A short-sighted left-handed person who is color blind marries a short-sighted right-handed person who is not color blind. In this family, one short-sighted daughter was born, the second daughter is left-handed and the son is dal-blind. Determine the genotypes of all members of this family, if it is known that myopia and the ability to better control the right hand are dominant autosomal signs, and color blindness is a recessive trait linked to the X chromosome.
Let’s designate the gene that causes the development of myopia as B, then the gene that causes the development of normal vision will be in.
Let’s designate the gene that determines the predominant possession of the right hand as C, then the gene for left-handedness will be c.
Let’s designate the gene for color blindness d, and the chromosome to which it is linked – Xd, then the chromosome with the gene for normal color discrimination will be XD.
Since in the offspring of the family under consideration there are children with normal vision (cc), left-handedness (cc), it should be concluded that both parents are heterozygous for the gene for visual acuity, as well as that the mother is heterozygous for the gene for right-handedness.
The son’s color blindness is due to the Xd Y genotype. From the father, the son inherits a neutral Y chromosome, therefore the mother transmitting to him a chromosome with the pathological Xd gene is a carrier of color blindness.
Let’s write down the genotypes of the parents.
Myopic right-handed mother – ВвСсXD Xd. It produces eggs bcXd, bCXD, bcXd, bCXd, bcXD, bcXd, bcXD, bCXD.
The short-sighted color-blind left-handed father – BwccXd Y. He produces spermatozoa – allY, BcXd, BcY, sun Xd.
A short-sighted daughter born in this family may have the genotype BBCcXD Xd or BbCcXD Xd.
A left-handed daughter will have the genotype bccXD Xd.