A blue-eyed man, both of whose parents had brown eyes, married a brown-eyed woman, whose father
A blue-eyed man, both of whose parents had brown eyes, married a brown-eyed woman, whose father was homozygous for dominant eye color, and whose mother was blue-eyed. Determine the possible genotypes of all 3 generations
Let the gene for the development of blue eyes in humans be denoted as m, then the gene for brown eyes will be denoted as M.
A blue-eyed male has the mm genotype because homozygosity is a prerequisite for the development of blue eyes, which are a recessive trait. Such a man has inherited recessive genes from both of his parents. They have brown eyes, therefore, in order to be the parents of a blue-eyed son. Must have heterozygous Mm genotypes and produce both M germ cells and m germ cells.
The woman’s brown-eyed father, homozygous for the dominant gene, had the MM genotype and produced the same type of spermatozoa M. According to the condition of the problem, her mother with blue eyes had the mm genotype and produced only m eggs. Consequently, the brown-eyed woman is heterozygous – Mm. It is capable of producing two types of oocytes – M and m.
The offspring possible for a given married couple are represented by the following options:
heterozygous children with brown (hazel) eyes (Mm) – 50%;
homozygous children with blue eyes (mm) – 50%.
Answer: the woman’s parents are MM and mm; parents of a man – Mm; man – mm; woman – Mm; theoretically possible children of a man and a woman: Mm and mm.