Viability Selection in detail

    Suppose 100,000 newborn mice are born in a population for which f(A) = 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. Suppose that fitness, measured as viability (the expectation of survival to reproductive age), of the corresponding three phenotypes are 0.5, 0.4, and 0.3, respectively. 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 newborns surviving as reproductive 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)' in the adults by 0.7000 - 0.6417 = 0.0583, about 6%, with respect to newborns.

   
In this case, change in allele frequencies occurs in a population that declines in numbers between the parents and offspring. Allele frequency change is determined from relative rather than absolute viability. With more intense selection, 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 [HOMEWORK: Prove this].

    Viability selection may  occur while population size remains constant, iff the expected number of newborn mice of each genotype surviving to adulthood is proportional to a constant carrying capacity K of the environment, in this case 100,000. Expected numbers can be predicted by weighting by the ratio of newborns to that of the 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 newborns that survive to adulthood is now 100,000.

    Calculated allele frequency change would be the same as in the original model [HOMEWORK: Prove this]. This is a K-selection model that acts to maintain N near a constant carrying capacity K. Computer models (such as NatSel) simulate this by generating genotypes at random, applying the viability of each type, and continuing to add to the population in the same manner, until the total number of individuals again reaches the carrying capacity of 100,000.

    Finally, 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 [HOMEWORK: Prove this]. This is still a K-selection model, where K varies according to environmental quality. Although population size increases, this would not be due to an increase in the frequency of the allele that offers the greater fitness advantage. This again shows that relative Darwinian fitness and absolute survival are unrelated.


Text material © 2024 by Steven M. Carr