The shape of the lesson
By the end of the lesson, learners can
- Water potential: Set pure water on the left and 0.50 mol dm −3 sucrose on the right, press Start and read both water potentials.
- Animal and plant cells: Use the Pure water, Isotonic for blood and 0.8 mol dm −3 buttons, then drag the Sucrose slider slowly.
- Measuring tissue water potential: Press Soak for 24 hours, then Line of best fit.
- Pure water = 0 kPa; solutions are negative.
- Osmosis: water moves from higher to lower ψ across a partially permeable membrane.
- Animal cells burst or crenate; plant cells become turgid or plasmolysed.
- No change in mass = solution ψ equals tissue ψ.
One picture every idea returns to
Water potential (ψ) is the tendency of water to move out of a solution, measured in kPa. Pure water has the highest water potential, 0 kPa; dissolving solutes lowers it, so all solutions have negative values. Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential, through a partially permeable membrane. An animal cell in a solution of higher water potential gains water and may burst; in a lower one it loses water and shrinks. A plant cell's wall stops it bursting, so it becomes turgid; in a solution of lower water potential the protoplast pulls away from the wall: plasmolysis. Soaking plant tissue in a range of solutions and finding where there is no change in mass gives the water potential of the tissue.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Water runs downhill in potential.. Drop a red blood cell into pure water and it bursts. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Water potential | Set pure water on the left and 0.50 mol dm −3 sucrose on the right, press Start and read both water potentials. Project tab 5 of the lab (Cell membranes lab, osmosis) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Animal and plant cells | Use the Pure water, Isotonic for blood and 0.8 mol dm −3 buttons, then drag the Sucrose slider slowly. Project tab 6 of the lab (Cell membranes lab, cells in solutions) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Measuring tissue water potential | Press Soak for 24 hours, then Line of best fit. Project tab 7 of the lab (Cell membranes lab, potato practical) and let a student drive it while the class predicts. | Anticipation / Reaction Guide |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 18 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Measuring tissue water potential, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Say "from a higher water potential to a lower water potential", not "from a high concentration to a low concentration". "Concentration" is ambiguous (of water or of solute?), and AS mark schemes expect osmosis to be explained in terms of water potential. | Return to Water potential: ask a pair to state the correct version and show it on the lab before moving on. |
| A cylinder that gains mass was in a solution of higher water potential than its cells. | Do not say the potato "absorbed sucrose": the membrane is partially permeable, and the mass change is due to water moving by osmosis. |
Osmosis and water potential: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
Define osmosis in terms of water potential.
Three plant cells, X, Y and Z, are next to each other in a row. Their water potentials are X −300 kPa, Y −550 kPa and Z −700 kPa. Describe and explain the movement of water between them.
Red blood cells and onion epidermis cells are each placed in a concentrated sucrose solution. Describe and explain what happens to each type of cell.
A potato cylinder had a mass of 2.40 g before soaking and 2.13 g after soaking in a sucrose solution. Calculate the percentage change in mass.
Potato cylinders were soaked in sucrose solutions. Percentage change in mass: 0.0 mol dm−3 +12%; 0.2 mol dm−3 +5%; 0.4 mol dm−3 −3%; 0.6 mol dm−3 −9%. A table gives the water potential of 0.30 mol dm−3 sucrose as −850 kPa and of 0.35 mol dm−3 as −990 kPa. Estimate the water potential of the potato tissue.
Suggest three ways to improve the reliability or accuracy of the potato cylinder investigation.
Answer key and marking notes
- The net movement of water molecules from a region of higher water potential to a region of lower water potential.
- Through a partially permeable membrane.
- Water moves from X to Y, because −300 kPa is higher than −550 kPa.
- Water moves from Y to Z, and also from X to Z, because Z has the lowest (most negative) water potential.
- The movement is by osmosis, through the partially permeable cell surface membranes.
- The solution has a lower water potential than the cell contents, so water leaves the cells by osmosis.
- Red blood cells shrink and become crenated (wrinkled), because there is no cell wall.
- In the onion cells the vacuole and cytoplasm shrink and the protoplast pulls away from the cell wall: plasmolysis.
- The cell wall is fully permeable, so sucrose solution fills the space between the wall and the membrane.
- Change = 2.13 − 2.40 = −0.27 g.
- Percentage change = −0.27 ÷ 2.40 × 100 = −11.25%, so −11% (a loss).
- Between 0.2 and 0.4 mol dm−3 the change falls from +5% to −3%, a fall of 8% over 0.2 mol dm−3.
- Zero is reached 5 ÷ 8 of the way: 0.2 + 0.125 = 0.325 mol dm−3 (or read from the line of best fit).
- No change in mass means the solution and tissue have the same water potential.
- 0.325 mol dm−3 lies halfway between the table values, so the tissue water potential is about −920 kPa.
- Cut all cylinders from the same potato, with the same cork borer and to the same length, and remove the skin.
- Blot every cylinder in the same way before each weighing, so surface liquid does not add mass; keep the time and temperature constant.
- Use three or more cylinders per concentration and calculate a mean; use more concentrations close to where the line crosses zero.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which of these has the lowest water potential?
- A1.0 mol dm−3 sucrose solution
- Bpure water
- C0.1 mol dm−3 sucrose solution
- D0.5 mol dm−3 sucrose solution
Red blood cells, with contents of water potential −750 kPa, are placed in a salt solution with a water potential of −1200 kPa. What happens to the cells?
- Athey lose water by osmosis, shrink and become crenated
- Bthey gain water by osmosis, swell and burst
- Cthey gain water and become turgid
- Dnothing happens, because salt cannot cross the membrane
An onion epidermal cell in a concentrated sucrose solution is plasmolysed. What fills the space between the cell wall and the cell surface membrane?
- Athe external sucrose solution
- Bair that has entered through the cell wall
- Ccytoplasm that has leaked out of the protoplast
- Dpure water that has left the vacuole
A potato cylinder had a mass of 3.20 g before and 3.52 g after soaking in a sucrose solution. What is the percentage change in mass?
- A+10.0%
- B+9.1%
- C+0.32%
- D−10.0%
A Visking tubing bag containing 0.5 mol dm−3 sucrose solution is placed in a beaker of 0.2 mol dm−3 sucrose solution. What happens?
- Awater moves into the bag by osmosis, so the bag gains mass
- Bwater moves out of the bag by osmosis, so the bag loses mass
- Csucrose moves out of the bag by diffusion until both are 0.35 mol dm−3
- Dthere is no net movement, because both liquids contain sucrose
The table shows results for carrot cylinders in sucrose solutions. In this model the water potential of sucrose solution is −2476 kPa per 1 mol dm−3. What is the best estimate of the water potential of the carrot tissue?
- A−620 kPa
- B−740 kPa
- C−500 kPa
- D+620 kPa
Fluid given to patients through a drip into a vein is 0.9% sodium chloride solution, not pure water. Why?
- Ait has about the same water potential as blood plasma, so red blood cells neither gain nor lose water
- Bit has a higher water potential than blood plasma, so cells gain water slowly and safely
- Cit supplies sodium ions, which red blood cells need for active transport of water
- Dpure water would have a lower water potential than the cells, so they would become crenated
Strips of onion epidermis were placed in sucrose solutions and the percentage of cells showing plasmolysis was counted: 0.30 mol dm−3 0%; 0.40 mol dm−3 20%; 0.50 mol dm−3 50%; 0.60 mol dm−3 85%. Which concentration gives the best estimate of the solution with the same water potential as the average cell?
- A0.50 mol dm−3, where half of the cells are plasmolysed
- B0.30 mol dm−3, the strongest solution in which no cells are plasmolysed
- C0.40 mol dm−3, where plasmolysis is first seen
- D0.60 mol dm−3, where most of the cells are plasmolysed
Quiz key
Adding solute lowers the water potential, so the most concentrated solution, 1.0 mol dm−3, has the most negative and therefore the lowest ψ. Pure water has the highest possible water potential, 0 kPa. The 0.1 mol dm−3 solution is only slightly below zero. The 0.5 mol dm−3 solution is lower than 0.1 but still higher (less negative) than 1.0 mol dm−3. Do not confuse a high solute concentration with a high water potential: they are opposites.
−1200 kPa is lower (more negative) than −750 kPa, so water leaves the cells by osmosis into the solution; with no cell wall, the cells shrink and their surface becomes crenated. Swelling and bursting (haemolysis) happens in a solution of higher water potential, such as pure water. Turgidity needs a cell wall to push against, which red blood cells do not have. Whether or not the salt crosses, water crosses the partially permeable membrane down its water potential gradient, so the cells do change.
The cellulose cell wall is fully permeable, so the external sucrose solution passes through it and fills the space left as the protoplast shrinks. Air cannot enter, because the tissue is surrounded by solution. The cytoplasm stays inside the cell surface membrane; it shrinks with the vacuole rather than leaking out. Water does leave the vacuole, but it mixes into the external solution, so the space contains that solution, not pure water.
Change = 3.52 − 3.20 = +0.32 g; percentage change = 0.32 ÷ 3.20 × 100 = +10.0%, a gain. +9.1% divides by the final mass (3.52 g) instead of the initial mass. +0.32% is the change in grams written as a percentage, without dividing by the initial mass. −10.0% has the right size but the wrong sign: the mass went up, so the cylinder gained water.
The 0.5 mol dm−3 solution inside has a lower water potential than the 0.2 mol dm−3 solution outside, so water moves into the bag by osmosis and it gains mass. Water moving out would mean moving from lower to higher water potential, which is the wrong direction. Sucrose molecules are too large to pass through the pores of Visking tubing, which is partially permeable. Both liquids contain sucrose, but at different concentrations, so there is a water potential gradient and net movement.
The change goes from +2.0% at 0.20 mol dm−3 to −2.0% at 0.30 mol dm−3, so zero change is halfway, at 0.25 mol dm−3. At that concentration there is no net osmosis, so the solution and the tissue have the same water potential: 0.25 × −2476 = −619, about −620 kPa. −740 kPa is the value for 0.30 mol dm−3, the first solution in which the tissue lost mass, not the point of no change. −500 kPa is the value for 0.20 mol dm−3, the last solution in which it gained mass. +620 kPa has the wrong sign: cells contain solutes, so their water potential must be below 0 kPa.
0.9% sodium chloride has about the same water potential as plasma and red blood cell contents, so there is no net osmosis and the cells keep their normal shape. Pure water would raise the plasma water potential above that of the cells, so water would enter them and they could burst (haemolysis). A higher water potential than plasma is exactly what is being avoided; cells without walls have nothing to stop them swelling. Water is never moved by active transport; it moves by osmosis. Pure water has the highest water potential possible, 0 kPa, so it would cause swelling and bursting, not crenation.
Cells in a tissue have slightly different water potentials. Where 50% are plasmolysed, the average cell is just at incipient plasmolysis, where its water potential equals that of the solution, so 0.50 mol dm−3 is the best estimate. At 0.30 mol dm−3 no cell is plasmolysed, which only shows that every cell has a water potential at or below that of this solution. At 0.40 mol dm−3 only the 20% of cells with the highest water potential have plasmolysed, so it reflects the extreme cells, not the average. At 0.60 mol dm−3 the solution is below the water potential of most cells, so it overestimates the concentration.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.