Lecture 8 - BCMB 3100 Lecture 8 Horton Chapter 5 Enzyme...

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BCMB 3100 – Lecture 8 Horton Chapter 5 Enzyme Kinetics Test Monday Feb. 7 Why study kinetics? To fully understand what is going on in a reaction we need to know If the reaction will occur (Thermodynamics) How long will the reaction take (Kinetics) How will the reaction occur (Mechanism) Kinetics is an important aspect of how enzymes work enzymes don’t create new chemistry they just make chemistry occur faster Living systems are not at equilibrium (parts may be) – they are often at steady state – a balance of reactions creating and destroying compounds
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A + B C + D K eq Substrates Products Free energy Δ G = -RT ln K eq [A][B] [C][D] K eq = [Substrates] [Products] = Tells you a reaction is favorable, but nothing about the rate Thermodynamics is about stability ( Δ G) Δ G E a A + B C + D Catalysts: enzymes increase reaction rates Urease urea Rate enhancement of 10 15 Test tube - 2,642 yrs/molecule Vs Cell - 37 μ sec/molecule Thermodynamics says nothing about rate Δ G E a A + B C + D Δ G ~ 25 kcal mol -1 or 104.5 kJ mol -1
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k = Ae E a The Arrhenius Equation shows that the rate of a reaction (k) is inversely proportional to activation energy (E a ). Δ G E a A+B C+D Kinetics is about Rates. A + B C + D k +1 k -1 K eq = k 1 k -1 = [A][B] [C][D] Gibb’s Free Energy can be calculated from the rates. Δ G = -RT ln K eq k = reaction rate Rates are about mechanisms Rate is Limited by the stability of the Transition State relative to the reactants. Enzymes increase forward and reverse reaction rates, thus equilibrium does not change. Enzyme Mechanism is a chemical model describing how an enzyme reduces E A .
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Zero-Order: Rate does not depend on [S] ν 0 = k 1 ν 0 = k 1 [S 1 ] 1 st -Order: Rate does depends on [S] Order of Chemical Reactions The forward rate constant (k 1 ) dependence on [S]. S P k 1 k -1 S 1 + S 2 k 1 k -1 P 1 + P 2 2 nd -Order: Rate does depends on both [S 1 ] AND [S 2 ] ν 0 = k 1 [S 1 ][S 2 ] K eq = k 1 k -1 [P] [S] = = 1.8 If you know the forward rate constant and the equilibrium constant, you can calculate the reverse rate constant First-Order Reaction 64% 36% k 1 k -1 Initial velocity = ν 0 = k 1 [S] Isomerization
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1) ν 0 (‘initial velocity’) is the initial slope of the progress curve 2) Get ν 0 at different [S] 3) Plot ν 0 against [S] 4) Slope is the forward rate constant (k 1 ) 1 Progress Curves Increasing k -1 influence --> S P k 1 k -1 First-Order Reaction ν 0 = k 1 [S] [S] First-Order Reaction 1 Progress Curves Increasing k -1 influence --> S k 1 k -1 First-Order Reaction ν 0 = k 1 [S] [S] First-Order Reaction Q. Why do the progress curves plateau?
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This note was uploaded on 02/07/2011 for the course BCMD 3100 taught by Professor Rose during the Spring '11 term at University of Georgia Athens.

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Lecture 8 - BCMB 3100 Lecture 8 Horton Chapter 5 Enzyme...

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