Revision Library

revisionlibrary.co.uk

GCSE · Biology

B1c Transport in and out of Cells

AQA 8464 · Calculator allowed · about 95 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Getting substances into and out of cells

Revision Library

Every cell needs a constant supply of substances such as oxygen and glucose, and needs to remove waste products such as carbon dioxide. Three processes move substances across the cell membrane: diffusion, osmosis and active transport. Diffusion and osmosis are passive, meaning they need no energy input, and both move particles from a region of higher concentration to a region of lower concentration. Active transport is different: it moves substances against a concentration gradient, from a lower to a higher concentration, so it requires energy released by respiration. How quickly these processes happen depends on factors such as concentration gradient, temperature, and the surface area to volume ratio of the cell or organism. This last idea explains why single-celled organisms can rely on diffusion alone, while larger, multicellular organisms need specialised exchange surfaces and transport systems. The required practical in this pack investigates osmosis using potato tissue placed in a range of sucrose solutions, measuring the resulting percentage change in mass.

Diffusion is one of the ways that substances move into and out of cells.

(a)Define diffusion.(2)

(b)Give one example of diffusion occurring in the human body.(1)

(Total for Question 1 is 3 marks)

As well as temperature, other factors affect the rate of diffusion.

(Total for Question 2 is 2 marks)

A student investigates how temperature affects the rate of diffusion. They place a small crystal of purple potassium manganate(VII) at the bottom of a test tube of water, and measure how far the purple colour has spread after 10 minutes, at different water temperatures.

Temperature (deg C) -- Distance diffused in 10 minutes (mm)
10 -- 4
20 -- 9
30 -- 15
40 -- 22

(a)Describe the pattern shown by the results.(2)

(b)Explain, in terms of particles, why increasing the temperature increases the rate of diffusion.(3)

(c)Use the pattern in the table to predict the distance diffused in 10 minutes at 50 deg C.(1)

(Total for Question 3 is 6 marks)

Osmosis is a special case of diffusion.

(a)Define osmosis.(3)

(b)State one similarity and one difference between osmosis and diffusion.(2)

(Total for Question 4 is 5 marks)

Root hair cells absorb mineral ions, such as nitrate ions, from the soil. The concentration of mineral ions is usually higher inside the root hair cell than in the soil around it.

(a)Name the process by which root hair cells take up mineral ions in this situation.(1)

(b)Explain why root hair cells need energy to take up mineral ions in this way.(2)

(c)Root hair cells contain a large number of mitochondria. Explain how this is an adaptation for active transport.(2)

(Total for Question 5 is 5 marks)

Complete the table below to compare diffusion, osmosis and active transport. For each process, state (i) whether energy from respiration is required, and (ii) the direction of movement relative to the concentration gradient.

Process -- (i) Energy from respiration required? -- (ii) Direction relative to concentration gradient
Diffusion -- ? -- ?
Osmosis -- ? -- ?
Active transport -- ? -- ?

(Total for Question 6 is 6 marks)

Required practical: a student investigates the effect of sucrose solution concentration on the mass of potato tissue.

(a)Describe a method the student could use to investigate this, including how to make the investigation a fair test.(6)

(b)Identify the independent variable and the dependent variable in this investigation.(2)

(Total for Question 7 is 8 marks)

Required practical: the table shows a student's results for potato chips left in different concentrations of sucrose solution for 24 hours.

Concentration (mol/dm³) -- Initial mass (g) -- Final mass (g) -- Percentage change in mass (%)
0.0 -- 1.52 -- 1.71 -- ?
0.2 -- 1.48 -- 1.58 -- +6.8
0.4 -- 1.50 -- 1.50 -- 0.0
0.6 -- 1.55 -- 1.46 -- ?
0.8 -- 1.49 -- 1.32 -- -11.4
1.0 -- 1.53 -- 1.24 -- ?

(a)Calculate the percentage change in mass for the potato chip in 0.6 mol/dm³ sucrose solution. Give your answer to 1 decimal place.(3)

(b)Complete the table by calculating the percentage change in mass at 0.0 mol/dm³ and at 1.0 mol/dm^3.(4)

(c)Using the results in the table, estimate the concentration of sucrose solution that has the same concentration as the liquid inside the potato cells. Explain your reasoning.(2)

(Total for Question 8 is 9 marks)

A student carrying out the potato osmosis practical did not blot the potato chips dry with a paper towel before weighing them at the start and end of the investigation.

(a)Explain the effect this could have had on the results.(2)

(b)Suggest one way the student could improve the reliability of the results, other than repeating the experiment.(1)

(c)The student's teacher suggests using several chips per concentration and calculating a mean percentage change in mass. Explain why this would improve the investigation.(2)

(Total for Question 9 is 5 marks)

As organisms increase in size, their surface area to volume ratio decreases.

(Total for Question 10 is 4 marks)

A teacher gives three groups of students agar cubes of different sizes, each containing sodium hydroxide and phenolphthalein indicator (pink in alkaline conditions, colourless in acidic conditions). Each cube is placed into hydrochloric acid, and the time taken for the pink colour to disappear completely from the cube is recorded.

Cube side length (mm) -- Surface area (mm²) -- Volume (mm³) -- Surface area to volume ratio -- Time for colour to disappear (s)
10 -- 600 -- 1000 -- 0.6:1 -- 240
20 -- 2400 -- 8000 -- 0.3:1 -- 580
30 -- 5400 -- 27000 -- 0.2:1 -- 1020

(Total for Question 11 is 6 marks)

A biologist models cells as cubes to compare their surface area to volume (SA:V) ratio.

(a)A cube-shaped cell has sides of length 2 mm. Calculate its surface area, volume, and SA:V ratio. Give your ratio in the form n:1.(3)

(b)A second cube-shaped cell has sides of length 5 mm. Calculate its SA:V ratio, in the form n:1.(3)

(c)Use your answers to (a) and (b) to explain why smaller cells are more efficient at exchanging substances, such as oxygen, with their environment.(2)

(Total for Question 12 is 8 marks)

A plant biologist investigates the uptake of nitrate ions by active transport in root hair cells at different temperatures, up to the plant's optimum temperature.

(a)Predict what would happen to the rate of active transport of nitrate ions if the temperature increased from 20 deg C to 30 deg C.(1)

(b)Explain your prediction from part (a) in terms of respiration.(3)

(Total for Question 13 is 4 marks)

The small intestine is adapted for the efficient diffusion of digested food molecules into the blood.

(a)Name two adaptations of the small intestine that increase the rate of diffusion of digested food into the blood.(2)

(b)Using ideas about surface area and concentration gradient, explain how each adaptation you named in part (a) increases the rate of diffusion.(2)

(Total for Question 14 is 4 marks)

A biologist places two identical strips of rhubarb tissue into different solutions for one hour: strip A into distilled water, and strip B into a concentrated salt solution.

(a)Predict what would happen to the mass of strip A. Explain your answer.(2)

(b)Predict what would happen to the mass of strip B. Explain your answer.(2)

(Total for Question 15 is 4 marks)

A cell has a surface area of 96 mm² and a surface area to volume ratio of 4:1.

(Total for Question 16 is 2 marks)

Two potato chips, cut from the same potato, are placed into distilled water for 30 minutes.

Chip 1: initial mass 2.84 g, final mass 3.20 g
Chip 2: initial mass 1.05 g, final mass 1.19 g

(a)Calculate the percentage change in mass for each chip. Give your answers to 1 decimal place.(4)

(b)A student concludes: 'Chip 2 absorbed more water than Chip 1, because its percentage change in mass is higher.' Evaluate this conclusion.(3)

(Total for Question 17 is 7 marks)

Mark scheme · B1c Transport in and out of Cells

Question 1

(a) B1 (net) movement of particles (molecules/ions) from an area of higher concentration to an area of lower concentration, oe
(a) B1 movement continues until the concentration is the same throughout / an equilibrium is reached, oe
(a) Answer: Diffusion is the net movement of particles from an area of higher concentration to an area of lower concentration, until the concentration is the same throughout.
(b) B1 any correct named example, e.g. oxygen diffusing from the alveoli into the blood; carbon dioxide diffusing from the blood into the alveoli; glucose diffusing from the small intestine into the blood, oe
(b) Answer: Oxygen diffuses from the air in the alveoli into the blood in the surrounding capillaries.

Question 2

B1 the concentration gradient (the difference in concentration between the two areas), oe
B1 the surface area (of the membrane or exchange surface) across which diffusion is happening, oe (accept: the distance/thickness of the membrane)
Answer: Any two from: the concentration gradient (difference in concentration); the surface area of the membrane; the distance (thickness of the membrane) across which particles must diffuse.

Question 3

(a) B1 as temperature increases, the distance diffused (rate of diffusion) increases, oe
(a) B1 the increase is not constant / the rate of increase itself gets bigger with each 10 deg C rise (e.g. the distance increases by 5 mm, then 6 mm, then 7 mm), oe with correct use of data
(a) Answer: As temperature increases, the distance diffused in 10 minutes increases. The increase is not constant: it rises by 5 mm, then 6 mm, then 7 mm as temperature increases by 10 deg C each time, so the rate of increase itself gets bigger.
(b) B1 particles have more kinetic energy at a higher temperature, oe
(b) B1 particles move faster (on average), oe
(b) B1 so particles move from the area of higher concentration to the area of lower concentration more quickly (spread out/mix faster), oe
(b) Answer: At a higher temperature, particles have more kinetic energy, so they move faster. This means they spread out from the area of higher concentration to the area of lower concentration more quickly, increasing the rate of diffusion.
(c) B1 an answer in the range 27-34 mm, showing correct extrapolation of the increasing difference pattern (ideally 30 mm), oe
(c) Answer: Approximately 30 mm (accept any value from 27-34 mm with correct reasoning).

Question 4

(a) B1 the diffusion of water (molecules), oe (cso: 'osmosis' is the term being defined and must not be confused with 'diffusion' alone)
(a) B1 from a dilute solution to a concentrated solution (from a region of higher water concentration to a region of lower water concentration), oe
(a) B1 through a partially permeable membrane
(a) Answer: Osmosis is the diffusion of water molecules from a dilute solution to a concentrated solution, through a partially permeable membrane.
(b) B1 similarity: any one of: both involve net movement of particles down a concentration gradient (from high to low concentration); both are passive processes and do not require energy, oe
(b) B1 difference: any one of: osmosis only involves the movement of water, while diffusion can involve any particle/solute; osmosis requires a partially permeable membrane, diffusion does not, oe
(b) Answer: Similarity: both are passive processes involving net movement down a concentration gradient. Difference: osmosis only involves water molecules, whereas diffusion can involve any type of particle.

Question 5

(a) B1 active transport
(a) Answer: Active transport.
(b) B1 mineral ions are absorbed against a concentration gradient (from a lower to a higher concentration), oe
(b) B1 energy (from respiration) is needed to move particles against a concentration gradient, oe
(b) Answer: The ions move against a concentration gradient, from a lower concentration in the soil to a higher concentration in the cell, so energy from respiration is needed to move them.
(c) B1 mitochondria (cso) are the site of (aerobic) respiration, oe
(c) B1 this releases (more) energy for active transport, allowing more ions to be moved against the concentration gradient, oe
(c) Answer: Mitochondria are the site of aerobic respiration. A large number of mitochondria releases more energy, allowing more mineral ions to be actively transported against the concentration gradient.

Question 6

B1 diffusion: energy not required (passive process)
B1 diffusion: down the concentration gradient (from high to low concentration)
B1 osmosis: energy not required (passive process)
B1 osmosis: down the (water) concentration gradient (dilute to concentrated solution)
B1 active transport: energy required (from respiration)
B1 active transport: against (up) the concentration gradient (from low to high concentration)
Answer: Diffusion: no energy needed, moves down the concentration gradient. Osmosis: no energy needed, moves down the water concentration gradient. Active transport: needs energy from respiration, moves against the concentration gradient.

Question 7

(a) B1 prepare a range of at least five sucrose solutions of different, known concentrations, e.g. by serial dilution from a stock solution, oe
(a) B1 cut potato chips of the same size using a cork borer (same diameter) and a scalpel/ruler (same length), oe
(a) B1 blot each chip dry with paper towel to remove surface moisture, then measure and record its starting mass using a balance, oe
(a) B1 place one chip into each concentration, using the same volume of solution for every chip, for the same fixed time and at the same temperature, oe
(a) B1 remove each chip, blot dry again, then measure and record its final mass, oe
(a) B1 repeat the whole investigation at least twice more for each concentration, and calculate a mean percentage change in mass, oe
(a) Answer: Prepare at least five sucrose solutions of known concentration by serial dilution. Cut identical potato chips using a cork borer. Blot and weigh each chip, then place one into each solution (same volume, time, temperature). Blot and re-weigh each chip. Repeat the whole investigation at least twice more and calculate a mean.
(b) B1 independent variable = concentration of the sucrose solution
(b) B1 dependent variable = (percentage) change in mass of the potato chip
(b) Answer: Independent variable: concentration of the sucrose solution. Dependent variable: percentage change in mass of the potato chip.

Question 8

(a) M1 change in mass = 1.46 - 1.55 = -0.09 (g), oe
(a) M1 percentage change = (change in mass / initial mass) x 100, correct substitution (ft from their change in mass), oe
(a) A1 -5.8% awrt, cao (allow -5.81% to -5.8%)
(a) Answer: -5.8% (a decrease of 5.8%).
(b) M1 0.0 mol/dm³: change in mass = 1.71 - 1.52 = +0.19 (g), oe
(b) A1 0.0 mol/dm³: +12.5% awrt, cao
(b) M1 1.0 mol/dm³: change in mass = 1.24 - 1.53 = -0.29 (g), oe
(b) A1 1.0 mol/dm³: -19.0% awrt, cao
(b) Answer: 0.0 mol/dm³: +12.5%. 1.0 mol/dm³: -19.0%.
(c) B1 approximately 0.4 mol/dm³ (accept 0.35-0.45 mol/dm³, ft from their table if internally consistent), oe
(c) B1 because there is 0% (no) change in mass at this concentration, so there is no net movement of water into or out of the potato cells (the solution is isotonic with the cell contents), oe
(c) Answer: Approximately 0.4 mol/dm³, because this is where the percentage change in mass is 0%, showing no net movement of water into or out of the cells.

Question 9

(a) B1 surface water/solution left on the chip would add to its measured mass, oe
(a) B1 this would make the calculated percentage change in mass inaccurate/unreliable (e.g. mass could appear to increase more, or decrease less, than the true value), oe
(a) Answer: Surface water or solution left on the chip would add to its measured mass, making the calculated percentage change in mass inaccurate.
(b) B1 any one from: use a larger sample size (more chips per concentration); use a narrower range of concentrations with smaller intervals; use a more precise balance; control temperature more closely (e.g. a water bath); ensure all chips are cut to the same size using a cork borer and template, oe
(b) Answer: Use a larger number of potato chips for each concentration.
(c) B1 an anomalous result can be identified and excluded from the calculation, oe
(c) B1 the mean is more representative of the true value, reducing the effect of random error and making the data more reliable, oe
(c) Answer: Using several chips allows anomalous results to be identified and excluded. The mean is then more representative of the true value, making the data more reliable.

Question 10

B1 single-celled organisms have a large surface area to volume ratio, oe
B1 so diffusion alone is fast enough to supply/remove substances at the rate needed by the cell, oe
B1 large multicellular organisms have a (relatively) small surface area to volume ratio, oe
B1 so diffusion alone would be too slow to meet the needs of all their cells, meaning they need specialised exchange surfaces and transport systems, oe
Answer: Single-celled organisms have a large surface area to volume ratio, so diffusion alone can supply their needs. Large multicellular organisms have a small surface area to volume ratio, so diffusion alone would be too slow, and they need specialised exchange surfaces and transport systems instead.

Question 11

Level 1 (1-2): Simple, isolated statements about surface area to volume ratio or diffusion, with little or no reference to the data provided. Limited linkage between ideas.
Level 2 (3-4): Some clear links are made between surface area to volume ratio, diffusion distance and time taken, with reference to some of the data in the table.
Level 3 (5-6): A detailed, logically structured explanation linking surface area to volume ratio, diffusion distance and time taken, supported throughout by correct use of the data provided.
Indicative content:
- As cube side length increases, volume increases faster than surface area (volume is cubed, surface area is squared), so the surface area to volume ratio decreases (from 0.6:1 to 0.2:1 as side length increases from 10 mm to 30 mm)
- Acid must diffuse from the surface to the centre of the cube; larger cubes have a greater distance for the acid to diffuse relative to their surface area
- Smaller cubes have a larger surface area to volume ratio, so acid can diffuse in relatively faster per unit volume, reaching the centre (and turning the whole cube colourless) more quickly
- This explains why the time taken increases from 240 s for the 10 mm cube to 1020 s for the 30 mm cube as the surface area to volume ratio falls
- Correct use of data, e.g. the 30 mm cube has a surface area to volume ratio 3 times smaller than the 10 mm cube, but takes over 4 times as long for the colour to disappear
- Rate of diffusion depends on surface area, concentration gradient and diffusion distance; a smaller surface area to volume ratio means the surface area is not large enough, relative to the volume needing supply, for diffusion to keep pace

Question 12

(a) M1 surface area = 6 x 2² = 24 mm², oe
(a) M1 volume = 2³ = 8 mm³, oe
(a) A1 ratio = 3:1 cao (ft from their surface area and volume)
(a) Answer: Surface area = 24 mm², volume = 8 mm³, SA:V ratio = 3:1.
(b) M1 surface area = 6 x 5² = 150 mm², oe
(b) M1 volume = 5³ = 125 mm³, oe
(b) A1 ratio = 1.2:1 cao (oe e.g. 6:5, ft from their surface area and volume)
(b) Answer: Surface area = 150 mm², volume = 125 mm³, SA:V ratio = 1.2:1.
(c) B1 the smaller cell has a larger SA:V ratio (3:1 compared with 1.2:1 for the larger cell), ft from (a) and (b), oe
(c) B1 so a greater surface area is available for diffusion/exchange per unit of volume, meaning substances can diffuse into/out of the cell more quickly relative to its needs, oe
(c) Answer: The smaller cell has a larger SA:V ratio (3:1 vs 1.2:1), meaning it has more surface area available for exchange relative to its volume, so substances can diffuse in and out more quickly relative to its needs.

Question 13

(a) B1 the rate of active transport would increase
(a) Answer: The rate of active transport would increase.
(b) B1 a higher temperature increases the rate of (aerobic) respiration, up to the optimum, oe
(b) B1 more energy is released/becomes available per unit time, oe
(b) B1 so more energy is available to actively transport nitrate ions against the concentration gradient, increasing the rate of active transport, oe
(b) Answer: A higher temperature increases the rate of respiration, releasing more energy per unit time. This extra energy allows nitrate ions to be actively transported against the concentration gradient more quickly.

Question 14

(a) B1 villi (and microvilli), which increase the surface area (for absorption), oe
(a) B1 any one other correct adaptation: a single layer of (thin) epithelial cells, giving a short diffusion distance; a good/rich blood supply, which maintains a steep concentration gradient, oe
(a) Answer: Villi (and microvilli), which increase the surface area, and a good blood supply, which maintains a steep concentration gradient.
(b) B1 a larger surface area (from villi/microvilli) means more molecules can diffuse across per unit time, ft from (a), oe
(b) B1 a good blood supply removes absorbed molecules quickly, maintaining a steep concentration gradient, which keeps the rate of diffusion high, ft from (a), oe
(b) Answer: The large surface area of the villi means more molecules can diffuse across at once. A good blood supply removes absorbed molecules quickly, maintaining a steep concentration gradient and keeping the rate of diffusion high.

Question 15

(a) B1 the mass increases, oe
(a) B1 water moves into the cells by osmosis, because the surrounding water is more dilute (has a higher water concentration) than the cell contents, oe
(a) Answer: The mass increases, because water moves into the cells by osmosis from the more dilute solution outside.
(b) B1 the mass decreases, oe
(b) B1 water moves out of the cells by osmosis, into the more concentrated salt solution (which has a lower water concentration than the cell contents), oe
(b) Answer: The mass decreases, because water moves out of the cells by osmosis into the more concentrated salt solution.

Question 16

M1 correct method to find volume from the ratio, e.g. volume = surface area / 4, oe
A1 24 mm³ cao
Answer: 24 mm^3.

Question 17

(a) M1 chip 1: change in mass = 3.20 - 2.84 = +0.36 (g), oe
(a) A1 chip 1: +12.7% awrt, cao
(a) M1 chip 2: change in mass = 1.19 - 1.05 = +0.14 (g), oe
(a) A1 chip 2: +13.3% awrt, cao
(a) Answer: Chip 1: +12.7%. Chip 2: +13.3%.
(b) B1 the conclusion is incorrect in terms of the actual mass of water gained: chip 1 gained more water in total (0.36 g compared with 0.14 g for chip 2), ft from (a), oe
(b) B1 percentage change is scaled to (accounts for) the different starting masses of the chips, so a higher percentage does not mean more water was absorbed in absolute terms, oe
(b) B1 percentage change is still useful because it allows a fair comparison between chips of different starting sizes/masses, oe
(b) Answer: The conclusion is incorrect: chip 1 actually gained more water in total (0.36 g vs 0.14 g), even though its percentage change is smaller, because chip 1 started with a larger mass. Percentage change accounts for the different starting masses, so it is not the same as the absolute amount of water absorbed.

Transport in and out of Cells - Worksheets, Questions and Revision