Cambridge International AS & A Level Biology 9700: Nucleic acids and protein synthesis – Study Guide

This study guide teaches topic 6, Nucleic acids and protein synthesis, from the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027. It covers sections 6.1 (Structure of nucleic acids and replication of DNA) and 6.2 (Protein synthesis). This is AS Level content, so it is examined on Paper 1 (Multiple Choice) and Paper 2 (AS Level Structured Questions). Paper 4 is based on the A Level content but the syllabus states that knowledge of AS material will be required there too.

Useful links: the Cambridge A Level Biology hub, the printable 9700 checklist, the condensed revision notes for this topic and the practice questions with marked answers. For a whole-course check, try the AS Level diagnostic. Condensation reactions and hydrogen bonds are covered in the biological molecules study guide.

What this topic covers

Syllabus ref.	What you must be able to do	Stage
6.1.1	Describe the structure of nucleotides, including ATP (no structural formulae)	AS
6.1.2	State which bases are purines (double ring) and which are pyrimidines (single ring)	AS
6.1.3	Describe DNA as a double helix: antiparallel strands, complementary base pairing, hydrogen bonding in A–T and C–G, phosphodiester bonds	AS
6.1.4	Describe semi-conservative replication in S phase: DNA polymerase, DNA ligase, leading and lagging strands	AS
6.1.5	Describe the structure of RNA, using mRNA as the example	AS
6.2.1–6.2.2	State what a gene is; describe the universal genetic code, including start and stop codons	AS
6.2.3–6.2.4	Describe transcription and translation; name the template and non-transcribed strands	AS
6.2.5	Explain how introns are removed and exons joined in eukaryotes	AS
6.2.6–6.2.7	State what a gene mutation is; explain substitution, deletion and insertion and their effects	AS

6.1 Nucleotides and nucleic acids

The nucleotide

A nucleotide has three parts:

• a pentose sugar (5-carbon): deoxyribose in DNA, ribose in RNA
• a phosphate group
• a nitrogenous base

The base and the phosphate are both attached to the sugar. You do not need structural formulae.

ATP is a phosphorylated nucleotide. It contains the base adenine, the sugar ribose and three phosphate groups. Adenine plus ribose is called adenosine, so ATP is adenosine triphosphate.

Purines and pyrimidines

Group	Ring structure	Bases
Purines	Double ring	Adenine (A), guanine (G)
Pyrimidines	Single ring	Cytosine (C), thymine (T), uracil (U)

Thymine is found only in DNA. Uracil is found only in RNA. A memory aid: “Pure As Gold” for the purines.

Building a polynucleotide

Nucleotides join by condensation reactions. The phosphate of one nucleotide bonds to the sugar of the next, forming a phosphodiester bond. This makes a sugar–phosphate backbone with the bases sticking out to one side.

Each strand has direction. One end has a free phosphate on carbon 5 of the sugar (the 5′ end). The other end has a free –OH on carbon 3 (the 3′ end).

The DNA double helix

• Two polynucleotide strands wind round each other to form a double helix.
• The strands are antiparallel: one runs 5′ to 3′, the other runs 3′ to 5′.
• Bases pair across the helix by complementary base pairing: A with T, C with G.
• Each pair is one purine plus one pyrimidine, so every rung is the same width and the helix has a constant diameter.
• A–T pairs are held by two hydrogen bonds. C–G pairs are held by three. A region rich in C–G is harder to separate.
• Hydrogen bonds are individually weak, but there are so many that the molecule is very stable. Because they are weak, the strands can still be separated for replication and transcription.

Worked example 1 – base proportions and hydrogen bonds

A sample of double-stranded DNA contains 18% adenine. A 20-base-pair section has 12 A–T pairs and 8 C–G pairs.

Base proportions
A pairs with T, so T = 18%
A + T = 36%, so G + C = 100 − 36 = 64%
G = C = 64 ÷ 2 = 32%

Hydrogen bonds in the 20 bp section
A–T: 12 × 2 = 24
C–G:  8 × 3 = 24
Total = 48 hydrogen bonds

This rule only works for double-stranded DNA. In single-stranded mRNA the bases are not paired, so A need not equal U.

Structure of mRNA

• Single polynucleotide strand, not a double helix.
• Sugar is ribose; base uracil replaces thymine.
• Much shorter than DNA: it is a copy of one gene (or a few genes), not a whole chromosome.
• Its bases are read in groups of three, called codons.

6.1 Semi-conservative replication

DNA is copied in the S phase of the cell cycle, before mitosis. Replication is semi-conservative: each new DNA molecule contains one original (parental) strand and one newly made strand.

Steps

1. Hydrogen bonds between the bases are broken, so the two strands separate. Both strands act as templates. (You do not need to name other enzymes involved, such as the one that unwinds the helix.)
2. Free DNA nucleotides in the nucleus pair with exposed bases by complementary base pairing.
3. DNA polymerase joins the new nucleotides to the growing strand by forming phosphodiester bonds.
4. DNA ligase joins separate sections of new DNA together, again forming phosphodiester bonds.
5. Two identical DNA molecules result, each with one old and one new strand.

Leading and lagging strands

DNA polymerase can only add nucleotides to the 3′ end of a growing strand. So a new strand is always built in the 5′ to 3′ direction.

• Leading strand: the template runs 3′ to 5′ in the direction the fork is opening. DNA polymerase follows the fork and builds this new strand continuously.
• Lagging strand: the template runs the other way. DNA polymerase has to work away from the fork, so this strand is built discontinuously in short fragments as more template is exposed.
• DNA ligase joins these fragments into one continuous strand.

The difference between the strands is a direct result of the 5′ to 3′ rule. Always state that link in an answer.

6.2 Genes and the genetic code

A gene is a sequence of nucleotides that forms part of a DNA molecule and codes for a polypeptide.

The genetic code is a triplet code. Each set of three DNA bases either codes for a specific amino acid or corresponds to a start or stop signal. On mRNA these triplets are called codons.

• Start codon: AUG on mRNA. It also codes for methionine.
• Stop codons: UAA, UAG and UGA on mRNA. They code for no amino acid and end translation.
• The code is universal: the same triplets code for the same amino acids in almost all organisms.
• Most amino acids have more than one codon. This matters when you explain mutations.

6.2 Transcription (in the nucleus)

1. The gene unwinds and hydrogen bonds between the two strands break in that region.
2. Only one strand is copied. This is the transcribed or template strand. The other is the non-transcribed strand.
3. RNA polymerase moves along the template strand. Free RNA nucleotides pair with it by complementary base pairing (A on DNA pairs with U on RNA).
4. RNA polymerase joins the RNA nucleotides with phosphodiester bonds.
5. At the end of the gene, the RNA molecule is released and the DNA re-forms its double helix.

Because the RNA is complementary to the template strand, its sequence matches the non-transcribed strand, except that it has U where the DNA has T.

Splicing in eukaryotes

The first RNA made is the primary transcript. It contains non-coding sequences called introns and coding sequences called exons. The introns are removed and the exons are joined together to form mRNA. The mRNA then leaves the nucleus through a nuclear pore.

6.2 Translation (in the cytoplasm)

1. mRNA attaches to a ribosome.
2. Each tRNA molecule carries one specific amino acid. It has a three-base anticodon that is complementary to one codon.
3. The first tRNA’s anticodon pairs with the start codon AUG. A second tRNA pairs with the next codon.
4. A peptide bond forms between the two amino acids.
5. The ribosome moves along one codon. The first tRNA leaves and can pick up another amino acid.
6. This repeats until a stop codon is reached. The polypeptide is released.

Worked example 2 – from template strand to polypeptide

Template strand, read 3′ to 5′: TAC AAG CCG TTG ATT

Template (3'→5'):  TAC  AAG  CCG  TTG  ATT
mRNA     (5'→3'):  AUG  UUC  GGC  AAC  UAA
Amino acid:        Met  Phe  Gly  Asn  STOP

Using a codon table: AUG = Met, UUC = Phe, GGC = Gly, AAC = Asn, UAA = stop. The polypeptide has 4 amino acids. The tRNA that brings phenylalanine has anticodon AAG, the same letters as the template triplet.

Worked example 3 – how long is the coding sequence?

A polypeptide has 300 amino acids. How many mRNA nucleotides code for it, from the start codon to the stop codon inclusive?

300 amino acids × 3 nucleotides = 900 (this includes AUG for the first Met)
Add the stop codon: 900 + 3 = 903 nucleotides

The gene in DNA will be longer still, because it also contains introns.

6.2 Gene mutations

A gene mutation is a change in the sequence of base pairs in a DNA molecule that may result in an altered polypeptide.

Type	What happens	Possible effect on the polypeptide
Substitution	One base pair replaced by another	No change if the new triplet codes for the same amino acid; one amino acid changed; or a new stop codon that makes the polypeptide shorter
Deletion	One base pair removed	Frame shift: every triplet after the mutation is read differently, so many amino acids change; a stop codon may appear early
Insertion	One base pair added	Frame shift, with the same range of effects as a deletion

A changed amino acid sequence may change the tertiary structure, so a protein such as an enzyme may not function. A deletion or insertion of three base pairs removes or adds one amino acid without a frame shift.

Common errors

• Writing that DNA contains ribose, or that RNA contains thymine.
• Saying A–T has three hydrogen bonds. It is C–G that has three.
• Describing replication as “conservative” or forgetting to say each new molecule has one old strand.
• Saying DNA polymerase breaks hydrogen bonds. Its role is joining nucleotides with phosphodiester bonds.
• Saying DNA ligase joins bases. It joins sections of the sugar–phosphate backbone.
• Explaining the lagging strand without linking it to DNA polymerase working only 5′ to 3′.
• Mixing up codon (on mRNA) and anticodon (on tRNA).
• Placing transcription in the cytoplasm or translation in the nucleus of a eukaryotic cell.
• Counting the stop codon as coding for an amino acid.
• Saying “a mutation changes the protein” without saying how the triplet and amino acid change.

Where next

• Definitions, tables and a quick self-test: nucleic acids and protein synthesis revision notes
• Original exam-style questions with marked answers: nucleic acids and protein synthesis practice
• Where ribosomes and the nuclear envelope fit in the cell: cell structure study guide

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.