CINÉTICA ENZIMÁTICA

A P

A PI1 I2

Chemical Kinetics

Reaction Order

A P

At constant temperature, the rate of an elementary reaction is proportional to the frequency with which the reacting molecules come together

I1 and I2: Intermediates in

the reaction

k : rate constant

The instantaneous rate of appearance of products or disappearance of reactant is called the velocity (v) of the reaction

k has units of (s-1)

First order reaction

Reaction kinetics: 1st order reactions•

•••

• • • • • • •

[A]

t

k1A B (+ C)

Decay reactions, like radio-activity;SN1 reactions

[A]k=dt

d[A]1Rate: -

Rewriting: - dtk=[A]

d[A]1

Integration gives: t

0

t

0

ktdtd[A][A]

1

So: ln[A]t – ln[A]0 = -kt or: =]A[

]A[ln

0

t -kt

t 1/2 = ln2/k = 0.693/k

The time for half of the reactant initially present to decompose, its half-time or half-life, t1/2 , is a constant and hence independent of the initial concentration of reactant.

By substituting the relationship [A] = [A0] / 2 when t = t1/2 into ln [A]=ln [A]0 - ktand rearranging:

Substances that are inherently unstable, such as radioactive nuclei, decompose through first order reactions

Rate Equations

Radionuclide Half-life Type of Radiationa

The half-time for a second order reaction is expressed t 1/2 = 1/k [A]0 and therefore, in contrast to a

first order reaction depends on the initial reactant concentration.

Second-order reaction 2A P

A+B P

Here, the reaction is said to be first order in A and first order in B.

Unimolecular and bimolecular reactions are common. Termolecular reactions are unusual because the simultaneous collision of three molecules is a rare event. Fourth and higher order reactions are unknown.

Reaction kinetics: 2nd order reactions

k2A + B P

dt

]A[dSo: - = - ]B][A[k=

dt

]P[d+=

dt

]B[d2

When [A] [B], this equation is mathematically rather complicated.A simplification reads as follows:take [P] = x, then [A] = [A]0 – x and [B] = [B]0 – x

The rate then becomes: = =]B][A[k=

dt

dx=

dt

]P[d2 k2([A]0-x)([B]0-x)

so dtk=dxx)-x)([B]-([A]

12

00

Integration gives: tk=]B][A[

]B[]A[ln

[A]-[B]

12

0

0

00

Plotting]A[

]B[ln against t gives a straight line with slope k2([B]0-[A]0)

Special cases:

• [A]0>>[B]0 (pseudo-first order kinetics)

k2CH3I + H2O CH3OH + HI

H2OExample:

- ]ICH['k=]OH][ICH[k=dt

]ICH[d3232

3 in which k'=k2[H2O]

This is a pseudo-first order reaction, since [H2O] is constant. The second-order rate constant k2 can be calculated from k' and [H2O]. In a dilute aqueous solution, [H2O]=55 M.

Reversible reactions

A Bk1

k-1Take the simplest possibility:

On t = 0: [A] = [A]0 [B] = 0 t = t: [A] = [A]0-x [B] = x

= k1[A] – k-1[B] = k1([A]0 – x) – k-1x = k1[A]0 – (k1 + k-1)x

Integration gives: (1)

At equilibrium, the net reaction rate = 0, so [B]t is constant(=[B]e = xe), so: k1[A]e = k-1[B]e = k-1xe

dt

dx

t)k+-(k=]A[k

x)k+(k-]A[kln 1-1

01

1-101

[A]0

xe

t

[A]t

[B]t

There is an equilibrium constant:

so: (2)

e0

e

e

e

1-

1

x-]A[

x=

]A[

]B[=

k

k=K

Combining eq (1) with (2) gives: t)k+-(k=x

x-xln 1-1

e

e

This is the rate equation for a first order process!

Determination of (k1 + k-1) by plotting against t e

e

x

x-xln

Eq (2) gives 1

1

k

k 2 equations, 2 unknowns

Individual values of k1 and k-1 can be determined

1

1e0 k

k1x[A]

Preequilibria

A + B A·Bk1

k-1

Ck2

Very complicated kinetics, unless you assume that [A·B] is constant during a large part of the reaction (steady state approach)

0=dt

d[A·B]k1[A][B] = k-1[A·B] + k2[A·B] = (k-1 + k2)[A·B]

So the rate equation now becomes:

= k2[A·B] = 21

21

kk

[A][B]kk

[A·B] = k1[A][B] / k-1 + k2

Two possibilities:

- rapid breakdown of A·B, k2>>k-1, so = k1[A][B]

- slow breakdown of the complex: k2<<k1,k-1, so:

= k2[A·B] = =

k2K[A][B]

[A·B]

[C]

A0

t

[A·B]

[C]

A0

t

xe

= k2[A·B] =21

21

kk

[A][B]kk

]B][A[k

kk

1-

12

A + B A·Bk1

k-1

Ck2

Sucrose + H2O glucose + fructose

Enzyme Kinetics

ß-fructofuranosidase:

When [S] » [E] : the rate is zero order with respect to sucrose. Initial rate no longer increases at S higher than S4

The Michaelis-Menten Equation

This equation cannot be explicitly integrated, however, without simplifying assumptions, two possibilities are:

1. Assumption of equilibrium. Leonor Michaelis and Maud Menten, building on the work of Victor Henri, assumed that k-1 » k2, so that the first step of the reaction reaches equilibrium.

Ks is the dissociation constant of the first step in the enzymatic reaction

v= Vmax = k2 ETET = E + ES

ES = E * S / Ks

v/Vmax = k2 ES / k2 ET = ES/ ET

v/Vmax = ES/ ET

v/Vmax = ES/ E + ES ¿How to know ES?

The Michaelis-Menten Equation

1. Assumption of steady-state. Figure illustrates the progress curves of the various participants in reaction

under the physiologically common conditions that substrate is in great excess over Enzyme ([S] » [E]).

ES maintains a steady state and [ES] can be treated as having a constant value:

The so called steady state assumption, a more general condition than that of equilibrium, was first proposed in 1925 by G. E. Briggs and B. S. Haldane

The Michaelis constant, KM , is defined as

The Michaelis-Menten Equation

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

Therefore: ES = E * S / KM

The Michaelis-Menten Equation

The expression of the initial velocity (v0) of the reaction, the velocityat t=0, thereby becomes

The maximal velocity of a reaction, Vmax occurs at high substrate concentrations when the enzyme is saturated, that is, S>> Km, and ET is entirely in the ES form v= Vmax when

This expression, the Michaelis-Menten equation, is the basic equation of enzyme kinetic.

v/Vmax = ES/ (E + ES)

v/Vmax = (E*S)/Km/ (E + (E*S)/Km )

v/Vmax = S/ Km / (1 + S/Km)

v/Vmax = S / Km + S

Significance of the Michaelis Constant

The Michaelis constant, KM, has a simple operational definition. At the substrate concentration at which [S] = KM, this equation

yields v0 = Vmax/2 so that

KM is the substrate concentration at which the reaction velocity is half maximal

Significance of the Michaelis Constant

The magnitude of KM varies widely with the identity of the enzyme and the nature of the substrate. It is also a function of temperature and pH. The Michaelis constant can be expressed as

Since Ks is the dissociation constant of the Michaelis complex, as Ks decreases, the enzyme’s affinity for substrate increases. KM in therefore also a measure of the affinity of the enzyme for its substrate, provided k2/k1 is small compared to Ks, that is k2 ‹ k-1 so that the ES P reaction proceeds more slowly than ES reverts to E + S

kcat/KM Is a Measure of Catalytic Efficiency

We can define the catalytic constant, kcat, of an enzyme as

This quantity is also known as the turnover number of an enzyme because it is the number of reaction processes (turnovers) that each active site catalyzes per unit time.

Turn Over Numbers of Enzymes

Catalase H2O2

Carbonic anhydrase HCO3-

Acetylcholinesterase Acetylcholine

40,000,000

400,000

140,000

-Lactamase Benzylpenicillin 2,000

Fumarase Fumarate 800

RecA protein (ATPase) ATP 0.4

Enzymes Substrate kcat (s-1)

The number of product transformed from substrate by one enzyme molecule in one second

Adapted from Nelson & Cox (2000) Lehninger Principles of Biochemistry (3e) p.263

kcat/KM Is a Measure of Catalytic Efficiency

When [S] « KM, very little ES is formed. Consequently, [E] ≈ [E]T, so

reduces to a second-order rate equation:

The quantity kcat/KM is a measure of an enzyme’s catalytic efficiency.

There is an upper limit to the value of kcat/KM : It can be not greater than k1; that is, the decomposition of ES to E + P can occur no more frequently than E and S come together to form ES. The most efficient enzymes have kcat/KM values near to the diffusion-controlled limit of 108 to 109 M-1.s-1

Chymotrypsin Has Distinct kcat / Km to Different Substrates

O R O

H3C–C–N–C–C–O–CH3

H H

= – =––

–HGlycine

kcat / Km

1.3 ╳ 10-1

–CH2–CH2–CH3Norvaline 3.6 ╳ 102

–CH2–CH2–CH2–CH3Norleucine 3.0 ╳ 103

–CH2–Phenylalanine 1.0 ╳ 105

(M-1 s-1)

R =

Adapted from Mathews et al (2000) Biochemistry (3e) p.379

Analysis of Kinetic Data

Lineweaver-Burk or double-reciprocal plot

S >> Kmvi=VmaxVmax= k2Et

- dS/dt = vi = So dX/dt

Al iniciar: t = 0, S = So

A cualquier tiempo:T = t S = S X = (So-S)/So

• As is the case with most reactions, an increase in temperature will result in an increase in kcat for an enzymatic reaction.

• From general principles, it can be determined that the rate of any reaction will typically double for every 10°C increase in temperature.

• Many enzymes display maximum temperatures around 40°C, which is relatively close to body temperature.

• There are enzymes that are isolated from thermophilic organisms that display maxima around 100°C, and some that are isolated from psychrophilic organisms that display maxima around 10°C.

Temperature Dependence of Enzymes

Kinetics and Transition State Theory

Consider a bimolecular reaction that proceeds along the following pathway

X‡ is the transition state

k is the ordinary rate constant of the elementary reaction and k’ is the rate constant for the decomposition of X‡ to products.Although X‡ is unstable, it is assumed to be in rapid equilibrium with the reactants; that is:

K‡ is an equilibrium constant

T is the absolute temperatureR is the gas constant (8.3145 J.K-1 mol -1)

Combining the three preceding equations yields

This equation indicates that the rate of a reaction not only depends on the concentration of its reactants, but also decreases exponentially with ∆G‡

Kinetics and Transition State Theory

The larger the difference between the free energy of the transition state and that of the reactants, that is, the less stable the transition state, the slower the reaction proceeds.

k’ is the rate at which X‡ decomposesv is the vibrational frequency X‡ of the bond that breaks as X‡ decomposes to productsk, the transmission coefficient, is the probability that the breakdown of X‡ will be in the direction of products formation rather than back to reactants. For most spontaneous reactions, k is assumed to be 1.0 (although this number, which must be between 0 and 1, can rarely be calculated with confidence)

Planck’s law states that

is the average energy of the vibration that leads to the decomposition of X‡

h is Planck’s constant (6.6261 x 10 -34 J.s) At temperature T, the classical energy of an oscillator iskB is the Boltzmann constant (1.3807 x 10 -23 J.K-1)

As the temperature rises, so that there is increased thermal energy available to drive the reacting complex over the activation barrier, the reaction speeds up.

Interpretation of rate constants:the Arrhenius equation

Eact

ES

P

Reaction coordinate

X‡ (TS)

Every reaction has to overcome an energy barrier: the transition state (TS, X‡).At higher temperature, more particles are able to overcome the energy barrier.

Arrhenius equation:

RT

E-A·expk a

obs

Ea can be determined by measuring kobs at two different temperatures:

21

a

2

1

T

1

T

1

R

Eexp

k

k

Idem, from statistical mechanics (collision theory)

RT

E-P·Z·expk a

obs

P = probability factor (not every collision is effective)Z = collision number (number of collisions per second)

Arrhenius:

RT

E-A·expk a

obs

Idem, from transition state theory:

k‡

A + B productsK‡

X‡

[A][B]

][X=K

‡‡ or [X‡] = K‡[A][B]

= k‡[X‡] = k‡K‡[A][B] = k[A][B], so k = k‡K‡

Statistical mechanics gives us the following relation:

h

Tk=k B‡ so ‡B K

h

Tk=k

kB = Boltzmann’s constant;h = Planck’s constant

Eact

ES

P

Reaction coordinate

X‡ (TS)

For all equilibria we can write: G0 = - RT ln K, so for our case we get: G‡ = - RT ln K‡

Expressing K‡ in terms of G‡ and RT gives the following equation for k:

RT

ΔGexp

h

Tkk

‡B

Since G‡ = H‡ - TS‡, we can also write:

R

ΔSexp

RT

ΔHexp

h

Tkk

‡‡B

(1)

(2)

Eq (1) and (2) are called the Eyring equations

‡B Kh

Tk=k

The Eyring and Arrhenius equations resemble each other:

Arrhenius:

RT

E-A·expk a

obs

so:R

E

d(1/T)

klnd a

Eyring:R

ΔHT

d(1/T)

klnd ‡

so Ea = H‡ + RT

In order to determine H‡ and S‡ it is easier to differentiate ln (k/T) to 1/T:

R

ΔH

d(1/T)

ln(k/T)d ‡

ln k

1/T

ln A

slope = R

Ea

1/T

slope =

T

kln

R

ΔH‡

So, the procedure to determine activation parameters is:

- determine k at different temperatures- plotting ln(k/T) against 1/T gives H‡

-

R

ΔSexp

RT

ΔHexp

h

Tkk

‡‡B then gives S‡

and when you have H‡ and S‡, you also have G‡ sinceG = H-TS

Interpretation of activation parameters

• G‡, the Gibbs free energy of activation, determines at which rate a certain reaction will run at a given temperature

• H‡ is a measure for the amount of binding energy that is lost in the transition state relative to the ground state (including solvent effects)

• S‡ is a measure for the difference in (dis)order between the transition state and the ground state– for monomolecular reactions: S‡ 0 J/mol.K

– for a bimolecular reaction: S‡ << 0 J/mol.K(two particles have to come together in the transition state to form one particle, demanding a much greater order)

Example:

N

CH2Ph

O

NH2

HH

N

S N+Me2Me2N

N

CH2Ph

O

NH2

H

N

S NMe2Me2N

H

MBH

+

BNAH(NADH model)

methylene blue(MB+)

+

BNA+

(NAD+ model)

ln(At-A)

t (sec)

xx

xx

xx

xx

xx

x

G‡ = 62.8 kJ/mol (very fast rx)H‡ = 33.0 kJ/mol (rel. low, compensation of C-H bond cleavage by hydration TS)S‡ = -100 J/mol.K (bimolecular rx)

Another example:

N

CH2Ph

O

NH2

CNH

N

CH2Ph

O

NH2

H

CH3CN

1

+ CN

H‡ = 85 kJ/mol (relatively high: no new bonds to beformed, no compensation for the partial cleavageof the C-C bond in the transition state; acetonitrileis aprotic, compensation of H‡ by solvation willbe less than in water

S‡ = 0 J/mol.K (monomolecular reaction)

Application of activation parameters for the elucidationof reaction mechanisms:

H3CO

O P

O

O

O

H3CO

O+ H2O + H2PO4

A S‡ of +12 J/mol.K was found monomolecular process

H3CO

O P

O

O

O

PhH3C

O

OHHO P

O

O

O

Ph+ H2O +

A S‡ of -117 J/mol.K was found bimolecular process; rate determining step in this case is the attack of water on the carbonyl group.

Look in your course book for the exact reaction mechanisms!

Solvation (solvent effects)

Influence of solvation on the reaction rate:

CH3I + Cl CH3Cl + I

k(H2O) = 10-7 l.mol-1.s-1; k(DMF) = 10-1 l.mol-1.s-1

so G‡ ~ 30 kJ/mol

H

O

NCH3

CH3

= DMF

H2O

DMF

G‡H2O

G‡DMF

E

reaction progressG‡

DMF < G‡H2O

What is the background of this strong solvent effect?

In H2O there is more solvation than in DMF, due to hydrogen bonds.

Note the changes in entropy: loss of S‡ because of orientation of the substrates, gain of S‡ because of the liberation of water (less solvated transition state). The balance is not easy to predict!In general, in case of ions, the ground state is more solvated than the transition state:

O2N ON

OO

CN

OO2N

O C O

O2N ON

OO

‡

+

TS (‡) is hardly solvated due to the spreading of charge.Again a strong solvent effect here: k(H2O)= 7.4x10-6 s-1; k(DMF) = 37 s-1

Cl

H

O H

HO H

HO HH

O HH

OH

HOH

HOH

HO

H

H

OH

HO H

HO

H

Cl

HO

H

HOH

HOH

HO

H

C I

H H

H

H

OH

HO H

HO

H

IH3C

HO H

+ x H2O

Solvation effects in (bio)polymers

Polymers or enzymes may have apolar pockets, which leads to:- less solvation and therefore higher reaction rates;- changes in pKa’s of acidic/basic groups:

N

NRH3C

N

NRH3C

H

+ H+ R = CH3 or compound with polymer

R = CH3: pKa = 9.7R = polymer: pKa = 7.7

Ka = [PyN][H+][PyNH+]

E.g. lysine, R-NH2 + H+ R-NH3+

pKa (H2O) = 10.4, in some enzymes pKa = 7 !

1/v = 1/ Vmax + Km/Vmax (1/S) + (1 / Ki Vmax) S

S pequeñas

1/v = 1/ Vmax + Km/Vmax (1/S) + (1 / Ki Vmax) S

S grandes

1/v = 1/ Vmax + Km/Vmax (1/S) + (1 / Ki Vmax) S

Enzyme Inhibition

Competitive Inhibition

Many substances alter the activity of an enzyme by reversibly combining with it in a way what influence the binding of substrate and/or its turnover number. Substances that reduce an enzyme’s activity in this way are known as inhibitors

A substance that competes directly with a normal substrate for an enzyme’s substrate-binding site is known as a competitive inhibitor.

Here it is assumed that I, the inhibitor, bind reversibly to the enzyme and is in a rapid equilibrium with it so that

And EI, the enzyme-inhibitor complex, is catalytically inactive. A competitive inhibitor therefore reduces the concentration of free enzyme available for substrate binding.

Enzyme Inhibition

This is the Michaelis-Menten equation that has been modified by a factor, , which is defined as

Competitive Inhibition

Is a function of the inhibitor’s concentration and its affinity for the enzyme. It cannot be less than 1.

Enzyme Inhibition

Competitive Inhibition

Recasting in the double-reciprocal form yields

A plot of this equation is linear and has a slope of KM/Vmax, a 1/[S] intercept of -1/ KM, and a 1/v0 intercept of 1/ Vmax

Enzyme Inhibition

Uncompetitive Inhibition

In uncompetitive inhibition, the inhibitor binds directly to the enzyme-substrate complex but not to the free enzyme

In this case, the inhibitor binding step has the dissociation constant

The uncompetitive inhibitor, which need not resemble the substrate, presumably distorts the active site, thereby rendering the enzyme catalytically inactive.

Enzyme Inhibition

Uncompetitive Inhibition

The double-reciprocal plot consists of a family of parallel lines with slope KM/Vmax, 1/v0 intercepts of ’/Vmax and 1/[S] intercept of -’/KM

Enzyme InhibitionMixed Inhibition (noncompetitive inhibition)

A mixed inhibitor binds to enzyme sites that participate in both substrate binding and catalysis. The two dissociation constants for inhibitor binding

Double-reciprocal plots consist of lines that have the slope KM/Vmax, with a 1/v0 intercept of ’/Vmax and 1/[S] intercept of -’/ KM

Steady State Kinetics Cannot Unambiguously Establish a Reaction Mechanism

Measurements of a multistep reaction can be likened to a “black box” containing a system of water pipes with one inlet and one drain

The steady state kinetics analysis of a reaction cannot unambiguously establish its mechanism

Bisubstrate Reactions

Almost all of these so called bisubstrate reactions are either transferase reactions in which enzyme catalyzed the transfer of a specific functional group, X, from one of the substrates to the other:

or oxidation-reduction reactions in which reducing equivalents are transferred between two substrates.

Sequential Reactions

Reactions in which all substrates must combine with the enzyme before a reaction can occur and products be released are known as Sequential reactions

Sequential Reactions

Ordered bisubstrate reaction

Random bisubstrate reaction

A and B : substrates in order that they add to the enzymeP and Q : products in order that they leave the enzyme

Group-transfer reactions in which one or more products are released before all substrates have been added are known as Ping Pong reactions

Ping Pong Reactions

Bisubstrate Reactions