lecture4 - 10.492 - Integrated Chemical Engineering (ICE)...

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10.492 - Integrated Chemical Engineering (ICE) Topics: Biocatalysis MIT Chemical Engineering Department Instructor: Professor Kristala Prather Fall 2004 Lecture # 5, 6 – Enzyme Inhibition and Toxicity Handout: Derivation of Inhibition Kinetics Now that we’ve considered enzyme kinetics, let’s talk about the phenomenon of enzyme inhibition. In this situation, either the substrate itself or a different molecule affects the ability of the enzyme to convert. In a “clean” system where the substrate is pure and only one product is formed, the inhibitor will be the substrate or the product. This is the case that we’ll consider; however, keep in mind that if the substrate or the reaction environment is mixed, other compounds may function as inhibitors. There are three types of inhibition – competitive, uncompetitive, and noncompetitive . Each kind of inhibition leads to a different form of the rate equation. It’s the impact on the kinetics that leads one to identify inhibition in an enzyme reaction. Let’s look at each of the three cases and how the rate equations are altered from the standard Michaelis- Menten form. In each case, we’ll assume that inhibition is reversible. We’ll consider the case of irreversible inhibition to be toxicity , which will be discussed later. 1. Competitive Inhibition In this case, the inhibitor binds to the active site and prevents binding of the substrate. The reaction equations are as follows: k 2 (1.1) [ E ] + [ S ] ←⎯→ [ S E ] K S ⎯→ [ E ] + [ P ] K I (1.2) [ E ] + [ I ] ⎯→ [ I E ] Note that we are assuming that formation of both enzyme complexes is in equilibrium with the respective substrate/inhibitor, and that K S and K I are dissociation constants. The final form of the equation will be the same if we don’t assume equilibrium and instead use the quasi-steady state assumption, except that K M =K S in this case. The rate equation is then obtained from a mass balance on product formation and the enzyme species: [ (1.3) P d ] = k [ S E ] dt 2 [ S E ] = [ E ][ S ] (1.4) K M = [ E ][ S ] [ S E ] K M [ I E ] = [ E ][ I ] (1.5) K I = [ E ][ I ] [ I E ] K I Dr. Kristala L. Jones Prather, Copyright 2004. MIT Department of Chemical Engineering
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Lecture #4, p.2 (1.6) [ E ] = [ E ] + S ] + I ] = [ E ] 1 + [ S ] K M + [ I ] [ E [ E 0 K I 0 S ] K M 1 + [ S ] K M + K I (1.7) [ E = [ S ] [ E ] [ I ] Plug this back into Eq (1.3): dP 0 v max [ S ] (1.8) = v = k 2 [ E ][ S ] = app dt K M 1 + [ I ] K I + [ S ] K M + [ S ] So from the final rate expression, you can see that the impact of a competitive inhibitor is to alter the Michaelis constant K M such that the enzyme would appear to have a lower affinity for the substrate (higher K M = lower affinity). This makes sense, since the inhibitor is binding to the same site as the substrate. So, as is the case with high K
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lecture4 - 10.492 - Integrated Chemical Engineering (ICE)...

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