# In cattle, the horniness gene dominates over the horniness gene, and the black gene dominates over

In cattle, the horniness gene dominates over the horniness gene, and the black gene dominates over red. Both pairs of genes are not linked. For several years, black hornless cows were crossed with black hornless buogai. So 896 heads of young animals were received; of these, 535 calves were black and 161 were red. How many horned calves were there and how many of them were red? b) On the farm, 984 calves were received from 1000 horned red cows, of which 472 were red, 483 hornless, 50 horned. Determine the genotype and percentage of black calves

If the hornless gene dominates over the horniness gene, we denote A – hornless, and – horned (recessive trait).

Black gene B (dominant), b- red.

If the offspring included both black and red calves, then the parents will be heterozygous. Let’s draw up a scheme for solving the problem.

A – hornless, horned, B – black, b – red.

P ♀AaBb X AaBb

G AB, Ab, ab, ab / Ab, Ab, ab, ab.

Let’s make a Punnet lattice.

We get the splitting:

9: 3: 3: 1.

9 hornless black.

3 hornless red.

3 horned black.

1 horned red.

Find the number of horned calves.

896 – 535 = 361 -161 = 200 (horned).

Z: 1 = 4 parts.

200: 4 = 50 (1 part – horned red),

200 – 50 = 150 – horned black.

B)

♀ AaBb x ♂ aabbb.

G AB, Ab, ab, ab / ab.

F1 AaBb, Aabbb, aaBb, aabbb

A – horned, B – red, a – rough, b – black.

Splitting by phenotype: 1: 1: 1: 1, 25% horned red, 25% horned black, 25% hornless black; 25% fine red.

Black calves – 50%.

By genotype: 1: 1: 1: 1.

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