How to calculate effective population size dominated by alpha male
Effective Population Size:
Here’s the general idea: Factors like genetic drift depend on population size. But the census size of a population—the raw number of members—may not tell you what’s really going on. In a population of living things, some members may not be reproducing (because they’re too young or too old, because there’s an excess of males or females, because a few “privileged“ males mate with multiple females. . . etc. etc.) The effective population size is the size of a hypothetical ideal population, all of whose members have an equal probability of breeding, that has as much inbreeding and genetic drift as the real population.
Exactly how you calculate this depends a lot on the situation and can be a little tricky, but here’s a sample: In a population of diploid individuals with separate sexes, Ne, the effective population size, is equal to (4NfNm) / (Nf Nm), where Nf is the number of females and Nm is the number of males. If Nf and Nm are equal, then Ne reduces to Nf Nm, which is just N (the actual population size). But suppose that this population is dominated by a single alpha male who does all the breeding (which is close to what happens in some animals, such as elephant seals). Then Nm = 1, and the formula becomes Ne = 4Nf / (Nf 1).
So if you have a population of 100 elephant seals, 50 males and 50 females, N = 100, but Ne = 200/51 = 3.9. In other words, a population of 100 elephant seals in which only one male mates will have as much genetic drift and inbreeding as a population of four elephant seals in which all members could mate! This has a lot of implications for conservation and breeding programs.
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