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Enzymes

Enzymes. Kinetics and Inhibition. Enzymes. How do we characterize an enzyme? Biological catalyst Highly specific Mostly proteins. Enzyme Classification. Common name Not always informative about reaction catalyzed Ex. Chymotrypsin Systematic name – Enzyme Commission 6 major classes

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Enzymes

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  1. Enzymes Kinetics and Inhibition

  2. Enzymes • How do we characterize an enzyme? • Biological catalyst • Highly specific • Mostly proteins

  3. Enzyme Classification • Common name • Not always informative about reaction catalyzed • Ex. Chymotrypsin • Systematic name – Enzyme Commission • 6 major classes • Ex. Transferases • Classification number • EC 2.7.3.2 2=transferase, 7=phosphate, 3=nitrogen group as acceptor, 2=unique

  4. Enzyme Classification

  5. Enzymes • What are enzyme cofactors? • Non-protein component • Organic group or metal • When is a cofactor a prosthetic group? • Holoenzyme versus apoenzyme

  6. Enzymes • How do metal ion cofactors function? • Bridging group, coordination complex • Stabilize conformation in active form • Vitamins are parts of coenzymes • Thiamine pyrophosphate • Nicotinamide adenine dinucleotide • Pyridoxal phosphate

  7. Enzymes and Free Energy • What does G tell us about the nature of a reaction? • What does it mean if G is = 0? • What is true if G is positive?

  8. Calculating Free Energy • A + B C + D • G = G0 + RT ln[C][D] / [A][B] • G0 = standard free energy change; A, B, C, D present at 1M • G0/ = at pH 7

  9. Calculating Free Energy • At equilibrium 0 = G0/+ RT ln[C][D]/ [A][B] • Since Keq = [C][D]/ [A][B] • We can get • G0/ = -2.303RT log10Keq

  10. Calculating Free Energy

  11. Enzymes • Do not change equilibrium of reactions, speed up both forward and reverse reactions • Equilibrium depends upon free energy difference between products and reactants

  12. Enzymes • How do enzymes speed up reactions? • Enzymes decrease activation energy

  13. Enzyme Kinetics • What evidence supports the existence of an enzyme-substrate complex?

  14. Enzyme Kinetics • Images from x-ray crystallography

  15. Enzyme Kinetics • Change in spectroscopic characteristics

  16. Enzyme Kinetics • What do we know about the active site of an enzyme? • where substrate binds • three-dimensional cleft • small part of enzyme • substrate attached by weak bonds

  17. Enzyme Kinetics • What does the lock and key model represent?

  18. Enzyme Kinetics • How is it different from the induced fit model?

  19. k k 1 2 k -1 Michaelis –Menten Equation • Enzyme E combines with substrate S to form enzyme substrate complex ES and ES breaks down to form E and product P P + E E + S ES

  20. Michaelis-Menten Equation • Initial velocity equals rate of breakdown of ES • V0 = k2 [ES] • Must define ES in other terms • Formation of ES = k1[E][S] • Breakdown of ES = (k-1 + k2 )[ES] • At steady state: k1[E][S] = (k-1 + k2 )[ES]

  21. Michaelis-Menten Equation • Rearrange equation • [E][S]/[ES] = (k-1 + k2 )/k1 • Define new term Km • Michaelis-Menten constant • Km = (k-1 + k2 )/k1 we now substitute in above and solve for [ES] • [ES] =[E][S]/Km • [E] = [E]T – [ES]

  22. Michaelis-Menten Equation • Substitute for[E] • [ES] =([E]T – [ES])[S]/Km • Solve for [ES] • [ES] = [E]T [S]/[S] + Km • Going back to previous equation v0=k2[ES] • V0 = k2 [E]T [S]/[S] + Km • Maximal rate Vmax is when [ES] =[E]T • V0 = Vmax [S]/[S] + Km

  23. Michaelis-Menten Equation • V0 = Vmax [S]/[S] + Km • When initial velocity = ½ maximal velocity • Km = [S]

  24. Michaelis-Menten Equation • For a particular enzyme, can determine Km by varying substrate concentration keeping enzyme concentration constant and measuring initial velocity of reaction

  25. Lineweaver –Burk Equation • Use double reciprocal plot

  26. Enzyme Kinetics • What do we know about Km? • Not a fixed value • Varies with structure of substrate • Varies with pH, temperature, ionic strength • Each substrate has characteristic Km • Km usually between 10-1 and 10-6 M • Derived from rate constants

  27. Enzyme Kinetics • If k-1 is considerable larger than k2 so that Km = k-1/k1 this represents the dissociation constant of ES complex Therefore, Km tells us about strength of ES complex high Km = weak bonding; low Km = strong bonding

  28. Enzyme Kinetics • What does Vmax represent? • Turnover number • What is the kinetic constant or kcat? • Kcat = Vmax/[E]0 [E]0 = initial conc. Enzyme • # of substrate molecules converted into product in unit time when enzyme is saturated • When [S]  Km, kcat/Km can be used as measure of catalytic efficiency

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