In Drosophila melanogaster, cut wings(ct) is recessive to normal wings(ct+), sab
ID: 6998 • Letter: I
Question
In Drosophila melanogaster, cut wings(ct) is recessive to normal wings(ct+), sable body(s) is recessive to gray body(s+), and vermilion eyes(v) is recessive to red eyes(v+). All three recessive mutations are X-linked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male homozygous for the normal body, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The progeny resulting from the testcross are listed below:v ct s 610
v+ ct s 2
v+ ct+ s 15
v+ ct+ s+ 607
v+ ct s+ 77
v ct s+ 17
v ct+ s 82
v ct+ s+ 2
a
A. determine order of these genes on the chromosome(now I understand that one of the genes moves to the middle and in both cases I think it is v and/or v+ but I still don't know what order I to put them in)
B. Calculate the map distances between the genes
Explanation / Answer
According to given data,
v ct s 610
v+ ct s 2
v+ ct+ s 15
v+ ct+ s+ 607
v+ ct s+ 77
v ct s+ 17
v ct+ s 82
v ct+ s+ 2
Total progeny = 1412
Recombinant frequency between v+ and ct+ = 2+ 77+ 82+ 2 = 163 = 163/ 1412 = 0.11
Recombinant frequency between ct+ and s+ = 15+ 77+ 17+ 82 = 191 = 191/1412 = 0.13
Recombinant frequency between v+ and ct+ = 2+ 15+ 17+ 2 = 36 = 36/ 1412 = 0.02
From the above map distances order of genes is
ct+----------------------v+---------s+
<-----0.11---------------><--0.02-->
<----------------0.13---------------->