Physics 380, 2011: Lecture 18

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Back to Physics 380, 2011: Information Processing in Biology. We are proceeding with the dynamical information processing block. In the next couple of lectures, we will follow the article by Detwiler et al., 2000.

Warm up question

How do we calculate the amount of information processed in the vertebrate photoreceptor?

Main lecture

1. We follow Detwiler et al., 2000, article in this lecture as well.
2. Let $\delta ,\Delta$ denote the noise and the signal in a quantity, respectively. Then ${\frac {d\Delta X^{*}}{dt}}=-k_{XX}\Delta x^{*}+k_{XE}\Delta E_{a}$. This can be rewritten as ${\frac {d\Delta X^{*}}{dt}}=-{\frac {1}{\tau }}(\Delta x^{*}-g_{0}\Delta E_{a})$.
3. With noise, this gives: ${\frac {d\delta X^{*}}{dt}}=-k_{XX}\delta x^{*}+k_{XE}\delta E_{a}+\eta (t)$. Here we use $<\eta (t)\eta (t')>=(\Gamma _{a}+\Gamma _{d})\delta (t-t')$. In reality, noises in complex reactions are not as simple. See Sinitsyn and Nemenman, 2007.
4. The noise clearly contributes to the variance of $\delta X*$, but so does the noise in the enzyme. Overall: $\delta X_{\omega }={\frac {k_{XE}\delta E_{a\omega }+\eta _{\omega }}{k_{XX}+i\omega }}$.
5. What is $\eta _{\omega }$? It's a random variable. Hence $\delta X_{\omega }$ is also a random variable. We need to calculate its mean and variance.
6. Let's calculate the mean and the variance of each Fourier component. The mean is zero.
7. To calculate variances, we will need the Fourier transforms of $\delta$ functions. $\delta _{\omega }=\int dt\delta (t)e^{-i\omega t}=1$. And hence $\delta (t)=\int e^{i\omega t}{\frac {d\omega }{2\pi }}$
8. Variance is the mean square of the distance of the random variable from its mean (zero in this case). Recalling that components are now complex, we need for the variance $<\eta _{\omega }\eta _{-\omega }>$. This is given by the Wiener-Khinchin theorem as $<\eta _{\omega }\eta _{-\omega }>=\int dte^{-i\omega t}<\eta (t)\eta (t')>=\int dte^{-i\omega t}(\Gamma _{a}+\Gamma _{d})\delta (t-t')=\Gamma _{a}+\Gamma _{d}\equiv \Omega$
9. The expression $<X_{\omega }X_{-\omega }>=S_{X}(\omega )$ for any quantity X is called the spectrum.
10. Thus: $<\delta X_{\omega }^{*}\delta X_{-\omega }^{*}>=S_{\delta X^{*}}(\omega )={\frac {k_{XE}S_{\delta E_{a}}(\omega )+\Omega }{k_{XX}^{2}+\omega ^{2}}}$
11. The signal and the noise get "filtered" by the system. But notice that all frequency components are independent from each other. So each component will have its own signal and noise.
12. We have studied information transmission in such systems: each component independently contributes to information. We have $I[\Delta X^{*};\Delta E_{a}]=\int {\frac {d\omega }{2\pi }}{\frac {1/2}{\log }}_{2}\left(1+SNR\right)=\int {\frac {d\omega }{2\pi }}{\frac {1/2}{\log }}_{2}\left(1+{\frac {S_{\Delta X^{*}}(\omega )}{S_{\delta X^{*}(\omega )}}}\right)$, where the spectrum of the signal is given by $S_{\Delta X^{*}}(\omega )={\frac {k_{XE}S_{\Delta E_{a}}(\omega )}{k_{XX}^{2}+\omega ^{2}}}$