Enzyme Kinetics

ENZYME KINETICS Go to lecture notes and/or supplementary handouts for the following: 1. Basic observations in enzyme kinetics 2. Michaelis-Menten treatment of enzyme kinetics 3. Briggs-Haldane treatment of enzyme kinetics Always remember the following: [Eo ] <<<< [So ] [St] ~ [So ], t = early times rate = k[ES] [Eo ] = [E] + [ES] What happens to S, P, E, ES? 1 Comparison of MM and BH treatments Michaelis-Menten Treatment Vmax[So] Vo = ----------Keq + [S o] Briggs-Haldane Treatment Vmax[So] Vo = ----------Km + [So] How does either equation explain the basic observations of enzyme kinetics? E + S 1. Vo %% k1 k -1 k2 ES P + E Keq = k -1 [E][S] -------- = -----------k1 [ES] Eo : linear dependence . . constant So Vo . . Vo % [Eo] Eo Vmax[So] Vo = ----------Km + [So] Vmax = k2 [Eo ] k2 [Eo][S o] Vo = --------------Km + [So] 2 How does either equation explain the basic observations of enzyme kinetics? E + S k1 k -1 k2 ES Keq = P + E k -1 [E][S] -------- = -----------k1 [ES] 2. Vo vs So: Hyperbolic Dependence constant Eo . . . . . . . Vo = V m ax Vo So Case 1: Low [So ] Case 2: High [So ] How does either equation explain the basic observations of enzyme kinetics? E + S k1 k -1 k2 ES Keq = P + E k -1 [E][S] -------- = -----------k1 [ES] 2. Vo vs So: Hyperbolic Dependence constant Eo Case 1: Low [So ] . . . . . .. Vo = V m ax Vo Vo % [So] So Vmax[So] Vo = ----------Km + [So] Choose [So ] << Km Vmax[So] Vo . ----------Km 3 How does either equation explain the basic observations of enzyme kinetics? E + S k1 k2 ES k -1 Keq = P + E k -1 [E][S] -------- = -----------k1 [ES] 2. Vo vs So: Hyperbolic Dependence constant Eo Case 2: High [So ] . . . . . . . Vo = V m ax Vo Vo = constant = V max = k2 [Eo ] So Vmax[So] Vo = ----------Km + [So] Choose [So ] >> Km Vo = Vmax What is the value of So when Vo = Vmax/2? Briggs-Haldane Treatment Vmax[So] Vo = ----------Km + [So] Vmax -----2 Vmax[So] = -------------Km + [So] 4 What is the value of So when Vo = Vmax/2? 1 -----2 [So] = -------------Km + [So] constant Eo Vo If Vo = Vmax 2 2 [So] = Km + [So] . . . . . .. So Km = So This is a numerical relationship only at a specific value of Vo = Vmax /2. This “Km = So” when Vo = Vmax/2 is not a conceptual idea! Km = [So] Method to approximates a value for Km Recall when we chose [So] << Km? E + S k1 k -1 k2 ES Keq = P + E k -1 [E][S] -------- = -----------k1 [ES] 2. Vo vs So: Hyperbolic Dependence constant Eo Case 1: Low [So ] . . . . . .. Vo = V m ax Vo Km = So Vo % [So] So Point of Comparison Vmax[So] Vo = ----------Km + [So] Choose [So ] << Km Vmax[So] Vo . ----------Km 5 Recall when we chose [So] >> Km?? E + S k1 k2 ES k -1 Keq = P + E k -1 [E][S] -------- = -----------k1 [ES] 2. Vo vs So: Hyperbolic Dependence constant Eo Case 2: High [So ] . . . . . . . Vo = V m ax Vo Vo = constant = V max = k2 [Eo ] Km = So So Point of Comparison Vmax[So] Vo = ----------Km + [So] Choose [So ] >> Km Vmax[So] Vo = ----------Km + [So] How can we determine if experimental data fits these equations? Michaelis-Menten Treatment Vmax[So] Vo = ----------Keq + [S o] Briggs-Haldane Treatment Vmax[So] Vo = ----------Km + [So] 6 Transform the equations to a linear form: Lineweaver-Burk MICHAELIS-MENTEN EQUATION: V max [So] Vo = K m + [So] LINEWEAVER-BURK EQUATION: 1 / Vo = Km/Vmax (1 / So) + 1 / Vmax y = mx + b y = 1 / Vo 1 / Vo Y = y-intercept Y = 1 / Vmax x = 1 / So m = Km / Vmax b = 1 / Vmax extrapolation of data line to x axis NOTE: If data points fit a straight line, the data is said to "fit" the theoretical equation 1 / So Vmax = 1 / Y X = - 1 / Km Km = 1 / X X = x-intercept (X is a negative number) What is the interpretation of K m? Case #IV E + S k1 k -1 k2 ES k 2[Eo ][So ] P + E Keq = Recall V max [So ] Vo = k -1 [E][S] -------- = -----------k1 [ES] = k2 + k -1 + [So ] k1 [So ] >> [Eo ] at all [So ] K m + [So] CASE IV: k -1 ~ k2 ; Km = (k1 + k2)/k -1 CLASSICAL BRIGGS-HALDANE, OR MICHAELIS-MENTEN KINETICS [ES] does not rapidly reach equilibrium [ES] builds up to a small level depending upon the magnitude of Km initial rate = k2[E o]/(Km + So ) = Vmax/(Km + So ) [ES] or [P] [P] [ES] x 10x TIME . . . . . .. = Vmax Vo [ES]max k2 is called “kcat” S 7 What is the interpretation of K m? Case #I E + S k1 k -1 ES k2 Recall K m + [So ] [So ] >> [Eo ] at all [So ] = k2 + k -1 + [So] k1 k -1 [E][S] -------- = -----------k1 [ES] V max [So] k 2[Eo][So] Vo = Keq = P + E CASE I: k2 >> k -1 and k2 << k1 so k2 is rate determining step [ES] builds up to maximum before any P forms rate = complex function of [Eo] and [So] k1 k2 E + S k ES P + E -1 [ES] or [P] [Eo] = [ES] max [P] [ES] Under these conditions, the reaction appears to be independent of So and only dependent upon Eo x 10x TIME What is the interpretation of K m? Case #II E + S k1 k -1 ES k2 k 2[Eo][So] Vo = Keq = P + E k -1 [E][S] -------- = -----------k1 [ES] V max [So] Recall K m + [So ] [So ] >> [Eo ] at all [So ] = k2 + k -1 + [So] k1 CASE II: k2 >> k -1 and k2 >> k1 so k1 is rate determining step [ES] does not build up to any extent rate = k1[Eo][So] E + S k1 ES [P] k2 P + E Under these conditions, the reaction appears to be first order in Eo and So and the Vo vs So plot would be linear (not hyperbolic) TIME 8 What is the interpretation of K m? Case #III E + S k1 k -1 ES k2 P + E k 2[Eo][So] Vo = Keq = k -1 [E][S] -------- = -----------k1 [ES] V max [So] Recall K m + [So ] [So ] >> [Eo ] at all [So ] = k2 + k -1 + [So] k1 CASE III: k -1 >> k2; Km = Kd = k1/k -1; k2 is rate determining step [ES] rapidly reaches equilibrium concentration before any product builds up [ES] depends upon the magnitude of Kd; rate = k2[ES] = k2[Eo][So]/Kd if Kd >> So (Note Kd << 1) Under these [ES] or [P] [P] [ES] x 10x .. .. . . . = Vm ax Vo [ES]max TIME conditions, Km is the dissociation constant for the ES complex: k-1 Km= ----------more general k1 S Interpretation of Km Conceptually, think of Km as the dissociation constant! # x 10-3 M Higher Km values = lower affinity of E for S, weaker binding Lower Km values = higher affinity of E for S, tighter binding 9 The Catalytic Constant, kcat or k 2 Conceptually, think of kcat as the ability of E to convert S to P kcat is called the turnover number Higher kcat values = higher ability of E to convert S to P Lower kcat values = lower ability of E to convert S to P The Meaning of the “Specificity Constant” kcat/Km Conceptually, think of kcat/Km as the 2nd order rate constant if E reacted with S without forming ES k E+SàE + P 10 Values for the “Specificity Constant” kcat/Km Conceptually, think of kcat/Km as the 2nd order rate constant if E reacted with S without forming ES ([So ]<