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

This study guide follows the OCR A Level Biology A specification (H420, Version 4.1, April 2026, first assessment 2025). It covers module 2 section 2.1.3, learning outcomes (a) to (e): nucleotides, phosphodiester bonds, ADP and ATP, the structure of DNA, the DNA purification practical, and semi-conservative replication. Outcomes (f) and (g), the genetic code and protein synthesis, are separate topics and are not taught here. Pair this page with the revision notes and the practice questions. The wider module is in Foundations in Biology. Course pages: OCR A Level Biology hub and the printable checklist. To find weak spots first, try the free diagnostics.

What this unit covers

Outcome	What you must be able to do
2.1.3(a)	Describe a nucleotide as the monomer of nucleic acids; state the differences between RNA and DNA nucleotides, including purines, pyrimidines and the pentose sugar
2.1.3(b)	Explain synthesis and breakdown of polynucleotides by forming and breaking phosphodiester bonds
2.1.3(c)	Describe ADP and ATP as phosphorylated nucleotides: ribose, adenine and inorganic phosphates
2.1.3(d)(i)	Describe DNA: hydrogen bonding between complementary bases (A to T, G to C) on two antiparallel polynucleotides, and the twisting that gives the double helix
2.1.3(d)(ii)	Carry out and interpret a DNA purification by precipitation
2.1.3(e)	Explain semi-conservative replication, the roles of helicase and DNA polymerase, accuracy in conserving genetic information, and random spontaneous mutation

The specification links outcome (a) to an ICT-based practical, investigating DNA structure using RasMol (a molecular-modelling program), and outcome (d) to a qualitative-testing practical group. Expect to meet a model of DNA on screen as well as in diagrams.

Nucleotide structure (2.1.3(a))

A nucleotide is the monomer from which nucleic acids are built. Every nucleotide has three parts joined by covalent bonds:

• a pentose sugar (a five-carbon sugar)
• a phosphate group
• a nitrogenous base

The bases fall into two families. Purines have a double-ring structure: adenine (A) and guanine (G). Pyrimidines have a single ring: cytosine (C), thymine (T) and uracil (U).

Feature	DNA nucleotide	RNA nucleotide
Pentose sugar	Deoxyribose	Ribose
Bases present	A, G, C, T	A, G, C, U
Pyrimidines	C and T	C and U
Purines	A and G	A and G

Ribose has a hydroxyl group on the carbon next to the one that carries the base; deoxyribose has only a hydrogen atom there. That is the origin of the name deoxy-ribose, “ribose without an oxygen”. Describe the difference like this: deoxyribose has one fewer oxygen atom than ribose.

In diagrams the carbons of the sugar are numbered 1′ to 5′ (“one prime” and so on). The base attaches to carbon 1′, the phosphate to carbon 5′, and a hydroxyl group sits on carbon 3′. You need these numbers to understand bonding and direction.

Phosphodiester bonds (2.1.3(b))

Nucleotides join in a condensation reaction. The phosphate on the 5′ carbon of one nucleotide bonds to the hydroxyl on the 3′ carbon of the next. The new link is a phosphodiester bond: a phosphate group joined by two ester bonds, one to each sugar. One molecule of water is released for every bond formed.

The repeated sugar and phosphate units form the sugar-phosphate backbone. The bases project sideways from it. A polynucleotide therefore has two different ends: a free phosphate at the 5′ end and a free hydroxyl at the 3′ end. This gives the chain a direction.

Breakdown is the reverse. Hydrolysis adds one water molecule to each phosphodiester bond, breaking it and releasing the nucleotides.

Worked example 1: counting bonds

A single-stranded polynucleotide has 22 nucleotides. A double-stranded DNA fragment has 22 base pairs.

1. Joining 22 nucleotides in one chain needs 22 − 1 = 21 bonds, so 21 water molecules are released.
2. The double-stranded fragment has two chains of 22 nucleotides: 2 × 21 = 42 phosphodiester bonds.
3. Completely hydrolysing the fragment to separate nucleotides needs one water molecule per bond: 42 water molecules.

Always count the bonds between units, not the units themselves. Hydrogen bonds between bases are separate, and are not phosphodiester bonds.

ADP and ATP (2.1.3(c))

ATP (adenosine triphosphate) is a phosphorylated nucleotide. It comprises:

• the pentose sugar ribose
• the base adenine
• three inorganic phosphates in a row

ADP (adenosine diphosphate) is identical except that it has two phosphates. Adding a phosphate to ADP is phosphorylation; the reverse, ATP + water → ADP + inorganic phosphate (Pi), is hydrolysis and releases energy that cells can use for work such as active transport or muscle contraction. Because ATP contains ribose and adenine, it is closely related to the adenine nucleotide of RNA, which has one phosphate.

Worked example 2: ATP and a cell process

A reaction needs 244 kJ per mole of product. Hydrolysing one mole of ATP supplies 30.5 kJ (value supplied for this example).

Minimum moles of ATP per mole of product = 244 ÷ 30.5 = 8.0 mol.

The calculation is a minimum because some energy is always lost as heat.

DNA structure (2.1.3(d)(i))

A DNA molecule has two polynucleotide strands. Complementary base pairs hold them together by hydrogen bonds: adenine pairs with thymine (two hydrogen bonds) and guanine pairs with cytosine (three hydrogen bonds). Each pair is one purine and one pyrimidine, so every pair has the same width.

The strands are antiparallel. One runs 5′ to 3′ and its partner runs 3′ to 5′. The strands then twist around each other to form a double helix, with the sugar-phosphate backbones on the outside and the base pairs inside.

Why this structure works:

• Many hydrogen bonds together make the molecule stable, yet each bond is weak, so the strands can be separated for copying.
• The covalent backbone is strong, which protects the base sequence.
• The base sequence along a strand carries information.

Worked example 3: hydrogen bonds and base counts

A 40 base-pair fragment has 104 hydrogen bonds in total. How many A–T and G–C pairs does it have?

Let a = number of A–T pairs and g = number of G–C pairs.

• a + g = 40
• 2a + 3g = 104

Substituting a = 40 − g gives 80 − 2g + 3g = 104, so g = 24 and a = 16.

The fragment has 16 A–T pairs and 24 G–C pairs: 16 adenine, 16 thymine, 24 guanine, 24 cytosine. The GC content is 24 ÷ 40 = 60%.

Worked example 4: length of a helix

Data supplied for this example: 10 base pairs per turn and 0.34 nm between neighbouring base pairs. A segment has 2,040 base pairs.

• Turns = 2,040 ÷ 10 = 204 turns
• Length = 2,040 × 0.34 = 693.6 nm

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

A typical class method uses a plant tissue such as strawberry, onion or split pea.

1. Break up the tissue by mashing or blending to separate cells and break cell walls.
2. Add a salt and detergent solution. The detergent dissolves the phospholipids of the cell surface membrane and nuclear envelope, so DNA is released. Sodium ions help the DNA clump by masking the negative charge of the phosphate groups.
3. Warm gently (about 60 °C for a few minutes, then cool). This helps the detergent work and inactivates enzymes that would digest DNA. Do not overheat.
4. Filter to remove cell debris.
5. Optionally add a protease to digest proteins that are bound to the DNA.
6. Pour ice-cold ethanol slowly down the side of a tilted tube so it forms a layer on top of the filtrate. DNA is soluble in water but not in ethanol, so it precipitates at the boundary as white, stringy threads.
7. Spool the DNA onto a glass rod.

Interpreting results: the product is crude. It will contain some RNA and protein, so you would use a DNA-specific test to confirm identity. To compare tissues, use the same starting mass, the same volumes and the same timings, and record the dried mass of DNA.

Worked example 5: comparing yields

Three tissues, each 20.0 g, gave dried DNA of 5.6 mg (strawberry), 3.1 mg (onion) and 9.4 mg (split pea).

• Strawberry: 5.6 ÷ 20.0 = 0.28 mg per g
• Onion: 3.1 ÷ 20.0 = 0.155 mg per g
• Split pea: 9.4 ÷ 20.0 = 0.47 mg per g

Split pea yields (0.47 − 0.155) ÷ 0.155 × 100 = 203% more than onion. With one run per tissue you cannot say the difference is reliable; repeating and calculating a mean would help.

Semi-conservative replication (2.1.3(e))

Semi-conservative means each new DNA molecule has one original strand and one newly made strand. Replication happens before cell division and is described in steps:

1. Helicase breaks the hydrogen bonds between complementary bases, unwinding and separating the two strands.
2. Each separated strand acts as a template.
3. Free DNA nucleotides line up by complementary base pairing: A with T, G with C.
4. DNA polymerase joins the aligned nucleotides by forming phosphodiester bonds. It adds nucleotides to the 3′ end of the growing strand.
5. Each product molecule has one old and one new strand.

Replication is important because it conserves genetic information with accuracy: the template and base pairing rules mean the base sequence is copied, so daughter cells receive the same information.

Evidence from labelling

Bacteria grown for many generations on heavy nitrogen (¹⁵N) have all-heavy DNA. Moved to light nitrogen (¹⁴N), their new strands are light. A separation by density shows a band position for each type.

Worked example 6: switching back

Bacteria with all-heavy DNA (HH) are grown for one generation in ¹⁴N, then moved back to ¹⁵N for one generation. What DNA results?

• After the ¹⁴N generation, every molecule is hybrid (HL): one old heavy strand, one new light strand.
• In ¹⁵N, each hybrid molecule separates. The heavy template builds a new heavy strand, giving HH. The light template builds a new heavy strand, giving LH (hybrid).
• So the result is 50% HH and 50% hybrid, with no all-light molecules.

A conservative model would have kept a fully original molecule intact and produced a fully new one; the observed single hybrid band after one generation does not fit it.

After n rounds from one molecule, 2ⁿ molecules exist but only two contain an original strand. After 4 rounds: 16 molecules, 2 with original strands (12.5%), and 2 of 32 strands are original (6.25%).

Spontaneous mutation (2.1.3(e))

Copying is accurate but not perfect. A mutation is a change in the order of the bases along a DNA molecule. Spontaneous mutations arise without a mutagen, for example when a wrong nucleotide is paired and joined during replication. They are random: you cannot predict which base, which gene or which cell. The specification notes that at AS Level learners are not required to distinguish between different types of mutation, so here the idea matters more than the classification.

Worked example 7: expected mutations

Data supplied: a bacterial genome of 4.6 × 10⁶ base pairs, with 5 × 10⁻¹⁰ errors per base pair per replication.

• Per replication: 4.6 × 10⁶ × 5 × 10⁻¹⁰ = 2.3 × 10⁻³ errors
• In 10⁹ replications: 2.3 × 10⁻³ × 10⁹ = 2.3 × 10⁶ errors

A tiny rate per replication still produces many mutations across a large population, and these give variation.

Common errors

• Writing that hydrogen bonds join nucleotides in one strand. Phosphodiester bonds do that.
• Saying DNA polymerase “joins bases”. It joins nucleotides and forms phosphodiester bonds.
• Saying helicase breaks “the DNA”. It breaks hydrogen bonds between bases only.
• Calling the strands “parallel” or “identical”.
• Saying ATP is “an energy store”. It is a molecule whose hydrolysis releases energy.
• Using “ribose” for DNA or “thymine” for RNA.

Official syllabus

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