Refer to the diagram below that shows a pedigree of a sex-linked recessive trait. In generation II, female number 5 marries a man and has three children. One daughter and one son do not have the recessive trait; the other son does have the recessive trait. Based on these results, one can conclude that:
a. the mother is heterozygous for the trait.
b. the mother is homozygous for the trait.
c. a mutation has occurred.
d. the father is heterozygous for the trait.
e. the father is homozygous for the trait
The answer is a, that the mother was heterozygous for the trait.
b is incorrect, because if the mother had been homozygous for a recessive trait, she would have had the trait herself and therefore her circle would have been shaded in on the chart (shaded-in shapes have the trait)
c could be possible, but the chances of that particular mutation occurring are rare, and a is totally possible, as I'll explain below
d is impossible because a male cannot be heterozygous for a sex-linked trait. Since males only have one X chromosome, they are what is called "hemizygous" because they can only have one allele for a trait
a is the correct answer because if a female is heterozygous, since not having the trait is dominant, she would still appear as a white circle (which she did). However, she would carry the recessive allele and could therefore pass it on to her children.
Her genotype would be XAXa, whereas her husband would be XAY. All the girls would receive at least one copy of the dominant allele (from the father), but the boys would only receive alleles from their mother, and could thsu receive either one, which explains why one boy could have the trait while the other would not, as seen in the pedigree.
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