Cambridge Notes, Past Papers, Revision Questions

Movement into and out of Cells – Cambridge IGCSE/A‑Level (9700)

Learning Objective

Students will be able to illustrate and explain why the surface‑area‑to‑volume (SA:V) ratio decreases as a cell (or any 3‑D object) becomes larger, and to apply this principle to biological situations such as diffusion, osmosis and cellular transport.

Assessment Objectives Covered

AO1 – Knowledge: recall definitions, formulas and the biological significance of SA:V, diffusion, facilitated diffusion, osmosis, active transport, endocytosis, exocytosis and water potential.

AO2 – Application: calculate SA, V and SA:V for simple shapes; convert microscope measurements (mm → µm) and interpret the results.

AO3 – Analysis: evaluate how SA:V influences the rate of exchange of substances and link the mathematical result to real‑world cellular adaptations.

1. Key Biological Concepts

Process	Definition	Key Features / Examples
Simple diffusion	Passive movement of particles from an area of higher concentration to an area of lower concentration.	Gases (O₂, CO₂), small non‑polar molecules; rate ↑ with temperature, concentration gradient, and SA:V.
Facilitated diffusion	Passive transport of specific molecules through membrane‑bound proteins.	Channel proteins – form pores; e.g., aquaporins for water, ion channels for Na⁺/K⁺. Carrier proteins – bind a molecule, change shape, release it; e.g., GLUT transporters for glucose. No energy required; rate limited by protein number and concentration gradient.
Osmosis	Diffusion of water across a semi‑permeable membrane driven by a water‑potential (Ψ) gradient.	Water moves from higher Ψ (less negative) to lower Ψ (more negative). Influenced by solute concentration and pressure.
Active transport	Movement of substances against their concentration gradient using metabolic energy (ATP).	Examples: Na⁺/K⁺‑ATPase, proton pumps in plant root hairs.
Endocytosis / Exocytosis	Bulk‑phase transport involving membrane invagination (endocytosis) or vesicle fusion (exocytosis).	Uptake of large particles, secretion of hormones, neurotransmitters.
Water potential (Ψ)	Potential energy of water in a system: Ψ = Ψ_s + Ψ_p (solute + pressure components).	Direction of water movement is from higher to lower Ψ. No calculations required for the exam, but students must describe the trend.

2. Practical Context – Measuring Size

Measure the linear dimension (diameter, edge length, radius, height) on a stage‑micrometer in mm.

Convert to micrometres (1 mm = 1000 µm) – the unit required by the syllabus.

Insert the value into the appropriate surface‑area or volume formula.

Tip: Keep track of units throughout. Surface area will be in µm², volume in µm³ and SA:V in µm⁻¹.

3. Surface‑Area‑to‑Volume Calculations

3.1 General formulas

Sphere (radius r): SA = 4πr² V = (4/3)πr³

Cube (edge a): SA = 6a² V = a³

Cylinder (radius r, height h, closed at both ends): SA = 2πr² + 2πrh V = πr²h

3.2 Numerical examples (µm)

Shape	Size parameters	Surface Area (µm²)	Volume (µm³)	SA : V (µm⁻¹)
Sphere	r = 5	4π(5)² ≈ 314	(4/3)π(5)³ ≈ 524	3/5 = 0.60
	r = 10	4π(10)² ≈ 1 257	(4/3)π(10)³ ≈ 4 189	3/10 = 0.30
	r = 20	4π(20)² ≈ 5 027	(4/3)π(20)³ ≈ 33 510	3/20 = 0.15
Cube	a = 5	6(5)² = 150	5³ = 125	6/5 = 1.20
	a = 10	6(10)² = 600	10³ = 1 000	6/10 = 0.60
	a = 20	6(20)² = 2 400	20³ = 8 000	6/20 = 0.30
Cylinder (closed)	r = 5, h = 10	2π(5)² + 2π·5·10 ≈ 471	π·5²·10 ≈ 785	3/5 = 0.60
	r = 10, h = 20	2π(10)² + 2π·10·20 ≈ 1 885	π·10²·20 ≈ 6 283	3/10 = 0.30
	r = 20, h = 40	2π(20)² + 2π·20·40 ≈ 7 540	π·20²·40 ≈ 50 265	3/20 = 0.15

3.3 Key observations

For each shape, SA:V ∝ 1 / linear dimension. Doubling the characteristic size roughly halves the ratio.

When the height of a closed cylinder equals twice the radius (h = 2r), the ratio simplifies to SA:V = 3/r, identical to the sphere.

If a cylinder is open at one end, subtract the area of that base: SA = πr² + 2πrh. The inverse relationship with size still holds.

4. Interpretation of the Results

Smaller objects possess a larger surface area relative to their volume, allowing a higher rate of exchange of gases, nutrients and waste by diffusion.

As a cell grows, its SA:V falls, limiting the speed at which substances can reach the interior by simple diffusion alone.

Consequently, most cells stay small (≈10–30 µm) or develop specialised adaptations (flattening, extensions, transport systems) to maintain an effective exchange rate.

5. Biological Relevance – Real‑World Examples

Animal cells – typically ≤30 µm in diameter to keep SA:V favourable for O₂, CO₂ and metabolite diffusion.

Plant cells – a large central vacuole pushes the cytoplasm into a thin peripheral layer, effectively increasing membrane SA relative to cell volume.

Alveoli (lungs) – millions of tiny sacs give a huge total SA while each sac remains small enough for rapid gas exchange.

Intestinal villi & microvilli – fold the surface into many narrow projections, maximising SA for nutrient absorption.

Phagocytes (e.g., macrophages) – extend pseudopodia, increasing the effective SA that contacts pathogens.

6. Experimental Investigations (Syllabus Requirement)

6.1 Factors affecting the rate of simple diffusion

Factor	How it influences the rate	Typical investigation where it is varied
Temperature	Higher temperature increases kinetic energy → faster diffusion.	Diffusion in gelatin/agar blocks kept at different water‑bath temperatures.
Concentration gradient	Larger gradient → greater net movement per unit time.	Vary the dye concentration on one side of a gelatin block.
Surface‑area‑to‑volume ratio (size of the object)	Higher SA:V provides more area for molecules to cross per unit volume.	Use blocks of different dimensions (e.g., 1 cm³ vs 2 cm³) in the same solution.
Medium viscosity	More viscous media slow molecular movement.	Add glycerol to agar and compare diffusion rates.

6.2 Typical set‑ups

Diffusion in gelatin/agar blocks – place a coloured dye on one face; measure distance travelled at regular time intervals. Variables: temperature, block size, dye concentration, medium viscosity.

Osmosis using dialysis tubing – fill tubing with a sucrose solution, immerse in distilled water (or vice‑versa); record mass change over time. Variables: external solute concentration, temperature.

Plant tissue (potato cores) in sucrose solutions – weigh cores before and after immersion; plot % mass change against solution concentration. Variables: external concentration, temperature.

For each investigation students must be able to:

Identify independent, dependent and controlled variables.

Record data in a clear table (initial/final mass, distance, time, temperature).

Sketch a simple graph (e.g., distance vs time or % mass change vs concentration) and interpret the trend in terms of diffusion or osmosis.

7. Connections to Other Syllabus Topics

Connections box

Transport in plants – root‑hair uptake of water is limited by SA:V; xylem and phloem provide bulk flow to overcome this limitation.

Gas exchange in animals – alveolar walls and capillary networks maximise SA while keeping diffusion distances short.

Immune response – large SA of macrophage membranes aids phagocytosis and signalling.

Cellular adaptations – flattening of epidermal cells, formation of mycelial hyphae, multinucleated muscle fibres and the presence of microvilli all serve to preserve a favourable SA:V.

8. Summary

Mathematical derivations for spheres, cubes and cylinders demonstrate that SA:V ∝ 1 / linear dimension. This geometric constraint explains why most living cells are small, why specialised tissues increase effective surface area, and why larger organisms have dedicated transport systems (circulatory, vascular) to overcome the low SA:V ratio of bulk tissue.

Suggested diagram: three spheres (or cubes) of increasing size with arrows indicating surface area (blue) and volume (red); the SA:V ratio is shown decreasing from left to right.

illustrate the principle that surface area to volume ratios decrease with increasing size by calculating surface areas and volumes of simple 3-D shapes (as shown in the Mathematical requirements)