Why is DNA replication described as semi-conservative?

Each new DNA molecule contains one original parent strand and one newly synthesised strand. This was demonstrated by the Meselson-Stahl experiment in 1958.

What is a frameshift mutation and why is it so damaging?

A frameshift mutation occurs when a base is inserted or deleted, shifting every subsequent codon out of alignment. This changes almost every amino acid downstream and usually produces a non-functional protein.

What does it mean that the genetic code is degenerate?

Most amino acids are coded for by more than one codon. With 64 possible codons but only 20 amino acids, this redundancy means some substitution mutations are silent.

What is the role of tRNA in protein synthesis?

tRNA carries specific amino acids to the ribosome during translation. Its anticodon base-pairs with the complementary codon on mRNA, delivering the correct amino acid to the growing polypeptide chain.

What are introns and exons?

Introns are non-coding sequences removed by splicing from pre-mRNA. Exons are coding sequences joined together to form mature mRNA. Alternative splicing enables one gene to code for multiple polypeptides.

How do I describe DNA replication for a 6-mark question?

Include: helicase breaks hydrogen bonds; each strand acts as template; free nucleotides align by complementary base pairing; DNA polymerase forms phosphodiester bonds; two identical molecules produced each with one original and one new strand (semi-conservative).

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Why DNA Replication and Protein Synthesis Decides Your Grade

DNA replication and protein synthesis is the molecular backbone of A-Level Biology. Every exam board allocates substantial marks to this topic, and the questions demand precision. You cannot bluff your way through a “describe semi-conservative replication” question – either you use the exact terminology that earns marks, or you do not.

Having examined this topic for both WJEC/Eduqas and Edexcel, I can tell you that the single most common reason students lose marks is not a lack of knowledge. It is using the wrong verb. Students write “DNA polymerase forms hydrogen bonds between bases” and lose a mark they should have gained. The correct statement is that DNA polymerase forms phosphodiester bonds between adjacent nucleotides. Hydrogen bonds between complementary bases form spontaneously. Getting that distinction right is worth marks every single year.

On this page I am going to take you through every aspect of DNA replication and protein synthesis that your exam board requires. More importantly, I am going to show you exactly which phrases examiners accept, which they reject, and how to construct your answers so that you earn maximum credit. This is information I have from standardisation meetings, not from reading a textbook.

Board check: DNA replication and protein synthesis appears in AQA 3.1.5 & 3.4.2, OCR A Modules 2.1.3 & 6.1.1, Edexcel A Topic 2 & 6, Edexcel B Topic 1 & 7, and WJEC/Eduqas Unit 1 Section 1.5. OCR A is the most demanding board for this topic – it uniquely requires the lac operon, Hox genes, AND post-translational modification. OCR B is the most restrictive – it explicitly excludes post-transcriptional modification.

Key Terminology – The Words That Earn Marks

Before you tackle any question on this topic, you need to know these definitions precisely. Examiners have specific accept and reject criteria for every term.

Gene A sequence of DNA nucleotides that codes for a polypeptide. Note: not “codes for a protein” – some proteins contain multiple polypeptides coded by different genes.

Codon A sequence of three adjacent nucleotide bases (triplet) on mRNA that codes for a specific amino acid or a stop signal.

Anticodon A sequence of three bases on tRNA that is complementary to a specific codon on mRNA.

Template strand (antisense strand) The strand of DNA that is read by RNA polymerase during transcription. It runs 3′→5′ and is complementary to the mRNA produced.

Coding strand (sense strand) The strand of DNA that has the same base sequence as the mRNA (with thymine instead of uracil). It is NOT read during transcription.

Semi-conservative replication DNA replication in which each new DNA molecule contains one original (parent) strand and one newly synthesised strand.

Degenerate (redundant) genetic code The genetic code in which most amino acids are coded for by more than one codon. This is because there are 64 possible codons but only 20 amino acids (plus 3 stop signals).

Examiner reject list: Do NOT write “DNA reproduces” or “DNA multiplies” – the correct term is “replicates.” Do NOT use single letter abbreviations (A, T, G, C) where full base names are needed. Do NOT write “phosphorus” when you mean “phosphate.” Do NOT say “pentose” alone when the question requires “deoxyribose” (for DNA) or “ribose” (for RNA).

DNA Replication – Semi-Conservative Mechanism

DNA replication occurs during the S phase of interphase, before a cell divides. The purpose is to produce two identical copies of each DNA molecule so that each daughter cell receives a complete genome. The mechanism is described as semi-conservative because each new DNA molecule retains one original strand.

The Five Steps That Earn Full Marks

Examiners typically award 3–7 marks for a DNA replication answer. Here is the sequence of marking points, in the order you should present them:

DNA helicase unwinds the double helix The enzyme DNA helicase breaks the hydrogen bonds between complementary base pairs, separating the two strands at the replication fork.

Each strand acts as a template The exposed bases on each separated strand serve as a template for the assembly of a new complementary strand.

Free nucleotides align by complementary base pairing Free DNA nucleotides from the nucleoplasm align opposite their complementary bases: adenine pairs with thymine (A–T), guanine pairs with cytosine (G–C).

DNA polymerase joins adjacent nucleotides The enzyme DNA polymerase catalyses the formation of phosphodiester bonds between adjacent nucleotides, building the new strand in the 5′→3′ direction.

Two identical DNA molecules are produced Each new molecule contains one original strand and one newly synthesised strand – this is why the process is called semi-conservative.

The mark-losing mistake: Every year, students write “DNA polymerase forms hydrogen bonds between complementary bases.” This is wrong and loses a mark. DNA polymerase forms phosphodiester bonds in the sugar-phosphate backbone. Hydrogen bonds between complementary bases form spontaneously. This single error appears in AQA examiner reports almost every year.

Edexcel B only: Edexcel B (9BI0) is the only UK A-Level board that requires you to name DNA ligase – the enzyme that joins Okazaki fragments on the lagging strand. No other UK board requires ligase, leading strand, lagging strand, or Okazaki fragments. If you are on AQA, OCR, WJEC, or Edexcel A, you do not need this detail.

The Meselson-Stahl Experiment – Evidence for Semi-Conservative Replication

The Meselson-Stahl experiment (1958) provided the definitive evidence that DNA replication is semi-conservative rather than conservative or dispersive. Several boards name this experiment explicitly: Edexcel A (2.11), WJEC (1.5h), and Eduqas (CC5h). AQA implies it through “evaluate the work of scientists in validating the Watson–Crick model.”

Grow E. coli in heavy nitrogen (¹⁵N) medium Bacteria were cultured for many generations in a medium containing only ¹⁵N (heavy nitrogen). All their DNA incorporated ¹⁵N and became “heavy.”

Transfer to light nitrogen (¹⁴N) medium Bacteria were then transferred to a medium containing only ¹⁴N (light nitrogen) and allowed to replicate.

Extract DNA and centrifuge DNA was extracted at intervals and separated by density gradient centrifugation in caesium chloride (CsCl).

After one generation: single intermediate band All DNA appeared at an intermediate density – ruling out conservative replication (which would show one heavy + one light band).

After two generations: two bands One band at intermediate density and one at light density – consistent with semi-conservative replication and ruling out dispersive replication.

Why this rules out each model: Conservative replication predicts one heavy band + one light band after generation 1. The intermediate band proves this wrong. Dispersive replication predicts a single band that gets progressively lighter with each generation – it would never produce two distinct bands. The appearance of two separate bands after generation 2 rules out dispersive replication.

Transcription – DNA to mRNA

Transcription is the process by which the genetic information in a gene is copied from DNA into messenger RNA (mRNA). It occurs in the nucleus in eukaryotes and is catalysed by the enzyme RNA polymerase.

The Mark-Earning Steps

DNA strands separate at the gene Hydrogen bonds between complementary base pairs break. Only the section of DNA containing the relevant gene unwinds – not the entire chromosome.

RNA polymerase binds to the template strand RNA polymerase reads the template strand (antisense strand) in the 3′→5′ direction, synthesising mRNA in the 5′→3′ direction.

Free RNA nucleotides align by complementary base pairing Adenine on the template pairs with uracil (not thymine – RNA has uracil). Thymine on the template pairs with adenine. Guanine pairs with cytosine.

RNA polymerase joins RNA nucleotides Phosphodiester bonds form between adjacent RNA nucleotides, producing a single-stranded pre-mRNA molecule.

Pre-mRNA is processed (eukaryotes only) Introns (non-coding sequences) are removed by splicing. Exons (coding sequences) are joined together to produce mature mRNA, which then leaves the nucleus through a nuclear pore.

Template vs coding strand – the confusion that costs marks: The mRNA sequence is identical to the coding strand (with U replacing T), but it is complementary to the template strand. RNA polymerase reads the template strand. Example: if the coding strand is 5′-ATGCGA-3′, the template strand is 3′-TACGCT-5′, and the mRNA is 5′-AUGCGA-3′.

Board differences for pre-mRNA processing: AQA (3.4.1 + 3.4.2), OCR A (6.1.1b), Edexcel A (6.10), Edexcel B (7.2iii), WJEC (1.5k), and Eduqas (CC5k) all require intron/exon knowledge. OCR B explicitly excludes post-transcriptional modification – so OCR B students should not discuss splicing unless specifically asked.

Translation – mRNA to Polypeptide

Translation is the process by which the base sequence of mRNA is decoded to produce a specific sequence of amino acids, forming a polypeptide. It occurs at ribosomes in the cytoplasm (free ribosomes or on the rough endoplasmic reticulum).

The Steps Examiners Want to See

mRNA attaches to a ribosome The mRNA molecule binds to a ribosome. The ribosome reads the mRNA codons sequentially, starting at the start codon AUG (which codes for methionine).

tRNA brings the correct amino acid A transfer RNA (tRNA) molecule with the complementary anticodon binds to the first codon on the mRNA. Each tRNA carries a specific amino acid determined by its anticodon.

Anticodon pairs with codon by complementary base pairing The three bases of the anticodon on tRNA form hydrogen bonds with the complementary three bases of the codon on mRNA.

Ribosome moves along mRNA to the next codon A second tRNA brings the next amino acid. A peptide bond forms between adjacent amino acids (condensation reaction). The first tRNA is released and recycled.

Polypeptide chain grows This process repeats – the ribosome moves along the mRNA one codon at a time, and peptide bonds link successive amino acids into a growing polypeptide chain.

Stop codon terminates translation When the ribosome reaches a stop codon (UAA, UAG, or UGA), there is no corresponding tRNA. The polypeptide is released and folds into its functional 3D shape.

ATP is needed for translation. Energy from ATP is required for the attachment of amino acids to their specific tRNA molecules (a process called amino acid activation) and for the movement of the ribosome along the mRNA. This is a commonly missed marking point.

Common error: Students write “tRNA reads the codon.” This is imprecise and may not earn the mark. The correct statement is that the anticodon on the tRNA is complementary to the codon on the mRNA, and they bind by complementary base pairing. The ribosome facilitates this matching.

Properties of the Genetic Code

The genetic code has five key properties that you must be able to define and explain. These are tested regularly, often as short-answer questions worth 1–2 marks each.

Triplet Three consecutive bases (a codon) code for one amino acid. With four possible bases, this gives 4³ = 64 possible codons.

Degenerate (redundant) Most amino acids are coded for by more than one codon. For example, leucine is coded by six different codons. This is possible because 64 codons code for only 20 amino acids plus 3 stop signals. The third base position (wobble position) is often interchangeable without changing the amino acid.

Universal The same codons code for the same amino acids in almost all organisms – from bacteria to humans. This provides powerful evidence for a common evolutionary origin. Rare exceptions exist in mitochondrial DNA.

Non-overlapping Each base is part of only one codon. Codons are read sequentially without sharing bases with adjacent codons.

Non-punctuated (comma-free) There are no gaps, spaces, or punctuation marks between codons. They are read continuously from the start codon. This is why insertions and deletions cause frameshift mutations – they shift every subsequent codon out of alignment.

WJEC/Eduqas additional properties: WJEC and Eduqas teaching guidance adds two further properties: linear (read in one direction along mRNA) and unambiguous (each codon codes for only one specific amino acid, even though one amino acid may be coded by multiple codons). These are worth knowing for WJEC/Eduqas students.

Comparing mRNA, tRNA, and rRNA

Comparison questions between the types of RNA are common and require direct paired comparisons. Writing features in isolation scores zero – you must compare like with like.

Feature	mRNA	tRNA	rRNA
Function	Carries genetic code from DNA to ribosome	Carries specific amino acids to the ribosome	Structural and catalytic component of ribosomes
Structure	Linear, single-stranded	Cloverleaf shape with anticodon loop and amino acid binding site	Complex folded structure; combines with proteins
Size	Variable (depends on gene length)	Short (~75–90 nucleotides)	Large (forms ribosomal subunits)
Lifespan	Short-lived; degraded after translation	Reusable; recycled after each amino acid delivery	Long-lived; stable component of ribosomes
Made where	Nucleus (transcription)	Nucleus	Nucleolus

Gene Mutations – Types and Consequences

A gene mutation is a change in the base sequence of DNA. All UK A-Level boards require you to understand three types of point mutation and their effects on the polypeptide produced.

Types of Gene Mutation

Mutation type	What happens	Effect on reading frame	Likely severity
Substitution	One base is replaced by a different base	No frameshift – only one codon affected	Variable: silent, missense, or nonsense
Insertion	An extra base is added into the sequence	Frameshift – every downstream codon altered	Usually severe – most amino acids changed
Deletion	A base is removed from the sequence	Frameshift – every downstream codon altered	Usually severe – most amino acids changed

Why Substitutions Can Be Harmless

Because the genetic code is degenerate, a substitution may produce a different codon that still codes for the same amino acid. This is called a silent (synonymous) mutation. The primary structure of the protein is unchanged, so its tertiary structure and function are unaffected.

If the substitution changes the amino acid (missense mutation), the effect depends on where the change occurs. If the new amino acid has similar properties, the protein may still function. If the change disrupts key bonds in folding or occurs at the active site, the protein may lose function entirely.

A nonsense mutation occurs when a substitution creates a premature stop codon, resulting in a shortened, usually non-functional polypeptide.

The full answer chain examiners want (4 marks): (1) Only one base/codon is changed → (2) the genetic code is degenerate → (3) so the same amino acid may still be coded for → (4) primary structure is unchanged, therefore tertiary structure and protein function are unchanged. You need all four links.

Why Insertions and Deletions Are More Severe

Insertions and deletions cause a frameshift – because the genetic code is read in non-overlapping triplets with no punctuation, adding or removing a single base shifts every subsequent codon. The sentence analogy makes this clear:

Original: THE CAT ATE THE RAT
Substitution: THE CAT ATE THE BAT (one “word” changed)
Deletion of C: THE ATA TET HER AT (every “word” after the deletion is changed)

Sickle cell anaemia – the required example: A substitution in the β-globin gene changes codon GAG to GTG (mRNA: GAG→GUG), resulting in valine replacing glutamic acid at position 6. This single amino acid change causes haemoglobin S to polymerise under low oxygen conditions, distorting red blood cells into a sickle shape. This example is named in AQA (3.4.3), WJEC/Eduqas (Unit 4/Comp 2), and Edexcel B (1.4).

Exam Board Comparison – What Your Board Requires

This is the table no other revision site provides. Use it to check exactly what your board requires – do not waste time studying content your specification does not examine.

Subtopic	AQA	OCR A	OCR B	Edexcel A	Edexcel B	WJEC / Eduqas
DNA helicase named	✔	✔	✔	❌	✔	✔
DNA ligase named	❌	❌	❌	❌	✔	❌
Meselson-Stahl named	Implicit	❌	❌	✔	❌	✔
Pre-mRNA splicing	✔	✔	❌	✔	✔	✔
Genetic code properties	✔	✔	✔	✔	✔	✔
Gene mutations	✔	✔	Via examples	Via CF	✔	✔
Lac operon	❌	✔	❌	✔	❌	❌
Hox genes	❌	✔	❌	❌	❌	❌
Epigenetics	✔	Implicit	✔	✔	✔	✔
Post-translational modification	❌	✔	❌	❌	❌	✔

8 Common Mistakes from Examiner Reports

These are the errors I see repeatedly, both as an examiner and as a tutor. Every one of them costs marks.

#	The mistake	The correction
1	“DNA polymerase forms hydrogen bonds between bases”	DNA polymerase forms phosphodiester bonds between adjacent nucleotides in the backbone. Hydrogen bonds between bases form spontaneously.
2	“DNA helicase hydrolyses hydrogen bonds”	Helicase breaks hydrogen bonds (AQA explicitly rejects “hydrolyses” here – hydrolysis refers to covalent bond breaking with water).
3	Confusing replication, transcription, and translation	Replication = DNA→DNA. Transcription = DNA→mRNA. Translation = mRNA→polypeptide. Learn this as “ReTranTran.”
4	“mRNA is a copy of the whole DNA”	mRNA is a copy of one gene only – a tiny fraction of the genome.
5	Writing isolated features in comparison questions	Mark schemes require direct paired comparisons: e.g., “DNA has deoxyribose; mRNA has ribose.” Listing features separately scores zero.
6	“All mutations are harmful”	Many mutations are silent (due to code degeneracy) or neutral (amino acid change has no functional effect).
7	Failing to mention ATP in translation	ATP is required for amino acid activation (attachment of amino acid to tRNA) and ribosome movement. This is a commonly missed mark.
8	Saying mRNA leaves the nucleus “through the membrane”	mRNA exits the nucleus through nuclear pores – not “through the membrane.”

Struggling with DNA Replication or Protein Synthesis?

I have examined questions on this exact topic for WJEC/Eduqas and Edexcel. I know what earns marks and what does not. If you cannot confidently describe replication in the correct order, or if transcription and translation blur into one process in your mind, tutoring will fix that.

Tyrone John • CBiol MRSB • Former Examiner • 25+ Years Teaching

Book a Free Consultation

The Lac Operon – Gene Regulation in Prokaryotes

The lac operon is an example of inducible gene regulation in E. coli. It is required by OCR A (6.1.1b) and Edexcel A (3.12) only. If you are on AQA, Edexcel B, OCR B, or WJEC/Eduqas, this section is extension content.

Components of the Lac Operon

Structural genes (lacZ, lacY, lacA) Code for enzymes needed to metabolise lactose: β-galactosidase (lacZ) breaks lactose into glucose and galactose; lactose permease (lacY) transports lactose into the cell.

Operator A short DNA sequence adjacent to the structural genes. When the repressor protein is bound to the operator, RNA polymerase cannot transcribe the structural genes.

Promoter The binding site for RNA polymerase. Located upstream of the operator.

Regulatory gene (lacI) Located elsewhere on the chromosome. It continuously produces the repressor protein.

How the Lac Operon Works

Without lactose: The repressor protein (produced by lacI) binds to the operator. This physically blocks RNA polymerase from transcribing the structural genes. The enzymes are not produced – there is no point making lactose-digesting enzymes when there is no lactose present.

With lactose present: Lactose (or more precisely, allolactose – a metabolite of lactose) acts as an inducer. It binds to the repressor protein, causing a conformational change that prevents the repressor from binding to the operator. RNA polymerase can now transcribe the structural genes, and the lactose-metabolising enzymes are produced.

Expert note: At A-Level, stating “lactose” as the inducer is perfectly acceptable and will earn full marks. Technically, β-galactosidase converts lactose to allolactose, and it is allolactose that binds the repressor. This distinction is not required at A-Level but demonstrates the depth of understanding a Chartered Biologist brings to tutoring.

Epigenetics – Controlling Gene Expression Without Changing DNA Sequence

Epigenetics is required by most boards: AQA (3.8.2.2), OCR B (5.1.2), Edexcel A (3.14), Edexcel B (7.2iv), and WJEC/Eduqas. OCR A covers gene regulation through the lac operon rather than explicit epigenetics.

DNA Methylation

Methyl groups (CH₃) are added to cytosine bases in DNA, typically at CpG sites. Methylation of a promoter region prevents transcription factors from binding, so the gene is silenced (switched off). Methylation patterns can be passed to daughter cells during replication and, in some cases, inherited across generations.

Histone Modification

DNA wraps around histone proteins to form nucleosomes. Acetylation of histones reduces the positive charge on histone tails, weakening the attraction between histones and the negatively charged DNA. This causes chromatin to loosen (euchromatin), making the DNA more accessible to RNA polymerase and increasing transcription. Deacetylation has the opposite effect – chromatin condenses (heterochromatin) and transcription is reduced.

AQA additional content: AQA uniquely requires RNA interference (RNAi) as an epigenetic mechanism (3.8.2.2). Small interfering RNA (siRNA) molecules bind to complementary mRNA sequences, leading to mRNA degradation and preventing translation. AQA students must know all three mechanisms: methylation, histone acetylation, and RNAi.

Frequently Asked Questions – DNA Replication & Protein Synthesis

DNA helicase unwinds the DNA double helix by breaking the hydrogen bonds between complementary base pairs. This separates the two strands at the replication fork, exposing the bases so they can act as templates for the synthesis of new complementary strands. Helicase is required by all UK A-Level boards except Edexcel A (Salters-Nuffield), which only names DNA polymerase in its specification.

Transcription is the process of copying the genetic information from a gene in DNA into messenger RNA (mRNA). It occurs in the nucleus and is catalysed by RNA polymerase. Translation is the process of decoding the mRNA base sequence to assemble a polypeptide from amino acids. It occurs at ribosomes in the cytoplasm and involves transfer RNA (tRNA) molecules carrying specific amino acids.

DNA replication is semi-conservative because each new DNA molecule produced contains one original (parent) strand and one newly synthesised strand. The term “semi” means half – half of the original molecule is conserved in each daughter molecule. This was demonstrated experimentally by Meselson and Stahl in 1958 using density gradient centrifugation with heavy nitrogen isotopes.

A frameshift mutation occurs when a base is inserted into or deleted from the DNA sequence. Because the genetic code is read in non-overlapping triplets with no punctuation, adding or removing a single base shifts every subsequent codon out of its correct reading frame. This means that almost every amino acid downstream of the mutation will be incorrect, and a premature stop codon is often introduced. The result is usually a non-functional protein, making frameshifts far more damaging than substitutions.

A degenerate (or redundant) genetic code means that most amino acids are coded for by more than one codon. There are 64 possible codons (4³) but only 20 amino acids plus 3 stop signals. For example, leucine has six different codons. This redundancy means that some base substitution mutations are “silent” – they change the codon but not the amino acid, so the protein is unaffected. The third base position of a codon is often interchangeable.

Transfer RNA (tRNA) carries specific amino acids to the ribosome during translation. Each tRNA molecule has an anticodon – a sequence of three bases that is complementary to a specific codon on the mRNA. When the anticodon binds to the codon by complementary base pairing, the tRNA delivers its amino acid to the growing polypeptide chain. Each tRNA is specific to one amino acid, and the attachment of the amino acid to its tRNA requires energy from ATP (amino acid activation).

Introns are non-coding sequences within a gene that are transcribed into pre-mRNA but are then removed by splicing before the mRNA leaves the nucleus. Exons are the coding sequences that remain after splicing and are joined together to form the mature mRNA. This process allows alternative splicing – different combinations of exons can be joined together, enabling a single gene to code for multiple different polypeptides. This is required by most UK boards (AQA, OCR A, Edexcel A, Edexcel B, WJEC/Eduqas) but is explicitly excluded by OCR B.

A 6-mark extended response on DNA replication is marked using levels of response. For Level 3 (5–6 marks), your answer must be logically structured, use precise terminology, and include all key steps: DNA helicase breaks hydrogen bonds between complementary base pairs, separating the strands; each strand acts as a template; free nucleotides align by complementary base pairing (A–T, G–C); DNA polymerase catalyses the formation of phosphodiester bonds between adjacent nucleotides; two identical DNA molecules are produced, each containing one original and one new strand (semi-conservative). Mention that replication occurs during the S phase of interphase and is necessary so each daughter cell receives a complete genome.

Written by Tyrone John

CBiol MRSB • Former WJEC/Eduqas & Edexcel Examiner • BSc Immunology (King’s College London) • Research Degree in Molecular Pharmacology (Newcastle University) • PGCE (University of Wales)

Tyrone has over 25 years of A-Level Biology teaching experience, including 18 years at Gower College Swansea. As a former examiner, he has first-hand knowledge of how mark schemes are applied and what examiners look for in student answers. Learn more →

Biology Education is not affiliated with AQA, OCR, Edexcel (Pearson), WJEC, Eduqas, CIE, or the IB. Exam board names and specification references are used for identification purposes only. All content is original and written by Tyrone John, CBiol MRSB.