Question

A blue-eyed, left-handed woman marries a brown-eyed, right-handed man who is heterozygous for both of his traits. if blue eyes and left-handedness are recessive, how many different phenotypes are possible in their children? three

Answer 1

If you look at a Punnet square of this, you can determine the possible phenotypes. Lets assign B for brown eyes, b for blue eyes, H for right hand and h for left hand.

The mother is blue eyed and left handed which are both recessive traits. This means that her genotype would be bbhh. Now the father is heterozygous for both, which means he has a dominant and recessive allele for each trait. His genotype would then be BbHh

The Punnet square should look like the table below:

Mother:bbhh
Father:BbHh

bh bh bh bh
BH BbHh BbHh BbHh BbHh
Bh Bbhh Bbhh Bbhh Bbhh
bH bbHh bbHh bbHh bbHh
bh bbhh bbhh bbhh bbhh

Based on the Punnet square, you have 4 possible genotypic combinations and 4 possible phenotypic combinations.

BbHh: Brown-eyed, right-handed
Bbhh: Brown-eyed, left-handed
bbHh Blue-eyed, right handed
bbhh Blue-eyed, left-handed