A chicken is homozygous for the dominant P allele of the Pigment gene for pigmen

Question

A chicken is homozygous for the dominant P allele of the Pigment gene for
pigmented feathers, but it is also homozygous for the dominant I allele of the Inhibitor gene that inhibits pigmentation of the feathers. So, it has colorless feathers. A second chicken is homozygous for the recessive p allele of the Pigment gene and makes no pigment, but it is homozygous ii at the Inhibitor gene. So, it also has colorless feathers. Assume that P and I are unlinked. What fraction of the F2 progeny of these two chickens will have pigmented feathers? (put answer in Fraction Form)

Explanation / Answer

Given that,
One chicken is homozygous for the dominant P allele of the pigment gene for
Pigmented feathers, and homozygous for the dominant I allele of the Inhibitor gene, that inhibits pigmentation of the feathers. (PPII)

Second chicken is homozygous for the recessive p allele of the Pigment gene and makes no pigment and homozygous ii at the Inhibitor gene. (ppii)

Both P and I are unlinked genes.

Cross between PPII * ppii gives

PPII * ppii

PpIi - progeny

All F1 progeny show colourless feathers due to the presence of inhibitor gene.

Cross between F1 progeny give the following results

PpIi * PpIi

F2 progeny are

PPII - colourless feathers
PpII - colourless feathers
PPIi - colourless feathers
PpIi - colourless feathers
PpII - colourless feathers
ppII - colourless feathers
PpIi - colourless feathers
ppIi - colourless feathers
PPIi - colourless feathers
PpIi - colourless feathers
PPii - PIGMENTED FEATHERS
Ppii - PIGMENTED FEATHERS
PpIi - colourless feathers
ppIi - colourless feathers
Ppii - PIGMENTED FEATHERS
ppii - colourless feathers

The fraction of the F2 progeny of these two chickens with pigmented
feathers are 3/16 = 0.185

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