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This lesson covers transcription — the first stage of gene expression in which DNA is used as a template to synthesise messenger RNA (mRNA). Understanding transcription is essential for the Edexcel A-Level Biology specification (9BI0, Topic 7).
Transcription is the process by which the nucleotide sequence of the template strand of DNA is used to produce a complementary mRNA molecule. It takes place in the nucleus of eukaryotic cells.
Transcription is the first step in the central dogma of molecular biology:
DNA → mRNA → Protein
The central dogma can be visualised as follows:
graph LR
A["DNA"] -->|"Transcription<br/>(RNA polymerase)"| B["mRNA"]
B -->|"Translation<br/>(ribosomes)"| C["Protein"]
A -->|"Replication<br/>(DNA polymerase)"| A
The enzyme responsible for catalysing transcription is RNA polymerase. In eukaryotes, RNA polymerase II is the enzyme that transcribes protein-coding genes.
| Template base | mRNA base added |
|---|---|
| A | U (uracil) |
| T | A (adenine) |
| G | C (cytosine) |
| C | G (guanine) |
Exam Tip: Remember that RNA contains uracil (U) instead of thymine (T). When writing out base-pairing during transcription, A on the template pairs with U on the mRNA, not T.
In eukaryotes, the initial transcript (pre-mRNA) undergoes several modifications before it can leave the nucleus and be translated:
Different combinations of exons can be retained in the final mRNA, producing different proteins from the same gene. This is called alternative splicing and is a key mechanism for increasing protein diversity.
For example:
Exam Tip: Post-transcriptional modifications (splicing, capping, polyadenylation) only occur in eukaryotes. In prokaryotes, mRNA is not processed — translation can even begin while transcription is still occurring (coupled transcription-translation).
| Feature | Eukaryotes | Prokaryotes |
|---|---|---|
| Location | Nucleus | Cytoplasm (no nucleus) |
| RNA polymerase | Multiple types (I, II, III) | Single type |
| Transcription factors | Required for initiation | Not required (simpler initiation) |
| mRNA processing | Splicing, capping, poly-A tail | No processing |
| Coupled with translation? | No — mRNA must leave nucleus first | Yes — ribosomes can attach while transcription occurs |
| Introns | Present in most genes | Rare |
Transcription factors are proteins that regulate transcription by binding to specific DNA sequences. They can be:
The combination of transcription factors present in a cell determines which genes are expressed. This is why different cell types in the same organism — which all contain the same DNA — can produce different proteins and have different functions.
Multiple RNA polymerase molecules can transcribe the same gene simultaneously. This allows the cell to produce many copies of the same mRNA in a short time, which is important for genes whose products are needed in large quantities (e.g. ribosomal RNA genes, histone genes).
When multiple RNA polymerase molecules are transcribing the same gene, the resulting structure resembles a "Christmas tree" when viewed under an electron microscope — this is sometimes called a transcription unit or Miller spread.
The mature mRNA molecule in eukaryotes has the following structure:
| Region | Function |
|---|---|
| 5' cap | Protection from degradation; ribosome recognition |
| 5' untranslated region (5' UTR) | Regulation of translation efficiency |
| Start codon (AUG) | Beginning of the coding sequence |
| Coding sequence | Specifies the amino acid sequence |
| Stop codon (UAA/UAG/UGA) | End of the coding sequence |
| 3' untranslated region (3' UTR) | Regulation of mRNA stability and localisation |
| Poly-A tail | Protection from degradation; export from nucleus |
| Stage | Key Events |
|---|---|
| Initiation | Transcription factors + RNA polymerase bind to promoter; DNA unwinds |
| Elongation | RNA polymerase reads template 3'→5'; mRNA grows 5'→3' |
| Termination | RNA polymerase reaches terminator; pre-mRNA released |
| Processing | Introns spliced out; 5' cap and poly-A tail added |
Exam Tip: A common exam question asks you to explain why a single gene can produce multiple different proteins. The answer is alternative splicing — different exons are included or excluded to produce different mature mRNAs from the same pre-mRNA.
This material sits in Edexcel 9BI0 Topic 8 (Grey Matter — Coordination, Response and Gene Technology), which expects candidates to describe transcription as the synthesis of a complementary RNA copy of the template (antisense) strand of DNA, catalysed by RNA polymerase, in the 5'→3' direction, using ribonucleoside triphosphates (ATP, GTP, CTP, UTP — note U replaces T in RNA), and to describe post-transcriptional modification of pre-mRNA in eukaryotes (5' cap, 3' poly-A tail, splicing of introns by the spliceosome). Synoptic links run backwards to the previous lesson on gene structure (the promoter, exons, introns and terminator defined there are the operational substrates of transcription); to Topic 1 (Lifestyle, Health and Risk — biological molecules) for the antiparallel double helix and the chemistry of phosphodiester bond formation between ribonucleotides; to Topic 2 (Genes and Health — cell biology) for the location of transcription inside the nuclear envelope and the export of mature mRNA via nuclear pores to cytoplasmic ribosomes; to Topic 6 (Infection, Immunity and Forensics) for recombinant DNA technology, where reverse transcriptase copies mRNA into cDNA (a sliced, intron-free version of the gene that can be expressed in bacteria) and where PCR amplifies that cDNA; and forwards to the next lesson on translation (which decodes the mature mRNA produced here at cytoplasmic ribosomes). Refer to the official Pearson Edexcel 9BI0 specification document for exact wording.
Question (8 marks):
(a) Describe the events of transcription in a eukaryotic cell, from RNA polymerase binding to release of pre-mRNA. (4)
(b) The coding (sense) strand of a short region of a eukaryotic gene reads 5'-ATG GCA TTT TAA-3'. State (i) the sequence of the corresponding template (antisense) strand, (ii) the sequence of the mRNA produced by transcription, and (iii) one reason why pre-mRNA cannot leave the nucleus until it has been processed. (4)
Solution with mark scheme:
(a) M1 (AO1) — initiation. RNA polymerase (RNA polymerase II for protein-coding genes in eukaryotes) binds to the promoter region upstream (5' side) of the gene, assisted by general transcription factors that form a transcription initiation complex. The DNA double helix locally unwinds and hydrogen bonds between complementary bases break, exposing the bases of both strands.
A1 (AO1) — elongation. RNA polymerase moves along the template (antisense) strand in the 3'→5' direction, aligning free ribonucleoside triphosphates (ATP, GTP, CTP, UTP) by complementary base pairing — A pairs with U, T pairs with A, G pairs with C, C pairs with G — and catalysing the formation of phosphodiester bonds between adjacent ribonucleotides. The growing pre-mRNA is therefore extended in the 5'→3' direction.
A1 (AO1) — re-annealing. Behind the polymerase, the two DNA strands re-form the double helix; only a short transcription bubble of around 10–15 base pairs is unwound at any moment.
A1 (AO1) — termination. When RNA polymerase reaches a terminator sequence, it dissociates from the DNA and the pre-mRNA is released into the nucleoplasm.
(b) M1 (AO2) — template strand. The template strand is the antiparallel complement of the coding strand, so it reads 3'-TAC CGT AAA ATT-5' (or, written 5' to 3', 5'-TTA AAA TGC CAT-3').
A1 (AO2) — mRNA sequence. mRNA is identical to the coding strand with U replacing T: 5'-AUG GCA UUU UAA-3'.
A1 (AO3.1) — reason for nuclear retention. Pre-mRNA still contains introns, which must be removed by splicing, and lacks the 5' cap and 3' poly-A tail that protect mRNA from exonuclease degradation and license export through nuclear pores.
A1 (AO3.2) — extension. The cap is added co-transcriptionally as the 5' end emerges from the polymerase; the poly-A tail is added post-transcriptionally by polyA polymerase after a polyadenylation signal (AAUAAA); splicing is carried out by the spliceosome (snRNPs U1, U2, U4, U5, U6) recognising 5' GU and 3' AG splice sites.
Total: 8 marks (M2 A6).
Question (6 marks): A scientist measured the length of pre-mRNA produced from a particular eukaryotic gene and the length of the mature mRNA exported from the nucleus. The pre-mRNA was approximately 8,500 nucleotides long, while the mature mRNA was approximately 1,800 nucleotides long. Western blotting showed that the same gene gave rise to three distinct polypeptides in different tissues of the organism.
Using the data, explain how a single gene can produce a much shorter mRNA and several different polypeptides.
Mark scheme decomposition by AO:
| Mark | AO | Earned by |
|---|---|---|
| 1 | AO1.1 | Stating that pre-mRNA contains introns and exons; introns are non-coding and are removed before export |
| 2 | AO1.2 | Stating that splicing is carried out by the spliceosome and joins exons to form mature mRNA |
| 3 | AO2.1 | Calculating that ~6,700 nucleotides (≈79% of the pre-mRNA) are introns removed during splicing |
| 4 | AO2.7 | Linking the three polypeptides to alternative splicing — different combinations of exons retained in different tissues |
| 5 | AO3.1 | Concluding that one locus can therefore specify multiple polypeptide isoforms, expanding proteome diversity beyond gene number |
| 6 | AO3.2 | Justifying tissue specificity by tissue-specific splicing factors that bias spliceosome choice between alternative splice sites |
Total: 6 marks (AO1 = 2, AO2 = 2, AO3 = 2). Edexcel reliably tests transcription through "compare pre-mRNA with mature mRNA / explain how one gene gives multiple proteins" prompts; candidates who name splicing without quantifying the intron fraction or naming alternative splicing lose AO2 and AO3 marks.
Lesson 1 (gene structure) — the promoter, exons, introns and terminator defined there are the operational substrates of transcription. RNA polymerase binds the promoter; it reads through both exons and introns to produce pre-mRNA; the terminator releases it; splicing then resolves the exon/intron architecture into a continuous coding sequence. Without the architecture established in lesson 1, transcription has no defined start, no defined stop, and no rule for what to keep in the mature transcript.
Lesson 3 (translation) — the mature mRNA produced here is the template for translation. The 5' cap recruits the small ribosomal subunit; the 5' UTR is scanned for the first AUG; codons are decoded by tRNA; the 3' poly-A tail recruits poly-A binding proteins that loop the mRNA into a closed-loop "Christmas-cake" topology that increases translation efficiency. Faults in transcription (mis-spliced mRNAs, premature termination, missing cap) directly compromise translation downstream.
Topic 1 (DNA chemistry) — the chemistry of phosphodiester bond formation. RNA polymerase catalyses nucleophilic attack of the 3' OH of the growing chain on the α-phosphate of the incoming ribonucleoside triphosphate, releasing pyrophosphate (PPi) and forming a phosphodiester bond. This is the same chemistry as DNA replication, which is why both polymerases extend strictly in the 5'→3' direction. Note, however, that RNA polymerase does not need a primer — unlike DNA polymerase, it can initiate de novo on a single-stranded template.
Topic 2 (cell compartmentation) — transcription is nuclear, translation is cytoplasmic in eukaryotes. This compartmentation is what allows extensive post-transcriptional modification (capping, polyadenylation, splicing) to occur before translation begins. Prokaryotes, lacking a nucleus, cannot regulate this way — translation begins on the still-transcribing mRNA (coupled transcription–translation), so introns and splicing are largely absent.
Topic 6 (recombinant DNA) — reverse transcriptase reverses this reaction. HIV reverse transcriptase copies an RNA template into DNA (the opposite direction to RNA polymerase), producing cDNA. Biotechnologists exploit this to make intron-free copies of eukaryotic genes that can be expressed in bacteria. Antiretroviral drugs (e.g. zidovudine, AZT) are nucleoside analogues that target reverse transcriptase, exploiting the fact that the viral enzyme is more error-prone and lacks the proofreading of host DNA polymerases.
| AO | Typical share on transcription questions | Earned by |
|---|---|---|
| AO1 (knowledge) | 35–45% | Naming RNA polymerase, promoter, terminator, template strand, pre-mRNA, spliceosome; stating 5'→3' synthesis direction |
| AO2 (application) | 35–50% | Translating a coding strand into mRNA (with U replacing T); calculating intron fraction from pre-mRNA / mature mRNA lengths; predicting effects of mutations in promoter, splice sites, polyA signal |
| AO3 (analysis / evaluation) | 10–20% | Linking alternative splicing to proteome expansion; arguing why eukaryotes cannot couple transcription–translation; evaluating reverse transcription as an exception to the central dogma |
Examiner-rewarded phrasing: "RNA polymerase reads the template (antisense) strand in the 3'→5' direction and synthesises mRNA in the 5'→3' direction"; "U replaces T in RNA"; "introns are transcribed into pre-mRNA and then spliced out by the spliceosome before nuclear export"; "the 5' cap is added co-transcriptionally and the poly-A tail is added post-transcriptionally"; "alternative splicing allows one gene to produce multiple polypeptide isoforms".
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