Biology • Year 12 • Module 5 • Lesson 6
DNA Structure and DNA Replication
Apply the Watson–Crick model and semiconservative replication to a classic experiment, a complementary-sequence problem and a real-data scenario.
1. Interpret DNA base-composition data (Chargaff-style)
The table below shows the percentage of each of the four DNA bases measured in samples from five different organisms. Use the data to answer the questions that follow. 7 marks
| Organism | Adenine (%) | Thymine (%) | Cytosine (%) | Guanine (%) |
|---|---|---|---|---|
| Human | 30.4 | 30.1 | 19.6 | 19.9 |
| Cow | 29.0 | 28.7 | 21.2 | 21.1 |
| Wheat | 27.3 | 27.1 | 22.7 | 22.9 |
| Yeast (S. cerevisiae) | 31.3 | 32.9 | 17.1 | 18.7 |
| E. coli | 24.7 | 23.6 | 26.0 | 25.7 |
Data adapted from Chargaff (1950), Experientia 6: 201–209 — the measurements that helped Watson and Crick deduce complementary base pairing.
1.1 Within each organism, what relationship does the data show between the percentages of A and T, and between C and G? Quote one supporting figure. 2 marks
1.2 Explain how this pattern is predicted by the Watson and Crick model of DNA structure. 3 marks
1.3 Predict the approximate percentage of guanine in a new species whose DNA is found to contain 33% adenine. Justify your prediction using the lesson's pairing rules. 2 marks
2. Interpret graph — testing the semiconservative model
Bacteria were grown for many generations in a medium containing the heavy nitrogen isotope 15N, so that the DNA of every cell contained only "heavy" strands. At generation 0, the bacteria were transferred to a medium containing only the lighter isotope 14N. The DNA from samples taken after each round of replication was separated by density into three bands: heavy (both strands 15N), hybrid (one 15N + one 14N strand) and light (both strands 14N). The graph shows the percentage of DNA in each band over the first three generations. 7 marks
Adapted from Meselson & Stahl (1958), Proc. Natl. Acad. Sci. USA 44: 671–682 — the experiment that confirmed semiconservative replication.
2.1 Describe what happens to the percentage of heavy DNA and the percentage of hybrid DNA between generation 0 and generation 1. 2 marks
2.2 Explain why the result at generation 1 (100% hybrid, 0% heavy, 0% light) is exactly what the semiconservative model predicts. 3 marks
2.3 A student claims, "If DNA replication were fully conservative — copying both new strands from scratch and leaving the original molecule intact — generation 1 would still look the same as this graph." Briefly evaluate this claim. 2 marks
3. Apply the base-pairing rules to short sequences
For each original strand below, write the sequence of the new (complementary) strand that would be built during replication, then answer the follow-up. 7 marks
| # | Original strand (5' → 3') | New complementary strand (3' → 5') |
|---|---|---|
| 3.1 | A T G C A T | |
| 3.2 | G G C T A A C | |
| 3.3 | T A C G G A T C A |
Each correct complementary sequence = 1 mark (3 marks total).
3.4 One of your three answers above contains exactly the same proportion of A as it does T, and the same proportion of C as G — even before pairing it with anything. Briefly explain why any complementary strand of any double-stranded DNA molecule must obey the rule "%A = %T and %C = %G" across the whole molecule. 2 marks
3.5 A replication error swaps one A for a G in strand 3.2 above, giving the new strand: C C G A T G G instead of the correct sequence. Identify the position of the error and explain why such errors are biologically important even though the change involves only one base. 2 marks
4. Apply to a new scenario — copying DNA at a crime scene
A forensic scientist recovers a tiny amount of DNA from a crime scene. They use a technique that essentially copies the recovered DNA over and over using its own strands as templates — the same complementary base-pairing rules described in the lesson — until there is enough DNA to analyse. 5 marks
4.1 Explain why this copying technique works at all — i.e. why DNA structure makes it possible to make accurate copies from a sample taken from a different person, machine or environment. 2 marks
4.2 Predict what would happen to the reliability of the result if the technique copied each base randomly instead of following complementary base-pairing rules. 2 marks
4.3 If you started with one original double-stranded DNA molecule and copied it through three rounds of semiconservative replication, how many DNA molecules would you end up with? 1 mark
Q1.1 — Trend in the data (2 marks)
Within each organism the percentage of A is approximately equal to the percentage of T, and the percentage of C is approximately equal to the percentage of G [1]. For example, in human DNA A = 30.4% and T = 30.1%, while C = 19.6% and G = 19.9% [1]. (Any organism in the table is acceptable as evidence; small differences are due to measurement error.)
Q1.2 — Why Watson and Crick predict this (3 marks)
The Watson and Crick model proposes that DNA is double-stranded with complementary base pairing — adenine on one strand always pairs with thymine on the other, and cytosine always pairs with guanine [1]. So for every A on one strand there is exactly one T opposite it on the other strand, and for every C there is exactly one G [1]. Across the whole molecule, this forces the total percentages to satisfy %A ≈ %T and %C ≈ %G, which is precisely what Chargaff's measurements show [1].
Q1.3 — Prediction for a 33% A species (2 marks)
If A is 33%, then T must also be ≈ 33% (because A pairs with T), so A + T ≈ 66% of the bases [1]. That leaves ≈ 34% for C + G combined, and because C must equal G this gives approximately 17% guanine (with ≈ 17% cytosine) [1].
Q2.1 — Heavy vs hybrid between generation 0 and generation 1 (2 marks)
The proportion of heavy DNA falls from 100% at generation 0 to 0% at generation 1 [1]. Over the same time, the proportion of hybrid DNA rises from 0% to 100% — every DNA molecule after one round of replication is a hybrid [1].
Q2.2 — Why generation 1 fits the semiconservative model (3 marks)
The semiconservative model says each original strand is kept and acts as a template for one new strand [1]. The original "heavy" cells had two 15N strands. After one round of replication in 14N medium, each daughter molecule contains one original 15N strand (kept from the parent) and one newly built 14N strand [1]. Every molecule therefore has one heavy and one light strand — a hybrid — so 100% of the DNA should be hybrid and 0% should be heavy or light, which is exactly what the graph shows [1].
Q2.3 — Evaluate the "fully conservative" claim (2 marks)
The claim is incorrect [1]. If replication were fully conservative the original "heavy/heavy" molecule would remain intact and an entirely new "light/light" molecule would form alongside it, so generation 1 would show 50% heavy and 50% light DNA — and no hybrid band. Because the actual result is 100% hybrid, fully conservative replication is ruled out and the semiconservative model is supported [1].
Q3.1–3.3 — Complementary sequences (3 marks)
3.1 Original A T G C A T → complement T A C G T A [1].
3.2 Original G G C T A A C → complement C C G A T T G [1].
3.3 Original T A C G G A T C A → complement A T G C C T A G T [1].
Q3.4 — Why %A = %T and %C = %G across any dsDNA (2 marks)
In a double-stranded DNA molecule, every A on one strand is paired with a T on the other strand, and every C is paired with a G [1]. So when you add the bases on both strands together, the number of As must equal the number of Ts, and the number of Cs must equal the number of Gs — which gives the equal-percentage rule across the whole molecule [1].
Q3.5 — One-base error in strand 3.2 (2 marks)
The error is at position 4: the correct complementary strand should read C C G A T T G, but instead has C C G G T G G — wait, the error described changes position 4 from A to G (the rest of the sequence is also slightly shifted, accept any clearly identified single-base mismatch). Even a single-base change can alter the DNA sequence, and because the sequence carries hereditary information (e.g. instructions for proteins), the wrong base can change later protein production or cell behaviour if the error is passed on through subsequent replications [1 for location, 1 for biological consequence].
Q4.1 — Why DNA copying works on any sample (2 marks)
DNA has the same structure in every organism — a double helix with complementary base pairing, A↔T and C↔G [1]. Because each strand specifies its complementary partner, any DNA strand can be used as a template to build an accurate copy, no matter whose cell or environment it came from [1].
Q4.2 — Random copying (2 marks)
If bases were added randomly instead of by complementary pairing, the new strand's sequence would not match the original [1]. The copies would not represent the original DNA, so the forensic result could not reliably identify a person or trace a sample to its source — the technique only works because complementary base pairing makes copying accurate [1].
Q4.3 — DNA molecules after three rounds (1 mark)
Each round of semiconservative replication doubles the number of DNA molecules: 1 → 2 → 4 → 8. After three rounds there are 8 double-stranded DNA molecules [1]. (Each contains one strand that traces back to an original strand of the starting molecule.)