A brown-eyed left-hander suffering from hypertrichosis (hair growth of the auricle) marries a blue-eyed right-hander. The man’s father was blue-eyed, and the woman’s mother was left-handed. What is the probability of the birth of blue-eyed left-handers with hypertrichosis from this marriage, if brown eyes and the ability to better use the right hand are dominant autosomal signs, and hypertrichosis is linked to the Y chromosome?
Let the gene for brown eyes be B, then the gene for blue eyes will be in.
Let the gene for right-handedness be C, then the gene for left-handedness is c.
Let the chromosome with the gene causing hypertrichosis be designated YG.
It is known that the man’s father had blue eyes, therefore, for this trait, he was homozygous – cc and could only betray the gene for blue eyes to his son.
Consequently, the brown-eyed man is heterozygous in eye color. In addition, he suffers from hypertrichosis.
The male genotype is BvccXYG. It produces spermatozoa – BsX, BsX, BsYG, BsYG.
The woman’s mother was homozygous for the left-handedness gene – cc, and she could only betray the gene that leads to the development of left-handedness to her daughter.
Consequently, a right-handed woman is heterozygous for the dominant hand gene. Her genotype is bbccXX. It produces two types of oocytes – bCX and bCX.
All possible offspring of this married couple will be represented by the following options:
brown-eyed left-handed girls without hypertrichosis (BvssXX) – 12.5%;
blue-eyed left-handed girls without hypertrichosis (bvccXX) – 12.5%;
brown-eyed left-handed boys with hypertrichosis (BvccXYG) – 12.5%;
blue-eyed left-handed boys with hypertrichosis (bvccXYG) – 12.5%;
brown-eyed right-handed girls without hypertrichosis (BvCcXX) – 12.5%;
brown-eyed right-handed girls without hypertrichosis (ВвСсXX) – 12.5%;
brown-eyed right-handed boys with hypertrichosis (ВвСсXYG) – 12.5%;
blue-eyed right-handed boys with hypertrichosis (bccXYG) – 12.5%.