Question

2. A rose- combed rooster is mated with two rose-combed hens. Hen A produces 14 chicks, all rose-combed. Hen B produces 9 chicks, 7 of which are rose-combed and 2 single-combed. What are likely genotypes of the parent? Explain.

Answer 1

Step-by-step explanation:

Based on the given information, we can use Punnett squares to determine the genotypes of the parent birds.

Let's use "R" to represent the rose-combed allele and "r" to represent the single-combed allele.

From the problem statement, we know that:

The rooster is rose-combed, so it must have at least one copy of the R allele.

Hen A produces 14 chicks, all of which are rose-combed. This means that Hen A must be homozygous for the R allele (RR).

Hen B produces 9 chicks, 7 of which are rose-combed and 2 of which are single-combed. This means that Hen B must be heterozygous for the R allele (Rr).

Using this information, we can set up Punnett squares to show the possible genotypes of the offspring from each parent:

Rooster: R r

---- ----

Hen A: R R R R

Offspring: R R R R (all rose-combed)

Hen B: R r R r

Offspring: R R R R (rose-combed)

R r R r (rose-combed)

r R r R (single-combed)

r R r R (single-combed)

Based on the Punnett squares, we can see that the rooster must be heterozygous for the R allele (Rr), since both hens produce only rose-combed offspring. Hen A is homozygous for the R allele (RR), while Hen B is heterozygous (Rr). This means that the genotype of the rooster is Rr, and the genotypes of the hens are RR and Rr, respectively.