Answer:
The expected double recombinant frequency = 0.0402 = 4.02%
Explanation:
<u>Available data</u>:
- 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 <u>a few recombination events</u> 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) <em>The expected double recombinant frequency = recombination frequency in region I x recombination frequency in region II.</em>
The expected d. recombinant frequency = 32.2% x 12.5% = 0.322 x 0.125
The expected double recombinant frequency = 0.0402 = 4.02%
