Question

Мужчина-дальтоник (цветовая слепота, признак сцеплен с х-хромосомой)женился на женщине с нормальным зрением, но не имевшие отца дальтоника . составить схему решение задачи. определить генотипы мужчины и женщины, генотипы и фенотипы возможного потомства. кокова вероятность рождения дочерей и сыновей-дольтоников?

Answer 1

Как известно, дальтонизм сцеплен с Х - хромосомой, поэтому переносчики - женщины, а болеют. как правило, мужчины. Дальтонизм у мужчин встречается в 20 раз чаще.

 

 

В данной задаче мы имеем:

Мужчина, больной дальтонизмом      ХDY

Здоровая женщина, не являющаяся носителем гена дальтонизма - ХХ

 

Могут иметь следующее потомство:

Все девочки будут здоровы, но они будут носителями гена дальтонизма, и имеют генотип ХХD

Все мальчики здоровы и имеют генотип ХY

Вероятность рождения детей-дальтоников - 0

 

 

Answer 2

Problem Statement

A man with color blindness (a form of color vision deficiency linked to the X chromosome) marries a woman with normal vision, but whose father was not color blind. We need to determine the genotypes of the man and the woman, as well as the possible genotypes and phenotypes of their offspring. Additionally, we need to calculate the probability of having daughters and sons who are color blind.

Solution

To solve this problem, we need to understand the inheritance pattern of color blindness and the genotypes involved.

1. Inheritance of Color Blindness: - Color blindness, specifically red-green color blindness, is a sex-linked trait that is recessive and carried on the X chromosome. - The gene responsible for color blindness is denoted as "Xc," where the lowercase "c" represents the color blindness allele. - The normal vision allele is denoted as "X."

2. Genotypes of the Man and the Woman: - The man is color blind, which means he must have inherited the color blindness allele from his mother. - Since color blindness is a recessive trait, the man's genotype is "XcY". - The woman has normal vision, but her father was not color blind. This means she must have inherited a normal vision allele from her father. - Since the woman's father was not color blind, he must have had the genotype "XY". - Therefore, the woman's genotype is "XX."

3. Genotypes and Phenotypes of the Offspring: - The possible genotypes of the offspring can be determined by combining the genotypes of the parents. - Since the man is color blind (XcY) and the woman has normal vision (XX), the possible genotypes of their offspring are as follows: - Sons: 50% chance of inheriting the color blindness allele (XcY) from the father and 50% chance of inheriting the normal vision allele (XY) from the mother. - Daughters: 100% chance of inheriting the normal vision allele (XX) from the mother. - Therefore, the possible genotypes of the offspring are: - Sons: XcY (color blind) and XY (normal vision). - Daughters: XX (normal vision).

4. Probability of Having Color Blind Daughters and Sons: - To calculate the probability of having color blind daughters and sons, we need to consider the possible combinations of genotypes. - Since color blindness is a recessive trait, a daughter would need to inherit the color blindness allele (Xc) from both parents (XcY from the father and XX from the mother) to be color blind. - Therefore, the probability of having a color blind daughter is 0%. - On the other hand, a son would only need to inherit the color blindness allele (Xc) from the father (XcY) to be color blind. - Therefore, the probability of having a color blind son is 50%.

Summary

In summary, the man is color blind (XcY) and the woman has normal vision (XX). Their possible offspring genotypes are XcY (color blind sons) and XY (normal vision sons), as well as XX (normal vision daughters). The probability of having color blind daughters is 0%, while the probability of having color blind sons is 50%.