Give your answer separately for males and females.?

Question

One form of color blindness (c ) in human is caused by sex-linked recessive mutant gene located on X chromosome. A women with normal color vision (c+) and whose father is color-blind marries a man of normal vision whose father was also color-blind. What proportion of their offspring will be color-blind?

Answer 1

We generally write this a XC or Xc (the c's are superscript)
to show that this gene is carried on the X chromosome. The Y does not carry it.
The genotype and phentypes are
XCXC normal female
XCXc carrier female (but normal vision)
XCY normal male
XcY affected male.

Now just to point out, males give their Y chromosome to sons and X's to females, so a male cannot pass this condition onto any son, but will to all females IF he has the condition.
Females can pass it to both male and female offspring as she will give and X chromosome to both sexes (eg X(X) and X(Y))

Ok a normal vision means she MUST have at least XC allele, but if her father was colourblind, he will have passed the defective Xc chromosome to her, she is therefore XCXc

Her partner is normal, therefore he must have XCY

the punnett square will be (where the top line will show females and bottom will be male)

XC Xc

XC XCXC XCXc

Y XCY XcY

So it will be 1 normal female, 1 carrier female (so all females will be normal vision)
Males will be 50% affected and 50% normal.

Answer 2

the geno type for the mother would be *XX and the father would be XY. the mother is a carrier of the colorblindness because her father passed on the trait to her. it doesn't matter if the father of the father is color blind or not, he would have got his color blind gene from his mom.

if the first two peopl listed has children, using a punnett square, the percentages of offspring are as follows:

50% of females would be normal
50% of females would be carriers of the trait

50% of the males would be color blind
50% of the males would be normal

Answer 3

The woman is heterozygous because she got an X from her father with the colorblindness allele.

Mother is Xc+ Xc
Father is Xc+ Y.

If you put these gametes on a Punnett square, you'll see that the first row is Xc+ Xc+ and Xc+ Y.
The second row is Xc Xc+ and Xc Y.

Only the last box (Xc Y) is a colorblind offspring, a male.
1/4 or 25% of their offspring are expected to be colorblind.
-- 0% of their daughters would be colorblind.
-- 50% of their sons would be expected to be colorblind.