Huntington\'s disease is a rare autosomal dominant trait (lethal in the homozygo

Question

Huntington's disease is a rare autosomal dominant trait (lethal in the homozygous dominant condition). The mutation is an expansion of a nucleotide repeat in DNA that encodes the huntingtin protein. Normal individuals have between 6-35 copies of a CAG repeat within the gene. Affected individuals can have between 36-121 repeats. In addition, as the number of repeated triplets (CAG) increases, the age of onset in the patient decreases. Individuals with this disease suffer from progressive neurodegeneration eventually resulting in death.
A) A 20 year old single woman whose father died of Huntington's disease if concerned about her risk of eventually developing the disease. What is the probability that she will eventually get HD, if her mother is phenotypically normal?
B) The HD gene has been sequenced and its location on chromosome 4 has been well characterized. The entire DNA sequence of chromosome 4 has been sequenced. A genetic screen can definitively determine whether this woman has the HD mutant allele.
i. List components needed to PCR amplify the HD alleles from her genome?
ii. Once the alleles were amplified, what technique would you use to compare the size of her alleles to alleles from other individuals?
iii. if this woman was heterozygous at the HD lous, describe the sizes of the two PCR-amplified alleles and how they would compare to amplified alleles from both of her parents?

Explanation / Answer

B. (i) We can take samples from her cheek, hair, blood and skin. The alleles are isolated and then amplified using PCR. We use the PCR machine with the approriate enzymes and at the right temperature.
(ii) We can use VNTR technique to compare the alleles of this lady with the normal ones. VNTR stands for variable number tandem repeats.So the number of CAG repeats in this lady may be more than the normal individual.So we can find out whether the alleles of the lady are just a variation or the sign of this disease.
(iii) If this lady was heterozygous she would have one normal and one defective allele. The normal allele would be shorter than the defective one because the defective one will have more copies of the CAG repeat and hence will be longer.
The father will have both defective alleles and hence both will be longer than the standard length of the allele.
The mother will have both normal alleles which will show the normal length.
hope it helps

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