Transport in Cells: Key Takeaways for GCSE Revision
1. Diffusion
- Definition: Net movement of particles from an area of high concentration to low concentration (down a concentration gradient).
- Key Rules:
- Passive process (no energy required).
- Occurs in liquids and gases (e.g., oxygen in lungs, deodorant particles in air).
- Net movement means overall direction, as particles move randomly in both directions.
- Examples:
- Oxygen diffusing from alveoli into blood capillaries.
- Carbon dioxide diffusing from blood into alveoli to be exhaled.
- Adaptations for Efficient Diffusion:
- Lungs: Large surface area (alveoli), thin walls (one cell thick), moist surfaces, rich blood supply.
- Small intestine: Villi increase surface area for nutrient absorption.
- Tip: Use the phrase “High to Low is the way to go!” to remember direction.
2. Osmosis
- Definition: Net movement of water molecules across a partially permeable membrane from high water concentration (dilute solution) to low water concentration (concentrated solution).
- Key Terms:
- Hypotonic: Solution with higher water concentration (lower solute).
- Hypertonic: Solution with lower water concentration (higher solute).
- Isotonic: Equal concentration of water/solute on both sides.
- Examples:
- Water entering plant roots from soil.
- Red blood cells shrinking in salty water (hypertonic) or bursting in pure water (hypotonic).
- Experiment: Potato cylinders in salt solutions – mass changes indicate water gain/loss.
- Tip: “Water follows solute” – moves to where there’s more solute.
3. Active Transport
- Definition: Movement of particles against the concentration gradient (low → high) using energy from respiration.
- Key Examples:
- Plants: Root hair cells absorbing mineral ions (e.g., nitrates) from soil.
- Humans: Glucose absorption in the small intestine when gut concentration is low.
- Adaptations:
- Root hair cells have mitochondria to provide energy for active transport.
- Tip: Remember it’s “active” because it requires ATP (energy).
4. Factors Affecting Diffusion
- Concentration Gradient: Steeper gradient = faster diffusion.
- Example: Stronger deodorant smell spreads quicker.
- Temperature: Higher temperature increases kinetic energy → faster particle movement.
- Surface Area to Volume Ratio (SA:Vol):
- Formula:
SA:Vol=Surface AreaVolumeSA:Vol=VolumeSurface Area - Cube Example:
- Cube A (1 cm³): SA = 6 cm², Volume = 1 cm³ → SA:Vol = 6:1.
- Cube B (2 cm³): SA = 24 cm², Volume = 8 cm³ → SA:Vol = 3:1.
- Smaller organisms (high SA:Vol) rely on diffusion; larger organisms need specialised systems (e.g., lungs).
- Formula:
5. Adaptations for Exchange
- Lungs:
- Alveoli provide huge surface area (~half a tennis court).
- Thin walls (one cell thick) for short diffusion distance.
- Small Intestine:
- Villi increase surface area for nutrient absorption.
- Microvilli on epithelial cells further enhance absorption.
- Fish Gills: Filaments and lamellae increase surface area for oxygen uptake.
6. Experiments & Practical Skills
- Agar Cube Experiment:
- Variables: Controlled variables = dye concentration, temperature.
- Observation: Smaller cubes (higher SA:Vol) show deeper dye penetration.
- Potato Osmosis Practical:
- Method: Measure mass change in potato cylinders placed in salt solutions.
- Improvements: Use multiple repeats, control potato size, blot excess water.
- Visking Tubing: Models osmosis – water moves into tubing with concentrated solution.
7. Common Mistakes & Tips
- Diffusion vs. Osmosis:
- Diffusion = any particle; Osmosis = water only.
- Breathing vs. Respiration:
- Breathing = gas exchange (ventilation); Respiration = energy release in cells.
- SA:Vol Calculations:
- Always check units (e.g., cm² vs. cm³).
- Use SA=6×side2SA=6×side2 and Volume=side3Volume=side3 for cubes.
8. Exam-Style Questions
- Q: Explain why insects are small.
A: Small size = high SA:Vol → oxygen can diffuse efficiently without lungs. - Q: Describe active transport in root hairs.
A: Minerals move from low concentration (soil) to high (root) using energy from respiration.
Final Tip: Practice drawing labelled diagrams of alveoli, villi, and osmosis experiments – visuals boost marks! 🌟
Transport in Cells: 50 GCSE Questions & Answers
Diffusion
1. Define diffusion.
Answer: Diffusion is the net movement of particles from an area of higher concentration to lower concentration (down a concentration gradient). It is a passive process requiring no energy.
2. Give two examples of diffusion in everyday life.
Answer:
- Smelling deodorant: Particles diffuse from under the arm (high concentration) into the air (low concentration).
- Tea brewing: Tea particles diffuse from the teabag (high concentration) into hot water (low concentration).
3. Explain why diffusion is a ‘net’ movement.
Answer: Particles move randomly in all directions, but overall, more particles move from high to low concentration. Some may move back, but the net result is movement down the gradient.
4. Describe how oxygen diffuses in the lungs.
Answer: Oxygen in alveoli (high concentration) diffuses across the one-cell-thick membrane into capillaries where blood has lower oxygen concentration. Blood circulation maintains the gradient by carrying oxygenated blood away.
5. What is a concentration gradient?
Answer: The difference in concentration of a substance between two areas. Steeper gradients (larger differences) increase diffusion rate.
6. Why can’t diffusion occur in solids?
Answer: Particles in solids are fixed in position and cannot move freely. Diffusion requires particles to mix, which is only possible in liquids/gases.
7. Name the specialised cells in plant roots for water absorption.
Answer: Root hair cells.
8. Explain why ventilation maintains a steep concentration gradient in the lungs.
Answer: Breathing replaces air in alveoli, keeping oxygen concentration high and carbon dioxide low, ensuring continuous diffusion.
9. How do insects absorb oxygen?
Answer: Through spiracles (tiny tubes) connected to tracheae. Oxygen diffuses directly into cells due to their small size and high SA:Vol.
10. Why are insects limited in size?
Answer: Larger insects would have a lower SA:Vol ratio, making diffusion insufficient to meet oxygen demands.
Osmosis
11. Define osmosis.
Answer: The net movement of water molecules across a partially permeable membrane from a region of higher water concentration (dilute solution) to lower water concentration (concentrated solution).
12. What is a partially permeable membrane?
Answer: A membrane that allows small molecules (e.g., water) to pass through but blocks larger molecules (e.g., sucrose).
13. Describe what happens to a plant cell in pure water.
Answer: Water enters by osmosis, causing the cell to become turgid. The cell wall prevents bursting.
14. What happens to a red blood cell in a hypertonic solution?
Answer: Water leaves the cell by osmosis, causing it to shrink (crenate).
15. Explain why plants wilt in salty soil.
Answer: Soil water becomes hypertonic relative to root cells. Water leaves roots by osmosis, causing cells to become flaccid.
16. Calculate the SA:Vol ratio of a 3 cm³ cube.
Answer:
- Surface area: 6×(3)2=54 cm26×(3)2=54cm2
- Volume: 33=27 cm333=27cm3
- SA:Vol = 54:27=2:154:27=2:1
17. Why is osmosis important for transpiration?
Answer: Water absorbed by roots via osmosis is transported to leaves, where it evaporates (transpiration), creating a continuous water column.
18. Name the holes in leaves for gas exchange.
Answer: Stomata, controlled by guard cells.
19. What is the difference between osmosis and diffusion?
Answer: Osmosis involves water only and requires a partially permeable membrane; diffusion involves any particles and no membrane is required.
20. Describe an experiment to investigate osmosis in potatoes.
Answer:
- Cut potato cylinders of equal mass.
- Place in salt solutions of varying concentrations.
- Measure mass change after 24 hours.
- Calculate % change:
% change=Final mass – Initial massInitial mass×100% change=Initial massFinal mass – Initial mass×100
Active Transport
21. Define active transport.
Answer: Movement of particles against the concentration gradient (low → high) using energy from respiration.
22. Give an example of active transport in plants.
Answer: Root hair cells absorb mineral ions (e.g., nitrates) from soil (low concentration) into roots (high concentration).
23. Why do root hair cells contain many mitochondria?
Answer: Mitochondria produce ATP for energy to power active transport.
24. Where does active transport occur in humans?
Answer: In the small intestine to absorb nutrients (e.g., glucose) when gut concentration is lower than blood.
25. Explain why active transport is essential for mineral uptake in plants.
Answer: Soil minerals are often in lower concentrations than root cells. Active transport allows plants to accumulate essential ions.
26. How does active transport differ from diffusion?
Answer:
- Direction: Against vs. down gradient.
- Energy: Requires ATP vs. passive.
27. Why can’t osmosis be used to absorb minerals?
Answer: Minerals are often in lower concentrations in soil than roots. Osmosis only moves water, not solutes.
28. Name a human organ that uses active transport.
Answer: Kidneys during reabsorption of glucose and ions from filtrate.
Factors Affecting Diffusion
29. How does temperature affect diffusion?
Answer: Higher temperature increases kinetic energy of particles, speeding up diffusion.
30. Why does emphysema reduce gas exchange?
Answer: Damaged alveoli reduce surface area, slowing oxygen diffusion into blood.
31. Explain why villi are adapted for absorption.
Answer: Villi increase surface area, have thin walls (one cell thick), and contain capillaries for efficient nutrient uptake.
32. Calculate the SA:Vol ratio of a 2 cm³ cube.
Answer:
- SA: 6×(2)2=24 cm26×(2)2=24cm2
- Volume: 23=8 cm323=8cm3
- SA:Vol = 24:8=3:124:8=3:1
33. Why do fish gills have lamellae?
Answer: Lamellae increase surface area for faster diffusion of oxygen into blood.
34. What is the role of capillaries in alveoli?
Answer: Maintain steep concentration gradient by carrying blood away after oxygenation.
35. How does surface area affect diffusion rate?
Answer: Larger surface area = more space for particles to diffuse, increasing rate.
Experiments & Data
36. Identify two controlled variables in the agar cube experiment.
Answer:
- Dye concentration.
- Temperature.
37. Why are agar cubes cut into different sizes?
Answer: To compare how SA:Vol ratio affects diffusion depth.
38. What does the potato osmosis experiment measure?
Answer: Mass change indicates water uptake (hypotonic) or loss (hypertonic).
39. How would you improve the potato experiment?
Answer:
- Use multiple repeats for reliability.
- Control potato size/mass precisely.
- Blot excess water before weighing.
40. Explain the purpose of Visking tubing in osmosis experiments.
Answer: Acts as a partially permeable membrane, allowing water but not solute molecules to pass.
41. Why is a red mesh bag used to model alveoli?
Answer: Represents large surface area and network structure of alveoli.
42. What is the error in Table 3.4 (deodorant smell experiment)?
Answer:
- Units mixed (seconds vs. minutes).
- Independent variable (temperature) not ordered ascendingly.
43. Redraw Table 3.4 correctly.
Answer:
| Temperature (°C) | Time taken (seconds) |
|---|---|
| 10 | 105 |
| 15 | 90 |
| 20 | 60 |
| 25 | 54 |
| 30 | 42 |
44. Design a table for the Visking tubing experiment.
Answer:
| Beaker | Starting Mass (g) | Mass at 5 min (g) | Mass at 10 min (g) | Mass at 15 min (g) | Mass at 20 min (g) |
|---|---|---|---|---|---|
| A | |||||
| B |
45. Why is distilled water used in the Visking tubing experiment?
Answer: To ensure no solutes interfere with osmosis; only water movement is measured.
Adaptations
46. List three adaptations of alveoli.
Answer:
- Large surface area (millions of alveoli).
- Thin walls (one cell thick).
- Moist lining and rich blood supply.
47. How are villi adapted for absorption?
Answer:
- Microvilli on epithelial cells increase surface area.
- Capillaries transport absorbed nutrients.
48. Why do small organisms not need lungs?
Answer: High SA:Vol ratio allows sufficient gas exchange via diffusion through skin/spiracles.
49. Explain the term ‘excretion’.
Answer: Removal of metabolic waste (e.g., urea, carbon dioxide) from the body.
50. How are fish gills adapted for gas exchange?
Answer:
- Gill filaments and lamellae increase surface area.
- Countercurrent flow maintains steep concentration gradient.
Tip: Revise SA:Vol calculations and practise annotating diagrams of exchange surfaces! 🌟
