What is protein synthesis and why is it important in Biology?

Protein synthesis is the process by which cells build proteins, which are essential molecules for various cellular functions. It involves two main stages: transcription and translation. This process is crucial because proteins play vital roles in the structure, function, and regulation of the body's tissues and organs. Understanding protein synthesis is fundamental in Edexcel International A Levels Biology as it links to topics like cell biology, genetics, and health.

How does transcription differ from translation in protein synthesis?

Transcription and translation are two distinct stages of protein synthesis. Transcription occurs in the nucleus where the DNA sequence of a gene is copied into messenger RNA (mRNA). Translation, on the other hand, occurs in the cytoplasm where the mRNA is read by ribosomes to assemble amino acids into a polypeptide chain, forming a protein. These processes are integral to the Edexcel International A Levels Biology curriculum, highlighting the flow of genetic information from DNA to protein.

What role do ribosomes play in protein synthesis?

Ribosomes are the cellular structures where protein synthesis occurs. They facilitate the translation of mRNA into a polypeptide chain by reading the sequence of codons on the mRNA and matching them with the appropriate amino acids. This process is essential for the production of proteins, which are critical for cell function and health. Understanding the role of ribosomes is important for students studying Edexcel International A Levels Biology, as it connects to broader topics like cellular function and molecular biology.

Why is the genetic code considered universal in the context of protein synthesis?

The genetic code is considered universal because it is nearly the same in all organisms, meaning that the same codons specify the same amino acids across different species. This universality is a key concept in protein synthesis, as it underscores the shared evolutionary heritage of life on Earth. For Edexcel International A Levels Biology students, this concept is important as it relates to genetics, evolution, and molecular biology.

How do mutations affect protein synthesis and what are the potential implications for health?

Mutations are changes in the DNA sequence that can affect protein synthesis by altering the mRNA and, consequently, the amino acid sequence of a protein. This can lead to nonfunctional proteins or diseases such as cystic fibrosis or sickle cell anemia. Understanding mutations is crucial for Edexcel International A Levels Biology students, as it links to topics like genetics, molecular biology, and health, highlighting the impact of genetic changes on organismal function and disease.

Why is "Defining 'degenerate' as 'there are lots of codons' or 'the code is degenerate'" a common mistake in Protein Synthesis?

Why it happens: Students restate the word instead of explaining the relationship between codons and amino acids. How to avoid it: State the relationship: **more than one triplet/codon codes for the same amino acid**. A definition must not use the term being defined. Source: Edexcel IAL Biology Unit 1 examiner reports

Why is "Writing thymine in an mRNA sequence (or pairing A with T in RNA)" a common mistake in Protein Synthesis?

Why it happens: Habit from DNA base pairing carries over to transcription and translation. How to avoid it: RNA uses **uracil, not thymine**. When transcribing, pair the DNA base adenine with **uracil** on the mRNA (A-U), and remember the mRNA contains U everywhere DNA would have T. Source: Edexcel IAL Biology Unit 1 examiner reports

Why is "Saying tRNA carries a 'codon' instead of an 'anticodon'" a common mistake in Protein Synthesis?

Why it happens: The two terms sound similar and both refer to triplets of bases. How to avoid it: mRNA has **codons**; tRNA has **anticodons** that are complementary to the codons. The anticodon pairs with the codon during translation. Source: Edexcel IAL Biology Unit 1 examiner reports

Why is "Adding an amino acid for the stop codon when reading a codon table" a common mistake in Protein Synthesis?

Why it happens: Students translate every codon mechanically and forget that stop codons signal termination. How to avoid it: A **stop codon (e.g. UAA, UAG, UGA) codes for no amino acid** — it ends translation. Do not include it as an amino acid when writing out the polypeptide. Source: Edexcel IAL Biology Unit 1 examiner reports

Why is "Naming the bond between amino acids as a 'glycosidic' or 'phosphodiester' bond" a common mistake in Protein Synthesis?

Why it happens: Bond names from carbohydrates and nucleic acids get muddled together. How to avoid it: Amino acids in a polypeptide are joined by **peptide bonds** (condensation). Phosphodiester bonds join the nucleotides within the mRNA backbone — a different bond made during transcription. Source: Edexcel IAL Biology Unit 1 examiner reports

Edexcel International A Levels Biology (XBI11-YBI11)

Protein Synthesis

Q: Why is "Transcribing the coding (sense) strand instead of the template (antisense) strand" a common mistake in Protein Synthesis?

Why it happens: Questions often quote the coding strand because it matches the mRNA, and students copy it directly without thinking about which strand is read. How to avoid it: RNA polymerase reads the **template (antisense) strand**, and mRNA is **complementary** to it. The mRNA only LOOKS like the coding strand (with U for T) — build it complementary to the template. Source: Edexcel IAL Biology Unit 1 examiner reports

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Short Notes - Protein Synthesis

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Detailed Study Notes

Detailed notes on Molecules, Diet, Transport and Health for Edexcel International A Levels Biology, covering key concepts, explanations, examples, and exam-focused revision points.

Protein Synthesis — Pearson Edexcel International A Level Biology (XBI11/YBI11) Study Notes (Unit 1)

The genetic code (triplet, non-overlapping, degenerate), the definition of a gene, the two types of RNA (mRNA and tRNA) and how they differ from DNA, then transcription and translation step by step — including the roles of RNA polymerase, ribosomes, codons, anticodons and peptide bonds.

At a glance

Gene = a sequence of DNA bases that codes for the amino acid sequence of one polypeptide (and so a protein).
Genetic code is a TRIPLET code — three bases (one codon) code for one amino acid.
Non-overlapping — each base belongs to only one triplet; degenerate — most amino acids have more than one codon.
mRNA is a single-stranded copy of one gene; tRNA is a folded clover-leaf with an anticodon and an amino-acid attachment site.
RNA differs from DNA: single-stranded, ribose (not deoxyribose), uracil instead of thymine, much shorter.
Transcription — RNA polymerase copies the template (antisense) strand into mRNA in the nucleus.
Translation — ribosomes read mRNA codons; tRNA anticodons pair with them and bring amino acids; peptide bonds join them into a polypeptide.
Template/antisense strand is read; coding/sense strand is NOT — mRNA is complementary to the template and identical (U for T) to the coding strand.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XBI11-YBI11 syllabus (2018-onwards).

2.11 — State that a gene is a sequence of DNA bases that codes for the sequence of amino acids in a polypeptide.
2.11 — Describe the genetic code as triplet, non-overlapping and degenerate, and explain what each property means.
2.12 — Describe the structure of messenger RNA (mRNA) and transfer RNA (tRNA) and state how RNA differs from DNA.
2.13 — Describe transcription: the role of RNA polymerase, the template (antisense) strand and the synthesis of mRNA.
2.13 — Describe translation: the role of ribosomes, codons, anticodons, complementary codon–anticodon pairing and peptide-bond formation.
2.13 — Explain the roles of mRNA, tRNA and ribosomes in protein synthesis.

Genes and the genetic code

A gene codes for one polypeptide; the code is triplet, non-overlapping and degenerate.

A gene is a length of DNA — a sequence of bases — that codes for the sequence of amino acids in a polypeptide (and therefore for a protein). The order of bases in the gene determines the order of amino acids, which determines the protein's primary structure and ultimately its three-dimensional shape and function.

The genetic code is the set of rules linking DNA base sequences to amino acids. You must be able to state AND explain three properties:

Property	What it means	Why it matters
Triplet	Each amino acid is coded for by a sequence of three bases (a triplet / codon).	There are 4 bases; $4^1 = 4$ and $4^2 = 16$ are too few for 20 amino acids, but $4^3 = 64$ combinations are enough.
Non-overlapping	Each base is part of only one triplet; the code is read in fixed blocks of three from a start point.	A single base substitution changes only one triplet, so usually only one amino acid is affected.
Degenerate	Most amino acids are coded for by more than one triplet (e.g. leucine has six codons).	A base substitution in the third position often gives the same amino acid, so the protein is unchanged (a silent mutation).

There are 64 possible triplets but only 20 amino acids, plus there are stop codons that signal the end of a polypeptide and a start codon (AUG, which also codes for methionine) that signals the beginning.

Edexcel discipline. When a question says "explain what is meant by triplet / non-overlapping / degenerate", do not just restate the word. For degenerate you must say "more than one triplet/codon codes for the same amino acid" — "lots of codons" alone scores nothing.

The overall flow of genetic information is summarised by the central dogma: the triplet base sequence of a gene is copied into mRNA by transcription, then read three bases at a time during translation to build a polypeptide.

The central dogma: a gene's triplet base sequence is transcribed into mRNA in the nucleus, then translated at a ribosome into a polypeptide — each codon (three bases) specifying one amino acid.

Gene = DNA sequence coding for the amino acid sequence of one polypeptide.
Triplet: 3 bases = 1 amino acid.
Non-overlapping: each base is in only one triplet.
Degenerate: most amino acids have more than one triplet.

See the full worked example for protein synthesis →

RNA — mRNA, tRNA and how RNA differs from DNA

Two RNA molecules carry the message and the amino acids; RNA is single-stranded, uses ribose and uracil.

Protein synthesis uses ribonucleic acid (RNA). Like DNA, RNA is a polymer of nucleotides (pentose sugar + phosphate + base), but it differs in four ways:

Feature	DNA	RNA
Strands	Double-stranded helix	Single-stranded
Pentose sugar	Deoxyribose	Ribose
Bases	A, T, C, G	A, U (uracil), C, G — uracil replaces thymine
Length	Very long (whole chromosome)	Short (one gene / one amino acid)

Messenger RNA (mRNA) is a single-stranded molecule made in the nucleus during transcription. It is a copy of one gene and carries the coded message from the DNA in the nucleus to the ribosomes in the cytoplasm. It is read in codons (triplets of bases). It is relatively short-lived.

Transfer RNA (tRNA) is a small single strand that folds back on itself into a clover-leaf shape, held by hydrogen bonds between complementary bases within the molecule. It has:

an anticodon — three exposed bases at one end that pair with a complementary codon on mRNA;
an amino-acid attachment (binding) site at the other end (the —CCA / 3′ end) that carries the specific amino acid corresponding to its anticodon.

Each amino acid is brought to the ribosome by a tRNA whose anticodon matches the appropriate mRNA codon.

tRNA folds into a clover-leaf: an anticodon at one end reads the mRNA codon, while the matching amino acid is carried at the 3′ attachment site.

RNA: single-stranded, ribose, uracil not thymine, short.
mRNA: copy of one gene, read in codons, carries message to ribosome.
tRNA: clover-leaf, has an anticodon + amino-acid attachment site.

Transcription — copying the gene into mRNA

RNA polymerase reads the template (antisense) strand and builds a complementary mRNA in the nucleus.

Transcription makes a single-stranded mRNA copy of one gene. It happens in the nucleus.

Step by step:

The enzyme DNA helicase (or RNA polymerase itself) causes part of the DNA double helix to unwind and the hydrogen bonds between the two strands to break, exposing the bases of the gene.
Only one of the two DNA strands is copied — the template strand (also called the antisense strand). The other strand is the coding (sense) strand and is not transcribed.
RNA polymerase moves along the template strand. Free RNA nucleotides line up against the exposed template bases by complementary base pairing: A–U, T–A, C–G, G–C (note: the DNA base A pairs with the RNA base uracil).
RNA polymerase joins adjacent RNA nucleotides together by forming phosphodiester bonds in the sugar–phosphate backbone, building the mRNA molecule.
When RNA polymerase reaches a stop signal, the completed mRNA detaches, the DNA re-forms its double helix, and the mRNA leaves the nucleus through a nuclear pore to reach a ribosome.

Why the template strand matters. Because mRNA is built complementary to the template strand, the mRNA base sequence is identical to the coding (sense) strand except that uracil replaces thymine. This is a classic exam trap — see the worked example.

DNA coding (sense) strand	5′ — A T G C A G — 3′
DNA template (antisense) strand	3′ — T A C G T C — 5′
mRNA transcribed	5′ — A U G C A G — 3′

Notice the mRNA matches the coding strand (with U for T) but is actually built against the template strand.

RNA polymerase moves 3′→5′ along the template (antisense) strand. Free RNA nucleotides pair complementarily (A–U, T–A, C–G, G–C), so the mRNA is built complementary to the template — and therefore identical to the coding (sense) strand except that uracil replaces thymine.

Occurs in the nucleus; product is mRNA.
DNA unwinds; hydrogen bonds between strands break.
RNA polymerase reads the template (antisense) strand.
Free RNA nucleotides pair complementarily (A–U, T–A, C–G, G–C).
mRNA is complementary to template, identical to coding strand (U for T).

See the full worked example for protein synthesis →

Translation — building the polypeptide at the ribosome

Ribosomes read mRNA codons; tRNA anticodons pair with them and peptide bonds join the amino acids.

Translation converts the codon sequence on mRNA into a sequence of amino acids — a polypeptide. It happens at a ribosome in the cytoplasm (or on the rough endoplasmic reticulum).

Step by step:

The mRNA binds to a ribosome. The ribosome holds two codons (six bases) at a time within its two binding sites.
A tRNA molecule with an anticodon complementary to the first (start) codon binds. The first codon is AUG (start), so the tRNA with the anticodon UAC brings the amino acid methionine. Pairing follows the base-pairing rules (A–U, C–G).
A second tRNA, carrying its amino acid, binds to the next codon by complementary codon–anticodon pairing.
The two amino acids held side by side are joined by a peptide bond (a condensation reaction, catalysed by the ribosome).
The ribosome moves along the mRNA by one codon (translocation). The first tRNA, now empty, is released and can be reloaded with another amino acid. The next tRNA brings the next amino acid, a new peptide bond forms, and the chain grows.
This repeats until the ribosome reaches a stop codon, which codes for no amino acid. The completed polypeptide is released, and the ribosome separates from the mRNA.

The polypeptide then folds into its secondary, tertiary and (if relevant) quaternary structure to become a functional protein.

Roles to remember (a frequent 3-mark "state the roles" question):

mRNA — carries the coded sequence (codons) from DNA in the nucleus to the ribosome; the sequence of codons determines the order of amino acids.
tRNA — brings specific amino acids to the ribosome; its anticodon pairs with the complementary mRNA codon so the correct amino acid is added in the correct order.
Ribosome — the site of translation; holds the mRNA and tRNAs in place and catalyses peptide-bond formation between adjacent amino acids.

The ribosome holds two codons at a time. tRNA anticodons (UAC, GAA) pair complementarily with mRNA codons (AUG, CUU) and deliver specific amino acids (methionine, leucine); the ribosome catalyses a peptide bond between them, then moves one codon along so the polypeptide grows until a stop codon.

Occurs at a ribosome; product is a polypeptide.
Ribosome reads mRNA codons; tRNA anticodons pair with them.
Codon–anticodon pairing is complementary (A–U, C–G).
Adjacent amino acids joined by peptide bonds (condensation).
Stops at a stop codon; polypeptide released and folds.

See the full worked example for protein synthesis →

Quick recap

A gene codes for the amino acid sequence of one polypeptide.
The code is triplet (3 bases = 1 amino acid), non-overlapping (each base in one triplet) and degenerate (most amino acids have more than one triplet).
RNA is single-stranded, uses ribose and uracil, and is short; mRNA carries codons, tRNA carries amino acids via an anticodon.
Transcription: RNA polymerase copies the template (antisense) strand into mRNA in the nucleus.
mRNA is complementary to the template strand and identical to the coding strand (U replacing T).
Translation: ribosomes read codons, tRNA anticodons pair complementarily, peptide bonds join amino acids into a polypeptide.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Triplet code: three bases (one codon) code for one amino acid.
Non-overlapping: each base is part of only one triplet.
Degenerate: most amino acids are coded for by more than one triplet/codon.
RNA: single-stranded, ribose, uracil (not thymine), short.
Transcription: RNA polymerase reads the template (antisense) strand; mRNA complementary to it.
Translation: complementary codon–anticodon pairing; amino acids joined by peptide bonds.
Roles — mRNA carries the message; tRNA brings amino acids; ribosome is the site and catalyses peptide bonds.

How it’s examined

Protein synthesis is examined on Unit 1 (Lifestyle, Transport, Genes and Health). Typical questions: define a gene or explain the three properties of the genetic code (2–3 marks, AO1); describe transcription or translation as a sequence of steps (4–6 marks, often levels-based for the 6-markers); work out an mRNA, tRNA anticodon or amino-acid sequence from a given DNA strand or codon table (2–4 marks, AO2); and 'state the roles of mRNA, tRNA and ribosomes' (3 marks). The most common lost marks are confusing template vs coding strand, writing thymine instead of uracil in RNA, and defining 'degenerate' circularly.

Sources: Pearson Edexcel International Advanced Level Biology (YBI11/XBI11) specification (Issue 2, 2018 onwards); Edexcel IAL Biology Unit 1 examiner reports; Edexcel IAL Biology Unit 1 sample assessment materials and mark schemes. Last reviewed 2026-06-24.

Take this whole topic with you

Step-by-step worked examples — Protein Synthesis

Step-by-step solutions to past-paper-style questions on protein synthesis, written exactly the way a tutor would explain them at the board.

Question type:

1Defining a gene (2 marks)
Getting startedIdentify & classify• gene, Unit 1, AO1
Question
State what is meant by the term gene. (2 marks)
Step-by-step solution
1. Step 1
  Identify what a gene is made of: a gene is a sequence (length) of DNA bases.
2. Step 2
  State what it codes for: it codes for the sequence of amino acids in a polypeptide (and therefore for a protein).
Answer
A gene is a sequence of DNA bases that codes for the sequence of amino acids in a polypeptide.
Examiner tip
Marks: (1) sequence/length of DNA (bases); (2) codes for the amino acid sequence of a polypeptide. 'A section of DNA that makes a protein' is too vague — the link to the AMINO ACID SEQUENCE of a polypeptide is the credited point.
2How RNA differs from DNA (3 marks)
Getting startedIdentify & classify• RNA, DNA, mRNA, Unit 1
Question
State three ways in which the structure of an mRNA molecule differs from that of a DNA molecule. (3 marks)
Step-by-step solution
1. Step 1
  Number of strands: mRNA is single-stranded, whereas DNA is double-stranded (a double helix).
2. Step 2
  Pentose sugar: mRNA contains ribose, whereas DNA contains deoxyribose.
3. Step 3
  Bases: mRNA contains the base uracil in place of thymine; DNA contains thymine. (A fourth valid point: mRNA is much shorter — a copy of one gene.)
Answer
mRNA is single-stranded (DNA double); contains ribose (DNA deoxyribose); contains uracil instead of thymine; and is much shorter.
Examiner tip
Marks: any 3 of — single-stranded vs double; ribose vs deoxyribose; uracil replaces thymine; shorter. Each point must CONTRAST the two molecules (state the DNA feature too) to be safe.
3Explaining the three properties of the genetic code (3 marks)
Building confidenceIdentify & classify• genetic code, triplet, degenerate, Unit 1, AO1
Question
The genetic code is described as triplet, non-overlapping and degenerate. Explain what is meant by each of these three terms. (3 marks)
Step-by-step solution
1. Step 1
  Triplet: a sequence of three bases (one codon) codes for one amino acid.
2. Step 2
  Non-overlapping: each base is part of only one triplet/codon — the code is read in fixed, separate blocks of three.
3. Step 3
  Degenerate: most amino acids are coded for by more than one triplet/codon (e.g. leucine has six codons).
Answer
Triplet = 3 bases code for 1 amino acid; non-overlapping = each base belongs to only one triplet; degenerate = most amino acids have more than one codon.
Examiner tip
Marks: (1) triplet = three bases per amino acid; (2) non-overlapping = each base in one triplet only; (3) degenerate = more than one codon per amino acid. The most common 0-mark answer is defining 'degenerate' as 'there are lots of codons' — you must say MORE THAN ONE CODON CODES FOR THE SAME AMINO ACID.
4Template strand vs coding strand (4 marks)
Building confidenceWord problem• transcription, template strand, coding strand, Unit 1, AO2
Question
Part of the coding (sense) strand of a gene has the base sequence 5'-A T G C A G-3'. (a) Write the base sequence of the template (antisense) strand. (b) Write the base sequence of the mRNA transcribed from this gene. (c) Explain why the mRNA sequence is identical to the coding strand apart from one base. (4 marks)
Step-by-step solution
1. Step 1
  (a) The template strand is complementary and antiparallel to the coding strand, using DNA base pairing (A-T, C-G). So the template reads 3'-T A C G T C-5'.
2. Step 2
  (b) mRNA is built complementary to the TEMPLATE strand, with uracil pairing to adenine. Pairing T-A, A-U, C-G, G-C, T-A, C-G gives mRNA 5'-A U G C A G-3'.
3. Step 3
  (c) The mRNA is complementary to the template strand; the coding strand is ALSO complementary to the same template strand, so mRNA and coding strand have the same sequence.
4. Step 4
  The single difference is that RNA contains uracil where DNA (the coding strand) contains thymine.
Answer
Template: 3'-TACGTC-5'. mRNA: 5'-AUGCAG-3'. mRNA matches the coding strand because both are complementary to the template; the only difference is U replacing T.
Examiner tip
Marks: (1) correct complementary template strand; (2) correct mRNA with U not T; (3) both mRNA and coding strand complementary to the template; (4) U replaces T. The classic trap is transcribing the CODING strand by mistake (gives the wrong, complementary mRNA) — RNA polymerase reads the TEMPLATE strand.
5Working out a polypeptide from a codon table (4 marks)
StretchMulti-step problem• translation, codon table, polypeptide, Unit 1, AO2
Question
The template (antisense) strand of a short gene reads 3'-T A C A A A G G C A T T-5'. Using the mRNA codon table below, determine the sequence of amino acids in the polypeptide produced. Codon table: AUG = methionine (start); UUU = phenylalanine; CCG = proline; UAA = stop. (4 marks)
Step-by-step solution
1. Step 1
  Transcribe the mRNA complementary to the template (A-U, T-A, C-G, G-C). Template 3'-TAC AAA GGC ATT-5' gives mRNA 5'-AUG UUU CCG UAA-3'.
  $3'\text{-TAC AAA GGC ATT-}5' \;\to\; 5'\text{-AUG UUU CCG UAA-}3'$
2. Step 2
  Split the mRNA into codons (read in non-overlapping blocks of three from the start): AUG | UUU | CCG | UAA.
3. Step 3
  Use the codon table to translate each codon: AUG = methionine, UUU = phenylalanine, CCG = proline, UAA = STOP (codes for no amino acid).
4. Step 4
  The stop codon ends translation and adds no amino acid, so the polypeptide is methionine - phenylalanine - proline (three amino acids).
Answer
mRNA = AUG-UUU-CCG-UAA → polypeptide: methionine - phenylalanine - proline (the stop codon adds no amino acid).
Examiner tip
Marks: (1) correct mRNA from the template (U for T, complementary); (2) correctly split into the four codons; (3) correct three amino acids named; (4) recognise UAA is a STOP codon and adds no amino acid. Two frequent errors: translating the template directly without transcribing, and adding an amino acid for the stop codon.
6Codon-anticodon pairing and the role of tRNA (4 marks)
StretchWord problem• translation, anticodon, tRNA, Unit 1, AO2
Question
During translation an mRNA codon reads 5'-G C A-3'. (a) Give the base sequence of the tRNA anticodon that pairs with this codon. (b) Explain how this pairing ensures the correct amino acid is added to the growing polypeptide. (4 marks)
Step-by-step solution
1. Step 1
  (a) The anticodon is complementary to the codon (A-U, C-G, G-C). Codon 5'-GCA-3' pairs with anticodon 3'-CGU-5' (i.e. CGU).
2. Step 2
  Each tRNA has a specific anticodon and carries one specific amino acid bound to its attachment site, so a particular anticodon is always linked to a particular amino acid.
3. Step 3
  The anticodon pairs only with the complementary codon on the mRNA by hydrogen bonding, so a tRNA can only bind where its anticodon matches.
4. Step 4
  Because the codon sequence on the mRNA is fixed (transcribed from the gene), the order of codons determines the order in which tRNAs bind, and therefore the correct amino acids are added in the correct sequence to build the polypeptide.
Answer
Anticodon = CGU. tRNA carries a specific amino acid; its anticodon pairs only with the complementary mRNA codon, so the codon order sets the amino acid order.
Examiner tip
Marks: (1) correct complementary anticodon CGU; (2) each tRNA carries a specific amino acid; (3) anticodon pairs with complementary codon only; (4) codon order on mRNA determines amino acid order. Watch base-pairing direction — anticodon bases must be complementary to the codon, not the same.

Model Answers — Protein Synthesis

High-scoring sample answers for protein synthesis on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Edexcel IAL Biology Unit 1 style2 marks
Q (2 marks). State what is meant by a gene and explain what is meant by saying the genetic code is triplet.
Model answer
A gene is a sequence of DNA bases that codes for the sequence of amino acids in a polypeptide (and therefore for a particular protein).

The genetic code being triplet means that each amino acid is coded for by a sequence of three bases (one codon) in the DNA or mRNA.
Why this scores
Marks: (1) gene = DNA base sequence coding for the amino acid sequence of a polypeptide; (2) triplet = three bases code for one amino acid. Both definitions must avoid using the term being defined.
Question 2
Edexcel IAL Biology Unit 1 style3 marks
Q (3 marks). State the roles of mRNA, tRNA and ribosomes in protein synthesis.
Model answer
mRNA carries the coded genetic information, as a sequence of codons, from the DNA in the nucleus to the ribosomes in the cytoplasm; the order of its codons determines the order of amino acids.

tRNA brings a specific amino acid to the ribosome; its anticodon pairs with the complementary codon on the mRNA so that the correct amino acid is positioned in the correct place.

A ribosome is the site of translation: it holds the mRNA and the tRNAs together and catalyses the formation of peptide bonds between adjacent amino acids to build the polypeptide.
Why this scores
Marks: (1) mRNA carries the (codon) sequence from nucleus to ribosome / determines amino acid order; (2) tRNA brings specific amino acids, anticodon pairs with codon; (3) ribosome = site of translation / joins amino acids by peptide bonds. One clear role for each of the three molecules.
Question 3
Edexcel IAL Biology Unit 1 style5 marks
Q (5 marks). Describe how a molecule of mRNA is produced from a gene during transcription.
Model answer
Transcription takes place in the nucleus. The relevant part of the DNA double helix unwinds and the hydrogen bonds between the two strands break, exposing the bases of the gene.

Only one of the two DNA strands acts as a template — the template (antisense) strand; the other, the coding (sense) strand, is not transcribed.

Free RNA nucleotides line up alongside the exposed bases of the template strand by complementary base pairing: cytosine pairs with guanine, and the base adenine on the DNA pairs with uracil on the RNA (uracil replaces thymine).

The enzyme RNA polymerase moves along the template strand and joins adjacent RNA nucleotides together by forming phosphodiester bonds in the sugar-phosphate backbone, producing a single-stranded mRNA molecule.

When RNA polymerase reaches the end of the gene the completed mRNA detaches, the DNA double helix re-forms, and the mRNA passes out of the nucleus through a nuclear pore to a ribosome.
Why this scores
Marks (point-based, max 5): (1) occurs in the nucleus / DNA unwinds and hydrogen bonds break; (2) template (antisense) strand is used; (3) complementary base pairing with free RNA nucleotides, uracil pairs with adenine; (4) RNA polymerase joins nucleotides / phosphodiester bonds; (5) mRNA detaches and leaves the nucleus through a nuclear pore. Naming the template strand and using uracil (not thymine) are the most commonly dropped marks.
Question 4
Edexcel IAL Biology Unit 1 style6 marks
Q (6 marks). Describe how a polypeptide is assembled during translation. (This part of the question is marked using levels.)
Model answer
Translation takes place at a ribosome in the cytoplasm. The mRNA binds to the ribosome, which holds two codons (six bases) at a time.

A tRNA molecule whose anticodon is complementary to the first (start) codon of the mRNA binds to it. Pairing follows the base-pairing rules, with the start codon AUG paired by the anticodon UAC, bringing the first amino acid (methionine).

A second tRNA, carrying its own specific amino acid, then binds to the next mRNA codon by complementary codon-anticodon pairing.

The two amino acids now held next to each other are joined by a peptide bond, formed by a condensation reaction that is catalysed by the ribosome.

The ribosome then moves along the mRNA by one codon (translocation). The first, now empty, tRNA is released and can collect another copy of its amino acid, while the next tRNA brings the next amino acid and a further peptide bond forms.

This process repeats, codon by codon, so the polypeptide chain grows in the order set by the mRNA codons, until the ribosome reaches a stop codon, which codes for no amino acid. The completed polypeptide is then released and folds into its three-dimensional structure.
Why this scores
Levels-based (max 6). Level 3 (5-6): a clear, well-ordered sequence covering binding of mRNA to the ribosome, complementary codon-anticodon pairing bringing specific amino acids, peptide-bond formation, ribosome moving along by one codon, and termination at a stop codon. Level 2 (3-4): several correct steps but gaps or muddled order. Level 1 (1-2): a few isolated correct points. This answer reaches Level 3 because it is fully sequenced and uses precise terminology (anticodon, complementary pairing, peptide bond, translocation, stop codon) throughout. The A*-distinguishing detail is naming translocation and stating the stop codon adds no amino acid. Saying 'tRNA codons' instead of 'tRNA anticodons' is penalised.
Question 5
Edexcel IAL Biology Unit 1 style6 marks
Q (6 marks). Explain how the fact that the genetic code is degenerate and non-overlapping can reduce the effect of a base-substitution mutation on the protein produced.
Model answer
A base-substitution mutation changes a single base in the DNA, and therefore changes a single base in one mRNA codon.

Because the code is non-overlapping, each base belongs to only one triplet/codon. A substitution therefore alters only one codon, rather than shifting and changing every codon downstream, so at most one amino acid in the polypeptide can be affected.

Because the code is degenerate, most amino acids are coded for by more than one codon, and codons that differ only in their third base often code for the same amino acid. So a substitution — particularly in the third position of a codon — may produce a different codon that still codes for the same amino acid.

If the same amino acid is still coded for, the primary structure (amino acid sequence) of the polypeptide is unchanged. This is called a silent mutation, and the protein's tertiary structure and function are unaffected.

Even where the amino acid does change, only one amino acid is altered (because the code is non-overlapping), so the change to the protein is usually small. If that single amino acid is not at the active site or another critical position, the protein's overall shape and function may be little affected.

Together, the non-overlapping nature limits the change to one codon, and the degeneracy of the code means that change frequently has no effect on the amino acid sequence at all — both reduce the impact of the mutation on the protein.
Why this scores
Marks (point-based, max 6): (1) substitution changes one base, so one codon; (2) non-overlapping means each base is in only one triplet; (3) so only one codon/amino acid can be affected; (4) degenerate = more than one codon per amino acid; (5) the new codon may code for the same amino acid (silent mutation); (6) so the amino acid sequence / primary structure and protein function are unchanged. Linking 'third-base degeneracy' to a silent mutation is the strong AO2 point.
Question 6
Edexcel IAL Biology Unit 1 style4 marks
Q (4 marks). Describe the structure of a tRNA molecule and explain how its structure is related to its function in translation.
Model answer
A tRNA molecule is a short, single-stranded RNA that folds back on itself into a clover-leaf shape, held by hydrogen bonds between complementary bases within the molecule.

At one end it has an anticodon — a sequence of three exposed bases. This anticodon pairs with the complementary codon on the mRNA, so each tRNA binds only at the correct position on the mRNA.

At the other end it has an amino-acid attachment (binding) site that carries one specific amino acid corresponding to its anticodon.

Because a particular anticodon is always linked to a particular amino acid, the complementary codon-anticodon pairing ensures that the correct amino acid is delivered to the correct place on the growing polypeptide, so the amino acid order matches the codon order on the mRNA.
Why this scores
Marks: (1) single-stranded RNA folded into a clover-leaf shape held by hydrogen bonds; (2) anticodon of three bases / amino-acid attachment site; (3) anticodon pairs with complementary mRNA codon; (4) ensures the correct/specific amino acid is added in the correct order. A structure-function question: each feature must be linked to its role.
Question 7
Edexcel IAL Biology Unit 1 style3 marks
Q (3 marks). The genetic code is described as triplet, non-overlapping and degenerate. Explain what is meant by each of these three terms.
Model answer
Triplet: a sequence of three bases (one codon) codes for one amino acid.

Non-overlapping: each base is part of only one triplet/codon — the sequence is read in fixed, separate blocks of three from a start point, so consecutive triplets do not share bases.

Degenerate: most amino acids are coded for by more than one triplet/codon (for example, leucine is coded for by six different codons).
Why this scores
Marks: (1) triplet = three bases code for one amino acid; (2) non-overlapping = each base belongs to only one triplet; (3) degenerate = more than one codon codes for the same amino acid. The A*-distinguisher is defining 'degenerate' by the codon-to-amino-acid relationship rather than circularly ('lots of codons' scores zero). Each definition must avoid using the word being defined.
Question 8
Edexcel IAL Biology Unit 1 style5 marks
Q (5 marks). Part of the template (antisense) strand of a gene reads 3'-T A C G G A T A A-5'. (a) Write the base sequence of the mRNA transcribed from this strand. (b) Write the base sequence of the coding (sense) strand of this gene. (c) Explain why RNA polymerase uses the template strand and not the coding strand to make mRNA.
Model answer
(a) The mRNA is built complementary to the template strand, with uracil pairing with adenine and the usual C–G pairing. Template 3'-TAC GGA TAA-5' therefore gives mRNA 5'-AUG CCU AUU-3'.

(b) The coding (sense) strand is complementary to the template strand using DNA base pairing (A–T, C–G), so it reads 5'-ATG CCT ATT-3'. Notice it is identical to the mRNA except that thymine replaces uracil.

(c) RNA polymerase joins RNA nucleotides by complementary base pairing with the strand it reads. Only the template strand carries the base sequence that, when copied complementarily, produces an mRNA with the correct codon sequence to specify the right amino acids. If the coding strand were transcribed, the mRNA would be complementary to the coding strand and so would carry the wrong codons, producing the wrong polypeptide.
Why this scores
Marks: (1) correct mRNA 5'-AUGCCUAUU-3' (U for T, complementary); (2) correct coding strand 5'-ATGCCTATT-3'; (3) recognise coding strand = mRNA with T for U; (4) mRNA made by complementary base pairing with the template; (5) transcribing the coding strand would give the wrong codon/amino acid sequence. The A*-point is the cause→effect link: only the template gives an mRNA with the correct codons. The common error is to transcribe the template as if it were the coding strand.
Question 9
Edexcel IAL Biology Unit 1 style6 marks
Q (6 marks). The template (antisense) strand of a short section of a gene reads 3'-T A C C G A G G G A C T-5'. (a) Determine the mRNA codon sequence. (b) Give the tRNA anticodons, in order, that would pair with these codons. (c) Using the codon table, state the amino acid sequence of the polypeptide. Codon table: AUG = methionine (start); GCU = alanine; CCC = proline; UGA = stop.
Model answer
(a) mRNA codons. Transcribe complementary to the template (A–U, T–A, C–G, G–C), reading the template 3'→5'. Template 3'-TAC CGA GGG ACT-5' gives:

mRNA = 5'-AUG GCU CCC UGA-3', i.e. codons AUG | GCU | CCC | UGA.

(b) tRNA anticodons. Each anticodon is complementary to its codon (and so reads the same as the original template triplet). In order they are:

codon AUG → anticodon UAC

codon GCU → anticodon CGA

codon CCC → anticodon GGG

(UGA is a stop codon, so no tRNA/anticodon binds it.)

(c) Amino acid sequence. Using the codon table: AUG = methionine, GCU = alanine, CCC = proline, UGA = STOP (no amino acid). The polypeptide is therefore methionine – alanine – proline (three amino acids); the stop codon ends translation and adds no amino acid.
Why this scores
Marks: (1) correct mRNA with U for T, complementary to template; (2) mRNA correctly split into codons AUG/GCU/CCC/UGA; (3) correct anticodons UAC, CGA, GGG (complementary to codons); (4) recognise UGA binds no tRNA; (5) correct three amino acids named (Met-Ala-Pro); (6) stop codon adds no amino acid. The A*-distinguisher is realising the anticodons read the same as the original template triplets (both complementary to the codon). Common errors: translating the template directly, writing T in RNA, and adding an amino acid for the stop codon.
Question 10
Edexcel IAL Biology Unit 1 style7 marks
Q (7 marks). In a cell-free experiment, a synthetic mRNA was added to a mixture containing ribosomes, tRNAs, amino acids and ATP. The rate of polypeptide synthesis was measured at different temperatures. The results were: 10 °C, 4 amino acids min⁻¹; 25 °C, 11 amino acids min⁻¹; 37 °C, 23 amino acids min⁻¹; 45 °C, 9 amino acids min⁻¹; 55 °C, 0 amino acids min⁻¹. (a) Describe the trend shown by the data. (b) Explain the results in terms of the enzymes and bonding involved in translation.
Model answer
(a) Trend. As temperature increases from 10 °C to 37 °C the rate of polypeptide synthesis increases, rising from 4 to 23 amino acids min⁻¹ (a peak at 37 °C). Above 37 °C the rate falls sharply, to 9 amino acids min⁻¹ at 45 °C and to 0 at 55 °C. The optimum temperature is therefore 37 °C.

(b) Explanation. Translation depends on enzyme activity — the ribosome catalyses peptide-bond formation, and tRNAs are loaded with amino acids by enzymes. As temperature rises towards 37 °C, the kinetic energy of the molecules increases, so there are more frequent, successful enzyme–substrate collisions and more frequent collisions between tRNA anticodons and mRNA codons; the rate of complementary codon–anticodon pairing and peptide-bond formation therefore increases.

Above the optimum, the high temperature gives atoms enough energy to break the hydrogen bonds (and ionic bonds) that hold the ribosomal/enzyme proteins in their precise tertiary structure. The active site changes shape and is no longer complementary to its substrate, so the enzyme is denatured. At 55 °C the enzymes are fully denatured, so no peptide bonds can form and the rate is 0 amino acids min⁻¹.

The temporary hydrogen bonds between codons and anticodons are also disrupted at high temperature, so tRNAs can no longer pair stably with the mRNA, further reducing synthesis.
Why this scores
Levels/point-based (max 7). (a) Describe with direction AND quoted values: increases 4→23 to a peak at 37 °C, then decreases to 0 at 55 °C; optimum 37 °C [2]. (b) up to 37 °C: increased kinetic energy → more enzyme-substrate collisions → faster peptide-bond formation [2]; above optimum: hydrogen/ionic bonds break, tertiary structure/active site changes, enzyme denatured, no longer complementary [2]; at 55 °C rate is 0 / codon-anticodon hydrogen bonds also disrupted [1]. The A*-distinguisher is quoting exact figures with units and using 'complementary active site' and 'denatured' correctly rather than 'killed'.

Key Definitions and Keywords — Protein Synthesis

Definitions to memorise and the exact keywords mark schemes credit for protein synthesis answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Gene
Examiner keyword
A sequence of DNA bases that codes for the sequence of amino acids in a polypeptide (and therefore for a protein).
Genetic code
Examiner keyword
The set of rules by which sequences of three bases (codons) in DNA/mRNA specify the amino acids in a polypeptide. It is triplet, non-overlapping and degenerate.
Triplet code
Examiner keyword
A property of the genetic code in which a sequence of three bases (one codon) codes for one amino acid.
Non-overlapping code
Examiner keyword
A property of the genetic code in which each base is part of only one triplet/codon; the code is read in fixed, separate blocks of three.
Degenerate code
Examiner keyword
A property of the genetic code in which most amino acids are coded for by more than one triplet/codon.
Codon
Examiner keyword
A sequence of three adjacent bases on an mRNA molecule that codes for one amino acid (or, for AUG, also signals the start; some codons signal stop).
Anticodon
Examiner keyword
A sequence of three exposed bases on a tRNA molecule that is complementary to, and pairs with, a specific codon on mRNA during translation.
Messenger RNA (mRNA)
Examiner keyword
A single-stranded RNA molecule, made during transcription, that carries the coded base sequence (as codons) of one gene from the nucleus to a ribosome.
Transfer RNA (tRNA)
Examiner keyword
A short single-stranded RNA folded into a clover-leaf shape, with an anticodon at one end and an amino-acid attachment site at the other, that carries a specific amino acid to the ribosome during translation.
RNA (vs DNA)
Examiner keyword
Ribonucleic acid: a single-stranded polynucleotide containing the pentose sugar ribose and the base uracil in place of thymine, and shorter than DNA.
Transcription
Examiner keyword
The synthesis of a single-stranded mRNA copy of a gene in the nucleus, in which RNA polymerase assembles RNA nucleotides complementary to the template (antisense) DNA strand.
Translation
Examiner keyword
The synthesis of a polypeptide at a ribosome, in which the codon sequence of mRNA is read and tRNA anticodons deliver amino acids that are joined by peptide bonds.
Template (antisense) strand
Examiner keyword
The one DNA strand of a gene that is copied during transcription; RNA nucleotides pair complementarily with it to build the mRNA.
Coding (sense) strand
Examiner keyword
The DNA strand of a gene that is NOT transcribed; its base sequence is the same as the mRNA produced, except that thymine replaces uracil.
RNA polymerase
Examiner keyword
The enzyme that catalyses transcription: it moves along the template DNA strand and joins RNA nucleotides together by forming phosphodiester bonds.
Peptide bond
Examiner keyword
The covalent bond formed by a condensation reaction between the amino group of one amino acid and the carboxyl group of the next, joining amino acids in a polypeptide.

Common Mistakes and Misconceptions — Protein Synthesis

The traps other students keep falling into on protein synthesis questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Defining 'degenerate' as 'there are lots of codons' or 'the code is degenerate'
Edexcel IAL Biology Unit 1 examiner reports
Why it happens
Students restate the word instead of explaining the relationship between codons and amino acids.
How to avoid it
State the relationship: more than one triplet/codon codes for the same amino acid. A definition must not use the term being defined.
✕Writing thymine in an mRNA sequence (or pairing A with T in RNA)
Edexcel IAL Biology Unit 1 examiner reports
Why it happens
Habit from DNA base pairing carries over to transcription and translation.
How to avoid it
RNA uses uracil, not thymine. When transcribing, pair the DNA base adenine with uracil on the mRNA (A-U), and remember the mRNA contains U everywhere DNA would have T.
✕Transcribing the coding (sense) strand instead of the template (antisense) strand
Edexcel IAL Biology Unit 1 examiner reports
Why it happens
Questions often quote the coding strand because it matches the mRNA, and students copy it directly without thinking about which strand is read.
How to avoid it
RNA polymerase reads the template (antisense) strand, and mRNA is complementary to it. The mRNA only LOOKS like the coding strand (with U for T) — build it complementary to the template.
✕Saying tRNA carries a 'codon' instead of an 'anticodon'
Edexcel IAL Biology Unit 1 examiner reports
Why it happens
The two terms sound similar and both refer to triplets of bases.
How to avoid it
mRNA has codons; tRNA has anticodons that are complementary to the codons. The anticodon pairs with the codon during translation.
✕Adding an amino acid for the stop codon when reading a codon table
Edexcel IAL Biology Unit 1 examiner reports
Why it happens
Students translate every codon mechanically and forget that stop codons signal termination.
How to avoid it
A stop codon (e.g. UAA, UAG, UGA) codes for no amino acid — it ends translation. Do not include it as an amino acid when writing out the polypeptide.
✕Naming the bond between amino acids as a 'glycosidic' or 'phosphodiester' bond
Edexcel IAL Biology Unit 1 examiner reports
Why it happens
Bond names from carbohydrates and nucleic acids get muddled together.
How to avoid it
Amino acids in a polypeptide are joined by peptide bonds (condensation). Phosphodiester bonds join the nucleotides within the mRNA backbone — a different bond made during transcription.

Protein Synthesis — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Protein Synthesis

Short Notes - Protein Synthesis

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Defining a gene (2 marks)

2How RNA differs from DNA (3 marks)

3Explaining the three properties of the genetic code (3 marks)

4Template strand vs coding strand (4 marks)

5Working out a polypeptide from a codon table (4 marks)

6Codon-anticodon pairing and the role of tRNA (4 marks)

Gene

Genetic code

Triplet code

Non-overlapping code

Degenerate code

Codon

Anticodon

Messenger RNA (mRNA)

Transfer RNA (tRNA)

RNA (vs DNA)

Transcription

Translation

Template (antisense) strand

Coding (sense) strand

RNA polymerase

Peptide bond

✕Defining 'degenerate' as 'there are lots of codons' or 'the code is degenerate'

✕Writing thymine in an mRNA sequence (or pairing A with T in RNA)

✕Transcribing the coding (sense) strand instead of the template (antisense) strand

✕Saying tRNA carries a 'codon' instead of an 'anticodon'

✕Adding an amino acid for the stop codon when reading a codon table

✕Naming the bond between amino acids as a 'glycosidic' or 'phosphodiester' bond

Protein Synthesis — frequently asked questions