Question

Помогите пожалуйста с решением задач, очень нужны помощь. ПО ТЕМЕ СЦЕПЛЕННЫХ ПОЛОВ С ПРИЗНАКОМ 1. Гипертрихоз наследуется как признак, сцепленный с Y-xpoмосомой. Какова вероятность рождения детей с этой аномалией в се­мье, где отец страдает гипертрихозом? 2. У бронзовых индеек вибрация (дрожание) тела передается ге­ном, сцепленным с полом (с Z-хромосомой). При спаривании вибрирующих индюков и нормальных самок в F, все самки были вибрирую­щими, а самцы — нормальными. Определите, какой из двух признаков доминирует, и установите генотипы родителей.

Answer 1

1. так как наследуется признак с У-хромосомой, то все девочки будут здоровы, а все мальчики больны. вероятность рождения с гипертрихозом 50%

ХХ х ХУа

а1 ХХ : ХУа

2. у птиц - самка ZW, самец - ZZ

А-норма

а- дрожание

P ZАW x ZаZа

F1 ZAZa : ZaW

(самец норм : самка дрож)

Answer 2

Inheritance of Hypertrichosis

Hypertrichosis is a condition characterized by excessive hair growth. In this case, it is inherited as a trait linked to the Y-chromosome. The Y-chromosome is passed from father to son, so if the father has hypertrichosis, there is a possibility that his children may also have this anomaly.

The probability of having children with hypertrichosis in a family where the father has hypertrichosis can be determined using the principles of Mendelian genetics. Since hypertrichosis is linked to the Y-chromosome, it is considered a sex-linked trait.

In humans, males have one X-chromosome and one Y-chromosome, while females have two X-chromosomes. Since the Y-chromosome is only present in males, hypertrichosis is more commonly observed in males.

To calculate the probability of having children with hypertrichosis, we need to consider the genotypes of the parents. Let's assume that the father is affected by hypertrichosis and has the genotype XY (where X represents the normal allele and Y represents the hypertrichosis allele). The mother, being a female, does not have the Y-chromosome and has the genotype XX.

The possible genotypes of the offspring can be determined using a Punnett square:

| | X | Y | |-----|-----|-----| | X | XX | XY | | X | XX | XY |

From the Punnett square, we can see that all male offspring (XY) will inherit the hypertrichosis allele from their father and will have hypertrichosis. Female offspring (XX) will not inherit the hypertrichosis allele and will not have hypertrichosis.

Therefore, in this family, all male children will have hypertrichosis, while female children will not have the condition.

Please note that this is a simplified explanation and assumes that hypertrichosis is solely determined by the Y-chromosome. In reality, the inheritance of hypertrichosis can be more complex and involve other genetic factors.

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