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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 » k 2, so that the first step of the reaction reaches equilibrium. K s is the dissociation constant of the first step in the enzymatic reaction v= V max = k 2 E T E T = E + ES ES = E * S / Ks v/V max = k 2 ES / k 2 E T = ES/ E T v/V max = ES/ E T v/V max = ES/ E + ES ¿How to know ES?

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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

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The Michaelis constant, K M, is defined as The Michaelis-Menten Equation Solving for [ES]: ES = E * S /(k -1 +k 2 )/k 1 Therefore: ES = E * S / K M

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The Michaelis-Menten Equation The expression of the initial velocity (v 0 ) of the reaction, the velocity at t=0, thereby becomes The maximal velocity of a reaction, V max occurs at high substrate concentrations when the enzyme is saturated, that is, S>> Km, and E T is entirely in the ES form v= Vmax when This expression, the Michaelis-Menten equation, is the basic equation of enzyme kinetic. v/V max = ES/ (E + ES) v/V max = (E*S)/Km/ (E + (E*S)/Km ) v/V max = S/ Km / (1 + S/Km) v/V max = S / Km + S

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Significance of the Michaelis Constant The Michaelis constant, K M, has a simple operational definition. At the substrate concentration at which [S] = K M, this equation yields v 0 = V max /2 so that K M is the substrate concentration at which the reaction velocity is half maximal

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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 k 2= 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 Graficar por ejemplo : (So – S)/t vs (1/t) ln (So/S)

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O -CO -C Time (sec) Nitrophenol Two-Stage Catalysis of Chymotrypsin O CH 3 –C–O– –NO 2 Nitrophenol acetate OCOC O CH 3 –C HO– –NO 2 + H 2 O O-H C CH 3 COOH Kinetics of reaction Two-phase reaction Acylation Deacylation (slow step) Adapted from Dressler & Potter (1991) Discovering Enzymes, p.169

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DESVIACIONESA M&M dE/dt = -k 1 E*S + k -1 ES + k 3 ES dES/dt = k 1 E*S – ES (k -1 + k 2 ) dES/dt = k 2 ES – k 3 ES v = dP 1 /dt = k 2 ES v = dP 2 /dt = k 3 ES E T = E + ES + ES From the steady state assumption:: dE/dt = 0 -k 1 E*S + k -1 ES + k 3 ES dES/dt = 0 ES = k 1 E*S / (k -1 + k 2 ) dES/dt = 0 ES = k 2 ES / k 3 dP 2 /dt = v = k 3 ES v = k 3 k 2 ES / k 3 v = k 2 k 1 E*S / (k -1 + k 2 ) Vmax = k 3 E T Vmax = k 3 (E + ES + ES) Vmax = k 3 (E + k 1 E*S / (k -1 + k 2 ) + k 2 ES / k 3 Vmax = k 3 (E + k 1 E*S / (k -1 + k 2 ) + k 2 k 1 E*S / (k -1 + k 2 ) / k 3 ) Vmax = k 3 E + k 1 k 3 E*S / (k -1 + k 2 ) + k 2 k 1 E*S / (k -1 + k 2 )

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v / Vmax = k 2 k 1 E*S / (k -1 + k 2 ) k 3 E + k 1 k 3 E*S / (k -1 + k 2 ) + k 2 k 1 E*S / (k -1 + k 2 ) v / Vmax = k 2 k 1 S k 3 (k -1 + k 2 ) + k 1 k 3 S + k 2 k 1 S Vmax = k 3 E T v / Vmax = k 2 k 1 S k 3 (k -1 + k 2 ) + k 1 S (k 2 + k 3 ) v / Vmax = k 2 S / (k 2 + k 3 ) k 3 (k -1 + k 2 ) / (k 2 + k 3 ) + k 1 S v = k 2 k 3 E T S / (k 2 + k 3 ) k 3 (k -1 + k 2 ) / (k 2 + k 3 ) + k 1 S

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K-2+ Vmax f = k 2 k 3 E T k -2 + k 3 + k 2 Vmax r = k -1 k -2 E T k -1 + k 2 + k -2 Ks = k -1 k -2 + k -1 k 3 + k 2 k 3 k 1 ( k 2 + k -2 + k 3 ) Kp = k -1 k -2 + k -1 k 3 + k 2 k 3 k -3 ( k -1 + k 2 + 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

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