IGEM:IMPERIAL/2007/Projects/Modelling/Tutorial: Difference between revisions
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*Enzyme is not allosteric. | *Enzyme is not allosteric. | ||
*Reaction is in steady state | *Reaction is in steady state | ||
*Equilibrium favours product over transition complex. | |||
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1. Rate of formation of ES: '''k<sub>1</sub>[E][S] + k<sub>4</sub>[E][P]''' | 1. Rate of formation of ES: '''k<sub>1</sub>[E][S] + k<sub>4</sub>[E][P]''' | ||
However, we assume | However, we assume k<sub>3</sub> ≫ k<sub>4</sub>, i.e. the equilibrium favours product over transition complex, so we ignore the term with k<sub>4</sub> in it. | ||
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'''V = k<sub>3</sub>[ES]''' | '''V = k<sub>3</sub>[ES]''' | ||
If we substitute this into the last result, we get: | |||
'''V = k<sub>3</sub>[E]T[S] / ([S]+K<sub>m</sub>)''' | |||
6. When the enzyme is saturated, [ES]=[E]T, so we can define a maximum velocity, V<sub>m</sub>: | |||
'''V<sub>m</sub> = k<sub>3</sub>[E]T''' | |||
And this gives us our final result: | |||
'''V = V<sub>m</sub> [S] / K<sub>m</sub> + [S]''' | |||
===MM Parameters=== | |||
[[Image:icgems_mm0.png|frame|Michaelis-Menton Kinetics]] | |||
V<sub>m</sub> (or Vmax) is the maximum rate at which a certain mass of enzyme can work, and equals k<sub>3</sub> (the rate constant for [ES] degradation to product) times the total enzyme concentration [E]T. Fast enzymes have higher V<sub>m</sub>. V<sub>m</sub> is not a constant, because it is dependent on the amount of enzyme present in a solution, and on the volume of this solution. It is measured in mol s<sub>-1</sub> or more usually μmol min<sub>-1</sub>. | |||
K<sub>m</sub> is (k<sub>2</sub>+k<sub>3</sub>) / k<sub>1</sub> = [E][S] / [ES], i.e. K<sub>m</sub> is the K<sub>eq</sub> for the non-productive dissociation of the ES complex back to E and S. A low K<sub>m</sub> indicates the enzyme has a high affinity for its substrate. Therefore efficient enymes have lower K<sub>m</sub>. K<sub>m</sub> is a constant: it equals [S] at ½V<sub>m</sub>, and is measured in mol L<sub>-1</sub>, or more usually mM. |
Revision as of 17:10, 29 July 2007
Modelling: Tutorial
Michaelis-Menton Kinetics
Introduction
Michaelis-Menten kinetics describes the kinetics of many enzymes.
Assumptions of Model:
- [Enzyme] is much less than the [Substrate]
- Enzyme is not allosteric.
- Reaction is in steady state
- Equilibrium favours product over transition complex.
Derivation of MM
Consider the following reaction:
k1 and k3 are forward rate constants. k2 and k4 are backward rate constants.
1. Rate of formation of ES: k1[E][S] + k4[E][P]
However, we assume k3 ≫ k4, i.e. the equilibrium favours product over transition complex, so we ignore the term with k4 in it.
2. Rate of removal of ES: k2[ES] + k3[ES]
3. In the steady state, rate of formation of ES = rate of removal of ES:
k1[E][S] = k2[ES] + k3[ES]
If we rearrange this to get all the rate constants on one side and all the concentration terms on the other, we can define a constant called Km, which is the Michaelis constant:
Km = (k2+k3) / k1 = [ES] / [E][S]
4. We can’t measure [E] (free enzyme) or [ES] easily, but we do know the total enzyme concentration, as that is what we will have added when we made up our enzyme solution.
[E] = [E]T – [ES]
If we substitute this into our definition of Km and rearrange it, we get:
[ES] = [E]T[S] / ([S]+Km)
5. The overall (initial) reaction rate V is
V = k3[ES]
If we substitute this into the last result, we get:
V = k3[E]T[S] / ([S]+Km)
6. When the enzyme is saturated, [ES]=[E]T, so we can define a maximum velocity, Vm:
Vm = k3[E]T
And this gives us our final result:
V = Vm [S] / Km + [S]
MM Parameters
Vm (or Vmax) is the maximum rate at which a certain mass of enzyme can work, and equals k3 (the rate constant for [ES] degradation to product) times the total enzyme concentration [E]T. Fast enzymes have higher Vm. Vm is not a constant, because it is dependent on the amount of enzyme present in a solution, and on the volume of this solution. It is measured in mol s-1 or more usually μmol min-1.
Km is (k2+k3) / k1 = [E][S] / [ES], i.e. Km is the Keq for the non-productive dissociation of the ES complex back to E and S. A low Km indicates the enzyme has a high affinity for its substrate. Therefore efficient enymes have lower Km. Km is a constant: it equals [S] at ½Vm, and is measured in mol L-1, or more usually mM.