Male: XC YFemale: XC XcThe chances of this couple havinga color-blind daughter is 0.hope this helps!......
The Punnett square is:
_____C_____c_____←♀ female gametes
C___CC____Cc____← the daughters
Y___CY____cY___ ← the colorblind son
↑
♂ gametes
Therefore, the probability of a colorblind daughter is 0. Of the daughters, one will be homozygous normal, and the other will be heterozygous...
Color-blindness is a recessive trait. Remember this tip: Sex-linked is X-linked. The genes are carried on the X chromosome.
First find the genotypes of the parents:
Both have normal vision so they each must have at least one dominant allele (XC).
Mother: XCX_ (we need to fill in this blank)
Father: XCY (males only have one X, and we know he has normal vision)
We know females can only give their offspring X chromosomes. So the sex of the child is determined by the male because he can either give an X or a Y. They have a son (mom gave an X while dad gave a Y) who is color-blind, remember that color-blindness is recessive, so the mother must have gave her son an X with a recessive allele (Xc). This tells us the mother's genotype (XCXc).
Using the parent's genotypes, simply do a normal punnett square.
Male: XC Y
Female: XC Xc
The chances of this couple having a color-blind daughter is 0.
hope this helps!
