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I was reading this review. On page 11, left column, first paragraph, one can read:

[..] there is a Poisson distribution of the equilibrium number of mutations per individual, if fitness effects are multiplicative.

with no further explanations. In other words, let the variable $X$ be the number of (deleterious) mutations found in one individual (irrespectively to whether the mutation occurred in the parents gonads or is older back in the lineage). The variable $X$ is Poisson distributed.

Why is this statement true? I suppose that the statement also assumes that all mutations have the same effect on fitness, is that true?

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  • $\begingroup$ It might be worth reading the paper that is cited there, Kimura & Matuyama (1966). They seem to justify it via the process of free recombination (see page 2, though the derivation starts at the bottom of page 1). $\endgroup$
    – HDE 226868
    Commented Jul 30, 2015 at 16:07
  • $\begingroup$ Oh I missed this part where he first talk about it and accompany the statement with a reference. Ok. I'll read that then. The question still hold though. Thank you $\endgroup$
    – Remi.b
    Commented Jul 30, 2015 at 16:11

1 Answer 1

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A Poisson process follows these postulates:

  1. $\lim\limits_{h\to0+}\frac{P(N_h=1)}h=\lambda$

    i.e. the probability of occurrence one event in a very small interval of time is equal to the macroscopic rate or intensity ($\lambda\,$).

  2. $P(N_h\geqslant2)=o(h)$

    i.e. the probability of occurrence of more than one events in an infinitesimal interval is essentially zero.

  3. Events are independent.

If you consider a single individual (for simplicity assume a single cell), then the DNA will undergo mutations at some fixed rate (which we assume to be uniform for all loci). Now each mutation event is independent of the previous event and in a very small interval of time the chance of two or more mutations is negligible. Considering all these facts and assumptions, it can be said that mutation in a single cell would behave like a Poisson process.

From the Poisson postulates you can derive the expression for the Poisson distribution which describes the probability of $k$ number of events in a given time interval, $t$. Hence, the number of mutations in an individual for a fixed time window ($t\,$) follows a Poisson distribution.

$$P(N=k)=\frac{(\lambda t)^k e^{-\lambda t}}{k!}$$

You can find the derivation of Poisson distribution from the postulates from many sources. I referred to this book:

Hogg, Robert V., and Allen T. Craig. Introduction to mathematical statistics. New York: Macmillan, 1978.

EDIT

The effect of deleterious mutations, in the mentioned section of the linked paper talks about Muller's ratchet that describes the accumulation of deleterious mutations and its effect on the population (i.e. extinction). Like any mutation event, accumulation of deleterious mutations will also follow Poisson distribution. Muller's ratchet just says that beyond a tolerance limit, deleterious mutations will cause extinction of asexually reproducing organisms. Perhaps if each deleterious mutation had a strong effect on the fitness, then sampling from the population may lead to non-Poissonian estimates.

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  • $\begingroup$ Sorry, I think there is a missunderstanding. I was a bit unclear. The question is not why the number of new mutations per reproduction event is a Poisson distribution but why the number of deleterious mutations per individual follows a Poisson distribution. Thank you though for your answer, I'll try to clarify the question. $\endgroup$
    – Remi.b
    Commented Jul 30, 2015 at 17:28
  • $\begingroup$ @Remi.b probability of any mutation will have a Poisson distribution. Not specific to deleterious mutations. In fact for deleterious mutations there can be non Poisson observation when you sample the population because they may be selected out. $\endgroup$
    – WYSIWYG
    Commented Jul 30, 2015 at 17:37
  • $\begingroup$ I talk about deleterious because they assume multiplicative fitness effects. But what they say is that the number of mutations/segregating loci in any single individual is Poisson distributed. Of course the number of new mutations transmitted to the offspring is also Poisson distributed (and I understand why) but I don't understand why the number of mutations/segregating loci is Poisson distributed. Do I missunderstand something? $\endgroup$
    – Remi.b
    Commented Jul 30, 2015 at 18:54
  • $\begingroup$ Sorry, I will just repeat what I said above. I think you are not answering to my question here. I am NOT asking why does the number of new mutations per reproduction event follows a Poisson distribution. I am asking why does the number of segregating sites per individual follow a Poisson distribution. $\endgroup$
    – Remi.b
    Commented Jul 30, 2015 at 19:10
  • $\begingroup$ @Remi.b I did not say anything about reproduction. I also do not know what is a segregating site. I am just saying how the process of accumulation of mutations in the DNA will have a Poisson distribution. $\endgroup$
    – WYSIWYG
    Commented Jul 30, 2015 at 19:13

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