A woman with brown eyes marries a blue-eyed man whose parents both had blue eyes.

A woman with brown eyes marries a blue-eyed man whose parents both had blue eyes. One blue-eyed child was born from this marriage. Determine the genotypes of all three members.

If a brown-eyed woman and a blue-eyed man have a child with blue eyes. while the man’s parents were blue-eyed themselves, the brown-eyed woman was heterozygous for this gene, and the brown-eyed gene was dominant.

Let us prove this by the crossing scheme.

Let A be brown eyes and blue eyes.

RR g Aa (k) x ma aa (g)

m / f a a A Aa (k) Aa (k) a
aa (g)

aa (g)
By phenotype: 50% – brown eyes, 50% – blue eyes.

Answer: Mother is heterozygous – Aa, father and child are homozygous – aa.