DNA Structure and Replication

Deoxyribonucleic acid (DNA) is the universal molecule of inheritance. Every aspect of its function — coding, copying, mutation, repair, recombination — flows from its molecular architecture. This lesson examines that architecture in detail, traces the experimental work that established it, and dissects the mechanism of semi-conservative replication that allows the genetic information to be transmitted with extraordinary fidelity from one cell generation to the next.

Spec mapping: This lesson sits in AQA 7402 Section 3.4.1 — DNA, genes and chromosomes, and connects forwards to Section 3.4.2 (DNA and protein synthesis) and 3.4.3 (mutation). The relevant content covers nucleotide structure, the antiparallel double helix, complementary base pairing, the semi-conservative mechanism, and the enzymes catalysing replication. (Refer to the official AQA specification document for exact wording.)

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The Nucleotide — The Monomer of Nucleic Acids

A DNA nucleotide is a tripartite molecule formed by condensation of three components linked together by covalent bonds. Its structure dictates both the sequence-coding capacity of DNA and the directional nature of the polymer:

• A phosphate group (PO₄³⁻) — provides the acidic backbone and the linkages between successive nucleotides. The phosphate is negatively charged at physiological pH, making DNA an anion that migrates to the anode in electrophoresis and associates electrostatically with the positively-charged histone proteins of chromatin.
• A deoxyribose sugar — a five-carbon (pentose) sugar. The "deoxy" prefix reflects the absence of a hydroxyl group on carbon 2′; this lost –OH is what distinguishes deoxyribose from ribose (RNA). The chemical consequence is that DNA is more stable than RNA — the 2′ hydroxyl in RNA participates in autocatalytic cleavage of the backbone, which is why ancient DNA is recoverable from fossils but ancient RNA is not.
• A nitrogenous base — one of four heterocyclic aromatic bases: adenine (A), thymine (T), cytosine (C) or guanine (G). The bases are weakly basic owing to nitrogen atoms in the ring system, giving them their name.

The carbon atoms of the sugar are numbered 1′ through 5′ ("prime") to distinguish them from carbons in the base. The base is attached at the 1′ carbon by an N-glycosidic bond; the phosphate is attached at the 5′ carbon by a phosphoester bond. The free 3′-OH on one nucleotide and the 5′-phosphate on the next are what allow nucleotides to be chained together.

Type	Bases	Ring structure
Purines	Adenine (A), Guanine (G)	Fused double ring (six-membered + five-membered)
Pyrimidines	Thymine (T), Cytosine (C)	Single six-membered ring

Key Definition: A nucleotide is the monomer of nucleic acids, consisting of a phosphate group, a pentose sugar and a nitrogenous base linked by condensation reactions. A nucleoside is the same structure without the phosphate.

The choice of bases is not arbitrary. Two purines paired together would make the helix too wide; two pyrimidines too narrow. The "one purine + one pyrimidine" rule gives the helix its constant 2 nm diameter — a structural prerequisite for the proofreading machinery of DNA polymerase, which detects width irregularities as errors.

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The Watson-Crick Model — Structure and Evidence

In April 1953, James Watson and Francis Crick published a one-page paper in Nature proposing the double-helical structure of DNA. Their model rested on three streams of evidence, none of which they themselves generated:

• X-ray diffraction images taken by Rosalind Franklin and Maurice Wilkins at King's College London. Franklin's famous Photograph 51 showed the characteristic cross pattern of a helical structure and yielded the helix pitch (3.4 nm) and diameter (2 nm).
• Chargaff's base-composition data demonstrating that the molar ratio of A to T, and of G to C, was always 1:1 in any species.
• Astbury and Furberg's earlier crystallographic work on nucleotide structure, which established that the bases lay perpendicular to the sugar-phosphate backbone.

The model assembled these fragments into a single structure:

• DNA is a double helix — two antiparallel polynucleotide strands wound around a common axis in a right-handed twist.
• The two strands run in opposite directions: one 5′→3′, the other 3′→5′. The 5′ end terminates in a free phosphate group; the 3′ end terminates in a free –OH on the deoxyribose.
• The sugar-phosphate backbone lies on the outside; the bases project inwards into the hydrophobic core of the helix.
• Complementary base pairing holds the two strands together: adenine pairs with thymine via two hydrogen bonds (A=T), and guanine pairs with cytosine via three hydrogen bonds (G≡C).
• One complete turn of the helix spans 3.4 nm and contains 10 base pairs; each base pair is therefore separated vertically by 0.34 nm.
• The helix has a major groove (about 2.2 nm wide) and a minor groove (1.2 nm wide). Regulatory proteins typically read the base sequence from the major groove, where the chemical signatures of bases are most accessible without strand separation.

flowchart TD
  A["Sugar-phosphate backbone (5'→3')"] -->|"phosphodiester bonds"| B["Base 1 (A)"]
  B -->|"2 H-bonds"| C["Base T on opposite strand"]
  C --> D["Sugar-phosphate backbone (3'→5')"]
  A2["Backbone continues"] -->|"phosphodiester bonds"| B2["Base G"]
  B2 -->|"3 H-bonds"| C2["Base C on opposite strand"]
  C2 --> D2["Antiparallel strand"]

Exam Tip: Questions often ask why the double helix has a constant diameter. The answer: a purine (two rings, ~1.2 nm) always pairs with a pyrimidine (one ring, ~0.8 nm), giving a consistent three-ring span (~2.0 nm) across the helix. Two purines would bulge; two pyrimidines would pinch.

Acknowledging Franklin's contribution

Modern teaching is careful to credit Franklin's role. The X-ray diffraction data she generated were shown to Watson without her permission shortly before the 1953 paper, and the model could not have been constructed without them. She died of ovarian cancer in 1958, four years before the Nobel Prize was awarded to Watson, Crick and Wilkins; Nobel rules prohibit posthumous awards. The story is now a standard ethics case study in the history of molecular biology and a reminder that the construction of scientific knowledge is rarely the work of single individuals.

The Avery-MacLeod-McCarty and Hershey-Chase experiments

The Watson-Crick model answered the structural question, but the question of which molecule carried the genetic information was settled earlier. Avery, MacLeod and McCarty (1944) showed that the substance transforming non-virulent into virulent Streptococcus pneumoniae was DNA, not protein — they purified the transforming principle and demonstrated that DNase abolished its activity but protease did not. Hershey and Chase (1952) used differentially radio-labelled bacteriophage to show that only the DNA (³²P-labelled), not the protein coat (³⁵S-labelled), entered the bacterial cell to direct new phage synthesis. By 1953 the question was therefore: not "is DNA the genetic material?" but "how can a single molecule store, transmit and express genetic information?". The double-helix model answered all three.

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Chargaff's Rules — The Chemical Foundation

Erwin Chargaff's contribution preceded the Watson-Crick paper by several years. By hydrolysing DNA from a wide variety of organisms and quantifying the bases, he established two rules that ruled out the prevailing "tetranucleotide hypothesis" (which held that DNA was a repetitive A-T-G-C polymer too uniform to carry genetic information):

1. In any DNA sample, the amount of adenine equals the amount of thymine, and the amount of guanine equals the amount of cytosine. (A = T and G = C.)
2. The ratio of (A+T) to (G+C) varies between species but is essentially constant within a species, regardless of which tissue the DNA is extracted from.

Rule 1 was the direct chemical evidence for complementary base pairing. Rule 2 was the first hint that the sequence of bases — not their proportion — carries genetic information.

Worked example — applying Chargaff's rules

A sample of DNA is found to contain 22% cytosine. Calculate the percentage of each other base.

• If C = 22%, then G = 22% (Chargaff: G = C).
• A + T + G + C = 100%, so A + T = 100% − (22% + 22%) = 56%.
• Chargaff: A = T, therefore A = 28% and T = 28%.

Worked example — base ratio from sequence length

A linear DNA molecule contains 1,000,000 base pairs and the (A+T):(G+C) ratio is 3:2. How many of each base are present?

• Total bases = 2,000,000 (each base pair contributes two bases).
• A+T fraction = 3/5 of total → 1,200,000 bases of A+T.
• G+C fraction = 2/5 of total → 800,000 bases of G+C.
• By Chargaff: A = T = 600,000 and G = C = 400,000.

Such species would be classed as "AT-rich" — a property often associated with thermolabile DNA (fewer triple-hydrogen-bond G≡C pairs to stabilise the helix), seen in some thermophilic archaea where the genome shows the opposite GC bias as a thermal adaptation. GC content is a useful diagnostic tool in microbial taxonomy and can be calculated from the melting temperature (T_m) of DNA — the temperature at which half the helix has denatured into single strands. T_m increases linearly with GC content because of the extra hydrogen bond per G≡C pair.

Worked example — calculating numbers of hydrogen bonds

A double-stranded DNA fragment of 1000 base pairs has 40% GC content. How many hydrogen bonds hold the two strands together?

• G≡C pairs: 0.40 × 1000 = 400 pairs, each with 3 H-bonds → 1200 H-bonds.
• A=T pairs: 0.60 × 1000 = 600 pairs, each with 2 H-bonds → 1200 H-bonds.
• Total H-bonds: 2400.

In a 3.2 × 10⁹ bp human genome with ~41% GC content, the total number of H-bonds across all chromosomes runs into the trillions — illustrating both the magnitude of the stability and the precision with which the helix can still be locally unwound for replication and transcription.

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The Sugar-Phosphate Backbone and Directionality

Nucleotides are joined by phosphodiester bonds formed during condensation reactions. Each bond links the 3′-OH of one sugar to the 5′-phosphate of the next, with the elimination of water. The resulting backbone has the following properties:

• It is covalent and therefore strong — the helix is held together longitudinally by phosphodiester bonds and laterally by the much weaker hydrogen bonds between bases.
• It is negatively charged at physiological pH (one negative charge per phosphate), which makes DNA migrate to the anode in gel electrophoresis and which dictates its association with positively charged histone proteins in chromatin.
• It is directional. By convention, sequences are written 5′→3′. Enzymes that synthesise DNA (DNA polymerase) and RNA (RNA polymerase) extend the chain in the 5′→3′ direction only.

The antiparallel arrangement of the two strands is not a stylistic choice. It is dictated by the geometry of complementary base pairing: A pairs with T (and G with C) only when the two backbones run in opposite directions. This geometric constraint is what creates the asymmetric replication problem dealt with by leading and lagging strand synthesis.

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Semi-Conservative Replication — The Mechanism

DNA replication occurs during the S phase of interphase, before mitosis or meiosis. It is described as semi-conservative because each daughter DNA molecule contains one strand from the original (parental) molecule and one newly synthesised strand. The mechanism proceeds in a coordinated sequence:

1. Initiation at the origin of replication. In eukaryotes, replication starts simultaneously at many origins along each chromosome to keep total replication time manageable (~8 hours in a human cell). Initiator proteins load the replication machinery.
2. Helicase unwinds the helix. DNA helicase moves along the backbone, breaking the hydrogen bonds between complementary base pairs, generating a replication fork at which the two single-stranded templates diverge.
3. Single-strand binding proteins (SSBs) coat the separated strands to prevent them from snapping back together.
4. Primase synthesises an RNA primer. DNA polymerase cannot start a strand from nothing; it requires a free 3′-OH to extend. Primase lays down a short (~10-nucleotide) RNA primer to provide that 3′-OH.
5. DNA polymerase extends the primer. Free DNA nucleotides (as deoxynucleoside triphosphates, dNTPs) are added one at a time. The polymerase tests each incoming nucleotide for complementarity with the template; correct pairs are stabilised by the right number of hydrogen bonds. The triphosphate is cleaved (releasing pyrophosphate), driving the formation of a new phosphodiester bond.
6. Leading vs lagging strand. Because polymerase only synthesises 5′→3′, only one of the two new strands can be made continuously towards the replication fork — the leading strand. The other strand (the lagging strand) must be synthesised in short fragments (Okazaki fragments, ~150–200 nucleotides in eukaryotes) moving away from the fork. Each Okazaki fragment requires its own RNA primer.
7. Primer removal and ligation. RNA primers are excised and replaced by DNA (by a nuclease + polymerase activity). DNA ligase seals the nicks between adjacent Okazaki fragments by forming the missing phosphodiester bonds.
8. Proofreading and repair. DNA polymerase has a 3′→5′ exonuclease activity that detects and removes mispaired nucleotides before continuing synthesis. Post-replication, mismatch repair systems scan the new strand for any errors that escaped proofreading. The net replication error rate in human cells is ~1 mistake per 10⁹ nucleotides.

Key enzymes in replication

Enzyme	Role
Helicase	Unwinds the double helix; breaks hydrogen bonds between complementary bases at the fork
Single-strand binding proteins	Stabilise exposed single strands and prevent reannealing
Topoisomerase	Relieves the torsional supercoiling that builds up ahead of the replication fork
Primase	Synthesises a short RNA primer to provide a starting 3′-OH for DNA polymerase
DNA polymerase	Adds complementary deoxynucleotides in the 5′→3′ direction; proofreads via 3′→5′ exonuclease activity
DNA ligase	Joins Okazaki fragments on the lagging strand by forming phosphodiester bonds

Key Definition: Semi-conservative replication is the mechanism of DNA replication in which each new molecule consists of one strand from the original parental molecule and one newly synthesised strand.

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The Meselson-Stahl Experiment (1958) — The Definitive Evidence

Matthew Meselson and Franklin Stahl, working at Caltech, provided the experimental confirmation of semi-conservative replication. At the time, three models were viable: semi-conservative (each daughter has one old + one new strand), conservative (one daughter is entirely old, one entirely new), and dispersive (both daughters are patchwork mixtures of old and new fragments).

Method

1. Escherichia coli were cultured for many generations in a medium where the sole nitrogen source was ¹⁵NH₄Cl. Over time, every base in every DNA molecule contained the heavy ¹⁵N isotope.
2. Bacteria were rapidly transferred to a medium containing only the normal ¹⁴N isotope and allowed to undergo a controlled number of replication rounds.
3. DNA was extracted at successive generations and centrifuged in a caesium chloride (CsCl) density gradient. DNA migrates to the position where its density matches the local CsCl density, forming a sharp band.

Results and interpretation

• Generation 0 (before transfer): A single heavy band (¹⁵N/¹⁵N).
• Generation 1 (after one round in ¹⁴N): A single band of intermediate density (one ¹⁵N strand + one ¹⁴N strand per molecule). This ruled out the conservative model, which predicted two bands (one heavy, one light) at generation 1.
• Generation 2: Two bands in 1:1 ratio — intermediate density (50% ¹⁵N/¹⁴N hybrids) and light (50% ¹⁴N/¹⁴N). This ruled out dispersive replication, which predicted a continued shift towards lighter (but not yet pure ¹⁴N) density.
• Predicted generation 3: 25% intermediate, 75% light. The intermediate fraction halves each generation; the light fraction approaches but never reaches 100% (the original ¹⁵N strands persist indefinitely, just diluted across more daughter molecules).

Exam Tip: Be prepared to predict banding patterns for further generations. By generation n, the proportion of intermediate DNA is 1/2^(n−1) and the rest is light. This is a frequent calculation step in higher-mark questions.

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Synoptic links

This material cross-cuts strongly with three other parts of the AQA 7402 specification:

1. Section 3.1.5 — Nucleic acids: the chemistry of nucleotides is built here, and the chemical distinction between RNA (ribose, uracil) and DNA (deoxyribose, thymine) underpins the transcription/translation lessons later in this course.
2. Section 3.2.2 — Mitosis and the cell cycle: S phase is when replication occurs. Defects in replication generate mitosis-derived mutations and underpin the link between DNA damage and cancer.
3. Section 3.5.2 — Respiration: mitochondrial DNA replicates by its own machinery, independent of the nuclear cell cycle. Inherited mitochondrial diseases (Leber's hereditary optic neuropathy, MELAS) reflect mtDNA mutations.

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Specimen question modelled on the AQA paper format

Question (6 marks): Describe the structure of a DNA molecule and explain how this structure allows DNA to perform its function as the molecule of inheritance.

Mark scheme decomposition by AO:

Mark	AO	Awarded for
1	AO1	Naming the nucleotide as phosphate + deoxyribose + base, joined by phosphodiester bonds
2	AO1	Stating antiparallel double helix with complementary base pairing (A-T, G-C) and hydrogen bonds
3	AO2	Linking base sequence to genetic coding
4	AO2	Linking complementarity to accurate replication
5	AO2	Linking weak H-bonds + strong covalent backbone to "strand separation without unzipping the chain"
6	AO3	Synthesis — e.g. "the antiparallel, complementary architecture is simultaneously the storage and the copying mechanism"

Grade C model answer (~165 words)

DNA is made of nucleotides. Each nucleotide has a phosphate, a deoxyribose sugar and a base. The bases are A, T, C and G. A always pairs with T and G always pairs with C. Two strands of DNA are joined together by hydrogen bonds between the bases. The two strands twist around each other to make a double helix. The bases code for amino acids in groups of three called triplets, so the order of bases is the genetic code. When DNA replicates, the two strands separate and each one acts as a template. New nucleotides line up by complementary base pairing, so each new DNA molecule has one old strand and one new strand. The hydrogen bonds are weak so the strands can be separated, but the phosphodiester bonds are strong so the DNA chain does not break. This means DNA can be copied accurately when cells divide.

Examiner commentary: Awarded approximately 4/6 (M1, M1, M1, M1). The candidate names the nucleotide components, identifies the antiparallel double helix in essence (though the word "antiparallel" is missing — a discriminator), gives the base-pairing rules, and links base sequence to genetic coding. Mark losses: no use of "antiparallel"; no explicit AO3 synthesis sentence; the weak/strong bond comparison is correct but underexplained. Many candidates lose marks here by treating the H-bonds as "loose" rather than as a specific design feature enabling controlled strand separation.

Grade A* model answer (~230 words)

DNA is a polymer of nucleotides. Each nucleotide consists of a deoxyribose sugar, a phosphate group and one of four nitrogenous bases — adenine, thymine, cytosine or guanine. Adjacent nucleotides are joined by phosphodiester bonds between the 3′-OH of one sugar and the 5′-phosphate of the next, generating a directional 5′→3′ backbone. Two such strands wind around each other in an antiparallel double helix, with bases projecting inwards and paired by hydrogen bonding — adenine to thymine via two H-bonds, guanine to cytosine via three. The geometry requires a purine to pair with a pyrimidine, which keeps the helix diameter constant at 2 nm and provides a substrate that DNA polymerase can proofread for width irregularities.

This architecture is simultaneously the storage and the copying mechanism. The sequence of bases is the genetic message; the complementarity guarantees that each strand carries enough information to specify its partner. During replication, helicase breaks the weak hydrogen bonds — without disrupting the strong covalent backbone — and each parental strand templates the synthesis of a new complementary strand. The result is semi-conservative replication, demonstrated experimentally by Meselson and Stahl in 1958. The elegance lies in the dual function: the same antiparallel-complementary design that stores the information is what copies it.

Examiner commentary: Full marks (6/6). The candidate names the nucleotide components, gives the antiparallel double helix with the correct H-bond counts, explains the purine-pyrimidine geometric constraint, links sequence to coding and complementarity to replication, references the weak/strong bond hierarchy explicitly, and closes with an AO3-targeted "dual function" synthesis. Examiner-rewarded phrasing throughout.

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Common errors and mark-loss patterns

• "The chromosome unzips" when the candidate means "hydrogen bonds between bases break". Many candidates lose marks here by personifying the helix; mark scheme requires the molecular event.
• Confusing "antiparallel" with "asymmetric". Antiparallel means the two strands run in opposite 5′→3′ directions. Asymmetric means the leading and lagging strands are synthesised differently. Use both terms precisely.
• Stating that DNA polymerase "joins" Okazaki fragments. It does not; DNA ligase seals the nicks. Polymerase only extends from a free 3′-OH.
• Confusing semi-conservative with conservative. A common phrasing error is "each new DNA has one parent and one daughter strand". The correct phrasing is "each new molecule has one parental strand and one newly synthesised strand".
• Counting hydrogen bonds wrongly. A-T has 2 H-bonds; G-C has 3. Reversing this is a frequent careless error and an easy mark to lose.

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A-Level-depth misconceptions

• Replication is not a single continuous event. Eukaryotic chromosomes have multiple origins; each opens a bidirectional fork; the forks meet and terminate. Picturing replication as "one fork that runs the whole length" is GCSE-level thinking.
• Helicase does not "break the chain" — it breaks the hydrogen bonds between base pairs. The phosphodiester backbone remains intact throughout replication.
• The 5′→3′ rule is a property of the enzyme, not the DNA. DNA itself is read in both directions during transcription (depending on which strand is the template). Replication is unidirectional only because DNA polymerase is.

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Going further

• Undergraduate reading: Alberts et al., Molecular Biology of the Cell (Chapter 5 — DNA replication, repair and recombination). Lewin's GENES XII (Chapter 11 — the replisome). For the historical context, read Horace Freeland Judson, The Eighth Day of Creation — still the definitive popular history of the molecular biology revolution.
• Interview prompt: "If DNA polymerase only synthesises 5′→3′, why doesn't evolution simply select for a 3′→5′ polymerase to remove the need for lagging-strand synthesis?" (The answer involves the chemical asymmetry of the dNTP substrate — the high-energy phosphoanhydride bond is on the incoming nucleotide's triphosphate, not on the growing chain's 3′ end. A 3′→5′ polymerase would need an entirely different reaction chemistry, where the energy comes from the chain end rather than the substrate. This would compromise proofreading: removing an incorrect nucleotide would simultaneously remove the energy source for continued synthesis.)
• Further interview prompt: "Telomeres shorten with each cell division because of the end-replication problem. What is the end-replication problem, and how do germ-line cells avoid the resulting catastrophe?" (Discussion: the lagging strand cannot be primed at the very 5′ end of a linear chromosome; telomeres of repeated TTAGGG sequence buffer this loss in somatic cells; telomerase, a reverse transcriptase, extends telomeres in germ-line and stem cells; the consequences of telomere shortening connect to ageing and cancer.)

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Summary

• A DNA nucleotide is built from a phosphate group, a deoxyribose sugar, and one of four nitrogenous bases (A, T, C, G).
• DNA is an antiparallel double helix held longitudinally by phosphodiester bonds and laterally by hydrogen bonds between complementary base pairs (A=T, G≡C).
• Chargaff's rules (A=T, G=C) provided the chemical foundation for the Watson-Crick base-pairing model; Franklin's X-ray data provided the geometric foundation.
• Replication is semi-conservative: each daughter molecule consists of one parental and one newly synthesised strand.
• Helicase, primase, DNA polymerase, DNA ligase and topoisomerase together produce the replisome; leading-strand synthesis is continuous and lagging-strand synthesis proceeds via Okazaki fragments.
• The Meselson-Stahl experiment (1958) confirmed semi-conservative replication using ¹⁵N/¹⁴N density-gradient centrifugation.

Exam Tip: When describing replication, always state the role of each enzyme, the direction of synthesis (5′→3′), and the leading/lagging distinction. High-mark questions often require you to explain why replication is discontinuous on the lagging strand — the answer is the antiparallel geometry combined with the directional polymerase.

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Spec alignment: AQA 7402 Section 3.4.1 — DNA, genes and chromosomes; linked to 3.1.5, 3.2.2, 3.5.2. Refer to the official AQA specification document for exact wording.