1. In Drosophila, brittle (b), ebony body (e) and gaucho (g) are three recessive
ID: 281070 • Letter: 1
Question
1. In Drosophila, brittle (b), ebony body (e) and gaucho (g) are three recessive genes. If homozygous EBONY females are crossed with homozygous BRITTLE & GAUCHO males, the resulting F1 progeny are all wild-type. If heterozygous F1 females are mated with EBON)Y BRITTLE & GAUCHO males, the following 1,000 progeny appear: brittle, gaucho: 290 genes. ebony: 280 brittle, ebony: 4 gaucho: 16 brittle: 132 ebony, gaucho: 128 wild-type: 70 brittle, gaucho, ebony: 80 a) Derive a map for the three (3)Explanation / Answer
Calculating the map distances between e and b by using formula we get,
(Total Single crossovers + Double crossovers) X 100/ Total Number of progeny
= (132 + 128 + 4 + 16) X 100 /1000
= 0.28 *100
= 28 map units
Thus the distance between e and b is 28 map units.
Considering the recombination frequency between only the region b--g we get,
Now substituting the values in equation (1) we get,
(70+80+4+16)/1000 * 100
= 0.17*100 = 17 map units
The map distances are 28 and 17 map units between ebony and brittle, brittle and gaucho respectively.
The total distance between c and b is found to be 28+17= 45 map units.
Drawing the map we get,
e---------------------b--------------g
28 17
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