A. Suppose that the sequence
A. Suppose that the sequence belowwas the open reading frame of a gene. It’s not the entire mRNAmolecule, just the protein coding sequence and in this case thefirst three bases is the start codon. The sequence is given as DNAsense strand because DNA is where mutations happen.
5’-ATGTCTAAACTGCGTGAGTAA-3’
i. Give the sequence of thepolypeptide it encodes. Make sure to mark the carboxyl and aminoterminal ends.
ii. Give a mutant version of thesecond codon that has a single base substitution that is asynonymous mutation.
iii. Give a mutant version of thefourth codon that has a single base substitution that is a missensemutation.
iv. Write a mutant version of the entiresequence that is an example of an in frame insertion. Indicate whatpart of the sequence is different.
Answer:
Ans.
Given sense DNA strand-
5′-ATGTCTAAACTGCGTGAGTAA-3′
RNA POLYMERASE always transcribe Antisense strand means 3′—> 5′ .
Hence Antisense DNA Will be-
3′-TACAGATTTGACGCACTCATT-5′
mRNA-
5′-AUGUCUAAACUGCGUGAGUAA-3′
polypeptide-
NH2-met-ser-lys-leu-arg-glu-cooH.
ii. Synonymous mutation in second codon.
Antisense mutant DNA-
3′-TACTGATTTGACGCACTCATT-5′
Transversion mutation because A (purine) is replaced by T(pyrimidine)
mRNA-
5′- AUGACUAAACUGCGUGAGUAA-3′
polypeptide-
NH2-met-thr-lys-leu-arg-glu-cooH.
iii. Fourth codon missense mutation-
Antisense mutant DNA in fourth codon-
3′-TACAGATTTGCCGCACTCATT-5′
Transversion mutation because A is replaced by C(pyrimidine)
mRNA-
5′-AUGUCUAAACGGCGUGAGUAA-3′
polypeptide-
NH2-met-ser-lys-arg-arg-glu-cooH.
iv. Insertion-
Insertion of C in between second and 3rd codon then allremaining reading frame in right side will be changed as-
Mutated antisense DNA strand-
3′-TACAGACTTTGACGCACTCATT-5′
mRNA-
5′-AUGUCUGAAACUGCGUGAGUAA-3′
polypeptide-
NH2-met-ser-glu-thr-ala-cooH.
Because 6th codon is converted in stop codon.
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