Question 6

Realize that the father exhibited the dominant right-handedness trait and the recessive blue-eyed trait. Thus his genotype is best given as R_bb.

The first wife was right-handed and brown-eyed (dominant): R_B_.

The second wife was also right-handed and brown-eyed: R_B_

The first marriage yielded one right-handed, blue-eyed child (R_bb) and one left-handed, brown-eyed child (rrBb).

The only way to get a left-handed child is if both parents offered a left-handed allele, r. Therefore both parents must have been heterozygous with respect to handedness (i.e., father = Rr; and first wife = Rr).

The only way to get a blue-eyed child is if both parents offered a blue-eyed allele, b. Therefore the mother (first wife) must be heterozygous (Bb) with respect to eye color. The father's eye color genotype was bb.

Therefore the genotypes must have been as follows:

Father: Rrbb

First Wife: RrBb

The second marriage yield 10 right-handed and brown-eyed children. If she were heterozygous for either character, chances are some of the offspring would have exhibited the recessive trait for the character, because occasionally a recessive allele from the father would be matched with a recessive allele from the second wife in the zygote. However, none of the offspring exhibited the recessive trait for either character. Therefore, the second wife must be homozygous dominant for both characters (RRBB).