Genetic Cross in Drosophila Melanogaster: Phenotypes Calculation

What phenotypes would you expect in offspring from a genetic cross between wild-type female fruit flies and miniature-winged males with garnet eyes?

If 800 offspring were produced from the cross, in what numbers would you expect the following phenotypes?

a. wild type

b. miniature wings

c. garnet eyes

d. miniature wings, garnet eyes

Phenotypes Expectation from the Genetic Cross

For a genetic cross between phenotypically wild-type female fruit flies and miniature-winged males with garnet eyes in Drosophila, we would expect an equal number of wild-type and miniature-winged offspring, and a smaller, equal number of garnet eyes only and miniature wings with garnet eyes, based on an 8% recombination frequency.

Explanation:

Genetics of Drosophila Phenotypes

The student asks about a genetic cross in Drosophila melanogaster, where the genes for miniature wings (m) and garnet eyes (g) are located on chromosome 1 and are 8 map units apart. Wild-type females with the genotype (m+ g+ / mg+) are crossed with miniature-winged males with garnet eyes (mg/mg). Given the recombination frequency of 8%, we would expect 8% of the offspring to be recombinants and 92% to exhibit parental phenotypes.

Since there are 800 offspring, we would expect 64 (8% of 800) to be recombinant phenotypes and 736 to be parental phenotypes. The expected number of offspring for each phenotype would be:

Wild type: 368 (46%)

Miniature wings: 368 (46%)

Garnet eyes: 32 (4%)

Miniature wings, garnet eyes: 32 (4%)

This calculation assumes that the parental phenotypes will appear at an equal frequency and that recombinant phenotypes will be split evenly between the two possible combinations.