Presentation on theme: "The Michaelis-Menten Equation"— Presentation transcript:
1The Michaelis-Menten Equation ET = E + ESv= Vmax = k2 ETv/Vmax = k2 ES / k2 ET = ES/ ETv/Vmax = ES/ ETv/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 / KsKs is the dissociation constant of the first step in the enzymatic reaction1
2The Michaelis-Menten Equation Assumption of steady-state. Figure illustrates the progress curves of the various participants in reactionunder the physiologically common conditions that substrate is in great excess overEnzyme ([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. Haldane2
3ES = E * S /(k-1+k2)/k1 ES = E * S / KM The Michaelis-Menten Equation Solving for [ES]:The Michaelis constant, KM , is defined asES = E * S / KMTherefore:3
4The Michaelis-Menten Equation The expression of the initial velocity (v0) of the reaction, the velocityat t=0, thereby becomesv/Vmax = ES/ (E + ES)v/Vmax = (E*S)/Km/ (E + (E*S)/Km )v/Vmax = S/ Km / (1 + S/Km)v/Vmax = S / Km + SThis 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 when4
5Significance of the Michaelis Constant The Michaelis constant, KM, has a simple operational definition. At the substrate concentration at which [S] = KM, this equationyields v0 = Vmax/2 so thatKM is the substrate concentration at which the reaction velocity is half maximal5
8k2= 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 ETk2= 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 = SX = (So-S)/So- dS/dt = vi = So dX/dtGraficar por ejemplo :(So – S)/t vs (1/t) ln (So/S)9