Team:KU Leuven/Modeling/Internal
Internal Model
In the internal model we will try to simulate the dynamics of the processes in the cell. For this we use the Simbiology toolbox in Matlab. This allows us to model the production of leucine and AHL in cell A and the changing behavior of cell B due to the changing concentration. $$\frac{{\large d} m_{CI}}{d t} = \alpha_1 {\cdot} CI_{gene} - d_1 {\cdot} m_{CI}$$ $$\frac{{\large d}{CI}}{d t} = \beta_1 {\cdot} {CI} - d_{CI} {\cdot} {CI} $$ $$\frac{{\large d} m_{luxI}}{d t} = L_{lambda} + {\frac{V_{max}}{1 + ({\frac{CI}{K_{d1}}})^{n_{CI}}}} {\cdot} luxI_{gene} - d_{luxI} {\cdot} m_{luxI} $$ $$\frac{{\large d} m_{luxR}}{d t} = L_{lambda} + {\frac{V_{max}}{1 + ({\frac{CI}{K_{d1}}})^{n_{CI}}}} {\cdot} luxR_{gene} - d_{luxR} {\cdot} m_{luxR} $$ $$\frac{{\large d} luxI}{d t} = \beta_{luxI} {\cdot} {m_{luxI}} - d_{luxI} {\cdot}{luxI} $$ $$\frac{{\large d} luxR}{d t} = \beta_{luxR} {\cdot} {m_{luxR}} - d_{luxR} {\cdot}{luxR} $$ $$\frac{{\large d} AHL_{in}}{d t} = {k_{luxI}} {\cdot} {luxI} + ( {D_{IN}} {\cdot} {AHL_{out}} - {D_{OUT}} {\cdot} {AHL_{in}} ) - d_{AHL,in} {\cdot} {AHL_{in}} $$ $$\frac{{\large d} AHL_{out}}{d t} = ( {D_{OUT} {\cdot} {AHL_{in}} - {D_{IN}}{\cdot}{AHL_{out}} ) -d_{AHL,out}{\cdot}{luxR} $$ $$\frac{\partial R}{\partial t} = D_r \bigtriangledown^2 B + k_r A - k_{lossH} R $$ $$\frac{\partial H}{\partial t} = D_h \bigtriangledown^2 B + k_h A - k_{lossR} H . $$
Test test test test test.
Parameters
| Parameter | Value | Unit | Source | Comment |
|---|---|---|---|---|
| $D_a$ | $0.072 \cdot 10^{-3}$ | $cm^2/h$ | following [1] | |
| $D_b$ | $2.376 \cdot 10^{-3}$ | $cm^2/h$ | following [1] | |
| $D_r$ | $26.46 \cdot 10^{-3}$ | $cm^2/h$ | as found in [6] | $298.2 K$ |
| $D_h$ | $50 \cdot 10^{-3}$ | $cm^2/h$ | from [3] | |
| $K_{c}$ | $8.5 \cdot 10^{-3}$ | $cm^2/h$ | guessed | |
| $\gamma$ | $10^{-5}$ | $h^{-1}$ | from [1] | |
| $k_p$ | $1.0 \cdot 10^2$ | $cl^{-1}$ | from [1] | |
| $k_h$ | $17.9 \cdot 10^{-4}$ | $fmol/h$ | computed from [4] and [8] | |
| $k_r$ | $5.4199\cdot 10^{-4}$ | $fmol/h$ | computed from [7] and [8] | |
| $k_{lossH}$ | $1/48$ | $h^{-1}$ | from [5] | $ ph = 7$ |
| $k_{lossR}$ | $1/80$ | $h^{-1}$ | guessed |
References
