Question

Карий цвет глаз доминирует над голубым, ретинобластома(опухоль глаза) определяется другим доминантным аутосомным геном. Пенетрантность заболевания равна60%. Какова вероятность того, что здоровыми от брака гетерозиготных по обоим признакам родителей, будут кареглазые дети?

Answer 1

Кажется так....))) Вроде неплохо получилось)

Answer 2

Probability of Healthy Children with Heterozygous Parents

To calculate the probability of having healthy children with heterozygous parents, we need to consider the inheritance pattern and the penetrance of the disease.

Given the information provided, we know that brown eye color is dominant over blue eye color, and retinoblastoma (a tumor of the eye) is determined by a different autosomal dominant gene. The penetrance of the disease is stated to be 60%.

To determine the probability of having healthy children with heterozygous parents, we need to consider the possible genotypes of the parents and the inheritance pattern of the traits.

Let's assume that the dominant allele for brown eye color is represented by "B" and the recessive allele for blue eye color is represented by "b". Similarly, let's assume that the dominant allele for retinoblastoma is represented by "R" and the recessive allele for not having the disease is represented by "r".

The possible genotypes of the parents can be represented as follows: - Parent 1: BbRr (heterozygous for both eye color and retinoblastoma) - Parent 2: BbRr (heterozygous for both eye color and retinoblastoma)

To determine the probability of having a child with brown eyes and without retinoblastoma, we need to consider the probability of inheriting the dominant alleles for both traits from each parent.

The probability of inheriting the dominant allele for brown eye color (B) from each parent is 1/2, as both parents are heterozygous for this trait.

The probability of inheriting the dominant allele for retinoblastoma (R) from each parent is also 1/2, as both parents are heterozygous for this trait.

To calculate the overall probability, we multiply the probabilities of inheriting the dominant alleles for both traits from each parent:

Probability of inheriting brown eye color (B) from both parents: (1/2) * (1/2) = 1/4

Probability of inheriting retinoblastoma (R) from both parents: (1/2) * (1/2) = 1/4

To calculate the probability of having a child with both brown eyes and without retinoblastoma, we multiply the probabilities of inheriting the dominant alleles for both traits:

Probability of having a child with brown eyes and without retinoblastoma: (1/4) * (1/4) = 1/16

Therefore, the probability of having healthy children with heterozygous parents, who are carriers for both brown eye color and retinoblastoma, is 1/16.

Please note that this calculation assumes that the traits are inherited independently and that there are no other genetic factors or complications involved.