
Гетерозиготный бык голштинской породы несет рецессивный ген, вызывающий отсутствие шерсти. В
гомозиготном состоянии этот ген ведет к гибели теленка. Определите вероятность рождения нежизнеспособного теленка от скрещивания этого быка с одной из его дочерей от нормальной коровы.

Ответы на вопрос

Аа, аа-гибель
Р: Аа * АА
Г: А а А
дети : Аа АА
Р: Аа* Аа
г: а А А а
дети : АА Аа Аа аа(гибель)



Probability of a Nonviable Calf from Crossbreeding
To determine the probability of a nonviable calf (one that does not survive) from the crossbreeding of a heterozygous Holstein bull carrying a recessive gene for absence of hair with one of his daughters from a normal cow, we need to consider the inheritance patterns of the gene.
The bull is heterozygous for the recessive gene, which means he carries one copy of the gene for absence of hair (recessive allele) and one copy of the gene for hair (dominant allele). In the homozygous state, where both copies of the gene are recessive, the gene leads to the death of the calf.
When the bull is crossed with his daughter from a normal cow, there are two possible scenarios for the offspring:
1. The daughter does not carry the recessive gene: In this case, the offspring will receive one copy of the dominant allele for hair from the daughter and either the dominant or recessive allele from the bull. Since the dominant allele is present, the calf will have hair and will be viable.
2. The daughter carries the recessive gene: In this case, there is a 50% chance that the daughter will pass on the recessive allele to the offspring. If the calf receives the recessive allele from both parents, it will be homozygous for the recessive gene and will not survive.
Therefore, the probability of a nonviable calf from this crossbreeding is 50%.
Please note that the information provided is based on the assumption that the inheritance of the absence of hair gene follows simple Mendelian genetics. Other factors, such as genetic interactions or modifiers, could influence the outcome.


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