# A blue-eyed woman married a brown-eyed man, they had a child with brown eyes.

A blue-eyed woman married a brown-eyed man, they had a child with brown eyes. Determine the genotype and phenotype of the child.

Given:

A – brown eye color;

a – blue eye color.

Determine the phenotype and genotype of the child.

Decision:

Brown eye color is a dominant trait, but an organism can be either heterozygous or homozygous. Blue eye color is a recessive trait, the body is always homozygous, always aa.

If the brown-eyed father is heterozygous, then we denote Aa. It turns out the following:

F1 Aa x aa

/ \

Aa Aa    aa aa

Thus, we get a 50% probability of having a brown-eyed baby Aa and a 50% probability of having a blue-eyed baby aa. By condition, we are given that the child was born with brown eyes, that is, with the Aa genotype.

If the brown-eyed father is homozygous for this trait, then we denote him AA:

F1 AA x aa

/ \

Aa Aa   Aa Aa

All children in this case will be brown-eyed with the Aa genotype.

Answer: the child’s phenotype is brown eyes; the child’s genotype is Aa.

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