In chickens, a wrinkled comb dominates over a simple comb. A rooster with a wrinkled comb was crossed

In chickens, a wrinkled comb dominates over a simple comb. A rooster with a wrinkled comb was crossed with two chickens with wrinkled combs. The first bore 14 chicks with wrinkled combs, and the second – 7 with wrinkled and 2 normal ones. A. Describe the genotype of the rooster in words. B. Describe the genotype of the first chicken in words. Q. Describe in words the genotype of the second chicken D. What theoretical phenotype cleavage resembles the cleavage in the second case? E. Could a simple-combed bird be a recessive homozygous?

Let’s designate the gene that causes the development of the wrinkled ridge as B, then the simple ridge gene will be designated as b.

Considering the fact that all offspring from crossing a rooster with the first hen had combs, it should be concluded that the first hen was homozygous for the dominant gene that causes the development of the comb. Thus, the genotype of the first chicken will be BB.

Since when a rooster was crossed with a second hen, some of the chickens did not have a crest, and, therefore, had the aa genotype, it should be concluded that the rooster and the second hen carry the recessive gene in their genotype in a latent form. Their genotypes are Bb.

A. Rooster: heterozygous, Bb;

B. The first chicken is homozygous for the dominant gene, BB;

B. The second chicken is heterozygous, Bb;

D. The law of splitting 1: 2: 1 by genotype and 3: 1 by phenotype in F2 offspring of pure lines (2nd Mendelian law);

E. A bird with a simple crest is necessarily a recessive homozygote, since homozygosity is an indispensable condition for the phenotypic realization of a recessive trait.