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.