We begin by identifying the classes of mating males and their population frequencies. Here, we represent the proportion of the male population in each mating class as pj, where j represents the number of females in the jth mating class of males. There are three such classes: males who do not mate, p0 (=1/10 males = 0.1), males who mate once, pi (= 8/10 males = 0.8), and males who mate twice, p2 (= 1/10 males = 0.1). The sum of all male mating classes, X p}- = (0.1 + 0.8 + 0.1) = 1. Next, we use these values to identify the average offspring produced by males in each of the jth mating classes, O?j, as well as the average offspring produced by all males across all mating classes, O? . The average offspring that males in each mating class produce, O?, equal the average offspring per female, O,, multiplied by the number of mates, j, that males in each jth mating class obtain, or O?j=j(O,).

Clearly, the average number of offspring produced by males who do not mate, O? 0, equals (0)(1) = 0. The average number of offspring for males who mate once, O? 1, equals (1)(1) = 1, and for males who mate twice, O? 2 equals (2)(1) = 2. The average number of offspring produced by all males, across all mating classes, O?, is equal to the number of offspring produced by the average female, O,, multiplied by the number of females mated by males in each mating class, j(=jO,); then, each quantity is multiplied by the fraction of the males belonging to that jth mating class, pj, and summed over all j mating classes, so that

Using the values in our example above, we can easily see that O? = O, = 1. Thus, although females are distributed unevenly among the 10 males, relative to females, as well as to the initial case in which all 10 males have equal mate numbers, the average number of offspring produced by all males in this example again equals 1.

The distribution of females across all classes of mating males is equal to the population sex ratio, which we can call R. This value can be calculated as the number of females mated by males in each mating class, j, multiplied by the fraction of the males in each mating class, pj, and summed over all classes of males, or, R = jpj = 1. Because the distribution of all females with all males equals the average number of mates per male, R also equals N,/N? (=1). That is, R is the reciprocal of OSR (=N?/N,). But whereas OSR measures the apparent intensity of male-male competition, R measures a slightly more useful quantity for estimating how selection works; R measures the population-wide average in the number of mates per male (in sex-role-reversed species, 1/R = RO measures the analogous quantity). By substitution, we can see that the average offspring per male, O? , equals the average mates per male, R, multiplied by the average offspring per female, O,, or O? = RO, = 1. Furthermore, while the distribution of females is now uneven among males, the average mates per male, R, the average offspring per female, O,, and the average offspring per male, O? , all remain unchanged relative to our initial example.

We can now express the total variance in offspring numbers for males, V0?, in terms of the average number of mates per male and the average number of offspring per female. As in a standard analysis of variance (ANOVA) problem, the total variance in male fitness can be partitioned into the sum of two components: (1) the average variance in offspring numbers for males within the classes of males who sire offspring, and (2) the variance in the average number of offspring sired by males among these same categories.

The first component of variance in male offspring numbers is calculated in three steps. First, for each mating class of males, the variance in female offspring numbers, Vg?, is multiplied by the number of mates obtained by males in each jth mating class (=jVq?). Next, this product is multiplied by the proportion of males in the population, pj, that belong to each jth mating class (=pj(jVO,)). Finally, these products are summed over all j mating classes. Thus, the variance in offspring numbers within the classes of mating males equals

In this example, because all females produce exactly 1 offspring, there is no variance in offspring numbers among females ( Vq, = 0), and, consequently, the variance in offspring numbers within the classes of mating males is also zero (V0? (within) = 0). We will return to this point below.

The second component of variance in male offspring numbers equals the variance in the average number of offspring sired by males among these same categories. This quantity is calculated in four steps. First, for each jth mating class of males, we calculate the difference between the average number of offspring per male, 0?, and the average number of offspring produced by that mating class, 0?j (=[O? — 0?j]). Second, we square each difference (=[O? — 0?j]2). Third, we multiply each squared difference by the fraction of males belonging to each mating class, pj (=p[0? — O?j]2), and fourth, we sum across all classes to obtain

Substituting in the values from above, we have

The total variance in offspring numbers among males is the sum of the within and among male components in offspring numbers, or

Because there is no variance in offspring numbers for females, VO, = 0, the first term in eqn [4] drops out. Thus, V0? (among) = V0?, and we can easily see that the variance in fitness among males goes from 0 to 0.2 when a single male mates with 2 females instead of 1. Note too that the increase in fitness variance comes entirely from the among-male component of total fitness variance. Now, if 1 male mates with all 10 of the females, the mean and variance in offspring numbers for females remains unchanged (O, = 1; Vq, = 0), and there is no change in either the sex ratio, R = 1, or the average number of offspring per male, O? = 1. But, because 1 male mates 10 times, 9 males do not mate at all. Thus, p?o = 9/10 = 0.9, p? -p? = 0, and p?w = 1/10 = 0.1. When these values are substituted in eqn [4], we see that VO? now increases 45-fold, to 9.

This exercise shows three important relationships. First, when the sex ratio equals 1, both sexes must have equal, average fitnesses. This is the reason why in most sexual populations, the sex ratio remains at 1. Deviations in sex ratio cause the average fitness of individuals of the majority sex to decrease relative to the average fitness of the minority sex, a condition that favors population-wide production of individuals belonging to the minority sex. Secondly, when some individuals are excluded from mating, the variance in offspring numbers within that sex will increase. This is the source of sexual selection. Exclusion of some individuals from mating means that the traits of the individuals who do mate will be represented disproportionately in the next generation. Third, if the fraction of individuals excluded from mating is larger in one sex than it is in the other, a sex difference in the variance in offspring numbers will appear. Because fitness variance is proportional to selection intensity, the magnitude of this sex difference in fitness variance determines the actual intensity of sexual selection. The larger the sex difference in fitness is, the more the sexes will diverge in phenotype.

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