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SURVEY OF BIOCHEMISTRY Enzyme Kinetics and Inhibition. k. v = A P. Rates of Chemical Reactions. Enzyme kinetics is the study of rates of reactions catalyzed by enyzmes. The rxn rate (velocity, v) can be described in several ways: [1] disappearance of reactant, A
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k v= A P Rates of Chemical Reactions Enzyme kinetics is the study of rates of reactions catalyzedby enyzmes. • The rxn rate (velocity, v) can be described in several ways: • [1] disappearance of reactant, A • [2] appearance of product, P • These eqn’s relate velocity to concentration of reactants and products.
k v= A P Rate Laws Enzyme kinetics is the study of rates of reactions catalyzedby enyzmes. • A rate law is an equation describing the velocity of achemical reaction. • Differential Rate Laws • Integrated Rate Laws
Differential Rate Laws • Differential rate laws correspond to order of the reaction. Order of Reaction Rate Law 0 v = k 1 v = k [A] 2 v = k [A]2 or k[B]2 or v = k [A] x [B]
Rate of Disappearance of A - d[A] A products = k [A] first order rxn dt d[A] Rearranging… = - k dt [A] Integrated Rate Laws:First Order • Integrated rate laws express the rxn velocity in terms of time.
d[A] Rearranging… = - k dt [A] Integrated Rate Laws:First Order [A]t Integrate onboth sidesof eqn: dA = - k dt [A]0 (ln [A]t + constant) - (ln [A]0 + constant) = - kt ln [A]t - ln [A]0 = - kt
Integrated Rate Laws:First Order ln [A]t - ln [A]0 = - kt ln [A] = ln [A]0 - kt
[A]t [A]t ln [A]0 [A]0 = - kt [A]t = [A]0 e -kt Integrated Rate Law:Other Versions of First Order ln [A]t - ln [A]0 = - kt first order rxn Rearranging: Take exponentof both sides: = e -kt first order rxn
2A products Rate of Disappearance of A - d[A] = k [A]2second order rxn dt d[A] Rearranging… = - k dt [A]2 Integrated Rate Law: Second Order • How does the integrated rate law change if the order of the reaction is second order? Show result on board
k1 k2 E + S ES E + P k-1 Michaelis-Menten Equation Many enzymes obey Michaelis-Menten kinetics behavior: Rate limiting step Problem: [ES] is difficult to measure!What can we do?
k1 k2 E + S ES E + P k-1 Michaelis-Menten Equation Recall Assume equilibrium is maintained in 1st step Assume “steady state” k1 [E] [S] - k-1 [ES] - k2 [ES] = 0 Formation of ES Depletion of ES See notes on board…
Enzyme Inhibition • What is an inhibitor? • Modes of Inhibition • Competitivebinds to same site in E as S • Uncompetitive • Noncompetitive • Mixed bind to different site in E than S Note: Text does not distinguish “non” and “mixed”
Competitive Inhibition • Competitive inhibitors bind to the same site on E as S
Uncompetitive Inhibition Uncompetitive inhibitors bind directly to the ES complex but not to the free enzyme
Mixed Inhibition Mixed inhibitors can bind to E or ES complexS cannot bind if I is already bound!
Noncompetitive Inhibition Noncompetitive inhibitors can bind to E or ES complexS can bind even if I is already bound! + I See board for plot
Updates and Reminders • Exam #2 in two weeks (June 26) • Chapter 7: Protein Function • Chapter 11: Enzyme Catalysis • Chapter 12: Kinetics & Inhibition • Chapter 8: Carbohydrates • Chapter 14: Introduction to Metabolism • Suggested HW problems online this weekend • Resources: What You Should Know more coming soon