A cross is made between homozygous wild-type female Drosophila (a^+ a^+ b^+ b^+
ID: 86009 • Letter: A
Question
A cross is made between homozygous wild-type female Drosophila (a^+ a^+ b^+ b^+ c^+ c^+) and triple-mutant males (aa bb cc) (the order here is arbitrary). The F1 females are test crossed back to the triple-mutant males and the F2 phenotypic ratios are as follows: a^+ b c = 18 a b^+ c = 112 a b c = 308 a^+ b^+ c = 66 a b c^+ = 59 a^+ b^+ c^+ = 320 a^+ b c^+ = 102 a b^+ c^+ = 15 total = 1000 Map these gene to a chromosome in a correct order and determine the map distance between them. Show all your work.Explanation / Answer
The following is the data provided:
Genotype of the parents:
Female : wild type, (a+, b+, c+/a+,b+, c+)
Male: triple mutant (a, b, c / a, b, c)
Therefore, the genotype of the F1 progeny would be a+, b+, c+ / a, b, c.
The F1 Females are test crossed to the triple mutant male.
The following are the gametes produced by the F1 female
(a+, b+, c+), (a, b, c), (a+, b, c+), (a, b+, c), (a+, b+, c), (a, b, c+), (a, b+, c+), and (a+, b, c)
The following is the phenotypic rations of the F2 offspring:
Genotype
Ratio
Parental/recombinant/SCO/DCO
b+ a+ c+
320
Recombinant
b a c
308
Recombinant
b a+ c+
102
Recombinant/SCO
b+ a c
112
Recombinant/SCO
b+ a+ c
66
Recombinant/SCO
b a c+
59
Recombinant/SCO
b+ a c+
15
Recombinant/DCO
b a+ c
18
Recombinant/DCO
From the data given, it can be seen that the gene ‘a’ is flapped in the double recombinants. Hence, the gene is between the genes b and c.
The distance between genes a and b is calculated as follows:
(No. of SCO between genes a & b + No of DCO / Total number of progeny) * 100
=( 102 + 112 + 15 + 18 / 1000) * 100
= 24.7
The recombination frequency between the genes a and b is 0.247 and the map distance is 24.7 mu.
The distance between genes b and c is calculated as follows:
(66 + 59 + 15 + 18/1000) * 100
= 15.8
The recombination frequency between the genes b and c is 0.158 and the map distance is 15.8 mu.
Therefore, the map distance between the genes b and c = 24.7 + 15.8 = 40.5 mu
The expected frequency of DCO = 0.247 * 0.158 = 0.0390
The expected number of DCO = 0.0390 * 1000 = 39
The actual number of DCO observed = 33
Genotype
Ratio
Parental/recombinant/SCO/DCO
b+ a+ c+
320
Recombinant
b a c
308
Recombinant
b a+ c+
102
Recombinant/SCO
b+ a c
112
Recombinant/SCO
b+ a+ c
66
Recombinant/SCO
b a c+
59
Recombinant/SCO
b+ a c+
15
Recombinant/DCO
b a+ c
18
Recombinant/DCO
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.