When a colour blind woman marries a normal man, what is the probability of their son being colour blind?
25%
100%
50%
0%
Explanation
Video Explanation
Post your Contribution
Discussions (27)

Color blindness is a genetic condition that is typically inherited through a recessive gene on the X chromosome. In humans, females have two X chromosomes (XX), while males have one X and one Y chromosome (XY).
If a color blind woman (let's represent her genotype as XcX) marries a normal man (genotype as XY), the possible combinations of their offspring are as follows:
- Sons:
- 50% chance of inheriting the X chromosome with the normal color vision gene (X) from the father and the X chromosome with the color blind gene (Xc) from the mother. These sons will be unaffected carriers of the color blindness trait (XcY).
- 50% chance of inheriting the Y chromosome from the father, which means they will have normal color vision (XY).
- Daughters:
- 100% chance of inheriting one X chromosome with the normal color vision gene (X) from the father.
- 100% chance of inheriting one X chromosome with the color blind gene (Xc) from the mother.
Therefore, the probability of their son being color blind (XcY) is 50%.
I think my school is wrong

100%
This is because color blindness and other s*x-linked traits are located on the X chromosome and the only X chromosome that a son (boy) can inherit is from the mother
Remember XY make a boy
XX make a girl
s*x-linked trait are expressed more in male that in female because the only X chromosome of a male child is from the mother.
hope I'm able to convince and not to confuse you 

okayy, the probability of their the sons being colour blind not the whole children

correct answer is C
1. The mother is color blind, which means she has two mutated genes (xx).
2. The father is normal, which means he has one X chromosome with the normal gene and one Y chromosome (Xy).
3. Their son inherits one X chromosome from his mother, which has a 50% chance of carrying the mutated gene (X or x).
4. He also inherits one Y chromosome from his father, which carries the normal gene (Y).
5. If the son inherits the mutated gene from his mother (x), he will be color blind (Xy).
6. Since there's a 50% chance of inheriting the mutated gene from his mother, the probability of their son being color blind is 50% (C).
So, the key point is that the son inherits one X chromosome from his mother, which has a 50% chance of carrying the mutated gene, making him color blind.

The correct answer is B. 100%.
Explanation
Color blindness is typically an X-linked recessive trait. Let's denote the allele for color blindness as "c" and the normal allele as "C".
- A color-blind woman would have the genotype cc (homozygous recessive) on her X chromosomes.
- A normal man would have the genotype CY (where Y is the Y chromosome and C is the normal allele on his single X chromosome).
Their sons receive the Y chromosome from their father and the X chromosome from their mother. Since the mother is color-blind (cc), she can only pass on a c allele to her sons. Therefore, all sons will receive the Xc chromosome from their mother and the Y chromosome from their father, making them cY, which means they will be color-blind.
Thus, the probability of their son being color-blind is 100%.



