Question

Joanna has short fingers (brachydactyly). She has two olderbrothers who are identical twins; they both have short fingers.Joanna’s two younger sisters have normal fingers. Joanna’smother has normal fingers, and her father has short fingers.Joanna’s paternal grandmother (her father’s mother) has shortfingers; her paternal grandfather (her father’s father), who isnow deceased, had normal fingers. Both of Joanna’s maternalgrandparents (her mother’s parents) have normal fingers. Joanna marries Tom, who has normal fingers; they adopt a son namedBill who has normal fingers. Bill’s biological parentsboth havenormal fingers. After adopting Bill, Joanna and Tom producetwo children: an older daughter with short fingers and a younger son with normal fingers. Please answer each of the following questions with clear explanations

1. What is the most likely mode of inheritance for short fingers in this family?
2. If Joanna and Tom have another three biological child, what is the probability(based on you answer to part b) that these children will have series of short –normal –short fingers ?Please denote the genotype of Joanna and Toms’, and explain your answer.

Answer 1

Answer a: The most likely mode of inheritance of short fingers in this family is autosomal dominant , as 50% of the offspring are affected, even if one of the parents is affected.

It is not Y linked as male and female both are affected

It is not X linked recessive or X linked dominant since it can be seen that the male is able to transmit the trait to the male offspring (Joanna’s father). We know that male transmits only Y chromosome to the male offspring.

Bill an adopted son would not be able to contribute to the inheritance of this trait hence not considered in the pedigree.

Answer b: Let Joann’s genotype be Bb, she can not be homozygous BB, otherwise all her children will express the character which is not the case.

Let Tom’s genotype be ‘bb’ as he is having normal fingers which is a recessive character, it can be expressed only in homozygous condition.

The possibility of their offspring’s phenotype will be as follows

Gamates	B	b
b	Bb Short fingers	bb Normal fingers
b	Bb Short fingers	bb Normal fingers

This couple will always have the probability of the offspring being born with short fingers as 50%

And offspring born with normal fingers as 50%

The sequence of the offspring doesn’t matter, as it is the random union of the gamates, and for each child it will be randomly taking place.

Note: if you have doubt in any of the concepts discussed above, please ask it in the comments section.