Q: What does H2 Biology: DNA Replication, Transcription & Translation cover? A: H2 Biology molecular biology notes covering DNA replication, transcription, translation, gene expression regulation, and exam-style worked examples for 9477.
TL;DR Replication copies DNA, transcription makes RNA from DNA, and translation uses mRNA codons to build a polypeptide. Most exam marks come from directionality, enzyme roles, and how sequence changes affect protein structure.
Concrete example: During transcription, RNA polymerase reads the DNA template strand 3' to 5' and makes mRNA 5' to 3'. During translation, each mRNA codon is matched to a tRNA anticodon, adding one amino acid to the chain.
Directionality checkpoint
Before writing a process answer, separate the template being read from the product being built. Most sequence errors come from mixing these two directions.
Process
Template being read
Product being built
Check before moving on
DNA replication
Each parental DNA strand is read 3' to 5'.
New DNA strands are synthesised 5' to 3'.
Leading strand is continuous; lagging strand needs Okazaki fragments.
Transcription
One DNA template strand is read 3' to 5'.
Pre-mRNA is synthesised 5' to 3'.
mRNA matches the coding strand except U replaces T.
Translation
mRNA codons are read 5' to 3'.
The polypeptide grows from N-terminus to C-terminus.
Reviewed by
Ezekiel Tan·Academic Advisor (Biology)
Codon, anticodon, and amino acid must stay in the same reading frame.
Common trap: do not say "polymerase moves 5' to 3'" without saying what is moving and what is being built. The enzyme reads the template in one direction while extending the new nucleic acid in the opposite direction.
Coding, template, and mRNA checkpoint
When a question gives a short DNA sequence, label the strand before converting bases. This prevents the common error of making mRNA complementary to the coding strand.
Strand named in the question
What it means
What to write next
Coding DNA strand
Same base order as the mRNA, except DNA has T where RNA has U
Replace T with U to get the mRNA sequence.
Template DNA strand
The strand RNA polymerase reads 3' to 5'
Build the complementary mRNA 5' to 3'.
mRNA
The sequence read by the ribosome
Split into codons from the start codon and keep the reading frame fixed.
Worked check: if the coding strand is 5'-ATG GAA TTT-3', the mRNA is 5'-AUG GAA UUU-3'. The matching template strand is 3'-TAC CTT AAA-5'. During translation, the ribosome reads AUG, then GAA, then UUU in that order.
Misconception check: RNA polymerase reads the template strand, but the mRNA sequence matches the coding strand except for U replacing T. Do not complement the coding strand unless the question asks for the template strand.
Molecular biology sits at the heart of Core Idea 3 (Genetics) in the H2 Biology syllabus. DNA replication, transcription, and translation form a tightly connected sequence - the central dogma - that examiners revisit in almost every sitting. Students who can name every enzyme, explain directionality, and link mutations to protein consequences tend to pick up full marks on structured questions.
Status: SEAB's current H2 Biology (9477) syllabus PDF is labelled for 2026 and identifies 2026 as the first year of examination. This page aligns with Core Idea 3 - Genetics and the Molecular Basis of Inheritance. [1]
Quick revision box
What this topic tests: DNA structure, semi-conservative replication, transcription and post-transcriptional modification, translation, gene expression regulation, mutations, and their consequences for protein function.
Top mistakes to avoid: Writing that DNA polymerase reads 5' to 3' (it reads template 3' to 5' and synthesises 5' to 3'); omitting the role of primase; confusing the template strand with the coding strand during transcription.
20-minute sprint plan: 5 min enzyme table for replication; 10 min walk-through of transcription then translation with directionality; 5 min mutation type drill with worked example.
1 DNA Structure Review
Before studying the central dogma, ensure the following structural features are secure.
Double helix: Two polynucleotide strands wound around each other in a right-handed helix.
Antiparallel orientation: One strand runs 5' to 3' while the complementary strand runs 3' to 5'. This directionality is critical for understanding replication and transcription.
Complementary base pairing: Adenine pairs with thymine (two hydrogen bonds); guanine pairs with cytosine (three hydrogen bonds). This specificity underpins faithful replication.
Sugar-phosphate backbone: Phosphodiester bonds between the 3' carbon of one deoxyribose and the 5' carbon of the next sugar provide structural integrity. The bases project inward.
Major and minor grooves: Formed by the helical twist; proteins such as transcription factors interact with DNA primarily through the major groove.
2 DNA Replication
2.1 Overview
DNA replication is semi-conservative: each daughter molecule contains one original (parental) strand and one newly synthesised strand. Replication occurs during the S phase of interphase and ensures that every daughter cell receives a complete copy of the genome.
2.2 Evidence - the Meselson-Stahl experiment
Meselson and Stahl (1958) grew E. coli in a medium containing heavy nitrogen (15N), then transferred the bacteria to a 14N medium.
Generation 0: All DNA was heavy (15N-15N).
Generation 1: All DNA appeared at an intermediate density, consistent with one heavy strand and one light strand - ruling out conservative replication.
Generation 2: Half of the DNA was intermediate, half was light - ruling out dispersive replication and confirming semi-conservative replication.
2.3 Key enzymes and proteins
Enzyme / Protein
Role
Helicase
Unwinds the double helix by breaking hydrogen bonds between complementary base pairs
Single-strand binding (SSB) proteins
Stabilise the separated single strands and prevent them from re-annealing
Primase
Synthesises a short RNA primer complementary to the template strand, providing the free 3'-OH group that DNA polymerase III requires
DNA polymerase III
Adds deoxyribonucleotides to the 3' end of the growing strand (5' to 3' synthesis); also has 3' to 5' proofreading exonuclease activity
DNA polymerase I
Removes RNA primers and replaces them with DNA
DNA ligase
Seals the nicks (phosphodiester bond gaps) between Okazaki fragments on the lagging strand
2.4 Leading strand vs lagging strand
Because DNA polymerase III can only synthesise in the 5' to 3' direction:
Leading strand: Synthesised continuously in the same direction as the replication fork movement.
Lagging strand: Synthesised discontinuously as short Okazaki fragments (approximately 1,000 to 2,000 nucleotides in prokaryotes), each requiring its own RNA primer. Fragments are later joined by DNA ligase after primer removal.
Replication fork side checkpoint
When a replication-fork diagram is given, first decide which way the fork is opening. Then label the template directions before naming a strand as leading or lagging.
fork opens this way ->
template read 3' to 5' -> new strand can grow continuously -> leading
template read 5' to 3' -> new strand must be made in short pieces -> lagging
Diagram clue
What to label first
Answer move
Common trap
Fork arrow is shown
Direction the fork is opening
The leading strand grows toward the fork.
Calling the upper strand leading just because it is drawn on top.
Template strand direction is shown
Whether DNA polymerase can read that template 3' to 5' while moving toward the fork
If yes, synthesis is continuous.
Forgetting that synthesis always extends the new strand 5' to 3'.
Many short new fragments are shown
Each fragment needs a primer and later ligase action
Identify the lagging strand and name Okazaki fragments.
Saying fragments form because DNA polymerase is slow rather than because of antiparallel templates.
Worked check: if the fork opens to the right and the lower template runs 3' to 5' toward the fork, DNA polymerase can follow the fork while extending the new lower strand 5' to 3'. The lower new strand is leading. The upper new strand must be made as Okazaki fragments away from the fork, so it is lagging.
Misconception check: leading and lagging are not fixed to "top" and "bottom". They depend on fork direction and template polarity.
2.5 Proofreading and error correction
DNA polymerase III possesses 3' to 5' exonuclease activity. If an incorrect nucleotide is incorporated, the enzyme reverses, excises the mismatch, and replaces it with the correct base. This proofreading reduces the error rate to approximately one mistake per 109 base pairs.
3 Transcription
Transcription is the synthesis of a messenger RNA (mRNA) molecule using one strand of DNA as a template. It occurs in the nucleus of eukaryotic cells.
3.1 Initiation
Transcription factors bind to the promoter region upstream of the gene.
RNA polymerase is recruited to the promoter and binds, forming the transcription initiation complex.
The DNA double helix unwinds locally, exposing the template strand (also called the antisense strand), which runs 3' to 5'.
3.2 Elongation
RNA polymerase reads the template strand in the 3' to 5' direction and synthesises the mRNA in the 5' to 3' direction.
Ribonucleoside triphosphates (ATP, GTP, CTP, UTP) are added by complementary base pairing with the template strand. Note that uracil replaces thymine in RNA.
Unlike DNA replication, no primer is required - RNA polymerase can initiate synthesis de novo.
The coding strand (sense strand) has the same sequence as the mRNA, except with thymine instead of uracil.
3.3 Termination
RNA polymerase reaches a terminator sequence on the DNA. The mRNA transcript is released, and the DNA re-anneals.
The primary transcript (pre-mRNA) undergoes three major modifications before leaving the nucleus:
5' cap: A modified guanine nucleotide is added to the 5' end. The cap protects the mRNA from degradation and assists ribosome recognition during translation.
3' poly-A tail: A string of adenine nucleotides (typically 100 to 250) is added to the 3' end. The poly-A tail increases mRNA stability and aids nuclear export.
Splicing: Non-coding sequences called introns are removed by the spliceosome. The remaining coding sequences, called exons, are joined together to form the mature mRNA. Alternative splicing allows one gene to encode multiple protein variants.
4 Translation
Translation is the synthesis of a polypeptide chain from an mRNA template. It occurs at ribosomes in the cytoplasm.
4.1 Ribosome structure
Small subunit: Binds the mRNA and ensures correct codon-anticodon pairing.
Large subunit: Catalyses peptide bond formation (peptidyl transferase activity).
Three binding sites on the ribosome:
A site (aminoacyl): Incoming aminoacyl-tRNA binds here.
P site (peptidyl): Holds the tRNA carrying the growing polypeptide chain.
E site (exit): Deacylated tRNA exits the ribosome from this site.
4.2 tRNA structure
Cloverleaf shape with an anticodon loop at one end and an amino acid attachment site (3' CCA end) at the other.
Each tRNA is charged with its specific amino acid by an aminoacyl-tRNA synthetase enzyme (one synthetase per amino acid).
The anticodon base-pairs with the complementary codon on the mRNA in an antiparallel fashion.
4.3 Initiation
The small ribosomal subunit binds to the 5' cap of the mRNA and scans until it reaches the start codon (AUG).
The initiator tRNA (carrying methionine) binds to the start codon at the P site via its anticodon (UAC).
The large ribosomal subunit then joins, forming the complete ribosome.
4.4 Elongation
An aminoacyl-tRNA enters the A site, with its anticodon complementary to the mRNA codon.
A peptide bond forms between the amino acid in the A site and the growing polypeptide in the P site (catalysed by the ribosome's peptidyl transferase).
The ribosome translocates one codon along the mRNA in the 5' to 3' direction. The tRNA in the P site moves to the E site and exits; the tRNA in the A site moves to the P site.
Steps 1 to 3 repeat, elongating the polypeptide chain.
Translation site checkpoint
For translation diagrams, track what each tRNA is carrying before and after peptide-bond formation. The site names only make sense when linked to the cargo movement.
Ribosome site
What is there before peptide bond formation
What happens next
Common trap
A site
Incoming aminoacyl-tRNA carrying the next amino acid
Its amino acid is joined to the growing chain
Saying the A site already holds the full polypeptide.
P site
tRNA carrying the growing polypeptide chain
The chain is transferred to the amino acid in the A site
Forgetting that the P-site tRNA becomes uncharged after transfer.
E site
Empty until a deacylated tRNA moves there
Deacylated tRNA exits the ribosome
Treating the E site as another codon-reading site.
mRNA
Codons exposed in the ribosome
Ribosome moves one codon in the 5' to 3' direction
Moving the mRNA reading frame by two or four bases.
Worked check: after a peptide bond forms, the growing polypeptide is attached to the tRNA in the A site. During translocation, that tRNA moves into the P site, the empty tRNA moves to the E site, and the next codon enters the A site.
Misconception check: the ribosome does not choose amino acids directly. Codon-anticodon pairing selects the charged tRNA, and the charged tRNA carries the amino acid.
4.5 Termination
When a stop codon (UAA, UAG, or UGA) enters the A site, no tRNA can bind.
A release factor binds to the stop codon in the A site.
The completed polypeptide is released, and the ribosomal subunits dissociate from the mRNA.
5 Gene Expression Regulation
Not all genes are expressed at all times. Regulation occurs at multiple levels and differs between prokaryotes and eukaryotes.
5.1 Prokaryotic regulation - operons
Lac operon (inducible):
In the absence of lactose, a repressor protein (encoded by the lacI gene) binds to the operator, blocking RNA polymerase from transcribing the structural genes (lacZ, lacY, lacA).
When lactose is present, allolactose (an isomer of lactose) binds to the repressor and changes its shape, causing it to detach from the operator. RNA polymerase can then transcribe the structural genes.
Trp operon (repressible):
When tryptophan levels are low, the repressor is inactive and the structural genes for tryptophan biosynthesis are transcribed.
When tryptophan accumulates, it acts as a co-repressor, binding to the repressor and activating it. The activated repressor binds the operator and blocks transcription.
Operon logic checkpoint
For operon questions, decide the default transcription state before describing what the signal molecule does. Lac and trp operons use repressors, but the signal has the opposite effect on each system.
Operon
Default when the signal is absent
Signal molecule
Effect of signal on repressor
Transcription outcome
Common trap
Lac operon
Repressor binds the operator, so structural genes are not transcribed.
Lactose, through allolactose
Inactivates the repressor so it leaves the operator.
Structural genes can be transcribed.
Saying lactose is the enzyme instead of the inducer signal.
Trp operon
Repressor is inactive, so structural genes are transcribed.
Tryptophan
Activates the repressor as a co-repressor.
Structural genes are not transcribed.
Saying high tryptophan should increase tryptophan synthesis.
Worked check: if lactose is present, the lac repressor is inactivated and cannot block the operator, so transcription can occur. If tryptophan is abundant, tryptophan activates the trp repressor, so the operator is blocked and transcription decreases.
Misconception check: "repressor present" is not enough. State whether the repressor is active and whether it is bound to the operator.
5.2 Eukaryotic regulation
Eukaryotic gene expression is regulated at multiple levels:
Transcription factors: Proteins that bind to specific DNA sequences (e.g. enhancers or the promoter region) and recruit or block RNA polymerase. Activators increase transcription; repressors decrease it.
Enhancers and silencers: Regulatory DNA sequences that can be thousands of base pairs away from the gene they regulate. They function through DNA looping, bringing transcription factors into contact with the promoter.
Epigenetic modifications:
DNA methylation: Addition of methyl groups to cytosine bases (commonly at CpG sites). Methylation of a promoter region typically silences gene expression.
Histone modification: Acetylation of histone tails loosens chromatin (euchromatin), promoting transcription. Deacetylation or methylation of histones can compact chromatin (heterochromatin), repressing transcription.
6 Mutations
A mutation is a permanent change in the nucleotide sequence of DNA. Mutations can arise spontaneously during replication or be induced by mutagens (e.g. UV radiation, chemical agents).
6.1 Point mutations (single nucleotide changes)
Type
What happens
Effect on protein
Silent
Changed codon still codes for the same amino acid (degeneracy of the genetic code)
No change in protein
Missense
Changed codon codes for a different amino acid
May alter protein folding and function (e.g. sickle cell anaemia: GAG to GUG in the haemoglobin gene)
Nonsense
Changed codon becomes a premature stop codon
Truncated, usually non-functional protein
6.2 Frameshift mutations
Insertion or deletion of one or more nucleotides (not in multiples of three) shifts the reading frame. Every codon downstream of the mutation is altered, typically producing a non-functional protein. Frameshifts are generally more damaging than point mutations because they affect a larger portion of the polypeptide.
6.3 Consequences for protein function
A mutation in the active site of an enzyme may abolish catalytic activity.
A mutation affecting protein folding (e.g. altering a disulfide bond or hydrophobic core) can reduce stability.
Some mutations are neutral if they occur in non-coding regions or do not alter the protein's functional domains.
6.4 Mutation effect checkpoint
When a sequence-change question appears, decide the effect in this order before naming the mutation type.
First check
What to ask
Why it matters
Reading frame
Was a number of bases not divisible by three inserted or deleted?
A frameshift changes every downstream codon, so the effect is usually larger than one amino acid substitution.
Stop signal
Did the new codon become a stop codon?
A premature stop codon truncates the polypeptide and may remove entire functional regions.
Amino acid property
If one amino acid changed, did charge, polarity, or size change?
A conservative substitution may have little effect, while a change in an active site or binding site can be severe.
Location in the gene
Is the mutation in a coding exon, splice site, promoter, or non-coding region?
Protein sequence, mRNA processing, and expression level can be affected in different ways.
Misconception check: "point mutation" describes the size of the DNA change, not the severity of the phenotype. A single-base change can be silent, missense, nonsense, or disruptive if it affects regulation or splicing.
Worked check: compare the coding-strand sequence 5'-ATG GAA TTT CCG-3' with a mutant sequence 5'-ATG GAT TTC CG-3'. First split the normal coding strand into codons: ATG, GAA, TTT, CCG. The mutant has one base deleted after GAT, so the grouping becomes ATG, GAT, TTC, ... and every codon after the deletion is read in a new frame. This is a frameshift, not just a single amino acid substitution.
If the mutant were 5'-ATG TAA TTT CCG-3' instead, only the second codon changes from GAA to TAA. Since TAA is a stop codon in the coding DNA sequence, this is a nonsense mutation that would terminate translation early.
Misconception check: do not classify mutations by how similar the printed sequences look. Count inserted or deleted bases first, then check codons and amino acid consequences.
7 Comparison Table: Replication vs Transcription vs Translation
Feature
DNA Replication
Transcription
Translation
Template
Both DNA strands
One DNA strand (template / antisense strand)
mRNA
Product
Two identical DNA molecules
Pre-mRNA (then mature mRNA after processing)
Polypeptide
Key enzyme
DNA polymerase III
RNA polymerase
Ribosome (with peptidyl transferase activity)
Direction of synthesis
5' to 3'
5' to 3'
N-terminus to C-terminus (mRNA read 5' to 3')
Primer required?
Yes (RNA primer by primase)
No
No (initiator tRNA binds start codon)
Nucleotides used
dATP, dTTP, dGTP, dCTP
ATP, UTP, GTP, CTP
Amino acids (carried by tRNA)
Location (eukaryotes)
Nucleus
Nucleus
Cytoplasm (ribosomes)
Base pairing rule
A-T, G-C
A-U (template to mRNA), G-C
Codon-anticodon (mRNA to tRNA)
8 How Molecular Biology Appears in Exams
Paper 2 (structured): Expect diagrams of a replication fork or a ribosome at a mRNA, with blanks to label enzymes, sites, or directions. Data from the Meselson-Stahl experiment may be presented as a density gradient graph for interpretation.
Paper 3 (free-response): Essay prompts such as "Describe the process of translation" or "Compare and contrast DNA replication and transcription" are common. Full marks require precise enzyme names, directionality, and logical sequencing of steps.
Cross-topic links: Mutations connect to inheritance patterns and genetic disease (Core Idea 3). Gene regulation connects to cell differentiation (Core Idea 1) and evolution (Core Idea 4). Exam questions often span these boundaries.
Exam tip: When describing replication or transcription, always state the direction of synthesis (5' to 3') and the direction the template is read (3' to 5'). Omitting directionality is one of the most common reasons students lose marks.
Quick Retrieval Check
Name three enzymes involved in DNA replication and state the role of each.
Explain why the lagging strand is synthesised discontinuously.
Describe two post-transcriptional modifications and explain the function of each.
Outline the events that occur at the A site, P site, and E site during translation elongation.
Distinguish between a missense mutation and a nonsense mutation, giving one example of each.
Where can I find the full H2 Biology Notes series? Start at the H2 Biology Notes hub, then follow Core Ideas 1-4 and the Extension Topics.
Is this topic tested in Paper 4 (practical)? Molecular biology itself is rarely tested as a standalone practical. However, techniques such as gel electrophoresis and PCR - both of which rely on DNA replication and base-pairing principles - may appear as data-handling or planning questions in Paper 4.
Do I need to memorise every enzyme in the replication fork? Yes. SEAB expects you to name helicase, primase, DNA polymerase III, DNA polymerase I, DNA ligase, and SSB proteins, and to state each enzyme's specific role. Simply writing "DNA polymerase" without specifying III or I is insufficient for full marks. [1]
What is the difference between the template strand and the coding strand? The template strand (antisense strand) is the strand read by RNA polymerase during transcription (read 3' to 5'). The coding strand (sense strand) has the same base sequence as the mRNA (with T instead of U) and is not directly read during transcription.
How do prokaryotic and eukaryotic gene regulation differ? Prokaryotes primarily use operons (clusters of genes under one promoter, regulated by repressors and inducers). Eukaryotes regulate gene expression at multiple levels including chromatin remodelling, transcription factor binding to enhancers/silencers, mRNA splicing, and epigenetic modifications such as DNA methylation and histone acetylation. [1]
