Question

Drosophila melanogaster (the pumice fly, sometimes called a fruit fly) normally has gray bodies. When two normal-looking flies ( coded parent 1 and parent 2) were crossed, all 100 of their offspring have normal gray bodies. From among the 100 offspring, two mating pairs are picked at random. One pair is coded Male 1 and Female 1; the other pair is coded Male 2 and Female 2. The Male 1 -female 1 mating produced 74 offspring with gray bodies and 26 offspring with back bodies. The Male 2-Female 2 mating produced 100 offspring with gray bodies. What can you infer? ( select all correct answers, assume new mutations do not occur and that the body- color trait is controlled by single gene.

1) Male or female ( but not both) is heterozygous.

2) Either P1 or P2, but not both P1 and P2, is heterozygous.

3) Gray is dominant to back

4) Either Parent 1 or Parent 2 ( but not both ) is heterozygous

5) Either Male 2 or Female 2 ( but not both ) might be hetereozygous

Answer 1

first cross

normal fly * normal fly

all progenies have normal grey body colour

and male 1 and female 1 from F1 gave both grey bodied and black-bodied flies in the ratio grey body: black body=2.8:1 this is almost equal to 3:1 that is male 1 and female 1 were heterozygous and male 2 and female 2 produced all grey offspring this shows at least one of the male 2 and female 2 are homozygous dominant.

here dominant allele is grey body colour because the cross between male 1 and female 1 ( let alleles be G- grey and g- black)

Gg* Gg

	G	g
G	GG (grey)	Gg (grey)
g	Gg (grey)	gg ( black)

the ratio expected when 2 heterozygous flies are crossed

grey body: black body=3:1

to get heterozygous offsprings for 2 normal parents at least one of them has to be heterozygous so the parents in first cross are GG and Gg

GG*Gg

	G	g
G	GG	Gg

and the all the offsprings are phenotypically normal

and male 2* female 2

all offsprings are normal, this shows at least one parent is homozygous dominant

GG*Gg

	G	g
G	GG	Gg

so the answers are

2) Either P1 or P2, but not both P1 and P2, is heterozygous.

3) Gray is dominant to back

4) Either Parent 1 or Parent 2 ( but not both ) is heterozygous