Team:Paris Saclay/Modeling/BacterialPopulationGrowth

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(Difference between revisions)
(Modeling of the bacterial population growth)
(Modeling of the bacterial population growth)
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Let $N(t)$ denote the population size at time $t$.
Let $N(t)$ denote the population size at time $t$.
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Then in the subsequent ''small'' time interval of length $h$ the ncrease in population time due to a single organism is $\lambda\times h$ - i.e the rate $\itmes$ the time - so the increase in size due to all $N(t)$ organisms is $\lambda\times h\times N(t)$. Thus
+
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
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\[ N(t+h) \]
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\[ N(t+h) = N(t) + \lambda h N(t) \]
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which on dividing both sides by h gives
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\[ \frac{N(t+h)-N(t)}{h} = \lambda N(t) \]
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Letting $h$ approach zero then yields the differential equation
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\[ \frac{dN(t)}{dt} = \lambda N(t) \]
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which integrates to give
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\[ N(t) = N_0 \exp{\lambda t} \]
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where $N_0$ denotes the initial population size at time $t=0$.

Revision as of 10:20, 10 September 2014

Contents

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 $t=0$.



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.