1
answer
2
watching
1,375
views
28 Sep 2019
Drosophila females heterozygous for three recessive X-linkedmarkers, y (yellow body), ct (cut wings), and m (miniature wings),and their wild-type alleles were crossed to y ct m males. Thefollowing progeny were obtained:
1 YELLOW, CUT, MINIATURE = 30.
2. Wild-type = 33
3. yellow = 10
4. cut, miniature = 12
5. miniature = 8
6. yellow, cut = 5
7 yellow miniature = 1
8. cut = 1
1. Determine the order of these genes
2. Calculate the map distances between the genes
3.determine the coefficient of coincidence and the interferenceamong these genes
Drosophila females heterozygous for three recessive X-linkedmarkers, y (yellow body), ct (cut wings), and m (miniature wings),and their wild-type alleles were crossed to y ct m males. Thefollowing progeny were obtained:
1 YELLOW, CUT, MINIATURE = 30.
2. Wild-type = 33
3. yellow = 10
4. cut, miniature = 12
5. miniature = 8
6. yellow, cut = 5
7 yellow miniature = 1
8. cut = 1
1. Determine the order of these genes
2. Calculate the map distances between the genes
3.determine the coefficient of coincidence and the interferenceamong these genes
Liked by soujitkumar414
Sixta KovacekLv2
28 Sep 2019
Related textbook solutions
Related questions
| In Drosophila melanogaster, cut wings (ct) is recessive to normal wings (ct+), sable body (s) is recessive to gray body (s+), and vermilion eyes (v) is recessive to red eyes (v+). All three recessive mutations are X-linked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The following are the progeny resulting from the testcross. v ct s 510 v+ ct s 17 v+ ct+ s 4 v+ ct+ s+ 500 v+ ct s+ 56 v ct s+ 5 v ct+ s 61 v ct+ s+ 19 Total รขยย 1172 What is the correct genetic map with respect to gene order and distances (in cM) for these three genes? | |||||||||
|
