The PCR primers D1S80(F) and D1S80(R) amplify DNA from a specific region of huma
ID: 81972 • Letter: T
Question
The PCR primers D1S80(F) and D1S80(R) amplify DNA from a specific region of human chromosome 1. The target region contains a VNTR that is highly informative in DNA fingerprinting studies. The expected PCR product can vary from 340 - 650 base pairs in length. The sequence of the primers is given below: D1580 (F) 5' -GAA aCT GGC CAA ACA CTG CCC GCC GTC CA approximate annealing temperature D1580 (R) 5'-CTC TTG TTC GAG ATG CAC GTG CCC CTT GCA GG approximate annealing temperature Based on the data presented in the figure above what can you determine about the paternity of the four children? State your reasons for reaching your conclusion.Explanation / Answer
There are various formulae to calculate annealing temperature. In this answer, the following formula was used to calculate annealing temperature.
Tm= 64.9 +41*(yG+zC-16.4)/(wA+xT+yG+zC)
where w,x,y,z are the number of the bases A,T,G,C in the sequence, respectively.
D1S80F:
GAAACTGGCCTCCAAACACTGCCCGCCGTCCA
Total bases 32
w = 8
x = 4
y = 6
z = 14
Tm= 64.9 +41*(yG+zC-16.4)/(wA+xT+yG+zC)
= 64.9 +41*(6+14-16.4)/(8+4+6+14)
= 64.9 + (41 x 3.6)/32
=64.9 + 4.615 =69.515°C
Ta (°C) = Tm (°C) - 5°C
= 69.5-5 = 64.5°C
D1S80R:
GTCTTGTTGGAGATGCACGTGCCCCTTGCAGG
Total bases 32
w = 4
x = 9
y = 11
z = 8
Tm= 64.9 +41*(yG+zC-16.4)/(wA+xT+yG+zC)
= 64.9 +41*(11+8-16.4)/(4+9+11+8)
= 64.9 + (41 x 2.6)/32
=64.9 + 3.33 =68.23°C
Ta (°C) = Tm (°C) - 5°C
= 68.2-5 = 63.2°C
Mom and dad shown in the gel are biological parents of children a, b and c.
Mom has two different alleles (heterozygous) and dad also has two different alleles (heterozygous). Both f them have a common allele (the one with low molecular weight)
Child A inherited low molecular weight allele from both parents. It is homozygous for low molecular weight allele.
Child B inherited low molecular weight allele from mother and high molecular weight allele from father.
Child C inherited high molecular weight alleles from both parents.
Child D inherited high molecular weight allele. But she is homozygous for that allele. So, she also inherited the same allele from mother. However, the mother shown in the gel does not have that allele. So, we can conclude that D is child of dad shown in the gel with some other female. Mother shown in the gel is not the biological mother of child D.
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