Segregational Load

Kimura and Crow first put forward this argument, in 1964. We start with the standard selection scheme for heterozygous advantage:

The equilibrial gene frequencies at the locus are (p for gene A , q for a ):

p = t/(s + t)
q = s/(s + t)

The proportion of heterozygotes in the population at the locus is:

H = 2st/((s+t)²)

The mean fitness of the population (for this one locus) is:

mean(v) = 1 - (st/(s+t))

The formula for mean fitness can be further simplified if we substitute the formula for heterozygosity into the one for mean fitness. Then,

mean(v) = 1- H((s+t)/2)

With mean s = (s + t ) / 2 then:

mean(v) = 1 - H mean(s)

If H = 0.33, the average member of the population has a fitness of 0.967 compared with 1 for a heterozygote. Mean fitness is less than one, because s and t are both positive. With heterozygous advantage, the population mean fitness is lower than what it could be if all individuals were heterozygotes.

Mean v is lower if H is higher because when there is a higher proportion of heterozygotes, selection against the homozygotes must be stronger (higher s ). More homozygotes must die each generation and the population 'suffers' more reduction in mean fitness than when the homozygotes and heterozygotes have similar fitnesses.