Track template direction and processed transcript length

1. Frame the molecular question

Transcription makes RNA complementary to a DNA template read 3′→5′, with RNA extended 5′→3′. The coding strand, written in the same 5′→3′ orientation as the RNA, matches its sequence except DNA T corresponds to RNA U. A promoter helps define initiation and which strand is used for a particular gene; neither strand is the universal template for an entire chromosome.

2. Explain mechanisms and assumptions

For a short original example, template 3′–TACCTTAAACGAATT–5′ produces RNA 5′–AUGGAAUUUGCUUAA–3′. The corresponding coding strand is 5′–ATGGAATTTGCTTAA–3′. Reverse orientation and complementing are different operations; always write the ends before deriving a transcript.

Eukaryotic pre-mRNA processing commonly includes a 5′ cap, intron removal and a poly(A) tail. Retained exons can contain untranslated regions as well as coding sequence. Alternative splicing changes which segments remain; skipping an exon whose coding length is a multiple of three can preserve the downstream frame, but another skip may shift it. Not every retained RNA nucleotide encodes an amino acid.

The supplied length packet has exons 120, 90 and 150 nt, introns 80 and 60 nt, a 30 nt 5′UTR, a 27 nt 3′UTR and one three-base stop codon. The middle exon is an in-frame optional coding segment. The cap’s added nucleotide is excluded from the length convention, while the selected poly(A) tail is included in mature-RNA length but not protein coding.

3. Work a quantitative example

The primary transcript is 500 nt. Retaining all exons gives 360 nt before the tail; a 40 nt tail gives 400 nt by the stated convention. Coding length including stop is 360−30−27=303 nt, encoding (303−3)/3=100 residues. Skipping the 90 nt exon gives 70 residues, not a frameshift in this packet.

4. Predict and investigate

Predict first: Does adding 60 nt to a poly(A) tail add 20 amino acids to the protein?

Compare exon retention and skipping with the same tail, then vary tail without changing exon choice. Keep primary RNA, mature RNA, coding length and residue count separate. Independently derive the short template example with both end labels.

Retain the first prediction, settings and complete evidence ledger in the lesson notebook. All sequences and numerical packets are original bounded teaching examples, not real genes, observed samples or performed experiments.

5. Apply the reasoning to a changed case

Apply the mechanism to a changed case

A processed transcript has 420 retained exon nucleotides, 45 nt 5′UTR, 42 nt 3′UTR and one stop codon. It receives an 80 nt tail. Find mature length by the stated cap-excluding convention and encoded residues.

Retain your calculation, units and assumptions in the lesson notebook before opening this public practice response.

Compare a worked response

Mature length is 500 nt. Coding length including stop is 420−45−42=333 nt, so residues are (333−3)/3=110. UTRs and the added tail do not count as amino-acid codons.

6. Reflect and retain evidence

Which independent observation would separate the molecular mechanism from an assay, sample or model artifact? State orientation, units, controls, biological replication and omitted alternatives. Public practice and written plans do not certify independently demonstrated reasoning or actual supervised laboratory participation.

Sources and course continuation

Scientific supporting reference, checked 30 October 2026. Original prose, packets and diagrams; no source exercise or figure reproduced. Continue through General Biology I for mixed evidence and retained project review. No genetic engineering, biological culture or clinical procedure is requested.