A female Drosophila fly heterozygous for the traits c (curved wings) and bw (bro
ID: 252752 • Letter: A
Question
A female Drosophila fly heterozygous for the traits c (curved wings) and bw (brown eyes) was mated to a homozygous male who had curved wings and brown eyes. Both traits are recessive. 1000 progeny were scored with the following phenotypes: Wild type wings, wild type eyes Curved wings, wild type eyes Curved wings, brown eyes Wild type wings, brown eyes 350 flies 150 flies 360 flies 140 flies A. What do these results tell us about the c and bw genes? (1 point) B. Identify the recombinant phenotypes in the progeny. (1 point) Estimate the genetic map distance between the two genes. Show your calculations. (1 point; half credit if calculations are not shown) C.Explanation / Answer
Assume that the gene coding for curved wings is “c” and the gene coding for wild type wings is, “C.” The genes coding for brown eyes is, “bw” and the gene coding for wild type eyes is, BW.
The genotype of female fly is, “Cc BWbw,” whereas the genotype of male is, “cc bwbw.”
A). The results are indicating that the genes are linked and not following independent assortment
B). The recombinant phenotypes in the progeny are, “curved wings, wild type eyes, and wild type wings, brown eyes.”
C). The genetic map distance between the two genes is, = recombinant genotypes/ total progeny = 150+140/ 1000 = 0.29 or 29 map units
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