Osmosis

This lesson covers osmosis — a special case of diffusion — as required by the Edexcel GCSE Biology specification (1BI0), Topic 1: Key Concepts in Biology. You need to understand the definition of osmosis, its effects on plant and animal cells, and be able to describe Core Practical 2 — investigating the effect of different concentrations of sucrose solution on potato tissue.

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What Is Osmosis?

Osmosis is the movement of water molecules from a dilute solution (high water concentration) to a more concentrated solution (lower water concentration) through a partially permeable membrane.

Key Points

• Osmosis is a special case of diffusion — it involves only water molecules.
• Water moves from where there is a higher water potential (dilute solution, more water molecules) to where there is a lower water potential (concentrated solution, fewer free water molecules).
• The partially permeable membrane allows water molecules to pass through but prevents larger solute molecules from passing.
• Osmosis is a passive process — it does not require energy (ATP).
• Osmosis continues until the water concentration (water potential) is equal on both sides of the membrane.

Exam Tip: The key phrase is "through a partially permeable membrane". Without the membrane, it is just diffusion. The membrane is what makes osmosis different from ordinary diffusion of water.

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Understanding the Terms

Term	Meaning
Dilute solution	Contains a low concentration of solute and a high concentration of water molecules
Concentrated solution	Contains a high concentration of solute and a low concentration of water molecules
Partially permeable membrane	A membrane with tiny pores that allow small molecules (like water) through but block larger molecules (like sugar)
Water potential	A measure of the tendency of water molecules to move. Pure water has the highest water potential. Adding solute lowers the water potential.

Direction of Water Movement

Water always moves from:

• Dilute solution → Concentrated solution
• High water potential → Low water potential

Or equivalently:

• High water concentration → Low water concentration

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Osmosis and Plant Cells

Plant cells are surrounded by a cell membrane (partially permeable) and a cell wall (freely permeable — allows all molecules through). The behaviour of plant cells in different solutions is very important.

Plant Cell in a Dilute Solution (e.g. Pure Water)

1. The solution outside the cell is more dilute (higher water potential) than the cell sap inside the vacuole (lower water potential).
2. Water enters the cell by osmosis through the partially permeable cell membrane.
3. The vacuole swells and pushes the cytoplasm against the cell wall.
4. The cell wall prevents the cell from bursting — it stretches slightly but resists further expansion.
5. The cell becomes turgid (firm and swollen).
6. Turgor pressure is the pressure of the swollen contents pushing against the cell wall.

Turgid cells are important because they provide support for the plant. A plant with turgid cells stands upright. This is why plants wilt when they lose water.

Plant Cell in a Concentrated Solution

1. The solution outside the cell is more concentrated (lower water potential) than the cell sap inside the vacuole (higher water potential).
2. Water leaves the cell by osmosis.
3. The vacuole shrinks and the cytoplasm contracts.
4. The cell membrane pulls away from the cell wall — this is called plasmolysis.
5. The cell is described as plasmolysed (or flaccid before full plasmolysis).

A plasmolysed cell has lost so much water that the cell membrane has detached from the cell wall. The space between the membrane and wall fills with the external solution.

If a plant loses too much water from its cells, it wilts — the cells become flaccid and can no longer support the plant's stems and leaves.

Plant Cell in an Isotonic Solution

• The solution outside has the same concentration as the cell sap.
• There is no net movement of water (water moves equally in both directions).
• The cell is flaccid — neither turgid nor plasmolysed.

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Osmosis and Animal Cells

Animal cells do not have a cell wall — they only have a cell membrane. This makes them much more vulnerable to changes in water concentration.

Animal Cell in a Dilute Solution (e.g. Pure Water)

1. Water enters the cell by osmosis (from high water concentration outside to low water concentration inside).
2. The cell swells.
3. Because there is no cell wall to prevent expansion, the cell continues to swell until it bursts (lyses).
4. This is called lysis (or cytolysis).

Animal Cell in a Concentrated Solution

1. Water leaves the cell by osmosis.
2. The cell shrinks and becomes wrinkled.
3. This is called crenation.

Animal Cell in an Isotonic Solution

• No net movement of water.
• The cell remains its normal shape and size.

Summary Table: Effects of Osmosis on Cells

Solution	Plant Cell	Animal Cell
Dilute (high water potential)	Turgid — swells, firm (cell wall prevents bursting)	Lysis — swells and bursts (no cell wall)
Isotonic (equal water potential)	Flaccid — normal	Normal — no change
Concentrated (low water potential)	Plasmolysed — membrane pulls away from cell wall	Crenation — shrinks and wrinkles

Exam Tip: This comparison table is a favourite exam question. Make sure you know the correct terms: turgid, flaccid, plasmolysed for plant cells; lysis (burst) and crenation (shrink) for animal cells.

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Core Practical 2: Investigating Osmosis Using Potato Cylinders

This is one of the key Edexcel Core Practicals that you must be able to describe and analyse.

Aim

To investigate the effect of different concentrations of sucrose solution on the mass of potato tissue (and therefore demonstrate osmosis).

Equipment

• Potato
• Cork borer (to cut uniform cylinders)
• Ruler
• Electronic balance (accurate to at least 0.01 g)
• Sucrose solutions of different concentrations (e.g. 0.0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M, 1.0 M)
• Boiling tubes or beakers
• Paper towels (for blotting)
• Stopwatch / timer
• Labels

Method

1. Use a cork borer to cut potato cylinders of equal length (e.g. 3 cm) and equal diameter. This ensures they all have the same starting surface area and volume.
2. Blot each cylinder gently with a paper towel to remove excess moisture.
3. Weigh each potato cylinder on an electronic balance and record the initial mass.
4. Place each cylinder into a separate boiling tube containing a different concentration of sucrose solution (e.g. 0.0 M to 1.0 M).
5. Ensure the potato cylinders are fully submerged in the solution.
6. Leave for a set time (e.g. 30 minutes or 24 hours for better results).
7. Remove each cylinder, blot gently to remove surface liquid, and reweigh.
8. Record the final mass and calculate the change in mass.
9. Calculate the percentage change in mass for each concentration.
10. Repeat the experiment at least three times and calculate a mean percentage change for each concentration.

Calculating Percentage Change in Mass

Percentage change = ((final mass − initial mass) ÷ initial mass) × 100

This is used instead of just the change in mass because it accounts for any slight differences in the starting mass of the potato cylinders. It allows a fair comparison.

Worked Example:

A potato cylinder has an initial mass of 2.50 g and a final mass of 2.85 g.

Percentage change = ((2.85 − 2.50) ÷ 2.50) × 100 = (0.35 ÷ 2.50) × 100 = +14.0%

The positive value means the potato gained mass (water entered by osmosis).

Another potato cylinder has an initial mass of 2.60 g and a final mass of 2.20 g.

Percentage change = ((2.20 − 2.60) ÷ 2.60) × 100 = (−0.40 ÷ 2.60) × 100 = −15.4%

The negative value means the potato lost mass (water left by osmosis).

Variables

Variable	Details
Independent variable	Concentration of sucrose solution
Dependent variable	Percentage change in mass of potato cylinder
Control variables	Length and diameter of potato cylinder (same cork borer, same ruler length), time left in solution, temperature, volume of sucrose solution, type of potato, blotting technique

Expected Results