Step 1: Understanding the Concept:
The coding strand has the same sequence as the mRNA (with T replaced by U). The template strand is complementary to the coding strand. Splicing removes introns in eukaryotes but not in prokaryotes.
Step 2: Detailed Explanation:
(i) Strands:
- Template Strand: Complementary to the coding strand with reversed polarity.
\( 3'-\text{TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT AAT}-5' \)
- mRNA Strand: Matches coding strand but uses Uracil (U) instead of Thymine (T).
\( 5'-\text{AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA}-3' \)
(ii) Intron Processing:
- (1) Bacterium: Bacteria are prokaryotes and do not possess introns or the splicing machinery. Therefore, the entire sequence (including the GUA segment) remains in the final transcript.
- (2) Humans: Humans are eukaryotes. Introns (non-coding regions) are removed during splicing. If GUA is an intron, all instances of GUA will be removed.
(iii) Amino Acid Count:
- The original mRNA has \(39\) nucleotides, which equals \(13\) codons.
- The sequence has 3 GUA codons (at positions 3, 6, and 9).
- In humans, after splicing out the 3 GUA introns, \(13 - 3 = 10\) codons remain.
- Assuming none of the remaining are stop codons (as the segment is a "stretch"), the polypeptide will have \(10\) amino acids.
Step 3: Final Answer:
(i) mRNA is 5'--AUG...UUA--3'. (ii) Bacterium keeps the whole sequence; Humans splice out GUA. (iii) 10 amino acids in humans.