Biology • Year 12 • Module 5 • Lesson 6

DNA Structure and DNA Replication

Build HSC band 5–6 extended-response technique on DNA structure, complementary base pairing and semiconservative replication — using two pieces of real foundational research as your evidence.

Master · Extended Response

1. Stimulus-based extended response — Meselson and Stahl (1958) (Band 5–6)

8 marks   Band 5–6

Stimulus. Meselson and Stahl (1958) grew Escherichia coli for many generations in medium containing only the heavy nitrogen isotope 15N. Once every DNA molecule contained only 15N, the bacteria were transferred to medium containing only the lighter 14N isotope. DNA was extracted at intervals and separated by density into three possible bands: heavy (15N/15N), hybrid (15N/14N) and light (14N/14N). After exactly one round of replication in 14N medium, 100% of the DNA was hybrid. After two rounds, the DNA was approximately 50% hybrid and 50% light. After three rounds, the DNA was approximately 25% hybrid and 75% light. At the time, three competing models of replication were under serious consideration: conservative (the original molecule is preserved intact and a wholly new molecule is made alongside it), semiconservative (each daughter molecule contains one old strand and one new strand) and dispersive (pieces of old and new DNA are interspersed within both strands of each daughter molecule).

Q1. Analyse and evaluate, using the data above and the Watson–Crick model of DNA structure, how the Meselson–Stahl experiment supports the semiconservative model of DNA replication. In your response you must:

  • Outline the structure of DNA that makes copying by complementary base pairing possible (Watson and Crick model).
  • State the predictions each of the three competing models (conservative, semiconservative, dispersive) makes for generations 1 and 2.
  • Use the actual data (100% hybrid at gen 1; 50% hybrid + 50% light at gen 2) to rule out the two incorrect models.
  • Explicitly define semiconservative in your own words and link it to "one original strand + one new strand".
  • Reach an evidence-based overall judgement about which model best explains the data.
Stuck? Plan first: structure → predictions of all three models → use the data to eliminate two → define semiconservative → judgement. The "Big Idea" callout from Card 2 is your spine: structure explains the mechanism.

2. Stimulus-based extended response — Chargaff (1950) (Band 5–6)

7 marks   Band 5–6

Stimulus. In a series of papers published in the late 1940s and early 1950s, Erwin Chargaff measured the proportions of the four DNA bases in samples from many different species. Selected results:

Organism%A%T%C%G
Human30.430.119.619.9
Cow29.028.721.221.1
E. coli24.723.626.025.7
Yeast (S. cerevisiae)31.332.917.118.7

Adapted from Chargaff (1950), Experientia 6: 201–209. The %A:%T ratio is ≈ 1 and the %C:%G ratio is ≈ 1 in every organism measured, although the (%A+%T):(%C+%G) ratio differs between species.

Q2. Justify the claim that Chargaff's base-composition data is best explained by the Watson and Crick double-helix model with complementary base pairing. In your response you must:

  • State Chargaff's rule (%A = %T and %C = %G) using a supporting figure from the table.
  • Outline the Watson and Crick double-helix model, including nucleotide structure and which bases pair with which.
  • Explain mechanistically why a double-helix structure with complementary base pairing forces %A = %T and %C = %G across the whole molecule.
  • Comment on the fact that the (%A+%T):(%C+%G) ratio differs between species, and explain why this is consistent with the model rather than evidence against it.
  • Reach a justified judgement linking DNA structure to the biological function of storing and accurately copying hereditary information.
Stuck? Use the pair-grid from Card 1 (A↔T, C↔G) to motivate the rule, and the "Big Idea" callout from Card 2 to land the function point at the end.
Answers — Do not peek before attempting

Q1 — Explicit marking criteria (8 marks)

  • 1 mark — Outlines the Watson and Crick model: DNA is a double helix of two strands made of nucleotides (sugar + phosphate + base), held together by hydrogen bonds between complementary base pairs (A–T, C–G).
  • 1 mark — States the conservative prediction at generation 1: 50% heavy + 50% light (no hybrid band).
  • 1 mark — States the semiconservative prediction at generation 1 (100% hybrid) and at generation 2 (50% hybrid + 50% light).
  • 1 mark — Notes the dispersive prediction at generation 1 also predicts a single intermediate-density band, but at generation 2 dispersive predicts a band of even lighter intermediate density rather than two distinct bands (50% hybrid + 50% light).
  • 1 mark — Uses the gen-1 data (100% hybrid, 0% heavy, 0% light) to rule out the conservative model explicitly.
  • 1 mark — Uses the gen-2 data (50% hybrid + 50% light) to rule out the dispersive model explicitly.
  • 1 mark — Defines semiconservative in their own words and links it explicitly to "one original (template) strand + one newly synthesised strand" in every daughter molecule.
  • 1 mark — Reaches an explicit overall judgement: the data is consistent with only one of the three models — semiconservative replication — and connects this back to the Watson–Crick claim that complementary base pairing makes accurate copying possible.

Q1 — Sample Band 6 response (8 marks), annotated

The Watson and Crick model describes DNA as a double helix of two strands built from nucleotides — each nucleotide containing a sugar, a phosphate group and a nitrogenous base — held together by hydrogen bonds between complementary base pairs A–T and C–G. [1 — structure]

Three replication models were on the table in 1958. The conservative model predicts that the original "heavy/heavy" molecule is preserved intact and an entirely new "light/light" molecule is synthesised alongside it, so generation 1 should show 50% heavy and 50% light DNA — and no hybrid band. [1 — conservative prediction] The semiconservative model predicts that each parent strand is kept and templates one new strand, so generation 1 should show 100% hybrid DNA (one 15N strand + one 14N strand in every molecule), and generation 2 should show 50% hybrid + 50% light (because the hybrid molecules separate and template new 14N strands). [1 — semiconservative prediction at gen 1 and gen 2] The dispersive model predicts that old and new DNA are interspersed throughout both strands of every daughter molecule, so at generation 1 every molecule would be a single intermediate-density band (like semiconservative), but at generation 2 the molecules would have an even lighter average density forming a single intermediate band, not two distinct bands. [1 — dispersive prediction]

The observed result at generation 1 (100% hybrid, 0% heavy, 0% light) rules out the conservative model entirely — if conservative replication were occurring, half the DNA at generation 1 would still be heavy. [1 — eliminates conservative using data] The observed result at generation 2 (50% hybrid + 50% light, in two distinct density bands) rules out the dispersive model, because dispersive replication predicts a single band of progressively lighter average density rather than two clearly separated bands. [1 — eliminates dispersive using data]

The only remaining model is semiconservative replication, which means that during replication each old strand acts as a template for one new strand, so every daughter DNA molecule contains exactly one original strand and one newly synthesised strand. [1 — defines semiconservative in own words]

Therefore the Meselson–Stahl data is consistent with only the semiconservative model and directly supports the central claim of the Watson–Crick model: that the double helix is organised in a way that allows accurate copying because each existing strand serves as a complementary template for a new one. [1 — overall evidence-based judgement linked back to structure]

Q2 — Explicit marking criteria (7 marks)

  • 1 mark — States Chargaff's rule: within any DNA sample %A ≈ %T and %C ≈ %G, supported by at least one figure from the table (e.g. human A 30.4 / T 30.1; C 19.6 / G 19.9).
  • 1 mark — Outlines the Watson and Crick model: DNA is a double helix of two strands of nucleotides (sugar + phosphate + base).
  • 1 mark — Identifies the specific complementary pairing rules (A with T, C with G) and that paired bases are held together by hydrogen bonds.
  • 1 mark — Explains mechanistically why complementary pairing forces %A = %T (every A on one strand requires a T opposite on the other strand) and %C = %G across the whole molecule.
  • 1 mark — Comments correctly on the (%A+%T):(%C+%G) ratio differing between species: this reflects different sequences in different genomes; complementary pairing only forces A=T and C=G, not equality of all four bases.
  • 1 mark — Explains that species differences in base ratio are therefore consistent with the Watson–Crick model rather than evidence against it.
  • 1 mark — Reaches an explicit judgement linking DNA structure (double helix, complementary pairing) to function (accurate storage and copying of hereditary information).

Q2 — Sample Band 6 response (7 marks), annotated

Chargaff's data shows that within every DNA sample the percentage of adenine is approximately equal to the percentage of thymine, and the percentage of cytosine is approximately equal to the percentage of guanine — for example in human DNA A = 30.4% and T = 30.1%, while C = 19.6% and G = 19.9%. This relationship is now called Chargaff's rule (%A = %T and %C = %G). [1 — rule + supporting figure]

The Watson and Crick model describes DNA as a double helix made of two strands. Each strand is built from nucleotides — a sugar, a phosphate group and a nitrogenous base — with the bases projecting inwards from a sugar–phosphate backbone. [1 — structure] The two strands are held together by hydrogen bonds between complementary base pairs: A on one strand always pairs with T on the other, and C on one strand always pairs with G on the other. [1 — pairing rules]

This specific pairing is the mechanistic reason for Chargaff's rule. In a double-stranded molecule, every A on one strand has exactly one T opposite it on the other strand, so the total count of A bases must equal the total count of T bases. The same applies to C and G. When the bases on both strands are counted together, %A is forced to equal %T and %C is forced to equal %G across the whole molecule. [1 — mechanism for %A = %T and %C = %G]

The (%A+%T):(%C+%G) ratio is not forced to be the same in every species, because that ratio depends on the underlying sequence of each species' genome — some genomes simply contain a higher proportion of A–T-rich regions than others. The complementary pairing rules only constrain A to T and C to G, not all four bases to one another. [1 — interprets between-species variation] So differences in the (%A+%T):(%C+%G) ratio between species are consistent with the Watson–Crick model rather than contradictory to it — the model predicts equality of pairs, not equality of all bases. [1 — consistency, not contradiction]

Chargaff's data is therefore best explained by a double-stranded structure with complementary base pairing, and this same structural feature is what allows DNA to fulfil its biological function: each strand can serve as a template for the other, so hereditary information is both stored stably and copied accurately during replication. [1 — structure–function judgement]

HSCScience • Biology • Year 12 • Module 5 • Lesson 6 • Worksheet 3 (Master)