In chickens, the black plumage is determined by the E gene, brown – e, the presence of a crest – C, and the absence – c.

In chickens, the black plumage is determined by the E gene, brown – e, the presence of a crest – C, and the absence – c.
1) A brown crested rooster is crossed with a black hen without a crest. In their offspring, half of the chicks are black crested, the other half are brown crested. What are the genotypes of the rooster and chicken?
2) The hen and the rooster are black crested. From them 13 chickens were received: 7 black crested, 3 brown crested, 2 black without a crest and 1 brown without a crest. What are the genotypes of the rooster and chicken?
3) Crested chickens with black plumage – the result of crossing a brown crested hen with a black rooster without a crest. Determine the formula for F2 cleavage by genotype and phenotype. What offspring can be expected from backcrossing F1 hybrids with both parental forms?

1) Genotypes P: (women) Eess, (husband) herSS
2) genotypes P: EeCc;
3) a) the formula for splitting F2 by genotype – 1 (EECC): 2 (EECc): 1 (EECc): 2 (EeCC): 4 (EeCc): 2 (Eess): 1 (eeCC): 2 (eeCc): 1 (eess);
b) the formula for splitting F2 by phenotype – 9 black crested: 3 black without a crest: 3 brown crested: 1 brown without a crest;
c) from backcrossing F1 hybrids with the maternal form, black crested and black without a crest can be expected offspring in a ratio of 1: 1 (genotypes EECc, EEcc, EeCc, Eess in a ratio of 1: 1: 1: 1);
d) from backcrossing F1 hybrids with the paternal form, black crested and brown crested offspring can be expected in a 1: 1 ratio (genotypes EeCC, EeCc, eeCC, eeCc in a ratio of 1: 1: 1: 1).

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