D2.3 Water potential. Practice questions with markscheme.
68 original IB-style questions on D2.3, written from the 2025 guide: 30 multiple-choice, 24 short-answer, 9 data-based, 2 extended-response parts, 2 drawing, 1 labelling. Below are two ready-made practice papers built from them, one at standard level (20 marks) and one at higher level (20 marks); switch between them above the paper. Hand one out as a class quiz or homework, or sit it yourself and mark against the scheme. Print it, project it, or build a fresh one on the same topic.
What the guide asks for
7 statements at SL and HL, 4 additional higher level.
- D2.3.1SL / HL Solvation with water as the solvent
- D2.3.2SL / HL Water movement from less concentrated to more concentrated solutions
- D2.3.3SL / HL Water movement by osmosis into or out of cells
- D2.3.4SL / HL Changes due to water movement in plant tissue bathed in hypotonic and those bathed in hypertonic solutions
- D2.3.5SL / HL Effects of water movement on cells that lack a cell wall
- D2.3.6SL / HL Effects of water movement on cells with a cell wall
- D2.3.7SL / HL Medical applications of isotonic solutions
- D2.3.8HL Water potential as the potential energy of water per unit volume
- D2.3.9HL Movement of water from higher to lower water potential
- D2.3.10HL Contributions of solute potential and pressure potential to the water potential of cells with walls
- D2.3.11HL Water potential and water movements in plant tissue
In the bank for D2.3
- 30 multiple-choice
- 24 short-answer
- 9 data-based
- 2 extended-response parts
- 2 drawing
- 1 labelling
- 23 higher level only
Every question is original and tagged to a guide statement. See the whole bank →
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The practice paper
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Red blood cells are placed in three solutions. Which row correctly shows how they respond?
| Row | Hypotonic solution | Isotonic solution | Hypertonic solution |
|---|---|---|---|
| A. | swell / burst | no net change | shrink (crenate) |
| B. | shrink | swell | no change |
| C. | no change | burst | swell |
| D. | burst | shrink | no change |
- Row A
- Row B
- Row C
- Row D
A gardener tips a large dose of soluble fertiliser into the pot of a non-woody house plant and waters it in. Within a few hours the plant wilts, even though the soil is still wet. Which explanation is correct?
- The soil solution is now hypertonic to the root cells, so the plant loses water and its cells lose turgor
- The roots absorb so much water from the wet soil that their cells swell until they burst and stop working
- Fertiliser ions enter the root cells, and these extra solutes then pull water out of the cells by osmosis
- Fertiliser ions block the xylem vessels, so that water can no longer be carried up the stem to the leaves
Blood plasma is isotonic with a 0.9 % (w/v) solution of sodium chloride. A hospital pharmacy prepares an intravenous fluid by diluting a sterile 3.6 % (w/v) sodium chloride solution with sterile water. Which volumes should be mixed to make 1000 cm³ of a fluid that can be infused without harming the patient's red blood cells?
- 90 cm³ of the 3.6 % solution and 910 cm³ of water
- 500 cm³ of the 3.6 % solution and 500 cm³ of water
- 750 cm³ of the 3.6 % solution and 250 cm³ of water
- 250 cm³ of the 3.6 % solution and 750 cm³ of water
Tomato (Solanum lycopersicum) fruits store carbohydrate as starch while they are green. As a fruit ripens, the starch in the cells of its flesh is broken down to glucose and fructose. A starch molecule is made of thousands of glucose units joined together and does not dissolve, whereas glucose and fructose are very soluble in water. Water lying on the surface of a fruit can pass slowly through the skin into the flesh. Ripe tomatoes often split open after a few days of rain, but green tomatoes on the same plants rarely do.
Explain why turgor pressure develops in plant cells but not in animal cells placed in a hypotonic solution.
To estimate the solute concentration of potato tissue, a student cut cylinders of potato of the same length and diameter, blotted and weighed them, and immersed each in a different concentration of sucrose solution for 24 hours. The cylinders were then blotted and reweighed, and the percentage change in mass was calculated. Three cylinders were used at each concentration; means are shown.
| Sucrose concentration / mol dm⁻³ | Mean change in mass / % |
|---|---|
| 0.0 | +18.4 |
| 0.2 | +8.9 |
| 0.4 | +0.6 |
| 0.6 | −7.2 |
| 0.8 | −13.5 |
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One mark per point; / separates alternative wording within a point, OR separates alternative answers, words in brackets are not required, underlined words are essential. OWTTE = or words to that effect.
- A: water enters in hypotonic, leaves in hypertonic, no net movement in isotonic solutions;
- A: the soil solution is more concentrated than the cell sap of the root cells, so there is net movement of water out of the roots by osmosis instead of in; the cells lose water and turgor pressure and become flaccid, so the non-woody plant wilts; the cell walls stop plant cells from bursting, and extra solutes inside a cell would draw water in, not out;
- D: 250 cm³ × 3.6 % ÷ 1000 cm³ = 0.9 %, so the fluid is isotonic with plasma and there is no net movement of water into or out of the blood cells; B (1.8 %) and C (2.7 %) are hypertonic, so red blood cells would lose water and shrink (crenate); A (0.32 %) is hypotonic, so they would take in water, swell and could burst;
- (a) [3 max]
- glucose and fructose dissolve: they form hydrogen bonds with water molecules / are each surrounded by a shell of water molecules, whereas starch stays undissolved;
- one starch molecule gives thousands of dissolved sugar molecules, so the solute concentration of the cell sap rises (greatly);
- the cell sap becomes hypertonic to the solution outside the cells / in the cell walls;
- water moves into the cells by osmosis, from the less concentrated to the more concentrated solution;
Do not accept 'water moves from high to low water concentration' as the direction point.
- (b) [3 max]
- rain water is hypotonic to (much more dilute than) the cell sap of the ripe flesh;
- water that passes through the skin enters the flesh cells by osmosis;
- the protoplasts swell and press against their cell walls, so turgor pressure rises;
- the cell walls resist expansion, so the cells do not burst;
- the many swelling cells enlarge the flesh, which pushes outwards on the skin until the skin is stretched further than it can stand and splits, OWTTE;
Do not accept 'the cells burst' as the cause of splitting.
- (c) [1 max]
- their carbohydrate is still (mostly) undissolved starch, so their cell sap has fewer dissolved solute molecules / a lower solute concentration;
- so the sap is less hypertonic to rain water, less water enters and turgor pressure rises less, OWTTE;
- the skin of a green fruit may be stronger / more elastic than the skin of a ripe fruit, so it stretches without splitting, OWTTE;
Accept a suggestion about the skin as an alternative to the solute points.
- in both, water enters by osmosis (down the water-potential gradient);
- the plant cell's (cellulose) wall resists expansion, so (hydrostatic) pressure builds, turgor;
- (turgor) opposes further net water entry, reaching equilibrium;
- animal cells lack a wall, so no pressure develops, they keep swelling and may burst/lyse, OWTTE;
- (a) [1]
- ≈ 0.4 mol dm⁻³ (accept 0.40–0.45), where the change in mass is (closest to) zero;
- (b) [3 max]
- the solution is hypotonic / has a lower solute concentration than the potato cells;
- water enters the cells by osmosis (across the partially permeable membranes);
- (so) the tissue gains mass;
- cells become (more) turgid, OWTTE;
Accept ψ notation.
- (c) [1]
- to remove surface liquid, so only the water inside the tissue is weighed / a fair comparison, OWTTE;
- (d) [2]
- the cylinders' initial masses differed (slightly);
- percentages allow fair comparison / control for initial size, OWTTE;
Content pivot toward experimental skills.
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A unicellular organism living in fresh water has a contractile vacuole that fills and empties every few seconds. When the organism is transferred to a slightly salty solution, the vacuole empties less often. Why?
- Less water enters the cell by osmosis, because the external solution is less hypotonic to the cytoplasm
- The salt enters the cell through the membrane and is pumped out by the contractile vacuole instead of water
- The organism absorbs more water in the salty solution and stores it in the vacuole for a longer time
- The cell wall of the organism resists the entry of water when the external solution contains salt
Three adjacent plant cells lie in a row. Cell X: ψs = −1000 kPa, ψp = +300 kPa. Cell Y: ψs = −800 kPa, ψp = +250 kPa. Cell Z: ψs = −600 kPa, ψp = +100 kPa. Predict the direction of net water movement along the row, showing your reasoning.
Cubes of storage tissue were cut from a turnip (Brassica rapa) root. At the start, the cells had a solute potential (ψs) of −1000 kPa and a pressure potential (ψp) of +200 kPa. The cubes were bathed in a sucrose solution with a water potential of −1200 kPa until there was no further change in their mass. The table shows a student's account of what happened in the cells.
| Statement | Student's account |
|---|---|
| 1 | At the start, the water potential of the cells was −800 kPa. |
| 2 | The solution had a lower water potential than the cells, so water left the cells by osmosis. |
| 3 | As the cells lost water, their cell sap became more dilute, so the solute potential rose towards zero. |
| 4 | When the protoplasts stopped pressing on the cell walls, the pressure potential fell to zero. |
| 5 | As more water left, the plasma membranes pulled away from the walls and the pressure potential became negative, because the shrinking protoplasts were pulling inwards on the walls. |
| 6 | Net movement of water stopped when the water potential of the cells had fallen to −1200 kPa. |
Strips of red onion epidermis were mounted in sucrose solutions of different concentrations. After 15 minutes, 50 cells in each strip were examined and the number that were plasmolysed was recorded.
| Sucrose concentration / mol dm⁻³ | Cells plasmolysed (out of 50) | Percentage plasmolysed / % |
|---|---|---|
| 0.0 | 0 | 0 |
| 0.2 | 3 | 6 |
| 0.3 | 12 | 24 |
| 0.4 | 26 | 52 |
| 0.5 | 41 | 82 |
| 0.6 | 48 | 96 |
Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
Rebuild or edit this exact paper (and its markscheme): biologybybradford.com/exam-maker?code=BbB-EAEAAAAAAhQACFSv
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Markscheme BbB-EAEAAAAAAhQACFSv
One mark per point; / separates alternative wording within a point, OR separates alternative answers, words in brackets are not required, underlined words are essential. OWTTE = or words to that effect.
- A: in fresh water the cytoplasm is strongly hypertonic to the surroundings, so water enters continuously and must be expelled; adding salt reduces the difference in solute concentration, so less water enters and the vacuole needs to empty less often; B fails because the contractile vacuole expels water, not salt; C reverses the direction of change; D fails because such organisms have no cell wall, which is why they need a contractile vacuole;
- ψX = −1000 + 300 = −700 kPa;
- ψY = −800 + 250 = −550 kPa;
- ψZ = −600 + 100 = −500 kPa;
- water moves from higher (less negative) to lower (more negative) ψ, so from Z → Y → X (net movement toward cell X), OWTTE;
Award ECF for a direction consistent with the candidate's calculated values.
- (a) [1]
- statements 3 and 5;
Both are needed for the mark.
- (b) [4 max]
- (statement 3) water leaves but the solutes stay inside the cell / the same amount of solute is now in less water;
- so the cell sap becomes more concentrated and ψs falls / becomes more negative, not towards zero;
- (statement 5) once the plasma membrane has pulled away from the wall, the protoplast no longer pushes on the wall and does not pull on it either, so ψp stays at zero;
- the wall keeps its shape while the protoplast shrinks away from it / the space between the wall and the membrane fills with the external solution, OWTTE;
- negative pressure potentials occur where sap is transported under tension, in xylem vessels, not in living cells with walls, OWTTE;
- ψp cannot fall below zero in a plasmolysed cell, so the cell's water potential equals its solute potential, OWTTE;
Award [2 max] if only one statement is corrected. Accept ψ notation or the words solute and pressure potential.
- (c) [2]
- ψp = 0 kPa, because the cells are plasmolysed / the protoplasts no longer press on the walls;
- ψs = −1200 kPa, because the water potential of the cells equals that of the solution and ψs = ψw − ψp = −1200 − 0;
Allow ECF for ψs from a candidate's stated ψp. Do not accept −1000 kPa, the starting value.
- (a) [2]
- percentage plasmolysed increases with sucrose concentration;
- slowly at first then steeply / steepest between 0.3 and 0.5 mol dm⁻³ (S-shaped), OWTTE;
- (b) [3]
- concentrated sucrose is hypertonic to the cells;
- water leaves the (protoplast of the) cell by osmosis (down its gradient);
- the protoplast/cytoplasm shrinks and (the membrane) pulls away from the wall;
- (c) [1]
- ~0.39 to 0.40 mol dm⁻³; read between 0.3 (24 %) and 0.4 (52 %);
- (d) [2]
- individual cells vary (in solute concentration/response);
- a larger sample gives a more reliable estimate of the true proportion / reduces the effect of chance/anomalies, OWTTE;
More in Theme D · Continuity and change
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- D1.2 Protein synthesis 64
- D1.3 Mutation and gene editing 82
- D2.1 Cell and nuclear division 80
- D2.2 Gene expression 77
- D3.1 Reproduction 70
- D3.2 Inheritance 109
- D3.3 Homeostasis 63
- D4.1 Natural selection 64
- D4.2 Stability and change 66
- D4.3 Climate change 66
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