Questions on diffusion and osmosis

Question 1

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.

Accepted Answer

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

Question 2

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.

Accepted Answer

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; for Pearson Edexcel Biology A, accept a higher concentration of free water molecules in the distilled water than in 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; for Pearson Edexcel Biology A, accept a lower concentration of free water molecules in the solution than in the cell contents B1 the cell shrinks and puckers into a spiky shape, which is crenation

Question 3

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.

Accepted Answer

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

Question 4

A mammal's lungs have a gas-exchange surface area of 60 m² and a diffusion barrier 0.3 µm thick. A fish's gills have a surface area of 3 m² and a barrier 15 µm thick. Assuming the same concentration difference for both, calculate the value of area ÷ thickness for each, and calculate how many times greater the mammal's rate of diffusion would be than the fish's.

Accepted Answer

M1 for the mammal, area ÷ thickness = 60 ÷ 0.3 = 200 A1 for the fish, area ÷ thickness = 3 ÷ 15 = 0.2 M1 divide the mammal's value by the fish's: 200 ÷ 0.2 A1 1000 times greater

Question 5

Compare simple diffusion with facilitated diffusion, referring to the route taken, the substances each carries, and what happens to the rate at a very high concentration gradient.

Accepted Answer

B1 simple diffusion occurs directly between the phospholipids of the bilayer, whereas facilitated diffusion occurs through a channel or on a carrier protein B1 simple diffusion carries small non-polar molecules such as oxygen and carbon dioxide, whereas facilitated diffusion carries substances that cannot cross the bilayer unaided, such as ions and glucose B1 at a very high concentration gradient the rate of simple diffusion keeps rising roughly in proportion, whereas facilitated diffusion levels off once the available channels or carriers are saturated B1 neither process requires ATP, since both move a substance down its own gradient

Question 6

A plant cell has a water potential of −300 kPa and a solute potential of −900 kPa. Calculate its pressure potential, and calculate its new pressure potential if, after further water uptake, its water potential rises to −50 kPa while its solute potential remains −900 kPa.

Accepted Answer

M1 rearrange ψ = ψs + ψp to give ψp = ψ − ψs, and substitute the first pair of values: −300 − (−900) A1 600 kPa, the pressure potential M1 substitute the second pair of values: −50 − (−900) A1 850 kPa, the new pressure potential

Question 7

Two agar cubes set with the same alkali-phenolphthalein mixture are cut to different sizes, a 5 mm cube and a 20 mm cube, and both are placed in the same dilute acid at the same time. Suggest which cube loses its colour first, and suggest the reason in terms of surface area to volume ratio.

Accepted Answer

B1 the smaller, 5 mm cube loses its colour first B1 the smaller cube has a larger surface area to volume ratio than the larger cube, so relatively more of its volume is close to a face through which acid can diffuse in B1 no point inside the small cube is far from a surface, so acid reaches the centre sooner, whereas in the large cube the centre is much further from any face B1 the acid and the jelly are the same in both cases, so the only thing that differs between the two cubes is this geometric ratio

Question 8

A nurse mistakenly infuses a patient with distilled water directly into a vein instead of isotonic (0.9%) saline. Suggest the effect this would have on the patient's red blood cells, and suggest why isotonic saline is used instead in normal practice.

Accepted Answer

B1 distilled water has a water potential of 0 kPa, which is much higher, less negative, than the roughly −800 kPa of the red blood cells' contents; for Pearson Edexcel Biology A, accept a higher concentration of free water molecules in distilled water than in the cells B1 water would enter the red blood cells by osmosis, down this water potential gradient, through their partially permeable cell surface membranes B1 with no cell wall to resist the rising pressure, the cells would swell and burst, which is haemolysis B1 isotonic saline is formulated to have almost the same water potential as blood plasma, so there is no net osmotic movement of water into or out of the red blood cells, keeping them at their normal volume

Question 9

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.

Accepted Answer

M1 rearranges the relationship to give solute potential = water potential − pressure potential M1 substitutes correctly: −1150 − (+450) A1 −1600 kPa, negative as a solute potential must always be

Question 10

Explain how the alveoli of the lungs and the blood flowing past them maintain a high rate of diffusion of oxygen into the blood.

Accepted Answer

B1 the alveoli give an enormous total surface area, about 70 m² in human lungs, and rate is proportional to surface area B1 the barrier is only about 0.3 µm thick, one flattened epithelial cell and one capillary endothelial cell, and rate is inversely proportional to thickness B1 ventilation and continuous blood flow keep replacing the air and the blood, which maintains a steep concentration difference that would otherwise even out

Question 11

Explain what is meant by incipient plasmolysis, and explain why a plant cell at incipient plasmolysis has a water potential equal to its solute potential.

Accepted Answer

B1 incipient plasmolysis is the point at which a plant cell's pressure potential has just fallen to zero, so its contents are just touching the cell wall but exerting no pressure on it B1 water potential is the sum of solute potential and pressure potential, ψ = ψs + ψp B1 at incipient plasmolysis ψp = 0, so this relationship reduces to ψ = ψs, meaning the two are equal at that specific point only

Question 12

Describe how Visking tubing can be used to show that starch cannot cross a partially permeable membrane while glucose can.

Accepted Answer

B1 a length of Visking tubing is filled with a mixture of starch and glucose solutions, sealed, rinsed on the outside, and suspended in a boiling tube of water B1 samples of the water outside the tubing are tested at intervals with iodine and with Benedict's solution B1 the iodine stays orange-brown, showing no starch has crossed, while Benedict's solution turns from blue to a positive colour, showing glucose has crossed the tubing's pores

Question 13

Explain why percentage change in mass, rather than the raw change in mass, is used to compare potato cylinders in the water potential investigation.

Accepted Answer

B1 the cylinders are not all cut to exactly the same starting mass, even when cut with the same cork borer and trimmed to the same length B1 a raw change in mass in grams would be affected both by the water potential of the solution and by how large the cylinder happened to be to start with, mixing the two together B1 dividing the change in mass by the initial mass and multiplying by 100 removes the effect of the cylinder's starting size, so cylinders of different starting mass can be fairly compared

Question 14

State the water potential of pure water at atmospheric pressure, and state what this tells you about the sign of the water potential of a solution at atmospheric pressure.

Accepted Answer

B1 pure water at atmospheric pressure has a water potential of 0 kPa B1 a solution at atmospheric pressure therefore has a negative water potential, because adding solute lowers the solute potential and no pressure is raising it

Question 15

State the term for the shrinking and puckering of an animal cell as water leaves it by osmosis, and state the term for a plant cell whose contents are pressing against its wall.

Accepted Answer

B1 crenation, the shrinking and puckering B1 turgid, pressing against the wall