Team:RHIT/Modeling.html

We created a mathematical model to verify that obligate mutualism was theoretically possible. Since the S. cerevisiae and E. coli will depend on each other to survive, if one population dies the other will follow. From the model, we developed predictions about the initial concentrations of S. cerevisiae and E. coli needed for co-dependence, and devised more informed laboratory experiments. See the technical section for the model derivation or either section for the model's predictions.

The following ODEs describe our system $$ \frac{dE}{dt} = -k_1 E + k_3 L E $$ $$ \frac{dY}{dt} = -k_2 Y + k_4 E Y $$ $$ \frac{dL}{dt} = k_5 Y - k_6 E L$$ where
• E represents E. coli
• Y represents Yeast
• L represents Lactate
• t represents time
• Parameters $k_1$ through $k_6$ describe the population dynamics

To simplify the system further we assume that the concentration of lactate equilibrtates rapidly compared to the bacteria and yeast concentrations so that $$\frac{dL}{dt} = 0 \\ k_5 Y - k_6 E L = 0 \\ L = \frac{k_5 Y}{k_6 E}$$ Substituting this into the original model gives the following two by two ODE system $$\frac{dE}{dt} = -k_1 E + k_3 \frac{k_5}{k_6} Y \\ \frac{Y}{dt} = -k_2 Y + k_4 E Y \\$$ Before we can analyze the model, we must make it dimensionless. Let $k_1 = \alpha$, $k_3 \frac{k_5}{k_6} = \gamma$, $k_2 = \sigma$, and $k_4 = \phi$ so that the model becomes $$\frac{dE}{dt} = -\alpha E + \gamma Y \\ \frac{dY}{dt} = -\sigma Y + \phi E Y $$ Setting the dimensionless parameters to be $$ e = \frac{E}{\frac{\sigma}{\phi}} \\ y = \frac{Y}{\frac{\alpha \sigma}{\phi \gamma}} \\ s = \frac{t}{\frac{1}{\alpha}} $$ simplifies the model to \begin{align} \frac{de}{ds} = y - e \\ \frac{dy}{ds} = \theta y (e - 1) \end{align} where $\theta = \frac{\sigma}{\alpha} = \frac{k_2}{k_1}$ which is the ratio of the decay rate of S. cerevisiae to E. coli. We can now qualitivately analyze the model and make predictions for behavior. The dimensionless model $$ \frac{de}{ds} = y - e \\ \frac{dy}{ds} = \theta y (e - 1)$$ has steady states at ($0$, $0$) and ($1$, $1$). We get the system's Jacobian to be $$ J = \begin{bmatrix} -1 & 1 \\ \theta y & \theta e - \theta \end{bmatrix} \\ $$ At the steady state ($0$, $0$) the Jacobian, trace, and determinant are \begin{align} J(0, 0) &= \begin{bmatrix} -1 & 1 \\ 0 & -\theta \end{bmatrix} \\ tr(J(0,0)) &= -1 - \theta \\ det(J(0,0))&= \theta \end{align} Since the determinant is positive and the trace is negative, this steady state is always stable. This means that initial concentrations close to (0, 0) are predicted to die. $\\$ The other steady state is ($1,1$), and its Jacobian, trace, and determinant are \begin{align} J(1, 1) &= \begin{bmatrix} -1 & 1 \\ \theta & 0 \end{bmatrix} \\ tr(J(1, 1) &= -1 \\ det(J(1, 1) &= -\theta \end{align} Since both the trace and determinant are negative, this steady state is a saddle point. Therefore, depending on the initial conditions, the population will either die or survive as shown in the next section.

Our model has three unknown parameters: the initial concentration of yeast, the initial concentration of E. coli, and the dimensionless parameter theta. In wet lab, we can control the initial concentrations used, but we cannot control theta. In the figures below, we graphed the phase plane of our model and the solution to the system of differential equations. Drag and drop the blue dot to adjust the initial concentrations used and the slider to adjust theta to determine if the populations will survive or not!