Huntington’s disease is an autosomal dominant disorder caused by an abundance of
ID: 212007 • Letter: H
Question
Huntington’s disease is an autosomal dominant disorder caused by an abundance of trinucleotide (CAG) repeats inside the Huntingtin gene. People with less than 36 repeats of (CAG) do not suffer any negative effects, but anyone with more than 40 repeats will suffer neurodegeneration starting around 35-44 years old. We took a random sampling of people and did PCR to determine their number of trinucleotide repeats in this gene and their risk for developing Huntington’s disease. The primers we used give a band of 200 bp if a person has no trinucleotide repeats. The size of each band is listed on the side of the gel. Each lane is a different person.
A: (8 pts) Tell me how many repeats each person has in their genes.
Patient 1: __________________
Patient 2: __________________
Patient 3: __________________
Patient 4: __________________
B: (2 pts) Which patient(s) will develop
Huntington’s disease? ______________________
Explain your answer
Explanation / Answer
As mentioned in the question that primer generates 200bp band IF Patient Doesn't have any trinucleotide.
it means the Additional nucleotides after 200bp, are all of the trinucleotide.
so 200 band size should be deducted from the patient band size. the answer of that deduction should be devided by 3 (as the trinucleotide repeats are of three nucleotide CAG). and answer will be your repeat no. having the patien
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A
Patient 1: Band Size - 230
so 230 - 200 = 30
30/3 = 10
this patient has 10 repeats.
Note : Single band means that patient has two allels of same size. patient 2 and 3 have two bands in the gel. two band means they have two allele which are different in their size. one allele having different no. of reapeats and another having the different no of repeats. but this disease is AUTOSOMAL DOMINANT, so if even a single allele crossing the limit of repeat, then he or she will develop a disease.
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Patient 2: allele 1 - band size - 260 so, 260 - 200 = 60 / 3 = 20 repeats
allele 2 - band size - 305 so, 305- 200 = 105 / 3 = 35 repeats
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Patient 3: Allele 1 - band size 290 so, 290 - 200 = 90 / 3 = 30 repeats
Allele 2 - band size 335 so, 335 - 200 = 135 / 3 = 45 repeats.
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Patient 4: Band size - 350 so, 350 - 200 = 150 / 3 = 50 repeats
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B
Patient No 3 and 4 will develop the disease because 3 has a one allele having 45 repeats where the 4 has a both alleles having 50 repeats.
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