A brown-eyed man and a blue-eyed woman had three blue-eyed girls, one brown-eyed boy.

A brown-eyed man and a blue-eyed woman had three blue-eyed girls, one brown-eyed boy. The brown-eye gene is dominant. Record the results of the cross and explain them.

A brown-eyed person can have blue-eyed children only if he is heterozygous for this trait, and in the second parent, both alleles are either the same or recessive.

The gene for brown eyes is dominant, so let’s designate A – brown eyes, and – blue.

Let’s write down the genotypes of the parents, as well as the gametes that they form.

Parents: (mother) aa x Aa (father).

Gametes: a x A, a.

Offspring: Aa (brown eyes), aa (blue eyes), ratio 1: 1.

Among children, the splitting turned out to be 3: 1, because the meeting of two gametes is random, and it is impossible to accurately predict which alleles they contain.