On chromosome #2 of the fruit fly there are three linked genes: vestigial, black
ID: 6582 • Letter: O
Question
On chromosome #2 of the fruit fly there are three linked genes: vestigial, black, brown or bl-vg-br. In my parental cross there a vestigial black brown fly is mated with a wild type fly. All F1 are wild type. The F2 are the following:1675 +++
23 bl vg br
176 bl ++
13 + vg br
57 ++ br
52 bl vg +
93 + vg +
17 bl + br
Based on this data, could you help me construct a map of the genes? In addition, the data should fit a 3:1 ratio, based on a chi square analysis, does it? Thank you so much for any help.
Explanation / Answer
Given data is
The order of the genes is bl(black)-vg(vestigial)-br(brown)
vestigial black brown fly is mated with a wild type fly, all F1 are wild type and F2 progeny are as follows,
1675 + + +
23 bl vg br
176 bl + +
13 + vg br
57 + + br
52 bl vg +
93 + vg +
17 bl + br
Total progeny = 2106
Recombinant frequency between bl and vg = 176+13+93+17 = 299 / 2106
= 0.141
Recombinant frequency between vg and br = 57+52+93+17 = 219 / 2106
= 0.103
Recombinant frequency between bl and br = 176+13+57+52 = 298 / 2106
= 0.141
The map of genes can be in the following manner
+ -----------------bl-------------------+
Vg---------------+--------------------br
because the recombinant frequency between the bl, vg and bl, br are the same. So the bl gene will be located in between of vg and br genes.
The data fits the 3:1 ratio.
The number of wild type progeny is 1675.
The number of recombinants is 431
Ratio of 1675: 431 is 3:1
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