In humans, brown eyes (b) are dominant over blue (b)*. a brown-eyed man marries a blue-eyed woman and they have three children, two of whom are brown-eyed and one of whom is blue-eyed. draw the punnett square that illustrates this marriage. what is the man’s genotype? what are the genotypes of the children?

Man' s genotype= Bb

Children's genotype= Bb, Bb and bb for the three children

Explanation:

The gene involved here is that coding for eye color in humans. The allele for brown eyes (B) is dominant over that of blue eyes (b) i.e. brown eyes allele (B) will mask the expression blue eyes allele (b) in an heterozygous state.

A brown-eyed man is crossed with a blue-eyed woman (bb). The woman will be homozygous recessive since the recessive trait can only be expressed in a homologous state. They produce 2 brown-eyed offsprings and 1 blue-eyed offspring.

Note that, for the man who possess the dominant trait (brown eyes), he can either possess a homozygous or heterozygous genotype. Since he produced a recessive phenotyped (blue eyes) offspring when crossed with a blue-eyed woman, it means he donated a recessive allele (b) which combined with that of the woman. Hence, the man has a heterozygous genotype (Bb).

Having known the man's genotype, a cross (see attached image) between Bb and bb will produce four possible offsprings; 2 will be heterozygous brown-eyed (Bb) while the other two will be homozygous blue-eyed (bb).

Since two of their children were brown-eyed, they will have a (Bb) genotype while the blue-eyed child will have a (bb) genotype.

i believe it would be c: tons.