In Drosophila, the genes for eye color, wing shape, and wing length are located on chromosome II. Purple eyes (pr), arc bent wings (a), and vestigial wings (vg) are the mutant forms of the wild type traits red eyes, straight wings, and long wings, respectively. You've discovered some data in your genetics laboratory which indicates that the distance between vg and pr is 12.5 m.u., the distance between a and pr is 44.7 m.u., and the distance between a and vg is 32.2 m.u. what is the chance of a double crossover between these three genes

genes for eye color, wing shape, and wing length are located on the same chromosome
Purple eyes (pr) is the mutant form of the wild type traits red eyes (pr+)
Arc bent wings (a) is the mutant form of the straight wings (a+)
Vestigial wings (vg) are the mutant forms of the long wings (vg+)
the distance between vg and pr is 12.5 m.u
the distance between a and pr is 44.7 m.u
the distance between a and vg is 32.2 m.u

→ Two genes that are very close will have a few recombination events and are strongly bounded.

→ The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.

1) First, we need to know their order in the chromosome, and to do so, we need to compare the distances between them.

The most distant pair is a-pr

----a---------------------------------------------pr---

║-------------------44,7 MU-------------║

From gene a to vg there are 32.2 MU

----a-----------------------------vg--------------pr---

║-------------------44,7 MU-------------║

║----------32.2 MU------║

Finally, de distance from vg to pr is 12.5 MU

----a-----------------------------vg--------------pr---

║-------------------44,7 MU---------------║

║----------32.2 MU------║--12.5 MU--║

We can name region I to the distance between a-vg, and region II to the distance between vg-pr.

2) To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cM. And that the maximum recombination frequency is always 50%.

1 MU -----------1% of recombination

32.2 MU -----X = 32.2 % of recombination

12.5 MU-------X = 12.5 % of recombination

3) The expected double recombinant frequency = recombination frequency in region I x recombination frequency in region II.

The expected d. recombinant frequency = 32.2% x 12.5% = 0.322 x 0.125

The expected double recombinant frequency = 0.0402 = 4.02%