BiologyCBSE Class 12 · Molecular Basis of Inheritance

Transcription: Copying DNA into RNA

Updated 15 October 2026 · Fact-checked

Transcription copies the information of one DNA strand into RNA. A transcription unit has a promoter, a structural gene and a terminator. RNA polymerase reads the 3'→5' template strand and builds RNA 5'→3', so the RNA matches the coding strand with U in place of T. In eukaryotes, hnRNA is spliced, capped and tailed into mRNA.

Understand Transcription

What and why only one strand. Transcription is the copying of genetic information from one strand of DNA into RNA, using A-U, T-A, G-C, C-G pairing (adenine in DNA pairs with uracil in RNA). Only a segment of DNA, and only one strand, is copied, because:
1. If both strands were copied, the two RNAs would have different sequences, so one DNA segment would code for two different proteins, complicating information transfer.
2. The two RNAs would be complementary and would pair into double-stranded RNA, which cannot be translated, making transcription futile.

The transcription unit has three parts:

Promoter: towards the 5' end (upstream) of the structural gene (with respect to the coding strand); the binding site for RNA polymerase. Its position decides which strand is the template.
Structural gene: the part that is copied.
Terminator: towards the 3' end (downstream) of the coding strand; marks the end of transcription.

Template and coding strands. RNA polymerase also works only 5'→3', so the strand with polarity 3'→5' is the template strand. The other strand (5'→3') has the same sequence as the RNA, except T in place of U; it is the coding strand (it does not itself code for anything, but all reference points are given on it).
Example: template 3'-TACGGATCC-5' → coding 5'-ATGCCTAGG-3' → RNA 5'-AUGCCUAGG-3'.
If the promoter and terminator swapped positions, the template and coding strands would also swap.

Gene and cistron. A cistron is a DNA segment coding for one polypeptide. Structural genes are monocistronic (mostly eukaryotes) or polycistronic (mostly bacteria). Eukaryotic genes are split genes: coding sequences, exons, appear in the mature RNA; they are interrupted by introns (intervening sequences), which do not.

Three types of RNA: mRNA (template for protein), tRNA (brings amino acids and reads the code), rRNA (structural and catalytic roles in translation).

Transcription in bacteria. One DNA-dependent RNA polymerase makes all types of RNA.
1. Initiation: RNA polymerase, associated with the initiation factor sigma (σ), binds the promoter.
2. Elongation: using nucleoside triphosphates as substrates, it polymerises RNA complementary to the template; it helps open the helix; only a short stretch of RNA stays bound to the enzyme.
3. Termination: at the terminator, with the termination factor rho (ρ), the new RNA and the polymerase fall off.
RNA polymerase itself catalyses only elongation; σ and ρ associate with it transiently. Bacterial mRNA needs no processing and there is no nuclear membrane, so translation can begin before transcription ends (transcription and translation are coupled).

Transcription in eukaryotes: two extra complexities.
1. Three RNA polymerases in the nucleus:

RNA polymerase I: rRNAs (28S, 18S, 5.8S);
RNA polymerase II: the precursor of mRNA, hnRNA (heterogeneous nuclear RNA);
RNA polymerase III: tRNA, 5S rRNA and snRNAs (small nuclear RNAs).

2. Processing of hnRNA (in the nucleus):

Splicing: introns are removed and exons are joined in a defined order.
Capping: an unusual nucleotide, methyl guanosine triphosphate, is added to the 5' end.
Tailing: 200-300 adenylate residues (poly-A tail) are added to the 3' end, template-independently.

The processed hnRNA, now mRNA, is transported out of the nucleus for translation. Split genes are thought to be an ancient feature, and splicing reflects the dominance of the RNA world.

Flow chart (eukaryote): promoter → RNA polymerase II → hnRNA (exons + introns) → splicing (remove introns) → capping (5') → tailing (3', poly-A) → mRNA → leaves nucleus → translation.

Draw and label 1: transcription unit. Checklist: two parallel DNA strands with polarity marked; promoter box at the left (upstream), structural gene in the middle, terminator at the right; an arrow showing the direction of transcription. Labels: promoter, structural gene, terminator, template strand (3'→5'), coding strand (5'→3'), 5' and 3' ends.

Draw and label 2: processing of hnRNA. Checklist: a long bar with alternating shaded exons and unshaded introns; below it the spliced bar with exons joined; a cap at the 5' end and a tail of A's at the 3' end. Labels: exon, intron, hnRNA, splicing, cap (5'), poly-A tail (3'), mRNA.

Transcription: key facts

Strand relationships: $template 3^{'} \to 5^{'}; coding 5^{'} \to 3^{'}; RNA = coding strand with T \to U$; RNA is synthesised 5'→3'.
Transcription unit: $promoter (5’, upstream) - structural gene - terminator (3’, downstream)$; Positions are given with respect to the coding strand.
Bacterial factors: $σ (initiation), ρ (termination)$; RNA polymerase alone catalyses elongation.
Eukaryotic RNA polymerases: $I: rRNA (28S, 18S, 5.8S); II: hnRNA; III: tRNA, 5S rRNA, snRNA$; All three are in the nucleus.
hnRNA processing: $splicing + 5’ cap (methyl GTP) + 3’ poly-A tail (200-300 A) \to mRNA$; Happens in the nucleus.

How to answer Transcription questions

Sequence questions are the most common. Work strand by strand and always write polarity.

1Identify what is given: coding strand, template strand, or mRNA. Note the 5' and 3' ends.
2From the coding strand: mRNA has the same sequence and polarity, with U in place of T.
3From the template strand: write the complement (A→U, T→A, G→C, C→G) with opposite polarity.
4For a transcription unit, put the promoter at the 5' end (upstream) of the coding strand and the terminator at its 3' end; the template is the 3'→5' strand.
5For split genes, remove the intron sequences from the transcript before counting codons or amino acids.
6For processing, give each step with its location on the RNA: splicing (introns out), cap at 5', poly-A tail at 3', all in the nucleus.

Transcription in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Coding strand, template strand and splicing

Suppose a student in a genetics club is given this coding strand of a tiny eukaryotic structural gene, 5'-ATG GCT GTA CCA GTA TTC GGA TAA-3', and told that each GTA triplet is an intron.

Write the template strand, the primary transcript and the mature mRNA coding sequence. How many amino acids will the mature mRNA code for?

Try it first, then show the solutionSolution

Template (complement, antiparallel): 3'-TAC CGA CAT GGT CAT AAG CCT ATT-5'.
Primary transcript = coding strand with U for T: 5'-AUG GCU GUA CCA GUA UUC GGA UAA-3'.
Splice out both GUA introns: mature 5'-AUG GCU CCA UUC GGA UAA-3' (six codons).
AUG, GCU, CCA, UUC, GGA code for amino acids; UAA is a stop codon. So 5 amino acids.

Answer: Template 3'-TACCGACATGGTCATAAGCCTATT-5'; mature mRNA AUG GCU CCA UUC GGA UAA; 5 amino acids.

Memory hook: Read the coding strand, swap T for U, cut the introns, stop at the stop.

Tricks to remember Transcription

Short memory aids. Say each one aloud once, then use it on a question the same day.

Coding = Copy
The Coding strand is what the RNA Copies (with U for T); the template is its mirror.
Use it when: Writing mRNA from a given strand.
Promoter at the Prologue
Promoter comes first (5', upstream); terminator at the end (3', downstream).
Use it when: Drawing a transcription unit.
1-2-3 = Ribosome, Messenger, Transfer
Pol I → rRNA, Pol II → mRNA (hnRNA), Pol III → tRNA (and 5S, snRNA).
Use it when: Eukaryotic polymerases.

Transcription: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: mRNA from a coding strand · 1 mark · MCQBoard level

The coding strand of a transcription unit reads 5'-ATGAATG-3'. Write the RNA transcript.

Try it first, then show the solutionSolution

RNA has the same sequence and polarity as the coding strand.
Replace T with U.

Answer: 5'-AUGAAUG-3'.

Example 2: Why one strand · 2 marks · Very short answerBoard level

Give two reasons why both strands of DNA are not copied during transcription.

Try it first, then show the solutionSolution

The two RNAs would have different sequences, so one DNA segment would code for two different proteins, complicating the information-transfer machinery (1 mark).
The two RNAs would be complementary and form double-stranded RNA, which cannot be translated, so transcription would be futile (1 mark).

Answer: Two different proteins from one segment; complementary RNAs would pair into untranslatable dsRNA.

Example 3: Processing hnRNA · 3 marks · Short answerHarder board level

Explain the processing of hnRNA into functional mRNA in eukaryotes. Where does this take place?

Try it first, then show the solutionSolution

Splicing: introns are removed and exons joined in a defined order (1 mark).
Capping: methyl guanosine triphosphate is added to the 5' end; tailing: 200-300 adenylate residues are added to the 3' end, template-independently (1 mark).
This occurs in the nucleus; the mature mRNA then moves to the cytoplasm for translation (1 mark).

Answer: Splicing, 5' capping, 3' poly-A tailing, in the nucleus.

Example 4: Transcription unit from a template · 5 marks · Long answerHarder board level

A hypothetical template strand reads 3'-TACGTACGTACG-5'. (a) Write the coding strand and the mRNA. (b) Draw a transcription unit, placing the promoter and terminator correctly. (c) What would happen to the template and coding strands if the promoter were placed at the other end?

Try it first, then show the solutionSolution

(a) Coding: 5'-ATGCATGCATGC-3' (1 mark). mRNA: 5'-AUGCAUGCAUGC-3' (1 mark).
(b) Promoter at the 5' end (upstream) of the coding strand, structural gene in the middle, terminator at the 3' end (downstream); template strand 3'→5' below (1 mark for positions, 1 mark for labelled polarity).
(c) The promoter decides the template; if promoter and terminator swapped, the other strand would become the template and the definitions of coding and template strands would be reversed (1 mark).

Answer: Coding 5'-ATGCATGCATGC-3'; mRNA 5'-AUGCAUGCAUGC-3'; promoter upstream, terminator downstream; swapping reverses the strands' roles.

Example 5: Coupled transcription and translation · 2 marks · Very short answerHarder board level

Explain why transcription and translation can be coupled in bacteria but not in eukaryotes.

Try it first, then show the solutionSolution

Bacterial mRNA needs no processing and there is no nuclear membrane, so ribosomes can start translating mRNA while it is still being transcribed (1 mark).
In eukaryotes transcription occurs in the nucleus and hnRNA must be spliced, capped and tailed before the mRNA is exported to the cytoplasm, so the two processes are separated in space and time (1 mark).

Answer: Bacteria: no processing, one compartment. Eukaryotes: nucleus vs cytoplasm and processing.

Example 6: Which polymerase? · 3 marks · Short answerHOTS / competency

Suppose a drug selectively blocks eukaryotic RNA polymerase II in a cell. Which RNAs stop being made, which continue, and what happens to protein synthesis after a while? Explain.

Try it first, then show the solutionSolution

RNA polymerase II makes hnRNA, the precursor of mRNA, so no new mRNA is produced (1 mark).
RNA polymerase I (rRNA 28S, 18S, 5.8S) and III (tRNA, 5S rRNA, snRNA) still work, so these RNAs continue (1 mark).
Existing mRNA is used up and not replaced, so protein synthesis gradually stops even though ribosomes and tRNAs are present (1 mark).

Answer: mRNA synthesis stops; rRNA, tRNA, snRNA continue; protein synthesis declines as mRNA runs out.

Where students lose marks in Transcription

The slips examiners see most often, and what to write instead.

Mistake: Writing mRNA as the complement of the coding strand
Students complement whatever strand is given.
Wrong working
Coding 5'-ATGAATG-3' → mRNA 3'-UACUUAC-5'.
Right working
Coding 5'-ATGAATG-3' → mRNA 5'-AUGAAUG-3'.
Do this: Coding strand = mRNA sequence (T → U); complement only the template.
Usually costs: 1 mark.
Mistake: Dropping polarity
Students write sequences without 5' and 3'.
Wrong working
mRNA: AUGCAUGC.
Right working
mRNA: 5'-AUGCAUGC-3'.
Do this: Mark both ends on every strand.
Usually costs: ½-1 mark.
Mistake: Promoter at the 3' end
Students think 'start' means 3'.
Wrong working
The promoter lies at the 3' end of the coding strand.
Right working
Promoter at the 5' end (upstream); terminator at the 3' end (downstream) of the coding strand.
Do this: Promoter at the Prologue.
Usually costs: 1 mark.
Mistake: Saying the tail is added at the 5' end
Cap and tail are swapped.
Wrong working
Poly-A tail at 5' end; cap at 3' end.
Right working
Cap (methyl guanosine triphosphate) at 5'; poly-A tail (200-300 A) at 3'.
Do this: Cap goes on the head (5'), tail on the tail (3').
Usually costs: 1 mark.

Exam tips

In sequence questions write the strand name, the polarity and the bases, line by line.
Use 'DNA-dependent RNA polymerase' and name σ and ρ for bacterial initiation and termination.
Learn the eukaryotic polymerases with their RNAs; Pol III's list (tRNA, 5S rRNA, snRNA) is often tested.
When an intron sequence is given, delete it before counting amino acids, and do not count the stop codon.
Mention that hnRNA processing occurs in the nucleus.

Practice questions on Transcription

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The upper strand of a DNA fragment, 3'-ATTGCC-5', is the template. The mRNA will be:
(a) 5'-UAACGG-3' (b) 3'-UAACGG-5' (c) 5'-AUUGCC-3' (d) 5'-TAACGG-3'
Hint
Complement with opposite polarity, U for T.
Answer
(a) 5'-UAACGG-3'.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Ribosomal RNA is synthesised in the nucleus.
Reason (R): rRNA is transcribed by RNA polymerase III.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Which polymerase makes 28S, 18S, 5.8S rRNA?
Answer
(c). The large rRNAs are made by RNA polymerase I (only 5S rRNA by Pol III).
Q3 · 2 marks · Very short answerBoard level
Define exons and introns. Why do genes with introns need splicing before translation?
Hint
Which appears in mature RNA?
Answer
Exons are coding (expressed) sequences that appear in mature RNA; introns are intervening sequences that do not. The primary transcript (hnRNA) contains both and is non-functional; introns must be removed and exons joined so that the mRNA carries an uninterrupted coding sequence.
Q4 · 4 marks · Case-basedHarder board level
Read and answer.
Suppose a diagram of bacterial transcription shows: a DNA region X where an enzyme Z binds with the help of factor Y; at the other end of the gene, a region where factor W causes the enzyme and RNA to fall off.
(a) Identify X and Z. (1)
(b) Identify Y and W. (1)
(c) Which of the steps does Z catalyse on its own? Why can a bacterial ribosome start translating this RNA before it is complete? (2)
Hint
σ and ρ.
Answer
(a) X = promoter; Z = DNA-dependent RNA polymerase. (b) Y = sigma (σ) initiation factor; W = rho (ρ) termination factor. (c) Z alone catalyses elongation. Bacterial mRNA needs no processing and there is no nucleus separating DNA from ribosomes, so translation can begin on the growing mRNA.
Q5 · 3 marks · Short answerHOTS / competency
A eukaryotic gene has four exons of 120, 90, 150 and 60 nucleotides separated by three introns of 400, 250 and 300 nucleotides. Ignoring regions outside the exons and introns, find the length of (a) the primary transcript, (b) the spliced sequence before capping and tailing. If 250 A residues are added as the tail, what is the length of this part of the mature mRNA?
Hint
Primary transcript contains exons and introns.
Answer
(a) Exons 120 + 90 + 150 + 60 = 420; introns 400 + 250 + 300 = 950; primary transcript = 1370 nucleotides. (b) Spliced = 420 nucleotides. With a 250-A tail: 420 + 250 = 670 nucleotides (plus the cap nucleotide at the 5' end).
Q6 · 3 marks · Short answerHarder board level
Name the three RNA polymerases of the eukaryotic nucleus and the RNAs each transcribes.
Hint
I, II, III.
Answer
RNA polymerase I: rRNAs (28S, 18S, 5.8S). RNA polymerase II: hnRNA, the precursor of mRNA. RNA polymerase III: tRNA, 5S rRNA and snRNAs.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What are the three parts of a transcription unit?Show answer
Promoter, structural gene, terminator.
2.Which strand is the template?Show answer
The strand with polarity 3'→5'.
3.Which factor initiates bacterial transcription?Show answer
Sigma (σ) factor.
4.What does RNA polymerase II transcribe?Show answer
hnRNA (precursor of mRNA).
5.What is added at the 5' end of hnRNA?Show answer
Methyl guanosine triphosphate (cap).
6.How many adenylate residues form the tail?Show answer
About 200-300.

Transcription: frequently asked questions

What is transcription?

The copying of genetic information from one strand of DNA (the template) into RNA by DNA-dependent RNA polymerase.

What is the difference between the template and coding strands?

The template strand (3'→5') is read to make RNA. The coding strand (5'→3') has the same sequence as the RNA, with T instead of U, and is used as the reference.

What is a split gene?

A eukaryotic gene whose coding sequences (exons) are interrupted by non-coding introns that are removed by splicing.

What happens during capping and tailing?

Methyl guanosine triphosphate is added to the 5' end (capping) and 200-300 adenylate residues to the 3' end (tailing) of hnRNA.

Why are transcription and translation coupled in bacteria?

Bacterial mRNA needs no processing and there is no nucleus, so translation can begin before transcription ends.