D1.2 Protein synthesis

Protein synthesis links DNA information to functional proteins through transcription, RNA processing, translation, genetic-code reading and post-translational modification in cells.

Syllabus
First assessment 2025
Topic
D1.2
Level
SL

Transcription Copies a Gene into mRNA

Transcription is the synthesis of an RNA strand using one DNA strand as the template.

RNA polymerase binds the selected DNA region, separates a short section, matches RNA nucleotides to exposed template bases and joins the nucleotides into RNA.

DNA template exposed → complementary RNA nucleotides pair → RNA polymerase forms the RNA backbone → RNA separates and DNA re-forms its double helix.

For a protein-coding gene, RNA polymerase produces an mRNA transcript whose base sequence can later be translated.

Transcription synthesizes RNA, not DNA or polypeptide. Promoter/transcription-factor detail belongs to the later HL initiation objective.

Transcription exam focus

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through structured response, commonly using State / Outline / Explain.

Command terms

State / Outline / Explain / Describe / Distinguish

What earns marks

Build the answer around this relationship: Transcription copies one DNA strand into an RNA molecule.

Watch for

Confusing transcription with translation by naming ribosomes or amino acids instead of RNA polymerase and mRNA.

Representative question

Question 1

[Maximum number: 8]

Explain the process of transcription in prokaryotes.

Hydrogen Bonds Let the Template Open and Close

Temporary hydrogen bonds between complementary bases allow an RNA sequence to be specified from a DNA template during transcription.

DNA hydrogen bonds separate locally while the covalent sugar–phosphate backbones remain intact. RNA nucleotides form complementary hydrogen bonds to exposed template bases before RNA polymerase joins them.

DNA template A pairs with RNA U; DNA template T pairs with RNA A; DNA C pairs with RNA G; DNA G pairs with RNA C.

A DNA template segment 3′-TACG-5′ specifies RNA 5′-AUGC-3′: template adenine is represented by uracil in RNA, not thymine.

Complementary does not mean identical. Hydrogen bonds guide base choice; phosphodiester bonds form the stable RNA backbone.

Hydrogen bonding in transcription

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Determine.

Command terms

Determine

What earns marks

Build the answer around this relationship: RNA uses uracil instead of thymine during base pairing.

Watch for

Writing thymine in an mRNA sequence instead of uracil.

Representative question

Question 1

[Maximum number: 1]

The sequence of bases on a short section of the antisense strand of a gene undergoing transcription is shown:

5 CATG 35^{\prime} \text { CATG } 3^{\prime}

What is the sequence of bases on the resulting mRNA?

A

33^{\prime} CATG 55^{\prime}

B

5' GUAC 3'

C

33^{\prime} GUAC 55^{\prime}

D

3GTAC3^{\prime} \mathrm{GTAC} 5'

DNA Stays Stable While Genes Are Read

A single DNA strand can act repeatedly as a transcription template without its base sequence changing.

RNA polymerase reads the template and joins separate RNA nucleotides; it does not consume or replace the DNA bases. After the transcription bubble passes, the DNA strands pair again.

DNA opens locally → RNA copy is synthesized → RNA leaves → original DNA sequence remains conserved and available for later transcription.

A non-dividing neuron may transcribe the same essential gene many times while conserving that gene's DNA sequence throughout the life of the cell.

Stable does not mean inaccessible: local opening permits transcription. The RNA transcript can change or degrade while the DNA template sequence remains unchanged.

Transcription Turns Gene Information into a Message

Transcription is the first stage of gene expression and a key point at which expression can be switched on or off.

Cells contain many genes but do not express all of them at the same time. If a gene is not transcribed, its mRNA is unavailable for translation; starting transcription makes expression possible.

Regulatory state → transcription on/off → mRNA absent/present → translation possible/not possible. Different cell types express different subsets of the same genome.

A cell can switch on transcription of an enzyme gene when the enzyme is needed while other genes remain untranscribed.

Transcriptional control is a key switch, not the only possible control. More mRNA does not guarantee proportionally more protein if later steps are limiting.

Transcription for gene expression

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Determine / Outline / Compare.

Command terms

Determine / Outline / Compare / Deduce / Evaluate

What earns marks

Build the answer around this relationship: mRNA presence indicates that a gene has been transcribed.

Watch for

Treating a visible mRNA band as protein evidence rather than evidence of transcription.

Representative question

Question 1

[Maximum number: 2]

The scientists concluded that auxin activates the transcription of the GH 3 gene. Using the information on the auxin concentration in the stem base in the graph on page 4 and the Northern blot, evaluate whether this conclusion is supported.

Translation Builds a Polypeptide from mRNA

Translation uses a ribosome to read mRNA codons and join amino acids in the encoded order.

tRNA molecules pair anticodons with codons and deliver specific amino acids. The ribosome forms peptide bonds as it moves from start toward stop.

Trace: start codon; tRNA matching; peptide bond; codon movement; stop codon.

A ribosome reading three codons adds three specified amino acids before a stop signal ends the chain.

Translation reads RNA; it does not copy mRNA back into DNA. The ribosome assembles the chain; later folding and modification determine whether the polypeptide becomes functional.

Translation exam focus

Assessment in practice

1–8 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Explain / Describe.

Command terms

Outline / Explain / Describe

What earns marks

Build the answer around this relationship: Translation occurs on ribosomes in the cytoplasm or on rough endoplasmic reticulum.

Watch for

Naming RNA polymerase as the enzyme for translation instead of locating translation at ribosomes.

Representative question

Question 1

[Maximum number: 8]

Explain how polypeptides are produced by the process of translation.

Ribosome, tRNA and mRNA Divide Translation Work

During translation, mRNA supplies codon order, tRNAs deliver specific amino acids and the ribosome positions the molecules and catalyses peptide-bond formation.

mRNA first binds to the small ribosomal subunit. The large subunit forms the catalytic complex and can hold two tRNAs simultaneously so the amino acid on one can be linked to the growing chain on the other.

mRNA: ordered codons. tRNA: anticodon plus specific amino acid. Small subunit: binds/positions mRNA. Large subunit: binds two tRNAs and supports peptide-bond formation.

One tRNA holds the growing chain while a second matching tRNA places the next amino acid beside it; a peptide bond extends the polypeptide.

No component determines the protein alone: mRNA sets sequence, tRNAs match codons, and the ribosome coordinates assembly.

Roles in translation

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Outline / Explain.

Command terms

Identify / Outline / Explain

What earns marks

Build the answer around this relationship: mRNA supplies codons that are read by the ribosome.

Watch for

Swapping the roles of codons and anticodons between mRNA and tRNA.

Representative question

Question 1

[Maximum number: 7]

Explain the role of RNA in translation, resulting in the formation of polypeptide chains.

Complementary Pairing Transfers Information

A three-base tRNA anticodon forms complementary base pairs with a three-base mRNA codon during translation.

Correct codon–anticodon pairing positions the tRNA carrying the amino acid specified by that codon, transferring mRNA sequence information into polypeptide sequence.

Read the mRNA codon 5′→3′; write the antiparallel complementary anticodon; pair A–U and C–G; identify the amino acid from the mRNA codon, not the anticodon.

mRNA codon 5′-AUG-3′ pairs with tRNA anticodon 3′-UAC-5′, positioning the tRNA carrying methionine.

Codon and anticodon are complementary and antiparallel, not identical. DNA–RNA pairing during transcription is a different Objective.

Complementary base pairing

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Explain.

Command terms

Outline / Explain

What earns marks

Build the answer around this relationship: mRNA codons pair with complementary tRNA anticodons.

Watch for

Using DNA base-pairing letters instead of RNA uracil when writing anticodons.

Representative question

Question 1

[Maximum number: 3]

Outline how translation depends on complementary base pairing.

The Genetic Code Maps Codons to Amino Acids

The genetic code is a triplet, degenerate and nearly universal mapping from mRNA codons to amino acids or stop signals.

Four bases taken two at a time give only 16 combinations, fewer than the 20 amino acids; triplets give 4³ = 64 codons, enough for all amino acids plus stop signals.

Degenerate means more than one codon can specify the same amino acid. Universal means the same codon usually specifies the same amino acid across organisms, with limited exceptions.

UUU and UUC both code for phenylalanine, showing degeneracy; UUU has one defined meaning, so degeneracy is not ambiguity.

Read codons in a fixed triplet frame. Universality is a broad biological pattern, not a claim that no exceptions exist.

Genetic code features

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Outline / Suggest.

Command terms

Identify / Outline / Suggest / Describe

What earns marks

Build the answer around this relationship: A codon is a triplet of bases on mRNA.

Watch for

Counting amino acids as if one base rather than three bases codes for each residue.

Representative question

Question 1

[Maximum number: 5]

Describe the genetic code and its relationship to polypeptides and proteins.

A Code Table Converts Codons into a Sequence

A genetic-code table is used by reading mRNA codons 5' to 3' and matching each to its amino acid.

The first, second and third bases select positions in the table. Reading-frame errors shift every later codon, so strand labels and grouping matter.

Use: locate start; split triplets; read each codon; stop at a stop codon; report amino-acid order.

For 5'-AUG-GCU-UAA-3', AUG starts, GCU adds alanine and UAA terminates. If one base is omitted before AUG, every later triplet may be regrouped, so the table must be used from the correct start frame.

Do not use the DNA template directly as if it were mRNA; transcribe first.

Using genetic code table

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Determine.

Command terms

Determine

What earns marks

Build the answer around this relationship: Genetic code tables use mRNA codons.

Watch for

Using DNA triplets directly in the mRNA code table without converting thymine to uracil.

Representative question

Question 1

[Maximum number: 1]

Which sequence of mRNA bases and amino acids could be produced by transcription and translation of the DNA molecule shown?

3' AAAGTGGCACGTATATTT 5'
5' TTTCACCGTGCATATAAA 33^{\prime}

\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline \multirow{6}{*}{} & \multicolumn{6}{|c|}{2nd base in codon} & \multirow{14}{*}{3rd base in codon} \\
\hline & & U & C & A & G & & \\
\hline & \multirow{4}{*}{U} & Phe & Ser & Tyr & Cys & U & \\
\hline & & Phe & Ser & Tyr & Cys & C & \\
\hline & & Leu & Ser & STOP & STOP & A & \\
\hline & & Leu & Ser & STOP & Trp & G & \\
\hline \multirow[t]{8}{*}{} & C & Leu & Pro & His & Arg & U & \\
\hline & \multirow{3}{*}{A} & lle & Thr & Asn & Ser & U & \\
\hline & & lle & Thr & Lys & Arg & A & \\
\hline & & Met & Thr & Lys & Arg & G & \\
\hline & \multirow{4}{*}{G} & Val & Ala & Asp & Gly & U & \\
\hline & & Val & Ala & Asp & Gly & C & \\
\hline & & Val & Ala & Glu & Gly & A & \\
\hline & & Val & Ala & Glu & Gly & G & \\
\hline
\end{tabular}

Sequence of mRNA bases

Sequence of amino acids

UUU-GAG-GCU-CGA-UAU-UUU

Phe-Glu-Ala-Arg-Tyr-Phe

AAA-CUC-CGA-GCU-AUA-UUU

Lys-Leu-Arg-Ala-lle-Phe

UUU-CAC-CGU-GCA-UAU-AAA

Phe-His-Arg-Ala-Tyr-Lys

AAA-GUG-GCA-CGU-AUA-UUU

Lys-Val-Ser-Arg-Ile-Phe

Elongation Repeats a Three-Step Ribosome Cycle

During elongation, a matching tRNA enters, a peptide bond forms and the ribosome moves to the next codon.

Repeating this cycle extends the chain in a fixed direction while mRNA advances through the ribosome. Energy from charged tRNAs and factors supports the process.

Track each cycle: codon recognition; chain transfer; translocation; empty tRNA exit.

After one cycle the chain has one additional amino acid and the next codon is exposed.

Elongation is not random amino-acid addition; codon recognition controls the order. The cycle stops at a termination signal rather than continuing until the physical end of the mRNA.

Elongation of polypeptide

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Explain.

Command terms

Explain

What earns marks

Build the answer around this relationship: Elongation lengthens a polypeptide one amino acid at a time.

Watch for

Describing several ribosomes on one mRNA as several genes rather than one transcript being translated many times.

Representative question

Question 1

[Maximum number: 2]

Six polypeptides are shown in the diagram. Explain the different lengths of these polypeptides.

A Mutation Can Change Protein Structure

A point mutation can change one mRNA codon, replace one amino acid and alter how a polypeptide folds or functions.

The effect depends on code degeneracy, mutation position and the chemical properties of the original and replacement amino acids. Some substitutions are synonymous; others alter interactions stabilizing protein structure.

In sickle-cell disease, a point substitution changes a β-globin codon from GAG to GUG in mRNA, replacing glutamic acid with valine. The hydrophobic replacement promotes abnormal haemoglobin association and changes red-cell shape under low oxygen.

DNA base substitution → changed mRNA codon → possible amino-acid replacement → altered side-chain interactions → altered protein structure/function.

A mutation is not automatically harmful or structure-changing: degeneracy can make it synonymous, and some amino-acid replacements have little effect.

Mutations changing protein structure

Assessment in practice

1–6 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Explain.

Command terms

Outline / Explain

What earns marks

Build the answer around this relationship: A base substitution can change one mRNA codon.

Watch for

Stopping at the DNA mutation without tracing the change through mRNA codon and amino acid sequence.

Representative question

Question 1

[Maximum number: 8]

Explain the cause of sickle cell anemia and how this disease affects humans.

Core Protein Synthesis

  • Transcription: RNA polymerase builds complementary mRNA from the DNA template; A pairs with U and C with G.
  • Translation: ribosomes read mRNA codons 5′ to 3′ while tRNA anticodons deliver specific amino acids for peptide-bond formation.
  • Genetic code: codons are triplets; the code is degenerate and almost universal, with start and stop signals. Use mRNA—not DNA—when reading a code table.
  • Information flow: codon order determines amino-acid sequence, which determines protein folding and function.
  • Expression and variation: cells regulate which genes are transcribed. A mutation may change a codon, primary structure and phenotype, as in sickle-cell haemoglobin.

Objective notes

11 learning objectives
D1.2.1Transcription• Transcription makes mRNA as a mobile copy of gene information• RNA polymerase synthesizes RNA complementary to the DNA template strand4% of analysed papers 6 papers · 6 questionsViewD1.2.2Hydrogen bonding in transcription• Free RNA nucleotides align by complementary base pairing and hydrogen bonding• DNA adenine pairs with RNA uracil, while cytosine pairs with guanine2% of analysed papers 3 papers · 3 questionsViewD1.2.3DNA template stability• DNA template strands are transcribed without altering the base sequence• Sugar-phosphate backbone and base pairing preserve genetic information0% of analysed papers ViewD1.2.4Transcription for gene expression• Transcription is the first stage of gene expression• Cells regulate which genes are transcribed according to tissue, stage, and signals1% of analysed papers 1 paper · 4 questionsViewD1.2.5Translation• Translation decodes mRNA at ribosomes to synthesize polypeptides• mRNA codon order determines amino acid sequence5% of analysed papers 7 papers · 7 questionsViewD1.2.6Roles in translation• mRNA provides codons; tRNA carries activated amino acids with anticodons• Ribosomes hold mRNA and tRNAs so peptide bonds can form4% of analysed papers 5 papers · 5 questionsViewD1.2.7Complementary base pairing• tRNA anticodons pair with complementary mRNA codons by hydrogen bonding• Specific tRNA-amino acid attachment helps ensure correct amino acid addition3% of analysed papers 4 papers · 4 questionsViewD1.2.8Genetic code features• The genetic code is triplet, degenerate, and almost universal• Codons specify amino acids, a start signal, or stop signals6% of analysed papers 8 papers · 10 questionsViewD1.2.9Using genetic code table• Genetic code tables use mRNA codons, not DNA triplets• Convert template DNA to mRNA first, then read codons 5' to 3'4% of analysed papers 5 papers · 5 questionsViewD1.2.10Elongation of polypeptide• Ribosomes move along mRNA one codon at a time from start to stop• Peptide bonds join amino acids; multiple ribosomes can form a polysome1% of analysed papers 1 paper · 1 questionViewD1.2.11Mutations changing protein structure• Mutations can change codons and therefore amino acid sequence• Changed primary structure may alter folding and function, such as sickle-cell haemoglobin4% of analysed papers 5 papers · 5 questionsView