# What does it mean for a distribution to be "consistent with a two rate-limiting stochastic steps"?

I'm reading a study (full text here) that examine the dynamic of nuclear translocation of a transcription factor in budding yeast, in response of calcium stress. They found that it occurs in bursts, which distribution of durations is plotted on this normalized histogram: In the text they say:

Normalized histograms, $h(t)$, of total burst duration at two calcium concentrations are both well-fit by $h(t)=te^{-t/\tau}$, with $\tau$ = 70 sec (black line).

And also

This distribution was consistent with two rate-limiting stochastic steps, each with a timescale of ~70 sec.

Can someone explain me the second sentence cited?

WYSIWYG is almost there, but you need one more piece of information to make this explicit.

The distribution cited in the paper is $h(t)\propto te^{-t/\tau}$ (we're going to ignore normalizing constants today). We can recognize this as a particular case of the Gamma distribution, with PDF:

$$f_{\mathrm{Gamma}}(t\,\big|\,k,\theta)\propto t^{k-1}e^{-t/\theta}.$$

In particular, $h(t)$ looks like the PDF of a $\mathrm{Gamma}(2,\tau)$ random variable, $$h(t)\propto f_{\mathrm{Gamma}}(t\,\big|\,2,\tau)\propto te^{-t/\tau}.$$

So the authors of the paper are claiming that the burst duration $\tau_{\mathrm{burst}}$ is a random variable with a $\mathrm{Gamma}(2,70\,\mathrm{ sec})$ distribution.

How do we get from that to "This distribution was consistent with two rate-limiting stochastic steps?" If we recall from our stats class (or look up properties of the Gamma distribution on Wikipedia), a Gamma distribution with integer values of $k$ is the distribution of the waiting time for $k$ events to occur in a Poisson process. The setup of a Poisson process is that you are tracking the occurrence of unlikely/infrequent events in time, so the fact that the distribution $h(t)$ of burst times looks like a $\mathrm{Gamma}(2,\tau)$ suggests that the burst duration $\tau_{\mathrm{burst}}$ is determined by the occurrence of 2 stochastic events (with the same "frequency" $1/\tau$). In other words, this distribution is consistent with the following scenario: a nuclear localization occurs, and then will stop after two particular events, where the timing of the events is governed by a Poisson process (which, as WYSIWYG pointed out, is a reasonable model for the occurrence of chemical reactions with reasonably slow kinetics, i.e. you can count individual reactions in time). If this is true, the distribution of $\tau_{\mathrm{burst}}$ will look like $te^{-t/\tau}$.

The authors generalize that statement somewhat by saying that technically there could be more events/reactions required to terminate a localization burst, but all but 2 of those reactions happen extremely rapidly, i.e. there are 2 rate-limiting steps in the decision process (both of which happen to have the same value for $\tau$).

Edit: I realized that one more mathematical point might make this more clear: The statement about the Gamma distribution and Poisson processes above is equivalent to the statement that the sum of $k$ iid random variables with distribution $\mathrm{Exponential}(\tau)$ is a random variable with a $\mathrm{Gamma}(k,\tau)$ distribution. Thus $h(t)$ is literally (up to normalization) the distribution of a sum of 2 independent exponential random variables. If you read up on the connection between the exponential distribution and waiting times that might make the claim in the paper seem more transparent.

http://en.wikipedia.org/wiki/Exponential_distribution

• A very thorough answer, now is truly clear. Thank you! Nov 5 '13 at 13:11
• Do you have any speculations about the identity of the two rate determining steps? Is it plausible that one of them is the binding of the molecule at the DNA (it's a transcription factor) and the other is the translocation process itself? Nov 5 '13 at 13:56
• The article mentions that Crz1 nuclear localization is controlled by phosphorylation and dephosphorylation. If I had to speculate I'd hazard that at least one of the steps is based on phosphorylation (see well-studied examples like the Pho2/Pho4 system in yeast). But I would caution against reading too much into this statement in the paper; without more rigorous tests of the properties of the empirical distribution, it's quite possible that this distribution that fits well "by eye" is not at all related to the process generating the empirical distribution. Nov 5 '13 at 23:27

The rate limiting step of a reaction is the slowest component of the reaction. So if there are two steps in a reaction, each of these steps will have a rate, and the slowest of those two rates will determine the overall rate of the whole reaction.

As a simple example, consider the equation

$A + B + C\rightarrow AB+C\rightarrow ABC$

If the rate of $A+B \rightarrow AB$ is fast, but the rate of $AB+C \rightarrow ABC$ is slow, then the overall reaction will still be slow.

The figure in your question is showing the distribution of the burst durations, and highlighting the fact that the calcium level has no effect on the burst duration. The second quote is a bit confusing, this is my interpretation:

The functional form which they fitted to the distribution indicates that the expected burst duration is about 70 seconds. I'm not completely sure which stochastic steps they are referring to, maybe translation and subsequent localisation (I think transcription would happen on a different timescale and so could be ignored in this case).

By the way, this paper is a really excellent example of an organism harnessing stochasticity to its advantage in order to synchronise its response to a stimulus (calcium in this case). I just think that it is so cool that the mean response to the signal level can be different to the response to the mean signal level - so counter-intuitive to me. And so much fun!!

Well.. First assume the chemical reactions to be a Poisson's process (which is a right stochastic approximation for processes such as chemical reactions and radioactive decay etc).

While the event of a burst follows a Poisson's distribution, the time interval between two reaction events follows an exponential distribution. (See wikipedia page on Poisson's distribution for details)

Since the both two mentioned steps have approximately equal time scale of ~70 sec we can expect that the mean delay between two reaction events to be equal to 70 sec.

The exponential distribution with a mean τ is:

p(t)=e^(-t/τ)