Segregrational load at more than one locus

The load exists because with Mendelian heredity it is impossible to create a population of pure-breeding heterozygotes.

Even in the extreme cases of s=t=1, when all homozygotes die before reproduction, the matings between heterozygotes will still generate the inferior homozygous forms; in practice, s and t are less than 1 and the homozygote crosses contribute too.

The population inevitably produces homozygotes every generation, and the average fitness of its members will therefore be below that of a heterozygote.

So far we have considered only one locus. But in a fruitfly population, there are about 3000 polymorphic loci, with heterozygosity averaging 0.33 each. If mean s = 0.1 for each of them, then at every locus the fitness of an average individual is about 0.967 compared with 1 for a heterozygote. We shall assume for now that natural selection operates independently on each locus; homozygotes at each of the 3000 loci contribute independently to the loss of average fitness in the population. The average fitness for all loci is found by multiplying up the average fitness of each of the 3000 polymorphic loci

mean v = (1 - 0.33 x 0.1)³⁰⁰⁰
= 0.967³⁰⁰⁰ = 10^-43

This is an absurdly low number. It means that if we compare the fitness of an imaginary individual who was heterozygous at all 3000 loci (who has w_opt = 1), with that of a random member of the population, the full heterozygote would be 10⁴³ times fitter than the randomly picked individual.