Team:Groningen/Modelling/Arsenic

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(Equilibrium: Equilibrium computations (without overexpression).)
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==Equilibrium==
==Equilibrium==
-
For many purposes, like determining the total amount of accumulated arsenic, it can be quite useful to consider the system at equilibrium. That is, when the derivatives of all variables to time are zero.
+
For many purposes, like determining the total amount of accumulated arsenic, it can be quite useful to consider the system at equilibrium. That is, when the derivatives of all variables to time are zero:
-
The following derives an equation for the concentration of free Operators (which is directly related to the amount of ArsR and ArsD produced) based on ArsR and ArsD (utilizing the same kind of equilibrium assumption as in the derivation of equation A.1.4 in [[Team:Groningen/Literature#Alon2007|Alon2007]]):
+
<pre>
 +
0 = - (k1on ArsR+k2on ArsD) As(III) + k1off ArsRAs + k2off ArsDAs
 +
0 = - (k3on ArsR+k4on ArsD) Operator + k3off ArsRop + k3off ArsDop
 +
0 = β1 Operator - (ln(2)/τ1+k1on As(III)+k3on Operator) ArsR + k1off ArsRAs + k3off ArsRop
 +
0 = β2 Operator - (ln(2)/τ2+k2on As(III)+k4on Operator) ArsD + k2off ArsDAs + k4off ArsDop
 +
0 = k1on ArsR As(III) - k1off ArsRAs
 +
0 = k2on ArsD As(III) - k2off ArsDAs
 +
0 = k3on ArsR Operator - k3off ArsRop
 +
0 = k4on ArsD Operator - k4off ArsDop
 +
</pre>
 +
 
 +
By eliminating the last four equations from the rest and dividing the last four by kNon we are left with:
 +
 
 +
<pre>
 +
0 = β1 Operator - (ln(2)/τ1) ArsR
 +
0 = β2 Operator - (ln(2)/τ2) ArsD
 +
0 = ArsR As(III) - K1d ArsRAs
 +
0 = ArsD As(III) - K2d ArsDAs
 +
0 = ArsR Operator - K3d ArsRop
 +
0 = ArsD Operator - K4d ArsDop
 +
</pre>
 +
 
 +
Using the fact that the total amount of operators remains constant the last two equations can be used to derive an equation for Operator:
 +
 
 +
<pre>
 +
    0 = ArsR Operator - K3d ArsRop
 +
    0 = ArsR Operator - K3d (OperatorT - Operator - ArsDop)
 +
    0 = (ArsR + K3d) Operator - K3d OperatorT + K3d ArsDop
 +
    0 = (ArsR/K3d + 1) Operator - OperatorT + ArsDop
 +
ArsDop = OperatorT - (ArsR/K3d + 1) Operator
 +
 
 +
      0 = ArsD Operator - K4d ArsDop
 +
      0 = ArsD Operator - K4d (OperatorT - (ArsR/K3d + 1) Operator)
 +
      0 = (ArsD/K4d) Operator - OperatorT + (ArsR/K3d + 1) Operator
 +
      0 = (ArsR/K3d + ArsD/K4d + 1) Operator - OperatorT
 +
Operator = OperatorT/(ArsR/K3d + ArsD/K4d + 1)
 +
</pre>
 +
 
 +
This leads to the following:
<pre>
<pre>
-
ArsR*Op=Kd3*ArsROp
+
  0 = β1 Operator - (ln(2)/τ1) ArsR
-
ArsD*Op=Kd4*ArsDOp
+
  0 = β1 OperatorT - (ln(2)/τ1) ArsR (ArsR/K3d + ArsD/K4d + 1)
-
OpT = Op + ArsROp + ArsDOp
+
  0 = β1 OperatorT - (ln(2)/τ1) ArsR&sup2;/K3d - (ln(2)/τ1) ArsR ArsD/K4d - (ln(2)/τ1) ArsR
 +
  0 = K3d (τ1/ln(2)) β1 OperatorT - ArsR&sup2; - (K3d/K4d) ArsD ArsR - K3d ArsR
 +
  0 = &frac12; ArsR&sup2; + &frac12; K3d (ArsD/K4d + 1) ArsR - &frac12; K3d (τ1/ln(2)) β1 OperatorT
 +
ArsR = -b1 &plusmn; &radic;(b1&sup2; + c1)
 +
  b1 = &frac12; K3d (ArsD/K4d + 1)
 +
  c1 =  K3d (τ1/ln(2)) β1 OperatorT
-
ArsD*Op=Kd4*(OpT-Op-ArsROp)
+
  0 = β2 OperatorT - (ln(2)/τ2) ArsD (ArsR/K3d + ArsD/K4d + 1)
-
ArsD*Op/Kd4=OpT-Op-ArsR*Op/Kd3
+
ArsD = -b2 &plusmn; &radic;(b2&sup2; + c2)
-
(1+ArsR/Kd3+ArsD/Kd4)*Op=OpT
+
  b2 = &frac12; K4d (ArsR/K3d + 1)
-
Op/Opt = 1/(1+ArsR/Kd3+ArsD/Kd4)
+
  c2 =  K4d (τ2/ln(2)) β2 OperatorT
</pre>
</pre>
-
And similarly for As(III):
+
As b1 and b2 are positive (concentrations are always non-negative) only the plus signs in the two plus-minuses above are valid choices. In addition, since c1 and c2 are non-negative the square roots are always larger than or equal to b1/b2, leading to non-negative values for ArsR/ArsD. The equations for ArsR and ArsD derived above can be used in a fixed point iteration as follows:
<pre>
<pre>
-
ArsR*As3=Kd1*ArsRAs3
+
f(ArsR) = -b1 + &radic;(b1&sup2; + c1)
-
ArsD*As3=Kd2*ArsDAs3
+
  with ArsD = -b2 + &radic;(b2&sup2; + c2)
-
As3T = As3 + ArsRAs3 + ArsDAs3
+
-
As3/As3T = 1/(1+ArsR/Kd1+ArsD/Kd2)
+
  f' = -b1' + (2 b1 b1')/&radic;(b1&sup2; + c1)
 +
  b1' = &frac12; (K3d/K4d) ArsD'
 +
ArsD' = -b2' + (2 b2 b2')/&radic;(b2&sup2; + c2)
 +
  b2' = &frac12; (K4d/K3d)
</pre>
</pre>
-
Finally the bound forms of ArsR and ArsD can be eliminated from the equations completely:
+
Assuming that the total amount of arsenic in the cell is also constant the following equation can be derived for arsenic, analogous to the derivation of the equation for Operator:
<pre>
<pre>
-
0 = &beta;1*Op - (ln(2)/&tau;1+k1on*As3+k3on*Op)*ArsR + k1off*ArsRAs3 + k3off*ArsROp
+
ArsR As(III) = K1d ArsRAs
-
  = &beta;1*Op - (ln(2)/&tau;1+k1on*As3+k3on*Op)*ArsR + k1off*ArsR*As3/Kd1 + k3off*ArsR*Op/Kd3
+
ArsD As(III) = K2d ArsDAs
-
  = &beta;1*Op - (ln(2)/&tau;1+k1on*As3+k3on*Op)*ArsR + (k1on*As3+k3on*Op)*ArsR
+
As(III)T = As(III) + ArsRAs + ArsDAs
-
  = &beta;1*Op - ln(2)/&tau;1*ArsR
+
-
0 = &beta;2*Op - ln(2)/&tau;2*ArsD
+
As(III)/As(III)T = 1/(ArsR/K1d + ArsD/K2d + 1)
</pre>
</pre>

Revision as of 12:13, 14 July 2009

Bold text

Igemhomelogo.png

Our initial ideas on how and what to model can be found at Brainstorm/Modelling.

Usage of graphs in wiki: Graphs

The raw model

Note: Math support is currently not enabled on this Wiki... (I've asked hq if they can enable it.)

The following variables play an important role in our system (these can be concentrations of substances, the density of the cell, etc.):

  • Extracellular:
    • As(III)
    • As(V)
  • Intracellular:
    • As(III)
    • Operator (concentration of unbound operators)
    • As(V)
    • ArsC
    • ArsR
    • ArsD
    • ArsRAs (bound to As(III))
      • At equilibrium: ArsR As(III) = (k1off/k1on) ArsRAs
    • ArsDAs (bound to As(III))
    • ArsRop (bound to operator)
    • ArsDop (bound to operator)

The variables above can be related to each other through the following "reactions" and/or equations:

  • As(V)ex → As(V), using phosphate transporters? (Summers2009)
  • As(V)ex → As(III), using ArsC (Summers2009)
  • As(III) → As(III)ex, using ArsAB (helped by ArsD) (Summers2009)
  • As(III)in + ArsR ↔ ArsRAs
  • As(III)in + ArsD ↔ ArsDAs
  • Operator + ArsR ↔ ArsRop
  • Operator + ArsD ↔ ArsDop
  • Operator → Operator + ArsR + ArsD (transcription + translation)
  • ArsR → null (degradation)
  • ArsD → null (degradation)

Resulting in the following differential equations (please note that the first two can be formed by linear combinations of the other six):

  • (d/dt) As(III) = - (k1on ArsR+k2on ArsD) As(III) + k1off ArsRAs + k2off ArsDAs
  • (d/dt) Operator = - (k3on ArsR+k4on ArsD) Operator + k3off ArsRop + k3off ArsDop
  • (d/dt) ArsR = β1 Operator - (ln(2)/τ1+k1on As(III)+k3on Operator) ArsR + k1off ArsRAs + k3off ArsRop
  • (d/dt) ArsD = β2 Operator - (ln(2)/τ2+k2on As(III)+k4on Operator) ArsD + k2off ArsDAs + k4off ArsDop
  • (d/dt) ArsRAs = k1on ArsR As(III) - k1off ArsRAs
  • (d/dt) ArsDAs = k2on ArsD As(III) - k2off ArsDAs
  • (d/dt) ArsRop = k3on ArsR Operator - k3off ArsRop
  • (d/dt) ArsDop = k4on ArsD Operator - k4off ArsDop

Using the following constants/definitions:

  • K1d = k1off/k1on
  • K2d = k2off/k2on = 60µM (Chen1997)
  • K3d = k3off/k3on = 0.33µM (Chen1997, suspect as the relevant reference doesn't actually seem to give any value for this)
  • K4d = k4off/k4on = 65µM (Chen1997)
  • degradation rate = ln(2)/τ
  • ArsR half-life time = τ1
  • ArsD half-life time = τ2
  • β1 = β2 ??? (and either value is unknown)

See Chen1997 for the interplay between ArsR and ArsD.

TODO Figure out relevant equations for metallochaperone function of ArsD?

TODO Make sure all the multiplicities are correct (and/or taken care of in constants). E.g. does 1 mol ArsR (if it is bound) bind 1 mol As(III)?

Equilibrium

For many purposes, like determining the total amount of accumulated arsenic, it can be quite useful to consider the system at equilibrium. That is, when the derivatives of all variables to time are zero:

0 = - (k1on ArsR+k2on ArsD) As(III) + k1off ArsRAs + k2off ArsDAs
0 = - (k3on ArsR+k4on ArsD) Operator + k3off ArsRop + k3off ArsDop
0 = β1 Operator - (ln(2)/τ1+k1on As(III)+k3on Operator) ArsR + k1off ArsRAs + k3off ArsRop
0 = β2 Operator - (ln(2)/τ2+k2on As(III)+k4on Operator) ArsD + k2off ArsDAs + k4off ArsDop
0 = k1on ArsR As(III) - k1off ArsRAs
0 = k2on ArsD As(III) - k2off ArsDAs
0 = k3on ArsR Operator - k3off ArsRop
0 = k4on ArsD Operator - k4off ArsDop

By eliminating the last four equations from the rest and dividing the last four by kNon we are left with:

0 = β1 Operator - (ln(2)/τ1) ArsR
0 = β2 Operator - (ln(2)/τ2) ArsD
0 = ArsR As(III) - K1d ArsRAs
0 = ArsD As(III) - K2d ArsDAs
0 = ArsR Operator - K3d ArsRop
0 = ArsD Operator - K4d ArsDop

Using the fact that the total amount of operators remains constant the last two equations can be used to derive an equation for Operator:

     0 = ArsR Operator - K3d ArsRop
     0 = ArsR Operator - K3d (OperatorT - Operator - ArsDop)
     0 = (ArsR + K3d) Operator - K3d OperatorT + K3d ArsDop
     0 = (ArsR/K3d + 1) Operator - OperatorT + ArsDop
ArsDop = OperatorT - (ArsR/K3d + 1) Operator

       0 = ArsD Operator - K4d ArsDop
       0 = ArsD Operator - K4d (OperatorT - (ArsR/K3d + 1) Operator)
       0 = (ArsD/K4d) Operator - OperatorT + (ArsR/K3d + 1) Operator
       0 = (ArsR/K3d + ArsD/K4d + 1) Operator - OperatorT
Operator = OperatorT/(ArsR/K3d + ArsD/K4d + 1)

This leads to the following:

   0 = β1 Operator - (ln(2)/τ1) ArsR
   0 = β1 OperatorT - (ln(2)/τ1) ArsR (ArsR/K3d + ArsD/K4d + 1)
   0 = β1 OperatorT - (ln(2)/τ1) ArsR²/K3d - (ln(2)/τ1) ArsR ArsD/K4d - (ln(2)/τ1) ArsR
   0 = K3d (τ1/ln(2)) β1 OperatorT - ArsR² - (K3d/K4d) ArsD ArsR - K3d ArsR
   0 = ½ ArsR² + ½ K3d (ArsD/K4d + 1) ArsR - ½ K3d (τ1/ln(2)) β1 OperatorT
ArsR = -b1 ± √(b1² + c1)
  b1 = ½ K3d (ArsD/K4d + 1)
  c1 =   K3d (τ1/ln(2)) β1 OperatorT

   0 = β2 OperatorT - (ln(2)/τ2) ArsD (ArsR/K3d + ArsD/K4d + 1)
ArsD = -b2 ± √(b2² + c2)
  b2 = ½ K4d (ArsR/K3d + 1)
  c2 =   K4d (τ2/ln(2)) β2 OperatorT

As b1 and b2 are positive (concentrations are always non-negative) only the plus signs in the two plus-minuses above are valid choices. In addition, since c1 and c2 are non-negative the square roots are always larger than or equal to b1/b2, leading to non-negative values for ArsR/ArsD. The equations for ArsR and ArsD derived above can be used in a fixed point iteration as follows:

f(ArsR) = -b1 + √(b1² + c1)
  with ArsD = -b2 + √(b2² + c2)

   f' = -b1' + (2 b1 b1')/√(b1² + c1)
  b1' = ½ (K3d/K4d) ArsD'
ArsD' = -b2' + (2 b2 b2')/√(b2² + c2)
  b2' = ½ (K4d/K3d)

Assuming that the total amount of arsenic in the cell is also constant the following equation can be derived for arsenic, analogous to the derivation of the equation for Operator:

ArsR As(III) = K1d ArsRAs
ArsD As(III) = K2d ArsDAs
As(III)T = As(III) + ArsRAs + ArsDAs

As(III)/As(III)T = 1/(ArsR/K1d + ArsD/K2d + 1)

Kinetic Laws

TODO Add references.

TODO Find out how to determine experimentally which is applicable (and if you know, what the parameters are).

Mass Action
Molecules randomly interact, the reaction rate is simply the product of the concentrations of the reactants (multiplied by a constant).
Michaelis-Menten
Applicable to situations where there is a maximum reaction rate (due to needing a catalyst/transporter/binding site of which there is only a limited amount for example) under the assumption that there is much more of the "main" reactant than of the catalyst/transporter. Has two constants, the maximum reaction rate and the concentration at which the reaction rate is half the maximum reaction rate.
Michaelis-Menten reversible
TODO
Hill
Generalization of Michaelis-Menten. More detail.