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Diffusion and osmosis: movement that costs nothing questions
Simple and facilitated diffusion, the factors in Fick's law and the relationship between them, osmosis defined by water potential in kilopascals, the equation relating water potential to solute and pressure potential, plasmolysis and incipient plasmolysis in plant cells, haemolysis and crenation in animal cells, and the potato serial dilution investigation.
6 original questions · 20 marks · the diffusion and osmosis: movement that costs nothing notes · Membranes and transport across cells
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A potato cylinder of initial mass 4.20 g is left in a sucrose solution for thirty minutes, blotted dry and reweighed at 3.78 g. Calculate the percentage change in mass and state the direction of net water movement.
Mark scheme
- M1 change in mass = 3.78 − 4.20 = −0.42 g
- M1 percentage change = change in mass divided by initial mass, multiplied by one hundred: (−0.42 ÷ 4.20) × 100
- A1 −10.0 per cent, that is a loss of a tenth of the initial mass
- B1 water moved out of the cylinder into the solution, so the solution had the lower, more negative, water potential
Explain what happens to a red blood cell placed in distilled water, and what happens to an identical red blood cell placed in a concentrated sodium chloride solution.
Mark scheme
- B1 distilled water has a water potential of 0 kPa, which is higher, that is less negative, than the roughly −800 kPa of the cell contents
- B1 water therefore enters by osmosis through the partially permeable cell surface membrane, and with no cell wall to resist the pressure the cell bursts, which is haemolysis
- B1 the concentrated sodium chloride solution has a lower, more negative, water potential than the cell contents, so water leaves the cell by osmosis
- B1 the cell shrinks and puckers into a spiky shape, which is crenation
The uptake of glucose by red blood cells rises as the external glucose concentration rises and then levels off, while the uptake of oxygen by the same cells continues to rise across the whole range tested. Suggest an explanation for both results.
Mark scheme
- B1 glucose is a large polar molecule and cannot cross the hydrophobic core unaided, so it enters by facilitated diffusion through carrier proteins
- B1 the rate levels off once every carrier protein is occupied, so a steeper concentration gradient cannot raise it any further
- B1 oxygen is small and non-polar, so it dissolves in the bilayer and crosses by simple diffusion between the phospholipids
- B1 simple diffusion uses no proteins, so there is nothing to become saturated and the rate keeps rising with the concentration difference
A plant cell has a water potential of −1150 kPa and a pressure potential of +450 kPa. Calculate the solute potential of the cell, giving your answer in kPa.
Explain how the alveoli of the lungs and the blood flowing past them maintain a high rate of diffusion of oxygen into the blood.
State the water potential of pure water at atmospheric pressure, and state what this tells you about the sign of the water potential of any solution.
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