Team:British Columbia/Modeling

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

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

The model in it's entirety involves a system of ten linear first order ordinary differential equations, subject to ten initial conditions and one boundary condition. When introducing an expression, a general form will be expressed leaving our defined system to be shown at the end.

To start we make a few assumptions to do with bacterial growth: 1. Bacteria is grown in a batch culture and follows the Monod growth equation. 2. Bacteria with the CRISPR are nearly 100% immune to the specific phage infection. 3. Bacteria without CRISPR system are completely susceptible to phage infection.

We begin by showing the rate of change in the total bacterial growth for a given strain $\phi$ \begin{align} \frac{d\phi}{dt} = \frac{d\phi_i}{dt} + \frac{d\phi_u}{dt} + \delta_\phi \end{align} The infected bacterial is represented by $\phi_i$ and uninfected by $\phi_u$, the $d_\phi$ represents the decrease in population during lysis and is defined as a delta function smeared over a lysis period using a normal distribution. \begin{align} \delta_\phi &= N(\phi_i), &during \ lysis\\ &= 0, &otherwise \end{align}

The infected and uninfected cells are described by (4), however the constants differ. \begin{align} \frac{d\phi_n}{dt} = \big(1 - e^{-\alpha t}\big)\big( \mu_{\phi, n} - k_{d_{\phi,n}} \big) \phi_n \end{align}

$\mu_{\phi_n}$ $k_{d_{\phi, n}}$ represents the growth rate of strain $\phi$ with health $n$. Modelling of the lag phase is handled by a lag phase term: $(1- e^{-\alpha t})$, note that this function is identically 0 at the initial start time, and as time progresses the term tends to 1 (as the dampening effect on growth disappears.

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 infects 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 after lysis is expected to be: \begin{align} E(X_i) = X_u(1 - e^{-MOI}) \end{align}

We must also consider that during growth, the bacterial culture must use substrate to convert to biomass. In our growth experiments, the concentration of substrate limits the growth of bacteria, so we use Monod Kinetics to describe the growth rate of bacteria with respect to substrate concentration.

\begin{align} \mu_i = \mu_{i, max}\frac{S}{K_s S}\\ \mu_u = \mu_{u, max}\frac{S}{K_s S} \end{align}

Where $\mu_{i, max}$ and $\mu_{u, max}$ are the maximum growth rates of infected and uninfected bacteria respectively, $S$ denotes the substrate concentration and $K_s$ is the concentration of substrate when the growth rate is half of it's maximum. Substrate concentration also changes with time and this rate is depended on the concentration available and the concentration of bacteria.

\begin{align} \frac{dS}{dt} = \frac{\mu_u(S)}{Y}X_u + \frac{\gamma_i(S)}{Y}X_i \end{align} Where $Y$ is a yield term, and $\gamma_i$ is the utilization rate of substrate by the infected cells. We choose to represent $\gamma_i$ as a utilization instead of a growth because the growth is very difficult to measure since infected cells do not appear to grow. (15) is subject to the boundary conditions: \begin{align} S(0) = S_0\\ S(t) = S_t \end{align}

Thus we have a system of 4 ODE's with 8 eighen \begin{align*} &\frac{dX}{dt} = \frac{dX_u}{dt} + \frac{dX_i}{dt} + \delta\\ &\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\\ &\frac{dS}{dt} = \frac{\mu_u(s)}{Y}X_u + \frac{\gamma_i(s)}{Y}X_i \end{align*} Which are subject 5 boundary conditions: \begin{align*} &X(0) = X_0\\ &X(t) = X_t\\ &X_i(t) = X_u(1 - e^{-MOI})\\ &S(0) = S_0\\ &S(t) = S_t \end{align*} Since we have 5 boundary conditions and 4 ODE's we can use an optimization to determine our eigenvalues.