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1.2.5Unit 1 — Cell Biology

Practical: Osmosis and Diffusion — SQA National 5 Biology Revision Notes & Topical Questions

5 min read 6 parts Lesson 10 of 28

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Part 1 of 6Osmosis in Potato Cylinders

1.2.5 Practical: Osmosis and Diffusion

Potato cells behave like plant cells, so they gain water in dilute solutions and lose water in concentrated ones.

1
Cut several potato cylinders with a cork borer and trim them to the same length, then blot each one dry and record its mass.
2
Place one cylinder in each of several beakers holding the same volume of different sugar solutions, including distilled water as a zero-concentration solution.
3
Leave all the beakers for the same time at the same temperature, such as thirty minutes.
4
Remove each cylinder, blot it dry in the same way, and measure its final mass.
5
Calculate the percentage change in mass for each cylinder and plot it against sugar concentration.

Part 2 of 6Interpreting the Results

Interpreting the Results
Mass increases
  • The cylinder gained water by osmosis because the solution outside had a higher water concentration than the potato cells.
Mass decreases
  • The cylinder lost water because the solution outside had a lower water concentration than the cells.
No change
  • The point where the line crosses zero shows a solution with the same water concentration as the potato cells.
Reliability
  • Repeating with several cylinders per solution and averaging reduces the effect of anomalous results.

Part 3 of 6Diffusion Through a Membrane

Visking tubing
  • A partially permeable tubing models a cell membrane by letting small molecules through but not large ones.
Starch and iodine
  • Starch solution inside the tubing and iodine solution outside;
    • iodine passes in and turns the contents blue-black.
Starch stays
  • Starch molecules are too large to pass out, so the water outside does not turn blue-black.
Conclusion
  • Small molecules can diffuse through a selectively permeable membrane while large molecules cannot.

Part 4 of 6Fair Testing

Fair Testing
Same size
  • All potato pieces must be the same size and shape so the surface area is the same.
Same volume
  • Equal volumes of solution and equal time in each beaker keep the comparison fair.
Blotting
  • Drying every piece in the same way removes surface liquid that would falsely affect the mass.

Part 5 of 6Worked Example

Question

  • A potato cylinder has a mass of 2.50 g at the start.
  • After 30 minutes in a sugar solution its mass is 2.80 g.
  • Calculate the percentage change in mass.

  1. Change in mass = final mass − starting mass = 2.80 − 2.50 = 0.30 g.
  2. Divide the change by the starting mass: 0.30 ÷ 2.50 = 0.12.
  3. Multiply by 100 to give a percentage: 0.12 × 100 = 12%.

Answer +12%, an increase, because the cylinder gained water by osmosis.

Part 6 of 6Exam Checklist & Tips

1.2.5 – Key Points for the Exam

  • Percentage change The change in a value divided by the starting value, multiplied by 100.
  • Visking tubing A partially permeable tubing used to model a cell membrane.
  • Anomalous result A result that does not fit the pattern and may be due to error.
  • Reliable Giving similar results when the experiment is repeated.

Common Mistakes

  • Forgetting to blot the potato before weighing, so surface water gives a false mass.
  • Comparing actual mass changes rather than percentage changes when starting masses differ.
  • Dividing the change by the final mass instead of the starting mass.
  • Saying that iodine did not move when the contents inside turned blue-black; iodine did diffuse in.

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