Team:British Columbia/Modeling

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(Difference between revisions)
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Where $\mu_u$ and $\mu_i$ are the growth rate of uninfected and infected cells respectively. And $k_{d_u}$, $k_{d_i}$ are the decay rates of the uninfected and infected bacteria. To create a more complete model, we have chosen to model the bacteriophage population as well.
Where $\mu_u$ and $\mu_i$ are the growth rate of uninfected and infected cells respectively. And $k_{d_u}$, $k_{d_i}$ are the decay rates of the uninfected and infected bacteria. To create a more complete model, we have chosen to model the bacteriophage population as well.
\begin{align}
\begin{align}
-
P = X_i \beta ^{\frac{\tau}{LT}}
+
V = X_i \beta ^{\frac{\tau}{LT}}
\end{align}
\end{align}
Where $\beta$ is the burst size, and $LT$ is latency time. Due to the nature of the phages in our experiment (in that they have a very low or non-existent phage secretion rate), we can redefine our phage population as a stepwise function.
Where $\beta$ is the burst size, and $LT$ is latency time. Due to the nature of the phages in our experiment (in that they have a very low or non-existent phage secretion rate), we can redefine our phage population as a stepwise function.
\begin{align}
\begin{align}
-
\Delta P &= E(X_i)\beta, \ \ \ \ for \ lysis\\
+
\Delta V &= E(X_i)\beta, \ \ \ \ for \ lysis\\
&= 0,  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ otherwise
&= 0,  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ otherwise
\end{align}
\end{align}
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The multiplicity of infection (MOI) is defined as the ratio of bacteriophages to bacterial cells.
The multiplicity of infection (MOI) is defined as the ratio of bacteriophages to bacterial cells.
\begin{align}
\begin{align}
-
MOI = \frac{P}{X}
+
MOI = \frac{V}{X}
\end{align}
\end{align}
We use the Poisson distribution to statistically determine the expected infected bacterial population.
We use the Poisson distribution to statistically determine the expected infected bacterial population.
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\begin{align}
\begin{align}
P(n > 0) = 1 - P(0) = 1 - e^{-MOI}
P(n > 0) = 1 - P(0) = 1 - e^{-MOI}
 +
\end{align}
 +
Hence, number of infected bacteria is after lysis is expected to be:
 +
\begin{align}
 +
E(X_i) = X_u(1 - e^{-MOI})
\end{align}
\end{align}

Revision as of 22:23, 31 August 2013

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Reaction Kinetics: - Modelling the production of cinnamaldehyde, vanillin and caffeine production


Population dynamics: - Modelling the population of a bacteria in a batch reactor with phage dependency

\begin{align} \frac{dX}{dt} = \frac{dX_u}{dt} + \frac{dX_i}{dt} + \delta \end{align} Where $\frac{dX}{dt}$ is rate of change in bacteria population over time, $\frac{dX_u}{dt}$ and $\frac{dX_i}{dt}$ are the rates of change in the uninfected and infected populations respectively. $\Gamma$ is defined as a stepwise function. \begin{align} \delta &= - E(X_i), \ \ \ during \ phage \ caused \ lysis\\ &= 0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ otherwise \end{align} Where $E(X_i)$ is the statistically expected infected bacterial populations. We define the rate of change in population in () and (). \begin{align} &\frac{dX_u}{dt} = \big(\mu_u + k_{d_u}\big)X_u\\ &\frac{dX_i}{dt} = \big(\mu_i + k_{d_i}\big)X_i \end{align} Where $\mu_u$ and $\mu_i$ are the growth rate of uninfected and infected cells respectively. And $k_{d_u}$, $k_{d_i}$ are the decay rates of the uninfected and infected bacteria. To create a more complete model, we have chosen to model the bacteriophage population as well. \begin{align} V = X_i \beta ^{\frac{\tau}{LT}} \end{align} Where $\beta$ is the burst size, and $LT$ is latency time. Due to the nature of the phages in our experiment (in that they have a very low or non-existent phage secretion rate), we can redefine our phage population as a stepwise function. \begin{align} \Delta V &= E(X_i)\beta, \ \ \ \ for \ lysis\\ &= 0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ otherwise \end{align}

The multiplicity of infection (MOI) is defined as the ratio of bacteriophages to bacterial cells. \begin{align} MOI = \frac{V}{X} \end{align} We use the Poisson distribution to statistically determine the expected infected bacterial population. \begin{align} P(n) = \frac{MOI^n e^{-MOI}}{n!} \end{align} Where $n$ is the number of bacteriophages attacking a cell. If one or more than one virus collides with a cell, the cell will be infected. Therefore, we calculate the fraction of cells to not be infected. \begin{align} P(n > 0) = 1 - P(0) = 1 - e^{-MOI} \end{align} Hence, number of infected bacteria is after lysis is expected to be: \begin{align} E(X_i) = X_u(1 - e^{-MOI}) \end{align}