Question

Given below is a stretch of DNA showing the coding strand of a structural gene of a transcription unit?

5’-ATG ACC GTA TTT TCT GTA GTG CCC GTA CTT CAG GCA TAA-3’

1. Write the corresponding template strand and the mRNA strand that will be transcribed, along with its polarity.
2. If GUA of the transcribed mRNA is an intron, depict the sequence involved in the formation of mRNA/the mature processed hnRNA strand.
   i. In a bacterium
   ii. In humans
3. Upon translation, how many amino acids will the resulting polypeptide have? Justify.

Answer 1

5’-ATG ACC GTA TTT TCT GTA GTG CCC GTA CTT CAG GCA TAA-3’ = CODING

1. 3’-TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT ATT-5’ = TEMPLATE
   5’-AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UAA-3’
2. i. In a bacterium
   5’-AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UAA-3’
   ii. In humans
   5’-mGpppAUG ACC UUU UCU GUG CCC CUU CAG GCA UAA - Poly A tail-3’
3. 9 amino acids in the polypeptide because UAA is a stop/terminator codon and does not code for any amino acid.