or2. Assume steady state:

d[ES]/dt = 0

(Michaelis and Menten, 1913)

(Briggs and Haldane, 1925)

E = enzyme, S = substrate, ES = enzyme-substrate complex,P = product

The Steady State in Enzyme Kinetics

Michaelis-Menten Kinetics Continued (2)

Rate of formation of ES complex = k1[E][S]Rate of breakdown of ES complex = k-1[ES] + k2[ES]

Because of steady state assumption:k1[E][S] = k-1[ES] + k2[ES]

Rearranging: [ES] = (k1/(k-1 + k2))[E][S]

Substituting Michaelis constant = KM = (k-1 + k2)/k1) = KS + k2/k1: [ES] = ([E][S])/KM

So: KM[ES] = [E][S]

Michaelis-Menten Kinetics Continued (3)

Substituting [E] = [E]t - [ES]: KM[ES] = [E]t[S] - [ES][S]

Rearranging: [ES](KM + [S]) = [E]t[S]

So: [ES] = [E]t[S]/(KM + [S])

Michaelis-Menten Kinetics Continued (4)

Now we can substitute for [ES] in the rate equationvo = k2[ES].

But first note that the velocity in v = k2[ES] we use is the initial velocity, vo, the velocity of the reaction after the pre-steady state and in the early part of the steady state, i.e., before ~10% of substrate is converted to product. This is because at this stage of the reaction, the steady-state assumption is reasonable ([ES] is still approximately constant). Also, since not much P has yet accumulated, we can approximate the kinetics for even reversible reactions with this equation if we limit ourselves to vo.

The Michaelis-Menten Equation

vo = k2[E]t[S]/(KM + [S])

or

vo = Vmax[S]/(KM + [S])

(since Vmax = k2[E]t when [S] >> KM)

A Lineweaver-Burk (Double Reciprocal) Plot

An Eadie-Hofstee Plot

Multistep Reactions

E + S ES ES E + P

vo = kcat[E]t[S]/(KM + [S])

k2 k3 k1

k-1

kcat = empirical rate constant that reflects rate-determining component. Mathematically, for the reaction above,kcat = k2k3/(k2 + k3).However, k2 and k3 often very hard to establish with precision as individual rate constants.

Catalytic Efficiency (kcat/KM )

“Perfect enzyme”

Diffusion-controlled limit: 108-109 M-1s-1

Substrate preferences of chymotrypsin

pH-Dependence of Enzyme Activity

Enzyme-Catalyzed Bisubstrate Reactions: Two Examples

Bisubstrate Reactions

S1 + S2 P1 + P2

A-X + B A + B-X (in transferase reactions)

• Sequential binding of S1 and S2 before catalysis:– Random substrate binding - Either S1 or S2

can bind first, then the other binds.– Ordered substrate binding - S1 must bind

before S2.• Ping Pong reaction - first S1 P1, P1 released

before S2 binds, then S2 P2.

E

E

Ping Pong reaction

Sequential binding

Ternarycomplex

Indicative of ternary complex formation and a sequential mechanism

Indicative of a Ping Pong mechanism

Enzyme Inhibition

Types of Enzyme Inhibition

• Reversible inhibition(Inhibitors that can reversibly bind and dissociate from enzyme; activity of enzyme recovers when inhibitor diluted out; usually non-covalent interaction.)– Competitive– Mixed (noncompetitive)– Uncompetitive

• Irreversible inhibition(Inactivators that irreversibly associate with enzyme; activity of enzyme does not recover with dilution; usually covalent interaction.)

Competitive Inhibition

Effects of Competitive Inhibitor on Enzyme Kinetics

KappM = KM(1 + [I]/KI) > KM

Vappmax = Vmax

KI (inhibitor dissociation constant) = koff/kon

= 1 + [I]/KI

A Substrate and Its Competitive Inhibitor

HIV Protease Inhibitors

Relationship of KI to Half-Maximal Inhibitory Concentration (IC50)

For a competitive inhibitor of an enzyme that follows Michaelis-Menton kinetics:

vI/v0 = (Vmax[S]/(KM + [S]))/(Vmax[S]/(KM + [S])) = (KM + [S])/(KM + [S])vI = initial velocity with inhibitorv0 = initial velocity without inhibitor= 1 + [I]/KI

When vI/v0 = 0.5, [I] = IC50 = KI(1 + [S]/KM)

If measurement made when [S] << KM, IC50 = KI

Uncompetitive Inhibition

Effects of Uncompetitive Inhibitor on Enzyme Kinetics

KappM = KM/(1 + [I]/KI) < KM

Vappmax = Vmax/(1 + [I]/KI) < Vmax

= 1 + [I]/KI

Mixed (Noncompetitive) Inhibition

Effects of Mixed (Noncompetitive) Inhibitor on Enzyme Kinetics

KappM = (1 + [I]/KI)KM/(1 + [I]/KI)

(= KM, when KI = KI, which is often the case.)

Vappmax = Vmax/(1 + [I]/KI) < Vmax

k1k-1

•Not the same as uncompetitive inhibition.•In mixed inhibition, inhibitor can bind E or ES.

= 1 + [I]/KI

= 1 + [I]/KI

= 1 + [I]/KI

= 1 + [I]/KI

(For mixed inhibitor, generally, ~ KM)

Irreversible Inhibition

k1

k-1

k2E + I E·I E-I Plot:ln(residual enzyme activity) vs. time

If [I]>>[E], conditions are pseudo-first order and slope is -kobs (pseudo-first order inactivation rate constant)

kinact (second-order inactivation constant) = k1k2/k-1 = kobs/[I]

Slope = -kobs

Irreversible Inhibition by Adduct Formation

(diisopropylfluorophosphate)

Irreversible Inhibition of Chymotrypsin by TPCK

(N-tosyl-L-phenylalanine chloromethylketone)