In humans, the gene for farsightedness dominates over the gene for normal vision.

In humans, the gene for farsightedness dominates over the gene for normal vision. In the family, the husband and wife suffer from farsightedness, but the mothers of both spouses had normal vision. How many types of gametes does a wife have? How many different genotypes can children in a given family have? How many phenotypes can children in a given family have? What is the probability that a child with normal vision will be born in this family? What is the probability of the birth of a child suffering from farsightedness in this family?

Let’s designate the gene that causes a person to develop hyperopia as D, then the gene for normal vision will be designated as d.

The mothers of both spouses have normal visual acuity. Let’s write them dd, because homozygosity is a prerequisite for the phenotypic realization of a recessive trait. The mothers of the spouses produced eggs exclusively of type d and could pass only the gene for normal vision to the offspring.

Consequently, farsighted men and women are heterozygous: Dd. They both produce eggs and sperm of two types – D and d.

The offspring of this married couple are represented by the following options:

children – homozygotes suffering from hyperopia (DD) – 25%;

children – homozygotes with normal vision (dd) – 25%;

children – heterozygotes suffering from hyperopia (Dd) – 50%.

Gametes in the mother are formed of 2 types.

Children can have 3 different genotypes.

Children can have 2 phenotypes.

The probability of having a child with normal vision in a family is 25%.

The probability of having a farsighted child in the family is 75%.