Viability Selection in detail

Suppose 100,000 newborn mice are  produced by a group of parents among which the frequency of allele A is 0.3. The newborns are distributed according to Hardy-Weinberg expectations, such that there are 9,000 AA, 42,000 AB, and 49,000 BB mice. The fitness of the three types measured as viability (the probability of survival to reproductive age) of each of the three phenotypes corresponding to genotypes AA, AB, and BB is 0.5, 0.4, and 0.3, respectively. Note that this is an additive model, in which each B allele decreases viability by 0.1. Therefore, the expected numbers of survivors to reproductive age are:

AA: 0.5 x 9,000 newborns = 4,500 AA adults
AB: 0.4 x 42,000 newborns = 16,800 AB  adults
BB: 0.3 x 49,000 newborns = 14,700 BB adults
The total number of surviving adults is (4,500 + 16,800 + 14700) = 36,000.

Following established principles, f(A) = [(2)(4,500) + 16,800] / (2)(36,000) = 0.3583, and f(B) = [(2)(14,700) + 16,800] / (2)(36,000) = 0.6417 = 1 - f(A). Viability selection has therefore decreased the relative frequency f(B)' by 0.7000 - 0.6417 = 0.0583, about 6%.

    Allele frequency change in the model just given is determined from relative rather than absolute viability. For example, if relative viabilities were 0.05, 0.04, and 0.03, the total number of mice surviving to adulthood would be only 3,600, but the calculated allele frequency change would be the same as before [PROVE this to yourself]. This also shows that evolution, defined as changes in allele frequencies, can occur in a population that is declining in numbers.

    Alternatively, viability selection may  allow population size to remain constant is, if the expected number of newborn mice of each genotype surviving to adulthood is weighted by the ratio of newborns to adults, in this case 100,000/36,000 = 2.778. Then

AA: 0.5 x 9,000 newborns x 2.778  = 12,500 AA adults
AB: 0.4 x 42,000 newborns  x 2.778 = 46,670 AB adults
BB: 0.3 x 49,000 newborns x 2.778 = 40,830 BB adults
The expected total number of adults is now 100,000.

    Calculated allele frequency change would be the same as in the original model [PROVE this to yourself]. [The original FORTRAN form of the Matlab NatSel program achieved the same result by repeatedly sampling from the post-selection adult population pool until the pre-selection population size is recovered. This is essentially K-selection, which maintains N near a constant carrying capacity].

    Finally, note that weighting by a larger factor of (say 3.0) would result in an increase in the population size, but once again calculated allele frequency changes would be the same. Population increase would not be due to an increase in the frequency of the allele at a fitness advantage. This again shows that relative Darwinian fitness and absolute survival are unrelated.


Example modified after Box 07-2 by Nielsen & Slatkin © 2013 by Sinauer; Text material © 2020 by Steven M. Carr