OCR A Level Biology A: Gene mutations and control of gene expression (H420) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions are for OCR A Level Biology A (H420), written to the OCR specification Version 4.1 (April 2026), for first assessment in 2025. They cover module 6.1.1 Cellular control, learning outcomes (a) to (d). All data are invented. Theory is in the study guide and recall prompts are in the revision notes. See also the OCR A Level Biology hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. State what is meant by a gene mutation, and name two types of change that can occur in the nucleotide sequence of a gene. [3]

2. The mRNA for a short polypeptide is AUG AAG GCC CUA GAU UUC CAC UGA. Data supplied: AUG is Met; AAG is Lys; GCC is Ala; CUA and CUG are Leu; GAU is Asp; UUC is Phe; CAC is His; UGA and UAG are stop codons.

(a) The fourth codon changes from CUA to CUG. Explain the effect on the polypeptide. [2] (b) The second codon changes from AAG to UAG. Explain the effect on the polypeptide. [2]

3. Use the mRNA and data in question 2.

(a) One C is deleted from the run of three consecutive C nucleotides (bases 8 to 10). Explain the effect on the polypeptide. [3] (b) Instead, the whole codon GAU is deleted. Describe and explain the effect on the polypeptide. [2]

4. A bacterium acquires a substitution mutation in a gene for a protein that an antibiotic normally binds.

(a) Explain how a substitution could stop the antibiotic from affecting the protein. [3] (b) Explain why the same mutation could be beneficial in one environment but neutral or harmful in another. [3]

5. Describe how the lac operon of E. coli stops the bacterium from making lactose-digesting enzymes when no lactose is present. [4]

6. Beta-galactosidase activity (arbitrary units) was measured in four strains of E. coli grown without and with lactose.

Strain	No lactose	Lactose present
W	3	900
X	850	870
Y	4	5
Z	0	0

(a) Calculate how many times greater the activity of strain W is with lactose than without. [1] (b) Explain the results for strain X. [2] (c) Suggest two possible explanations for the results for strain Y. [2] (d) State one disadvantage to strain X of growing without lactose. [1]

7. Explain why a liver cell and a muscle cell, both with identical DNA, make different proteins. [4]

8. A eukaryotic gene has four exons of 105, 192, 147 and 66 nucleotides. The introns between them are 640, 295 and 405 nucleotides long.

(a) Calculate the length of the primary mRNA. [1] (b) Calculate the percentage of the primary mRNA that is exons. Give your answer to 1 decimal place. [2] (c) Assume that all of the mature mRNA is translated and that the final codon is a stop codon. Work out how many amino acids the polypeptide contains. [2] (d) State where in the cell the introns are removed. [1]

9. Explain how a hormone can activate an enzyme using cyclic AMP, without new protein being made. [4]

10. Homeobox sequences are found in plants, animals and fungi.

(a) Explain how a gene containing a homeobox can regulate gene expression. [2] (b) A homeodomain from a fungus and the matching region in a plant protein have 44 identical amino acids out of 60. Calculate the percentage identity. Give your answer to 1 decimal place. [1] (c) Suggest why homeobox sequences have changed so little. [2] (d) A mutation in a Hox gene of an insect makes one body segment form the structures normally found on the next segment. Explain this. [2]

11. In a developing mouse limb, the cells between the future digits die. The number of apoptotic cells per square millimetre of this tissue was 48 in a normal embryo and 6 when an inhibitor of caspase enzymes was added.

(a) Calculate the percentage decrease in apoptotic cells caused by the inhibitor. [2] (b) Predict the appearance of the digits when the inhibitor is present. [1] (c) Explain how apoptosis separates the digits. [2] (d) Explain the role of mitosis in the growth of the limb. [2]

12. Cells from a culture were examined. Of 300 cells without a growth factor, 27 were in mitosis. Of 300 cells with the growth factor, 51 were in mitosis.

(a) Calculate the mitotic index for each culture. [2] (b) The growth factor is an external stimulus. Suggest how it causes the change in the mitotic index. [2] (c) Explain how DNA damage inside a cell could lead to apoptosis. [2] (d) Predict what could happen to a damaged cell if the genes controlling apoptosis no longer respond to internal stimuli. [1]

Answers

1. A permanent alteration to the order of nucleotides in a gene [1]. Any two of: substitution, insertion, deletion [2]. [3] Examiner insight: Name two different types; two sizes of the same type (insertion of one, insertion of two) are one type.

2. (a) CUA and CUG both code for Leu [1]. The polypeptide is unchanged (a silent change, because the code is degenerate) [1]. [2] (b) UAG is a stop codon [1]. Translation stops after Met, so only a very short polypeptide is made [1]. [2] Examiner insight: Use the data given: name the amino acid each codon codes for, not only “silent” or “stop”.

3. (a) Deleting one nucleotide shifts the reading frame [1]. The codons after the deletion are read differently, so GCC UAG is read and UAG is a stop codon [1]. The polypeptide is Met-Lys-Ala only, much shorter [1]. [3] (b) Three nucleotides are lost, so the reading frame is kept [1]. One amino acid (Asp) is missing: Met-Lys-Ala-Leu-Phe-His [1]. [2] Examiner insight: Link the number of nucleotides (one or three) to the frame; “deletions cause frameshifts” alone is incomplete, because a deletion of three does not.

4. (a) The substitution changes a codon, so a different amino acid is in the primary structure [1]. This can change interactions between R groups, altering the tertiary structure [1]. The antibiotic can no longer bind to the changed protein [1]. [3] (b) With the antibiotic present, the mutant survives and reproduces; others do not [1]. Without it there is no advantage [1]. If the changed protein works less well, the mutation may be harmful, so the effect depends on the environment [1]. [3] Examiner insight: Argue beneficial, neutral or harmful for a named environment; calling a mutation good or bad with no conditions is weak.

5. The regulatory gene lacI codes for the repressor protein [1]. With no lactose, the repressor binds to the operator [1]. This stops RNA polymerase transcribing lacZ and lacY [1]. No mRNA is made, so the enzymes are not made, saving energy and amino acids [1]. [4] Examiner insight: Keep the operator (a DNA sequence) and the repressor (a protein) separate.

6. (a) 900 / 3 = 300 times [1]. [1] (b) Activity is high without lactose [1]. The repressor is not working, or cannot bind the operator, so the genes are always transcribed [1]. [2] (c) Any two of: a faulty promoter, so RNA polymerase cannot bind [1]; a repressor that cannot be released by lactose, so the operator stays covered [1]; (a third possibility is a faulty permease, so lactose cannot enter the cell). [2] (d) It wastes energy and amino acids making enzymes that are not needed [1]. [1] Examiner insight: In the “suggest” part, give two different faults; the same fault in different words is one idea.

7. Transcription factors bind to specific base sequences near genes [1]. Activators help RNA polymerase bind to the promoter and start transcription [1]. The two cell types have different transcription factors active [1]. So different genes are transcribed, giving different mRNA and proteins from the same DNA [1]. [4] Examiner insight: Say the DNA is the same and the difference is which genes are transcribed; you need not name a transcription factor.

8. (a) 105 + 192 + 147 + 66 = 510 nucleotides of exons; 640 + 295 + 405 = 1340 of introns; primary mRNA = 1850 [1]. [1] (b) 510 / 1850 × 100 [1] = 27.6% [1]. [2] (c) Mature mRNA has 510 nucleotides, so 510 / 3 = 170 codons [1]. One is the stop codon, so 169 amino acids [1]. [2] (d) In the nucleus [1]. [1] Examiner insight: Show the working for each stage so your reasoning is clear. Remember to subtract the stop codon.

9. The hormone binds to a surface receptor [1]. The cell then makes cyclic AMP [1]. Cyclic AMP binds to an inactive enzyme and changes its shape, so it becomes active [1]. The enzyme was already present, so no new transcription or translation is needed [1]. [4] Examiner insight: Say that the protein already exists and is activated, not made.

10. (a) The gene codes for a protein with a DNA-binding domain [1]. The protein binds to specific base sequences of other genes and switches their transcription on or off (it is a transcription factor) [1]. [2] (b) 44 / 60 × 100 = 73.3% [1]. [1] (c) These genes control the body plan, so a mutation is likely to be harmful [1]. Such individuals are less likely to survive and reproduce, so selection keeps the sequence unchanged [1]. [2] (d) The Hox gene codes for a transcription factor that switches on the genes for structures in one segment [1]. The mutation changes which genes are switched on in that segment, so the wrong structures form [1]. [2] Examiner insight: Keep homeobox genes (plants, animals and fungi) separate from Hox genes (a subset in animals).

11. (a) (48 − 6) / 48 × 100 [1] = 87.5% [1]. [2] (b) The digits are joined by webbing (not separated) [1]. [1] (c) Genes trigger programmed cell death of the cells between the digits [1]. Caspase enzymes break the cells down and the fragments are removed, leaving separate digits [1]. [2] (d) Mitosis increases cell number, so the limb grows [1]. The new cells are genetically identical, so each has all the genes needed to specialise [1]. [2] Examiner insight: Percentage decrease is the change divided by the original value.

12. (a) Without: 27 / 300 × 100 = 9.0% [1]. With: 51 / 300 × 100 = 17.0% [1]. [2] (b) The growth factor binds to a receptor on the cell [1]. This activates the genes that drive the cell cycle, so more cells enter mitosis [1]. [2] (c) The damage is detected inside the cell (an internal stimulus) [1]. The genes for apoptosis are switched on, so the cell is broken down instead of dividing [1]. [2] (d) The damaged cell survives and may keep dividing, forming a tumour [1]. [1] Examiner insight: Quote the two mitotic index values when comparing, and name each stimulus as internal or external.

Where marks are usually lost

• Writing “the gene is faulty” without naming the codon, amino acid or polypeptide affected.
• Saying all deletions cause frameshifts; a deletion of three keeps the frame.
• Mixing up the operator, promoter and regulatory gene, or saying lactose binds the operator.
• Saying introns are removed from DNA, or removed in the cytoplasm.
• Calling apoptosis damage-caused death rather than programmed death.
• Dividing by the wrong value in a percentage change, or forgetting the stop codon.

Next steps

• Recall: the revision notes and the study guide.
• Course hub: OCR A Level Biology and the printable checklist.
• The free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Source: OCR, A Level Biology A (H420), specification Version 4.1 (April 2026), first assessment 2025, module 6.1.1 Cellular control, learning outcomes (a) to (d).