AQA A-level Biology: Nucleic Acids, DNA Replication and ATP (7402) – Revision Notes

These are condensed revision notes for the AQA AS and A-level Biology specification (7401 and 7402, first teaching 2015), sections 3.1.5.1, 3.1.5.2 and 3.1.6. The content is part of both the AS and the A-level. Full explanations and worked examples are in the study guide. Test yourself with the practice questions.

Wider topic: Biological Molecules revision notes. Course page: AQA A-level Biology. Tick off outcomes on the checklist. Not sure where to start? Take a free diagnostic.

3.1.5.1 DNA and RNA at a glance

Feature	DNA	RNA
Job	holds genetic information in all living cells	transfers genetic information from DNA to the ribosomes
Pentose	deoxyribose	ribose
Bases	adenine, cytosine, guanine, thymine	adenine, cytosine, guanine, uracil
Structure	double helix, two polynucleotide chains	relatively short polynucleotide chain
Chains linked by	hydrogen bonds between specific complementary base pairs	none: a single chain

Ribosomes are formed from RNA and proteins.

Key definitions

• Nucleotide: a pentose, a nitrogen-containing organic base and a phosphate group.
• Polynucleotide: a polymer of nucleotides.
• Phosphodiester bond: the bond formed by a condensation reaction between two nucleotides. It makes the sugar-phosphate backbone.
• Complementary base pairs: A with T (or U in RNA), C with G.

Must-know distinctions

Often confused	Difference
Phosphodiester bond and hydrogen bond	phosphodiester bonds join nucleotides along one strand; hydrogen bonds join bases between two strands
Ribose and deoxyribose	ribose in RNA and ATP; deoxyribose in DNA
Thymine and uracil	thymine in DNA; uracil in RNA
Nucleotide and base	a nucleotide contains the base plus a pentose and a phosphate group

Counting rules

• n nucleotides in one chain give n − 1 phosphodiester bonds and n − 1 water molecules released.
• A double-stranded molecule of b base pairs has 2b nucleotides and 2(b − 1) phosphodiester bonds.

Base-frequency method

1. Write down what you know. In double-stranded DNA, A = T and C = G.
2. If you know one value (say A), the matching base has the same value.
3. Subtract both from 100 and halve the remainder to get the other pair.
4. For one strand only: the other strand’s A equals this strand’s T, and so on. The whole molecule is the average of the two strands.
5. Check the four values total 100.

Worked reminder. Double-stranded DNA has 36% adenine. T = 36%. A + T = 72%. C + G = 28%, so G = C = 14%.

Why DNA’s simplicity caused doubt

• Only four different bases, so DNA looked too simple to carry the code.
• Proteins have many different amino acids, so they looked more able to carry information.
• The answer: the information is in the order and length of the bases.

3.1.5.2 Semi-conservative replication

Each new molecule has one original strand and one new strand. This gives genetic continuity between generations of cells.

Method in steps

1. The double helix unwinds.
2. DNA helicase breaks hydrogen bonds between complementary bases and unwinds the DNA.
3. Each strand acts as a template.
4. Free DNA nucleotides are attracted to the exposed bases and base-pair (A–T, C–G).
5. DNA polymerase catalyses the condensation reaction joining adjacent nucleotides.
6. Two identical molecules result, each half old and half new.

Enzyme	Reaction type	Bond affected
DNA helicase	breaks bonds	hydrogen bonds between bases
DNA polymerase	condensation	phosphodiester bonds in the new strand

Evaluating the Watson-Crick validation work

The Meselson-Stahl experiment grew bacteria in ¹⁵N, moved them to ¹⁴N and used a density gradient to separate DNA by density.

Generation in ¹⁴N	Result seen	Meaning
0	one heavy band	all DNA contains ¹⁵N
1	one intermediate band	rules out conservative
2	intermediate and light bands, equal amounts	rules out dispersive; fits semi-conservative

Predictions to learn. Conservative at generation 1: heavy and light bands. Dispersive at generation 2: one band, lighter than intermediate.

Fractions to learn. After n generations: 2ⁿ molecules, always 2 with a ¹⁵N strand. After 5 generations that is 2 of 32 molecules, 6.25%.

To evaluate, say:

• Strength: a clear prediction that could have failed, tested at several time points.
• Limit: shows distribution of strands, not the enzymes or the sequence accuracy.
• Limit: done in bacteria, so applying it to other organisms needs more evidence.

3.1.6 ATP

Feature	Detail
Structure	nucleotide derivative: ribose, adenine and three phosphate groups
Hydrolysis	ATP + water → ADP + Pi, catalysed by ATP hydrolase
Use of the energy released	coupled to energy-requiring reactions in cells
Use of the Pi	phosphorylates other compounds, often making them more reactive
Resynthesis	ADP + Pi → ATP by condensation, catalysed by ATP synthase, during photosynthesis or respiration

ATP compared with an RNA nucleotide

Part	RNA nucleotide	ATP
Pentose	ribose	ribose
Base	one of four (A, C, G, U)	adenine
Phosphate groups	one	three

ATP is called a nucleotide derivative because it is built like a nucleotide but with extra phosphate groups.

Coupling and phosphorylation in an answer

• Say the energy released by ATP hydrolysis is coupled to the reaction that needs energy.
• Say the Pi is transferred to another compound, which is phosphorylated.
• Say the phosphorylated compound is often more reactive, so the reaction goes ahead more easily.
• Name ATP hydrolase for hydrolysis, and say water is used.

Worked reminder (data supplied). A reaction needs 25.0 kJ per mole. ATP hydrolysis releases 30.5 kJ per mole. Net release = 30.5 − 25.0 = 5.5 kJ per mole, so the coupled pair can proceed.

Remember

• Hydrolysis uses water; condensation releases water.
• ATP hydrolase breaks down ATP. ATP synthase makes it.
• Phosphorylation means adding a phosphate group.

Worked reminder (data supplied). Hydrolysis of 0.20 mol ATP releases 30.5 kJ per mole. Energy = 0.20 × 30.5 = 6.1 kJ.

Model answer frames

Describe semi-conservative replication. Use this order and you will meet most points:

• The helix unwinds as DNA helicase breaks hydrogen bonds between complementary bases.
• Each separated strand acts as a template.
• Free nucleotides pair with the exposed bases (A with T, C with G).
• DNA polymerase joins adjacent nucleotides by condensation, forming phosphodiester bonds.
• Each new molecule has one original and one new strand.

Describe how a polynucleotide is built. Nucleotides join by condensation, a molecule of water is released, a phosphodiester bond forms and a sugar-phosphate backbone results with the bases projecting from it.

Explain the use of ATP in a reaction. ATP is hydrolysed by ATP hydrolase to ADP and Pi. The energy is coupled to the reaction, or the Pi phosphorylates a compound and makes it more reactive. ATP is later resynthesised from ADP and Pi.

Quick self-test

1. Name the pentose sugar in RNA.
2. Name the four bases found in RNA.
3. A single polynucleotide chain has 150 nucleotides. How many phosphodiester bonds does it have?
4. A DNA fragment is 40 base pairs long. How many phosphodiester bonds does it contain?
5. Double-stranded DNA has 18% thymine. What is the percentage of guanine?
6. One strand has 22% A, 30% T, 28% G and 20% C. State the percentage of A and of G in the whole molecule.
7. Which enzyme breaks hydrogen bonds between bases during replication?
8. Which enzyme catalyses the condensation reaction joining adjacent nucleotides?
9. After 5 generations in ¹⁴N, starting with all ¹⁵N DNA, what percentage of molecules contain ¹⁵N?
10. Hydrolysis of ATP releases 30.5 kJ per mole. How much energy is released from 0.20 mol?
11. Which enzyme resynthesises ATP, and what type of reaction is it?
12. Give one reason why DNA’s structure made scientists doubt it carried the genetic code.

Answers

1. Ribose.
2. Adenine, cytosine, guanine, uracil.
4. 2 × 39 = 78.
5. A = T = 18%, so A + T = 36%, C + G = 64%, G = 32%.
6. A = (22 + 30) ÷ 2 = 26%. G = (28 + 20) ÷ 2 = 24%.
7. DNA helicase.
8. DNA polymerase.
9. 2 of 32 molecules = 6.25%.
10. 0.20 × 30.5 = 6.1 kJ.
11. ATP synthase; a condensation reaction (ADP + Pi).
12. It has only four different bases, so it seemed too simple compared with proteins.

Where marks are usually lost

• Writing that helicase breaks “the DNA” or “bonds” without saying hydrogen bonds.
• Saying DNA polymerase “joins the bases” rather than joining adjacent nucleotides by condensation.
• Counting a double-stranded fragment as b nucleotides instead of 2b.
• Giving A = T but then forgetting that G and C must share what is left.
• Saying “ATP stores energy” without linking release to hydrolysis and to coupling.
• Using “ATP synthetase” or swapping ATP hydrolase and ATP synthase.
• Saying that Meselson-Stahl “proved” the model without describing the bands.
• In an evaluation, listing only strengths. Give at least one limitation too.

Official syllabus

AQA, AS and A-level Biology (7401 and 7402), specification for first teaching 2015, sections 3.1.5.1, 3.1.5.2 and 3.1.6.