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The mutation is a change in the nucleotidic sequence. In the exposed example, Patient 1: nucleotide 143, T<u>T</u>C⇒T<u>G</u>C. Patient 2: nucleotide 143, T<u>T</u>C⇒T<u>C</u>C. Patient 3: nucleotide 147, TT<u>C</u> ⇒ TT<u>G</u>.
<h3>What is a mutation?</h3>
A mutation is a change or alteration in DNI sequences that introduce new variants.
Many of these are eliminated, but some of them might succeed and be incorporated into each individual.
These mutations are the ones that have been selected by natural selection.
<h3>Solving the problem</h3>We know that each sequence initiates with nucleotide number 103 and ends in nucleotide 162.
So first, we will number the nucleotides, from 103 to 162. Each nucleotide has a number in increasing order.
Now, we will identify the mutations in each of the strands by comparing them with the wild-type sequence. The mutation occurs in one of the nucleotides, so we must look for the change in the bases.
Finally, we will identify the nucleotide location of each mutation.
nucleotide wild-type mutated
<u> location nucleotide nucleotide </u><u> </u>
Patient 1 143 TTC TGC
Patient 2 143 TTC TCC
<u>Patient 3 147 TTC TTG </u>
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In the attached files you will find an image for a better understanding.
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The immense amount of information stored in DNA is coded with just four base nucleotides. These four nucleotides are paired by twos to form <u>complementary </u>DNA chains.
The makeup of DNA is limited to two <u>pairs </u>of nucleotides bases. One of these pairs is a pyrimidine base whilst the other forms a purine base. The four nucleotides that will make up the entire DNA code are:
- Adenine (A)
- Guanine (G)
- Cytosine (C)
- Thymine (T).
A and G will pair to form the purine base, meanwhile, Cytosine pairs with Thymine to form the <em>pyrimidine base</em>. As per the entire DNA code, A will always pair with G.
Whilst C always pairs with T. This is due to the very nature of the nucleotides themselves and will not change throughout the synthesis or duplication of DNA. Therefore, to complete the complementary strand described in the assignment we must match the letter given with its corresponding base pair, using this rule of DNA.
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