Question

12. A Drosophila female is heterozygous for two differently marked X chromosomes. One chromosome carries the recessive white (w eye color mutation and a recessive lethal mutation The w and genes are 20 map units apart. The other chromosome carries the wildtype alleles of both genes. This female was crossed to a male whose Xchromosome also carried the w mutation. A. What fraction of daughters will be white-eyed? B. What fraction of sons will be white-eyed? C. What will be the proportion of daughters to sons among this female's progeny? D. Overall, what phenotypes (eye color and sex) do you expect to see and what will be their proportions?

I know drosophila only has recombination among females. So since 20 map units=20% recombination,

I HAVE THE ANSWERS! I need help understanding why the answers are what they are. PLEASE don't just copy the answers I will write down

a) 1/2

b) (.1)/(.4)(.1) = 20%

c) 2:1

d)

1/3 w female

1/3 w+ female

1/15 w male

4/15 w+ male

I NEED AN EXPLANATION OF THE ANSWERS. NOT THE ANSWERS THEMSELVES. THANK YOU IN ADVANCE.

Answer 1

A) 1/2 answer so in experiment we check two chromosomes so chances of any chromosome to show is equal hence it is 1/2.

B) 20% answer as you show in your image it is as per mendles law of mutation chances of son with white eyes is only 20%.

C) among female progeny it ratio is 2:1 because female it self contain it's own chromosome that is responsible for daughter progeny hence ration obviously higher side on daughter.

D) you already wirte answer and that is come on basis of cross link probability chances that is explained by mendles law hence that result can come as you show.