Answer:
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
WB WwBb WwBb WwBb WwBb Straight wing, Gray body
Wb Wwbb Wwbb Wwbb Wwbb Straight wing, Ebony body
wB wwBb wwBb wwBb wwBb Curved wing, Gray body
wb wwbb wwbb wwbb wwbb Curved wing, Ebony Body
The phenotypic ratio is: 1:1:1:1 for the four different phenotypes
