\"Antenna less (al), brown eyes (bw) and ebony eyes (e) are recessive autosomal
ID: 65242 • Letter: #
Question
"Antenna less (al), brown eyes (bw) and ebony eyes (e) are recessive autosomal mutants of drosophila melanogaster. A male expressing antenna less and ebony was mated to a brown female producing a Wildtype F1. And F1 female was then test crossed to produce the following progeny: total = 1000 al bw e=18, + + e=21, al bw +=17, + + +:16, al + e=232, + bw e=229, al + +=237, + bw +=230What is the map distance between the antenna less locus and the ebony locus? A. 7.2 B. 3.6 C. 8.9 D. 14.4 E. Antenna less and ebony are not linked
(I think the answer is E. Is this correct?) "Antenna less (al), brown eyes (bw) and ebony eyes (e) are recessive autosomal mutants of drosophila melanogaster. A male expressing antenna less and ebony was mated to a brown female producing a Wildtype F1. And F1 female was then test crossed to produce the following progeny: total = 1000 al bw e=18, + + e=21, al bw +=17, + + +:16, al + e=232, + bw e=229, al + +=237, + bw +=230
What is the map distance between the antenna less locus and the ebony locus? A. 7.2 B. 3.6 C. 8.9 D. 14.4 E. Antenna less and ebony are not linked
(I think the answer is E. Is this correct?) al bw e=18, + + e=21, al bw +=17, + + +:16, al + e=232, + bw e=229, al + +=237, + bw +=230
What is the map distance between the antenna less locus and the ebony locus? A. 7.2 B. 3.6 C. 8.9 D. 14.4 E. Antenna less and ebony are not linked
(I think the answer is E. Is this correct?)
Explanation / Answer
Map distance between antenna less (al) and ebony (e)
Total number of progeny = 1000
Number of 'al' containing progeny = 18 + 237 + 232 + 17 = 504
Number of 'e' containing progeny = 18 + 21 + 232 + 229 = 500
Frequency of 'al' = 504/1000 = 0.504
Frequency of 'e' = 500/1000 = 0.50
The distance between 'al' and 'e' = 0.504 - 0.50 = 0.004
Since, the distance is much small (only 0.004 m.u) between the gene loci and hence no recombination and linkage. Option E is correct.
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