Cambridge International AS & A Level Biology 9700: Nucleic acids and protein synthesis – Revision Notes

Condensed for the final weeks. For full explanations and worked examples, use the nucleic acids and protein synthesis study guide.

These notes cover topic 6 of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027: sections 6.1 (Structure of nucleic acids and replication of DNA) and 6.2 (Protein synthesis). It is AS Level content, examined on Paper 1 and Paper 2, and the syllabus says AS knowledge will also be required on Paper 4. Other links: the Cambridge A Level Biology hub, the printable 9700 checklist, the practice questions and the AS Level diagnostic.

6.1 Definitions to write word for word

• Nucleotide: a pentose sugar, a phosphate group and a nitrogenous base.
• ATP: a phosphorylated nucleotide – adenine, ribose and three phosphate groups.
• Purine: double-ring base – adenine, guanine.
• Pyrimidine: single-ring base – cytosine, thymine, uracil.
• Phosphodiester bond: the bond between the phosphate of one nucleotide and the sugar of the next, formed by condensation.
• Antiparallel: the two DNA strands run in opposite directions, one 5′ to 3′ and the other 3′ to 5′.
• Semi-conservative replication: each new DNA molecule has one parental strand and one new strand.

Three nucleotides compared

	DNA nucleotide	RNA nucleotide	ATP
Sugar	Deoxyribose	Ribose	Ribose
Base	A, T, C or G	A, U, C or G	Adenine only
Phosphate groups	One	One	Three
Role	Monomer of DNA	Monomer of RNA	Energy transfer in cells

No structural formulae are expected for any of these, or for the bases.

DNA vs mRNA

Feature	DNA	mRNA
Sugar	Deoxyribose	Ribose
Bases	A, T, C, G	A, U, C, G
Strands	Two, as a double helix	One
Length	Very long (whole chromosome)	Short (copy of one gene)
Base pairing within the molecule	A–T (2 H-bonds), C–G (3 H-bonds)	Not a double helix
Where found in eukaryotes	Nucleus	Made in nucleus, used in cytoplasm

Numbers to know

Fact	Value
Hydrogen bonds in an A–T pair	2
Hydrogen bonds in a C–G pair	3
Nucleotides per codon	3
Start codon (mRNA)	AUG, also codes for methionine
Stop codons (mRNA)	UAA, UAG, UGA
In double-stranded DNA	%A = %T and %C = %G

Method in steps: replication

1. S phase of the cell cycle.
2. Hydrogen bonds between bases break; strands separate; both act as templates.
3. Free nucleotides pair with exposed bases (A–T, C–G).
4. DNA polymerase forms phosphodiester bonds between new nucleotides, working only 5′ to 3′.
5. Leading strand made continuously; lagging strand made in fragments.
6. DNA ligase joins the fragments.
7. Result: two identical molecules, each one old strand + one new strand.

Direction rules (5′ and 3′)

• The 5′ end of a strand has the free phosphate; the 3′ end has the free –OH on the sugar.
• New nucleotides are only ever added to the 3′ end, so new strands grow 5′ → 3′.
• A template read 3′ → 5′ gives a new strand made 5′ → 3′.
• Always label both ends of any sequence you write. An unlabelled sequence can be read backwards and lose the mark.
• mRNA codons are read 5′ → 3′, starting at AUG.

Heavy and light DNA: predicting the bands

This is the standard way semi-conservative replication is tested with data. DNA made with heavy nitrogen (¹⁵N) is denser than DNA made with light nitrogen (¹⁴N), so the two form different bands after centrifuging. Cells start with all-heavy DNA and are then moved to ¹⁴N.

Generations in ¹⁴N	Heavy	Hybrid	Light
0	100%	0	0
1	0	100%	0
2	0	50%	50%
3	0	25%	75%
4	0	12.5%	87.5%

The number of hybrid molecules stays the same (the two original strands are never lost), while the total doubles every generation. Conservative replication would have given a heavy band and a light band after generation 1, with no hybrid band.

Method in steps: from gene to polypeptide

Transcription (nucleus)

1. Gene unwinds; hydrogen bonds break.
2. RNA polymerase uses the template (transcribed) strand.
3. RNA nucleotides pair: DNA A → RNA U, T → A, C → G, G → C.
4. RNA polymerase forms phosphodiester bonds.
5. Primary transcript: introns removed, exons joined → mRNA.
6. mRNA leaves through a nuclear pore.

Translation (cytoplasm)

1. mRNA binds to a ribosome.
2. tRNA with a complementary anticodon pairs with each codon, bringing a specific amino acid.
3. Peptide bond forms between adjacent amino acids.
4. Ribosome moves one codon along; used tRNA leaves.
5. Stops at UAA, UAG or UGA; polypeptide released.

Small worked reminders

Base percentage. DNA has 24% cytosine. So G = 24%, A + T = 100 − 48 = 52%, A = T = 26%.

Direction-safe conversion. Template 3′-TAC GGT-5′ → mRNA 5′-AUG CCA-3′ → Met–Pro. The non-transcribed strand reads 5′-ATG CCA-3′: the same as the mRNA with T for U.

Codon count. A polypeptide of 210 amino acids needs 210 codons + 1 stop codon = 211 codons = 633 nucleotides of coding mRNA.

Must-know distinctions

• Codon vs anticodon. Codon: three bases on mRNA. Anticodon: three bases on tRNA, complementary to the codon.
• Template vs non-transcribed strand. RNA polymerase reads the template strand. The mRNA is complementary to the template and has the same sequence as the non-transcribed strand (U for T).
• DNA polymerase vs RNA polymerase. DNA polymerase makes DNA in replication. RNA polymerase makes RNA in transcription.
• DNA polymerase vs DNA ligase. Polymerase adds nucleotides one at a time. Ligase joins whole sections of new strand.
• Leading vs lagging strand. Continuous vs fragments. The cause is that DNA polymerase adds only to the 3′ end.
• Intron vs exon. Introns are non-coding and are removed. Exons are coding and are joined.
• Substitution vs frame shift. A substitution affects one codon at most. A single deletion or insertion shifts every codon after it.

Mutation effects at a glance

Mutation	Frame shift?	Possible outcomes
Substitution	No	Same amino acid; one different amino acid; early stop codon
Deletion (1 base pair)	Yes	Many amino acids changed from that point; often an early stop codon
Insertion (1 base pair)	Yes	Same as deletion

Link it to function: altered primary structure → altered tertiary structure → for an enzyme, altered active site.

Phrases that carry the marks

• “Complementary base pairing, A with T and C with G, held by hydrogen bonds.”
• “Each new molecule contains one parental strand and one new strand.”
• “DNA polymerase forms phosphodiester bonds between adjacent nucleotides.”
• “DNA polymerase adds nucleotides only in the 5′ to 3′ direction.”
• “RNA polymerase uses the template (transcribed) strand.”
• “The anticodon on tRNA is complementary to the codon on mRNA.”
• “Introns are removed and exons are joined to form mRNA.”
• “Every triplet after the deletion is read differently (frame shift).”

Quick self-test

1. Name the two purine bases.
2. State the three components of ATP.
3. A DNA sample has 31% guanine. What percentage is adenine?
4. A 20-base-pair section contains 9 C–G pairs. How many hydrogen bonds hold it together?
5. Which bond joins nucleotides within one strand?
6. In which phase of the cell cycle does DNA replication happen?
7. Why is the lagging strand made in fragments?
8. Bacteria grown for many generations on heavy nitrogen (¹⁵N) are moved to light nitrogen (¹⁴N). After four generations, what fraction of DNA molecules is hybrid (¹⁵N/¹⁴N)?
9. Give the mRNA sequence for the template strand 3′-TAC GGT-5′.
10. A polypeptide has 210 amino acids. How many nucleotides are in its coding mRNA, including the stop codon?
11. What is removed from the primary transcript to form mRNA?
12. Why can a substitution have no effect on the polypeptide?

Answers

1. Adenine and guanine.
2. Adenine, ribose and three phosphate groups.
3. C = 31%, so A + T = 38% and A = 19%.
4. 9 × 3 + 11 × 2 = 49.
5. Phosphodiester bond.
6. S phase.
7. DNA polymerase can only add nucleotides in the 5′ to 3′ direction, so on this template it must work away from the fork in short sections, later joined by DNA ligase.
8. After each generation the number of hybrid molecules stays at 2 per original molecule while the total doubles: 2 ÷ 16 = 1/8.
9. 5′-AUG CCA-3′.
10. 211 codons × 3 = 633.
11. Introns (non-coding sequences); the exons are joined.
12. Most amino acids have more than one codon, so the new triplet may code for the same amino acid.

Where marks are usually lost

• Writing “deoxyribose” for RNA or “thymine” in an mRNA sequence.
• Giving A–T three hydrogen bonds instead of two.
• Describing the double helix without the word “antiparallel” when the question asks about strand orientation.
• Saying DNA polymerase “joins bases” – it forms phosphodiester bonds in the backbone.
• Explaining the lagging strand without the 5′ to 3′ reason.
• Writing mRNA with T instead of U, or reading the codon table with DNA triplets.
• Counting the stop codon as an amino acid when finding polypeptide length.
• Forgetting that introns are removed before mRNA leaves the nucleus.
• Stating “the protein changes” without linking mutation → codon → amino acid sequence → shape.
• Calling tRNA’s triplet a codon.

Official syllabus

Cambridge International, Cambridge International AS & A Level Biology 9700 syllabus for 2025, 2026 and 2027 (Version 1, published September 2022), Cambridge University Press & Assessment. Topic 6, Nucleic acids and protein synthesis: sections 6.1 and 6.2.