Practice Questions
OCR GCSE Biology A: Microscopy, eukaryotic and prokaryotic cells and electron microscopy (J247) – Practice Questions
Original practice questions with worked answers on OCR GCSE Biology A B1.1: microscopes, stains, magnification, cell structures and electron microscopy.
- Subject
- Biology
- Level
- GCSE
- Topic
- Microscopy, eukaryotic and prokaryotic cells and electron microscopy
- Author
- Marlbridge Academic Team
- Updated
Aligned to OCR GCSE Biology (J247), Version 4.0 (August 2026), for first assessment in 2018. Official specification .
Syllabus page (what it covers and how it is assessed): OCR GCSE Biology.
Syllabus points this page covers
J247
- B1.1 Cell structures
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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 cover statements B1.1a to B1.1c of OCR GCSE (9-1) Gateway Science Suite Biology A (J247), Version 4.0, first assessment in 2018. No statement in B1.1 is Higher Tier only, so every question suits both tiers. Learn the content first from the study guide and revision notes. The course hub is OCR GCSE Biology, with a printable checklist. Not sure where you stand? Try a free 10-minute diagnostic.
Questions
1. A student uses a light microscope with a ×10 eyepiece lens and a ×40 objective lens. (a) State the total magnification. [1] (b) State the job of the stage. [1] (c) State the job of the lamp. [1]
2. A student makes a slide of cheek cells and adds a drop of methylene blue. (a) Explain why a stain is used. [2] (b) The cover slip is lowered at an angle with a mounted needle. Give the reason. [1]
3. A leaf cell is 18 mm long on a light micrograph. Its real length is 45 µm. Calculate the magnification of the micrograph. [3]
4. A chloroplast is 54 mm long on a transmission electron micrograph (TEM) of magnification ×6000. (a) Calculate the real length of the chloroplast in µm. [2] (b) Give your answer to (a) in metres, in standard form. [2]
5. Copy and complete the table with a tick if the structure is present and a cross if it is absent.
| Structure | Palisade cell of a leaf | Cheek cell | Bacterium |
|---|---|---|---|
| Nucleus | |||
| Chloroplast | |||
| Mitochondrion | |||
| Plasmid |
[4]
6. A sperm cell has far more mitochondria than a skin cell. Explain why. [3]
7. (a) Describe the position and form of the DNA that makes up the main genome of a bacterium. [2] (b) State what a plasmid is and give one gene that it may carry. [2]
8. A hormone affects liver cells but has no effect on skin cells. (a) Explain, in terms of the cell membrane, why only liver cells respond. [2] (b) State one other role of the cell membrane. [1] (c) A liver cell makes large amounts of protein. Name the structure where proteins are made and suggest what you would expect to see a lot of in this cell. [2]
9. Ribosomes are about 25 nm across. They cannot be seen with a light microscope but are visible with a TEM. (a) Explain why. [3] (b) On a TEM image of magnification ×200 000, calculate the image size of a ribosome in mm. [2]
10. A TEM image has a scale bar of 20 mm, labelled 2 µm. A mitochondrion on the same image is 30 mm long. (a) Work out the magnification. [2] (b) Work out the actual length of the mitochondrion in µm. [2] (c) Give this length in nm. [1] (d) Suggest one reason why a TEM image gives more information about the mitochondrion than a light micrograph. [1]
11. A student examines onion epidermis with iodine solution. The field of view is 1.8 mm across and about 6 cells fit in a row across it. (a) Estimate the length of one cell in µm. [2] (b) The student draws one cell 60 mm long. Calculate the magnification of the drawing, assuming the cell is 0.3 mm long. [2] (c) The student sees nuclei but no chloroplasts. Suggest why chloroplasts are absent, and why iodine helps. [2] (d) Evaluate the use of a TEM rather than a light microscope for studying a living onion cell. [2]
Answers
1. (a) Total magnification = 10 × 40 = ×400 [1] (b) The stage holds the slide in position, with the specimen over the hole [1] (c) The lamp provides light that passes through the specimen [1] Examiner insight: For a total magnification, show the multiplication and include the “×” sign; a bare number gives no evidence of method if you slip.
2. (a) Cells are nearly colourless, so there is little contrast with the background [1]; the stain colours structures such as the nucleus so they can be seen [1] (b) To avoid trapping air bubbles under the cover slip [1] Examiner insight: A two-mark “explain why” answer is best written as a reason (colourless) plus a consequence (structures visible); “to see it better” alone is vague.
3. magnification = image size ÷ actual size [1]; convert 18 mm to 18 000 µm [1]; 18 000 ÷ 45 = ×400 [1] Examiner insight: Convert units before dividing. Computing 18 ÷ 45 = 0.4 skips the conversion and gives a figure that cannot be right for a micrograph.
4. (a) actual size = 54 ÷ 6000 = 0.009 mm [1]; 0.009 × 1000 = 9 µm [1] (b) 9 µm = 0.000009 m [1], which is 9 × 10⁻⁶ m [1] Examiner insight: Both parts of a standard-form answer count: the number and the power. A length well under 1 m needs a negative power.
5. Palisade cell: nucleus ✓, chloroplast ✓, mitochondrion ✓, plasmid ✗. Cheek cell: nucleus ✓, chloroplast ✗, mitochondrion ✓, plasmid ✗. Bacterium: nucleus ✗, chloroplast ✗, mitochondrion ✗, plasmid ✓. One mark for each structure (row) completely correct: nucleus [1], chloroplast [1], mitochondrion [1], plasmid [1]. Examiner insight: Fill every box, checking each row against what the cell type contains; a blank box tells the examiner nothing about what you know.
6. Mitochondria contain enzymes for cellular respiration [1]; respiration releases energy from glucose [1]; a sperm cell needs a lot of energy to swim, so it needs many mitochondria [1] Examiner insight: Link the structure, its process and the cell’s need. “Mitochondria make energy” is a common slip: energy is released in respiration.
7. (a) The genetic material is a single circular chromosome [1] lying free in the cytoplasm, because a bacterium has no nucleus [1] (b) A plasmid is a small ring of DNA separate from the main chromosome [1]; it may carry a gene for antibiotic resistance [1] Examiner insight: “DNA in the nucleus” for a bacterium is a very common error. The absence of a nucleus is what the question is testing.
8. (a) Liver cell membranes have receptor molecules [1] that bind the hormone, whereas skin cells lack the matching receptor [1] (b) The membrane is a selective barrier, controlling which molecules enter and leave [1] (c) Ribosomes are the site of protein synthesis [1]; you would expect many ribosomes in the cytoplasm [1] Examiner insight: Match your answer to the number of marks in each part. In (b) one clear role earns the mark, and the word “selective” is the key one.
9. (a) Resolution is the smallest gap at which two points still look separate [1]; a light microscope can only separate points more than about 200 nm apart, so 25 nm ribosomes are not resolved [1]; a TEM uses electrons, which give much higher resolution, so the ribosomes can be seen [1] (b) image size = 25 nm × 200 000 = 5 000 000 nm [1] = 5 mm [1] Examiner insight: Use the word resolution, not “clearer” or “magnifies more”. Higher magnification alone would not make the ribosomes visible.
10. (a) 20 mm = 20 000 µm [1]; 20 000 ÷ 2 = ×10 000 [1] (b) 30 mm = 30 000 µm; 30 000 ÷ 10 000 = 3 µm [2] (c) 3 × 1000 = 3000 nm [1] (d) The TEM has a higher resolution, so it shows internal detail such as the folds of the inner membrane [1] Examiner insight: Show each conversion and write units on every line; a clear layout makes slips easy to spot and lets the reader follow your reasoning.
11. (a) 1.8 mm ÷ 6 [1] = 0.3 mm = 300 µm [1] (b) magnification = 60 ÷ 0.3 [1] = ×200 [1] (c) Onion bulb tissue is underground and does not photosynthesise, so its cells do not contain chloroplasts [1]; iodine solution stains the nucleus so it stands out against the cytoplasm [1] (d) A TEM needs dead, very thin and treated specimens, so it cannot show a living cell [1]; a light microscope can be used on living cells, although its resolution is lower [1] Examiner insight: An “evaluate” answer is best when it gives a limit of one method and a benefit of the other, both tied to the living cell in the question.
Where marks are usually lost
- Dividing the actual size by the image size instead of the other way round.
- Skipping the conversion between mm and µm, or between µm and nm.
- Writing standard form with a positive power for a tiny size.
- Saying a TEM “magnifies more” instead of explaining that it has higher resolution.
- Placing DNA of a bacterium in a nucleus.
- Saying mitochondria “make” energy rather than release it during respiration.
- Calling the cell membrane a barrier that keeps everything in, without the word selective.
- Giving chloroplasts to cheek cells or to cells in non-green plant tissue.
- Leaving out the unit in a final answer, or using the symbol µ for mm.
- Describing a stain only as “to see it better”, without saying it adds contrast.
Next steps
- Re-read the study guide for any question you found hard.
- Condense with the revision notes.
- Return to the OCR GCSE Biology hub and tick off statements on the checklist.
- Take one of our free 10-minute diagnostics.
- Book a free trial class if you want help with magnification sums.
Official syllabus
Aligned to OCR GCSE (9-1) Gateway Science Suite Biology A (J247), Version 4.0, published by OCR (Oxford Cambridge and RSA), first assessment in 2018, Topic B1, statements B1.1a to B1.1c.
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Revision notes
OCR GCSE Biology A: Microscopy, eukaryotic and prokaryotic cells and electron microscopy (J247) – Revision Notes
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