The effective population size ($N_e$) is a central concept of evolutionary biology and is influenced by several parameters. For example: sex ratio bias affects $N_e$ $\left(N_e = \frac{4N_mN_f}{N_m+N_f}\right)$ and varying population size over time influences $N_e$ $\left(N_e = \frac{n}{\sum_{i=1}^n\frac{1}{N_i}}\right)$. There is a post on how overlapping generations influences population size.


Selection also influences the effective population size. Intuitively, I'd expect that the higher is the fitness variance, the lower is the effective population size as fewer individuals contribute to the next generation. Am I right? How (what is the mathematical formulation) does fitness variance influences the effective population size?


1 Answer 1


From Conner and Hartl's A primer of ecological genetics:

"Any variance in reproductive success among individuals greater than random expectations, a commonplace concurrence in natural populations, reduces effective population size."

So yes, selection does reduce the effective population size and for the reason you suggest - it removes some individuals from the mating pool/ reduces their contribution to the next generation. Mathematically it can be derived as

$N_e \approx \frac{ 8N_a}{V_m + V_f + 4 }$

where $V_m$ and $V_f$ are sex-specific variances of offspring production (males are often more variable in reproductive success than females).

They give an example of selection affecting $N_e$, in a population of deer 33 males had four times the variance ($V_m$ = 41.9) in reproductive success of 35 females where variance was $V_f$ = 9.1. Thus effective population size was

$N_e \approx \frac{ 8 * (33 + 35)}{41.9 + 9.1 + 4 } = 9.9$

The notation $N_a$ is the actual number of individuals in the population. This can be seen in the calculation of effective population size when the number of males and females are not equal

$N_e = \frac{4 N_m N_f}{N_m + N_f} = \frac{4 N_m N_f}{N_a}$

An example is a population of 80 males and 80 females compared to a population of 70 males and 90 females. Both populations have equal sizes but effective population size is reduced by unequal sex ratios

$N_e = \frac{4 * 80 * 80}{160} = 160 \neq \frac{4 * 70 * 90}{160} = 157.5$

  • $\begingroup$ +1 Nice! $N_a$ is just the population size I guess. What does the $a$ in index stands for? $\endgroup$
    – Remi.b
    Oct 14, 2014 at 16:32
  • $\begingroup$ I'll take a look tomorrow, just got home, can't remember why it's $N_a$ as opposed to anything else. @remi.b $\endgroup$
    – rg255
    Oct 14, 2014 at 17:16
  • $\begingroup$ edits done @Remi.b $\endgroup$
    – rg255
    Oct 15, 2014 at 8:29
  • $\begingroup$ Wow, usually when people say that they "I'll have a look at that tomorrow", it means that they won't never give any more attention to the post. But you actually came back and answered my question in the comments! Thank you $\endgroup$
    – Remi.b
    Oct 17, 2014 at 12:40
  • $\begingroup$ @Remi.b I'm guilty of that a lot, I think I still owe you an answer on evolution of the G-matrix!!! $\endgroup$
    – rg255
    Oct 17, 2014 at 12:47

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