Question

У овец ген белой масти (В) доминирует над черной (в), а ген длинных ушей U — над безухостью (u). Гетерозиготные по этому при­знаку животные имеют короткие уши. Гомозиготный белый длинноухий баран скрещивался с гетерозиготными белыми короткоухими овцемат­ками. Было получено 40 ягнят. Сколько разных генотипов может быть получено в результате этого скрещивания? Сколько будет ягнят гомо­зиготных по обеим признакам? Сколько ягнят будет иметь короткие уши?

Answer 1

B- белая масть

b- черная масть

U-длинные

u-безухие

Uu-короткие

Проводим скрещивание

P:♀BbUu*♂BBUU

G:BU/Bu/bU/bu ....BU

F: BBUU--бел.длин/25%

BBUu--бел.коротк. 25%

BbUU--бел.длин/25%

BbUu--бел .коротк/25%

Различных генотипов- 4
2) Составим пропорцию

40 ягнят-100%

х ягнят- 25%  следовательно гомогозиготных ягнят40*25/100=10

3) Составим пропорцию :

40 ягнят- 100%

х ягнят -50%  следовательно гетерозиготных ягнят40*50/100=20

Answer 2

Understanding the Problem

The problem describes a breeding experiment involving sheep with different traits. The dominant gene for white coat color (V) is dominant over the gene for black coat color (v), and the dominant gene for long ears (U) is dominant over the gene for no ears (u). Heterozygous animals have short ears due to this trait. A homozygous white, long-eared ram is crossed with heterozygous white, short-eared ewes, resulting in 40 lambs. The questions are: 1. How many different genotypes can be obtained from this cross? 2. How many lambs will be homozygous for both traits? 3. How many lambs will have short ears?

Answering the Questions

1. To determine the number of different genotypes that can be obtained from this cross, we need to consider the possible combinations of the genes for coat color and ear length. Since the ram is homozygous for both traits (VVUU), all of its offspring will inherit one copy of each gene. The ewes are heterozygous for both traits (VvUu), so they can pass on either the dominant or recessive gene for each trait.

To calculate the number of different genotypes, we can use the formula 2^n, where n is the number of heterozygous traits. In this case, there are two heterozygous traits (coat color and ear length), so the number of different genotypes is 2^2 = 4.

Therefore, there can be four different genotypes obtained from this cross.

2. To determine the number of lambs that will be homozygous for both traits, we need to consider the possible combinations of the genes for coat color and ear length. The homozygous ram will always pass on the dominant genes (VU) for both traits. The heterozygous ewes can pass on either the dominant or recessive gene for each trait.

Since the ram is homozygous for both traits, all of the lambs will inherit one copy of each gene from the ram. To calculate the number of lambs that will be homozygous for both traits, we can use the formula 1/4 * total number of lambs. In this case, there are 40 lambs, so the number of lambs that will be homozygous for both traits is 1/4 * 40 = 10.

Therefore, there will be 10 lambs that are homozygous for both coat color and ear length.

3. To determine the number of lambs that will have short ears, we need to consider the possible combinations of the genes for ear length. The homozygous ram will always pass on the dominant gene (U) for ear length. The heterozygous ewes can pass on either the dominant or recessive gene for ear length.

Since the ram is homozygous for long ears, all of the lambs will inherit the dominant gene (U) for ear length from the ram. To calculate the number of lambs that will have short ears, we can use the formula 1/2 * total number of lambs. In this case, there are 40 lambs, so the number of lambs that will have short ears is 1/2 * 40 = 20.

Therefore, there will be 20 lambs that have short ears.

Conclusion

In summary, the breeding experiment between a homozygous white, long-eared ram and heterozygous white, short-eared ewes can result in four different genotypes, with 10 lambs being homozygous for both coat color and ear length, and 20 lambs having short ears.