On the planet Strogus the equivalent of the earth drosophila

On the planet Strogus, the equivalent of the earth drosophila (fruit fly) has the Morty phenotype, in which those with the trait love bagels. The wild- type flies do not. In a cross between a Morty female and wild type male, of the 1000 offspring (F1 generation), 507 are wild-type females and 493 are Morty males. A complementation cross is performed on the parental generation. Which of the following counts would be most likely in the F1 generation, if there are 1000 offspring?. 751 Morty males, 249 female wild types. 751 male wild types, 249 female wild types. All 1000 F1 files are Morty flies. 512 male wild-types, 488 female wild-types. 512 male wild-types, 488 Morty females.

Solution

Wild type flies do not have the trait.

Let the symbol for wild type be “+”, and that having the trait be “t”. Initially let us consider that the morty characteristic is autosomal.

Morty female “tt” is crossed with wild type “+t” male. Remember in Drosophila males only one X-chromosome is there, and crossing over is absent.

Results should be “tt” (morty =50%) and +t (wild type=50%). This result can be with either males or females. But it is given that out of 1000, 507 are wild type females and 493 are morty males.

This means that the characteristic was X-linked. And the cross was tt female cross to + male, resulting in “t (morty)” male and “+t (wild type)” female.

This is same as the expected result. This means that morty characteristic is located on the X-chromosome. It is autosomal.

Now, a complementation cross is performed. A complementation test is one in which two parents having homozygous recessive mutation are crossed.

Morty male (t) crossed to morty female (tt)

Results: All will be morty. So, (C) is the correct answer.