Question

В семье, где оба родителя кареглазые, имеется четверо детей. Двое голубоглазых имеют I и IV группы крови, двое кареглазых II и III группы крови. Определите вероятность рождения следующего ребенка кареглазого с I группой крови, если известно, что карий цвет глаз доминирует над голубым обусловлен аутосомным признаком

Answer 1

Если анализировать детей, то можно понять, что генотипы родителей CcIAI0 и CcIBI0(C, c - правша\левша). 
Вероятность рождения кареглазого ребёнка с первой группой=3\16=18,75%

Answer 2

Problem Analysis

In this problem, we are given a family where both parents have brown eyes. There are four children in the family, two of whom have blue eyes and two have brown eyes. We need to determine the probability of the next child being born with brown eyes and blood type I, given that brown eye color is dominant over blue eye color and blood type is an autosomal trait.

Solution

To solve this problem, we need to consider the inheritance patterns of eye color and blood type separately.

# Eye Color Inheritance

In this case, we are given that brown eye color is dominant over blue eye color. This means that if a person has at least one brown eye color allele, they will have brown eyes. Only individuals with two blue eye color alleles will have blue eyes.

Based on the information provided, we know that both parents have brown eyes. This means that both parents must have at least one brown eye color allele. Since they have two children with blue eyes, we can infer that both parents are heterozygous for eye color (Bb), where B represents the brown eye color allele and b represents the blue eye color allele.

Now, let's consider the probability of the next child having brown eyes. Since both parents are heterozygous (Bb), there are three possible genotypes for the child: BB, Bb, or bb. Out of these three genotypes, only BB and Bb will result in brown eyes. Therefore, the probability of the next child having brown eyes is 2/3.

# Blood Type Inheritance

Next, let's consider the inheritance of blood type. Blood type is determined by multiple alleles, including A, B, and O. In this problem, we are given the blood types of the four children: two have blood type I and two have blood type II or III.

Based on the information provided, we can infer that both parents must be heterozygous for blood type, with one parent having genotype IAi and the other parent having genotype IBi. This is because blood type I is a recessive trait and can only be expressed when an individual has two copies of the recessive allele (i).

Now, let's consider the probability of the next child having blood type I. Since both parents are heterozygous (IAi and IBi), there are four possible genotypes for the child: IAIA, IAi, IBIB, or IBi. Out of these four genotypes, only IAIA and IAi will result in blood type I. Therefore, the probability of the next child having blood type I is 2/4 or 1/2.

# Combined Probability

To determine the combined probability of the next child having both brown eyes and blood type I, we need to multiply the probabilities of each trait independently. Therefore, the probability of the next child having brown eyes and blood type I is:

(Probability of brown eyes) * (Probability of blood type I) = (2/3) * (1/2) = 1/3.

Therefore, the probability of the next child being born with brown eyes and blood type I is 1/3 or approximately 33.33%.

Conclusion

In a family where both parents have brown eyes and there are four children (two with blue eyes and two with brown eyes), the probability of the next child being born with brown eyes and blood type I is approximately 33.33%. This calculation is based on the assumption that brown eye color is dominant over blue eye color and blood type I is a recessive trait.