IGCSE Biology Paper 4 (Theory / Extended)

Topic 3: Movement Into and Out of Cells -- Challenge Exam 2

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0610

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

This paper covers the whole of Topic 3. Like a real Cambridge paper, the seven questions range across every sub-topic — diffusion, osmosis and active transport — and they are deliberately mixed rather than grouped. All three Topic 3 papers do; they differ in the angle they come at it from, not in what they cover.

Question 1 – Water Uptake in Salty Soil

Total: 12 marks

A farmer near the coast in Gujarat finds that his wheat plants wilt during the day even though the soil is wet. Sea water has flooded the field in the past and the soil water now contains a high concentration of dissolved salts. A researcher measures the water potential of the soil water and of the cell sap of the root hair cells of two crops growing in the same field.

	relative water potential (arbitrary units, 0 = pure water)
soil water in the flooded field	−18
root hair cell sap of wheat	−12
root hair cell sap of a salt-tolerant grass growing beside it	−26

(a) [3]

Explain what is meant by water potential, and state how dissolving a solute in water changes it.

Model Answer – 1(a)

water potential is a measure of the tendency of water molecules to move out of a solution, or of the concentration of free water molecules in it [1]

pure water has the highest water potential of all [1]

dissolving a solute lowers the water potential, so the more concentrated a solution is, the lower its water potential [1]

⚠ If you missed marks here: The scale runs downwards from pure water, so a more negative figure means a more concentrated solution. Candidates who think “more solute means more water potential because there is more stuff” get every subsequent direction wrong. Anchor the idea with one sentence you never change: more solute, less free water, lower water potential.

1 mark -- water potential describes the tendency of water molecules to move out of a solution 1 mark -- pure water has the highest water potential 1 mark -- adding solute lowers the water potential

(b) [3]

Use the figures in the table to explain why the wheat plants wilt even though the soil is wet.

Model Answer – 1(b)

the soil water has a water potential of −18, which is lower than the −12 of the wheat root hair cell sap [1]

so there is a net movement of water out of the root hair cells, by osmosis, into the soil water [1]

the cells lose water, become flaccid and lose turgor pressure, so the plant wilts even though liquid water surrounds the roots [1]

⚠ If you missed marks here: Both figures must be quoted and compared — simply writing “the soil is too salty” does not use the data. The counter-intuitive part is that wet soil can still dehydrate a plant, and the explanation is entirely about the direction of the water potential gradient, never about how much liquid is present.

1 mark -- compares soil water potential of minus 18 with wheat sap of minus 12 1 mark -- net movement of water out of the root hair cells by osmosis 1 mark -- cells lose turgor and become flaccid so the plant wilts

Explain how the salt-tolerant grass is able to take up water from the same soil.

Model Answer – 1(c)

the cell sap of the grass has a water potential of −26, which is lower than the −18 of the soil water [1]

so the water potential gradient still runs into the root, and water enters the root hair cells by osmosis [1]

the grass achieves this by keeping a very high concentration of dissolved solutes in its cell sap [1]

⚠ If you missed marks here: The trap is to assume the tolerant plant must somehow keep salt out. It does the opposite: it makes its own contents even more concentrated so that its water potential stays below that of the soil. Note also that water is never taken up by active transport — that phrase appears in weak answers every year and cannot be credited.

1 mark -- grass cell sap at minus 26 is lower than the soil water at minus 18 1 mark -- water still enters by osmosis down the water potential gradient 1 mark -- achieved by a high concentration of dissolved solutes in the cell sap

(d) [3]

The farmer floods the field with fresh water to wash the salt away. Explain, in terms of water potential, why the wheat recovers.

Model Answer – 1(d)

removing dissolved salts raises the water potential of the soil water above that of the root hair cell sap [1]

so there is now a net movement of water into the root hair cells by osmosis, through their partially permeable cell membranes [1]

the cells become turgid again and the pressure of water pressing outwards on the cell walls restores support to the stem and leaves [1]

⚠ If you missed marks here: Answers here often stop at “the plant gets water back”. Three separate ideas earn the marks: the change to the soil’s water potential, the reversal of the gradient, and the return of turgor pressure as the mechanism of support. Use the word turgid, not “full” or “stiff”.

1 mark -- removing salt raises the water potential of the soil water 1 mark -- net movement of water into the cells by osmosis through partially permeable membranes 1 mark -- cells become turgid and turgor pressure restores support

Question 2 – Osmometer Readings

Total: 12 marks

A student builds an osmometer. A bag of dialysis tubing containing 1.0 mol dm⁻³ sucrose solution is fitted to a capillary tube and stood in a beaker of distilled water. The height of the liquid in the capillary tube is recorded every 10 minutes.

time / min	0	10	20	30	40	50	60
height of liquid in capillary tube / mm	0	26	48	64	73	77	78

(a) [2]

Calculate the mean rate of rise of the liquid over the first 20 minutes. Give the unit.

Model Answer – 2(a)

48 ÷ 20 [1]

= 2.4 mm/min (or mm min⁻¹) [1]

⚠ If you missed marks here: Rate questions are marked for the working and the unit as well as the number. Reading the height at 20 minutes but dividing by 10, or forgetting that the liquid started at 0 mm, both produce wrong answers, and a rate quoted without a unit will not gain the second mark however good the arithmetic is.

1 mark -- working shown dividing 48 mm by 20 min 1 mark -- answer 2.4 mm per minute with correct unit

(b) [3]

Explain why the liquid rises up the capillary tube.

Model Answer – 2(b)

the sucrose solution has a lower water potential than the distilled water outside [1]

water molecules move down the water potential gradient into the tubing by osmosis, through the partially permeable dialysis tubing [1]

the volume of liquid inside the bag increases, and because the bag cannot expand freely the extra liquid is pushed up the narrow capillary tube [1]

⚠ If you missed marks here: The third mark is the physical one and is often left out: the extra volume has to go somewhere. Sucrose cannot leave the bag because its molecules are too large to pass through the tubing, so an answer describing sucrose moving out has misread the whole apparatus.

1 mark -- sucrose solution has a lower water potential than the distilled water 1 mark -- water enters by osmosis through the partially permeable tubing 1 mark -- increased volume inside the bag pushes liquid up the capillary tube

Describe the shape of the results and explain why the rate of rise falls after 30 minutes.

Model Answer – 2(c)

the height rises quickly at first and then more and more slowly, levelling off at about 78 mm [1]

as water enters, the sucrose solution inside becomes more dilute, so its water potential rises and the water potential gradient becomes less steep [1]

the column of liquid in the capillary tube also exerts an increasing downward pressure, opposing further entry of water [1]

⚠ If you missed marks here: Two independent causes are available and either pair of them earns the marks, but the answer must explain why the gradient changes rather than simply saying “osmosis slows down”. Note that the liquid never stops moving through the membrane; when the line flattens, entry and exit have simply become equal.

1 mark -- describes a rapid rise that slows and levels off at about 78 mm 1 mark -- dilution of the sucrose raises its water potential so the gradient falls 1 mark -- pressure of the liquid column opposes further entry of water

(d) [2]

Predict how the results would differ if the beaker had contained 0.5 mol dm⁻³ sucrose solution instead of distilled water. Explain your prediction.

Model Answer – 2(d)

the liquid would rise more slowly and would level off at a lower height [1]

because the water potential difference between the inside and the outside of the bag is smaller, so the gradient driving water inwards is less steep [1]

⚠ If you missed marks here: A prediction has to state both the direction and the reason. Answers that say the liquid would not rise at all have overreached: 0.5 mol dm⁻³ is still more dilute than the 1.0 mol dm⁻³ inside the bag, so a gradient remains, just a shallower one.

1 mark -- liquid rises more slowly and levels off lower 1 mark -- smaller water potential difference gives a less steep gradient

(e) [2]

Suggest why the student rinsed the outside of the dialysis tubing with distilled water before placing it in the beaker.

Model Answer – 2(e)

rinsing removes any sucrose solution spilt on the outside of the tubing [1]

sucrose left on the outside would dissolve in the beaker water, lowering its water potential and reducing the gradient, so the results would not be valid [1]

⚠ If you missed marks here: This is a validity question, not a hygiene question. The credited answer explains the effect of the contamination on the water potential gradient. Answers such as “to keep it clean” or “so the tubing does not stick” describe an action without linking it to the measurement being affected.

1 mark -- removes sucrose spilt on the outside of the tubing 1 mark -- contamination would lower the water potential outside and reduce the gradient

Question 3 – Counting Plasmolysed Cells

Total: 12 marks

A student places strips of red onion epidermis in a range of sucrose solutions for 20 minutes, then examines each strip under a microscope and counts the percentage of cells in which the cell membrane has pulled away from the cell wall.

sucrose concentration / mol dm⁻³	0.2	0.3	0.4	0.5	0.6	0.7
plasmolysed cells / %	0	8	32	68	92	100

(a) [2]

State what is meant by plasmolysis and explain why red onion epidermis is a suitable tissue for this investigation.

Model Answer – 3(a)

plasmolysis is the pulling away of the cell membrane from the cell wall when a plant cell loses water by osmosis [1]

the epidermis is only one cell thick so it can be viewed directly, and the coloured cell sap in the vacuole makes the shrunken contents easy to see against the wall [1]

⚠ If you missed marks here: The definition must mention the membrane leaving the wall; “the cell shrinks” describes a flaccid cell just as well and is too vague. For the suitability mark, both the single cell layer and the pigment are creditable, but an answer that says only “onion is easy to get” explains nothing about the observation being made.

1 mark -- plasmolysis defined as the cell membrane pulling away from the cell wall 1 mark -- single layer of cells with coloured sap makes plasmolysis easy to see

(b) [3]

The concentration at which 50 % of the cells are plasmolysed is called the point of incipient plasmolysis. Use the results to estimate this concentration and explain what it tells you about the cells.

Model Answer – 3(b)

approximately 0.45 mol dm⁻³, between the 32 % at 0.4 and the 68 % at 0.5 [1]

at this concentration about half the cells have just begun to lose contact between membrane and wall, so on average the solution and the cell sap have about the same water potential [1]

it therefore gives an estimate of the water potential of the cell sap of this onion tissue [1]

⚠ If you missed marks here: Reading between two data points is expected, so an answer of exactly 0.4 or exactly 0.5 loses the first mark. The interpretation mark is for realising that the cells are not identical — each has a slightly different sap concentration, which is why the change happens over a range of concentrations rather than all at once.

1 mark -- estimate of about 0.45 mol per dm cubed by interpolation 1 mark -- at this point the solution and cell sap have about the same water potential 1 mark -- gives an estimate of the water potential of the cell sap

Explain, in terms of water potential and osmosis, why the percentage of plasmolysed cells increases as the sucrose concentration increases.

Model Answer – 3(c)

as the sucrose concentration rises, the water potential of the solution falls further below that of the cell sap [1]

the water potential gradient out of the cells becomes steeper, so more water leaves each cell by osmosis through the partially permeable cell membrane [1]

more cells lose enough water for the membrane to separate from the wall, so the percentage plasmolysed rises [1]

⚠ If you missed marks here: The link that examiners look for is steeper gradient means more water lost. Answers that say the sucrose “pulls” or “sucks” water out do not describe a gradient and are not credited, and neither is any suggestion that sucrose enters the cells — sucrose cannot cross the cell membrane, which is what makes the experiment work.

1 mark -- higher concentration means a lower water potential outside the cells 1 mark -- steeper gradient so more water leaves each cell by osmosis 1 mark -- more cells lose enough water for the membrane to separate from the wall

(d) [2]

The cells in 0.7 mol dm⁻³ sucrose are returned to distilled water for 20 minutes. Predict and explain what happens to them.

Model Answer – 3(d)

water re-enters the cells by osmosis because distilled water has a higher water potential than the cell sap, so the cells become turgid again [1]

plasmolysis is reversible in living cells, so the protoplast expands and presses on the cell wall once more [1]

⚠ If you missed marks here: The word examiners want is reversible. Many candidates assume plasmolysed cells are dead and predict no change; they are not dead, and the recovery is the standard demonstration of that. The cell cannot burst on the way back, because the cell wall stops the swelling once the cell is turgid.

1 mark -- water re-enters by osmosis and the cells become turgid again 1 mark -- plasmolysis is reversible in living cells

(e) [2]

Suggest two ways in which the student could improve the accuracy of the percentage figures obtained.

Model Answer – 3(e)

count a much larger number of cells in each strip, for example at least 100, and count several fields of view [1]

use a fixed rule for deciding whether a cell counts as plasmolysed, and count each field twice or have a second person count independently [1]

⚠ If you missed marks here: Counting is a sampling process, so accuracy improves with larger samples and consistent criteria. Suggestions such as “use a better microscope” do not address the source of error, and “leave the cells longer” changes the experiment rather than improving the measurement.

1 mark -- count a larger number of cells and several fields of view 1 mark -- use a consistent criterion or repeat counts to reduce counting error

Question 4 – Temperature and Diffusion in Agar

Total: 12 marks

A student investigates the effect of temperature on diffusion. Six identical dishes of agar jelly each have a well cut in the centre. A fixed volume of the same coloured solution is placed in each well, and the dishes are kept in water baths at different temperatures. The diameter of the coloured circle is measured after 40 minutes.

temperature / °C	10	20	30	40	50	60
diameter of coloured circle after 40 min / mm	11	15	19	24	28	33

(a) [3]

Describe and explain the effect of temperature on the diffusion of the coloured solution through the agar.

Model Answer – 4(a)

as the temperature increases the diameter of the coloured circle increases, from 11 mm at 10 °C to 33 mm at 60 °C [1]

at higher temperatures the particles have more kinetic energy [1]

so they move faster and spread out more quickly, giving a faster rate of diffusion [1]

⚠ If you missed marks here: Describe first with figures, then explain with kinetic energy — two different skills, each carrying marks. A frequent error is to claim that heating steepens the concentration gradient; heating adds no extra particles, so the gradient is unchanged and only the speed of movement alters.

1 mark -- describes increasing diameter with supporting figures from the table 1 mark -- particles have more kinetic energy at higher temperature 1 mark -- particles move faster so the rate of diffusion increases

(b) [2]

Calculate how far the coloured solution travelled outwards from the edge of the well at 40 °C, in mm/min. The well has a diameter of 4 mm.

Model Answer – 4(b)

distance from the edge of the well = (24 − 4) ÷ 2 = 10 mm [1]

rate = 10 ÷ 40 = 0.25 mm/min [1]

⚠ If you missed marks here: Two steps are hidden in this calculation and both are commonly missed. The measurement is a diameter, so the distance travelled outwards is only half of it; and the well itself already had a diameter of 4 mm, so that must be subtracted before halving. Forgetting either step gives a rate that is twice or more the correct value.

1 mark -- distance of 10 mm obtained by subtracting the well diameter and halving 1 mark -- rate of 0.25 mm per minute

State three variables that must be kept the same in this investigation, and for each explain why.

Model Answer – 4(c)

the volume and concentration of the coloured solution added to each well, because these set the concentration gradient, which is itself a factor affecting the rate of diffusion [1]

the depth and composition of the agar and the diameter of the well, because these affect the distance the particles travel and the surface area available [1]

the time before measuring, because the circle continues to grow, so measuring at different times would make the diameters incomparable [1]

⚠ If you missed marks here: Each mark requires a variable and a reason, and the strongest reasons name one of the four syllabus factors affecting diffusion. “Keep everything the same” earns nothing, and neither does naming a variable that has already been changed deliberately — temperature is the independent variable here, so it must not appear in this list.

1 mark -- volume and concentration of solution controlled to fix the concentration gradient 1 mark -- depth of agar and size of well controlled to fix distance and surface area 1 mark -- same time allowed before measuring so the diameters are comparable

(d) [2]

A classmate concludes: “This shows that diffusion gets faster and faster however hot it gets.” Explain why this conclusion goes beyond the evidence.

Model Answer – 4(d)

the results only cover temperatures from 10 °C to 60 °C, so the conclusion extrapolates beyond the range of the data [1]

at much higher temperatures other effects would occur, such as the agar softening or melting and the solution evaporating, so the trend cannot be assumed to continue [1]

⚠ If you missed marks here: Conclusions must stay inside the range investigated. This is a routine evaluation mark on Paper 4, and the way to secure it is to state the range tested and then name one specific reason the pattern might not continue outside it.

1 mark -- conclusion extrapolates beyond the range of temperatures tested 1 mark -- other effects such as agar melting or evaporation could change the pattern

(e) [2]

Suggest why measuring the diameter of the coloured circle is difficult to do accurately, and describe one way to reduce the error.

Model Answer – 4(e)

the edge of the coloured circle is not sharp but fades gradually, so deciding where it ends is a matter of judgement [1]

photograph each dish against a ruler and measure the images later using the same criterion, or take two measurements at right angles and use the mean [1]

⚠ If you missed marks here: The credited improvement must address the stated source of error. Answers such as “use a better ruler” do not, because the difficulty is not the scale but the fuzziness of the boundary. Measuring twice at right angles also corrects for a circle that has spread unevenly.

1 mark -- the edge of the coloured circle is fuzzy so judging the boundary is subjective 1 mark -- photograph and measure consistently or take two measurements at right angles and average

Question 5 – Glucose Uptake by Gut Lining Cells

Total: 10 marks

Pieces of the lining of the small intestine of a mammal are placed in solutions containing glucose. The uptake of glucose into the cells is measured under two conditions: with oxygen present and with oxygen absent.

external glucose concentration / arbitrary units	1	2	4	8	16
uptake with oxygen / arbitrary units	14	26	42	50	51
uptake without oxygen / arbitrary units	2	4	8	15	29

(a) [3]

Compare the two sets of results and explain the difference between them.

Model Answer – 5(a)

uptake is much greater with oxygen than without it at every concentration, for example 42 compared with 8 units at a concentration of 4 [1]

with oxygen the cells can respire aerobically, releasing much more energy [1]

this energy is used for active transport of glucose into the cells, which is why removing oxygen reduces uptake so severely [1]

⚠ If you missed marks here: A comparison mark needs a matched pair of figures from the same column, not two numbers picked at random. The explanation must then join oxygen to respiration to energy to active transport; missing out the middle of that chain is the difference between one mark and three.

1 mark -- compares uptake with and without oxygen using a matched pair of figures 1 mark -- oxygen allows aerobic respiration which releases more energy 1 mark -- energy is used for active transport of glucose

(b) [2]

Explain why uptake still occurs when there is no oxygen, and why it keeps rising as the external glucose concentration increases.

Model Answer – 5(b)

without oxygen, glucose can still enter by diffusion, which needs no energy from the cell [1]

as the external concentration rises the concentration gradient into the cell becomes steeper, so the rate of diffusion increases [1]

⚠ If you missed marks here: The pattern in the lower row is the fingerprint of simple diffusion: uptake rising steadily with concentration and no plateau. Recognising a process from the shape of the data, rather than from what you were told, is exactly the skill this question is testing.

1 mark -- glucose can still enter by diffusion which needs no energy 1 mark -- steeper concentration gradient at higher external concentration increases diffusion

With oxygen present, the uptake levels off above a concentration of 8 units. Suggest an explanation for this.

Model Answer – 5(c)

uptake by active transport depends on the number of protein carriers in the cell membrane [1]

at high external concentrations every carrier is already working as fast as it can, so it is the carriers, not the glucose supply, that limit the rate [1]

adding more glucose outside therefore cannot increase the rate any further, so the line levels off [1]

⚠ If you missed marks here: A plateau always means something other than the substance being supplied has become limiting. Saying “the cells are full” is not credited, because uptake is a rate rather than a total, and cells go on using glucose as fast as it arrives. The contrast with the no-oxygen row, which keeps climbing, is the evidence that carriers are involved.

1 mark -- uptake depends on the number of protein carriers in the membrane 1 mark -- all carriers are saturated so they limit the rate 1 mark -- adding more glucose cannot increase the rate further so the line levels off

(d) [2]

Explain why the cells lining the small intestine contain large numbers of mitochondria and are covered in microvilli.

Model Answer – 5(d)

mitochondria release energy by aerobic respiration for the active transport of the products of digestion into the cells [1]

microvilli greatly increase the surface area of the cell membrane, so more diffusion and more active transport can occur per second [1]

⚠ If you missed marks here: Two features, two different reasons — answers that give one reason for both lose a mark. Note that microvilli increase surface area at the level of a single cell, whereas villi do the same for the gut wall as a whole; using the two words interchangeably is a small error that examiners do notice.

1 mark -- mitochondria release energy by aerobic respiration for active transport 1 mark -- microvilli increase the surface area of the membrane for absorption

Question 6 – Gas Exchange in Four Organisms

Total: 12 marks

The table gives information about how four organisms obtain oxygen.

organism	how oxygen reaches the respiring cells	maximum distance from a gas exchange surface to a respiring cell
Amoeba, a single-celled protoctist	directly across the cell surface membrane	0.05 mm
a flatworm 0.4 mm thick	across the flattened body surface	0.2 mm
a locust	along fine air tubes that end beside the muscle cells	0.01 mm
a human	across the alveolus wall, then carried in the blood	0.001 mm from a capillary

(a) [3]

Explain why Amoeba and the flatworm can obtain enough oxygen without any gas exchange organ or transport system.

Model Answer – 6(a)

both are very small or very thin, so no respiring cell is more than a fraction of a millimetre from the surface [1]

diffusion is fast enough over such short distances to supply every cell [1]

both also have a large surface area to volume ratio, so there is plenty of surface for the amount of respiring tissue that must be supplied [1]

⚠ If you missed marks here: Two of the four factors do all the work here — distance and surface area — and a full answer names both. Quoting the flatworm’s shape matters: it is not simply small, it is flat, which is a way of staying thin while growing larger.

1 mark -- no cell is more than a fraction of a millimetre from the surface 1 mark -- diffusion is fast enough over such short distances 1 mark -- large surface area to volume ratio supplies the respiring tissue

(b) [3]

A human has a far larger total gas exchange surface than a locust, yet still needs a blood system while the locust does not. Explain why.

Model Answer – 6(b)

in the locust, fine air tubes carry air directly to the respiring cells, so oxygen only has to diffuse a very short distance at the end of the journey [1]

a human is far too thick for diffusion from the lungs to reach the innermost cells in time [1]

so blood carries oxygen most of the way by mass flow, leaving diffusion to cover only the last fraction of a millimetre from a capillary [1]

⚠ If you missed marks here: The distinction being tested is between mass flow and diffusion. Both organisms rely on diffusion for the final step; they differ in how oxygen is brought close to the cells. Answers that say humans need blood “because we are more complex” describe rather than explain and gain nothing.

1 mark -- air tubes bring air directly to the cells so diffusion distance is tiny 1 mark -- a human is too thick for diffusion alone to supply inner cells 1 mark -- blood carries oxygen by mass flow leaving diffusion for the final short distance

A cube-shaped organism has sides of 2 mm. Calculate its surface area to volume ratio, and calculate how many times greater the ratio would be if it were divided into eight separate cubes each with sides of 1 mm.

Model Answer – 6(c)

2 mm cube: surface area = 6 × 4 = 24 mm², volume = 8 mm³, ratio = 3 : 1 [1]

1 mm cube: surface area = 6 mm², volume = 1 mm³, ratio = 6 : 1 [1]

the ratio is 2 times greater [1]

⚠ If you missed marks here: The eight small cubes have the same total volume as the original, but twice the total surface area, so the ratio doubles. A common error is to compare total surface areas (24 against 48) and answer that the ratio is four times greater, or to forget that surface area to volume ratio is calculated per cube rather than for the whole set.

1 mark -- ratio of 3 to 1 for the 2 mm cube with working 1 mark -- ratio of 6 to 1 for the 1 mm cube with working 1 mark -- correct conclusion that the ratio is twice as great

(d) [3]

State the four factors that affect the rate of diffusion, and for each one give a feature of a human alveolus that makes diffusion faster.

Model Answer – 6(d)

surface area — there are millions of alveoli, giving a very large total surface area; distance — the alveolus wall is one cell thick, so the diffusion distance is very short [1]

concentration gradient — breathing replaces the air in the alveoli and the blood flow removes oxygen, so a steep gradient is maintained [1]

temperature — the body is kept at about 37 °C, which is warm enough for the particles to have plenty of kinetic energy [1]

⚠ If you missed marks here: The four factors must be given in the syllabus wording: surface area, temperature, concentration gradient and distance. The one candidates forget is temperature, and the one they describe loosely is the gradient — say what maintains it, namely ventilation on one side and blood flow on the other.

1 mark -- surface area and distance factors matched to alveolar features 1 mark -- concentration gradient maintained by ventilation and blood flow 1 mark -- temperature named as a factor with body temperature as the feature

Question 7 – Getting the Words Right

Total: 10 marks

A student has written five statements in her revision notes. Each one contains an error of biological vocabulary or reasoning of the kind that regularly loses marks in examinations.

(a) [4]

For each of the following two statements, identify the error and write a corrected version.

Statement 1: “Diffusion is the movement of particles from a high concentration to a low concentration.”
Statement 2: “Osmosis is the movement of water through a semi-permeable membrane.”

Model Answer – 7(a)

statement 1 omits the word net; particles move randomly in all directions and only the overall movement runs down the gradient [1]

corrected: diffusion is the net movement of particles from a region of their higher concentration to a region of their lower concentration, down a concentration gradient [1]

statement 2 uses semi-permeable, which is not the accepted term, and omits both net and the direction of movement [1]

corrected: 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 [1]

⚠ If you missed marks here: These are the two most expensive words in the whole topic. Net and partially permeable appear on the mark scheme almost every year, and their absence is one of the commonest reasons for losing marks that the candidate genuinely understood. Write both definitions out in full until they are automatic.

1 mark -- identifies that the word net is missing from the diffusion definition 1 mark -- gives a corrected diffusion definition including net and the gradient 1 mark -- identifies semi-permeable as the wrong term and notes the missing direction 1 mark -- gives a corrected osmosis definition with water potential and partially permeable

(b) [3]

Statement 3: “Water moves by osmosis to where there is more solute, because the solute attracts it.”
Explain what is wrong with this statement and give a correct explanation of the direction of osmosis.

Model Answer – 7(b)

the direction described happens to be correct, but the reason is wrong: solutes do not attract water molecules [1]

water molecules move at random, and a solution with more solute has fewer free water molecules, so it has a lower water potential [1]

the net movement of water is therefore down the water potential gradient, from higher to lower water potential [1]

⚠ If you missed marks here: This is the misconception that survives longest, because it always gives the right prediction. It fails as soon as a question asks why, and it makes the water potential ideas of later topics much harder. Replace “attracts” with “lower water potential” every time you use it.

1 mark -- states that solutes do not attract water molecules 1 mark -- more solute means fewer free water molecules and a lower water potential 1 mark -- net movement of water is down the water potential gradient

Statement 4: “A plant cell in distilled water bursts.”
Statement 5: “Active transport uses energy from the concentration gradient.”
Identify the error in each and give the correct biology.

Model Answer – 7(c)

statement 4 is wrong because a plant cell has a strong, inelastic cell wall which resists the swelling; the cell becomes turgid instead of bursting [1]

it is the animal cell, which has no cell wall, that bursts in distilled water [1]

statement 5 is wrong because a concentration gradient is not an energy source; the energy for active transport is released by respiration, mostly in the mitochondria [1]

⚠ If you missed marks here: Both errors come from the same habit — borrowing an idea from one part of the topic and using it where it does not belong. The cell wall is the entire reason plant and animal cells behave differently in dilute solutions, and active transport works against a gradient, so the gradient cannot possibly be paying for it.

1 mark -- plant cell becomes turgid because the cell wall resists swelling 1 mark -- it is the animal cell without a wall that bursts 1 mark -- energy for active transport comes from respiration not from the gradient

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