Pedigree and risk CER

Claim: There is a 50% chance that this couple's next son will be colorblind, and a 0% chance that a daughter will be colorblind.

Evidence: Red-green colorblindness is an X-linked recessive trait, so the allele sits on the X chromosome. The pedigree shows that the mother has a colorblind father, which means she inherited one affected X and is a carrier (X^A X^a), while the father has normal color vision (X^A Y). A Punnett square of X^A X^a mother by X^A Y father gives four equally likely children: X^A X^A, X^A X^a, X^A Y, and X^a Y. Among the two possible sons (X^A Y and X^a Y), one is affected, and neither of the two possible daughters is affected because each daughter still inherits at least one normal X^A from the father.

Reasoning: Because the trait is recessive and carried on the X chromosome, a son is affected whenever his single X carries the recessive allele, since he has no second X to mask it, giving 1 of 2 sons affected, or 50%. A daughter would need two recessive alleles to be affected, but the father can only pass a normal X^A, so no daughter in this cross can be colorblind, which is why the daughter risk is 0%. This reasoning assumes the mother is truly a carrier, which the pedigree supports because her father was colorblind and each daughter of a colorblind man must receive his X^a. A limitation is that the pedigree relies on reported vision status rather than a formal color-vision test, so mild cases could be missed or misreported, and confirming the mother's carrier status with genetic testing would make the 50% son estimate more reliable.