Question

In 1905, scientist Lucien Cuénot carried out multiple crosses with yellow and non- yellow colored mice, discovering that the yellow allele is dominant to all non-yellow colors, and that mice that are homozygous for yellow always die during embryonic development. In his laboratory population of mice, Cuénot started out with 72% non-yellow mice and 28% yellow mice. Calculate the allele frequencies for this color locus.

Answer 1

It is given that the yellow allele (Y) is dominant and non-yellow (y) is recessive. It is also given that the mice having YY (homozygous yellow) die during embryonic development.

According to Hardy-Weinberg equation,

p + q = 1

and, p² + q² + 2pq = 1

where p and q are the frequencies of the two alleles.

q² = 0.72 (because it is given that 72% are non-yellow)

q = 0.848

p = 1-q = 1- 0.848 = 0.152

Thus, frequency of allele Y = 0.152

frequency of allele y = 0.848