Problem 16.1: Dynamics of noise

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In the chapter, we worked out that the dynamics of the mean copy number of unregulated gene expression follow the same ODE we introduced at the very beginning,

$$\begin{align} \frac{\mathrm{d}\langle n \rangle}{\mathrm{d}t} = \beta - \gamma \langle n \rangle. \end{align}$$

We also worked out that the steady state distribution is Poisson;

$$\begin{align} P(n) = \frac{1}{n!}\left(\frac{\beta}{\gamma}\right)^n\mathrm{e}^{-\beta / \gamma}. \end{align}$$

Here, we will work out the dynamics of the variance in $n$ and demonstrate that the steady state variance and noise are indeed what we get from a Poisson distribution.

a) Derive a differential equation for the variance $\sigma^2 = \left\langle n^2 \right\rangle - \langle n \rangle ^2$. Your ODE should only have $\langle n \rangle$ and $\sigma^2$ as variables. Hint: First write down an expression for $\mathrm{d}\left\langle n^2 \right\rangle / \mathrm{d}t$, and then use the fact that

$$\begin{align} \frac{\mathrm{d}\sigma^2}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\left\langle n^2 \right\rangle - \langle n \rangle^2\right) = \frac{\mathrm{d}\left\langle n^2 \right\rangle}{\mathrm{d}t} - \frac{\mathrm{d}\left\langle n \right\rangle^2}{\mathrm{d}t} = \frac{\mathrm{d}\left\langle n^2 \right\rangle}{\mathrm{d}t} - 2\langle n \rangle\,\frac{\mathrm{d}\left\langle n \right\rangle}{\mathrm{d}t}. \end{align}$$

b) Assuming we start at $t = 0$ with no gene product ($n = 0$), solve for $\langle n \rangle(t)$ and $\sigma^2(t)$ to show that the mean and variance are equal for all time, and therefore the Fano factor is always unity. Hint: We have already solved for $\langle n \rangle(t)$ in the notes. The resulting ODE for $\sigma^2(t)$ may be solved by integrating factor.

c) Take the limit of $t \to \infty$ of the above solutions to show that the steady state noise is $\eta = \sqrt{\gamma/\beta}$, as we have shown in the chapter.