VCE Biology (Year 12) · Nucleic Acids and Proteins · 12m read

Transcription — From DNA to mRNA

Have you ever wondered how your cells read the instruction manual without damaging the original? That's transcription — the process where your cells copy a gene from DNA into mRNA, a temporary messenger molecule that carries instructions to make proteins. In eukaryotes (like your cells), this mRNA gets processed and cleaned up before it leaves the nucleus, ensuring only the useful information reaches the protein-making machinery. This is the crucial first step that turns your genetic code into the actual proteins your body needs to survive and function.

Learning objective: Describe transcription as the synthesis of mRNA from a DNA template, including RNA processing in eukaryotes.

By the end of this lesson you can:

• I identify RNA polymerase as the enzyme of transcription.
• I distinguish pre-mRNA from mature mRNA (5' cap, poly-A tail, splicing).
• I explain the roles of introns and exons.

Transcription is photocopying one recipe. You don't take the whole cookbook to the kitchen — you copy one recipe (gene) onto mRNA.

Transcription in the nucleus

Transcription produces a messenger RNA (mRNA) copy of a gene. RNA polymerase binds the promoter of a gene, unwinds the DNA, and reads the template strand 3'→5' while building mRNA 5'→3'. RNA uses uracil (U) instead of thymine; otherwise base pairing rules are the same.

In eukaryotes, the initial transcript (pre-mRNA) is processed before leaving the nucleus. A 5' cap (modified guanosine) protects against degradation and helps ribosome binding. A poly-A tail (many adenines) adds stability. Splicing removes non-coding introns, leaving only coding exons joined together. Alternative splicing allows one gene to produce multiple protein variants.

Prokaryotes have no nucleus and no introns — transcription and translation can occur simultaneously in the cytosol.

Transcription essentials

• Enzyme: RNA polymerase.
• Template read direction: 3'→5' on DNA.
• mRNA build direction: 5'→3'.
• Base pairing: A–U (instead of A–T), C–G as normal.
• RNA processing: 5' cap + poly-A tail + splice out introns.

Quick reference

| Step | Event |
|---|---|
| Initiation | RNA polymerase binds promoter |
| Elongation | mRNA built (U replaces T) |
| Termination | Terminator sequence |
| Processing | 5′ cap, poly-A tail, splicing introns |

Worked example (Medium)

Give the mRNA sequence transcribed from the DNA template strand 3'-TACGGCAT-5'.

1. Identify template direction. Read 3'→5'.
2. Pair each base (U, not T). T→A, A→U, C→G, G→C. Template: 3'-TACGGCAT-5'
3. Write mRNA 5'→3'. Antiparallel to template. mRNA: 5'-AUGCCGUA-3'

Answer: mRNA: 5'-AUGCCGUA-3'.

Worked example (Hard)

A eukaryotic gene has 5 exons and 4 introns. The pre-mRNA is 8000 nucleotides long, and the introns total 5000 nucleotides. How long is the mature mRNA (excluding the poly-A tail and 5' cap)?

1. Calculate coding length. Total − introns = 8000 − 5000 = 3000 nucleotides of exon sequence.
2. Mature mRNA. After splicing, the 5 exons are joined: mature mRNA ≈ 3000 nucleotides (plus 5' cap and poly-A tail, excluded here).
3. Protein prediction. 3000 ÷ 3 = 1000 codons. Subtract 1 stop codon = up to 999 amino acids.

Answer: Mature mRNA ≈ 3000 nucleotides after splicing removes 5000 nt of introns. This could encode a protein of up to 999 amino acids.

Common mistakes

• Using thymine (T) in RNA sequences instead of uracil (U) RNA replaces thymine with uracil — this is a fundamental difference between DNA and RNA. Fix: When writing any RNA sequence, use U wherever T would appear in DNA. A in DNA template → U in mRNA.
• Confusing the template strand with the coding strand The template strand is read by RNA polymerase; the coding strand has the same sequence as the mRNA (but with T instead of U). Fix: Template strand = read by enzyme (3'→5'). Coding strand = same sequence as mRNA (but T not U). VCAA may give either — read carefully.
• Forgetting RNA processing steps in eukaryotes Pre-mRNA must be processed before translation — splicing, capping, and polyadenylation are examinable. Fix: Three processing steps: (1) 5' cap added, (2) poly-A tail added, (3) introns spliced out, exons joined. Only mature mRNA leaves the nucleus.

Alternative splicing — one gene, many proteins

The human genome contains approximately 20,000 protein-coding genes — yet the human body produces over 100,000 different proteins. How? Alternative splicing allows a single gene to produce multiple different mRNA molecules and therefore multiple different proteins.

Mechanism: During RNA processing, the spliceosome can include or exclude different combinations of exons. An exon that is included in one tissue may be excluded in another. A gene with 10 exons could theoretically produce hundreds of different mRNA variants, each encoding a slightly different protein.

Example: The *Drosophila* DSCAM gene can produce over 38,000 different mRNA variants through alternative splicing — more than twice the total number of genes in the fruit fly genome. In humans, the calcitonin gene produces calcitonin in thyroid cells but CGRP (a neuropeptide) in neurons — same gene, different splicing, different protein, different function.

VCE relevance: Alternative splicing explains why the number of genes does not determine organismal complexity. Humans have roughly the same number of genes as a roundworm (~20,000) but far greater protein diversity due to alternative splicing, post-translational modifications, and regulatory complexity.

This concept also connects to disease — errors in splicing cause conditions including spinal muscular atrophy and some cancers. Understanding splicing is essential for both molecular biology and medical genetics.

Practice questions

1. The enzyme responsible for transcription is:

   1. RNA polymerase
   2. DNA polymerase
   3. Helicase
   4. Ligase

   (2 marks)

2. Which is NOT part of pre-mRNA processing in eukaryotes?

   1. Translation at the ribosome
   2. Addition of a 5' cap
   3. Addition of a poly-A tail
   4. Removal of introns

   (3 marks)

3. Explain the purpose of alternative splicing.

   (3 marks)

Summary

• Identify RNA polymerase as the enzyme of transcription.
• Distinguish pre-mRNA from mature mRNA (5' cap, poly-A tail, splicing).
• Explain the roles of introns and exons.