Molecular Genetics: Problem 4

A wild-type fruit fly (heterozygous for gray body color and normal wings was mated with a black fly with vestigial wings. The offspring had the following phenotypic distribution: wild type, 778; black-vestigial, 785; black-normal, 158; gray-vestigial, 162. What is the recombination frequency between these genes for body color and wing type.

First count the total number of offspring 778+785+158+162 = 1883
In all dihybrid test crosses (a cross between a known heterozygote for two wild type traits and a homozygous recessive individual for both traits) the expected ratio of phenotypes if the genes are on separate chromosomes must be:

wild type, 25%; black-vestigial, 25% black-normal, 25%; gray-vestigial, 25%. These results do not fit the experimental data above (778+785+158+162).

In fact the black-normal (158) and gray-vestigial (162) offspring represent recombinant individuals.

Calculation of recombination frequency:

778 - wild type

785 - black-vestigial

778/1883 = 41.3%

785/1883 = 41.7%

83% are non-recombinant

158 - black-normal

162 - gray-vestigial

158/1883 = 8.4%

162/1883 = 8.6%

17% are due to recombination

Recombination frequency = 17%

The generally accepted method of symbolizing the genotypes for a dihybrid cross of linked genes is as follows (pg. 264 - Campbell):

b⁺ = wild-type (gray body)

b = black body

vg⁺ = normal wing shape

vg = vestigial wings

The testcross is symbolized as follows:

If no cross over occurs then only two phenotypes should be seen. That is 50% of the offspring should be dominant for both traits ---- and the other 50% should be homozygous recessive ---- just as in the parents above.

Cross over in the heterozygous parent results in 50% recombinants: