In Drosophila cherub wings ch black body b and cinnabar eyes

In Drosophila, cherub wings (ch), black body (b), and cinnabar eyes (cn) result from recessive alleles that are all located on chromosome 2. A homozygous wild-type fly was mated with a cherub, black, and cinnabar fly, and the resulting F1 female were test-crossed with cherub, black, and cinnabar males. The following progeny were produced from the testcross:

ch        b+        cn        105

ch+      b+        cn+      750

ch+      b          cn        40

ch+      b+        cn        4

ch        b          cn        753

ch        b+        cn+      41

ch+      b          cn+      102

ch        b          cn+      5

                        -----------------

Total                            1800

(a) Determine the linear order of the genes on the chromosome (which gene is in the middle?) (b)Calculate the recombinant distance between the three loci.(c)Determine the coefficient of coincidence and the interference for these three loci

Solution

a)

Total number of progeny = 1800

Frequency of ch+ = 750 + 40 + 4 + 102 = 896 / 1800 = 0.497

Frequency of b+ = 105 + 750 + 4 + 41 = 900 / 1800 = 0.5

Frequency of cn+ = 750 + 41 +102 +5 = 898 / 1800 = 0.498

The order of genes is ch+ , cn+ and b

The gene cn+ is in the middle.

b)

Recombination distance between ch+ and b+ = 40 + 102 + 105 + 41 = 288

Recombination distance between b+ and cn+ = 105 + 750 + 102 + 5 = 962

Expected rate of double recombinants = (288 / 1800 ) (962 / 1800) = 0.085

Actual double recombinants = 750 + 102 = 852

The coefficient of coincidence = 852 / 85 = 0.021

Interference = 1 - 0.021 = 0.979