Population Dynamics by Probabilities
A Population Dynamics Model can be fitted in our system. Theoretically, we had first planned a probabilistic model:
When some new plasmids are genetically introduced into the cytoplasm of E.Coli bacteria, this doesn't garantee that the future copies will contain it. Indeed, these plasmids can be lost after cell division or replication, so it is interesting to study a model based on the different possibilities of plasmid combinations in bacteria, like in the studies of mutations in E.Coli.
Some crasy cases exist in theorical biology, like the Fibonnacci sequence, where the (n) term depends of the previous (n-1) and (n-2) terms. We will contruct here an original new recurrence sequence based on the same idea, but with five modifiable parameters.
$Set$ $of$ $Parameters$
Our parameters are $\hspace{0.12cm}\alpha$, $\beta$, $\gamma$, $\delta$, $\epsilon .\hspace{0.12cm}$ The values of these parameters are between 0 and 1 because we work now with probabilities. The three first parameters correspond to the evolution of bacteria with T$\scriptsize\&$A plasmids and the two last correspond to the evolution of bacteria which only get Antotixin plasmids.
$Hypothesis$
A Probabilistic Model is useful because easily undertsood, but necessits some assumptions to make sense.
Please discover the hypothesis of a new theorem after the following example.
$A$ $First$ $Example$
What happens when four kinds of plasmids are ingered at the same rate by E.Coli?
A first example [Fig m2] opens the game of probabilities.
Figure m2 : A first example of bacteria generations at same acceptation plasmid rate $\hspace{0.12cm}(\alpha = \beta = \gamma$, $\hspace{0.08cm}\delta = \epsilon )\hspace{0.12cm}$
for a Toxin (T), an Antitoxin (A), the two (T&A) or no plasmid (-).
The number on the right of the letter G indicates the generation number.
We remark that we will converge to finally obtain bacteria without Toxin or Antitoxin.
Now appears the following: And what happens when four kinds of plasmids are ingered at the different rates by E.Coli?
To study this, we have build our initial five-to-five generations theorem.
$Theorem$
$\small If \hspace{0.12cm}we\hspace{0.12cm} define$
\begin{array}.
i , j , n \hspace{0.12cm} \subset \hspace{0.12cm} \mathbb{N} \\
\alpha , \beta , \gamma , \delta , \epsilon \hspace{0.12cm} \subset \hspace{0.12cm} \mathcal{Q}^{+}_{0}\\
\alpha + \beta + \gamma = \mathtt{1} = \delta + \epsilon
\end{array}
$\small Then$ $(\small \forall$ $\small n > 2$$)$
\begin{equation}
\left( \sum_{i=0}^{n-1} \gamma^{i} \right)\hspace{0.08cm} \alpha \hspace{0.12cm}+\hspace{0.12cm}
\left( \sum_{i,j | (i+j) \leq (n-2)}^{n-2} \gamma^{i} \epsilon\hspace{0.02cm}^{j} \right) \hspace{0.08cm} \delta \hspace{0.06cm}\beta \hspace{0.12cm}+\hspace{0.12cm}
\left( \sum_{i,j | (i+j) = (n-1)}^{n-1} \gamma^{i} \epsilon\hspace{0.02cm}^{j} \right) \hspace{0.08cm} \beta \hspace{0.12cm}+\hspace{0.12cm}
\gamma^{n} \hspace{0.12cm}=\hspace{0.12cm} \mathtt{1}
\end{equation}
Interested to prove this? Please, practice first, and if you have some difficulty to do it,
the DigitalStudentStaff will be ready to send you the soluce.
$Control$ $on$ $Antitoxin$ $plasmids$
When we modify the parameters, we can bifurcate between three classes of dynamics for the Antitoxin plasmids {$monotonic-$, $constant$ $generation$, $monotonic+$}. There's a convergence for T$\scriptsize\&$A plasmids
and the (n) serie containing the sum of A and T$\scriptsize\&$A terms is always crescent.
Some choices of parameters are now illustrated, but not all the possibilities are here described and we hope the reader will invent himself new combinations, respecting the previous rules (cf Theorem).
One example by case:
{$\alpha , \beta , \gamma , \delta , \epsilon \hspace{0.03cm}$} $\hspace{0.1cm}=\hspace{0.1cm}$
{$\hspace{0.03cm} 1/6,1/3,1/2,1/4,3/4 \hspace{0.03cm}$} $\hspace{0.25cm}$ ($monotonic-$),
{$\alpha , \beta , \gamma , \delta , \epsilon \hspace{0.03cm}$} $\hspace{0.1cm}=\hspace{0.1cm}$
{$\hspace{0.03cm} 0,1/2,1/2,1/16,15/16 \hspace{0.03cm}$} $\hspace{0.25cm}$ ($constant$ $n=3$ $generation$),
{$\alpha , \beta , \gamma , \delta , \epsilon \hspace{0.03cm}$} $\hspace{0.1cm}=\hspace{0.1cm}$
{$\hspace{0.03cm} 1/12,1/6,3/4,1/10,9/10 \hspace{0.03cm}$} $\hspace{0.25cm}$ ($monotonic+$).
$\underline{Ex:}$ Show that
{$\alpha , \beta , \gamma , \delta , \epsilon \hspace{0.03cm}$} $\hspace{0.1cm}=\hspace{0.1cm}${$\hspace{0.03cm} 1/4,1/2,1/4,1/32,31/32 \hspace{0.03cm}$} and
{$\alpha , \beta , \gamma , \delta , \epsilon \hspace{0.03cm}$} $\hspace{0.1cm}=\hspace{0.1cm}${$\hspace{0.03cm} 0,1/2,1/2,1/32,31/32 \hspace{0.03cm}$}
$\hspace{0.25cm}$ are respectively ($constant$ $n=2$ $generation$) $\hspace{0.1cm}\scriptsize\&\hspace{0.1cm}$ ($constant$ $n=4$ $generation$).
$Steps$ $\scriptsize \&$ $example$
To be sure our modelling is understood, a solution will be detailed.
Now we will see that the following set of parameters corresponds to a $\hspace{0.1cm}monotonic-\hspace{0.1cm}$ plasmid dynamics:
{$\alpha , \beta , \gamma \hspace{0.04cm}$} $\hspace{0.1cm}=\hspace{0.1cm}$
{$\hspace{0.03cm} 1/6,1/3,1/2 \hspace{0.03cm}$}$\hspace{0.1cm}$
signify that if you have $\hspace{0.1cm}M\hspace{0.1cm}$ bacteria with T$\scriptsize\&$A plasmids in a generation $G_{m}$ in the colony, then you will obtain $\hspace{0.1cm}\alpha M$ (no plasmid A or T), $\beta M$ (only A plasmids)$\hspace{0.04cm}$ and $\hspace{0.06cm}\gamma M\hspace{0.1cm}$ (T$\scriptsize\&$A plasmids) bacteria in the generation $G_{m+1}$ in proportion to {$\hspace{0.03cm} 1/6,1/3,1/2 \hspace{0.03cm}$} $\hspace{0.25cm}$
& $\hspace{0.25cm}$
{$\hspace{0.03cm} \delta , \epsilon \hspace{0.03cm}$} $\hspace{0.1cm}=\hspace{0.1cm}$
{$\hspace{0.03cm} 1/4,3/4 \hspace{0.03cm}$}$\hspace{0.1cm}$
signify that if you have $\hspace{0.1cm}N\hspace{0.1cm}$ bacteria with Antoxin plasmids in a generation $G_{m}$, then you will obtain $\hspace{0.1cm}\delta N$ (no plasmid A or T)$\hspace{0.04cm}$ and $\hspace{0.06cm}\epsilon N$ (only A plasmids) in the generation $G_{m+1}$ in proportion to {$\hspace{0.03cm} 1/4,3/4 \hspace{0.03cm}$}.
You verify that the sum of the three first parameters and the sum of the two last parameters equals one and satisfy the hypothesis of our theorem, so you can apply it [You suppose that you begin initially with a "perfect population" in the petri box or in the bioreactor (i.e. that initially, there is 100% of bacteria with T$\scriptsize\&$A plasmids)].
This must gives what you will see in the following table (until $G\small 5$):
$\hspace{1.12cm}$Table m1 : An example of population dynamics results using our theorem for probabilities.
You write the elements of each column in decimal notation, in a sequence, like this (cf yellow/gold column):
0.00, 0.33, 0.42, 0.40, 0.34, ...
The Antitoxin plasmids column is important for the population dynamics because the first column always is crescent and the last always is decreasing in our five-to-five parameters system, but the yellow column can be crescent, temporary stationnary or decreasing, depending of the values assigned to the parameters.
You can remark that the two first elements of the sequence depend (burn-in) on the initial conditions (100% of bacteria with T$\scriptsize\&$A plasmids) and it is why you will truncate the sequence, to obtain correctly:
0.42 ($G\small 3$), 0.40, 0.34, 0.27, 0.22, 0.17, 0.13, 0.10, 0.07 ($G\small 11$), ...,
0.01 ($G\small 18$), etc.
This corresponds to a decreasing ($\equiv$ $monotonic-$) sequence in mathematical terms.
Finally, the interpretation is that in a long time, the bacteria that only have Antitoxin plasmids will disappear in the population and $the$ $system$ $ends$ $with$ $a$ $majority$ $of$ $bacteria$ $without$ $plasmids$ (because bacteria only with T plasmids die, and the proportions of bacteria with only A plasmids & with T$\scriptsize\&$A plasmids are decreasing with the time evolution of the population).
This particulary case corresponds well to what we had obtained in July on lab': a lot of bacteria without plasmids, so without the $\MyColi$ mechanim working in the E.Coli population.
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- Université Libre de Bruxelles -
