Calculating Risks in Pedigree Analysis
When a disease allele is known to be present in a family, knowledge of simple gene
transmission patterns can be used to calculate the probability of prospective parents'
having a child with the disorder. For example, a married couple finds out that each had an
uncle with Tay-Sachs disease (a severe autosomal recessive disease). The pedigree is as
follows:
What is the probability that their child will have Tay-Sachs?
Answer is worked out on next page--
The probability of their having a child with Tay-Sachs can be calculated in the following
way. The question is whether or not the man and woman are heterozygotes (it is clear that
they do not have the disease) because if they are both heterozygotes then they stand a
chance of having an affected child.
1. The man's grandparents must have both been heterozygotes T/t because they produced
a t/t child (the uncle). Therefore, they effectively constituted a monohybrid cross. The
man's father could be T/T or T/t, but we know that the relative probabilities of these
genotypes must be 1/4 and 1/2, respectively (the expected progeny ratio in a monohybrid
cross is 1/4 T/T, 1/2 T/t, and 1/4 t/t). Therefore, there is a 2/3 probability that the father is
a heterozygote (that is, 2/3 of the individuals displaying the dominant trait are
heterozygotes).
2. The man's mother must be assumed to be T/T, since she married into the family and
disease alleles generally are rare. Thus if the father is T/t, then the mating to the mother
was a cross T/t × T/T and the expected progeny proportions are 1/2 T/T and 1/2 T/t.
3. The overall probability of the man's being a heterozygote must be calculated using a
statistical rule called the product rule, which states that the probability of two independent
events both occurring is the product of their individual probabilities. Hence the
probability of the man's being a heterozygote is the probability of his father's being a
heterozygote times the probability of the father having a heterozygous son, which is 2/3 ×
1/2 = 1/3.
4. Likewise the probability of the man's wife being heterozygous is also 1/3.
5. If they are both heterozygous (T/t), then the probability of their having a t/t child is 1/4,
so overall the probability of the couple having an affected child is 1/3 × 1/3 × 1/4 = 1/36;
in other words, a 1 in 36 chance.