As far as I understand, in sexual reproduction, there is a mix of 50/50 genes from parents. In how many generations of asexual reproduction, will offspring reach differentiation level that of sexual one ?


closed as unclear what you're asking by AliceD, James, rg255, anongoodnurse, March Ho Aug 25 '16 at 14:00

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Clarification of the question

It is unclear what you mean by differentiation? I will assume that the question is

For what value of $n$. is the number of pairwise differences between any two siblings under sexual reproduction is equal to the number of pairwise differences between any two $n^{th}$ cousins under asexual reproduction?


The answer to this question mainly depend on (1) the mutation rate and (2) the amount of genetic variation present in the sexual population (which itself is a function of the mutation rate).

Assuming a panmictic population of constant size $N$, assuming equal sex-ratio and non-overlapping generation (and a few other assumptions), we can make some estimations. Let's assume that the sexual and asexual populations that we are comparing have the same mutation rate $\mu$.

Sexual population

Using coalescent theory, one can show that the expected number of pairwise difference between any two individual in a sexual population is $4 N \mu$. It would be off-topic to explain why this is true. I let you open a population genetics book and/or open a different post.

Asexual population

Assuming an infinite site model (just like coalescent theory does), the number of pairwise differences between $n^{th}$ cousins is $2 n \mu$.


Let's set the equation

$$2 n \mu = 4 N \mu$$

and solve for $n$, we get

$$n = 2 N$$

In words, $2N$ cousins in asexual populations have the same number of pairwise differences (on average) than two sibling in a sexual population of size $N$.


Please note that the conclusion is drawn from important (but common) assumptions. Note also that drawing conclusions from the above calculation such as "sex helps retain genetic diversity" would be wrong, as we are comparing sibling to $n^{th}$ cousins and not randomly chosen individuals in the population.

  • $\begingroup$ excellent answer, thank you. The reformulation is good $\endgroup$ – PetoU Aug 4 '16 at 8:01

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