Revision as of 08:21, 27 September 2013

Modelling Objectives:

Predict the growth of recombinant E. coli cultures under phage cultures under phage predation.
Predict caffeine production based on initial starting number of viruses, (ie. multiplicity of infection; MOI).

Assumptions:

Bacteria is grown in a batch culture and follows the Monod growth equation.
Bacteria with the CRISPR are nearly 100% immune to the specific phage infection.
Bacteria without CRISPR system are completely susceptible to phage infection.
Yield coefficients and phage attachment coefficients are constants.
Bacterial phages do not decrease significantly in number during the course of an experiment.
All considered bacterial phages are lytic.

Bacteria Growth:

We begin with a populations balance for a given strain $X$, \begin{align} \frac{dX}{dt} = \frac{dX_i}{dt} + \frac{dX_u}{dt} \end{align} The infected bacteria is represented by $X_i$, uninfected bacteria by $X_u$ and $X = X_i + X_u$. We use the Monod equation to model the substrate limited bacterial growth. \begin{align} \mu_{X} = \mu_{max, X} \frac{S}{K_{S, X} + S} \end{align}

Where $\mu_{X}$ is the bacterial growth rate, $\mu_{max, X}$ is the maximum bacterial growth rate, $S$ is the growth limiting substrate and $K_s$ is the amount of substrate remaining when the growth rate is half of maximum. Due to environmental factors, some bacteria will die; the death rate ($k_d$) is assumed to be directly proportional to the bacteria population. A lag phase is modelled by multiplying a dampening term to the growth rate, namely $(1 - e^{-\alpha t})$, where $\alpha$ is defined as a proportionality constant. The slowing of growth from exponential to stationary is modelled with yet another dampening term to be multiplied to the growth: $exp\big[{-\big(\frac{X}{X_c}\big)^m\big]}$. This dampening term is related to the quorom sensing in bacteria, once the bacteria reach some characteristic concentration ($X_c$) the bacteria begin to slow down growth. Thus the uninfected cell growth is described by: \begin{align} \frac{dX_u}{dt} = \big( \mu_{X, u}e^{-\big(\frac{X}{X_c}\big)^m} - k_{d_{X,u}} \big) X_u\big(1 - e^{-\alpha t}\big) \end{align} For clarity, in this model, $m$, $X_c$, and $\alpha$ are all to be empirically determined and assumed constant. In reality, each constant is dependent on many variables, for example: $\alpha(temperature, pressure, [substrate], environmental \ conditions, bacterial \ strain, more)$. Similarly for infected cells: \begin{align} \frac{dX_i}{dt} = \big( \mu_{X, i}e^{-\big(\frac{X}{X_c}\big)^m} - k_{d_{X,i}} \big) X_i\big(1 - e^{-\alpha t}\big) - \delta \end{align}

However, a major difference is that the decay rate for the infected cells is dependent on the latency time of the virus and thus we introduce $\delta$. The function $\delta$ is defined as follows, \begin{align} \delta = &N(\tau)X_i , \ \ during \ lysis\\ & 0, \ \ \ \ \ \ \ \ \ \ \ \ \ otherwise \end{align} Here $N(\tau)$ is a normal distribution over some time $\tau$, in other words, it is expected that the infected population will lyse according to a normal distribution over the domain $(-\tau, \tau)$.

Viral Growth

It is now necessary to consider the phage population $V$ as it directly affects the amount of bacteria that are infected at any given time. To describe the phage population it is necessary to consider how many phages are released per cell lysis, this number is referred to as the burst size $(\beta)$. \begin{align} V = V_0 + X_i \beta ^{\frac{t}{LT}} \end{align} Where $V_0$ is the initial phage populations and $LT$ is the latency time (they time between infection and lysis). However, since lab techniques are tricky to measure the exact phage at various time points as well as the exact amount of infected bacteria, it is desired to simplify the model. It is reasonable to use a stepwise function that describes the amount of free phage present. Also, since the burst size is large, on the order of $100$, we see that the loss of current phage due to infection of cells is small (roughly 1%), thus to simplify computations we consider this to be negligible. With these deductions we arrive at the following: \begin{align} V = V_0 + E(X_i)\beta^{\big[\frac{t}{LT}\big]} \end{align} Where $E(X_i)$ is the expected infected population and $\big[\frac{t}{LT}\big]$ is the largest integer equal or less than $\frac{t}{LT}$. To find the expected infected population, the Poisson distribution is used based on the MOI. However, there is an efficiency of viral attachment $\epsilon \in (0, 1)$ (in fact the efficiency has been found to be a function of certain proteins [reference]). \begin{align} MOI = \frac{V}{X} \end{align}

\begin{align} E(X_i) = \epsilon X(1 - e^{-MOI}) \end{align}

Substrate utilization

The bacteria will need to use a substrate, and the depletion of this substrate is proportional to the growth of the uninfected bacteria and the amount of bacteria. Although the infected bacteria do not multiply when infected, the infected cells may use up substrate to gain the energy needed to replicate the phage. Although it may be the case that the bacteria simply recycle intracellular material, a substrate utilization term ($\gamma$) for the infected cells is utilized (if the bacteria do in fact recycle intracellular material, it follows that this value will be 0). Moreover, when cells lyse, the cell materials can be metabolized by other bacteria and will add to the amount of nutrients available. Thus, the equation for substrate utilization rate is:

\begin{align} \frac{dS}{dt} = - \big[ \frac{\mu_u}{Y_u}X_u + \frac{\gamma}{Y_i}X_i \big] + \phi \delta \end{align} Where $Y_u$ and $Y_i$ are yield constants and $\phi$ is the average amount of substrate released per infected cell during lysis. Note that $\delta$ is as described above.

We use the Poisson distribution to statistically determine the expected infected bacterial population based on the multiplicity of infection (MOI), where MOI is defined as the ratio of virus to bacteria. The multiplicity of infection (MOI) is defined as the ratio of bacteriophages to bacterial cells. \begin{align} MOI = \frac{V_{X}}{X} \end{align}

\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 (or after initial phage is added) is expected to be: \begin{align} E(X_i) = X(1 - e^{-MOI}) \end{align} Given an initial bacterial population, we can use the above equations to determine the population at a later time, however, the infected population is dependent on the viral population. Thus, it is necessary to understand how the phage population changes with time. Fundamentally, we expect viral growth to follow:

V(t) = \phi_i (t) \beta ^{\frac{\t}{LT}}

Where $LT$ is the latency time, $\beta$ is the burst size (average number of phage that are released per infected cell). However, to simplify experimentation, we model the viral population as follows:

In this case, we only see the phage population increase during lysis, and much like the $\delta_{\phi}$ function, the step is smeared using a normal distribution over time. The only remaining factor in our growth model is substrate utilization. For uninfected cells, substrate utilization is expected to be a function of the growth rate and population size of bacteria. For infected cells the apparent growth rate is low since the cells are not multiplying, however, they will be using the nutrients and energy sources to produce the phage, therefore we use another function $\gamma$ to explain substrate utilization of infected cells. The substrate utilization then becomes,

\begin{align} \frac{dS}{dt} = \sum_{k = 1}^m \Big[\frac{\mu_{X_k , i}}{y_{i, X_k}}X_{k, i} + \frac{\mu_{X_{k, i} , u}}{y_{X_k , u}}X_{k, u}\Big] + \sum_{k = 1}^m \Big[\frac{\gamma_{X_k , i}}{y_{i, X_k}X_{k, i} }\Big] \end{align} where m is the number of strains

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(1-e^{-\alpha t})\\ &\frac{dX_i}{dt} = \big(\mu_i - k_{d_i}\big)X_i(1-e^{-\alpha t}) \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.

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.

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.

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.

Team:British Columbia/Modeling

From 2013.igem.org