1. 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 1

2. The Michaelis-Menten Equation
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 2

3. ES = E * S /(k-1+k2)/k1 ES = E * S / KM The Michaelis-Menten Equation
Solving for [ES]:
The Michaelis constant, KM , is defined as
ES = E * S / KM
Therefore: 3

4. 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 4

5. 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 5

8. k2= 10/5 = 2 moles/mg seg Vmax= 10 M/seg Km=10 x10-5 M
Si 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?

9. - dS/dt = vi = So dX/dt Al iniciar: t = 0, S = So A cualquier tiempo:
T = t S = S
X = (So-S)/So
- dS/dt = vi = So dX/dt
Graficar por ejemplo :
(So – S)/t vs (1/t) ln (So/S) 9

10. Two-Stage Catalysis of Chymotrypsin
Acylation O
CH3–C–O– –NO2
Nitrophenol acetate O C
CH3–C
HO– –NO2
O - C
Kinetics of reaction
Deacylation (slow step)
CH3COOH
+ H2O
Time (sec)
Nitrophenol
O-H C
現在 已經知道 chymotrypsin 的催化機制分成 acylation 及 deacylation 兩個階段，其研究是利用 nitrophenol acetate 為基質所進行的；在早期的觀察中，發現催化生成 nitrophenol 的速率有兩個 phases，起先相當快放出 nitrophenol，然後變慢並一直維持此一速度。
利用兩階段式催化機制，可說明為何開始速率會比較快。 這是因為第一階段的 acylation 反應很快，而剛開始酵素都空閒著，一抓到基質就水解並放出 nitrophenol，因此黃色產物迅速上升。但是因為醋酸根還留在酵素上，準備進行第二步的 deacylation，而此一步驟相當慢速，即為 速率決定步驟 rate-limiting step；酵素要在加入水分子，釋出醋酸後，才能繼續吸入 nitrophenol acetate 繼續下一輪反應。因此出現了兩相的反應速率。
Two-phase
reaction
Adapted from Dressler & Potter (1991) Discovering Enzymes, p.169
E5-21

11. DESVIACIONESA M&M From the steady state assumption::
dE/dt = -k1 E*S + k-1 ES + k3 ES’
From the steady state assumption::
dE/dt = 0
-k1 E*S + k-1 ES + k3 ES’
dES/dt = k1 E*S – ES (k-1 + k2)
dES’/dt = k2 ES – k3 ES’
dES/dt = 0
ES = k1 E*S / (k-1 + k2)
v = dP1/dt = k2 ES
v = dP2/dt = k3 ES’
dES’/dt = 0
ES’ = k2 ES / k3
ET= E + ES + ES’
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) 11

12. 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
v / Vmax = k2 k1 S
k3 (k-1 + k2) + k1 S (k2 + k3)
Vmax = k3 ET
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

13. 13

14. K-2+ Keq = Vmax f * Kp Vmax f = k2 k3 ET k-2 + k3 +k2 Vmax r * Ks
Vmax r = k-1 k-2 ET
k-1 + k2 +k-2
v = Vmax f Kp S – Vmax r Ks P
KsKp + KpS + KsP
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)
v = Vmax f S – P / Keq
Ks + S + (Ks/Kp) P 14

15. 15

16. 16

17. 17

18. 18

19. 19