Repressilator

A repressilator is a feedback loop in which each gene suppresses the next one. The following diagram shows an implementation of a repressilator.

Questions about the repressilator

1. When does the repressilator oscillate (or not)?
2. What are the design requirements?

Model

\[\begin{align*} x &\equiv \text{mRNA} \\ y &\equiv \text{Protein} \end{align*}\]

The reactions are the following:

\[\begin{align*} x_1 &\rightarrow y_1 \ \ \ \ \ \ y_1 \dashv x_2 \\ x_2 &\rightarrow y_2 \ \ \ \ \ \ y_2 \dashv x_3 \\ x_3 &\rightarrow y_3 \ \ \ \ \ \ y_2 \dashv x_1 \\ \end{align*}\] \[\dot{x}_j = f(y_{j-1}) - \mu x_j \\ \dot{y}_j = k x_j - \nu y_j \\ j \in \{1, 2, 3\}\] \[f(y) = g_0 + g \frac{K_d^n}{K_d^n+y^n}\]

We will simplify these equations by writing them in terms of dimensionless variables. So instead of having the derivatives in terms of \(d/dt\) let’s use \(d/d(\mu t)\).

After some algebra we get the following dimensionless rate equations

\[\begin{align*} \frac{d}{d(\mu t)} \frac{k}{\nu} \frac{x_j}{K_d} &= \frac{k}{\nu} \frac{g_0}{\mu K_d} + \frac{k}{\nu} \frac{g}{\mu K_d} \frac{1}{ \left[ 1 + \left(\frac{y_{j-1}}{K_d} \right)^n \right] } - \frac{k}{\nu} \frac{x_j}{K_d} \\ \frac{d}{d(\mu t)} \frac{y_j}{K_d} &= \frac{\nu}{\mu} \left[ \frac{k}{\nu} \frac{x_j}{K_d} - \frac{y_j}{K_d} \right] \end{align*}\]

We can make these equations a bit cleaner my making the following definitions

\[\tau \equiv \mu t \ , \ \ \ \ \beta \equiv \frac{\nu}{\mu} \\ m_j \equiv \frac{k}{\nu} \ , \ \ \ \ p_j \equiv \frac{y_j}{K_d} \\ \alpha_0 \equiv \frac{kg_0}{\nu\mu K_d} \ , \ \ \ \ \alpha \equiv \frac{kg}{\nu\mu K_d}\]

So the reduced system is…

\[\Rightarrow \begin{cases} \frac{dm_j}{d\tau} &= \alpha_0 + \frac{\alpha}{1+p_{j-1}^n} - m_j \\ \frac{dp_j}{d\tau} &= \beta \left( m_j - p_j \right) \end{cases}\]

Assume that the mRNA dynamics are faster than the protein dynamics.

\[\begin{align*} \mu >> \nu \ &\Rightarrow \beta << 1 \\ &\Rightarrow \frac{dm_j}{d\tau} \approx 0 \\ &\Rightarrow m_j = \alpha_0 + \frac{\alpha}{1+p_{j-1}^n} \end{align*}\] \[\therefore \underbrace{ \frac{dp_j}{d( \underbrace{ \beta\tau) }_{\tau'} } }_{ \dot{p}_j} = \underbrace{ \alpha_0 + \frac{\alpha}{1+p_{j-1}^n} }_{g(p_{j-1})} - p_j\]

Let p, q, and r represent the rescaled protein concentrations.

\[p_j \in \{ p, q, r\}\] \[\Rightarrow \begin{cases} \dot{p} &= g(r) - p \\ \dot{q} &= g(p) - q \\ \dot{r} &= g(q) - r \end{cases}\]

Like in previous lectures, we will now look for fixed points in the reduced system.

\[\frac{dp_j}{d\tau'} = 0 \\ \Rightarrow g(r^*) = p^* \ , \ \ g(p^*) = q^* \ , \ \ g(q^*) = r^*\]

By symmetry all the fixed points are equal to one another.

\[p^* = q^* = r^* \equiv s\] \[g(s) = s = \alpha_0 + \frac{\alpha}{1+s^n}\]

Assume that \(\alpha_0 \to 0\).

\[\Rightarrow s = \frac{\alpha}{1+s^n} \ \ \text{or} \ \ \ \alpha = s ( 1+s^n )\]

Is the fixed point stable or unstable?

First, assume a small deviation from s in order to linearize the system.

\[p = s + \delta p \ , \ \ q = s + \delta q \ , \ \ r = s + \delta r\] \[\begin{align*} \Rightarrow \delta\dot{p} &= g(s+\delta r) - (s + \delta r) \\ \delta\dot{p} &\approx g(s) + \delta r g'(s) -s - \delta p \\ \delta\dot{p} &= g'(s) \delta r - \delta p \end{align*}\] \[g'(s) = \frac{ - n s^n}{1+s^n}\] \[g'(s) < 0 \ \text{ so let } \ g'(s) \equiv -\gamma \ \text{ with } \ \gamma > 0 \ .\] \[\underbrace{ \frac{d}{d\tau'} \begin{bmatrix} \delta p\\ \delta q \\ \delta r \end{bmatrix} }_{ \dot{\vec{\delta}}} = \underbrace{ \begin{bmatrix} -1 &0 &-\gamma \\ -\gamma &-1 &0 \\ 0 &-\gamma &-1 \\ \end{bmatrix} }_J \ \underbrace{ \begin{bmatrix} \delta p\\ \delta q \\ \delta r \end{bmatrix} }_{ \vec{\delta}}\] \[\dot{ \vec{ \delta} } = J \vec{ \delta} \Rightarrow \vec{ \delta }(\tau') = \vec{ \delta}(0) e^{J \tau'}\]

Let’s find the eigenvalues of the matrix J. Since J is a 3 x 3 matrix it will have three eigenvalues.

\[\text{eig}(J) = \{ \lambda_1 , \lambda_2 , \lambda_3 \}\]

The eignevalues must have the form described below, where \(a\), \(b\), and \(c\) are real numbers.

\[\lambda_1 = a \ , \ \ \lambda_{2,3} = b \pm ic\]

If \(c = 0\)

\[\begin{align*} a, b < 0 &\rightarrow \text{stable} \\ \text{else} &\rightarrow \text{unstable} \end{align*}\]

If \(c \neq 0\)

Two of the eignevalues are complex.

\[\begin{align*} a, b < 0 &\rightarrow \text{stable with decaying oscillations} \\ \text{else} &\rightarrow \text{unstable} \end{align*}\]

The unstable case can exhibit several different behaviors.

• Growing oscillations
• Limit cycle
• Chaos

In the case of the repressilator we get a limit cycle.

Let’s find the eigenvalues for this case.

\[\det(J-\lambda I) = 0 \\ \Rightarrow (\lambda+1)^3 + \gamma^3 = 0 \\ \lambda + 1 = \gamma \ (-1)^{1/3}\]

We can write the factor of negative one in terms of a complex exponential.

\[-1 = e^{(2n+1)i\pi} \ \ \text{with} \ n = 0,1,2 \\ \Rightarrow \lambda + 1 = \gamma \ e^{ \left( \frac{2n+1}{3} \right) i\pi} \\ \lambda + 1 = \left\{ -\gamma, \frac{\gamma}{2} \pm i \gamma \frac{ \sqrt{3}}{2} \right\}\] \[\lambda_1 = -(\gamma+1) < 0 \\ \lambda_{2,3} = \frac{\gamma}{2} - 1 \pm i \gamma \frac{ \sqrt{3}}{2}\] \[\begin{align*} \text{If} \ \ \frac{\gamma}{2} - 1 < 0 &\Rightarrow \text{stable} \\ \text{If} \ \ \frac{\gamma}{2} - 1 > 0 &\Rightarrow \text{unstable (oscillations)} \end{align*}\]

For the case of oscillations

\[\frac{\gamma}{2} - 1 > 0 \ \rightarrow \ \gamma > 2\] \[\frac{ n s^n }{1+s^n} > 2 \\ (n-2)s^n > 2\]

Therefore \(n>2\) is required, i.e. we require cooperative repression. Recall that the expression for \(\alpha\) is

\[\alpha = s(1+s^n)\]

So the above constraint leads to the following inequalities which will constrain the possible values of \(\alpha\).

\[s^n > \frac{2}{n-2} \ , \ \ s > \left( \frac{2}{n-2} \right)^{1/n}\] \[\Rightarrow \alpha > \alpha_c = \left( \frac{2}{n-2} \right)^{1/n} \left( \frac{n}{n-2} \right)\]

This means that repression must be sufficiently strong in order for oscillations to occur.

\[\alpha \approx \text{height of } f(y)\]

So the hill function must be sufficiently steep and tall in order to get oscillations.

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