Question

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

5'- ATG ACC GTA TTT TCT GTA GTG CCC GTA CTT CAG GCA TTA 3'

1. Write the corresponding template strand and m-RNA strand that will be transcribed along with its polarity.     [2]
2. If GUA of transcribed mRNA is an intron, then depict the sequence involved in formation of mRNA/mature/processed hnRNA strand:
   1. In a bacterium     [1]
   2. In humans     [1]
3. How many amino acids the resulting polypeptide will have after the process of translation in humans?     [1]

Answer 1

1. DNA Template Strand and mRNA Strand:
   • Template Strand: This is complementary to the coding strand with reversed polarity.
     3'- TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT AAT -5'
   • mRNA Strand: This is identical to the coding strand, but Thymine (T) is replaced by Uracil (U).
     5'- AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA -3'
2. Processing if "GUA" is an Intron: If the sequence GUA is considered an intron (non-coding region), it must be removed to form mature mRNA. In the given sequence, GUA appears three times.
   1. In a Bacterium (Prokaryote) In bacteria, genes are usually monocistronic and do not contain introns. Transcription and translation are often coupled.
      Sequence: The mRNA remains exactly as transcribed because splicing does not occur in bacteria.
      5'- AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA -3'
   2. In humans, the primary transcript (hnRNA) undergoes splicing, where introns are removed and exons are joined together.
      Processed mRNA Sequence: (Removing all three GUA sequences)
      5'- AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA -3'
3. Translation in Humans:
   To find the number of amino acids, we look at the processed human mRNA and group them into codons (triplets):
   1. AUG (Start)
   2. ACC
   3.  UUU
   4. UCU
   5. GUG
   6. CCC
   7. CUU
   8. CAG
   9. GCA
   10. UUA
       Total Amino Acids: There are 10 codons in the processed sequence. Since none of these resulting triplets are “Stop” codons (UAA, UAG, UGA), all 10 will code for amino acids.
       Result: The polypeptide will have 10 amino acids.