Separate sexes

We'll start by assuming that $\hat f_t = 0$ to make the calculations simple. So we know that

$$N_e^{(f)} = \frac{1}{2\hat f_{t+1}} \quad .$$

The first thing to do is to calculate $\hat f_{t+1}$. To do this we have to break the problem down into pieces.²²

• We assumed that $\hat f_t = 0$, so the only way for two alleles to be identical by descent is if they are identical copies of the same allele in the immediately preceding generation.

• Even if the numbers of reproductive males and reproductive females are different, every new offspring has exactly one father and one mother. Thus, the probability that the first gamete selected at random is female is just 1/2, and the probability that the second gamete selected is male is just 1/2.

• The probability that the second gamete selected is female given that the first one we selected was female is $(N-1)/(2N-1)$, because $N$ out of the $2N$ alleles represented among offspring came from females, and there are only $N-1$ out of $2N-1$ left after we've already picked one. The same logic applies for male gametes.

• Finally, the probability that one particular female gamete was chosen is $1/2N_f$, where $N_f$ is the number of females in the population. Similarly the probability that one particular male gamete was chosen is $1/2N_m$, where $N_m$ is the number of males in the population.

With those facts in hand, we're ready to calculate $\hat f_{t+1}$.

$$\begin{eqnarray*} f_{t+1} &=& \left(\frac{1}{2}\right) \left(\frac{N-1}{2N-1}\r... ...rac{1}{4}\right) \left(\frac{1}{2N_f} + \frac{1}{2N_m}\right) \end{eqnarray*}$$

So,

$$N_e^{(f)} \approx \frac{4N_fN_m}{N_f + N_m} \quad .$$

What does this all mean? Well, consider a couple of important examples. Suppose the numbers of females and males in a population are equal, $N_f = N_m = N/2$. Then

$$\begin{eqnarray*} N_e^{(f)} &=& \frac{4(N/2)(N/2)}{N/2 + N/2} \\ &=& \frac{4N^2/4}{N} \\ &=& N \quad . \end{eqnarray*}$$

The effective population size is equal to the actual population size if the sex ratio is 50:50. If it departs from 50:50, the effective population size will be smaller than the actual population size. Consider the extreme case where there's only one reproductive male in the population. Then

$$N_e^{(f)} = \frac{4N_f}{N_f + 1} \quad .$$ (7)

Notice what this equation implies: The effective size of a population with only one reproductive male (or female) can never be bigger than 4, no matter how many mates that individual has and no matter how many offspring are produced.