Question

First, draw out the pedigree. Label everyone as affected/unaffected. Write in each genotype you KNOW for sure from the information you are given in the problem. Then use Mendelian genetics & statistical rules to figure out the chance the unaffected people are carriers. Note: in order to figure out the probability that a couple will have an affected child, you must first determine the probability that each parent is a carrier of the mutant gene in question. Then you must determine the probability that they would pass on the mutant gene if they were a carrier. All of this must happen in order for them to have an affected child.

Question: John’s paternal uncle died of a rare autosomal recessive disease, Disease X. “Wow, “ says John’s fiancée, “that is the same condition that my daughter from my first marriage died from.” What is the probability that a child born to John and his fiancée will be affected?

Answer 1

Hi,
The disease is autosomal and recessive. Let us denote it as 'a'. The person with AA or Aa is normal, the diease occurs only if the person has 'aa' genotype.
The John's paternal uncle had the disease, so his genotype = aa. This means each of the grandparents of John, had donated one 'a' allele to the uncle. That means the grandparents were heterozygotes, Aa genotye.
The John's father is normal, but he can be a carrier. So when two heterozygotes mate, Aa x Aa, out of 3 normal kids, 2 will be heterozygotes with Aa genotype. So probability that the John's father is a heterozygote = 2/3
Now, we must assume that John's mother who came from unrelated family is perfectly normal with AA genotype. So when they mate, the probability to have a heterozygous John = Probabiity of Johns father as a heterozygote x probability of heterozygote when (Aa x aa)
= 2/3 x 1/2
= 1/3 ; chance that John is a carrier.

Now coming to fiancee, she already had a diseased child, that means the fiancee is a heterozygote with Aa genotype. So when two heterozygotes mate, i.e John and his fiancee, Aa x Aa, we get AA, Aa,Aa, and aa genotypes. So only 1 out of 4 possible child will be affected. This means probability = 1/4
So total probability to have affected child = 1/3 x 1/4 = 1/12