Biology  by Bradford
support
IB Biology 2025 · Theme D · Continuity and change

D1.2 Protein synthesis. Practice questions with markscheme.

64 original IB-style questions on D1.2, written from the 2025 guide: 30 multiple-choice, 24 short-answer, 6 data-based, 2 extended-response parts, 1 drawing, 1 labelling. Below are two ready-made practice papers built from them, one at standard level (20 marks) and one at higher level (20 marks); switch between them above the paper. Hand one out as a class quiz or homework, or sit it yourself and mark against the scheme. Print it, project it, or build a fresh one on the same topic.

What the guide asks for

11 statements at SL and HL, 8 additional higher level.

  1. D1.2.1SL / HL Transcription as the synthesis of RNA using a DNA template
  2. D1.2.2SL / HL Role of hydrogen bonding and complementary base pairing in transcription
  3. D1.2.3SL / HL Stability of DNA templates
  4. D1.2.4SL / HL Transcription as a process required for the expression of genes
  5. D1.2.5SL / HL Translation as the synthesis of polypeptides from mRNA
  6. D1.2.6SL / HL Roles of mRNA, ribosomes and tRNA in translation
  7. D1.2.7SL / HL Complementary base pairing between tRNA and mRNA
  8. D1.2.8SL / HL Features of the genetic code
  9. D1.2.9SL / HL Using the genetic code expressed as a table of mRNA codons
  10. D1.2.10SL / HL Stepwise movement of the ribosome along mRNA and linkage of amino acids by peptide bonding to the growing polypeptide chain
  11. D1.2.11SL / HL Mutations that change protein structure
  12. D1.2.12HL Directionality of transcription and translation
  13. D1.2.13HL Initiation of transcription at the promoter
  14. D1.2.14HL Non-coding sequences in DNA do not code for polypeptides
  15. D1.2.15HL Post-transcriptional modification in eukaryotic cells
  16. D1.2.16HL Alternative splicing of exons to produce variants of a protein from a single gene
  17. D1.2.17HL Initiation of translation
  18. D1.2.18HL Modification of polypeptides into their functional state
  19. D1.2.19HL Recycling of amino acids by proteasomes

In the bank for D1.2

  • 30 multiple-choice
  • 24 short-answer
  • 6 data-based
  • 2 extended-response parts
  • 1 drawing
  • 1 labelling
  • 28 higher level only

Every question is original and tagged to a guide statement. See the whole bank →

Make your own

The practice paper

20 marks at SL · 20 at HL · each has its own code
Take it online → Build a fresh paper Paper code BbB-EAAAAAAgABQAO8O_
Biology · topic quiz
Standard level · topic practice, not an exam format
30 minutes20 marks

Paper code: this paper was generated, so the code is its recipe. Enter it at biologybybradford.com/exam-maker to rebuild this exact paper and its markscheme.

Covers D1.2 Protein synthesis · SL
Name:
1.

A student was asked to find the amino acid sequence coded by part of a gene whose template strand reads 3′-TAC CGA GTT-5′. The table shows the student's working. In the genetic code, AUG codes for methionine, GCU for alanine and CAA for glutamine; UAC codes for tyrosine, CGA for arginine and GUU for valine. Which statement identifies the mistake in the working?

[1]
StepStudent's working
1The mRNA is complementary to the template strand: 5′-AUG GCU CAA-3′
2The tRNA anticodons pair with these codons: 3′-UAC-5′, 3′-CGA-5′ and 3′-GUU-5′
3Looking up UAC, CGA and GUU in the table of the genetic code gives tyrosine, arginine, valine
  1. Step 2: each anticodon has the same base sequence as the codon that it pairs with
  2. Step 3: the table lists mRNA codons, so AUG, GCU and CAA should be looked up
  3. Step 1: the mRNA has the same bases as the template, so it is 5′-UAC CGA GUU-3′
  4. Step 3: each anticodon must be looked up backwards in the table, from 3′ to 5′
2.

A template (antisense) strand of DNA reads 3′-TAC GGA-5′. What is the sequence of the mRNA transcribed from it?

[1]
3′ 5′ 5′ 3′ T A C G G A template (antisense) strand of DNA mRNA
  1. 5′-AUG CCU-3′
  2. 5′-ATG CCT-3′
  3. 5′-UAC GGA-3′
  4. 3′-AUG CCU-5′
3.

What is transcription?

[1]
  1. Synthesis of a polypeptide using mRNA
  2. Conversion of RNA back into DNA
  3. Synthesis of RNA from a DNA template
  4. Copying the whole genome before division
4.

Several ribosomes are often attached to one mRNA at the same time (a polysome). What is the benefit?

[1]
  1. The ribosomes repair the mRNA molecule
  2. Many copies are made from one mRNA at once
  3. Each ribosome makes a different protein
  4. The mRNA is protected from translation
5.

An mRNA coding region (excluding the stop codon) is 450 bases long. How many amino acids does the polypeptide contain?

[1]
  1. 150
  2. 450
  3. 1350
  4. 225
6.

Sickle-cell anaemia results from a mutation of the gene for the β-chain of haemoglobin. What is the immediate cause of the change in the protein?

[1]
  1. Complete failure of the gene to be transcribed
  2. Loss of the whole β-chain gene from the chromosome
  3. A single base substitution changing one codon
  4. A frameshift changing every amino acid after it
7.

A neuron in an adult never divides, yet it transcribes the same gene many times a day for decades. Which statement explains why the polypeptide produced from this gene stays the same throughout the life of the cell?

[1]
  1. The mRNA transcribed from the gene is stored permanently and re-used for every round of translation
  2. The gene is replicated each time it is transcribed, so a fresh, unused copy is always available
  3. The template strand is rebuilt from the mRNA after each round of transcription, restoring the original base sequence
  4. The template strand of the gene is used repeatedly without its base sequence being altered by transcription
8.

The antibiotic chloramphenicol blocks the formation of peptide bonds by bacterial ribosomes. Predict, with reasons, its effect on protein synthesis in bacteria.

[3]
9.

Outline the process of transcription.

[3]
10.

Outline the process of translation.

[4]
11.

Distinguish between transcription and translation.

[3]

Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
Rebuild or edit this exact paper (and its markscheme): biologybybradford.com/exam-maker?code=BbB-EAAAAAAgABQAO8O_

Show the markscheme

Markscheme BbB-EAAAAAAgABQAO8O_

One mark per point; / separates alternative wording within a point, OR separates alternative answers, words in brackets are not required, underlined words are essential. OWTTE = or words to that effect.

1. D1.2.7, D1.2.9 [1]
  • B: the table of the genetic code is a table of mRNA codons, so the amino acids come from the codons AUG, GCU and CAA (methionine, alanine, glutamine), not from the anticodons, which are complementary to them; steps 1 and 2 are correct, because the mRNA is complementary to the template (A) and an anticodon is complementary, not identical, to its codon (B); reading the anticodons backwards still does not give the codons (D);
2. D1.2.2 [1]
  • A: mRNA is complementary and antiparallel to the template, with U replacing T: 5′-AUG CCU-3′;
3. D1.2.1 [1]
  • C: RNA polymerase makes an RNA copy of one (template/antisense) strand of a gene; translation is polypeptide synthesis;
4. D1.2.6 [1]
  • B: polysomes multiply output per transcript, many identical polypeptides in parallel;
5. D1.2.8 [1]
  • A: three bases per codon: 450 ÷ 3 = 150 amino acids;
6. D1.2.11 [1]
  • C: the GAG → GTG substitution changes a single codon and hence one amino acid; the altered haemoglobin (HbS) polymerizes at low oxygen, deforming red blood cells;
7. D1.2.3 [1]
  • D: a single DNA strand serves as a template for RNA synthesis by complementary base pairing, and the DNA sequence is read but not changed, so a non-dividing somatic cell can conserve the same sequence for its whole life; B fails because replication occurs only before cell division, not during transcription; C fails because mRNA is short-lived and is broken down, not stored permanently; D fails because DNA is never rebuilt from mRNA in the cell, and there would be no need since the template is unchanged;
8. D1.2.10 [3]
  • amino acids can no longer be linked (by condensation) to the growing polypeptide chain;
  • (so) the chain cannot grow / no (complete, functional) polypeptides are made, even though mRNA and tRNAs can still bind to the ribosome;
  • (so) bacterial protein synthesis stops, halting growth of/killing the bacteria, OWTTE;
9. D1.2.1-D1.2.3 [3 max]
  • RNA polymerase binds (to the promoter) and unwinds/separates the DNA strands;
  • free (RNA) nucleotides pair with complementary bases on the template/antisense strand (U with A);
  • RNA polymerase links the nucleotides (covalently), building mRNA in the 5′ → 3′ direction;
  • the mRNA detaches and the DNA (helix) re-forms / mRNA moves to the cytoplasm (for translation), OWTTE;
10. D1.2.6-D1.2.9 [4 max]
  • mRNA binds to a ribosome (small subunit, then large);
  • tRNAs carry specific amino acids (attached by tRNA-activating enzymes);
  • the tRNA anticodon pairs with the (complementary) mRNA codon (at the A site);
  • peptide bonds form between adjacent amino acids (catalysed by the ribosome);
  • the ribosome moves along the mRNA (5′ → 3′), codon by codon;
  • translation starts at a start codon (AUG) and ends at a stop codon, releasing the polypeptide, OWTTE;
11. D1.2.5 [3 max]
  • transcription produces RNA/mRNA, whereas translation produces a polypeptide;
  • the template for transcription is (the antisense strand of) DNA, whereas translation uses the (codon sequence of) mRNA;
  • transcription is carried out by RNA polymerase, whereas translation occurs on ribosomes (with tRNAs supplying amino acids);
  • (in eukaryotes) transcription occurs in the nucleus, whereas translation occurs in the cytoplasm, OWTTE;

Credit only explicitly comparative statements; award converse.

More in Theme D · Continuity and change

All 40 IB Biology subtopics →  ·  Command terms →  ·  AP Biology units →  ·  Open the exam maker →