In Drosophila melanogaster , cut wings ( ct ) is recessive to normal wings ( ct
ID: 79522 • Letter: I
Question
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 1172
What is the correct genetic map with respect to gene order and distances (in cM) for these three genes? ct 8.7 s 6.2 v s 4.5 v 15.6 ct s 12.5 ct 2.6 v v 3.8 s 10.8 ct 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 1172
What is the correct genetic map with respect to gene order and distances (in cM) for these three genes? 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 1172
What is the correct genetic map with respect to gene order and distances (in cM) for these three genes? ct 8.7 s 6.2 v s 4.5 v 15.6 ct s 12.5 ct 2.6 v v 3.8 s 10.8 ct
Explanation / Answer
Answer:
v 3.8 s 10.8 ct
Explanation:
Usually parental combinations will be more than the recombinants, so the arrangement of genes in heterozygous condition is v+ ct+ s+ / v ct s
1. If single cross over (SCO) occurs between v+ & ct+ and v & ct
Normal order= v+---------ct+ & v-----ct
After cross over= v+-----ct & v------ct+
v+-----ct recombinants are 17+56 = 73
v------ct+ recombinants are 61+19 = 80
Total recombinants = 153
RF = (153/1172.)*100 =13.05 %
2. If single cross over (SCO) occurs between ct+ & s+ and ct & s
Normal order= ct+---------s+ & ct----s
After cross over= ct+-----s & ct------s+
ct+-----s recombinants are 4+61 = 65
ct------s+ recombinants are 5+56= 61
Total recombinants = 126
RF = (126 / 1172)*100 = 10.75%
3. If single cross over (SCO) occurs between v+ & s+ and v & s
Normal order= v+---------s+ & v------s
After cross over= v+-----s & v------s+
v+-----s recombinants are 17+4 = 21
v------s+ recombinants are 19+5 = 24
Total recombinants = 45
RF = (45/1172)*100 =3.84%
% RF = Map unit distance
The order of genes is -----
v---3.8 m.u.---s----------10.8m.u.--------ct
Order of genes is v-s-ct