ENZYME KINETICS. What happens to S, P, E, ES?

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ENZYME KINETICS Go to lecture notes and/or supplementary handouts for the following: 1 Basic observations in enzyme inetics 2 Michaelis-Menten treatment of enzyme inetics 3 Briggs-Haldane treatment of enzyme inetics Always remember the following: [E o ] <<<< [ ] [S t ] ~ [ ], t = early times rate = [ES] [E o ] = [E] + [ES] What happens to S, P, E, ES? http://deptphysicsupennedu/courses/gladney/mathphys/subsection4_1_6html 1

Comparison of MM and BH treatments Michaelis-Menten Treatment K eq + [ ] Briggs-Haldane Treatment How does either equation explain the basic observations of enzyme inetics? 2 1 % E o : linear dependence constant E o Vmax = 2 [Eo] % [E o ] 2 [E o ][ ] ---- 2

How does either equation explain the basic observations of enzyme inetics? 2 2 Vo vs So: Hyperbolic Dependence Vo = Vmax Case 1: Low [ ] Case 2: High [ ] How does either equation explain the basic observations of enzyme inetics? 2 Case 1: Low [ ] 2 Vo vs So: Hyperbolic Dependence Vo = Vmax % [ ] Choose [ ] << K m ----------- K m 3

How does either equation explain the basic observations of enzyme inetics? 2 Case 2: High [ ] 2 Vo vs So: Hyperbolic Dependence Vo = Vmax = constant = V max = 2 [E o ] Choose [ ] >> K m = V max What is the value of when = V max /2? Briggs-Haldane Treatment V max ------ = -------------- 2 4

What is the value of when = V max /2? 1 [ ] ------ = -------------- 2 2 [ ] = If Vo = V max 2 K m = This is a numerical relationship only at a specific value of = V max /2 K m = [ ] This K m = when = V max /2 is not a conceptual idea! Method to approximates a value for K m Recall when we chose [So] << Km? 2 2 Vo vs So: Hyperbolic Dependence Case 1: Low [ ] Vo = Vmax % [ ] K m = Point of Comparison Choose [ ] << K m ----------- K m 5

Recall when we chose [So] >> Km?? 2 Case 2: High [ ] 2 Vo vs So: Hyperbolic Dependence Vo = Vmax = constant = V max = 2 [E o ] K m = Point of Comparison Choose [ ] >> K m How can we determine if experimental data fits these equations? Michaelis-Menten Treatment K eq + [ ] Briggs-Haldane Treatment 6

Transform the equations to a linear form: Lineweaver-Bur MICHAELIS-MENTEN EQUATION: = LINEWEAVER-BURK EQUATION: 1 / = K m /V max (1 / ) + 1 / V max Y = y-intercept 1 / y = mx + b y = 1 / m = K m / V max Y = 1 / V max x = 1 / b = 1 / V max extrapolation of data line to x axis X = - 1 / K m 1 / X = x-intercept (X is a negative number) NOTE: If data points fit a straight line, the data is said to "fit" the theoretical equation V max = 1 / Y K m = 1 / X What is the interpretation of K m? Case #IV 2 2 [E o ][ ] = = 2 + -1 + [ ] Recall [ ] >> [E o ] at all [ ] CASE IV: -1 ~ 2; K m = ( + 2)/ -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 K m initial rate = 2[E o]/(k m + ) = V max/(k m + ) [ES] or [ES] max [ES] x 10 x TIME = Vmax 2 is called cat S 7

What is the interpretation of K m? Case #I 2 2 [E o ][ ] = = 2 + -1 + [ ] Recall [ ] >> [E o ] at all [ ] CASE I: 2 >> -1 and 2 << 1 so 2 is rate determining step [ES] builds up to maximum before any P forms rate = complex function of [E o ] and [ ] E + S ES -1 2 [ES] or [E o ] = [ES] max [ES] x 10 x P + E Under these conditions, the reaction appears to be independent of and only dependent upon E o TIME What is the interpretation of K m? Case #II 2 2 [E o ][ ] = = 2 + -1 + [ ] Recall [ ] >> [E o ] at all [ ] CASE II: 2 >> -1 and 2 >> 1 so 1 is rate determining step [ES] does not build up to any extent rate = [E o ][ ] 2 E + S ES P + E Under these conditions, the reaction appears to be first order in E o and and the vs plot would be linear (not hyperbolic) TIME 8

What is the interpretation of K m? Case #III 2 2 [E o ][ ] = = 2 + -1 + [ ] Recall [ ] >> [E o ] at all [ ] CASE III: -1 >> 2; Km = Kd = 1/ -1; 2 is rate determining step [ES] rapidly reaches equilibrium concentration before any product builds up [ES] depends upon the magnitude of Kd; rate = 2 [ES] = 2 [E o ][ ]/K d if K d >> So (Note K d << 1) [ES] or [ES] max [ES] x 10 x TIME = Vmax ṁore general S Under these conditions, K m is the dissociation constant for the ES complex: -1 K m = ----------- Interpretation of K m Conceptually, thin of K m as the dissociation constant! # x 10-3 M Higher K m values= lower affinity of E for S, weaer binding Lower K m values = higher affinity of E for S, tighter binding 9

The Catalytic Constant, cat or 2 Conceptually, thin of cat as the ability of E to convert S to P cat is called the turnover number Higher cat values = higher ability of E to convert S to P Lower cat values = lower ability of E to convert S to P The Meaning of the Specificity Constant cat /K m Conceptually, thin of cat /K m as the 2 nd order rate constant if E reacted with S without forming ES E + S E + P 10

Values for the Specificity Constant cat /K m Conceptually, thin of cat /K m as the 2 nd order rate constant if E reacted with S without forming ES ([ ]<<K m ) Units of cat /K m are the same as a second order rate constant: ( cat = s -1 )/(K m = M) = M -1 s -1 -dp/dt = [E o ][ ] Ms -1 = (M -1 s -1 )(M)(M) Upper limit for 2 nd order rxn = diffusion controlled = 10 8 to 10 9 M -1 s -1 11