Vermilion eyes (v), crossveinless wings (cv). and singed bristles (m) are three
ID: 324475 • Letter: V
Question
Vermilion eyes (v), crossveinless wings (cv). and singed bristles (m) are three X-linked recessive traits in Drosophila. A testcross between a female who was heterozygous for all three traits with a male showing all three recessive traits produced the following progeny: singed, crossveinless, vermilion 3 crossveinless, vermilion 392 vermilion 34 crossveinless 61 singed, crossveinless 32 singed, vermilion 65 singed 410 wild type 3 Total 1000 Determine the chromosomal arrangement of alleles in the heterozygous female, including gene order and which allele is on each homolog. Calculate the genetic distances between these genes and draw a map showing their arrangement and distances.Explanation / Answer
Answer:
2. a. These classes tell us that the heterozygous female had the cv and v mutations on one of her X chromosomes and the sn mutation on the other. Her genotype was therefore (cv + v)/(+ sn +), with the parentheses indicating uncertainty about the gene order.
For the double cross overs, two exchanges in the genotype will produce gametes that are either cv sn v or + + +, which correspond to classes 1 and 8. Thus, the proposed gene order is:
cv + v / + sn +
b. Calculating the map distance:
recombinant classes represent crossovers between cv and sn, and which represent crossovers between sn and v.
Cross over between cv and sn = 34 + 32 + 3 + 3 = 72
Cross over between sn and v = 61 + 65 + 3 + 3 = 132
So, distance between cv-sn = 72/1000 = 7.2 cM, and
distance between cv-sn = 132/1000 = 13.2 cM
cv----------7.2--------sn---------13.2----------v
cv-v = 20.4 cM
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