Lecture 15 (10/20/17) Lecture 15 (10/20/17)

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1 Reading: Ch6; Ch6; Box 6- Lecture 5 (0/20/7) Problems: Ch6 (text); 8, 9, 0,, 2, 3, 4, 5, 6 Ch6 (study guide-facts); 6, 7, 8, 9, 20, 2 8, 0, 2 Ch6 (study guide-applying); NEXT Reading: Ch6; Ch6; 92-93, 95-96, Problems: Ch6 (text); 8, 9, 20, 2, 22 Ch6 (study guide-facts); 9, Ch6 (study guide-applying); 2 ENZYMES: & Catalysis A. Kinetics-review B.. Rate vs. [S] for enzyme catalyzed reaction a. initial rate (v 0 ) 2. ES complex a. Reaction i. reaction ii. Catalytic reaction Lecture 5 (0/20/7) b. Meaning of rate curve: hyperbolic curve 3. Rate expression; Michaelis-Menten Kinetics (M-M) a. Assumptions b. M-M equation derivation 4. Meaning of rate expression (M-M equation) a. [S] = Km b. [S] >> Km c. [S] << Km 5. Collection and manipulation of data

2 Substrate (S) Apply Chemical Kinetics to Enzymes Enzyme Substrate Complex (ES) Enzyme (E) Chemistry Rate (!) Product [E] (P) The rate that an individual enzyme can go around this cycle is called the turnover number, or k cat And, the rate depends on the [E] (recall) Enzyme Product Complex (EP) E + S ES EP E + P -But, Recall, this this is has ONLY units of if a there is AMPLE substrate. -What st order if rate [S] constant is low? -The (sec - ) velocity will now have a dependence on [S], or v [S] -If the velocity has a dependence on [S], or v [S], Then what is the proportionality constant in the rate equation? v = k [S] The curve for v versus [S] is not linear, like it is for [E]: OMG!! The relationship between reaction velocity and [S] is the same as between fraction bound (Y) and [L]... Hyperbolic Curve So, what is the equation that describes this behavior, i.e., the rate equation? But first, what is this initial velocity? 2

3 How to Do Kinetic Measurements Experiment:. Mix enzyme + substrate. 2. Record rate of substrate disappearance and/or product formation as a function of time (the velocity of reaction). 3. Plot initial velocity (v 0 ) at that substrate concentration [S]. 4. Change substrate or enzyme concentration and repeat to get the best slopes. rate The rate is getting slower and slower [S] Why is this curving off? As S goes to P, the [S] decreases. OK, back to the question: what is the equation that describes this behavior? How to Do Kinetic Measurements Experiment:. Mix enzyme + substrate. 2. Record rate of substrate disappearance and/or product formation as a function of time (the velocity of reaction). 3. Plot initial velocity (v 0 ) at that substrate concentration [S]. 4. Change substrate or enzyme concentration and repeat to get the best slopes. rate The rate is getting slower and slower [S] Why is this curving off? As S goes to P, the [S] decreases. OK, back to the question: what is the equation that describes this behavior? 3

4 So, what is the equation that describes this behavior, i.e., the rate equation? Lets focus on binding: Catalysis K eq = This is just like any other binding reaction, which we already discussed K d = What about catalysis? [ES] [E][S] OR [E][S] [ES] = k = - k Remember: K d = k What about the rate expression at ANY [S]? What is the rate equation for enzyme kinetics? Lets first look at the hyperbolic curve in more detail: 2 3 At low [S] nearly linear function of rate vs. [S] At this asymptote, what is the rate expression? Called V max At high [S] no dependence of rate on [S] 3 V max Catalysis Asymptote V max = [ES], but since [ES] [E] T 2 V max = [E] T Rate of the reaction does not change because all the E is in ES complex, so there is no free [E] in the reaction; [E] Total = [ES]. This is sometimes called zero order b/c there is no dependence on S. v 0 [S] [E] v 0 = k [S][E] This is the pseudo second order reaction with k having units of M - s Remember: - V max = [E] T 4

5 What about the rate expression at ANY [S]? What is the rate equation for enzyme kinetics? Lets first look at the hyperbolic curve in more detail: 2 3 At low [S] nearly linear function of rate vs. [S] At this asymptote, what is the rate expression? Called V max At high [S] no dependence of rate on [S] 3 V max Catalysis Asymptote V max = [ES], but since [ES] [E] T 2 V max = [E] T Rate of the reaction does not change because all the E is in ES complex, so there is no free [E] in the reaction; [E] Total = [ES]. This is sometimes called zero order b/c there is no dependence on S. v 0 [S] [E] v 0 = k [S][E] This is the pseudo second order reaction with k having units of M - s Remember: - V max = [E] T This question was answered in 93 by Michaelis & Menten They used these principles and derived the equation that describes the relationship between v 0 and [S] for enzymecatalyzed reactions that show hyperbolic behavior... The famous Michaelis-Menten Equation 5

6 Derivation of Michaelis-MentenEquation:. Start with a model mechanism. 2. Identify constraints and assumptions. 3. Carry out algebra. Simplest Model Mechanism: one reactant, one product, no inhibitors Derivation of Michaelis-MentenEquation: 2. Identify constraints and assumptions. There are 4 important ASSUMPTIONS for this derivation.. Use only INITIAL rate (v 0 ). This means [P] = 0, so there is no back reaction and you can ignore. The mechanism simplifies further: E + S ES E + P 2. The slow step is AFTER binding. In other words, binding is a rapid equilibrium. This is the so-called RAPID EQUILIBRIUM ASSUMPTION. v 0 = [ES] 3. The substrate is in vast excess of the enzyme (a true catalyst). This means you can ignore the amount of substrate in the ES complex, and [S] free = [S] Total. We can get everything in easily measured quantities: [E] T, [S] T, and v 0, but not [ES]. 4. The rate of formation of ES and the rate of breakdown are are equal. This is the so-called STEADY-STATE ASSUMPTION. Therefore, the [ES] does not change. 6

7 Derivation of Michaelis-MentenEquation: E + S ES E + P 3. Carry out the algebra. Starting with the Steady state assumption d[ ES] = rateof formationof ES- rateof breakdownof ES = 0 dt [E][S] = [ES] + [ES] Collect [ES] terms: [E][S] = ( + )[ES] Substitute the expression [E] = [E] T [ES], because free [E] is difficult to determine: ( ) ([E] T [ES])[S] = + [ES] Collect [ES] terms: [E] T [S] [ES][S] = ( + )[ES] ( ) [E] T [S] = + + [S] [ES] Derivation of Michaelis-MentenEquation: E + S ES E + P 3. Carry out the algebra. Solve for [ES]: Divide by / = ): + [S] [E] T [S] = ( + )[ES] [E] T [S] ( + + [S]) ( [E] T [S] + ( + [S] = [ES] = [ES] Define K m as this collection or rate constants: Substitute in K m : [E] T [S] = [ES] K m = ( Rate constant governing ES FORMATION + ( Rate constants governing ES BREAKDOWN Recall Assumption #2: is slow (small number) Km = K k d 7

8 Derivation of Michaelis-Menten Equation: 3. Carry out the algebra. [E] T [S] Use RAPID EQILIBRIUM ASSUMPTION (#2): [E] T [S] Solve for v 0 and make the rate equation: Recall definition of V max : V max = [E] T [E] T [S] E + S ES E + P = [ES] = v 0 / = v 0 v 0 = [ES] Michaelis-Menten Equation v 0 = V max [S] [ L] Y = KD + [ L] Recall: hyperbola à y = x/(b+x)! 0 V max [S] = What is the rate equation for enzyme kinetics? Michaelis-Menten Equation v 0 = V max [S] ½ V max Catalysis Special cases: What is v 0 when [S] = K m? When K M = [S], v 0 = ½ V max. Thus, K M is the substrate concentration that yields ½ V max. K m 8

9 What is the rate equation for enzyme kinetics? Michaelis-Menten Equation v 0 = V max [S] V max Catalysis 2 Special cases: What is v 0 when [S] >> K m? When K M << [S], its value is negligible and v 0 = V max. Thus, at high [S], the equation tells us the obvious; we are at V max. [S] What is the rate equation for enzyme kinetics? Michaelis-Menten Equation v 0 = V max [S] 3 Special cases: What is v 0 when [S] << K m? When K M >> [S], v 0 = (V max /K m ) [S]. Thus, the rate is strictly dependent on the [S]. But, remember that it also depends on [E]. [S] Catalysis k cat /K m is known as the Specificity Constant, or a measure of enzyme EFFICIENCY Recall that V max = [E] T = k cat [E] T. So, v 0 = (k cat /K m ) [E] T [S] K m Also recall: This is the same as: v 0 = k [S][E] This pseudo second-order rate constant we now know is k cat /K m with units of M - s - 9

10 Enzyme Efficiency is Limited by Specificity: k cat /K M Diffusion from the active site limits the maximum value for specificity. Can gain efficiency by having high velocity or affinity for substrate catalase vs. acetylcholinesterase TABLE 6-8 Enzymes for Which k cat /K m is Close to the Diffusion-Controlled Limit (0 8 to 0 9 M s ) Enzyme Substrate kcat (s ) Km (M) kcat/km (M s ) Acetylcholinesterase Acetylcholine.4 x x x0 8 Carbonic anhydrase CO2 x x x 0 7 HCO3 4 x x x 0 7 Catalase H2O2 x x x 0 8 Crotonase Crotonyl-CoA 5.7 x0 3 2 x x 0 8 Fumarase Fumarate Malate 8 x x x x x x 0 7 β-lactamase Benzylpenicillin 2.0 x x 0 5 x 0 8 Source: A. Fersht, Structure and Mechanism in Protein Science, p. 66, W. H. Freeman and Company, 999. SUMMARY: The final form in case of a single substrate is the Michaelis-Menten equation: v = v 0 = kcat[e tot][s] = K + [S] m V K [S] + [S] k cat (turnover number): how many substrate molecules one enzyme molecule can convert per second K m (Michaelis constant): an approximate measure of a substrate s affinity for an enzyme During steady state, the maximum velocity (V max ) occurs when all of the enzyme is in the ES complex and is dependent on the breakdown of that complex (k[es]). The microscopic meaning of K m and k cat depends on the details of the mechanism. m max 0

11 Two enzymes play a key role in the metabolism of alcohol. Some people respond to alcohol consumption with facial flushing and rapid heart beat, symptoms caused by excessive amounts of acetaldehyde in the blood. There are two different acetaldehyde dehydrogenases in most people, one with a low K M and one with a high K M. The low K M enzyme is genetically inactivated in some individuals. The enzyme with the high K M cannot process all of the acetaldehyde, and so some acetaldehyde appears in the blood. So, if knowing the values of the constants, K m and V max, for enzymes and their substrates is important, how are they determined? Determination of Kinetic Parameters A nonlinear Michaelis-Menten plot should be used to calculate parameters K m and V max. Lineweaver-Burke derived a linear form of the M-M equation by taking the reciprocal of both sides. This is called the linearized double-reciprocal plot. Its good for analysis of enzyme kinetic data to get these kinetic parameters.

12 Lineweaver-Burk Plot: Linearized, Double-Reciprocal The Michaelis-Menten equation can be manipulated into one that yields a straight-line plot. v 0 = V max [S] K v 0 = m + [S] Vmax [S] y = v 0 y = bx + a x = [S] K m v 0 = Vmax [S] + [S] V max [S] This double-reciprocal equation is called the Lineweaver-Burke equation. K v 0 = m + Vmax [S] V max 2