Question

Bill has come to you for genetic counseling.  He is concerned that he and his sister (Alice) are at risk for Huntington disease. Bill and Alice are too young to show symptoms.  Their grandmother, (Mary) had Huntington disease.  Her three children, Shirley, Tom, and Ed so far show no symptoms.  Shirley is the oldest (45 years old) and is Bill and Alice’s mother. A study by Adams (American Journal of Human Genetics 43;695 – 704, 1988) indicates that 68% of individuals heterozygous for the Huntington allele show symptoms by age 45.   

a. Draw a pedigree of this family.

b. What is the probability that Bill is heterozygous for Huntington disease?

c. What test(s) would you recommend for Bill and Alice to determine if they are heterozygous?

d. Explain why you recommend this test, what results it can give, and how these results can answer Bill’s question.

Answer 1

Answers:

(A). Mary had the disease. But we cannot be sure of the genotype of Mary as the disease is a dominant one. Hence, two different pedegree charts for the family is attached with this answer, one representing Mary with heterozygous genetic makeup for the trait and the other representing Mary with homozygous dominant genotype.

Nothing is said about the genotype of Mary's partner and hence should be considered disease free and hence homozygous recessive for the trait. Also, nothing is said about Shirley's partner and should be considered disease free and hence homozygous recessive. Keeping these things in mind, go through the pedegree charts attached.

(B). Even if Bill's mother, Shirley has the disease, she can only be heterozygous as her father was normal. Also, the genotype of Bill's father is homozygous recessive and hence if bill actually has the disease, he can only be heterozygous and not homozygous dominant.

Let p be the probability of Bill to be heterozygous incase if his mother, Shirley has the disease. In such case, there are 2 possible genotypes for Bill i.e. Hh (heterozygous) and hh (homozygous recessive ) for the trait of Huntington's disease.

Hence, p = no. Of possible heterozygous genotype ÷ tota no. Of possible genotypes

p = 1÷2 = 0.5 = 50%

But it is not certain that Shirley will have the disease.

Let probability of Shriley to have the disease be q

q = (number of times out of two that Shirley would have the disease if Mary was heterozygous + No.of times Shirley would have the disease if Mary was homozygous dominant)÷4

4 is the total number of daughter genotypes that can be produced by both the genetic makeup of Mary

q = [(1/2)+(2/2)]÷ 4

q = 3/8 Or 3÷8

Hence the actual probablity of Bill to have the heterozygous genetic make up is the product of p and q = 1/2 * 3/8 = 1/16 = 6.25%

(C). I would recommend DNA blood test ( More particularly the Huntington Disease DNA Test)

In people having the Huntington's disease, there is a repetitive sequence of 'CAG' nucleotide sequence that can extend even upto 120 times, found in the chromosome number 4 of the affected individual. And, DNA blood test examines exactly that.

(D). I prefer this test because, this test directly verifies the suspected nucleotide sequence repeats in an individual. Moreover, this test can give 100% accuracy since we know that one of his close relatives has the disease.

Possible Results of the test:

(i). Negetive result:

Both the alleles will have only the normal number of CAG repeats, that is below 40.

(ii). Positive result:

Both or one of the alleles will have CAG repeates over 40 in number, and it may even go upto or slightly beyond 120.

In case of Bill, since he can either be heterozygous or homozygous recessive, in case if he has the disease in his genotype, he will show high CAG repeat number in one homologue of chromosome 4 and a normal reoeat requence of CAG in the other.

Answering Bill's doubts:

If an allele of Bill's 4th chromosome show high CAG repeats that exceeds 40 in number, he can be 100% sure that he has the disease, although clinical symptoms take years to appear.

If he has a normal repeat sequence of CAG in his 4th chromosome, he can be 100% sure that he is free from the disease and that he will not be carrier for the trait in his successive generations.