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.


<br clear="all">
<br clear="all">
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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 k3 k4, i.e. the equilibrium favours product over transition complex, so we ignore the term with k4 in it.
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.




Line 69: Line 69:


'''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.

Reaction at Steady State

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

Michaelis-Menton Kinetics

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.