Paris/Continuous model

From 2007.igem.org

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The fixed points are (g = 0, s = 0) and the solution of :  
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The fixed points are (g = 0, s = 0) and the solution of : <br clear="all" />
[[Image:eq6.jpg|right]]
[[Image:eq6.jpg|right]]
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<br clear="all" />
==Analysis of stability ==
==Analysis of stability ==
=== Fixed point at the origin ===
=== Fixed point at the origin ===
Let us linearise the equations’ system (5a) close to the origin. For small perturbations  
Let us linearise the equations’ system (5a) close to the origin. For small perturbations  
around (g = 0, s = 0) (5a) is equivalent to :  
around (g = 0, s = 0) (5a) is equivalent to :  
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+
<br clear="all" />
[[Image:eq7.jpg|right]]
[[Image:eq7.jpg|right]]
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+
<br clear="all" />
We look for a solution of the form [[Image:inline7.jpg]] , with [[Image:inline8.jpg]] an arbitrary vector.  The values of λ  are the eigenvalues of the matrix in (7) and can be obtain here straightforwardly by  
We look for a solution of the form [[Image:inline7.jpg]] , with [[Image:inline8.jpg]] an arbitrary vector.  The values of λ  are the eigenvalues of the matrix in (7) and can be obtain here straightforwardly by  
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the characteristic polynomial : λ1 = −b, λ2 = −1, with respectively the eigenvectors
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the characteristic polynomial : λ1 = −b, λ2 = −1, with respectively the eigenvectors [[Image:inline9.jpg]].
 +
 
 +
The two eigenvalues are always negative : the origin is therefore
 +
always an attractive fixed point. For too weak initial concentrations of the two cellular
 +
types, the system is always going to die out.
 +
 
 +
===Second fixed point, out of the origin ===
 +
Let (g0 := β s0 , s0 := k/(α−1) ) be the non trivial solution of (6a). We can simplify the (5a) by
 +
dividing the two equations respectively by g0 et s0 and by taking as new variables G := g/g0
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and S := s/s0:
 +
<br clear="all" />
 +
[[Image:eq8.jpg|right]]
 +
<br clear="all" />
 +
 
 +
and with μ := α−1 = α1/α2− 1 :
 +
<br clear="all" />
 +
[[Image:eq9.jpg|right]]
 +
<br clear="all" />
 +
 
 +
The fixed point is now (G0=1,S0=1). We linearise system (9a) around this point and
 +
look for a solution close to it. Take x := G−G0 and y := S−S0 , close to (G0,S0) we can
 +
write by keeping only the first order terme of (9a) :
 +
 
 +
<br clear="all" />
 +
[[Image:eq10.jpg|right]]
 +
<br clear="all" />
 +
 
 +
Looking for a solution of the form [[Image:inline7.jpg]], we need to find the eigenvalues of the matrix in
 +
(10). Solving the characteristic polynomial gives :
 +
 
 +
<br clear="all" />
 +
[[Image:eq1112.jpg|right]]
 +
<br clear="all" />

Revision as of 21:22, 23 October 2007


Contents

Well mixed population model

We present here a theoretical approach based on population dynamics. We consider here the case of a well mixed, homogenous, culture of the SMB organism, i.e. there is no space in this analysis and we follow only the variation of the different cell lines concentrations in the culture volume.


Derivation of the model

Let the variables g, s and d describe respectively the concentrations of germinal and so- matic cells and the conentration of DAP, then we can write for our system : Eq1.jpg

Equation (1a) describes the growth of the germinal cell population in presence of sufficient DAP (interaction represented by the Michaelis-Menten function) ; the term proportional to α2 is the differentiation into somatic cells by recombination of the CRE/LOX box, and the last term proportional to α3 is the germinal cells’ death. Equation (1b) is the variation of the somatic cells’ population, with the term proportional to α4 for somatic cells’ death. The last line describes DAP production by the somatic cells, and includes a degradation term. In absence of any quantitative details on assimilation of DAP by germinal cells and response to DAP levels, n and k are are to be considered as arbirary phenomenological parameters. We take however in the following n = 1 neglecting potential saturation related non-linearities for high DAP concentrations. The value of k corresponds to the DAP concentration for half-maximal growth rate, and could set experimenntally. We simplify the previous system by assuming that the evolution of d is rapid compared to the cellular growth, so that at this time scale we can take d' = 0 and write d = s α5/ α6 . This gives the two-variable system :

Eq2.jpg

Redefinition of parameters k → kα6 /α5 and α3 → α2 + α3 leads to the simpler writing :

Eq3.jpg

Let us do some rewriting :

Eq4.jpg

and by redefining the time and most of the parameters we get :

Eq5.jpg


The fixed points are (g = 0, s = 0) and the solution of :

Eq6.jpg


Analysis of stability

Fixed point at the origin

Let us linearise the equations’ system (5a) close to the origin. For small perturbations around (g = 0, s = 0) (5a) is equivalent to :

Eq7.jpg


We look for a solution of the form Inline7.jpg , with Inline8.jpg an arbitrary vector. The values of λ are the eigenvalues of the matrix in (7) and can be obtain here straightforwardly by the characteristic polynomial : λ1 = −b, λ2 = −1, with respectively the eigenvectors Inline9.jpg.

The two eigenvalues are always negative : the origin is therefore always an attractive fixed point. For too weak initial concentrations of the two cellular types, the system is always going to die out.

Second fixed point, out of the origin

Let (g0 := β s0 , s0 := k/(α−1) ) be the non trivial solution of (6a). We can simplify the (5a) by dividing the two equations respectively by g0 et s0 and by taking as new variables G := g/g0 and S := s/s0:

Eq8.jpg


and with μ := α−1 = α1/α2− 1 :

Eq9.jpg


The fixed point is now (G0=1,S0=1). We linearise system (9a) around this point and look for a solution close to it. Take x := G−G0 and y := S−S0 , close to (G0,S0) we can write by keeping only the first order terme of (9a) :


Eq10.jpg


Looking for a solution of the form Inline7.jpg, we need to find the eigenvalues of the matrix in (10). Solving the characteristic polynomial gives :


Eq1112.jpg