Question

Пробанд - нормальная женщина имеет 5 сестер, две из которых однояйцевые близнецы , две двухяйцевые близнецы. Все имеют шесть пальцев на руке . Мать пробанда нормальная , отец шестипалый . Со стороны матери все предки нормальные . Дедушка по линии отца и все родственники пятипалые . Определите вероятность рождения в семье пробанда шестипалых детей при условии , если она выйдет замуж за нормального мужчина .

Answer 1

Обозначим:
А-норм.(пятипалый)
а- шестипалость
Генотип пробанда Аа т.к. ее отец был шестипалый и имел генотип аа, а мать Аа( такой вывод мы сделали потому как в семье имеются шестипалые сёстры пробанда).
Вероятность рождения шестипалых детей в семье пробанда зависит от того какой генотип будет у её нормального мужа. Он может быть либо Аа либо АА.
1) Если генотип мужа пробанда Аа, тогда:
Р Аа х Аа
G A a   A a
F1 AA Aa Aa aa
то есть вероятность рождения шестипалого ребёнка в этом случае составит 25%.
2) Если генотип мужа пробанда АА, тогда:
Р Аа х АА
G A a    A
F1 AA Aa 
то есть вероятность рождения шестипалого ребёнка в этом случае составит 0%, но 50% будут носителями гена шестипалости.

Answer 2

Probability of having six-fingered children in the Proband's family

To determine the probability of having six-fingered children in the Proband's family, we need to consider the information provided about the Proband, their siblings, parents, and grandparents.

According to the information given, the Proband is a normal woman who has five sisters. Two of her sisters are identical twins (monozygotic) and two are fraternal twins (dizygotic). All of them have six fingers on their hands. The Proband's mother is normal, and her father has six fingers. On the mother's side, all the ancestors are normal, while on the father's side, the grandfather and all other relatives have five fingers.

To calculate the probability, we need to consider the genetic factors involved. The presence of six fingers is a result of a dominant gene. Let's break down the probability step by step:

1. The Proband's father has six fingers, which means he is heterozygous for the six-fingered gene. Let's represent the gene as "S" for six fingers and "s" for five fingers. Therefore, the father's genotype can be represented as "Ss".

2. The Proband's mother is normal, which means she has two copies of the five-fingered gene. Her genotype can be represented as "ss".

3. The Proband's sisters include two identical twins and two fraternal twins, all of whom have six fingers. Since the Proband is also normal, it can be inferred that the Proband's genotype is "ss".

4. When the Proband marries a normal man, the probability of having six-fingered children depends on the genotype of the husband. Since the husband is normal, his genotype can be represented as "ss".

5. When the Proband and her husband have children, the probability of each child having six fingers can be calculated using Punnett squares. In this case, since both parents have the genotype "ss", the probability of having a six-fingered child is 0%.

Therefore, based on the information provided, the probability of the Proband and her normal husband having six-fingered children is 0%.

Please note that this calculation assumes that the information provided is accurate and complete.