3_Enzymatic%20reaction

3_Enzymatic%20reaction - Enzymatic Reactions S v Vmax Vmax...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Enzymatic Reactions ] S [ K ] S [ V dt ] P [ d v M max + = = S P enzyme Note: Without the enzyme, the reaction can still happen but at a slower rate. Question: How to model this reaction? Question: Can we assume it is an elementary reaction? [S] 0 V max V max 2 K M Expt Data v How to model biological systems Modeling Nature’s laws -1 st law -2 nd law - Mass conservation Experimental data - Model constants - Estalishing new relationships - Verify model predictions - Establish empirical Eqn in the model Hypotheses - e.g., law of mass action in this course Michaelis - Menten Model E, S, ES, and P are the enzyme, substrate, enzyme-substrate complex, and product. E + S k -1 k 1 ES k 2 ⎯→ P + E Assumptions : (1) Enzyme is not consumed in the reaction, i.e., the total concentration of enzyme [E] 0 is a constant. (2) [E] 0 << [S] and the reverse reaction from P to S is negligible. (This is only true initially since [S] 0 near the end of reaction.)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 The initial rate of the reaction: ] ES [ k dt ] P [ d v 2 0 = = How to determine [ES] ? d [ ES ] dt = k 1 [E][S] – k -1 [ES] - k 2 [ES] = k 1 [E][S] - (k -1 +k 2 )[ES] E + S k -1 k 1 ES k 2 ⎯→ P + E Special case 1. The first two reactions are at quasi-equilibrium Two special cases E + S k -1 k 1 ES k 2 P + E What is the rationale to make such an assumption? How to make an assumption? Assuming k -1 >> k 2 => the first reaction is at the equilibrium state. k 1 [S][E] k -1 [ES] The contribution of k 2 [ES] is negligible. j i [products] ] [reactants If all other reactions are very slow => Quasi-equilibrium A Sufficient Condition
Background image of page 2
3 K D : equilibrium dissociation constant of the first reaction. To cancel [E] in the equation, use the mass conservation law for E [E] 0 = [E] + [ES] [E] 0 = K D [ES]/[S] + [ES] const K k k ] ES [ ] S ][ E [ D = = 1 1 ] S [ K ] S [ ] E [ ] ES [ D 0 + = or ] S [ K ] S [ ] E [ k ] ES [ k dt ] P [ d v D 0 2 2 0 + = = = Thus, Let V max = k 2 [E] 0 ] [ ] [ ] [ max 0 S K S V dt P d v D + = = (Original Michaelis-Menten Eq) v 0 [S] 0 V max V max 2 K D Expt Data 2. k
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 12

3_Enzymatic%20reaction - Enzymatic Reactions S v Vmax Vmax...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online