Comparison with the data of the Wetlab team

In previous sections, we have built a descriptive model of the biological processes, and estimated the requested constants through biological considerations. We observed that the solution of the model (for the obtained constants) was qualitatively corresponding to the results of the biological reasoning.

Although, it is difficult to experimentally measure the amount of each protein (Gam, Bet, Exo, Flp, RepA101TS and CI857) in the different conditions (arabinose, $30\circ$C, $42\circ$C, ?), it is interesting to compare the data generate by our model with that obtain experimentally by the Wetlab team. In this section, we will study what we really can measure, and we will compare our predictions with the data of the Wetlab team.

pINDEL dilution at $42^\circ$C

It is possible to determine what proportion $p_1(t)$ of bacteria is resistant to ampicillin, that is possessing at least one pINDEL plasmid. Let us see what our model predicts in that case.

We define the random variable $\tilde{P}(t)$ as the amount of pINDEL plasmids in a given bacterium, over time. Our model only allows us to predict its mathematical expectation: \begin{equation} P(t)=\mathbb{E}[\tilde{P}(t)]. \end{equation}

Having no information about the distribution of $\tilde{P}(t)$ besides its expectation $P(t)$, the most natural and simple solution is to postulate that it follows a Poisson distribution of parameter $P(t)$ \begin{equation} \tilde{P}(t) \sim \operatorname{Poiss}(P(t)) \quad\Leftrightarrow\quad \mathbb{P}[\tilde{P}(t)=k]=e^{-P(t)}\frac{P(t)^k}{k!}, \quad k\in\mathbb{N}. \end{equation} Our model then predicts for $p_1(t)$: \begin{equation} p_1(t)=\mathbb{P}[\tilde{P}(t)\ge1]=1-\mathbb{P}[\tilde{P}(t)=0]=1-e^{-P(t)}. \end{equation}

Since we know $P(t)$, we can draw, for every phase, the evolution graph of that proportion $p_1(t)$. The phase at $30^\circ$C does not bring us any useful information: we have indeed $p_1(t)\approx1$ as expected; we can draw the graph (fig(\ref{graphe12})).
$N_{max}=2\cdot10^9$bact/ml, $N_0=10^8$bact/ml, $k_N=\frac{\log2}{20\cdot60}\mbox{s}^{-1}$, $k_P=\frac{\log2}{14.4}\mbox{s}^{-1}$, $P_ 0=19$ and $P_{max}=20$.}

For the phase at $42^\circ$C, however, the result is not so trivial (fig(\ref{graphe13})).

For the dilution experiment, we obtain (fig(\ref{graphe14})).
This is obtained for $k_N=\frac{\log2}{20\cdot60}\mbox{s}^{-1}$, $C_E=\frac1{18}\mbox{s}^{-1}$, $A_E=\frac{\log{2}}{30}\mbox{s}^{-1}$, $k_P=\frac{\log2}{14.4}\mbox{s}^{-1}$, $E_0=5\cdot10^3$, $P_{max}=20$ and $P_0=19$.} If we compare this to the observations of the Wetlab team (fig(\ref{graphe14+})), we notice the global concordance between predictions and observation.
This is obtained for $k_N=\frac{\log2}{20\cdot60}\mbox{s}^{-1}$, $C_E=\frac1{18}\mbox{s}^{-1}$, $A_E=\frac{\log{2}}{30}\mbox{s}^{-1}$, $k_P=\frac{\log2}{14.4}\mbox{s}^{-1}$, $E_0=5\cdot10^3$, $P_{max}=20$ and $P_0=19$.}

Deletion of the chloramphenicol resistance gene by FLP-mediated site-specific recombination

Similarly as in the last sub-section (\ref{ampiciline}), we define the random variables $\tilde{F}(t)$ and $\tilde{G}_i(t)$, corresponding respectively to the amount of active FLP and proteins $i$ in one bacterium, over time. Our model only allows us to predict the mathematical expectation of those random variables: \begin{equation} F(t)=\mathbb{E}[\tilde{F}(t)] \quad \mbox{and} \quad G_i(t)=\mathbb{E}[\tilde{G}_i(t)]. \end{equation} Having no information about the distribution of those random variables besides their expectation, the most natural and simple solution is to postulate that they follow a Poisson distribution with their expectation as parameter: \begin{align} \tilde{F}(t) \sim \operatorname{Poiss}(F(t)) &\quad\Leftrightarrow\quad \mathbb{P}[\tilde{F}(t)=k]=e^{-F(t)}\frac{F(t)^k}{k!}, \quad k\in\mathbb{N}\\ \mbox{and}\quad\tilde{G}_i(t) \sim \operatorname{Poiss}(G_i(t)) &\quad\Leftrightarrow\quad \mathbb{P}[\tilde{G}_i(t)=k]=e^{-G_i(t)}\frac{G_i(t)^k}{k!}, \quad k\in\mathbb{N}. \end{align}

During the insertion step, the three Red recombinase proteins insert the X gene and the chloramphenicol resistance cassette in the bacterial DNA, while the FLP excise the resistance cassette. Over time, it is possible to measure the proportion $p_2(t)$ of bacteria that are resistant to chloramphenicol. Let us see what our model predicts: \begin{align} p_2^{(42)}(t)&\approx\mathbb{P}[\mbox{$\exists\tau\in[0,t]$: $\tilde{G}_i(\sigma)=0\forall\sigma<\tau$, $\tilde{G}_i(\tau)\ge1$ and $\tilde{F}(\sigma)=0\forall\sigma>\tau$}] \end{align}

This probability is too hard to compute exactly. However, it is reasonable to assume that $\tilde{F}(t)$ and $\tilde{G}_i(t)$ are almost surely increasing, as we know their respective expectations are increasing. We then obtain: \begin{align} p_2(t)&\approx\mathbb{P}[\mbox{$\exists\tau\in[0,t]$: $\tilde{G}_i(\tau^-)=0$, $\tilde{G}_i(\tau^+)\ge1$ and $\tilde{F}(t)=0$}]\\ &=\mathbb{P}[\tilde{F}(t)=0]\cdot\int_0^t{\mathbb{P}[\tilde{G}_i(\tau)=0,\tilde{G}_i(\tau+d\tau)\ge1]}\\ &=e^{-F(t)}\cdot\int_0^t{\mathbb{P}[\tilde{G}_i(\tau)=0,\tilde{G}_i(\tau+d\tau)\ge1]}\\ &=e^{-F(t)}\cdot\int_0^t{\mathbb{P}[\tilde{G}_i(\tau)=0]\mathbb{P}[\tilde{G}_i(\tau+d\tau)\ge1| \tilde{G}_i(\tau)=0]}\\ &=e^{-F(t)}\cdot\int_0^t\underbrace{\mathbb{P}[\tilde{G}_i(\tau)=0]}_{=e^{-G_i(\tau)}}(1-\underbrace{\mathbb{P}[\tilde{G}_i(\tau+d\tau)=0| \tilde{G}_i(\tau)=0]}_{=1-G_i'(\tau)d\tau})\\ &=e^{-F(t)}\cdot\int_0^te^{-G_i(\tau)}G_i'(\tau)d\tau\\ &=e^{-F(t)}\cdot\left(1-e^{-G_i(t)}\right)\\ &\overset{t\rightarrow\infty}{\longrightarrow}\exp{\left(-\frac{10\%p_{simul}C_FP_{max}}{D_F}\right)}\left(1-\exp{(-\frac{C_iP_{max}}{D_i})}\right)\\ &\qquad\approx0.013 \end{align}

Let us plot the graphic for realistic values of the parameters (fig(\ref{graphe15})). We observe that, for long times, that proportion is almost $0$, as the Wetlab team have observed. In order to be sure being able to find some bacteria with the resistance cassette, they should look around the time for which $p_2$ is maximum, i.e. $t=t_\star\approx10000$, at which $p_2(t_\star)\approx0.75$, so that a good deal of the bacteria then have received the X gene without having lost the resistance cassette.
This is obtained for $N_{max}=2\cdot10^9$bact, $N_0=10^8$bact, $k_N=\frac{\log2}{20\cdot60}\mbox{s}^{-1}$, $k_P=\frac{\log2}{14.4}\mbox{s}^{-1}$, $P_ 0=19$, $P_{max}=20$, $D_i=\frac{\log2}{60\cdot60}\mbox{s}^{-1}$, $C_i=\frac1{40}\mbox{s}^{-1}$, $D_F=\frac{\log2}{60\cdot60}\mbox{s}^{-1}$, $C_F=\frac1{24}\mbox{s}^{-1}$ and $p_{simul}=0.01$.}

Early on in the excision step, all our bacteria include in their genome the X gene followed by the resistance gene to chloramphenicol, inserted previously. Later, the active FLP will excise that resistance. In good approximation we can suppose that $\tilde F(t)$ is decreasing, as FLP degradation is more important than FLP production; the proportion of bacteria resistant to chloramphenicol can thus be estimated to be \begin{align} \mathbb{P}[\mbox{$\forall\tau\in[0,t]$: $\tilde{F}(\tau)=0$}]&=\mathbb{P}[\tilde{F}(0)=0]\\ &=e^{-F(0)}=\exp{\left(-\frac{10\%p_{simul}C_FP_{max}}{D_i}\right)}\\ &\approx0.013 \end{align}