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

Vmax= 10 M/seg Km=10 x10-5 MSi en el ensayo se usaron 5mg/L de preparación enzimática, entonces:v= Vmax = k2 ET k2= 10/5 = 2 moles/mg seg

¿Qué predicciones podemos hacer a partir de esta información?

- dS/dt = vi = So dX/dt

Al iniciar: t = 0, S = So

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

Graficar por ejemplo :

(So – S)/t vs (1/t) ln (So/S)

O -

C

Time (sec)N

itrop

hen

ol

Two-Stage Catalysis of Chymotrypsin

O

CH3–C–O– –NO2

Nitrophenol acetate

OC

O

CH3–C HO– –NO2

+ H2O

O-HC

CH3COOH

Kinetics of reaction

Two-phasereaction

Acylation

Deacylation (slow step)

Adapted from Dressler & Potter (1991) Discovering Enzymes, p.169

DESVIACIONESA M&M

dE/dt = -k1 E*S + k-1 ES + k3 ES’

dES/dt = k1 E*S – ES (k-1 + k2)

dES’/dt = k2 ES – k3 ES’

v = dP1/dt = k2 ES

v = dP2/dt = k3 ES’

ET= E + ES + ES’

From the steady state assumption::

dE/dt = 0 -k1 E*S + k-1 ES + k3 ES’

dES/dt = 0 ES = k1 E*S / (k-1 + k2)

dES’/dt = 0 ES’ = k2 ES / k3

dP2/dt = v = k3 ES’ v = k3 k2 ES / k3

v = k2 k1 E*S / (k-1 + k2)

Vmax = k3 ET Vmax = k3 (E + ES + ES’) Vmax = k3 (E + k1 E*S / (k-1 + k2) + k2 ES / k3 Vmax = k3 (E + k1 E*S / (k-1 + k2) + k2 k1 E*S / (k-1 + k2) / k3 ) Vmax = k3 E + k1 k3 E*S / (k-1 + k2) + k2 k1 E*S / (k-1 + k2)

v / Vmax = k2 k1 E*S / (k-1 + k2) k3 E + k1 k3 E*S / (k-1 + k2) + k2 k1 E*S / (k-1 + k2)

v / Vmax = k2 k1 S k3 (k-1 + k2) + k1 k3 S + k2 k1 S

Vmax = k3 ET

v / Vmax = k2 k1 S k3 (k-1 + k2) + k1 S (k2 + k3)

v / Vmax = k2 S / (k2 + k3) k3 (k-1 + k2) / (k2 + k3) + k1 S

v = k2 k3 ET S / (k2 + k3)

k3 (k-1 + k2) / (k2 + k3) + k1 S

K-2+

Vmax f = k2 k3 ET k-2 + k3 +k2

Vmax r = k-1 k-2 ET k-1 + k2 +k-2

Ks = k-1 k-2 + k-1 k3 + k2 k3 k1( k2 + k-2 + k3)

Kp = k-1 k-2 + k-1 k3 + k2 k3 k-3( k-1 + k2 + k-2)

Keq = Vmax f * Kp Vmax r * Ks

v = Vmax f Kp S – Vmax r Ks P

KsKp + KpS + KsP

v = Vmax f S – P / Keq

Ks + S + (Ks/Kp) P