AQA AS Level Biology Paper 1, June 2025: Question 3

8 marks · Medium difficulty · Extended Answer

Describe the semi-conservative replication of DNA and explain the results of the Meselson-Stahl density gradient centrifugation experiment.

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Question

Question 3 displays three parts. Question 03.1 asks to describe the semi-conservative replication of DNA for 5 marks. Following this is a passage detailing an experiment where bacteria are grown in heavy nitrogen (15N, sample 1), then transferred to light nitrogen (14N) and grown for one division (sample 2) and two divisions (sample 3). Figure 5 shows three centrifuge tubes: sample 1 shows a band near the bottom; sample 2 shows an intermediate band in the middle; and sample 3 is empty. Question 03.2 asks to explain the position of the DNA band in sample 2 for 2 marks. Question 03.3 asks to complete sample 3 in Figure 5 to show the expected bands after two divisions for 1 mark.
Question text

03.1 Describe the semi-conservative replication of DNA.

[5 marks]

Scientists designed an experiment to investigate DNA replication.

They grew bacteria for several generations in a nutrient solution containing a heavy

form of nitrogen (15N). They obtained DNA from a sample of these bacteria

(sample 1).

The scientists then transferred the bacteria to a nutrient solution containing a light

form of nitrogen (14N). They allowed the bacteria to grow and divide twice. After each

division, they obtained DNA from a sample of bacteria (sample 2 and sample 3).

They suspended DNA from each sample of bacteria in a solution in separate tubes.

They spun these tubes in a centrifuge at the same speed and for the same time.

Based on its density, DNA settled as bands in each tube.

Figure 5 shows the scientists’ results.

Figure 5

03.2 Explain the position of the DNA band in sample 2 in Figure 5.

[2 marks]

03.3 Complete sample 3 in Figure 5 to show the result you would expect after two

divisions.

[1 mark]

Mark scheme

Show the mark scheme Mark scheme table for parts 03.1 to 03.3. For 03.1 (max 5 marks): DNA helicase unzips/separates strands or breaks hydrogen bonds; both strands act as templates; complementary nucleotides pair up; DNA polymerase joins nucleotides together forming phosphodiester bonds; each new DNA consists of one original and one new strand. For 03.2 (2 marks): semi-conservative replication occurs where one strand is old and one is new; DNA contains one light 14N strand and one heavy 15N strand. For 03.3 (1 mark): one band level with sample 2 and one band higher up in the sample 3 tube.

Question Marking Guidance Mark Comments

Max 5 from:

1. DNA helicase unwinds DNA/double helix 1. Reject hydrolyses

hydrogen bonds

OR 1. Accept ‘unzip’ for

unwind

DNA helicase separates strands

OR

DNA helicase breaks hydrogen bonds;

5 max

03.1 (5 x

2. Both strands act as templates;

AO1)

3. (Free DNA) nucleotides line up in

complementary pairs/A-T and G-C; 2. Accept description

of ‘template’, eg

4. DNA polymerase joins nucleotides (of new exposed bases on

strand); single (polynucleotide)

strands

5. Forming phosphodiester bonds; 4. Reject forms

hydrogen bonds/joins

6. (Each new DNA molecule consists of) one bases

old/original/template strand and one new strand;

1. (Produced by) semi-conservative replication

OR

03.2 One old/original strand and one new strand; (2 x

14 15 AO2)

2. (Contains) N/light (strand) and N/heavy

(strand);

One band level with band in sample 2 and one Allow lines of any

band higher, drawn onto the sample 3 tube; thickness

Ignore lines drawn on

03.3 (1 x

the other tubes

AO2)

Both lines needed for

mark

How to answer it

Semi-Conservative DNA Replication & Density Centrifugation

📋 What This Question Tests
  • Knowledge of biological mechanism (AO1): The detailed enzymatic steps of semi-conservative DNA replication, including the roles of DNA helicase, complementary base pairing, and DNA polymerase forming phosphodiester bonds.
  • Data interpretation & experimental analysis (AO2): Interpreting the classical Meselson-Stahl experiment involving isotope density centrifugation (¹⁵N and ¹⁴N).
  • Prediction of scientific results (AO2): Applying knowledge of exponential replication cycles to predict intermediate and light DNA bands after two rounds of division.
Question 03.1 • 5 Marks

Describing Semi-Conservative DNA Replication

Detailed step-by-step description of how DNA replicates in cells

✅ Mark Scheme Key Points (Any 5 from 6)

  1. DNA helicase unwinds the double helix / breaks hydrogen bonds / separates strands.
  2. Both strands act as templates.
  3. Free (DNA) nucleotides align via complementary base pairing (A to T, C to G).
  4. DNA polymerase joins adjacent nucleotides together (in the new strand).
  5. Forms phosphodiester bonds (via condensation reactions).
  6. Each new DNA molecule consists of one original (conserved) strand and one newly synthesised strand.

🧠 Exam Technique & Terminology Precision

This is a classic 5-mark recall question where accuracy in biological terminology is paramount:

  • State the enzymes clearly: DNA helicase unwinds/unzips; DNA polymerase joins nucleotides.
  • Be precise about what is joined: DNA polymerase joins nucleotides together into a sugar-phosphate backbone, NOT bases together across the helix.
  • Don't forget the definition: Finish with the definition of "semi-conservative" to secure mark point 6 easily.

❌ Common Errors & Misconceptions

  • Incorrect enzyme role: Saying "DNA polymerase forms hydrogen bonds between complementary bases". Hydrogen bonds form automatically due to chemical attraction; DNA polymerase catalyses phosphodiester bonds between phosphate and deoxyribose. (Rejected in mark scheme!)
  • Imprecise cleavage phrasing: Writing "DNA helicase hydrolyses hydrogen bonds". Hydrogen bonds are broken, not hydrolysed (hydrolysis breaks covalent bonds using water).
  • Vague references: Stating "bases line up" instead of free DNA nucleotides.

💡 Key Knowledge: The Two Enzymes

  • DNA Helicase: Breaks weak hydrogen bonds holding complementary base pairs together, exposing unpaired bases on both template strands.
  • DNA Polymerase: Synthesises the polynucleotide strand in a 5' to 3' direction by catalysing condensation reactions to build the sugar-phosphate backbone (phosphodiester bonds).
Mark distribution: 5 marks total (5 × AO1). 6 marking points available, capped at 5.
Question 03.2 • 2 Marks

Explaining the Position of the DNA Band in Sample 2

Accounting for the intermediate density band formed after one division

✅ Correct Answer Breakdown

  1. Mechanism: DNA has replicated by semi-conservative replication / each new molecule contains one original strand and one new strand [1 mark].
  2. Isotope Composition: Each DNA molecule contains one heavy strand (¹⁵N) and one light strand (¹⁴N), giving it an intermediate density [1 mark].

💡 Understanding Sample 2 (Generation 1)

  • Sample 1: Grown exclusively in ¹⁵N → Double-stranded heavy DNA (¹⁵N/¹⁵N), settling near the bottom.
  • Sample 2: Grown in ¹⁵N then divided once in ¹⁴N → Both original ¹⁵N strands separate and serve as templates; new strands are made using ¹⁴N.
  • All resulting DNA molecules are hybrid ¹⁴N/¹⁵N duplexes, creating a single band exactly midway between heavy and light.

❌ Common Errors

  • Omitting the specific isotopes: Simply saying "it has a mix of heavy and light" without identifying that one strand is ¹⁵N and the other strand is ¹⁴N.
  • Failing to link the two strands directly to semi-conservative replication.

🧠 Exam Technique

Always pair cause with molecular composition: state that replication is semi-conservative (cause), which results in a hybrid molecule made of one ¹⁵N strand and one ¹⁴N strand (composition).

Mark distribution: 2 marks total (2 × AO2).
Question 03.3 • 1 Mark

Completing Sample 3 (After Two Divisions)

Predicting the banding pattern in the centrifuge tube

✅ Required Drawing in Sample 3 Tube

You must draw two horizontal bands:

  • Band 1 (Intermediate): Drawn at the exact same level as the band in Sample 2 (representing hybrid ¹⁴N/¹⁵N DNA).
  • Band 2 (Light): Drawn higher up the tube than the Sample 2 band (representing light ¹⁴N/¹⁴N DNA).

📐 Visualising the Strand Tracking (Generation 2)

Start with 1 hybrid molecule entering 2nd division in ¹⁴N:

  1. Strand 1 (¹⁵N heavy): paired with new ¹⁴N strand → ¹⁴N/¹⁵N hybrid (intermediate density band).
  2. Strand 2 (¹⁴N light): paired with new ¹⁴N strand → ¹⁴N/¹⁴N light (low density band, sits higher).

Total ratio in Gen 2: 50% Intermediate : 50% Light (2 hybrid molecules : 2 fully light molecules).

❌ Common Traps

  • Drawing only one band (forgetting that the original ¹⁵N strands are still conserved and present).
  • Drawing a band near the bottom (at the level of Sample 1). There is NO fully heavy (¹⁵N/¹⁵N) DNA remaining.
  • Misaligning the intermediate band so it does not line up horizontally with Sample 2.

🧠 Examiner Guidance

Lines can be of any thickness, but both lines are required to earn the 1 mark. Any lines drawn mistakenly in Sample 1 or Sample 2 are ignored.

Mark distribution: 1 mark total (1 × AO2).

Topics

Biology · 3.1 Biological molecules

Question and mark scheme from the AQA AS Level Biology examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.