Colorium is an autosomal dominant trait in nutonian flies (identical to earth fruit flies in every way). there are two alleles at this locus: f = dominant allele; flies with this allele cannot observe the color fuschia; these flies have the colorium phenotype.+ = wild type allele; flies who are homozygous for this allele can observe the color fuschia; these flies do not have the colorium phenotype. a cross of a true breeding male with colorium and a true breeding female without colorium produces f1 offspring that all have the colorium phenotype. when a complementation cross is performed (with true breeding flies), we expect the following results: a. half of the f1 offspring will have the colorium phenotype. b. none of the f1 offspring will have the colorium phenotype. c. all the f1 offspring will have the colorium phenotype. d. half of the female f1 offspring will have the colorium phenotype. e. only fuschia-colored nutonian flies will have the colorium phenotype.

Half of the F1 offspring will have the Colorium phenotype

Explanation:

The Colorium has an autosomal dominant trait, so we can say the genotype of colorium phenotype is either OO or Oo and the non-colorium is oo.

In the initial cross(F1), the cross between male with colorium (OO or Oo) and female without colorium (oo) will be:

Let use the cross between Oo × oo

we will have: Oo, Oo, oo, oo (check the document below to view the punnet square of the cross)

From the results above, only half of the individuals get colorium, but in the question it is stated that the breeding produces F1 offspring that all have the Colorium phenotype.

Let's look at another cross between OO × oo

we will have: Oo,Oo,Oo,Oo  (check the document below to view the punnet square of the cross)

From the above cross, all the F1 generation having colorium phenotype.

This implies that the genotype of colorium phenotype is OO

True breeding implies  that the parents are homozygous and not heterozygous . As such If we make a complementation cross with true breeding flies.(i.e a true breeding female without Colorium) and the product of the F1 Ggeneration, we wil have:

the complementary cross between Oo × oo

which are: Oo,Oo,oo,oo  (check the document below to view the punnet square of the cross)

Therefore, we conclude that Half of the F1 offspring will have the Colorium phenotype

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