A fruit fly of genotype Aa Bb parent 1 is crossed to another

A fruit fly of genotype A/a. B/b (parent 1) is crossed to another fruit fly of genotype a/a. b/b (parent 2). The progeny of this cross was: What gametes were produced by parent 1 and in what proportions? Do these proportions demonstrate independent assortment of the two genes? What can be deduced from these proportions? Draw chromosomal diagram(s) to illustrate the arrangement of the genes on the chromosomes in parents 1 and 2. Draw a diagram or diagrams to explain the origin of the two rarest progeny genotypes.

Solution

(a) Parent1 has genotype ‘AaBb’. So, the gametes produced by this parent will be AB, ab, Ab, and aB.

Proportions are:

(32+33+17+18) = 100

Proportion of AB = 32/100 = 0.32

Proportion of ab = 0.33

Proportion of Ab=0.17

Proportion of aB =0.18

b)

These ratios are not determining independent assortment.

c) There is linkage in the two genes, \'A\' and \'B\'; because parental types (32+33 = 65) > recombinants (17+18 = 35)

d) 35 map units is the distance between the two genes. Please note that there is 35% recombination; and 1% recombination = 1 map unit.

A--------------35-----------------B (Parent 1)

a---------------35-----------------b (Parent 2)