Team:Paris Bettencourt/Modeling

Ferment It Yourself

iGEM Paris-Bettencourt 2O15

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Notebook

Vitamin A	Vitamin B2	Vitamin B12
Phytase	Riboswitch	Differentiation on E. coli
Differentiation on S. cerevisiae	Manufacturing	Idli and Micro-organisms

Kinetics Module

Introduction

Based on a set of ordinary differential equations (ODE) describing the kinetics of the cells differentiation, we design a model to find the best differentiation rate ie find the constant reaction \(k_{1}\) to optimize the vitamin production.

We find out that a stochastic algorithm is an other solution to solve our problem.
For system involving large cell counts, ODE model give an accurate representation of the behavior. But with small cell counts , the stochastic and discrete method has a significant influence on the observed behaviour.
These reasons led us to write a stochastic programm based on the Gillespie’s stochastic simulation algorithm (SSA). With this programm we obtain an accurate analysis.

Equations

\[MC \xrightarrow[]{k_{1}} DC\] \[MC \xrightarrow[]{k_{2}} 2.MC\] \[DC \xrightarrow[]{k_{3}} 2.DC\] \[DC \xrightarrow[]{k_{4}} DC + Vitamin\]

\[\frac{d[MC]}{dt}(t) = (k_{2} - k_{1}).[MC](t)\] \[\frac{d[DC]}{dt}(t) = k_{1}.[MC](t) + k_{3}.[DC](t)\] \[\frac{d[Vitamin]}{dt}(t) = k_{4}.[DC](t)\] \[[Vitamin](t) = \int_{0}^{t}{[DC](t').dt'}\]

\[[MC](t) = [MC]_{0}.e^{(k_{2} - k_{1}).t}\]
\[ [DC](t) = \begin{cases} [DC]_{0}.e^{k_{3}.t} + [MC]_{0}.\frac{k_{1}}{k_{2}-k_{1}-k_{3}}.(e^{(k_{2} - k_{1}).t} - e^{k_{3}.t}) & \mbox{if } k_{1} \ne k_{3} - k_{2} \\ \\ ([MC]_{0}.(k_{2} - k_{3}).t + [DC]_{0}).e^{k_{3}.t} & \mbox{if } k_{1} = k_{3} - k_{2} \end{cases} \]
\[ [Vitamin](t) = \begin{cases} [DC]_{0}.\frac{k_{4}}{k_{3}}.(e^{k_{3}.t} - 1) + [MC]_{0}.\frac{k_{4}.k_{1}}{(k_{2}-k_{1}-k_{3}).(k_{2} - k_{1})}.(e^{(k_{2} - k_{1}).t} - 1) + [MC]_{0}.\frac{k_{4}.k_{1}}{(k_{2}-k_{1}-k_{3}).k_{3}}.(1 - e^{k_{3}.t}) & \mbox{if } k_{1} \ne k_{3} - k_{2} \\ \\ [DC]_{0}.\frac{k_{4}}{k_{3}}.(e^{k_{3}.t} - 1) + [MC]_{0}.\frac{k_{4}}{k_{3}}.(k_{3}.t - 1).(\frac{k_{2}}{k_{3}} - 1).(e^{k_{3}.t} - 1) & \mbox{if } k_{1} = k_{3} - k_{2} \end{cases} \]