(2) If this Drosophila melanogaster is mated with wild normal-eyed male Drosophila melanogaster, the offspring of Drosophila melanogaster are xbxbxb, XbY, Y, which is fatal, so the offspring of Drosophila melanogaster are normal-eyed females, Roddy-eyed females and normal-eyed males, accounting for 1/3. If the genotype of wild-type normal-eyed male Drosophila melanogaster is represented by XbY after X-ray treatment, it will be crossed with female Drosophila melanogaster, XbXb, XbY. If the recessive lethal mutation did not occur, XbY survived, and the male offspring accounted for 1/3.
(3) According to the meaning of the question, gray is dominant to yellow, and the genotypes of X+X+XXYY offspring are X+Xy and X+Y, both of which are gray.
(4)Xyxy×X+Y, female Drosophila melanogaster appeared in the offspring, and then the female individuals appeared as composite chromosomes, and then the females produced gametes of XyXy and o, and when they combined with the X+ and y gametes produced by X+Y, they were followed by individuals of XyXy, X+Xy Xy, X+O and OY, so the number of dead individuals accounted for 50%, and the genotype of the surviving individuals was XyXy (orpiment).
So the answer: (1)2/3.
(2)0? 1/3
(3) Gray
(4) 50% women? XyXyY (female Drosophila melanogaster) X+O (male Drosophila melanogaster)