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Modeling of the bacterial population growth

This part aims to predict the bacterial population growth on an ellipsoidal object - a fake lemon in practice - over time.

Our work will focus on two different aspects of the bacterial population growth : first, we will study the overall growth, exprimed in concentration of bacteria, then we will have a look on the growth of a single bacterial colony, coming from one bacterium.

Overall growth

We are here considering a bacterial population uniformly spread on a surface in the euclidian space, in crowd-free conditions and with unlimited food resource.

Pure birth process

In this part, we assume that :

• organisms do not die,
• they develop without interacting with each other,
• the birth rate, $\lambda$, is the same for all the organisms, regardless of their age and does not change with time.

Determininistic model

Let $N(t)$ denote the population size at time $t$.

Then in the subsequent small time interval of length $h$ the increase in population time due to a single organism is $\lambda\times h$ - i.e the rate $\times$ the time - so the increase in size due to all $N(t)$ organisms is $\lambda\times h\times N(t)$. Thus \[ N(t+h) = N(t) + \lambda h N(t) \] which on dividing both sides by h gives \[ \frac{N(t+h)-N(t)}{h} = \lambda N(t) \] Letting $h$ approach zero then yields the differential equation \[ \frac{dN(t)}{dt} = \lambda N(t) \] which integrates to give \[ N(t) = N_0 \exp{\lambda t} \] where $N_0$ denotes the initial population size at time $t=0$. This form for $N(t)$ is known as the Malthusian expression for population development, and shows that the simple rules we used rises to exponential growth.

Stochastic model

A deterministic model only give us an average solution of the problem. In order to take into account the unpredictability of biology, we need a stochastic approach. Usualy, the stochastic model converge to a limit which is similar to the deterministic model. A stochastic model is all the more pertinent as the initial bacterial population is small. It give us an array of probabilities, describing each possible state for the population at time t.

As in the deterministic model, we call $\lambda$ the birth rate : in a short time interval of length $h$ the probability that any particular cell will divide is $\lambda h$. Then for the population to be of the size $N$ at the time $t+h$, either it is of the size $n$ at time $t$ and no birth occurs in the subsequent short interval $(t,t+h)$, or else it is of size $N-1$ at time $t$ and exactly one birth occurs in $(t,t+h)$. In fact, by choosing $h$ sufficiently small we may ensure that the probability of more than one birth occuring is negligible. Since the probability of $N$ increasing to $N+1$ in $(t,t+h)$ is $(\lambda h N$, it follows that the probability of no increase in $(t,t+h)$ is $1- \lambda h N$. Similarly, the probability of $N-1$ increasing to $N$ in $(t,t+h)$ is $\lambda(N-1)h$. Thus on denoting \[ p_N(t)(t) = \mathbb{P}(population~is~of~size~N~at~time~t) \] we have \[ p_N(t+h) = p_N(t)\times\mathbb{P}(no~birth~in~(t,t+h)) + p_{N-1}(t)\times\mathbb{P}(one~birth~in~(t,t+h)) \] i.e \[ p_N(t+h) = p_N(t)\times(1-\lambda N h)+p_{N-1}(t)\times\lambda(N-1)h \] On dividing both sides by $h$ \[ \frac{p_N(t+h)-p_N(t)}{h} = - \lambda N p_N(t)+\lambda(N-1)p_{N-1}(t) \] and as he approaches zero this becomes \[ \frac{dp_N(t)}{dt} = -\lambda p_N(t)+\lambda(N-1)p_{N-1}(t) \] for $N=N_0,~N_0+1,...$.

We use the following phenomenological law suggest by Adolphe Fick in 1855:

In an homogeneous and isotropic environment, containing particles distributed inhomogeneously,appears spontaneously a volumetric flow density vector particle $\overrightarrow{J}(M,t) $. In any point $M$ in space, this vector is proportional to the gradient of the particle density $n(M,t)$. Mathematicaly, this relationship take the form: \[ \overrightarrow{J}(M,t) = - D \times \overrightarrow{grad} n(M,t) \qquad (1) \] where $D$ is the diffusion coefficient.

Referring to the article [1], the diffusion coefficient of oxygen in agarose is $ D = 0{,}256 \times 10^{-8} m^2 s^{-1} $.

• To simplify the problem, we consider that the diffusion of oxygen particle occurs only in one direction. So $\overrightarrow{J}(M,t) = J(x,t) \overrightarrow{e}_x $.
• Spatial variations in the density of particles are connected to spatial variations of the vector $\overrightarrow{J}(M,t)$ by the material's equation of conservation in presence of volume distribution of particle source $\sigma (x,t)$ (device which injects or subtracted particles to the system) :

\[ \frac{\partial n}{\partial t} (x,t) = - \frac{\partial J}{\partial x} (x,t) + \sigma (x,t) \qquad (2) \]

By replacing $(2)$ in $(1)$, we obtain the following equation of diffusion \[ \forall t, \forall x, \bigg( \frac{\partial}{\partial t} - D \frac{\partial^2}{\partial x^2} \bigg) n(x,t) = \sigma (x,t) \qquad (3) .\]

As our lemon is exposed to the ambient air, we stay in steady state where the source $ \sigma (x,t) $ is equal to $N_0$ the quantity of $O_2$ in the air.

To solve this equation, we use Fourier's analysis (+ d'explications)

\[ \forall x, \forall t>0, n(x,t) = \frac{N_0}{\sqrt{4 \pi D t}} exp \bigg(- \frac{x^2}{4 D t} \bigg) + \int_{0}^{t} \underbrace{N_0 * exp \bigg(- \frac{|x|^2}{4 D \tau} \bigg)}_{= 0 \text{ by symmetry of the gaussian distribution }} \frac{d\tau}{\sqrt{4 \pi D \tau} } \]

The average dispersion particle is given by the variance $\Delta x = \sqrt{2Dt}$. Using this formula, we deduct that oxygen will penetrate $3 \times 10^{-3} m$ in $1956.522 s = 32.6082 $ minutes.

References:

[1] A.C. Hulst, H.J.H. Hens, R.M. Buitelaar and J. Tramper, Determination of the effective diffusion coefficient of oxygen in gel materials in relation to gel concentration, Biotechnology Techniques Vol 3 No 3 199-204 (1989).

[2] Vincent Renvoizé, Physique PC-PC*, Cap Prepas, Pearson Education, 2010.