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**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

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**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

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**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

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**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

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**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

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**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?

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**- 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

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**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

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**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

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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

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**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

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