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Team:British Columbia/Modeling
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\begin{align} | \begin{align} | ||
| - | \frac{dS}{dt} = \frac{\mu_u( | + | \frac{dS}{dt} = \frac{\mu_u(S)}{Y}X_u + \frac{\gamma_i(S)}{Y}X_i |
\end{align} | \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: | 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: | ||
| Line 66: | Line 66: | ||
S(t) = S_t | S(t) = S_t | ||
\end{align} | \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 U 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*} | ||
Revision as of 03:53, 8 September 2013
iGEM Home
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. $\delta$ 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 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 U 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*}
