IB DP Biology: Water potential and osmosis (DP Biology) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover section D2.3 (Water potential) of the IB Biology guide, first assessment 2025, for DP Biology. Questions 1 to 9 test statements D2.3.1 to D2.3.7 (SL and HL). Questions 10 to 12 test D2.3.8 to D2.3.11 and are labelled HL only.

Read the study guide first and use the revision notes for recall. See also the IB DP Biology hub, the printable checklist and Membranes and membrane transport. Not sure where to start? Take a free 10-minute diagnostic.

Questions

1. Explain how water molecules dissolve sodium chloride. [2]

2. A cell has cytoplasm with a solute concentration of 0.35 mol dm⁻³. It is placed in a solution of 0.15 mol dm⁻³ of a solute that cannot cross its membrane.

(a) State the term for the solution compared with the cytoplasm. [1] (b) State the direction of net water movement and explain it in terms of solute concentration. [2]

3. A student writes: “In an isotonic solution, water stops moving across the membrane.” Correct this statement. [3]

4. Red blood cells have no cell wall. Describe and explain what happens to them when they are placed in:

(a) a hypotonic solution [2] (b) a hypertonic solution [2]

5. A freshwater protist has one contractile vacuole. Each time it discharges, it expels 0.2 nL of fluid. In pond water, the vacuole discharges every 12 seconds. After the protist is moved into a salt solution, it discharges every 45 seconds.

(a) Calculate the volume expelled per minute in pond water. [1] (b) Calculate the volume expelled per minute in the salt solution, to 2 significant figures. [1] (c) Calculate the percentage decrease in the volume expelled per minute. [1] (d) Explain the change. [2]

6. Cylinders of turnip were weighed, left in sucrose solutions for one hour, blotted and reweighed. Mean percentage changes in mass (five cylinders each):

Sucrose / mol dm⁻³	0.1	0.2	0.3	0.4
Mean change in mass / %	+9.6	+3.2	−4.4	−9.0

(a) One cylinder fell from 3.40 g to 3.12 g. Calculate its percentage change in mass, to 1 decimal place. [2] (b) Estimate the sucrose concentration isotonic with the turnip cells, using straight-line interpolation. [2] (c) Explain why percentage change is used rather than change in grams. [1] (d) State what the answer to (b) shows about the cell contents. [1]

7. Strips of celery were left in sucrose solutions and their lengths measured before and after. There were six strips for each concentration.

Sucrose / mol dm⁻³	Mean change in length / mm	SD
0.1	+4.8	1.8
0.3	−1.2	2.4

(a) Calculate the standard error for each mean. Give answers to 2 decimal places. [2] (b) Error bars of mean ± SE are added to a graph. Deduce what the bars suggest about the two means. [2] (c) Suggest one way to reduce the standard error. [1]

8. Epidermis from a red onion was mounted in a strong sucrose solution and examined under a microscope. Describe and explain what is seen, and state what fills the space between wall and membrane. [4]

9. Explain why hospitals use isotonic saline for intravenous fluids rather than distilled water, and why a donor organ is bathed in isotonic solution before transplant. [4]

10. (HL only) Plant cell A has ψs = −750 kPa and ψp = +300 kPa. The adjacent cell B has ψs = −620 kPa and ψp = +150 kPa.

(a) Calculate ψw for cell A. [1] (b) Calculate ψw for cell B. [1] (c) State the direction of net water movement and give the reason. [2]

11. (HL only) A plant cell has ψs = −650 kPa and ψp = +350 kPa. It is placed in an open beaker of solution with ψw = −700 kPa. Treat ψs of the cell as unchanged.

(a) Calculate ψw of the cell. [1] (b) State the direction of net water movement and the size of the gradient. [2] (c) Explain what happens to ψp and whether the cell plasmolyses. [2]

12. (HL only) Water potentials in a plant are shown in this table.

Location	ψs / kPa	ψp / kPa
Soil water	−100	0
Root cortex cell	−400	+150
Xylem sap	−50	−400
Leaf mesophyll cell	−900	+300

(a) Calculate ψw at each location, including soil water. [3] (b) Deduce the path of net water movement and justify it. [2] (c) Explain the movement in terms of potential energy. [1] (d) Suggest why ψp is negative in the xylem. [1]

Answers

1. Sodium ions (Na⁺) are attracted to the slightly negative oxygen of water molecules, and chloride ions (Cl⁻) to the slightly positive hydrogens [1]. Water molecules surround each ion and separate it from the crystal lattice, so the ions disperse [1] [2] Examiner insight: Name the charge on each part of the water molecule and link each ion to the correct end; “water is polar so it dissolves salt” alone does not explain the attraction.

2. (a) The solution is hypotonic to the cytoplasm [1]. (b) Net movement of water is into the cell [1], because the solution has a lower solute concentration than the cytoplasm, and water moves towards the higher solute concentration [1] [3] Examiner insight: Give the direction and then the reason. A reason phrased in terms of “water concentration” does not match the required wording.

3. Water does not stop moving [1]. Molecules cross the membrane in both directions at equal rates [1], so there is no net movement: a dynamic equilibrium [1] [3] Examiner insight: A correction needs both halves: movement continues, and the net change is zero. Repeating “no net movement” alone earns only part of the answer.

4. (a) Water enters the cells by osmosis [1] because the solution has lower solute concentration than the cytoplasm; with no wall to resist, the cells swell and burst [1]. (b) Water leaves the cells by osmosis [1]; they shrink and the membrane becomes puckered (crenation) [1] [4] Examiner insight: Use the term crenation for the animal cell. Plasmolysis refers to a cell with a wall, so using it here suggests the structures have been confused.

5. (a) 60 ÷ 12 = 5 discharges; 5 × 0.2 = 1.0 nL per minute [1] (b) 60 ÷ 45 = 1.33 discharges; × 0.2 = 0.27 nL per minute [1] (c) (1.0 − 0.267) ÷ 1.0 × 100 = 73% [1] (d) In the salt solution the medium has a higher solute concentration, so less water enters the protist by osmosis [1]; less excess water has to be removed, so the vacuole discharges less often [1] [5] Examiner insight: Show the number of discharges per minute before multiplying. Using the unrounded value in (c) avoids an error from rounding 0.267 twice.

6. (a) Change = 3.12 − 3.40 = −0.28 g [1]; −0.28 ÷ 3.40 × 100 = −8.2% [1] (b) The mean changes from +3.2 to −4.4 across 0.1 mol dm⁻³, a difference of 7.6 [1]; 0.2 + 0.1 × (3.2 ÷ 7.6) = 0.24 mol dm⁻³ [1] (c) Pieces start at slightly different masses, so a percentage allows fair comparison [1] (d) The cell contents have a solute concentration equal to about 0.24 mol dm⁻³ sucrose [1] [6] Examiner insight: Divide by the starting mass and keep the negative sign. In an interpolation, show the two points used.

7. (a) SE = SD ÷ √n. For 0.1: 1.8 ÷ √6 = 0.73 [1]. For 0.3: 2.4 ÷ √6 = 0.98 [1] (b) The bars span +4.07 to +5.53 mm and −2.18 to −0.22 mm [1]. They do not overlap, which suggests the two means differ in a real way (but does not prove it) [1] (c) Increase the number of repeats [1] [5] Examiner insight: Say “suggests” rather than “proves”. Quote the ranges the bars cover so your comparison is checkable.

8. The sucrose solution is hypertonic to the cell contents, so water leaves the vacuole by osmosis [1]. The protoplast shrinks and the plasma membrane pulls away from the cell wall: plasmolysis [1]. The wall is freely permeable, so the gap fills with the external sucrose solution [1]. The cell loses turgor [1] [4] Examiner insight: The space contains the external solution, not water or air. Mention turgor to show the consequence for the cell.

9. Distilled water is hypotonic to blood plasma [1], so red blood cells would take in water by osmosis and burst [1]. Isotonic saline causes no net water movement, so blood cells keep their normal volume [1]. A donor organ in isotonic solution has no net water movement into or out of its cells, so they are not damaged by swelling or shrinking [1] [4] Examiner insight: Link each medical use to the cells concerned. “Isotonic is safe” does not say what the cells would otherwise do.

10. (a) ψw = −750 + 300 = −450 kPa [1] (b) ψw = −620 + 150 = −470 kPa [1] (c) Net movement is from cell A to cell B [1], because A has the higher (less negative) water potential, so water moves down the gradient of 20 kPa [1] [4] Examiner insight: Compare water potentials, not the solute potentials alone: cell B has the higher ψs but the lower ψw.

11. (a) ψw = −650 + 350 = −300 kPa [1] (b) Water leaves the cell [1], because the cell has the higher water potential; the gradient is 400 kPa [1] (c) As water leaves, ψp falls [1]. When ψp reaches zero the cell’s ψw equals ψs, −650 kPa, which is still higher than −700 kPa, so water continues to leave and the cell plasmolyses [1] [5] Examiner insight: Show the ψp = 0 point explicitly: the comparison between −650 and −700 kPa is what supports the conclusion.

12. (a) Soil water −100 kPa; root cortex −250 kPa; xylem sap −450 kPa; leaf −600 kPa [1] [1] [1] (b) Water moves soil, root cortex, xylem, leaf [1], down a gradient of falling water potential at each step [1] (c) Water moves from where it has higher potential energy per unit volume to where it has lower [1] (d) Sap in the xylem is pulled upwards under tension, so the pressure is below atmospheric pressure [1] [7] Examiner insight: In a calculation table, state each ψw as its own line. Writing the pathway as a list of values in descending order makes the justification clear.

Where marks are usually lost

• Giving direction of water movement in terms of “water concentration” rather than solute concentration.
• Saying water stops moving in an isotonic solution.
• Using plasmolysis for animal cells or crenation for plant cells.
• Dividing by the final mass rather than the initial mass for percentage change.
• Describing standard error bars as proof of a difference.
• Forgetting that the space in a plasmolysed cell contains external solution.
• Writing a positive ψs, or adding ψs and ψp with the wrong signs (HL).
• Comparing ψs where the question asks for ψw (HL).

Next steps

• Revise with the revision notes.
• Re-read the study guide.
• Go to the IB DP Biology hub and the printable checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Section D2.3 (Water potential), under Theme D: Continuity and change.