Homework Answers
Answer :)
To find out the map of those six genes, we need to calculate recombination frequencies of each allele pair.
In 2 point cross 1:
Recombination frequency between Short legs (d) and purple eyes (pr) = [Total progenies of recombinants /Total offsprings] x100
Recombination frequency between Short legs (d) and purple eyes (pr) = [(470+494) /(470+494 +146+146)] x100
Recombination frequency between Short legs (d) and purple eyes (pr) = [964 /1256] x100
Recombination frequency between Short legs (d) and purple eyes (pr) = 0.77 x100
Recombination frequency between Short legs (d) and purple eyes (pr) = 77 cM
That is how we get:
Recombination frequency between very short wings (vg) and purple eyes (pr) =13 cM
Recombination frequency between stunted wings (dp) and purple eyes (pr) =41 cM
Recombination frequency between black bodies (bl) and purple eyes (pr) = 94 cM
Recombination frequency between black bodies (bl) and very short wings (vg) =19 cM
Therefore the map of these genes are:
bl------d---------vg-----------pr----------al----------dp
The allele of very short wings (vg) is must be placed near the centromere of chromosome #2 because there is a large distance in between black bodies (bl) and purple eyes (pr). The allele of short legs (d) is also placed in the side of bl allele because allele d and pr are far than 50cM. Recombination frequency more than 50cM shows alleles either present on the different chromosome or on the different chromatids of the same chromosome separated by the centromere. The allele of aristaless (al) must present in between pr and dp because both are 41 cM apart, in which a big space is present.
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