A blue-eyed right-hander married a brown-eyed left-hander, they had eight children, all brown-eyed

A blue-eyed right-hander married a brown-eyed left-hander, they had eight children, all brown-eyed right-handers. What are the genotypes of parents and children?

Let the gene that determines the priority of the right hand over the left in a person be designated as K, then left-handedness will be determined by the gene in the homozygous genotype, designated as k.

Let the gene that determines the development of blue eyes in humans, provided the genotype is homozygous for it, be designated as m, then the gene for brown eyes will be designated as M.

Since all children of a left-handed woman with genotype kk homozygous for the leading left hand were born right-handed, and the woman could only pass on the recessive left-handedness gene k to them, it should be concluded that the father transmits only the right-handedness gene K, therefore, he is homozygous for this dominant trait – KK … Then all right-handed children are heterozygous Kk.

Since all children have brown eyes, and they could have inherited only the m blue eyes gene from the father mm homozygous for the recessive gene, it should be concluded that the brown-eyed woman is homozygous for the gene for the dominant trait – MM. She only produces M oocytes, and all children are heterozygous Mm.

Answer: the mother is homozygous for the dominant gene for eye color and for the recessive gene for the dominant hand (MMkk), the father is homozygous for the recessive gene for eye color and for the dominant gene for the dominant hand (mmKK). All 8 children are heterozygous for both traits (MmKk).