body charactenistiCS, y Case 4: Upon visual observation, it appears the gene for
ID: 209748 • Letter: B
Question
body charactenistiCS, y Case 4: Upon visual observation, it appears the gene for certain color, and hair length are linked. In the research data a cross in Drosophila involving the X- linked recessive mutations yellow (v). white (w), and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes a whose wings were cut. The F1 females were wild type for all three traits, while the F1 males expressed the yellow body and white-eye traits. The F1 progeny were interbred to produce the following results: nd body were normal but Y + ct male 9 + w +male 6 + w+ female 7 y w ct male 90 + ++male 95 +++female 519 + +ct male 424 y w + male 376 y w + female 466 y+ +female 9 A. Diagram the genotypes of the original parents and the F1 progeny B. Construct a map of the X chromosome, showing the correct gene order and distance between adjacent genes (show your work). C. Based on your evaluation of the above data, what are the potential linkage groups?Explanation / Answer
Part A-
The female is Yellow, white eyed and normal wings i.e. yy ww ++
The male is normal body, normal eye and cut wings i.e. + + ct
The F1 progeny would be-
+y +w +ct [normal female]
y w [Yellow, white eyed]
Here males have one genes in genotype because males in drosophila do not have extra sex chromosome, they have only X chromosome.
Part B-
The double cross over progenies are having the lowest progenies i.e. 9, 6, and 7.
We do not have any genotype for + w ct; therefore, the gene order must be w y +.
Gene distance between y and w
= 100 x [(Progenies of single cross over occurred between w and y) + Double Crossover Progenies / Total Progenies]
=[9+6+7+9 /2001] x100
= 1.5 cM
Between y and +
= [90+95+519+9+6+7+9 /2001] x100
=36.7 cM
Part c- the two linkage groups are y and w, y and +
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