A female animal with genotype A/a B/b is crossed with a double recessive male (a/a b/b). Their progeny include 442 A/a B/b, 458 a/a b/b, 46 A/a b/b and 54 a/a B/b. This problem is explained.© BrainMass Inc. brainmass.com June 22, 2018, 3:48 am ad1c9bdddf
The ratio of genotypes observed for the progeny is not the typical 1:1:1:1 that one would normally expect (ie, the law of independent assortment states that the segregation of alleles of ...
An example of a two allele determination of linkage is included.