The following is data from a cross of a heterozygous drosophila female and a hom
ID: 66711 • Letter: T
Question
The following is data from a cross of a heterozygous drosophila female and a homozygous recessive male. The data is examining three linked genes. Use the data to determine:
The order of the genes on the chromosome.
The pattern of the genes on the female’s pair of chromosomes.
The map distances between the genes.
Table 1 Drosophila Cross data 1000 F1 progeny scored for 3 linked loci
Trait
number
Smooth abdomen(sa)
3
Lobe Eye(le) and withered wings (ww)
2
Lobe eye (le)
87
withered wings and smooth abdomen
76
Wild type(+, +, +)
390
Lobe eyes, withered wings and smooth abdomen
344
Withered wings
45
Lobe eyes and smooth abdomen
53
I know the double cross over is between the two offspring classes with the fewest number and the parental are those with the highest. I dont know which genes are single crossing in the other progeny.
Trait
number
Smooth abdomen(sa)
3
Lobe Eye(le) and withered wings (ww)
2
Lobe eye (le)
87
withered wings and smooth abdomen
76
Wild type(+, +, +)
390
Lobe eyes, withered wings and smooth abdomen
344
Withered wings
45
Lobe eyes and smooth abdomen
53
Explanation / Answer
First you need to find the order of genes on the chromosomes. The important point is that a double-crossover event moves the middle allele from one sister chromatid to the other. This effectively places the non-parental allele of the middle gene onto a chromosome with the parental alleles of the two flanking genes.
Taking into consideration of parental phenotypes and double cross recombinants the order of genes on the chromosomes will be le---sa---ww
The map distance between le and sa
(87 and 76- single crossover between le and sa)
100 x (87+76+3+2)/1000
= 16.8 cM
The map distance between sa and ww
(45 and 53- single crossover between sa and ww)
100 x (45+53+3+2)/ 1000
= 10.3 cM
le-----16.8----sa-----10.3----ww
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