Team:ULB-Brussels/modeling/30

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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 amount of bacteria ($N(t)$) is $N_0:=N(0)\approx10^8$. It seems natural to use Verhulst's logistic model:
..math:: \dot N=k_NN\left(1-\frac N{N_{max}}\right)
..math:: \dot N=k_NN\left(1-\frac N{N_{max}}\right)
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:label: N30
+
:label:`N30`
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 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

Revision as of 12:18, 18 September 2011

Modeling : Phase at 30°C

Interference in the transcription: computer simulation

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 $100\%$).

...[Jo&Pierre]

Model

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 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.

Modelisation of the $30^\circ$C phase on arabinose

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: ..math:: \dot N=k_NN\left(1-\frac N{N_{max}}\right) :label:`N30` 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 E. Coli ideally duplicate every $20$min, if we are far of 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 \textit{low copy}). At initial time, the amount of Pindel plasmids per bacterium is $P_0:=P(0)\approx19$. Again, we naturally opted for a logistic model: \begin{equation} \dot P=k_PP\left(1-\frac{P}{P_{max}}\right). \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 anymore; 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 natural enough, 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 \textit{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-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 \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. Furthermore, if we take 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 $D_F\approx0.1$.

We thereby obtain the following system (see eqs :eq:N30 , (\ref{P30}), (\ref{Gi30}), (\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}

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} and 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)}\approx N_0e^{k_Nt} \end{equation} where the approximation stays 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 $t\approx2\mbox{h}30\mbox{min}$, like we can see on the following graph (established with realistic constants) \[\mbox{[insérer le graphique 1]}\]

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 can't be solved analytically. However, we can solve it numerically using \textit{Mathematica}: for realistic values of the parameters, \[\mbox{[insérer le graphique 2]}\]

The two last equations, for $F$ and $G_i$ (eqs (\ref{F30f}) et (\ref{Gi30f})), can also be solved via \textit{Mathematica}: for realistic constants, \begin{description} \item{for $F$:} \[\mbox{[insŽrer le graphique 3]}\] \item{for $G_i$:} \[\mbox{[insŽrer le graphique 4]}\] \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, like we can observe if we vary the values of the parameters a little around their estimation.

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