Question

The fitness for a particular species of South American bats is determined by a single gene locus with two segregating alleles, (W) and (w), that determines echo-location ability. The dominant allele W causes normal echo-location, whereas the recessive allele w impairs echo-location ability. In the large bat population that you are studying, you determine initial frequencies of the W and w alleles to be 0.6 (p) and 0.4 (q), respectively. If the genotypes are in Hardy–Weinberg equilibrium on fertilization, and the selection coefficient (s) is 0.36, determine the effects of natural selection on the allele frequencies after one generation.

Answer 1

1) Determine the initial status of the population.

For WW: relative frequency (p2) = 0.36; relative fitness = 1.0

For Ww: relative frequency (2pq) = 0.48; relative fitness = 1.0

For ww: relative frequency (q2) = 0.26; relative fitness = 1 – 0.36= 0.64

2) Determine contributions of each genotype to the next generation:

For WW: relative contribution to next generation = 0.36 × 1 (relative fitness) = 0.36

For Ww: relative contribution to next generation = 0.42 × 1 (relative fitness) = 0.48

For ww: relative contribution to next generation = 0.16×0.64 (relative fitness) = 0.10

(3) To determine the relative contributions of each genotype to the next generation, divide the individual relative contributions (just calculated) by the sum of all of their contributions (i.e., 0.36 + 0.48 + 0.10 = 0.94):

For WW: 0.36/0.94 = 0.383

For Ww: 0.48/0.94 = 0.511

For ww: 0.10/0.94 = 0.106

(4) It is now possible to calculate the frequency of the wallele after one generation of selection because all the genes transmitted by the ww homozygotes, and half the genes transmitted by the Wwheterozygotes, are w. In the next generation, the frequency of the wallele (q1) will be q1 = 0.106 + ½ (0.511) = 0.361. Note that the frequency of the w the bat population decreased from an initial frequency of 0.4 to 0.361 after one generation.