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.

In wet lab, we are planning on co-culturing our strains of yeast and E. coli. We envision the species forming a multicellular unit. We want the mathematical model to help inform us of the initial concentrations of yeast and E. coli that we should use in laboratory so that the populations will surivive.

In the deriving the model, we would like to account for the concentration of yeast, which we will represent with Y, the concentration of E. coli, which we will represent with E, and the concentration of lactate, which will we will represent with L. We will use the law of mass action to derive the mathematical model so we will use arbitrary constants to describe the population kinetics. We will denote them with K.

First, we will consider the interaction between the yeast and E. coli. Since to survive yeast needs alpha mating factor that the E. coli presents, we will assume that the yeast grow at rate proportional to the rate at which the yeast and E. coli meet. Mathematically, this can be described with the birth-term K Y E.

Likewise, since the E. coli needs lactate to survive, we will assume that the E. coli and lactate grow at a rate proportional to the rate at which the E. coli and lactate meet. This can be described with the birth term K E L.

Since the yeast or E. coli will die in the absence of each other, we will assume that the yeast and E. coli die at rate proportional to their own concentrations. These decay rates can be described with the terms -K Y and - K E respectively.

We must also consider what happens to the lactate. The lactate is secreted by the yeast, so we will assume that the lactate grow at a rate proportional to the concentration of yeast. This can be described with the birth term K Y L. On the other hand, the lactate is being consumed by the E. coli, so it will be degraded at a rate proportional to the rate at which the E. coli and lactate meet. This can be described with the death term - K E L.

The following ODEs describe this: $$\frac{dE}{dt} = -k_1 E + k_3 L E $$ $$ \frac{dY}{dt} = -k_2 Y + k_4 E Y $$ $$ \frac{dL}{dt} = k_5 Y - k_6 E L $$ where
• E represents the concentration of E. coli
• Y represents the concentration yeast
• L represents the concentration of 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 system of two ODE $$\frac{dE}{dt} = -k_1 E + k_3 \frac{k_5}{k_6} Y \\ \frac{Y}{dt} = -k_2 Y + k_4 E Y \\$$ 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 $$\frac{dE}{dt} = -\alpha E + \gamma Y \\ \frac{dY}{dt} = -\sigma Y + \phi E Y $$ Defining 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!

You will discover that there are certain thresholds that our initial concentrations of yeast andE. coli must be greater than so that the population survives.