2. In Drosophila , Lyra (ly) and Stubble (Sh) are dominant mutations located chr

ID: 177354 • Letter: 2

Question

2. In Drosophila, Lyra (ly) and Stubble (Sh) are dominant mutations located chromosome 3 at map locations 40 and 58 respectively. A recessive mutation with bright red eyes (br) was discovered and shown also be on chromosome 3. A map was obtained by crossing a female who was heterozygous for all three mutations with a male homozygous for bright red (but with none of the other mutations). There is no Map for this question.

The following data were obtained.

Phenotype
(1) Ly Sh br 404
(2) +++ 422

(3)Ly++ 18
(4) +Sb br 16

(5) Ly + br 75
(6)+Sb+ 56

(7) Ly Sh +                                4
(8) ++br                                     2
                                                ______
                                                1000

a. What is the arrangement of the alleles on the mother’s chromosomes?

b. What is the genotype of the father?

c. Determine the location of the bright red mutation on chromosome 3 relative to the other two mutations.

Explanation / Answer

Answer:

a). + ---br----+

b) Ly + br + Sh+

c). Bright red mutation present in between the Lyra and Stubble mutation

Ly----4mu----br------13.74 mu------Sh

Explanation:

Explanation

Shorthand

Number of progeny

Ly Sh br

404

+ + +

422

Ly + +

+ Sh br

Ly + br

+ Sh +

Ly Sh +

+ + br

Total

997

1. If single cross over (SCO) occurs between Ly & Sh

Normal order= Ly---------Sh & +-----+

After cross over= Ly-----+ & +------Sh

Ly-----+ recombinants are 18+75 =93

+------Sh recombinants are 16+56 = 72

Total recombinants = 165

RF = (165/997.)*100 =16.55%

2. If single cross over (SCO) occurs between Sh & br

Normal order= Sh---------br & +----+

After cross over= Sh-----+ & +------br

Sh-----+ recombinants are 56 + 4 = 60

+------br recombinants are 75 + 2 =77

Total recombinants = 137

RF = (137/997)*100 = 13.74%

3. If single cross over (SCO) occurs between Ly & br

Normal order= Ly---------br & +------+

After cross over= Ly-----+ & +------br

Ly-----+ recombinants are 18+4 = 22

+------br recombinants are 16 + 2 =18

Total recombinants = 40

RF = (40/997)*100 = 4%

% RF = Map unit distance

The order of genes is -----