Lecture 18

Autoregulation

We will continue our discussion of autoregulating genetic networks. Specifically, we will continue talking about autorepression.

Autorepression

Recall that there are two advantages of autorepression.

  1. Reduces response time.
  2. Buffers fluctuations.

Noise Reduction (buffer fluctuations)

Consider a small fluctuation \(\epsilon\) from the steady-state concentration \(x^*\).

\[x = x^* + \epsilon\] \[\dot{x} = \dot{\epsilon} = \underbrace{f_-(x^*+\epsilon)-r(x^*+\epsilon)}_{F(x^*+\epsilon)}\]

Let’s make a Taylor series expansion of the time derivative…

\[\dot{\epsilon}=\underbrace{F(x^*)}_{=0}+\epsilon F'(x^*)+\frac{1}{2}\epsilon^2F''(x^*)+...\]

We neglect terms of order \(\epsilon^2\).

\[\Rightarrow \dot{\epsilon}=F'(x^*)\epsilon \\ \therefore \epsilon(t)=\epsilon(0)e^{F'(x^*)t}\] \[\begin{align*} F'(x^*)<0 &\Rightarrow \text{ decay (stable)} \\ F'(x^*)>0 &\Rightarrow \text{ growth (unstable)} \end{align*}\]

In our case \(F'(x^*)<0\). So fluctuations in the number of proteins will decay exponentially leading to a stable stead-state concentration of protein.

Now lets derive the stochastic model for autorepression gene regulation.

Stochastic Model of Autorepression

\[\dot{p}_0 = rp_1-f_0p_0 \\ \dot{p}_n = f_{n-1}p_{n-1}+r(n+1)p_{n+1}-f_np_n-rnp_n \ \ \text{ for n}\geq\text{1}\] \[f_n = k \frac{n^h_0}{n^h_0+n^h} \\ \left[f_n\right] = \frac{1}{\text{time}}\]

Steady-State Solution

In steady the time derivative of the probabilities will be equal to zero.

For \(\ n=0\)

\[0= rp_1-f_0p_0 \\ \Rightarrow p_1=\frac{f_0}{r}p_0\]

For \(\ n=1\)

\[0 = f_{0}p_{0}+2rp_{2}-f_1p_1-rp_1 \\ p_2 = \frac{f_1}{2r}p_1 \\ \Rightarrow p_2 =\frac{1}{2}\frac{f_0f_1}{r^2}p_0\]

For \(\ n=2\)

\[0 = f_{1}p_{1}+3rp_{3}-f_2p_2-2rp_2 \\ p_3 = \frac{f_2}{3r}p_2 \\ \Rightarrow p_3 = \frac{1}{2\cdot 3}\frac{f_0f_1f_2}{r^3}p_0\]

A pattern emerges…

\[p_n = \frac{p_0}{n!r^n} \prod_{j=0}^{n-1}f_j\]

So we now have an exact expression for the probability distribution. However, this form is very unwieldy.

Autoactivation

This is the opposite of autorepression we can hypothesize that it has the following features:

  1. Increases the response time.
  2. Increases fluctuations due to noise.
  3. Can lead to bi-stability.

1. Increase Response Time

Lets examine how the response time increases for an autoactivation process.

\[\dot{x} = f_+(x)-rx \\ f_+(x)=\frac{k_0}{V}+\frac{x^h}{K_d^h+x^h}\] \[k_0 = \text{ Expression rate in the absence of the activator}\]

This differential equation is hard! So lets not make some approximations to make things easier to solve. We will assume non-cooperativity and that there is strong activation. This translates into the following contraints:

\[h = 1 \ , \ \ \ K_d>>x\] \[\Rightarrow \dot{x} = \frac{k_0}{V}+\frac{k}{K_dV}x-rx \\ \dot{x} = \frac{k_0}{V}- \underbrace{\left(r-\frac{k}{K_dV}\right)}_{r_\text{eff}}x\] \[\therefore x(t) = \left[x(0)-x^*\right]e^{-r_\text{eff}t}+x^*\] \[r_\text{eff}<r \ \Rightarrow \ \tau_\text{eff}>\tau\]

So the response time has increased in comparison with. the unregulated case. This increase in response time will lead to a broader probability distribution.

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Written on March 27, 2015