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: