Team:British Columbia/Modeling

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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}