Punnett square question?

Question

A sex-linked recessive gene produces red-green color blindness in a man. A normal appearing woman whose father was colour blind marries a color blind man.

1) What genotypes are possible for the mother of the color blind man?
2) What are the chances that the first child from this marriage will be a color blind boy?
3) Of all the girls produced by these parents, what percentage is expected to be color blind?

Thanks for any help

Answer 1

Denote the color blindness genes as c, recessive and C, normal and dominant, occurring only on the X chromosome.
The man is color blind, so his genotype is cY:

Man: cY

The woman's father is color blind and his genotype is also cY
Since she is a woman, she got the X chromosome with color blindness from him. Since she is not color blind, her genotype is cC.

Woman: Cc

The Punnett square looks like this:

.........woman
m.......C.....c
a...c...Cc...cc
n...Y...CY...cY

Now we can answer the questions.

1. The man got the Y chromosome from his father and the X chromosome with color blindness from his mother. The possible genotypes of the man's mother are Cc or cc. She has to have the gene for color blindness, but she is not necessarily color blind because it is recessive and she might only have 1 copy of it.

2. A color blind boy would have genotype cY. This occurs in only 1 of the 4 cells in the Punnett square, so the chance of that is1/4.

3. The girls occur twice in the square, with genotypes Cc and cc. The one with genotype cc is color blind and the other is unafflicted. 1 of the 2 cells that produce girls are color blind, so the chance that a girl will be color blind is 1/2.

Good question.

Answer 2

These alleles are on the X-chromosome, but not the Y.
So women have two alleles for color vision, but men only have one allele for color vision.

B - normal vision
b - color blindness

The woman is not color blind, but her father was. She had to get one X from her dad and one from Mom. She's XB Xb.

The color blind man she marries must be Xb Y.

Make the cross like this:
P XB Xb x Xb Y
gametes from Mom: XB and Xb. Put these down the side of the Punnett square.
gametes from Dad: Xb and Y. Put these across the top of the Punnett square.

Now fill in the four boxes of the square. The boxes on the top row should have: XB Xb, XB Y.
The second row should have: Xb Xb, Xb Y

Answer the three questions:
1. The mother of the color blind man must have had one Xb. We don't know about the other allele, so she has two possible genotypes: Xb XB or Xb Xb. The first one is more likely.
2. The chance of having a color blind boy? Look at the Punnett square and see how many of the four boxes show a color blind boy. Just the fourth box. So the chance is 1/4. (1/4 means one out of four)
3. Of all the girls, what percentage is expected to be color blind? Look at the girls in the boxes of the Punnett square. Note that only two of the boxes are girls. How many of the two boxes are XbXb? Just one of them. So 1/2, one out of two, or 50% of the girls would be expected to be color blind.