#3 Filled in Skeleton Notes

#3 Filled in Skeleton Notes - Hogan Chem 430 Lecture notes...

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Hogan, Chem 430 Lecture notes II. Enzyme catalyzed reactions in biological systems A. For a simple enzyme catalyzed reaction: DRAW: S = substrate and P = products B. Biological reactions are generally more complex 1. S 1 + S 2 P 2. S P 1 + P 2 all simple but complex in biological systems 3. S 1 + S 2 P 1 + P 2 Mathematically, this can get complicated in a hurry E + S must form ES complex and must interact for any product to form C. In the simple case of S being converted to P in an enzyme catalyzed reaction, the enzyme [E] must form a complex with substrate [S] to yield an enzyme- substrate complex [ES] in order to form product [P]. I. Michaelis and Menten (M & M): Their pioneering work on enzyme kinetics began around 1913 A. M & M began investigating the effects of [S] on the formation of the ES complex. They examined the effects by measuring the initial reaction velocity ( υ o ). they were observing: draw Question: What if we keep the [E] constant and vary the [S]? this effects ES complex note: product formation is related to ES formation Simple Concept Example: E + S ES P DRAW: -All enzymes have intrinsic rate of how much product they can form per unit time. -All E’s are unique and each ES has specific properties 1 Some transfer of functional group
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Hogan, Chem 430 Lecture notes II. Derivation of the M & M equation: Assumption 1 (assumption of equilibrium): Early in the reaction, little P has accumulated so k -2 can be ignored. -Because this is V-initial: The reaction does not have time to reach equilibrium where Some of the product that is formed has time to move back to ES complex This equilibrium could easily go back and forth easily w/o ever forming product. P/time = ( υ o ) = k 2 [ES] Eqtn. 1 Formation of ES: rate = k 1 [E] [S] bi-molecular Eqtn. 2 Breakdown of ES: rate = k -1 [ES] All rates could be different rate = k 2 [ES] Assumption 2 Steady State Assumption: Once reaction gets started, the [ES] remains constant. As a result, the formation of ES must equal the Breakdown of ES: DRAW: Eqtn. 3 Eqtn. 4 ES formation Now rewrite this expression in terms of ES: Eqtn. 5 Define K m -The Meckaelis Constant The Substrate Concentration at which reaction will proceed at ½ it’s Vmax DRAW: 2 *Once reaction gets going, the concentration of ES remains constant
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Hogan, Chem 430 Lecture notes *****Under saturating conditions of S (i.e. [S] = very high) then [E o ] = [ES]. Recall that Vmax is achieved when all E has S bound. IMPORTANT: know that Vmax is achieved when all E has an S bound Therefore, the Michaelis-Menten equation is: DRAW: Eqtn. 10 SO WHO CARES? Consider the special case when υ o = ½ Vmax Eqtn. 11 Eqtn. 12 Eqtn. 13 Eqtn. 14 Eqtn. 15 When υ o = ½ Vmax, K m = [S] that gives ½ Vmax So, when ½ Vmax = 1 2 1 K K K + - So, if K-1 goes up low affinity of E for S If K2 goes up low affinity of E for S 3 DRAW: So now we are able to find the V of an enzyme @ any [S] so long as we know Vmax and Km
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Hogan, Chem 430 Lecture notes If K1 goes up high affinity of E for S STUDY!
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