Free Cell structure questions for AS Level Biology (9700), part of Cells & molecules. Each question has the answer and the marking points an examiner would look for.
Generate a full Cell Structure worksheet →All Cells & molecules questionsIn a root tip squash, 200 cells were counted and 36 were in prophase, metaphase, anaphase or telophase. Of these, 12 were in prophase.
What is the mitotic index?
Answer: B
Which term is described by this statement?
“The ability to distinguish two points that are close together.”
Answer: D
Which row shows structures present in a typical palisade mesophyll cell and absent from a typical liver cell?
| chloroplast | centriole | large permanent vacuole | |
|---|---|---|---|
| A | absent | absent | present |
| B | present | absent | absent |
| C | present | present | present |
| D | present | absent | present |
Answer: D
A photomicrograph has a magnification of × 2500. A structure on it measures 48 mm. Calculate its actual length in μm.
| Part | Answer and marking points | Marks |
|---|---|---|
| 19.2 μm actual = image ÷ magnification C1 19.2 A1 | 2 |
A micrograph shows a mitochondrion with an image length of 60 mm at a magnification of ×1500.
Calculate the actual length of the mitochondrion in μm.
Describe how an eyepiece graticule is calibrated.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 40 μm actual = image ÷ magnification C1 40 μm A1 | 2 |
| (b) | stage micrometer; count divisions use a stage micrometer of known scale B1 line up and calculate the length of one eyepiece division at that magnification B1 | 2 |
A student calibrates an eyepiece graticule using a stage micrometer with divisions 10 µm apart. 40 eyepiece units line up with 15 stage divisions.
Explain why the eyepiece graticule must be calibrated at each magnification.
Calculate the actual length represented by one eyepiece unit.
A cell measures 18 eyepiece units. Use your answer to (b) to calculate its actual length.
State three structural differences between a prokaryotic cell and a eukaryotic cell.
Explain why an electron microscope has a much higher resolution than a light microscope, and give one disadvantage of electron microscopy.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | graticule fixed in eyepiece; apparent size of each division on specimen changes with objective the graticule scale stays the same size while the image is magnified differently by each objective B1 so the actual length represented by one eyepiece unit changes B1 | 2 |
| (b) | 3.75 µm 15 × 10 µm = 150 µm M1 ÷ 40 = 3.75 µm A1 | 2 |
| (c) | 67.5 µm 18 × 3.75 = 67.5 µm B1 | 1 |
| (d) | no nucleus (circular DNA); no membrane-bound organelles; 70S ribosomes; peptidoglycan wall; plasmids Any 3 of: prokaryotes: no nucleus (DNA free in cytoplasm, circular); no membrane-bound organelles (e.g. mitochondria); smaller (70S) ribosomes; cell wall of peptidoglycan; plasmids B3 | 3 |
| (e) | electrons have much shorter wavelength; resolution limited by wavelength; specimens dead / vacuum / artefacts electrons have a much shorter wavelength than light B1 resolution is limited to about half the wavelength, so finer detail is distinguished B1 disadvantage: specimens must be dead / in a vacuum / artefacts / no colour B1 | 3 |
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Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in