EnzymeKinetics_1

EnzymeKinetics_1 - Simple kinetic analysis Enzyme kinetics...

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Simple kinetic analysis Enzyme kinetics Simple kinetics analysis (continued) • Quasi steady state assumption (QSSA) • Quasi equilibrium assumption • Activation energy • Transition state theory Enzyme kinetics
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In chemical reactions, Steady state of a substance A [] 0 dA f dt = = Equilibrium state of a reversible chemical reaction Forward rate = Reverse rate Note: •Steady state is more “species-centric”. We often say some species is at a steady state, but not that some reaction is at steady state. •Equilibrium is more “process-centric”. We often say a reaction is at equilibrium but not that a species is at equilibrium. •For a system consisting of multiple species with multiple reactions, the system is at steady state if and only if each species is at steady state. •The system is at equilibrium if and only if each reaction (must be reversible) is at equilibrium. •A system at equilibrium must be at steady state, but a system at steady state is not necessarily at equilibrium.
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Sequential reactions AB C k 1 k 2 1 = - dA kA dt 12 = - dB kA kB dt 2 = dC kB dt At t = 0, A = 1, B = C = 0 At steady state 1 = - 0 dA dt = = - 0 dB dt = 2 = 0 dC dt = A ss = B ss = 0 C ss = 1 Starting from our specified initial condition, the system can only asymptotically approach the steady state
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Limiting cases 1. k 2 >> k 1, The first reaction A B is the rate-limiting process. There’s only one steady state for the system, when A is completely converted into C. For intermediate data points, changes in B is much slower than changes in A or C. To simplify the system, we can make the assumption that B is at a steady state. However, it’s not a genuine steady state – we call it “ quasi-steady state ”. That is, within a small time window, B appears to be at a steady state, when compared with changes in other components. A B C By making that assumption, we have dB/dt = k1 A – k2 B = 0 Thus, dC/dt = k2 B = k1 A The two reactions are reduced into one: A Æ C, with rate constant k1 The simplified solution is: ( ) 1 = exp A kt ( ) 1 = 1 exp Ck t −−
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Rate of changes (absolute values) in three species: a closer look |dA/dt| |dB/dt| |dC/dt|
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By making that assumption, we have dB/dt = k1 A – k2 B = 0 Thus, dC/dt = k2 B = k1 A The two reactions are reduced into one: A Æ C, with rate constant k1 The simplified solution is: ( ) 1 = exp A kt () 1 = 1 exp Ck t −− approximation Full solution As can be seen from numerical results, the assumption gives very good approximation to the full solution Quasi-steady state assumptions are extensively used to simplify complex kinetic models to simpler ones
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Notes on quasi-steady state QSS assumptions are used extensively for all types of mathematical models to simplify analysis. All kinetic models of biological systems used explicitly or implicitly the QSSA for some variables in the model.
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This note was uploaded on 08/03/2009 for the course BME 100 taught by Professor Yuan during the Spring '07 term at Duke.

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EnzymeKinetics_1 - Simple kinetic analysis Enzyme kinetics...

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