In fruit flies, curved wings (w) are recessive to the dominant straight wings (W), and ebony body (b) is recessive to the dominant gray body (B). A fly that has curved wings and ebony body is crossed with a fly that has straight wings and gray body. Of the offspring there were approximately equal numbers of straight gray, straight ebony, curved gray, curved ebony. What are the genotypes of the parents and offspring

The correct answer =

parents : WwBb x wwbb

offspring :

WwBb, Wwbb, wwBb, and wwbb

Explanation:

It is given that the straight wings (W) is dominant over curved wings (w) and the ebony body (b) is recessive to the dominant gray body (B) and there is a cross between curved wings and ebony body and straight wings and gray body is made. The phenotypic ratio is: 1:1:1:1 for the four different phenotypes straight gray, straight ebony, curved gray, curved ebony.

It is not given the genotype of parents, however, for the curved and ebony fly, it is clear that both alleles are recessive as recessive alleles only appear in absence of the dominant allele of the trait.

For another parent, the possible genotypes can be  WWBB or WwBb, however in the case of WWBB the phenotypic ratio would not be equal therefore,

crossing WwBb x wwbb

This can be represented in the form of a punnett square as below:

          wb   wb           wb           wb  

WBWwBbWwBbWwBbWwBbStraight wing, Gray body

WbWwbbWwbbWwbbWwbbStraight wing, Ebony body

wBwwBbwwBbwwBbwwBbCurved wing, Gray body

wbwwbbwwbbwwbbwwbbCurved wing, Ebony Body

The phenotypic ratio is: 1:1:1:1 for the four different phenotypes

hope it mark me as brainlieast.

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