A blue-eyed myopic man marries a brown-eyed woman with normal vision. They had a daughter with normal vision, blue-eyed. Brown dominates over blue. Nearsightedness over normal vision. 1. How many phenotypes can there be among the children of these spouses 2. How many genotypes can there be among the children of these spouses.
Let’s designate the gene that determines the development of the brown iris in humans as K, then the blue iris gene will be designated as k.
Let’s designate the gene that leads to the development of the pathology of visual acuity in humans by the type of myopia as L, then the gene for normal visual acuity will be l.
A daughter with blue eyes and normal visual acuity has a dihomozygous genotype, we will write it down as kkll. She inherits genes for recessive traits from both parents, therefore, the myopic father is heterozygous for visual acuity, and the brown-eyed mother is for eye color.
The mother has the Kkll genotype and produces eggs Kl, kl.
The father has the kkLl genotype and produces spermatozoa kL, kl.
The offspring of this married couple includes the following hypothetically possible options:
children with myopia and brown eyes (KkLl) – 25%;
children with myopia and blue eyes (kkLl) – 25%;
children with normal visual acuity and brown eyes (Kkll) – 25%;
children with normal visual acuity and blue eyes (kkll) – 25%.
4 phenotypes can be in children born to these spouses;
4 genotypes can be in the children of these spouses.