AS & A Level Biology · 9700 · Topic 1 · 2025–2027 Exam

Cell Structure

Foundations of cell biology: light and electron microscopy with magnification calculations, eukaryotic cell organelles in plants and animals, prokaryotic cell features, and the non-cellular nature of viruses.

Sub-sections 1.1–1.2 AS Level Papers 1–3 12 Learning Outcomes
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Topic 1.1 · AS

The microscope in cell studies

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Microscopy is the foundational practical skill for AS & A Level Biology. Whether examining tissue slides for Topic 1, photosynthetic pigments for Topic 13, or kidney sections for Topic 14, every candidate is expected to prepare slides, draw cells, calibrate measurements, and calculate magnifications and actual sizes using mm, µm, and nm.

Specimen preparation and drawing

For temporary mounts viewed under the light microscope, candidates should be able to perform the standard sequence: collect a thin specimen, place it on a clean slide with a drop of stain or water, lower a coverslip at an angle from one edge to exclude air bubbles, and blot excess fluid. Common stains include iodine (starch in plant cells), methylene blue (animal cell nuclei), and aceto-orcein (chromosomes).

Biological drawing rules
  • Use a sharp pencil — no shading or sketchy multiple lines
  • Lines must be single, continuous, and connect cleanly
  • Label every structure with horizontal label lines (no arrows)
  • Show only what is actually visible — do not add textbook detail
  • Include a scale bar OR magnification statement (never both)
  • Title with specimen + plane of section + objective lens used
Two drawing scales tested in Paper 3

Low-power plan: shows tissue layout and boundaries between regions. No individual cells drawn.

High-power detail: shows individual cells with visible organelles. Typically only 3–5 representative cells.

Magnification calculation

The fundamental relationship the syllabus expects candidates to apply fluently:

Core formula — rearrange in three directions
Magnification = Image size ÷ Actual size

Therefore: Actual size = Image size ÷ Magnification

And: Image size = Actual size × Magnification

Critical: Image size and actual size must be in the same unit before dividing. Convert first, then calculate.

Worked Example — Standard P3 calculation

An electron micrograph shows a mitochondrion measuring 28 mm long. The stated magnification is ×14 000. Calculate the actual length in µm.

Step 1 — Convert image size to µm: 28 mm = 28 000 µm

Step 2 — Apply formula: Actual size = 28 000 µm ÷ 14 000 = 2.0 µm

Mitochondria are typically 1–10 µm long, so 2.0 µm is biologically plausible — always sense-check against known organelle sizes.

Unit conversions for microscopy

UnitSymbolValue in metresTypical use
millimetremm10−3 mWhole organisms; small tissues
micrometreµm10−6 mCells (10–100 µm); organelles (1–10 µm)
nanometrenm10−9 mMembranes (~7 nm); ribosomes (~25 nm); viruses (20–300 nm)

Conversion shortcuts to memorise: 1 mm = 1000 µm = 1 000 000 nm. To convert mm to µm, multiply by 1000. To convert µm to nm, multiply by 1000.

Eyepiece graticule and stage micrometer

An eyepiece graticule is a small ruler etched onto a glass disc inside the eyepiece. Its divisions (epu — eyepiece units) appear superimposed on any specimen. But each epu has a different real-world value at each magnification — so it must be calibrated against a stage micrometer: a slide with a precise scale (typically 1 mm divided into 100 divisions of 10 µm each).

Calibration procedure
  1. Place stage micrometer on stage; focus until both scales are visible.
  2. Align the two scales so divisions line up (both ideally at left).
  3. Count how many epu match a known number of stage micrometer divisions.
  4. Calculate: 1 epu = (stage divisions × 10 µm) ÷ epu count
  5. Repeat for each objective lens used (×4, ×10, ×40).
Worked Example — Calibration calculation

Under the ×40 objective, 50 eyepiece divisions span 4 stage micrometer divisions. Calculate the value of 1 eyepiece division in µm.

Step 1: 4 stage micrometer divisions = 4 × 10 µm = 40 µm

Step 2: 50 epu = 40 µm, so 1 epu = 40 ÷ 50 = 0.8 µm

This calibration only applies to ×40. To use ×100 (oil immersion), recalibrate.

Magnification vs Resolution — the syllabus distinction

Candidates routinely confuse these terms. The syllabus requires defining each separately and explaining the difference — not just using them interchangeably.

TermDefinitionWhat changes it
MagnificationHow many times larger an image appears compared to the actual specimenLens system — can always be increased by adding lenses or projecting larger
ResolutionThe minimum distance between two points at which they can still be distinguished as separateWavelength of the radiation used to view the specimen
Why electron microscopes resolve more

Electrons have a much shorter wavelength than visible light. Light microscopes have a resolution limit of about 200 nm — structures closer than this appear as one blur, no matter how high the magnification. Electron microscopes resolve to about 0.2 nm, revealing organelle ultrastructure.

Empty magnification: Increasing magnification beyond the resolution limit only makes the blur bigger — it adds no biological detail. This is why a light microscope at ×2000 shows nothing more than at ×1000.

Light vs Electron Microscopes

FeatureLight microscopeTransmission EM (TEM)Scanning EM (SEM)
RadiationVisible lightElectron beam (transmitted)Electron beam (reflected)
Maximum magnification~×1500~×500 000~×100 000
Resolution~200 nm~0.2 nm~3–10 nm
Image typeColoured (with stain)2D black & white3D-appearance black & white
Specimen stateLive or deadDead, dehydrated, in vacuumDead, dehydrated, in vacuum
Best forWhole cells, tissues, processesInternal organelle detailSurface detail, 3D structure
Common mistakes — Microscopy

✕ "Electron microscopes magnify more, so they show more detail." — Wrong cause. They show more detail because of higher resolution, not magnification. A light microscope at ×2000 still has 200 nm resolution — just bigger blur.

✕ Confusing µm and nm in calculations — almost always loses the mark. Always state units in working.

✕ "Resolution" used to mean "image clarity" or "image sharpness" — the syllabus definition is specific: minimum separation distance.

MCQ · Topic 1.1 · Paper 1 style

Which statement correctly distinguishes resolution from magnification?

  • A. Resolution can be increased indefinitely by adding more lenses.
  • B. Magnification depends on the wavelength of radiation used.
  • C. Resolution is the minimum distance between two points at which they can still be distinguished as separate.
  • D. Light microscopes have higher magnification than electron microscopes.
Answer: C — This is the syllabus definition. (A) is wrong because resolution is fixed by wavelength, not lens count. (B) reverses cause: resolution depends on wavelength, not magnification. (D) is the opposite of reality — electron microscopes typically reach ×500 000 vs ~×1500 for light.
Structured · Topic 1.1 · Paper 2 style · 4 marks

A student observes a plant cell at ×400 magnification. The cell measures 32 eyepiece divisions across. Calibration showed that at ×400, 50 epu = 4 stage micrometer divisions, where each stage division is 10 µm.

(a) Calculate the value of 1 epu in µm. [2]
(b) Calculate the actual width of the cell in µm. [2]

(a) Value of 1 epu [2 marks]

4 stage divisions = 4 × 10 µm = 40 µm [1 mark for conversion]

1 epu = 40 µm ÷ 50 = 0.8 µm [1 mark for answer with unit]

(b) Actual cell width [2 marks]

Width = 32 epu × 0.8 µm [1 mark for setup]

= 25.6 µm [1 mark for answer with unit]

Mark scheme guidance
  • Unit (µm) is required for full marks — bare number loses 1
  • Working must show the calibration step explicitly; just stating "0.8 µm" without derivation typically scores half
  • Answer 25.6 or 26 (sig fig dependent) both accepted
Topic 1.2 · AS

Cells as the basic units of living organisms

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All living organisms are built of cells, and cells fall into two fundamentally different organisations: eukaryotic (with a true membrane-bound nucleus and complex organelles) and prokaryotic (without). Viruses are non-cellular — the syllabus uses them to challenge candidates to articulate what counts as life.

Eukaryotic cell organelles — structure and function

The syllabus expects candidates to recognise each organelle in photomicrographs and electron micrographs, and to outline its structure and function. Memorise the structure-function pairing for each:

OrganelleStructureFunctionIn plant?In animal?
NucleusSurrounded by double membrane (nuclear envelope) with pores; contains chromatin and one or more nucleoliStores genetic information; controls cell activity through transcription
NucleolusDense region within nucleus; not membrane-boundManufactures ribosomal RNA and assembles ribosomes
Rough ER (RER)System of flattened membrane sacs (cisternae) studded with ribosomesSynthesises and transports proteins (especially those for secretion)
Smooth ER (SER)Membrane sacs without attached ribosomesSynthesises lipids and steroids; detoxification
Golgi bodyStack of flattened membrane sacs (Golgi cisternae); vesicles bud from edgesModifies, sorts, and packages proteins and lipids; produces glycoproteins; forms lysosomes
Ribosomes (80S)Two subunits (large + small); not membrane-bound; ~25 nmSite of protein synthesis (translation)
MitochondrionDouble membrane; inner folded into cristae; matrix contains 70S ribosomes and circular DNASite of aerobic respiration; produces ATP
LysosomeSingle-membrane vesicle containing hydrolytic enzymesIntracellular digestion; destruction of worn-out organelles (autophagy)(rare)
CentriolesPair of cylindrical bundles of microtubules at right anglesOrganise microtubules during nuclear division (spindle formation)
MicrovilliFinger-like extensions of cell surface membraneIncrease surface area for absorption (e.g. small intestine epithelium)
CiliaHair-like projections; 9+2 arrangement of microtubules; basal body anchorsBeat to move fluid past cell surface (e.g. respiratory tract)
MicrotubulesHollow tubes of tubulin protein; ~25 nm diameterCytoskeleton; organelle movement; spindle fibres; cilia/flagella core
ChloroplastDouble membrane; stroma contains stacks of thylakoids (grana); 70S ribosomes; circular DNASite of photosynthesis
Permanent vacuoleLarge central sac surrounded by the tonoplast (single membrane); contains cell sapMaintains turgor; stores ions, sugars, pigments; some hydrolytic activity(small & temporary)
Cellulose cell wallRigid layer outside the cell surface membrane; cellulose microfibrilsMechanical support; prevents cell bursting under turgor
PlasmodesmataCytoplasmic strands through pores in adjacent plant cell wallsAllow movement of water, ions, and small molecules between cells
Cell surface membranePhospholipid bilayer with embedded proteins; ~7 nm thickControls movement of substances; cell signalling; recognition
Where ATP fits

The syllabus explicitly states (1.2 LO 4) that cells use ATP from respiration for energy-requiring processes. Examples to remember: active transport, protein synthesis, muscle contraction, DNA replication, mitosis, exocytosis. ATP itself is detailed in Topic 12 — here just connect "mitochondria → ATP → energy-requiring processes" as the chain.

Plant vs Animal cells — what differs

Plant and animal cells share most organelles. The syllabus expects candidates to compare structurally, focussing on what is present in one but not (or rarely) in the other:

FeaturePlant cellAnimal cell
Cellulose cell wall✓ Always present, outside membrane✕ Absent
Chloroplasts✓ In photosynthetic cells✕ Absent
Permanent large vacuole✓ Central, fluid-filled, with tonoplastSmall, temporary vacuoles only
Plasmodesmata✓ Present✕ Absent
CentriolesGenerally absent✓ Present
Cilia / flagella (typical somatic cells)Generally absentIn some cell types (e.g. respiratory epithelium)
ShapeFixed by cell wall (often box-like)Variable, often rounded
Storage carbohydrateStarch grainsGlycogen granules
Frequent errors — Plant vs Animal

✕ "Plant cells don't have mitochondria." — Wrong. All eukaryotic cells respire and need mitochondria. Photosynthesis produces glucose; respiration releases its energy.

✕ "Animal cells have no vacuoles." — Wrong. They have small, temporary vacuoles. The syllabus distinction is the permanent, large central vacuole, which is plant-specific.

✕ Lysosomes described as "plant organelles" because they sound similar to vacuoles — lysosomes are characteristic of animal cells.

Prokaryotic cells — the syllabus checklist

The 9700 syllabus (1.2 LO 5) gives an explicit list of features for "a typical bacterium". Memorise these six features:

Feature 1
Unicellular

Each prokaryote is a single cell. Some form colonies but each cell remains independent.

Feature 2
Generally 1–5 µm diameter

About 10× smaller than typical animal cells (~50 µm). Can be visualised under light microscope but ultrastructure requires EM.

Feature 3
Peptidoglycan cell walls

A polymer of sugars cross-linked by short peptides. Different chemistry from cellulose (plant) or chitin (fungi). Targeted by penicillin (Topic 10).

Feature 4
Circular DNA

One main circular chromosome floating freely in cytoplasm — not bounded by a membrane, so prokaryotes have no nucleus. Often plus small extra circular plasmids.

Feature 5
70S ribosomes

Smaller than 80S eukaryotic ribosomes. Same machinery for protein synthesis but different antibiotic sensitivity (e.g. streptomycin targets 70S).

Feature 6
No double-membrane organelles

No mitochondria, no chloroplasts, no nucleus. Some prokaryotes have internal membrane folds for respiration or photosynthesis, but no membrane-bound compartments.

Why the 70S / 80S distinction matters clinically

Many antibiotics (streptomycin, tetracycline, erythromycin) target 70S ribosomes specifically. Because mitochondria and chloroplasts contain 70S ribosomes, prolonged high-dose antibiotic use can cause mitochondrial side effects — an evolutionary echo of the endosymbiotic origin of these organelles. (Topic 10 develops this.)

Prokaryote vs Eukaryote — comparison table

FeatureProkaryotic (bacterium)Eukaryotic (plant / animal)
Typical size1–5 µm10–100 µm
Nucleus✕ Absent (DNA free in cytoplasm)✓ Membrane-bound nucleus
DNA arrangementCircular chromosome (+ plasmids)Linear chromosomes with histones
Ribosomes70S (smaller)80S (cytoplasm); 70S in mitochondria/chloroplasts
Cell wallPeptidoglycanCellulose (plants); none (animals); chitin (fungi)
Membrane-bound organelles✕ None with double membrane✓ Mitochondria, chloroplasts, nucleus, etc.
Cell divisionBinary fissionMitosis or meiosis
ExamplesE. coli, Mycobacterium tuberculosisOnion epidermis, human epithelium

Viruses — non-cellular structures

The syllabus carefully states (1.2 LO 7) that all viruses are non-cellular structures. A virus is not a cell — it has no cytoplasm, no ribosomes of its own, and cannot synthesise proteins or generate ATP outside a host cell. The syllabus expects candidates to know only three structural components:

1. Nucleic acid core

Either DNA or RNA (but not both). Single- or double-stranded depending on the virus. Carries the genetic instructions for making more virus particles.

2. Capsid

Protein shell surrounding the nucleic acid. Made of protein subunits (capsomeres) arranged in geometric patterns. Protects nucleic acid; provides surface proteins for host cell recognition.

3. Phospholipid envelope (some only)

Some viruses (e.g. HIV, influenza, SARS-CoV-2) have an outer envelope of phospholipids stolen from the host cell membrane during budding. Embedded glycoproteins help the virus bind to host receptors.

Are viruses alive?

This is precisely why the syllabus places viruses in this topic — to make candidates think critically about the cell theory. Viruses replicate, evolve, and possess heritable information, but they cannot do so independently. They lack the metabolic machinery of cells. Most biologists treat them as biological entities at the boundary of life rather than as living organisms in the strict sense. A safe phrasing for exam answers: "Viruses possess some features of life such as heritable nucleic acid, but lack independent metabolism and reproduction, so they are conventionally regarded as non-cellular biological agents."

MCQ · Topic 1.2 · Paper 1 style

Which feature is found in a prokaryotic cell but NOT in a eukaryotic cell?

  • A. A cell surface membrane
  • B. Ribosomes
  • C. A peptidoglycan cell wall
  • D. Circular DNA in mitochondria
Answer: C — Peptidoglycan is unique to bacterial cell walls. (A) all cells have membranes; (B) all cells have ribosomes (70S in prokaryotes, 80S + 70S in eukaryotes); (D) eukaryotic mitochondria do contain circular DNA — a remnant of their endosymbiotic ancestry — so it is not exclusively prokaryotic.
MCQ · Topic 1.2 · Paper 1 style

A student examines four structures under an electron microscope and lists their components. Which structure is a virus?

  • A. Cell surface membrane, cytoplasm with 70S ribosomes, circular DNA, peptidoglycan wall
  • B. Protein capsid surrounding RNA core, with phospholipid envelope and glycoprotein spikes
  • C. Double membrane bounding stroma containing thylakoids, 70S ribosomes, and circular DNA
  • D. Double membrane bounding cristae and matrix with 70S ribosomes and circular DNA
Answer: B — Capsid + nucleic acid core (here RNA) + optional phospholipid envelope = virus. (A) describes a typical bacterium; (C) is a chloroplast; (D) is a mitochondrion. The presence of cytoplasm/ribosomes/membrane (cellular features) is the giveaway that A, C, D are all cells or organelles, not viruses.
Structured · Topic 1.2 · Paper 2 style · 6 marks

Compare the structure of a typical bacterium with that of a typical animal cell. Give THREE differences. [6]

Any three of the following differences (2 marks each — 1 for the bacterial feature, 1 for the contrasting animal feature):

Acceptable comparison points
  • DNA arrangement: bacterium has circular DNA free in cytoplasm; animal cell has linear chromosomes inside a membrane-bound nucleus
  • Ribosomes: bacterium has 70S ribosomes only; animal cell has 80S ribosomes (plus 70S in mitochondria)
  • Cell wall: bacterium has a peptidoglycan cell wall; animal cell has no cell wall
  • Membrane-bound organelles: bacterium has none with double membranes; animal cell has nucleus, mitochondria, etc.
  • Size: bacterium is 1–5 µm; animal cell is typically 10–100 µm
  • Cell division: bacterium divides by binary fission; animal cell by mitosis

Mark scheme guidance: Each comparison must explicitly contrast both organisms — just stating "bacteria are smaller" without naming the animal cell scale typically scores 1 of 2. Use comparative connectives ("whereas", "in contrast", "but").

Exam Prep

Topic 1 Practice — Comprehensive

Mixed practice in 9700 P1/P2 style. Cover the answer; attempt fully; check; record where the rubric points are won or lost.

MCQ · Mixed · Magnification · Paper 1

A bacterial cell appears 12 mm long on an electron micrograph stated as ×6000. The actual length of the cell is

  • A. 0.2 µm
  • B. 2.0 µm
  • C. 20 µm
  • D. 200 µm
Answer: B (2.0 µm)
Image size 12 mm = 12 000 µm. Actual size = 12 000 ÷ 6000 = 2.0 µm. Sense check: typical bacteria are 1–5 µm, so 2.0 is plausible. (A) miscalculates; (C) forgets to convert mm; (D) is in the wrong order of magnitude.
MCQ · Cell type recognition · Paper 1

An electron micrograph shows a cell with: a true nucleus surrounded by a double membrane, mitochondria with cristae, chloroplasts containing grana, and a large central vacuole bounded by a single membrane. The cell is

  • A. an animal cell
  • B. a plant cell
  • C. a bacterial cell
  • D. a virus
Answer: B — Chloroplasts and a permanent large vacuole are diagnostic of plant cells. Animal cells lack both. Bacteria lack double-membrane organelles. Viruses are non-cellular.
MCQ · Resolution concept · Paper 1

Two ribosomes lie 20 nm apart in a cell. Which microscope can distinguish them as separate structures?

  • A. Light microscope at ×1000
  • B. Light microscope at ×3000 with oil immersion
  • C. Transmission electron microscope
  • D. None — ribosomes cannot be visualised
Answer: C — Light microscopes resolve to ~200 nm; ribosomes 20 nm apart appear as a single point regardless of magnification. TEM resolves to ~0.2 nm and easily distinguishes them. (B) is a trap — oil immersion improves a light microscope's resolution somewhat, but not enough to resolve 20 nm. (D) is false because TEM works.
Structured · Comprehensive · Paper 2 · 8 marks

Fig. 1 (not shown) is an electron micrograph of a single cell, magnification ×15 000.

(a) The cell shows a double-membrane bounded structure containing folded inner membranes (cristae) and a 70S ribosome. Identify this organelle and state TWO functions of the inner membrane folding. [3]
(b) The cell also contains chloroplasts. State TWO conclusions you can reach about this cell's identity, and justify each. [3]
(c) On Fig. 1, the cell measures 90 mm across at the widest point. Calculate the actual diameter of the cell in µm. [2]

(a) Mitochondrion + cristae function [3 marks]

Organelle: mitochondrion [1]

Two functions of inner membrane folding (cristae):

  • Increases surface area for the embedded electron transport chain proteins / ATP synthase — allowing more ATP production per mitochondrion [1]
  • Compartmentalises the intermembrane space from the matrix, enabling the proton gradient required for chemiosmosis [1]
(b) Cell identity from chloroplasts [3 marks]

Conclusion 1: The cell is eukaryotic [1]

Justification: chloroplasts are double-membrane organelles, only present in eukaryotic cells (prokaryotes have no double-membrane organelles) [1]

Conclusion 2: The cell is from a plant (or photosynthetic protoctist) [1]

Justification: chloroplasts are restricted to photosynthetic eukaryotes; animal cells do not contain them.

(c) Actual diameter [2 marks]

Image size: 90 mm = 90 000 µm [1 for unit conversion]

Actual size = 90 000 µm ÷ 15 000 = 6.0 µm [1 for answer with unit]

Examiner notes
  • (a) "More surface area for ATP production" alone scores 1 only. The question asks for TWO functions — chemiosmosis/proton gradient is the second mark
  • (b) Stating "plant cell" without justification scores at most 1 of 2 conclusions. Justify means link the structural feature to the inference
  • (c) Bare numerical answer "6.0" without µm scores half — units always count
Exam Prep

Topic 1 — Common Mistakes

Topic 1 strategy

Topic 1 underpins Topics 4 (membranes), 5 (mitosis), 6 (nucleic acids), 7 (transport in plants), and 11 (immunity). The highest-value items to over-prepare are: magnification calculations with unit conversion, the prokaryote 6-feature checklist, the plant-vs-animal comparison table, and the resolution-vs-magnification distinction. These appear in some form in nearly every Paper 1 series and are frequently embedded in Paper 2 structured questions on tissues from later topics.

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