Enzyme Kinetics

Enzyme Kinetics - BME100L:Topic6 EnzymeKinetics...

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1 BME 100L: Topic 6 Enzyme Kinetics Enzyme kinetics y Michaelis-Menten kinetics y Kinetic data analysis y Lineweaver-Burk plot y Dixon plot y Eadie-Hofstee plot y Specificity y Inhibition
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2 Enzymes: catalysts of life y Thousands of biochemical reactions proceed at any Thousands of biochemical reactions proceed at any given instant within living cells. These reactions are catalyzed by enzymes. y Enzymes are typically proteins. Ribozymes are enyzmes made of RNA. y Enzymes are the agents of metabolic function. Enzymes play key functions in controlling rate of reaction, coupling reactions, and sensing the momentary metabolic needs of the cell. For example ADP + P ATP + H O ATPase (1 Enzymes work by lowering the activation energy of a reaction i ATP + H 2 O ADP + P i ATP + H 2 O (1) (2) • The difference between the reactions is the rate! 0' 31 / Gk J m o l Δ= ATP + H2O ADP + Pi -ATPase +ATPase Recall ln ln , exp( ) a a E kA o r RT E RT =− Recall: ln k E a
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3 S P enzyme Product When t is small, dP/dt is constant This is the initial rat time Increasing Enzyme This is the initial rate S P enzyme v Initial Rate of P production Michaelis-Mentin Kinetics ] S [ K ] S [ V dt ] P [ d v M max + = = V max V max 2 Note: Without the enzyme, the reaction can still happen but at a slower rate [S] 0 K M but at a slower rate. How to model this reaction?
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4 Michaelis - Menten model E+S k f ES k 2 P+E E + S k r ES ⎯ → P + E E = enzyme ES = enzyme-substrate complex S = substrate P = product Assumptions : Enzyme is not consumed in the reaction, i.e., the total concentration of enzyme [E] 0 is a constant. Full model y 4 components, need four ODES ] [ ] [ ] ][ [ ] [ : 2 ES k ES k S E k E d E f + + = ] [ ] [ ] ][ [ ] [ : ] [ ] ][ [ ] [ : 2 P d ES k ES k S E k dt ES d ES ES k S E k dt S d S dt r f r f r = + = ] [ ] [ : 2 ES k dt P = The model can be numerically solved given initial conditions, e.g. [E] t=0 = [E] 0 [S] t=0 = [S] 0
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5 Case I: E 0 <<S 0 k f = 2, k r = 1, k 2 =0.5, E 0 = 0.2, S 0 =10 Substrate concentration is much higher than enzyme much higher than enzyme concentration Rate of changes in ES is much smaller than in S and P. We can apply QSSA. 1 st reaction not necessarily at equilibrium, but the forward reaction should be much faster than the 2 nd one. Assuming [ES] at steady state: Case I: Quasi steady state assumption 2 2 ] ][ [ ] [ 0 ] [ ] [ ] ][ [ ] [ k k S E k ES ES k ES k S E k dt ES d r f r f + = = = Since the total concentration of enzyme [E] 0 is [E] 0 = [E] + [ES]
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6 Substituting [E] = [E] 0 -[ES] V Case I: Quasi steady state assumption ] [ 1 ] [ ] [ ] ][ [ ] [ 2 0 2 S k k k E ES k k S E k ES f r r f + + = + = M E k V S K v 0 2 max max 0 ] [ ] [ 1 = + = Michaelis-Menten Equation Maximum rate ] [ 1 ] [ ] [ 2 0 2 2 0 S k k k E k ES k v f r + + = = f r M k k k K 2 + = Michaelis-Menten Constant When [S] = K M , v 0 = V max /2 Case II: k f , k r >> k 2 k f = k r = 50, k 2 =0.5, E 0 = 3, S 0 = 10 Apply Quasi equilibrium Linear scale log scale t t Note the lack of change in E, S, ES in shaded region
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7 Assuming k f , k r >> k 2 , => the first reaction is at the equilibrium state.
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This note was uploaded on 04/21/2011 for the course BME 100 taught by Professor Yuan during the Spring '07 term at Duke.

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Enzyme Kinetics - BME100L:Topic6 EnzymeKinetics...

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