A female fruit fly that is homozygous for ebony body color (on chromosome 3) and

Question

A female fruit fly that is homozygous for ebony body color (on chromosome 3) and brown eyes (on chromosome 2) is mated to a wild-type fly. All of the F1 offspring (both males & females) have wild-type bodies & eyes a) These wild type flies are mated among themselves to produce an F2. What are the expected phenotypic ratios among the F2 flies? b) One of the F1 male flies is test-crossed to a female with ebony body &brown; eyes. What are the expected phenotypic ratios among the offspring of this test cross?

Explanation / Answer

The genotype of homozygous ebony body colour and brown eyes fly will be: eebb and the genotype of wild type fly will be EEBB. Therefore the cross will be as follows:

eebb X EEBB

The gamete from eebb will be eb and from EEBB will be EB. Therefore all the F1 progeny will be EeBb (Similar phenotype with wild type).

a. These wild type mated among themselves. It means the cross will be as follows:

EeBb X EeBb

The gamete produced here will be: EB, Eb, eB, eb

Now according to punnet square:

It is clear from the Punnet square table, the phenotypic ratio will be 9:3:3:1 (9 wild, 3 nonebonybrown, 3 ebony nonbrown, 1 ebony brown).

b. The genotype of F1 male progeny is EeBb and the genotype of ebony brown fly is eebb. Therefore the cross will be as follows:

EeBb X eebb

The gamete produced by EeBb will be: EB, Eb, eB, eb and by eebb will be eb. Therefore according to punnet square table:

Here it is clear from punnet square table the phenotypic ratio will be 1:1:1:1

EB Eb eB eb EB EEBB EEBb EeBB EeBb Eb EEBb EEbb EeBb Eebb eB EeBB EeBb eeBB eeBb eb EeBb Eebb eeBb eebb

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