Consider a weird DNA polymerase that makes only specific biased mistakes during
ID: 63642 • Letter: C
Question
Consider a weird DNA polymerase that makes only specific biased mistakes during replication: a single nucleotide in tyrosine codons get specifically mutated (i.e. changed, not deleted) with 100% efficiency (no other mistakes are made). What could the different outcomes for the resulting protein sequence be if there were 2 tyrosines in the sequence (at site 1 and site 2)? What would the probability be of observing those outcomes? (Consider all possibly of codons for tyrosine in your calculation.)
Explanation / Answer
The amino acid Tyrosine is encoded by codons: UAU and UAC.
There is a single nucleotide change resulted in a mutant.
In a codon there are 3 bases. For example in UAU - U -1st base, A- 2nd base and U 3rd base
If the nucleotide change is resulted in 1st base (pyrimidine to pyrimidine transition), it results in CAU and CAC codons which code for histidine.
If the nucleotide change is resulted in 1st base (pyrimidine to purine transition), it results in AAU and AAC codons which code for aspargine.
If the nucleotide change is resulted in 1st base (pyrimidine to purine transition), it results in GAU and GAC codons which code for aspartic acid.
If the nucleotide change is resulted in 2nd base (purine to purine transition), it results in UGU and UGC codons which code for cysteine.
If the nucleotide change is resulted in 2nd base (pyrimidine to pyrimidine transition), it results in UCU and UCC codons which code for Serine.
If the nucleotide change is resulted in 2nd base (pyrimidine to pyrimidine transition), it results in UUU and UUC codons which code for phenylalanine.
If the nucleotide change is resulted in 3rd base (pyrimidine to purine transition), it results in UAA and UAG codons. These are stop codons that end translation.
Thus, the 2 codons of tyrosine can be changed to 8 other codons encoding 6 other amino acids and 2 stop codons.
The tyrosine is at 2 sites in the DNA, so there is possibility for 8 x 2 = 16 changes.
If the DNA polymerase is 100% efficient, then the probability of observing outcomes = 100 / 16 = 6.25%