Team:ULB-Brussels/modeling/30

From 2011.igem.org

(Difference between revisions)
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Modeling : Phase at 30°C </div>
Modeling : Phase at 30°C </div>
<div id="maint">
<div id="maint">
-
<h1>Transcriptional interference: computer simulation</h1>
+
<h1>Phase at $30^\circ$C on arabinose}</h1>
 +
 
 +
\label{Ph30}
 +
 
 +
<h2>{Transcriptional interference: computer simulation</h2>
 +
 
 +
\label{IntTranscr}
<p>
<p>
-
In this section, we study the interference in the transcription provoked by the simultaneous expression of the gene coding for the flippase (which is performed only for $\ldots\%$ at $30^\circ$C), and of the three genes $i$ for the insertion in the bacterial DNA (which is performed for $100\%$).
+
In this section, we study the interference in the transcription provoked by the simultaneous expression of the gene coding for the flippase (which is performed only for $\ldots\%$ at $30^\circ$C), and of the three genes $i$ for the Red recombinase (which is performed for $100\%$).
</p>
</p>
 +
<p>
<p>
-
...[Jo&Pierre]
+
\ldots[Jo\&Pierre]
</p>
</p>
-
 
-
 
-
<h1>Model</h1>
 
<h2>Preparation: electroporation and night culture</h2>
<h2>Preparation: electroporation and night culture</h2>
<p>
<p>
-
We electropore <emph>E. Coli</emph> with Pindel plasmids. Given that the plasmids include a resistance gene to ampiciline, we can see, by testing that resistance, which bacteria actually received a Pindel plasmid. One colony of those bacteria is then cultivated at $30^\circ$C in $10$ml, where she attains dew point (between $2\cdot10^9$ and $5\cdot10^9$ bacteria per ml). The solution is diluted $100$ to $1000$ times, then cultivated again, until we reach the optic density (OD) (at $600$nm) of $0.2$, which corresponds to approximatively $10^8$ bacteria. Those bacteria are then put in touch with arabinose at $30^\circ$C.
+
We electropore <em>E. coli</em> with Pindel plasmids. Given that the plasmids include a resistance gene to ampiciline, we can see, by testing that resistance, which bacteria actually received a Pindel plasmid. One colony of those bacteria is then cultivated at $30^\circ$C in $10$ml, where it attains saturation (between $2\cdot10^9$ and $5\cdot10^9$ bacteria per ml). The solution is diluted $100$ to $1000$ times, then cultivated again, until we reach an optic density (OD) (at $600$nm) of $0.2$, which corresponds to approximatively $10^8$ bacteria. Those bacteria are then put in touch with arabinose at $30^\circ$C.
</p>
</p>
-
<h2>Modelisation of the $30^\circ$C phase on arabinose</h2>
+
<h1>Model</h1>
 +
 
 +
<h2>Getting the equations</h2>
 +
\label{Mod30}
<p>
<p>
-
At the initial time ($t=0$), the amount of bacteria ($N(t)$) is $N_0:=N(0)\approx10^8$. It seems natural to use Verhulst's logistic model:
+
At the initial time ($t=0$), the concentration of bacteria ($N(t)$) is $N_0:=N(0)\approx10^8\mbox{bact}/\mbox{ml}$. It seems natural to use Verhulst's logistic model
\begin{equation}
\begin{equation}
\dot N=k_NN\left(1-\frac N{N_{max}}\right)
\dot N=k_NN\left(1-\frac N{N_{max}}\right)
\label{N30}
\label{N30}
\end{equation}
\end{equation}
-
where $N_{max}$ is the maximum amount of bacteria that the culture environment is able to contain and where $k_N$ corresponds to the growth rate one would observe in the limit where the saturation would be inexistent. In our case, the saturation density slightly exceeds $1$ OD (at $600$nm), that is approximatively $N_{max}\approx 2\cdot10^9$. On the other hand, since our <emph>E. Coli</emph> ideally duplicate every $20$min, if we are far of the saturation ($N_{max}=\infty$), we obtain
+
where $N_{max}$ is the maximum concentration of bacteria that is possible in the culture environment and where $k_N$ corresponds to the growth rate one would observe in the limit where the saturation would be inexistent. In our case, the saturation density slightly exceeds $1$ OD (at $600$nm), that is approximatively $N_{max}\approx 2\cdot10^9{\mbox{bact}/\mbox{ml}$. On the other hand, since our <em>E. coli</em> ideally duplicate every $20$min, if we are far from the saturation ($N_{max}=\infty$), we obtain
\begin{equation}
\begin{equation}
-
\dot N=k_NN \Rightarrow N_0e^{k_Nt}=N(t)=N_02^{t/20\mbox{min}} \quad\Rightarrow k_N\approx \frac{\log{2}}{20\cdot60}\mbox{s}^{-1}.
+
\dot N=k_NN \Rightarrow N_0e^{k_Nt}=N(t)=N_02^{t/20\mbox{\footnotesize{min}}} \quad\Rightarrow k_N\approx \frac{\log{2}}{20\cdot60}\mbox{s}^{-1}.
\label{k_N}\end{equation}
\label{k_N}\end{equation}
</p>
</p>
<p>
<p>
-
At this point, all the bacteria contain Pindel. At $30^\circ$C, RepA101 becomes active and is present in sufficient quantities to allow the plasmid's replication; the evolution of $E_{tot}$ and $E$ thus do not matter. However, we shall note that the amount of plasmids per bacterium cannot exceed a certain number $P_{max}\approx20$ (because the origin of replication of the plasmid is <em>low copy</em>). At initial time, the amount of Pindel plasmids per bacterium is $P_0:=P(0)\approx19$. Again, we naturally opted for a logistic model:
+
At this point, all the bacteria contain Pindel. At $30^\circ$C, RepA101 becomes active and is present in sufficient quantities to allow the plasmid's replication; the evolution of $E_{tot}$ and $E$ thus do not matter. However, we shall note that the amount of plasmids per bacterium cannot exceed a certain number $P_{max}\approx20$ (because the origin of replication of the plasmid is <em>low copy</em>). At initial time, the amount of Pindel plasmids per bacterium is $P_0:=P(0)\approx19$. Again, we naturally postulate a logistic model:
\begin{equation}
\begin{equation}
\dot P=k_PP\left(1-\frac{P}{P_{max}}\right).
\dot P=k_PP\left(1-\frac{P}{P_{max}}\right).
-
\end{equation}
+
\label{production}\end{equation}
</p>
</p>
<p>
<p>
-
By the same reasoning we used for $k_N$ (eq(\ref{k_N})), we compute $k_P\approx\frac{\log{2}}{11}\mbox{s}^{-1}$, since our plasmid replicates itself every $11$s (reference?). Moreover, we have to consider the dilution of those plasmids through the population, due to its increase. In that purpose, let us suppose for a moment that the plasmids don't replicate anymore; we then have $PN=\mbox{cst}$, thus
+
By the same reasoning we used for $k_N$ (eq(\ref{k_N})), we compute $k_P\approx\frac{\log{2}}{11}\mbox{s}^{-1}$, since our plasmid replicates itself every $11$s (reference?). Moreover, we have to consider the dilution of those plasmids through the population, due to its increase. In that purpose, let us suppose for a moment that the plasmids don't replicate any more; we then have $PN=\mbox{cst}$, thus
\begin{equation}
\begin{equation}
P=\frac{\mbox{cst}}N\quad\Rightarrow \dot P=-\mbox{cst}\frac{\dot N}{N^2}=-\frac{\dot N}NP.
P=\frac{\mbox{cst}}N\quad\Rightarrow \dot P=-\mbox{cst}\frac{\dot N}{N^2}=-\frac{\dot N}NP.
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\frac d{dt}(NP)=k_PNP\left(1-\frac P{P_{max}}\right),
\frac d{dt}(NP)=k_PNP\left(1-\frac P{P_{max}}\right),
\label{equNP}\end{equation}
\label{equNP}\end{equation}
-
which allows a convenient interpretation: $NP$, the total amount of Pindel plasmids, follows a logistic model but where the saturation is only due to $P$. This seems natural enough, as we will see. The evolution of the amount of plasmids has to be of the form
+
which allows a convenient interpretation: $NP$, the total amount of Pindel plasmids, follows a logistic model but where the saturation is only due to $P$. This seems quite natural, as we will see. The evolution of the amount of plasmids has to be of the form
\begin{equation}
\begin{equation}
\frac d{dt}(NP)=NP\cdot(b(N,P)-d(N,P))
\frac d{dt}(NP)=NP\cdot(b(N,P)-d(N,P))
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<p>
<p>
-
Arabinose activates Pbad (the promotor of the three-genes sequence $i$ on Pindel), in order that those $3$ genes are expressed. Keeping in mind that the expressed proteins naturally deteriorate, the easiest way to modelise the evolution of their quantity ($G_i$) is
+
Arabinose activates Pbad (the promotor of the three-gene sequence $i$ on Pindel), in order that those $3$ genes are expressed. Keeping in mind that the expressed proteins naturally deteriorate, the easiest way to modelise the evolution of their quantity ($G_i$) is
\begin{equation}
\begin{equation}
\dot{G_i}=C_iP-D_iG_i \quad (i=1,2,3)
\dot{G_i}=C_iP-D_iG_i \quad (i=1,2,3)
Line 678: Line 685:
<p>
<p>
-
The promotor of flippase is repressed by a thermo-sensible repressor, and, at $30^\circ$C, is only partially activated ($\ldots\%$); in addition, the transcription is hindered by a possible interference with the transcription of the genes $i$. By a computer simulation, we have been able to estimate $p_{simul}$, the probability that the flippase sequence gets entirely transcribed (see section \ref{IntTranscr}). Remark that at $30^\circ$C flippase is entirely active. Furthermore, if we take the natural deterioration of flippase in account, we can write
+
The promotor of flippase is repressed by a thermo-sensible repressor, and, at $30^\circ$C, is only partially activated ($\ldots\%$); in addition, the transcription is hindered by a possible interference with the transcription of the genes $i$. By a computer simulation, we have been able to estimate $p_{simul}$, the probability that the flippase sequence gets entirely transcribed (see section \ref{IntTranscr}). Remark that at $30^\circ$C flippase is entirely active. If we take furthermore the natural deterioration of flippase in account, we can write
\begin{equation}
\begin{equation}
\dot{F}=C_Fp_{simul}P-D_FF
\dot{F}=C_Fp_{simul}P-D_FF
\label{F30}\end{equation}
\label{F30}\end{equation}
-
where $C_F$ is the production rate of flippase by Pindel (in ideal conditions, at $100\%$ of its activity) and $D_F$ the natural deterioration rate of flippase. We estimated that $C_F\approx\ldots$ et $D_F\approx0.1$.
+
where $C_F$ is the production rate of flippase by Pindel (in ideal conditions, at $100\%$ of its activity) and $D_F$ the natural deterioration rate of flippase. We estimated that $C_F\approx\ldots$ et $D_F\approx0.01$.
</p>
</p>
<p>
<p>
-
We thereby obtain the following system (see eqs (\ref{P30}), (\ref{P30}), (\ref{Gi30}), (\ref{F30})):
+
We thereby obtain the following system (see eqs (\ref{N30}), (\ref{P30}), (\ref{Gi30}) and (\ref{F30})):
-
$$
+
\begin{numcases}{}
-
\left\{
+
-
\begin{array}{l}
+
\dot{N}=k_NN\left(1-\frac N{N_{max}}\right)\label{N30f}\\
\dot{N}=k_NN\left(1-\frac N{N_{max}}\right)\label{N30f}\\
\dot{P}=k_PP\left(1-\frac{P}{P_{max}}\right)-\frac{\dot N}NP\label{P30f}\\
\dot{P}=k_PP\left(1-\frac{P}{P_{max}}\right)-\frac{\dot N}NP\label{P30f}\\
\dot{G_i}=C_iP-D_iG_i \qquad (i=1,2,3)\label{Gi30f}\\
\dot{G_i}=C_iP-D_iG_i \qquad (i=1,2,3)\label{Gi30f}\\
\dot{F}=C_Fp_{simul}P-D_FF\label{F30f}
\dot{F}=C_Fp_{simul}P-D_FF\label{F30f}
-
\end{array}
+
\end{numcases}
-
\right.
+
-
$$
+
</p>
</p>
 +
 +
<h2>Solving the equations</h2>
<p>
<p>
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M(t)=\frac1{N_{max}}+(\frac1{N_0}-\frac1{N_{max}})e^{-k_Nt}
M(t)=\frac1{N_{max}}+(\frac1{N_0}-\frac1{N_{max}})e^{-k_Nt}
\end{equation}
\end{equation}
-
and thus
+
thus
\begin{equation}
\begin{equation}
-
N(t)=\frac{N_{max}N_0e^{k_Nt}}{N_0e^{k_Nt}+(N_{max}-N_0)}=N_0e^{k_Nt}\frac1{1+\frac{N_0}{N_{max}}\left(e^{k_Nt}-1\right)}\approx N_0e^{k_Nt}
+
N(t)=\frac{N_{max}N_0e^{k_Nt}}{N_0e^{k_Nt}+(N_{max}-N_0)}=N_0e^{k_Nt}\frac1{1+\frac{N_0}{N_{max}}\left(e^{k_Nt}-1\right)}\overset{\star}{\approx} N_0e^{k_Nt}
\end{equation}
\end{equation}
-
where the approximation stays valid for short times, that is  
+
where the approximation ($\star$) remains valid for short times, that is  
\begin{equation}
\begin{equation}
t\ll\frac1{k_N}\log{(\frac{N_{max}}{N_0}+1)}\approx5271\mbox{s}=1\mbox{h}27\mbox{min}51\mbox{s}.
t\ll\frac1{k_N}\log{(\frac{N_{max}}{N_0}+1)}\approx5271\mbox{s}=1\mbox{h}27\mbox{min}51\mbox{s}.
Line 719: Line 724:
<p>
<p>
-
Saturation is reached when $t\approx2\mbox{h}30\mbox{min}$, like we can see on the following graph (established with realistic constants)
+
Saturation is reached when $t\approx9000\mbox{s}=2\mbox{h}30\mbox{min}$, as we can see on the following graph (obtained for realistic values of the parameters)
-
\[\mbox{[insérer le graphique 1]}\]
+
\[\mbox{[insert graphic 1]}\]
</p>
</p>
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\dot P=k_PP\left(1-\frac{P}{P_{max}}\right)-k_N\frac{N_{max}-N_0}{N_0e^{k_Nt}+(N_{max}-N_0)}P
\dot P=k_PP\left(1-\frac{P}{P_{max}}\right)-k_N\frac{N_{max}-N_0}{N_0e^{k_Nt}+(N_{max}-N_0)}P
\end{equation}
\end{equation}
-
which can't be solved analytically.  However, we can solve it numerically using <em>Mathematica</em>: for realistic values of the parameters,
+
which cannot be solved analytically.  However, we can solve it numerically using <em>Mathematica</em>: for realistic values of the parameters,
-
\[\mbox{[insérer le graphique 2]}\]
+
\[\mbox{[insert graphic 2]}\]
 +
We observe that $P(t)\approx P_{max}$ as soon as $t\gtrsim50\mbox{s}$.
</p>
</p>
<p>
<p>
-
The two last equations, for $F$ and $G_i$ (eqs (\ref{F30f}) et (\ref{Gi30f})), can also be solved  via <em>Mathematica</em>: for realistic constants,
+
The two last equations, for $F$ and $G_i$ (eqs (\ref{F30f}) and (\ref{Gi30f})), can also be solved  via <em>Mathematica</em>: for realistic constants,
\begin{description}
\begin{description}
\item{for $F$:}
\item{for $F$:}
-
\[\mbox{[insŽrer le graphique 3]}\]
+
\[\mbox{[insert graphic 3]}\]
\item{for $G_i$:}
\item{for $G_i$:}
-
\[\mbox{[insŽrer le graphique 4]}\]
+
\[\mbox{[insert graphic 4]}\]
\end{description}
\end{description}
</p>
</p>
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<p>
<p>
-
It is important to point out that here, the solution of our model only presents a small sensitivity to the parameters around the estimated values: a small error on the parameters will only result in a small change in the solution, like we can observe if we vary the values of the parameters a little around their estimation.
+
It is important to point out that here, the solution of our model only presents a small sensitivity to the parameters around the estimated values: a small error on the parameters will only result in a small change in the solution, as we can observe if we vary the values of the parameters a little around their estimation. [mettre Žventuellement un lien vers une page avec tous les fichiers cdf]
</p>
</p>
-
 
</div>
</div>

Revision as of 18:13, 18 September 2011

Modeling : Phase at 30°C

Phase at $30^\circ$C on arabinose}

\label{Ph30}

{Transcriptional interference: computer simulation

\label{IntTranscr}

In this section, we study the interference in the transcription provoked by the simultaneous expression of the gene coding for the flippase (which is performed only for $\ldots\%$ at $30^\circ$C), and of the three genes $i$ for the Red recombinase (which is performed for $100\%$).

\ldots[Jo\&Pierre]

Preparation: electroporation and night culture

We electropore E. coli with Pindel plasmids. Given that the plasmids include a resistance gene to ampiciline, we can see, by testing that resistance, which bacteria actually received a Pindel plasmid. One colony of those bacteria is then cultivated at $30^\circ$C in $10$ml, where it attains saturation (between $2\cdot10^9$ and $5\cdot10^9$ bacteria per ml). The solution is diluted $100$ to $1000$ times, then cultivated again, until we reach an optic density (OD) (at $600$nm) of $0.2$, which corresponds to approximatively $10^8$ bacteria. Those bacteria are then put in touch with arabinose at $30^\circ$C.

Model

Getting the equations

\label{Mod30}

At the initial time ($t=0$), the concentration of bacteria ($N(t)$) is $N_0:=N(0)\approx10^8\mbox{bact}/\mbox{ml}$. It seems natural to use Verhulst's logistic model \begin{equation} \dot N=k_NN\left(1-\frac N{N_{max}}\right) \label{N30} \end{equation} where $N_{max}$ is the maximum concentration of bacteria that is possible in the culture environment and where $k_N$ corresponds to the growth rate one would observe in the limit where the saturation would be inexistent. In our case, the saturation density slightly exceeds $1$ OD (at $600$nm), that is approximatively $N_{max}\approx 2\cdot10^9{\mbox{bact}/\mbox{ml}$. On the other hand, since our E. coli ideally duplicate every $20$min, if we are far from the saturation ($N_{max}=\infty$), we obtain \begin{equation} \dot N=k_NN \Rightarrow N_0e^{k_Nt}=N(t)=N_02^{t/20\mbox{\footnotesize{min}}} \quad\Rightarrow k_N\approx \frac{\log{2}}{20\cdot60}\mbox{s}^{-1}. \label{k_N}\end{equation}

At this point, all the bacteria contain Pindel. At $30^\circ$C, RepA101 becomes active and is present in sufficient quantities to allow the plasmid's replication; the evolution of $E_{tot}$ and $E$ thus do not matter. However, we shall note that the amount of plasmids per bacterium cannot exceed a certain number $P_{max}\approx20$ (because the origin of replication of the plasmid is low copy). At initial time, the amount of Pindel plasmids per bacterium is $P_0:=P(0)\approx19$. Again, we naturally postulate a logistic model: \begin{equation} \dot P=k_PP\left(1-\frac{P}{P_{max}}\right). \label{production}\end{equation}

By the same reasoning we used for $k_N$ (eq(\ref{k_N})), we compute $k_P\approx\frac{\log{2}}{11}\mbox{s}^{-1}$, since our plasmid replicates itself every $11$s (reference?). Moreover, we have to consider the dilution of those plasmids through the population, due to its increase. In that purpose, let us suppose for a moment that the plasmids don't replicate any more; we then have $PN=\mbox{cst}$, thus \begin{equation} P=\frac{\mbox{cst}}N\quad\Rightarrow \dot P=-\mbox{cst}\frac{\dot N}{N^2}=-\frac{\dot N}NP. \label{dilution}\end{equation}

Combining both production (eq (\ref{production})) and dilution (eq(\ref{dilution})) effects, we get the evolution equation for $P$: \begin{equation} \dot P=k_PP\left(1-\frac P{P_{max}}\right)-\frac{\dot N}NP. \label{P30}\end{equation}

Remark that this equation can be written as follow: \begin{equation} \frac d{dt}(NP)=k_PNP\left(1-\frac P{P_{max}}\right), \label{equNP}\end{equation} which allows a convenient interpretation: $NP$, the total amount of Pindel plasmids, follows a logistic model but where the saturation is only due to $P$. This seems quite natural, as we will see. The evolution of the amount of plasmids has to be of the form \begin{equation} \frac d{dt}(NP)=NP\cdot(b(N,P)-d(N,P)) \end{equation} in term of a birth rate of new plasmids $b(N,P)$ and a death rate $d(N,P)$. The death rate is a priori constant and even zero in our case: $d(N,P)=d=0$. Regarding the birth rate, it has to diminish when $P$ increases, but is obviously unlinked with the amount of bacteria $N$; the easiest is then to postulate an affine function $b(N,P)=\alpha-\beta P$, so that we find \begin{equation} \frac d{dt}(NP)=NP(\alpha-\beta P) \end{equation} which is equivalent to (\ref{equNP}). This observation thus justifies our equation for $P$ (eq(\ref{P30})), initially obtained by heuristic reasoning.

Arabinose activates Pbad (the promotor of the three-gene sequence $i$ on Pindel), in order that those $3$ genes are expressed. Keeping in mind that the expressed proteins naturally deteriorate, the easiest way to modelise the evolution of their quantity ($G_i$) is \begin{equation} \dot{G_i}=C_iP-D_iG_i \quad (i=1,2,3) \label{Gi30}\end{equation} where $C_i$ is the production rate of the protein $i$ by the Pindel plasmid and $D_i$ the deterioration rate of that same protein. We estimated $C_i\approx\ldots$ and $D_i\approx\ldots$.

The promotor of flippase is repressed by a thermo-sensible repressor, and, at $30^\circ$C, is only partially activated ($\ldots\%$); in addition, the transcription is hindered by a possible interference with the transcription of the genes $i$. By a computer simulation, we have been able to estimate $p_{simul}$, the probability that the flippase sequence gets entirely transcribed (see section \ref{IntTranscr}). Remark that at $30^\circ$C flippase is entirely active. If we take furthermore the natural deterioration of flippase in account, we can write \begin{equation} \dot{F}=C_Fp_{simul}P-D_FF \label{F30}\end{equation} where $C_F$ is the production rate of flippase by Pindel (in ideal conditions, at $100\%$ of its activity) and $D_F$ the natural deterioration rate of flippase. We estimated that $C_F\approx\ldots$ et $D_F\approx0.01$.

We thereby obtain the following system (see eqs (\ref{N30}), (\ref{P30}), (\ref{Gi30}) and (\ref{F30})): \begin{numcases}{} \dot{N}=k_NN\left(1-\frac N{N_{max}}\right)\label{N30f}\\ \dot{P}=k_PP\left(1-\frac{P}{P_{max}}\right)-\frac{\dot N}NP\label{P30f}\\ \dot{G_i}=C_iP-D_iG_i \qquad (i=1,2,3)\label{Gi30f}\\ \dot{F}=C_Fp_{simul}P-D_FF\label{F30f} \end{numcases}

Solving the equations

In order to solve the first equation (eq(\ref{N30f})), we pose $M=1/N$; the equation then reads \begin{equation} \dot M=-\frac{\dot N}{N^2}=-k_N\left(\frac1N-\frac1{N_{max}}\right)=-k_NM+\frac {k_N}{N_{max}} \end{equation} and easily get solved to give \begin{equation} M(t)=\frac1{N_{max}}+(\frac1{N_0}-\frac1{N_{max}})e^{-k_Nt} \end{equation} thus \begin{equation} N(t)=\frac{N_{max}N_0e^{k_Nt}}{N_0e^{k_Nt}+(N_{max}-N_0)}=N_0e^{k_Nt}\frac1{1+\frac{N_0}{N_{max}}\left(e^{k_Nt}-1\right)}\overset{\star}{\approx} N_0e^{k_Nt} \end{equation} where the approximation ($\star$) remains valid for short times, that is \begin{equation} t\ll\frac1{k_N}\log{(\frac{N_{max}}{N_0}+1)}\approx5271\mbox{s}=1\mbox{h}27\mbox{min}51\mbox{s}. \end{equation}

Saturation is reached when $t\approx9000\mbox{s}=2\mbox{h}30\mbox{min}$, as we can see on the following graph (obtained for realistic values of the parameters) \[\mbox{[insert graphic 1]}\]

The equation for $P$ (eq(\ref{P30f})) then becomes \begin{equation} \dot P=k_PP\left(1-\frac{P}{P_{max}}\right)-k_N\frac{N_{max}-N_0}{N_0e^{k_Nt}+(N_{max}-N_0)}P \end{equation} which cannot be solved analytically. However, we can solve it numerically using Mathematica: for realistic values of the parameters, \[\mbox{[insert graphic 2]}\] We observe that $P(t)\approx P_{max}$ as soon as $t\gtrsim50\mbox{s}$.

The two last equations, for $F$ and $G_i$ (eqs (\ref{F30f}) and (\ref{Gi30f})), can also be solved via Mathematica: for realistic constants, \begin{description} \item{for $F$:} \[\mbox{[insert graphic 3]}\] \item{for $G_i$:} \[\mbox{[insert graphic 4]}\] \end{description}

As soon as $t\approx\ldots$, $G_i$ reaches its asymptotic maximum $C_iP_{max}/D_i\approx\ldots$ and $F$ reaches its asymptotic maximum $C_Fp_{simul}P_{max}/D_F\approx\ldots$.

It is important to point out that here, the solution of our model only presents a small sensitivity to the parameters around the estimated values: a small error on the parameters will only result in a small change in the solution, as we can observe if we vary the values of the parameters a little around their estimation. [mettre Žventuellement un lien vers une page avec tous les fichiers cdf]

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