A normalvisioned man marries a normalvisioned woman whose fa

A normal-visioned man marries a normal-visioned woman whose father is color-blind. They have two daughters who grow up and marry. The first daughter has five sons, all normal-visioned. The second daughter has two normal-visioned daughters and a color-blind son. Diagram the family history in a pedigree and indicate the genotypes of all family members.

Solution

Color vision is an X-linked recessive trait. In case of a woman (XX) to be color blind, both her X chromosomes must have the mutated gene. If she only gets one mutated gene, her other X chromosome can make up for it and she won\'t be colorblind, but she will be able to pass down the bad gene. If a man (XY) receives a defective gene on his X chromosome, he will express the trait since the Y chromosome can\'t compensate for the defect.

In this case, the woman\'s father was color blind, so his X chromosome contains the mutation. She is XX in which one X from her mother and one from her father. Therefore, she must have inherited her father\'s X chromosome, and she is a carrier of the color blind trait. If her husband is normal, he definitely doesn\'t have the defect.

So if the two were to have children, the girls will get the father\'s good X chromosome, and either the wild type or the mutated gene from their mother. Either way, they\'ll be phenotypically normal since they are guaranteed at least one good gene from their dad. They will, though, have a 50% chance of being a carrier like their mom.

The sons are at a higher risk. The get a Y from their father, and one of their mother\'s X chromosomes. If they get the mutated gene they\'ll be colorblind, so her boys have a 50% chance of being affected.

Woman (XX)                                                                                  Man (XY)

Girls (50% chance for color blindness)

Boys (50% chance for color blindness)

Mother\'s genotype (normal vision but whose father was color blind) X^NXⁿ

She is a carrier because of her heterozygous condition. For the recessive gene to be expressed one must be homozygous for it.

Father\'s genotype (color blind) XⁿY

From the Punnett square it is obvious that half of the children will be color blind and the other half will be normal. More specifically there will be 1 normal boy to 1 color-blind boy to 1 normal girl to 1 color- blind girl.

X^NXⁿ one normal (carrier) girl

X^NY one normal boy

XⁿXⁿ one color blind girl

XⁿY one color blind boy

The first daughters have 5 sons all are of normal visioned, Second daughter has 2 normal vision daughter and a colour blind son.

First daughter has

X^NY (normal visioned) son

X^NY (normal visioned) son

X^NY (normal visioned) son

X^NY (normal visioned) son

X^NY (normal visioned) son

Second daughter has

X^NXⁿ normal (carrier) girl

X^NXⁿ normal (carrier) girl

And, XⁿY one color blind Son