OCR A Level Biology A: Nucleotides, ATP, DNA and replication (H420) – Revision Notes

Condensed recall for OCR A Level Biology A (H420, Version 4.1, April 2026), section 2.1.3(a) to (e). For full explanations and worked examples use the study guide, then test yourself with the practice questions. Course pages: OCR A Level Biology hub and the printable checklist. If you are unsure where to start, take a free diagnostic. The genetic code, transcription and translation (2.1.3(f) and (g)) are not covered here.

2.1.3(a) Nucleotides: DNA against RNA

Each nucleotide is built from a five-carbon sugar, a phosphate and a base containing nitrogen. It is the monomer of nucleic acids.

	DNA	RNA
Sugar	Deoxyribose	Ribose
Pyrimidines	Cytosine, thymine	Cytosine, uracil
Purines	Adenine, guanine	Adenine, guanine
Strands	Two, antiparallel, twisted	Usually one

• Purines (A, G) have two rings. Pyrimidines (C, T, U) have one ring.
• Memory aid: “CUT the pyrimidines” (cytosine, uracil, thymine); “PURe As Gold” (adenine and guanine are purines).
• Deoxyribose has a hydrogen atom, not a hydroxyl group, on carbon 2′.
• Link to practical: OCR lists a modelling activity, investigating DNA structure using RasMol. Be able to read a molecular model.

2.1.3(b) Phosphodiester bonds

• Synthesis: condensation. 5′ phosphate of one nucleotide joins the 3′ hydroxyl of the next. One water is released per bond.
• Breakdown: hydrolysis. One water is used per bond.
• A chain of n nucleotides has n − 1 phosphodiester bonds.
• A double-stranded fragment of n base pairs has 2(n − 1) of them.

Step	Method
1	Count nucleotides in one strand: n
2	Bonds in one strand = n − 1
3	Multiply by 2 for a double helix
4	Water released or used = number of bonds

Reminder: a 30-nucleotide chain has 29 bonds; a 30 base-pair fragment has 58.

2.1.3(c) ADP and ATP

	ADP	ATP
Sugar	Ribose	Ribose
Base	Adenine	Adenine
Inorganic phosphates	2	3

• Both are phosphorylated nucleotides.
• ATP + water → ADP + Pi releases energy (hydrolysis). ADP + Pi → ATP is phosphorylation.
• Describe ATP as a molecule that releases energy when hydrolysed, not as an energy “store” or “source of energy” without the reaction.
• Reminder calculation: a process needing 183 kJ per mole, when each mole of ATP supplies 30.5 kJ, needs at least 183 ÷ 30.5 = 6.0 mol of ATP.

2.1.3(d)(i) DNA structure

• Two polynucleotides; sugar-phosphate backbone on the outside, bases inside.
• Complementary base pairing: A to T with two hydrogen bonds; G to C with three.
• Each pair = one purine + one pyrimidine, so the width is constant.
• Antiparallel: one strand runs 5′ to 3′, the other 3′ to 5′.
• The twisting of the paired strands gives the double helix.
• Stability: many hydrogen bonds together hold the strands; the backbone is covalent.
• Weak individual hydrogen bonds allow the strands to separate for copying.

Method for hydrogen-bond questions: let a = A–T pairs and g = G–C pairs. Write a + g = total pairs and 2a + 3g = total hydrogen bonds. Solve together.

2.1.3(d)(ii) Practical: DNA purification by precipitation

Reagent or step	Purpose
Mashing	Breaks up tissue and cell walls
Detergent	Dissolves phospholipids of membranes, releasing DNA
Salt	Masks the negative charge on phosphates so DNA clumps
Gentle warming	Helps detergent work; inactivates enzymes that digest DNA
Protease (optional)	Digests proteins bound to DNA
Filtering	Removes debris
Ice-cold ethanol layer	DNA is insoluble in ethanol and precipitates at the boundary
Glass rod	Spools the white strands

Controlling variables when comparing tissues: same starting mass, same volumes, same temperature and times, same ethanol layering. Dependent variable: dried mass or length of DNA. The product is crude, with RNA and protein present.

2.1.3(e) Semi-conservative replication

Semi-conservative: each new molecule = one original strand + one new strand.

Steps:

1. Helicase breaks hydrogen bonds between bases and unwinds the helix.
2. Each strand is a template.
3. Free nucleotides pair by complementary base pairing.
4. DNA polymerase forms phosphodiester bonds, adding nucleotides at the 3′ end of the new strand.
5. Two molecules, each hybrid of old and new.

• Accuracy: templates plus base pairing copy the sequence, conserving genetic information for daughter cells.
• Spontaneous mutation: random changes in base sequence that occur without a mutagen, for example a wrong nucleotide incorporated. They are random in place and time.
• At AS Level the specification does not require learners to distinguish between types of mutation.

Labelling reminder (¹⁵N heavy, ¹⁴N light): start with all-heavy; after one generation in light medium, all hybrid; after a second generation, half hybrid and half light. A conservative model predicts no hybrid band at generation one.

After n rounds from one molecule: 2ⁿ molecules, always only two containing an original strand.

Worked reminders

Yield from a purification

A sample of 15.0 g of tissue gives 4.5 mg of dried DNA.

• Yield = 4.5 ÷ 15.0 = 0.30 mg per g.
• Compare tissues by yield per gram, not by total mass, because the starting masses may differ.
• Percentage difference = (new − baseline) ÷ baseline × 100. State your baseline.

Hybrid DNA after switching medium

All-heavy bacteria (HH) are grown for one generation in light medium, then returned to heavy medium for one generation.

• After light medium: all HL.
• In heavy medium each HL splits into one HH and one HL.
• Result: half HH, half HL, no LL.

Expected mutations

Data supplied: 2.0 × 10⁶ base pairs and 1 × 10⁻⁹ errors per base pair per replication.

• Errors per replication = 2.0 × 10⁶ × 1 × 10⁻⁹ = 2.0 × 10⁻³.
• Across 10⁶ replications = 2.0 × 10³ errors.
• A rate that is tiny per replication still gives many errors in a large population.

Exam wording to use

If the question says	Write
Describe the structure of DNA	Two antiparallel polynucleotides, sugar-phosphate backbone, complementary base pairs joined by hydrogen bonds, twisted into a double helix
Explain why replication is semi-conservative	Each new molecule contains one original template strand and one new strand
Explain the role of DNA polymerase	Joins free nucleotides by forming phosphodiester bonds on the new strand, using the template and complementary base pairing
Explain the role of helicase	Breaks hydrogen bonds between complementary bases so the strands separate and act as templates
Describe a spontaneous mutation	A random change in the base sequence that arises during replication without a mutagen

Linking ideas across the unit

• ATP, ADP and the RNA adenine nucleotide share the same sugar and base, so you can compare them in one sentence.
• Phosphodiester bonds appear twice: in DNA synthesis by DNA polymerase and in the backbone of every polynucleotide.
• Hydrogen bonds appear twice: between complementary bases in the helix and, in replication, as the bonds helicase breaks.
• Precipitation in the practical works because of the same phosphate groups that make the backbone negatively charged.
• If a question gives numbers, write the equation you are using before you substitute, and give units.

Must-know distinctions

Often confused	Difference
Phosphodiester bond / hydrogen bond	Covalent link between nucleotides in a strand / weak link between complementary bases across strands
Helicase / DNA polymerase	Breaks hydrogen bonds between bases / forms phosphodiester bonds between nucleotides
ADP / ATP	Two phosphates / three phosphates
Purine / pyrimidine	Two rings / one ring
Condensation / hydrolysis	Water released / water used

Quick self-test

1. Name the three components of a nucleotide.
2. State one structural difference between ribose and deoxyribose.
3. How many phosphodiester bonds join the nucleotides in a single strand of 48 nucleotides?
4. How many phosphodiester bonds are there in a double-stranded fragment of 48 base pairs?
5. How many hydrogen bonds join a G–C pair?
6. State what is meant by antiparallel.
7. A 12 base-pair fragment has 30 hydrogen bonds. How many are G–C pairs?
8. Why is ice-cold ethanol added in DNA purification?
9. Name the enzyme that breaks hydrogen bonds during replication.
10. After four rounds of semi-conservative replication, how many molecules contain an original strand if you began with one molecule?

Answers

1. A pentose sugar, a phosphate group and a nitrogenous base.
2. Deoxyribose has hydrogen on carbon 2′ where ribose has a hydroxyl group (one fewer oxygen atom).
4. 2 × 47 = 94.
5. Three.
6. The two strands run in opposite directions, 5′ to 3′ and 3′ to 5′.
7. a + g = 12 and 2a + 3g = 30, so g = 30 − 24 = 6 G–C pairs (and 6 A–T pairs).
8. DNA is soluble in water but not in ethanol, so it precipitates as strands at the boundary.
9. Helicase.
10. Two (out of 16 molecules).

Where marks are usually lost

• Writing “hydrogen bonds” when the bond between nucleotides in one strand is a phosphodiester bond.
• Counting nucleotides instead of bonds between them.
• Saying “thymine in RNA” or “ribose in DNA”.
• Describing helicase as “unzipping the DNA” without saying it breaks hydrogen bonds between bases.
• Leaving out complementary base pairing when explaining accurate copying.
• Calling mutations “caused by” a mutagen when the question says spontaneous.
• Explaining the purification steps without saying what each reagent does to the cell or the DNA.
• Quoting percentage change without stating which value is the baseline.

Official syllabus

OCR, Level 3 Advanced GCE in Biology A (H420), Version 4.1 (April 2026), first assessment 2025, section 2.1.3(a) to (e).