These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: Ψ = Ψp + Ψs; Ψs = −iCRT; R = 0.0831 L bar mol⁻¹ K⁻¹; T in kelvin = °C + 273; i = 1 for sucrose, 2 for NaCl and 3 for CaCl₂. A calculator is assumed. Keep full calculator values until the last step. All data sets are fictional.
Question 1 (multiple choice · foundation)
A sample of red blood cells is placed in a solution. Within minutes the cells swell and many of them burst. Which statement best explains this?
- (A) The solution was hypotonic to the cells, so there was a net movement of water into the cells by osmosis.
- (B) The solution was hypertonic to the cells, so there was a net movement of water into the cells by osmosis.
- (C) The solution was hypotonic to the cells, so solute particles moved into the cells and the cells swelled with solute.
- (D) The solution was isotonic to the cells, so the cells took in water actively using ATP until they burst.
Answer and explanation
Answer: (A). A hypotonic solution has fewer solute particles than the cytoplasm, so it has a higher water potential. Water moves into the cells. With no wall to resist, the cells swell and lyse.
- (B) uses the wrong term: if the solution were hypertonic, water would leave the cells and they would shrivel.
- (C) describes solute movement. In osmosis, water moves; it is the extra water, not extra solute, that swells the cells.
- (D) is wrong twice: in an isotonic solution there is no net water movement, and osmosis is passive, so it uses no ATP.
Question 2 (multiple choice · core)
What is the solute potential of a 0.15 M calcium chloride (CaCl₂) solution at 27 °C?
- (A) −1.01 bar
- (B) −3.74 bar
- (C) −11.2 bar
- (D) +11.2 bar
Answer and explanation
Answer: (C). T = 27 + 273 = 300 K. Ψs = −iCRT = −(3)(0.15 mol L⁻¹)(0.0831 L bar mol⁻¹ K⁻¹)(300 K) = −11.2 bar.
- (A) uses 27 °C instead of 300 K: −(3)(0.15)(0.0831)(27) = −1.01.
- (B) uses i = 1, ignoring that each CaCl₂ unit gives three ions (one Ca²⁺ and two Cl⁻).
- (D) drops the minus sign. Solute potential is never positive.
Question 3 (multiple choice · core)
A plant cell has a solute potential of −8.0 bar and a pressure potential of +3.0 bar. It is placed in a solution with a water potential of −6.0 bar. Which prediction is correct?
- (A) Water leaves the cell, because the cell’s water potential (−5.0 bar) is higher than that of the solution.
- (B) Water enters the cell, because the cell’s solute potential (−8.0 bar) is lower than the water potential of the solution.
- (C) There is no net water movement, because the cell wall stops water entering or leaving.
- (D) Water leaves the cell, because water moves from lower water potential to higher water potential.
Answer and explanation
Answer: (A). Cell: Ψ = Ψp + Ψs = +3.0 + (−8.0) = −5.0 bar. The solution is at −6.0 bar, which is lower. Water moves from higher to lower Ψ, so it leaves the cell.
- (B) ignores the pressure potential. You must compare total water potentials, not one component.
- (C) is wrong because the cell wall is fully permeable to water; it resists expansion but does not stop water moving.
- (D) gives the right direction for the wrong reason: water moves from higher to lower water potential.
Question 4 (graph construction · core)
To estimate the water potential of onion cells, a student places strips of onion epidermis in sodium chloride solutions at 20 °C. After 30 minutes she counts 50 cells on each slide and records how many are plasmolysed.
| NaCl / M | 0.00 | 0.05 | 0.10 | 0.15 | 0.20 | 0.25 | 0.30 |
|---|---|---|---|---|---|---|---|
| Plasmolysed cells (out of 50) | 0 | 1 | 5 | 19 | 33 | 45 | 50 |
(a) Convert the counts to percentages and describe how you would construct a suitable graph. (b) Use the graph to estimate the concentration at which 50% of cells are plasmolysed. (c) At this concentration, the average cell is just starting to plasmolyse, so its pressure potential is about zero. Estimate the water potential of the onion cells. (d) Explain why few cells are plasmolysed at 0.10 M while all are plasmolysed at 0.30 M.
Worked solution and suggested Marlbridge rubric
(a) Percentages (count ÷ 50 × 100): 0, 2, 10, 38, 66, 90 and 100%. Plot NaCl concentration / M on the x-axis (independent variable) and percentage of cells plasmolysed / % on the y-axis (dependent variable), with even scales from 0 to 0.30 M and 0 to 100%. Plot the points and join them with a smooth curve or straight segments (the relationship is S-shaped, not a straight line). Give the graph a title.
(b) 50% lies between 0.15 M (38%) and 0.20 M (66%). Interpolating: 0.15 + (50 − 38) ÷ (66 − 38) × 0.05 = 0.17 M (accept 0.16–0.18 M read from a curve).
(c) With Ψp ≈ 0, the cells’ Ψ equals the solution’s Ψ. Ψs = −iCRT = −(2)(0.17)(0.0831)(293) = −8.3 bar.
(d) Cells in one tissue differ slightly in their solute content. At 0.10 M, the solution’s Ψ (about −4.9 bar) is higher than that of most cells, so few lose enough water to plasmolyse. At 0.30 M (about −14.6 bar) the solution is hypertonic to every cell, so all of them lose water from the vacuole and the membranes pull away from the walls.
| Point | What earns it |
|---|---|
| 1 | Percentages correct; concentration on x-axis and percentage on y-axis, with units and even scales |
| 1 | 50% concentration read as 0.16–0.18 M |
| 1 | Ψ ≈ −8.3 bar using i = 2 and T = 293 K (carry forward from (b)) |
| 1 | Variation between cells explained in terms of water potential |
Question 5 (constructed response · core)
A freshwater single-celled organism has a contractile vacuole. A student transfers cells into solutions of a solute that cannot cross the plasma membrane and counts how often the contractile vacuole empties.
| Solute concentration / mmol L⁻¹ | 0 | 25 | 50 | 75 | 100 |
|---|---|---|---|---|---|
| Contractions per minute (mean) | 12.0 | 9.0 | 6.5 | 4.0 | 2.0 |
(a) Describe the trend, using numbers. (b) Explain the trend in terms of water potential. (c) Predict what would happen to cells in pond water (0 mmol L⁻¹) treated with a substance that stops ATP production. Justify your prediction.
Model answer and suggested Marlbridge rubric
(a) As solute concentration rises from 0 to 100 mmol L⁻¹, the rate of contraction falls from 12.0 to 2.0 per minute, a decrease of about 83%. The time between contractions rises from 5 s to 30 s.
(b) Adding solute lowers the water potential of the outside solution, so it gets closer to the water potential of the cytoplasm. The difference in water potential is smaller, so less water enters by osmosis each minute. The vacuole has less water to remove, so it fills and empties less often.
(c) The cells would swell and burst. Pond water is hypotonic to the cytoplasm, so water keeps entering. Expelling water with the contractile vacuole needs energy; without ATP it stops working. The cell has no wall to resist swelling.
| Point | What earns it |
|---|---|
| 1 | Trend described with at least two quoted values |
| 1 | Higher external solute lowers outside Ψ, so the water potential difference is smaller |
| 1 | Less water enters by osmosis, so less needs to be pumped out |
| 1 | Prediction of swelling/lysis, linked to energy need and no cell wall |
Question 6 (constructed response · stretch)
Bony fish keep the solute concentration of their body fluids within a narrow range, whether they live in fresh water or in the sea.
(a) For a freshwater fish, state whether its body fluids are hypotonic or hypertonic to the water around it, and the direction of water movement across its gills. (b) Describe two ways a freshwater fish keeps its water and solute balance. (c) A marine bony fish lives in water that is hypertonic to its body fluids. Explain why it drinks seawater even though seawater is salty, and what it must then do. (d) Explain why both fish need a continuous supply of energy to osmoregulate.
Model answer and suggested Marlbridge rubric
(a) Its body fluids are hypertonic to fresh water, so water moves into the fish across the gills (from higher to lower water potential).
(b) It produces large volumes of dilute urine to remove the extra water, and it takes up salts from the water through its gills (and gets some from food).
(c) The fish loses water by osmosis to the hypertonic seawater, so it must replace it. Drinking seawater gains water but also salt, so it must remove the extra salt, for example through cells in its gills, and it produces only small amounts of urine.
(d) In both cases the fish keeps its internal solute concentration different from its surroundings. Ions must be moved against their concentration gradients (taken up in fresh water, removed in seawater), which needs active transport and so a supply of ATP. Without it, passive movement would bring internal and external concentrations closer and cells would swell or shrink.
| Point | What earns it |
|---|---|
| 1 | Freshwater fish hypertonic to surroundings, water enters |
| 1 | Two correct freshwater mechanisms |
| 1 | Marine fish loses water, so drinks, then removes excess salt |
| 1 | Energy needed because ions move against gradients to keep internal conditions different from outside |
Question 7 (calculation · stretch)
A plant cell with a solute potential of −9.0 bar is placed in an open beaker of 0.25 M sucrose at 25 °C. It reaches equilibrium with no change in its solute potential.
(a) Calculate the water potential of the sucrose solution. (b) Calculate the pressure potential of the cell at equilibrium. (c) The cell is then moved to an open beaker of 0.50 M NaCl at 25 °C. Calculate the water potential of this solution and predict what happens to the cell.
Worked solution
(a) T = 25 + 273 = 298 K. Ψ = Ψs = −(1)(0.25)(0.0831)(298) = −6.19 bar (open beaker, so Ψp = 0).
(b) At equilibrium the cell’s Ψ equals the solution’s Ψ: Ψp + (−9.0) = −6.19, so Ψp = −6.19 + 9.0 = +2.81 bar (about +2.8 bar). The cell is turgid.
(c) Ψ = −(2)(0.50)(0.0831)(298) = −24.8 bar. Even when Ψp falls to zero, the cell’s water potential can fall only to −9.0 bar, which is still far above −24.8 bar. Water leaves the cell, turgor is lost (Ψp falls to 0) and the cell plasmolyses.
Suggested mark points (3): 1 for −6.19 bar with T in kelvin; 1 for Ψp ≈ +2.8 bar from setting the two water potentials equal; 1 for −24.8 bar with i = 2 and the prediction of water loss and plasmolysis.
Common errors: leaving out i for NaCl gives −12.4 bar, which still predicts water loss but is the wrong value; using −9.0 as the cell’s Ψ in (b) forgets that pressure potential counts.
How did you do?
- Q1 or Q3 wrong: re-read “Tonicity: always a comparison” and “Water potential” in the study guide, and look again at Figure 1.
- Q2 or Q7 calculations wrong: rework Worked example 1; check i, the minus sign and kelvin.
- Q4 graph or reading wrong: re-read Worked example 2 and Figure 2 on choosing axes, scales and reading where a line crosses a value.
- Q5 or Q6 incomplete: re-read “Osmoregulation: keeping water balance within limits”.
Then tick off the topic checklist.