CINÉTICA ENZIMÁTICA
Chemical Kinetics
A
A
I1
P
I2
P
I1 and I2: Intermediates in
the reaction
Reaction Order
At constant temperature, the rate of an elementary reaction is proportional
to the frequency with which the reacting molecules come together
First order reaction
A
P
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)
Reaction kinetics: 1st order reactions
A
k1
•
B (+ C)
Decay reactions, like radio-activity;
SN1 reactions
d[A]
Rate: = k1[A]
dt
•
•
•
••
[A]
d[A]
Rewriting: = k1dt
[A]
t
t
t
1
Integration gives:
0 [A] d[A]  0 ktdt
So: ln[A]t – ln[A]0 = -kt
[ A ]t
or: ln
= -kt
[A ]0
••
•
•
•
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 - kt
and rearranging:
t 1/2 = ln2/k = 0.693/k
Rate Equations
Substances that are inherently unstable, such as radioactive nuclei, decompose through first order
reactions
Radionuclide
Half-life
Type of
Radiationa
Second-order reaction
2A
The half-time for a second order reaction is expressed
P
t 1/2 = 1/k [A]0
and therefore, in contrast to a
first order reaction depends on the initial reactant concentration.
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
d[A]
d[B]
d[P]
So: ==+
= k 2 [A][ B]
A+B
P
dt
dt
dt
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
k2
d[P] dx
The rate then becomes: n =
=
= k 2 [A ][ B] = k2([A]0-x)([B]0-x)
dt
dt
1
so
dx = k 2 dt
([A] 0 - x)([B] 0 - x)
[A]0 [B]
1
ln
= k 2t
Integration gives:
[B] 0 - [A]0 [A][ B]0
[B]
Plotting ln
against t gives a straight line with slope k2([B]0-[A]0)
[A]
Special cases:
• [A]0>>[B]0
Example:
(pseudo-first order kinetics)
CH3I + H2O
k2
H2O
CH3OH + HI
d[CH 3 I]
= k 2 [CH 3 I][ H 2 O] = k '[CH 3I] in which k'=k2[H2O]
dt
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
Take the simplest possibility:
A
On t = 0: [A] = [A]0 [B] = 0
t = t: [A] = [A]0-x [B] = x
k1
k-1
[A]0
B
[A]t
xe
[B]t
t
dx
= k1[A] – k-1[B] = k1([A]0 – x) – k-1x = k1[A]0 – (k1 + k-1)x
dt
k1[A]0 - (k 1 + k -1 ) x
= -(k1 + k -1 ) t
Integration gives: ln
k1[A]0
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
(1)
xe
k1 [B]e
There is an equilibrium constant: K =
=
=
k -1 [A]e [A]0 - x e
 k 1 

so: [A]0  x e 1 
k1 

(2)
Combining eq (1) with (2) gives:
xe - x
ln
= -(k1 + k -1 ) t
xe
This is the rate equation for a first order process!
xe - x
Determination of (k1 + k-1) by plotting ln
xe
Eq (2) gives
k 1
k1
against t
2 equations, 2 unknowns
Individual values of k1 and k-1 can be determined
Preequilibria
A+B
k1
k-1
A·B
k2
C
Very complicated kinetics, unless you assume that [A·B] is
constant during a large part of the reaction (steady state approach)
d[A·B]
=0
dt
k1[A][B] = k-1[A·B] + k2[A·B] = (k-1 + k2)[A·B]
[A·B] = k1[A][B] / k-1 + k2
So the rate equation now becomes:
k1k 2 [A][B]
n = k2[A·B] =
k 1  k 2
A+B
k1
k-1
A·B
k2
C
n = k2[A·B] =
Two possibilities:
k1k 2 [A][B]
k 1  k 2
A0
[C]
- rapid breakdown of A·B, k2>>k-1,
so n = k1[A][B]
[A·B]
t
- slow breakdown of the complex:
k2<<k1,k-1, so:
k1
n = k2[A·B] = k 2
[A ][ B] =
k -1
A0
xe
[A·B]
[C]
k2K[A][B]
t
Enzyme Kinetics
ß-fructofuranosidase:
Sucrose + H2O
glucose + fructose
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
ET = E + ES
v= Vmax = k2 ET
v/Vmax = k2 ES / k2 ET
=
ES/ ET
v/Vmax = ES/ ET
v/Vmax = ES/ E + ES ¿How to know ES?
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.
ES = E * S / Ks
Ks is the dissociation constant of the first step in the enzymatic reaction
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-Menten Equation
Solving for [ES]:
ES = E * S /(k-1+k2)/k1
The Michaelis constant, KM ,
is defined as
Therefore:
ES = E * S / KM
The Michaelis-Menten Equation
The expression of the initial velocity (v0) of the reaction, the velocity
at t=0, thereby becomes
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
This expression, the Michaelis-Menten
equation, is the basic equation of
enzyme kinetic.
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
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
kcat (s-1)
Enzymes
Substrate
Catalase
H2O2
Carbonic anhydrase
HCO3-
400,000
Acetylcholinesterase
Acetylcholine
140,000
b-Lactamase
Benzylpenicillin
Fumarase
Fumarate
RecA protein (ATPase)
ATP
40,000,000
2,000
800
0.4
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
kcat / Km
R=
Glycine
–H
1.3 ╳ 10-1
Norvaline
–CH2–CH2–CH3
3.6 ╳ 102
Norleucine
–CH2–CH2–CH2–CH3
3.0 ╳ 103
Phenylalanine –CH2–
1.0 ╳ 105
(M-1 s-1)
Adapted from Mathews et al (2000) Biochemistry (3e) p.379
Lineweaver-Burk or double-reciprocal plot
Analysis of Kinetic Data
S >> Km
vi=Vmax
Vmax= k2Et
Al iniciar:
t = 0, S = So
A cualquier tiempo:
T=t S=S
X = (So-S)/So
- dS/dt = vi = So dX/dt
Temperature Dependence of Enzymes
•
•
•
•
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.
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 temperature
R is the gas constant (8.3145 J.K-1 mol -1)
Combining the
equations yields
three
preceding
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‡ decomposes
v is the vibrational frequency X‡ of the bond that breaks as X‡ decomposes to products
k, 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 is
kB 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
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.
Eact
E
S
P
Reaction coordinate
 E 
k obs  A·exp - a 
 RT 
Ea can be determined by measuring kobs at two different
temperatures:
 E  1 1 
k
Arrhenius equation:
 exp   a    
k2
 R  T1 T2  
1
 Ea 
Arrhenius: k obs  A·exp 
 RT 
Idem, from statistical mechanics (collision theory)
k obs
 Ea 
 P·Z·exp 
 RT 
P = probability factor (not every collision is effective)
Z = collision number (number of collisions per second)
Idem, from transition state theory:
A+B
K‡
X‡
k‡
X‡ (TS)
products
Eact
E
‡
[X
]
‡
K =
[A][B]
or
S
[X‡] = K‡[A][B]
P
Reaction coordinate
n = k‡[X‡] = k‡K‡[A][B] = k[A][B], so k = k‡K‡
Statistical mechanics gives us the following relation:
k BT
k =
h
‡
so
k BT ‡
k=
K
h
kB = Boltzmann’s constant;
h = Planck’s constant
k BT ‡
k=
K
h
For all equilibria we can write: DG0 = - RT ln K,
so for our case we get: DG‡ = - RT ln K‡
Expressing K‡ in terms of DG‡ and RT gives the
following equation for k:
 ΔG ‡ 
k BT

k
exp  
h
 RT 
(1)
Since DG‡ = DH‡ - TDS‡, we can also write:
 ΔH ‡   ΔS‡ 
k BT
exp 

k
exp  
h
 RT   R 
(2)
Eq (1) and (2) are called the Eyring equations
The Eyring and Arrhenius equations resemble each other:
Arrhenius:
 Ea 
k obs  A·exp 
 RT 
Ea
d ln k

so:
d(1/T)
R
ln A
slope =
Ea
R
ln k
1/T
Eyring:
d ln k
ΔH‡
 T 
d(1/T)
R
so Ea = DH‡ + RT
In order to determine DH‡ and DS‡ it is
easier to differentiate ln (k/T) to 1/T:
d ln(k/T)
ΔH‡

d(1/T)
R
slope =
ln
k
T
1/T
ΔH ‡
R
So, the procedure to determine activation parameters is:
- determine k at different temperatures
- plotting ln(k/T) against 1/T gives DH‡
 ΔH ‡   ΔS‡ 
k BT
exp 

exp  
- k
h
 RT   R 
then gives DS‡
and when you have DH‡ and DS‡, you also have DG‡ since
DG = DH-TDS
Interpretation of activation parameters
•
DG‡, the Gibbs free energy of activation, determines at
which rate a certain reaction will run at a given
temperature
• DH‡ is a measure for the amount of binding energy that is
lost in the transition state relative to the ground state
(including solvent effects)
• DS‡ is a measure for the difference in (dis)order between
the transition state and the ground state
– for monomolecular reactions: DS‡  0 J/mol.K
– for a bimolecular reaction: DS‡ << 0 J/mol.K
(two particles have to come together in the transition state to form
one particle, demanding a much greater order)
Example:
H H
O
H
Me2N
NH2
N
CH 2Ph
BNAH
(NADH model)
S
O
N+Me2
Me2N
NH2
+
N
H
BNA+
(NAD+ model)
methylene blue
(MB+)
x
NMe2
+
N
CH2Ph
N
S
MBH
x
x
ln(At-A)
x
x
x
x
x
x
x
x
t (sec)
DG‡
= 62.8 kJ/mol (very fast rx)
DH‡ = 33.0 kJ/mol (rel. low, compensation
of C-H bond cleavage by hydration TS)
DS‡ = -100 J/mol.K (bimolecular rx)
Another example:
H CN O
H
O
NH 2
N
CH 2Ph
NH 2
CH 3CN
N
CH 2Ph
+ CN
1
DH‡ = 85 kJ/mol (relatively high: no new bonds to be
formed, no compensation for the partial cleavage
of the C-C bond in the transition state; acetonitrile
is aprotic, compensation of DH‡ by solvation will
be less than in water
DS‡ = 0 J/mol.K (monomolecular reaction)
Application of activation parameters for the elucidation
of reaction mechanisms:
O
H3C
O
+ H2O
O P O
O
O
H3C
+ H2PO4
O
A DS‡ of +12 J/mol.K was found  monomolecular process
O
H3C
O
+ H2O
O P O Ph
O
O
+ HO P O Ph
OH
O
O
H3C
A DS‡ 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 DDG‡ ~ 30 kJ/mol
DG‡DMF
O
H
CH 3
N
CH 3
= DMF
DG‡H2O
E
DMF
H2O
DG‡DMF < DG‡H2O
reaction progress
What is the background of this strong solvent effect?
H
O O H
H O
H H O
H
H
H O
H
H Cl
H
H
H H
O H
O
O H
H
O
H O
H
H

H O
H Cl
H
O
H
O H
H3C I H
H
O
H
H
O
H
O
H
O H
H H
 H
C
I
H
H
H
H O
O
H
H
+ x H2O
In H2O there is more solvation than in DMF, due to hydrogen bonds.
Note the changes in entropy: loss of DS‡ because of orientation of the substrates, gain of
DS‡ 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:
O

O
O
‡
O C O
O
+
C
N
O2N
O
N
N
O2N

O
O2N
O
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
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:
H3C
N
R
H3C
N
+ H+
N
R = CH3:
pKa = 9.7
R = polymer: pKa = 7.7
R
R = CH3 or compound with polymer
N
H
[PyN][H+]
Ka = [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
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
Competitive Inhibition
A substance that competes directly with a normal substrate for an enzyme’s substratebinding 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
Competitive Inhibition
This
is
the
Michaelis-Menten
equation that has been modified by a
factor, , which is defined as
 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 Inhibition
Mixed 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
Bisubstrate Reactions
Sequential Reactions
Ordered bisubstrate reaction
A and B : substrates in order that they add to the
enzyme
P and Q : products in order that they leave the
enzyme
Random bisubstrate reaction
Ping Pong Reactions
Group-transfer reactions in which one or
more products are released before all
substrates have been added are known as
Ping Pong reactions