Huntington’s disease is a rare autosomal dominant trait (lethal in the homozygou

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 the 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 is 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. What reaction components would be needed to PCR amplify the HD alleles from her genome?
ii. Once the alleles were amplified, how could you compare the size of her alleles to alleles from other individuals?
iii. If this woman was heterozygous at the HD locus, describe the sizes of the two PCR-amplified alleles and how they would compare to amplified alleles from both of her parents.

Explanation / Answer

a. As it is a dominant disorder, we dont know whether the father was homozygous dominant or heterozygous. As the mother is phenotypically normal, she is normal. So, this lady has 50% chance or probability of getting the HD allele. ------------------------------------------------------- b. i. To amplify through PCR we need oligonucleotide DNA primers specific to the sequence of the HD allele. We also need the PCR enzyme, Taq polymerase and nucleotides. ------------------------------------------------------ ii. We can compare the size of her alleles with alleles from other individuals by taking her mother's allele. As her mother is normal it would be a good standard to compare with. ---------------------------------------------------- iii. If this woman was heterozygous for the HD allele, she would have one normal and one HD allele. Her normal allele will be of normal size and would match with the normal allele from her mother. Her HD allele would be much longer than the normal allele. The size of the defective allele would help to identify the allele. This defective allele can be compared with the defective allele isolated from the father.

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