Potato in sugar solutions — answer key

1. PQ1 · Practice

Model answer. b. The cell membrane is partially permeable. The cell wall is fully permeable: water and dissolved substances pass straight through it.

A full answer includes

• I chose b.

Also accept

• b

Watch for

• “The cell wall controls what goes in and out.” — The wall is strong but fully permeable; the cell membrane is partially permeable.

2. PQ2 · Practice

Model answer. (2.16 − 2.40) ÷ 2.40 × 100 = −0.24 ÷ 2.40 × 100 = −10 %. The cylinder lost 10 % of its mass.

Checked automatically

• Percentage change in mass: -10 % (within ±0.2 %)
  • Common wrong answer 10 : The mass went down: the change is negative.
  • Common wrong answer -0.24 : That is the change in grams. Divide by 2.40 g and multiply by 100.
  • Common wrong answer -11.1 : Divide by the starting mass, 2.40 g, not the final mass.

A full answer includes

• (2.16 − 2.40) ÷ 2.40 × 100 = −10 %.

Also accept

• −10 %

Watch for

• “The change is 0.24 %.” — That is the change in grams. Divide it by 2.40 g and multiply by 100.
• “Divide by the final mass.” — Percentage change is always worked out from the starting mass.

3. PQ3 · Practice

Options as numbered on the pupil sheet: 1 — Cut cylinders of the same size from one potato; 2 — Blot each cylinder gently and weigh it; 3 — Put each cylinder into its own solution for 30 minutes; 4 — Lift each cylinder out and blot it the same way; 5 — Reweigh each cylinder and calculate the percentage change in mass.

Model answer. Cut cylinders of the same size from one potato; blot and weigh each; put each into its own solution for 30 minutes; lift out and blot each the same way; reweigh and calculate the percentage change in mass.

A full answer includes

• Cut cylinders the same size → blot and weigh each → put each in its own solution for 30 minutes → lift out and blot each → reweigh and calculate the percentage change.

Also accept

• cut, blot and weigh, soak, lift out and blot, reweigh and calculate

Watch for

• “Blotting is only needed after soaking.” — Blot before the first weighing too, so the surface liquid is removed the same way each time.

4. PQ4 · Practice

Model answer. The volume of solution, the temperature, the time in the solution and the size of the cylinders are controlled variables. The concentration of sucrose is the independent variable: it is changed on purpose.

A full answer includes

• I selected the volume of solution, the temperature, the time and the size of the cylinders, and not the concentration.

Also accept

• volume, temperature, time, size of cylinder

Watch for

• “Keep the concentration the same.” — The concentration is what is being tested: it is the independent variable.

5. PQ5 · Practice

Options as numbered on the pupil sheet: 1 — Turgid; 2 — Flaccid; 3 — Plasmolysis; 4 — Turgor pressure.

Model answer. Contents pressing on the cell wall: turgid. Contents no longer pressing on the wall: flaccid. Cytoplasm and membrane pulled away from the wall: plasmolysis. The pressure of the contents on the wall: turgor pressure.

A full answer includes

• Contents press on the wall: turgid.
• Contents no longer press on the wall: flaccid.
• Cytoplasm and membrane pulled away from the wall: plasmolysis.
• Pressure of the cell contents on the wall: turgor pressure.

Also accept

• turgid; flaccid; plasmolysis; turgor pressure

Watch for

• “A flaccid cell is plasmolysed.” — A flaccid cell’s contents still touch the wall; in plasmolysis they pull away after more water loss.

6. PQ6 · Practice

Model answer. (a) About 0.29 mol/dm³, where the line crosses 0 %. (b) Water has a higher water potential than the cell sap, so there was a net movement of water into the cells by osmosis through their partially permeable membranes, and the cylinder gained mass. (c) The cylinders did not all start with the same mass; percentage change compares the changes fairly.

A full answer includes

• (a) About 0.29 mol/dm³ (accept 0.25–0.35).
• (b) Water has a higher water potential than the cell sap, so water entered the cells by osmosis through their membranes.
• (c) The cylinders started with slightly different masses; percentage change allows a fair comparison.

Also accept

• (a) about 0.3 mol/dm³; (b) water entered the cells by osmosis; (c) different starting masses

Watch for

• “At 0.29 mol/dm³ water stops moving.” — Water still crosses both ways; there is no net movement.

7. PT1 · Apply

Model answer. In water: the water has a higher water potential than the cell sap, so water enters the cells by osmosis; the cells become turgid and press on their walls, and turgor pressure holds the stems and petals up. In the strong sugar solution: it has a lower water potential than the cell sap, so water leaves the cells by osmosis; they stay flaccid or become plasmolysed and the flowers stay limp or droop further.

A full answer includes

• Water: the flowers stand up again: water enters the cells by osmosis, they become turgid and turgor pressure supports the stems.
• Strong sugar solution: they stay limp or droop more: water leaves the cells by osmosis, so they stay flaccid (or become plasmolysed).

Also accept

• water: water enters, cells turgid, flowers revive; sugar solution: water leaves, cells flaccid, flowers droop

Watch for

• “Sugar is food for the flowers, so they recover faster.” — A strong sugar solution has a low water potential: the cells lose water by osmosis.