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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
HL

Transcription Copies a Gene into mRNA

Transcription uses one DNA strand as a template to make a complementary messenger-RNA strand.

RNA polymerase opens a local section, matches RNA nucleotides and joins them into a strand that can carry the gene sequence away from DNA.

Trace: promoter; template strand; complementary RNA; mRNA export for translation.

A gene's DNA sequence is copied into mRNA before a ribosome reads its codons. The RNA sequence preserves the information needed for a ribosome to assemble the same polypeptide.

Transcription is RNA synthesis, not DNA replication or protein assembly. RNA polymerase reads only the selected gene region, not the whole chromosome.

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

Hydrogen bonds between complementary DNA bases can separate locally during transcription and reform behind the polymerase.

The covalent backbone stays intact while temporary base-pair bonds open. This exposes one template strand without destroying the chromosome. Opening is local and temporary, so the rest of the chromosome remains paired and protected while the polymerase moves.

Identify: local opening; exposed template; new RNA pairing; re-formation behind the enzyme.

A short transcription bubble moves along a gene while the rest of the DNA remains double-stranded. The bubble closes behind polymerase, leaving the DNA double helix ready for later use.

Weak hydrogen bonds do not mean DNA is structurally unimportant; the backbone provides continuity.

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

DNA stability protects the information source while temporary local opening permits transcription or replication.

The backbone and most base pairs remain protected in the chromosome. Controlled unwinding exposes only the region needed, reducing damage and preserving sequence fidelity.

Separate global stability from local access: chromosome stays intact; one gene region opens; copying or reading proceeds.

A polymerase can transcribe one gene without unzipping the entire DNA molecule. A cell can transcribe one gene while unrelated genes remain closed and protected elsewhere on the same chromosome.

Stable does not mean permanently closed; regulated local opening is part of gene expression. Controlled access is a balance: excessive opening can increase damage, but no opening prevents gene expression.

Transcription Turns Gene Information into a Message

Transcription converts stored DNA information into an mRNA message that can be translated.

Regulatory signals determine whether transcription starts and how much mRNA is made. The message links a gene's sequence to the amount and timing of protein production.

Follow the chain: regulatory control; mRNA abundance; ribosome access; protein output.

If transcription of an enzyme gene increases, more mRNA can support more translation when later steps are not limiting.

More mRNA does not guarantee proportionally more protein; later controls can limit expression.

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

mRNA supplies codon order, tRNA matches codons and carries amino acids, and the ribosome positions molecules and catalyzes peptide bonds.

Translation is accurate because each component solves a different problem: information, matching and chemical assembly. Coordination converts nucleotide sequence into protein sequence.

For each step name: codon; anticodon; amino acid; ribosome site; peptide bond.

A matching tRNA enters, places its amino acid beside the growing chain and then leaves.

tRNA does not determine the whole protein alone; codon order and ribosome movement are also required.

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

Complementary base pairing allows DNA to specify an mRNA sequence and codons to be recognized by tRNA anticodons.

Hydrogen-bonding patterns and base size make particular pairs fit. The same principle connects template reading during transcription with codon decoding during translation.

Check strand direction and pair type before predicting the next nucleotide or amino acid. Check pairing; strand direction; codon reading frame; and the matching anticodon.

A DNA template produces complementary mRNA, whose codon pairs with a matching tRNA anticodon. A single template base determines its partner in mRNA, which then helps select the tRNA carrying the next amino acid.

Pairing is complementary, not identical: the message is related to the template rather than a literal copy.

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 assigns each three-base mRNA codon to an amino acid or stop signal.

Triplets provide enough combinations for all amino acids. The code is degenerate because several codons can specify one amino acid, but each codon has a defined meaning.

Read codons in threes, using the correct reading frame and checking start and stop signals. Check triplet grouping; reading frame; codon meaning; and stop signals.

Changing UUU to UUC may still specify phenylalanine because the code is degenerate. A codon table can show that two different triplets encode the same amino acid, reducing the effect of some substitutions.

Degenerate does not mean ambiguous for one codon; it means multiple codons can share an amino acid.

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 DNA mutation can alter an mRNA codon, amino-acid sequence and ultimately protein folding or function, although some changes have little effect.

Outcome depends on mutation type, code degeneracy, location and the chemical difference between residues. Folding translates sequence change into structure change.

Assess: DNA change; codon consequence; amino-acid change; structural or functional effect.

Replacing a charged residue in an enzyme's active site can alter binding and reduce catalytic activity.

A mutation is not automatically harmful; synonymous or distant changes may leave function largely unchanged.

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.

Nucleic-Acid Synthesis Has a Defined Direction

HL only

Polymerases extend nucleic acids 5' to 3', while ribosomes read mRNA codons 5' to 3'. Directionality fixes how information is copied and decoded.

The 3' hydroxyl chemistry fixes extension direction. Labels keep codons in a consistent reading frame and explain why template orientation matters.

Mark 5' and 3' ends first; then place extension or ribosome movement arrows.

A ribosome starts at the 5' side of mRNA and moves toward its 3' end while the polypeptide grows.

Saying left to right is meaningless without strand labels and diagram orientation.

Directionality exam focus

HL only

Assessment in practice

1 marks
How it is assessed

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

Command terms

Identify

What earns marks

Build the answer around this relationship: RNA synthesis proceeds in the 5 prime to 3 prime direction.

Watch for

Choosing a start codon without using the 5 prime to 3 prime direction shown in the diagram.

Representative question

Question 1

[Maximum number: 1]

Using the diagram, identify, with a reason, whether X or Y is the start codon.

Promoters Position Transcription Initiation

HL only

A promoter is a DNA region where transcription machinery binds to position RNA polymerase and define where transcription begins.

Promoter sequence and associated factors determine strand choice, start site and transcription level. Opening and early RNA synthesis proceed from that positioned complex.

Name: promoter; binding factor; start site; direction of RNA synthesis.

A promoter mutation can reduce mRNA production without changing the protein-coding sequence. RNA polymerase begins at the promoter and then moves along the template, so promoter position fixes which downstream sequence is copied.

A promoter is regulatory DNA, not the entire gene or translated region.

Initiation of transcription at promoter

HL only

Assessment in practice

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: Promoters are non-coding DNA sequences linked to transcription start sites.

Representative question

Question 1

[Maximum number: 2]

Explain the function of a promoter in DNA.

Non-Coding DNA Can Regulate the Genome

HL only

Non-coding DNA does not translate into protein, but it can contain regulatory sequences, introns, structural regions or genes for functional RNAs.

Its role depends on sequence and location. Binding sites can control transcription, while introns are removed from pre-mRNA and other regions organize chromosomes.

Classify a region as regulatory; transcribed but removed; structural; or functional RNA.

An enhancer can influence transcription even though its sequence is not translated into amino acids.

Non-coding does not mean useless, and it does not identify one single function.

Non-coding sequences in DNA

HL only

Assessment in practice

1 marks
How it is assessed

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

Command terms

State

What earns marks

Build the answer around this relationship: Non-coding DNA can regulate transcription.

Watch for

Equating non-coding DNA with having no function.

Representative question

Question 1

[Maximum number: 2]

DNA has regions that do not code for proteins. State two functions of these regions.
1.
2.

Post-Transcriptional Processing Makes Mature mRNA

HL only

Eukaryotic pre-mRNA is processed by adding a 5' cap and poly-A tail and removing introns so mature mRNA can be exported and translated.

Processing protects RNA, supports export and helps ribosomes recognize the message. Exons are joined in the order defining the coding sequence.

Trace: pre-mRNA; cap and tail; intron removal; exon joining; mature mRNA export.

Removing an intron prevents its sequence from being read as part of the final translated message.

Processing changes RNA, not the original DNA sequence; different processing choices can alter the final protein.

Post-transcriptional modification

HL only

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice.

What earns marks

Build the answer around this relationship: Eukaryotic pre-mRNA can contain both introns and exons.

Watch for

Saying introns are added during processing rather than removed.

Representative question

Question 1

[Maximum number: 1]

What happens to mRNA after transcription in eukaryotic cells?

A

Binding to the large subunit of the ribosome

B

Removal of introns

C

Addition of exons

D

Attachment of an amino acid

Alternative Splicing Produces Multiple Messages

HL only

Alternative splicing joins selected exons in different combinations, allowing one gene to produce different mature mRNAs and protein isoforms.

Tissue-specific splicing factors influence which exons remain. Different combinations change amino-acid sequence and can alter localization or function. Splicing factors recognize sequence features near splice sites and select an exon pattern appropriate to the cell type.

Compare isoforms by checking shared exons, included or skipped exon and resulting protein difference. Compare shared exons; skipped or included exon; reading frame; and resulting protein domain.

A neuron and muscle cell can splice the same pre-mRNA differently and produce proteins with distinct interaction domains. Including an exon that encodes a binding domain can let one isoform interact with a partner while another isoform cannot.

Alternative splicing increases output from a gene but does not create a new DNA gene for every isoform.

Alternative splicing

HL only

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice.

What earns marks

Build the answer around this relationship: Alternative splicing joins different exon combinations.

Watch for

Assuming one gene can produce only one protein product in all circumstances.

Representative question

Question 1

[Maximum number: 1]

The number of protein-coding genes in the human genome is estimated to be about 20000 , which is much less than the size of the proteome. What is one reason for this?

A

Exons are removed from RNA before translation.

B

There are more types of amino acids than nucleotides.

C

mRNA can be spliced after transcription.

D

Base substitutions occur during transcription.

Translation Initiation Sets the Reading Frame

HL only

Translation initiation positions the ribosome at a start codon and establishes the reading frame for all later codons.

Initiation factors and initiator tRNA recognize the start region. Once the frame is fixed, triplets are read sequentially until a stop codon.

Check: start codon; initiator tRNA; ribosome position; frame; downstream triplets.

If one base is inserted before the frame, every later triplet can change even though the start codon remains.

A start codon can also encode an amino acid internally; context determines initiation.

Translation initiation

HL only

Assessment in practice

1–3 marks
How it is assessed

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

Command terms

Outline

What earns marks

Build the answer around this relationship: The small ribosomal subunit binds mRNA early in initiation.

Watch for

Putting the initiator tRNA into the E site or A site instead of linking initiation to the start codon and P site.

Representative question

Question 1

[Maximum number: 4]

Outline the roles of the different binding sites for tRNA on ribosomes during translation.

Polypeptides Are Modified after Translation

HL only

New polypeptides can be folded, cleaved or chemically modified after translation before becoming functional proteins.

Processing can expose an active site, add a targeting group, join subunits or alter stability. The final protein depends on sequence and post-translational handling.

Name: initial chain; processing event; structural change; functional consequence.

A precursor protein can be cleaved to remove a signal segment and reveal the mature active form.

Translation ending does not guarantee a functional protein; folding and processing may still be required.

Polypeptide modification

HL only

Assessment in practice

2 marks
How it is assessed

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

Command terms

Suggest

What earns marks

Build the answer around this relationship: A newly translated polypeptide may be an inactive precursor.

Representative question

Question 1

[Maximum number: 2]

Insulin is produced by cutting C -peptide from the precursor molecule proinsulin. Suggest why group 1 has a greater level of C-peptide than group 2.

Proteasomes Recycle Unneeded Proteins

HL only

Proteasomes break selected proteins into peptides, allowing amino acids to be recycled for new protein synthesis or metabolism.

Damaged or short-lived proteins are tagged and fed into a proteasome, where peptide bonds are hydrolysed. Released amino acids return to the cellular pool.

Trace: protein tag; proteasome entry; peptide breakdown; amino-acid reuse or catabolism.

A misfolded protein can be removed and its amino acids reused to build a different protein.

Proteasomal degradation is controlled recycling, not random digestion of every protein. Proteasomes do not recycle every protein immediately; tagging and recognition determine which substrates are selected.

HL Protein Synthesis Details

HL only

RNA polymerase reads template DNA 3' to 5' and synthesizes RNA 5' to 3'; ribosomes translate mRNA codons in the 5' to 3' direction. Promoters mark transcription start regions and orientation; transcription factors help RNA polymerase bind and initiate in eukaryotes. Non-coding DNA does not code for polypeptide amino acid sequences and includes introns, regulatory sequences, telomeres, rRNA genes, and tRNA genes. Eukaryotic pre-mRNA is modified before export and translation by adding a 5' cap and poly-A tail and removing introns by splicing. Alternative splicing joins different exon combinations from one pre-mRNA, so one gene can produce multiple protein variants in different cells or stages. Translation initiation assembles ribosomal subunits at the start codon AUG; initiator tRNA enters the P site and A, P, and E sites organize tRNA movement. Newly made polypeptides may be folded, cleaved, or chemically modified; preproinsulin processing to active insulin is a key example. Proteasomes degrade tagged, damaged, or unneeded proteins; amino acid recycling supports new protein synthesis and proteome quality control.

ConceptIB Biology HL