Question

1. Four eye-color variants in Drosophila- brown, carnation, apricot, and purple - are inherited as recessive traits. Red is the dominant wild-type color. Eight crosses are made between parents of pure-breeding lines. Which of these traits are inherited as X-linked and which are autosomal? Explain how you distinguish between X-inked and autosomal inheritance. a. b. Predict the F2 ratios of crosses between the F1 progeny in A, B, D, and G Cross Parents F1 Progeny FemaleMale Red Red Red Red Red Red Red Female Male Red Red Purple Apricot A Apricot Apricot Red Red Red Carnation Red Red Red Brown Red Red Carnation Red D F Purple G Red H Red Red Brown Carnation Red

Answer 1

*Before start working with the problems the tips that have to be kept in mind are: a male progeny gets a X chromosome from his mother only. so X linked traits can be understand by examining the mother and the son.On reversing the male and female parents, the F1 ratio also change in X linked phenotype.

a) For answering the question a ,conisder the table and analyse the crosses Aand D,B and G,C and F,E and H as pairs.

A and D : Consider it is autosomal trait then 'a'(Apricot reccesive gene from mother) and 'R'(Red dominant gene from father) combines to give Ra F1 progeny.As R is dominant over a all the F1 will be red eyed.But that is not happening.

Now consider 'a' as X linked, so the genotype apricot mother will be X^aX^a and genotype of red father will be X^RY and the X^a is the gamete from the female and X^R and Y are the gametes from male.The F1 progeny will be X^aX^R(Female),redeyed female and X^aY(male) apricot eyed male. On reversing the parental phenotype(Cross D),X^RX^R(female) and X^aY(male) and the F1 progeny will be X^RX^a(female red eyed since a is reccessive) and X^RY(red eyed male). So it can be concluded that apricot is X linked.

Similarly, by considering the pairs B and G, C and F, and E and H it can be concluded that Brown and purple are autosomal and Carnation is X linked.

b)(A) F1 :red female and apricot male;female genotype(X^RX^a) and male genotype(X^aY). Gametes will be X^R, X^a (from female)and X^a,Y(from male) to give F2 progeny as X^aX^a,X^aY,X^RX^aand X^RY so the phenotypic ratio will be red :apricon=2:2 that is 1:1.Similarly solve for (D) to obatin the ratio red :apricot as 3:1

(B) brown(bb) and red(RR): it is autosomal,the F1 genotype will be Rb and Rb on crossing the F2 progenies will be RR,Rb,Rb,bb the phenotypic ratio will be red:brown=3:1.Similarly,solve for(G) to obtain the red : brown ratio as 3:1.