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⚡ Challenge Paper Preparation

Challenge Prep: Movement Into and Out of Cells

IGCSE Biology 0610 — Topic 3

Every exam paper in this topic is a challenge paper, so this is where the ramping happens. Topic 3 is the biggest mark-loser in the first half of the syllabus, and it is not because the science is hard — it is because three words go missing. Diffusion is net movement, not just movement. Membranes are partially permeable, never “semi-permeable”. And water moves down a water potential gradient, not “towards the solute”. On top of those sit the classics: plant cells said to burst, animal cells said to plasmolyse, mitochondria said to power osmosis, percentage change divided by the wrong mass, isotonic points read off the nearest data point instead of the crossing, and conclusions extrapolated far beyond the data. Twelve traps, six walkthroughs, six pairs to separate, six wrong answers to dissect and ten full challenge questions — all below.

⚠️ Common Traps & Misconceptions

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Twelve traps that cost marks on Topic 3 questions. The first three are the ones that cost the most, so they are dealt with in the most depth.

⚠️ TRAP 1
Trap 1: Leaving out the word “net”
The Trap“Diffusion is the movement of particles from a high concentration to a low concentration.” It sounds like a textbook sentence, it is what most students write, and on many mark schemes it earns nothing. The same trap has a second face: “at equilibrium the particles stop moving”.
The TruthParticles move randomly in every direction, all the time. They never form an orderly stream and they never stop. What runs downhill is the overall balance of those random journeys, because more particles happen to leave a crowded region than happen to enter it. That is why the definition must read: the net movement of particles from a region of their higher concentration to a region of their lower concentration, down a concentration gradient, as a result of their random movement. And at equilibrium the concentrations are equal, so equal numbers cross in each direction — the movement continues, the net movement is zero.
Why It Matters“Define diffusion” is close to a guaranteed question, and the word appears again in osmosis (“net movement of water molecules”), in the isotonic point (“no net movement”), in the fully turgid cell and in the isotonic drip. One word, at least five separate marks across a paper.
Example Question“Define diffusion and state where the energy for it comes from. [3]”
⚠️ TRAP 2
Trap 2: Writing “semi-permeable” instead of “partially permeable”
The Trap“Osmosis is the movement of water through a semi-permeable membrane.” The biology is right and the mark is gone. Older textbooks, older teachers and most of the internet still use the old term, so it feels completely safe.
The TruthCambridge 0610 uses partially permeable and mark schemes are written with that wording. The phrase means the membrane lets some substances through and not others — water molecules pass, larger solute molecules such as sucrose do not. That selectivity is the entire reason osmosis happens at all: if the solute could cross too, both would spread out and no water movement would be observed.
Why It MattersThe phrase belongs in every osmosis answer you will ever write — the definition, the potato explanation, the Visking tubing explanation, the plasmolysis explanation. Getting into the habit now converts a recurring lost mark into a recurring free one.
Example Question“Explain, in terms of water potential, why a potato cylinder loses mass in concentrated sucrose solution. [3]”
⚠️ TRAP 3
Trap 3: “Water moves to where there is more solute”
The TrapThis one is dangerous precisely because it works. It predicts the direction of osmosis correctly every single time, so it survives for years without ever being challenged — until a question says “explain in terms of water potential” and there is nothing to write. Its close relatives are “the salt sucks the water out” and “the sugar attracts the water”.
The TruthSolute particles do not reach out and grab water molecules. Water molecules move at random, exactly as in diffusion. A concentrated solution simply contains fewer free water molecules, and that is what “lower water potential” means. So the net movement of water runs from higher water potential to lower water potential — down the water potential gradient — through a partially permeable membrane. Pure water has the highest water potential of all; dissolving anything in it lowers the water potential.
Why It MattersEvery Extended osmosis question is marked in water potential language, and the same idea reappears in transport in plants and in the kidney. Students who never replace the “attraction” picture find water potential gradients baffling for the rest of the course.
Example Question“A student writes that water moves by osmosis because the solute attracts it. Explain what is wrong with this and give a correct explanation. [3]”
⚠️ TRAP 4
Trap 4: Saying a plant cell bursts, or an animal cell plasmolyses
The Trap“In distilled water the plant cell takes in water and bursts.” “The red blood cell in salt solution became plasmolysed.” The two cell types get swapped constantly, in both directions.
The TruthThe difference is one structure: the cell wall. A plant cell in a dilute solution swells until the strong, inelastic wall pushes back hard enough to stop further net entry — it becomes turgid, and it cannot burst. An animal cell has only a membrane, so nothing resists the swelling and it bursts. In a concentrated solution, the plant cell’s membrane peels away from its wall (plasmolysis), while the animal cell simply shrinks and crinkles — there is no wall for anything to peel away from.
Why It MattersThis comparison is examined every year, often as a two-part question with both cell types in the same solution. The mark always hangs on naming the cell wall and saying what it does mechanically, not on saying “plant cells are stronger”.
Example Question“Explain why a red blood cell bursts in distilled water but an onion cell does not. [2]”
⚠️ TRAP 5
Trap 5: Treating “flaccid” and “plasmolysed” as the same word
The Trap“The cells went flaccid, in other words plasmolysed.” Both describe a cell that has lost water, so they feel interchangeable — and a question asking for the difference then scores zero.
The TruthThey are consecutive stages of the same journey. Flaccid: enough water lost for turgor pressure to fall to zero, so the tissue is soft — but the membrane is still touching the wall. Plasmolysed: so much more water lost that the membrane has pulled away from the cell wall and the gap fills with the external solution. A wilted plant contains flaccid cells; only a cell in a strongly concentrated solution is plasmolysed. And plasmolysis is reversible in a living cell.
Why It Matters“Distinguish between flaccid and plasmolysed” is a standard two-mark question with a single dividing line, and the dividing line is whether the membrane still contacts the wall.
Example Question“Explain the difference between a flaccid plant cell and a plasmolysed plant cell. [2]”
⚠️ TRAP 6
Trap 6: “Mitochondria provide the energy for osmosis”
The Trap“Root hair cells have many mitochondria because they need energy to absorb water.” It sounds sensible — absorbing anything sounds like work — and it scores zero, because it is a biological error rather than an incomplete answer.
The TruthOsmosis is passive. It runs on the kinetic energy of the water molecules themselves and costs the cell nothing at all, which is why osmosis works perfectly well in dead potato tissue. The mitochondria are there for active transport of mineral ions: aerobic respiration releases energy, and that energy runs the protein carriers that drag nitrate, phosphate and potassium into the cell against a steep concentration gradient. There is also no such thing as “active transport of water”.
Why It MattersThe root hair cell is the syllabus’s showcase example for both processes at once, so it appears constantly — and the mitochondria question is always attached to the ions, never to the water.
Example Question“Root hair cells contain many mitochondria. Explain how this is related to their function. [2]”
⚠️ TRAP 7
Trap 7: Dividing by the final mass in a percentage change calculation
The TrapInitial 4.20 g, final 4.62 g, and the answer written down is +9.1 %. The arithmetic is flawless; the denominator is wrong. The twin error is dropping the minus sign when the cylinder has lost mass.
The TruthPercentage change = (final − initial) ÷ INITIAL × 100. Here (0.42 ÷ 4.20) × 100 = +10.0 %. Dividing by 4.62 gives 9.1 % and loses the accuracy mark. And the sign is information, not decoration: positive means water entered, negative means water left, and the examiner reads it as part of your answer.
Why It MattersPercentage change is the standard calculation of the whole topic and appears on almost every Paper 4. It is also the number you plot to find the isotonic point, so an error here propagates into the graph question that follows.
Example Question“A cylinder of initial mass 4.50 g has a final mass of 4.32 g. Calculate the percentage change in mass. [2]”
⚠️ TRAP 8
Trap 8: Reading the isotonic point off the nearest data point
The TrapThe table shows +2 % at 0.4 and −11 % at 0.6, so the answer written down is “0.4, because that is closest to zero”. Alternatively: “at that point osmosis stops.”
The TruthThe isotonic point is where the line crosses zero, which almost always lies between two tested concentrations — here around 0.43 mol dm⁻³. You are expected to interpolate. And what it means is that the solution and the cell sap have the same water potential, so there is no net movement of water. Water molecules still cross the membrane in both directions.
Why It MattersThis is the single most examined graph in Topic 3, and it usually carries three marks: the value, how you obtained it, and what it means. Each of the three has its own way of being lost.
Example Question“Estimate the concentration with the same water potential as the potato cells, and explain how you obtained your estimate. [3]”
⚠️ TRAP 9
Trap 9: Calling ion uptake “diffusion” without checking the gradient
The Trap“Nitrate ions are small and dissolved, so they diffuse into the root hair.” Size and solubility feel decisive, so the direction of the gradient never gets checked.
The TruthThe identity of the particle never tells you the process — only the direction of the gradient does. Root cells typically hold nitrate at around a hundred times the concentration of the soil water, so uptake runs against the gradient and must be active transport, using energy released by respiration and specific protein carriers. Glucose is the same story: it diffuses into a respiring muscle cell, but it is actively transported out of the gut and out of a kidney tubule.
Why It MattersData tables asking “name the process” are a Paper 2 and Paper 4 staple, and they are built precisely to punish answers based on the substance rather than the gradient.
Example Question“Name the process by which nitrate ions enter a root hair cell from very dilute soil water, and give a reason. [2]”
⚠️ TRAP 10
Trap 10: Explaining a maintained gradient by inventing a pump
The Trap“Oxygen keeps entering the muscle cell because the cell pumps it in to keep the gradient steep.” Anything that keeps going seems to need a machine driving it.
The TruthNo pump. Respiration continually uses the oxygen up, so its concentration inside the cell never rises to match the blood and the concentration gradient is permanently maintained. The mirror image works for carbon dioxide: respiration keeps making it, so it keeps diffusing out. The same sentence with different nouns explains carbon dioxide entering a photosynthesising leaf and glucose being absorbed from the gut while the blood carries it away.
Why It MattersThis is one of the most reusable two-mark answers in the entire syllabus, and it reappears in gas exchange, photosynthesis, transport and excretion. Learning it once pays out repeatedly.
Example Question“Explain how the concentration gradient for oxygen between the blood and a muscle cell is maintained. [2]”
⚠️ TRAP 11
Trap 11: “The big cube has more surface area, so it decolourises first”
The TrapPerfectly true and completely wrong. The 30 mm cube really does have the largest total surface area — 5400 mm² against 600 mm² — and it is still the last to change colour.
The TruthWhat matters is the surface area to volume ratio: how much surface there is for each unit of volume that has to be supplied. For a cube of side l the ratio is 6 ÷ l, so it falls as the object grows: 0.6 : 1, 0.3 : 1, 0.2 : 1. On top of that, the distance from surface to centre is 5 mm in the small cube and 15 mm in the large one. Two of the four factors, both favouring the small cube.
Why It MattersWhenever a question compares sizes it is asking about the ratio, and the same reasoning underlies why large organisms need transport systems and why gas exchange surfaces are always folded.
Example Question“Calculate the surface area to volume ratio of each cube and explain which becomes colourless first. [4]”
⚠️ TRAP 12
Trap 12: “It is a fair test” and “there could have been errors”
The TrapTwo phrases that feel like answers and are not. “Fair test” names no variable; “there could have been errors” names no error. Both are written under time pressure and both score zero.
The TruthFor a control variable, name the variable and say what it would otherwise change — and the strongest reasons name one of the four factors: “the same sucrose concentration was used, so the water potential gradient was the same for every cylinder and only the surface area varied”. For evaluation, name a specific weakness and its effect: “the edge of the coloured circle is fuzzy, so judging the diameter is subjective”. And remember the distinction: repeats and means improve reliability; better measurement improves accuracy.
Why It MattersChallenge papers put three or four of these marks in every paper, and they are the marks most often left on the table by students who did the science perfectly.
Example Question“State three variables that must be controlled and explain the effect of not controlling one of them. [3]”

🧩 Multi-Step Reasoning Walkthroughs

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Six challenge questions broken into steps. Try each step yourself before revealing the next — the reasoning is the point, not the answer.

Walkthrough 1 — The Potato Graph, From Raw Masses to Water PotentialA student cuts six cylinders from one potato. In 0.6 mol dm⁻³ sucrose a cylinder of initial mass 4.50 g has a final blotted mass of 4.32 g. The other results give percentage changes of +15.0, +8.0, +2.0, −10.0 and −15.0 % at 0.0, 0.2, 0.4, 0.8 and 1.0 mol dm⁻³. (a) Calculate the missing percentage change. [2] (b) Estimate the isotonic concentration and explain what it means. [3] (c) Explain what happened to the cells in 1.0 mol dm⁻³. [3]
1

Subtract in the right order

change = final − initial = 4.32 − 4.50 = −0.18 g. Subtracting the other way round gives +0.18 and reverses the entire biological conclusion, so always write final first. The negative sign is telling you water left the tissue.

2

The denominator that everyone gets wrong

(−0.18 ÷ 4.50) × 100 = −4.0 %. Dividing by the final mass of 4.32 gives −4.2 % and loses the accuracy mark. Circle the initial mass in the question before you pick up the calculator, every single time.

3

Interpolate — do not pick the nearest point

The values run +2.0 % at 0.4 and −4.0 % at 0.6, so the line crosses zero between them, roughly two thirds of the way across at about 0.47 mol dm⁻³. Answering “0.4” because it gives the smallest number loses the mark: the crossing point is not a tested concentration.

4

Water potentials equal, no NET movement

At this concentration the sucrose solution and the potato cell sap have the same water potential, so water molecules cross the membrane equally in both directions and there is no net movement. Never write “osmosis has stopped” — it has not, and that phrase is refused.

5

Use the template, word for word

The 1.0 mol dm⁻³ solution has a lower water potential than the cell sap, so there is a net movement of water molecules out of the cells by osmosis through their partially permeable cell membranes. The cells lose water and become flaccid; some become plasmolysed. The cylinder therefore loses mass.

Final Answer−4.0 %; isotonic point about 0.47 mol dm⁻³, where the water potentials are equal and there is no net movement; in 1.0 mol dm⁻³ water leaves the cells by osmosis and they become flaccid or plasmolysed.
Walkthrough 2 — Agar Cubes: Two Factors Working TogetherCubes of agar containing a pink indicator, with sides of 10 mm, 20 mm and 30 mm, are dropped into an excess of dilute acid. They become completely colourless after 240 s, 480 s and 720 s. (a) Calculate the surface area to volume ratio of each. [3] (b) Explain the order in which they change. [3] (c) Predict the time for a 25 mm cube and justify it. [2] (d) Explain why the acid was in excess. [2]
1

Six faces, and a cube of volume l cubed

Surface area = 6l², volume = l³, so the ratio is 6 ÷ l. That gives 0.6 : 1, 0.3 : 1 and 0.2 : 1. Forgetting the six faces divides every area by six; writing all three ratios the same assumes shape is what matters, when it is scale.

2

Surface area to volume ratio AND distance

The 10 mm cube has the largest surface area to volume ratio, so there is more surface for every unit of volume that must be reached. It also has the shortest distance from surface to centre — 5 mm against 15 mm — and rate of diffusion falls as distance increases. An answer naming only one factor scores one mark of three.

3

Equal steps in side, equal steps in time

240, 480 and 720 s rise in equal steps for equal steps of side, so time is directly proportional to the length of the side — 24 s per millimetre. For 25 mm: 25 × 24 = 600 s. Predictions are always marked as outcome plus justification, so quote the proportionality explicitly.

4

Name the factor being held constant

A large excess means the acid concentration barely falls during the experiment, so the concentration gradient stays high and constant. Using the same concentration for every cube makes that gradient a controlled variable, so any difference in time must be caused by the size of the cube. Do not write “to make it a fair test” — name the variable.

Final AnswerRatios 0.6 : 1, 0.3 : 1, 0.2 : 1; the smallest cube changes first because of its higher ratio and shorter diffusion distance; 600 s for 25 mm; the excess acid keeps the concentration gradient constant.
Walkthrough 3 — Why the Wheat Wilts in Wet SoilA coastal field has been flooded by sea water. Measurements of relative water potential (0 = pure water) give: soil water −18, wheat root hair cell sap −12, salt-tolerant grass cell sap −26. Explain why the wheat wilts although the soil is wet, and how the grass survives. [6]
1

More solute means lower water potential

Pure water is the highest water potential of all, at 0 on this scale. Every dissolved particle lowers it, so more negative means more concentrated. The soil water at −18 is therefore more concentrated than the wheat sap at −12.

2

Water moves towards the LOWER water potential

Water moves from higher water potential to lower. Wheat sap −12 is higher than soil water −18, so water moves out of the root hair cells into the soil. The plant is losing water to wet soil — physiological drought.

3

Turgor pressure is what holds the plant up

The cells lose water, turgor pressure falls and they become flaccid. Since a non-woody plant is supported by the pressure of water inside its cells pressing outwards on the cell walls, losing that pressure means the stem and leaves droop. That is wilting.

4

It out-concentrates the soil

The grass sap is −26, lower than the soil water at −18, so the gradient still runs into the root and water enters by osmosis. It achieves this by keeping a very high concentration of dissolved solutes in its cell sap. It does not keep the salt out, and it certainly does not actively transport water — there is no such process.

Final AnswerThe soil water has a lower water potential than the wheat sap, so water leaves the roots by osmosis and the cells go flaccid; the grass keeps its sap even more concentrated, so its water potential stays below the soil’s and water still enters.
Walkthrough 4 — Proving a Movement Is Active TransportBarley roots take up potassium ions from a dilute solution that contains about one hundredth of the concentration already inside the cells. Uptake is 4 units at 0 % oxygen, 22 at 5 %, 38 at 10 % and 50 at 21 %. (a) Describe the results using figures. [2] (b) Explain them. [3] (c) Suggest why uptake is not zero at 0 % oxygen. [2]
1

Rising, but not in a straight line

Uptake increases with oxygen concentration, from 4 to 50 arbitrary units. The increase is large at first (4 to 22 between 0 % and 5 %) and much smaller later (38 to 50 between 10 % and 21 %). Description marks almost always require quoted figures, and a second mark is often available for spotting that the relationship is not linear.

2

Uphill movement rules out the free processes

The ions are already about a hundred times more concentrated inside the cells, so they are moving against the concentration gradient. That eliminates diffusion and osmosis immediately and tells you energy must be involved. Get this sentence down first — without it, nothing that follows has a reason.

3

Oxygen, respiration, energy, carriers

More oxygen allows more aerobic respiration, which releases more energy, which is used by the protein carriers in the cell membrane to move more ions across per hour. Compressing this to “more oxygen means more uptake” scores one mark of three.

4

Respiration does not stop without oxygen

Two creditable ideas: anaerobic respiration still releases a small amount of energy, so a little active transport continues; and a small amount of movement may occur by diffusion, which needs no energy from the cell. “Experimental error” throws away a mark that reasoning would have earned.

Final AnswerUptake rises from 4 to 50 units with oxygen; the ions move against the gradient by active transport driven by aerobic respiration; the residual uptake without oxygen comes from anaerobic respiration and a little diffusion.
Walkthrough 5 — Two Movements Across One Piece of TubingVisking tubing containing a mixture of starch solution and glucose solution is rinsed, tied off and suspended in distilled water at 37 °C. After 30 minutes the water outside gives an orange-red precipitate with Benedict’s solution but stays orange-brown with iodine solution. The tubing and contents have also gained mass. Explain all three observations. [6]
1

The tubing is partially permeable

Visking tubing has tiny pores. Glucose molecules are small enough to pass through; starch molecules are far too large. So glucose diffuses out down its concentration gradient and gives a positive Benedict’s test outside, while starch stays inside and the iodine test outside stays negative.

2

Colours and conditions both earn marks

Benedict’s solution must be heated, and a positive result is an orange-red precipitate, not simply “it goes orange”. A negative iodine test is orange-brown — the colour it already was. Saying iodine “stays the same” is weaker than naming the colour.

3

This is osmosis, going the other way

The contents are a solution of starch and glucose, so they have a lower water potential than the distilled water outside. Water therefore moves into the tubing by osmosis through the partially permeable tubing, increasing the volume inside and so the mass.

4

Glucose out by diffusion, water in by osmosis

The single most common error here is explaining the mass gain with glucose — but glucose was leaving. Two different substances, two different processes, two opposite directions, all at the same time through the same barrier. Write them as two separate sentences so the examiner can see you have distinguished them.

Final AnswerGlucose diffuses out because it is small enough to cross the partially permeable tubing, giving a positive Benedict’s test; starch is too large so iodine stays negative; water enters by osmosis because the contents have the lower water potential, so the tubing gains mass.
Walkthrough 6 — Reading a Plateau Off a GraphUptake of a substance by gut lining cells is measured as its external concentration rises. With oxygen: 14, 26, 42, 50, 51 units. Without oxygen: 2, 4, 8, 15, 29 units. (a) Compare the two sets. [2] (b) Explain why the oxygen line levels off. [3] (c) Explain why the no-oxygen line keeps rising. [2]
1

Same column, both conditions

Uptake is far greater with oxygen at every concentration — for example 42 units against 8 units at the third concentration. Picking two numbers from different columns is the standard way of losing a comparison mark, so always take the pair from the same point on the scale.

2

A plateau means a fixed number of something

With oxygen, most of the uptake is active transport through protein carriers. There is a fixed number of carriers, so once every one of them is working flat out, adding more of the substance outside cannot increase the rate — the line flattens. Saying “the cells are full” is not credited: uptake is a rate, and the cells keep using the substance as fast as it arrives.

3

A straight climb means simple diffusion

Without oxygen there is little energy for active transport, so what remains is largely diffusion. The rate of diffusion depends on the steepness of the concentration gradient, so raising the outside concentration simply keeps raising the rate. No carriers involved, so no ceiling.

4

Graph shape identifies the process

Plateau plus oxygen dependence equals active transport. Straight rise with no oxygen dependence equals diffusion. You can identify the process from the shape of the data alone, without being told anything about the substance — which is exactly the skill challenge papers are testing.

Final AnswerUptake is much higher with oxygen (42 against 8 units); the oxygen line plateaus because all the carrier proteins are saturated; the no-oxygen line keeps rising because diffusion depends only on the concentration gradient.

🔍 Spot the Difference

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Six pairs of questions that look almost identical and have different answers. Find the distinction before you read the key difference.

Question A
A plant cell is placed in distilled water. What happens?
Water enters by osmosis and the cell becomes turgid. The strong, inelastic cell wall resists the swelling and its inward pressure stops further net entry.
Question B
A red blood cell is placed in distilled water. What happens?
Water enters by osmosis and the cell swells and bursts. There is no cell wall, so nothing resists the pressure.
Key DifferenceIdentical osmosis, opposite outcome, and the whole difference is the cell wall. Never write “the plant cell bursts” or “the red blood cell became turgid”.
Question A
Which cube has the greatest total surface area?
The largest cube. A 30 mm cube has 5400 mm² against the 10 mm cube’s 600 mm².
Question B
Which cube becomes completely colourless first in acid?
The smallest cube — largest surface area to volume ratio (0.6 : 1) and the shortest distance to its centre (5 mm).
Key DifferenceTotal area and area per unit volume are different quantities and they point to opposite cubes. Whenever a question compares sizes, it is asking about the ratio.
Question A
A cell is flaccid. What does that mean?
Enough water has been lost for turgor pressure to fall to zero, so the tissue is soft — but the cell membrane is still touching the cell wall.
Question B
A cell is plasmolysed. What does that mean?
So much more water has been lost that the cell membrane has pulled away from the cell wall, and the gap fills with the external solution.
Key DifferenceOne dividing line: is the membrane still in contact with the wall? A wilted plant has flaccid cells; only a cell in a strongly concentrated solution is plasmolysed — and it can recover.
Question A
Nitrate ions move into a root hair cell from very dilute soil water. Name the process.
Active transport — the ions are already far more concentrated inside, so this is movement against the gradient, using energy from respiration.
Question B
Oxygen moves into a root hair cell from air spaces in the soil. Name the process.
Diffusion — respiration keeps using oxygen up inside, so oxygen is less concentrated inside and moves down its gradient.
Key DifferenceSame cell, same membrane, same moment — but opposite gradients. The direction of the gradient, never the identity of the particle, decides the process.
Question A
What happens at the isotonic point on a potato graph?
The solution and the cell sap have the same water potential, so water crosses the membrane equally in both directions and there is no net movement.
Question B
What happens in a fully turgid plant cell?
Turgor pressure has risen until it balances the tendency of water to enter, so again there is no net movement — but here the water potentials are not equal; pressure is doing the balancing.
Key DifferenceBoth end in “no net movement”, but for different reasons: equal water potentials in one case, pressure from the cell wall in the other. In neither case has osmosis stopped.
Question A
How do you improve the reliability of a potato experiment?
Repeat each concentration at least three times and calculate a mean, so anomalies from uneven cutting or blotting matter less.
Question B
How do you improve the accuracy of the same experiment?
Use a balance reading to 0.01 g, cut the cylinders to identical dimensions, and blot every cylinder in exactly the same controlled way.
Key DifferenceMark schemes distinguish them sharply: reliability comes from repeats and means; accuracy comes from better measurement. Offering repeats as an accuracy improvement scores nothing.

🔗 Topic 3 Concept Map

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Click each node to see how the three processes connect to one another and to the rest of the syllabus.

⭐ CORE FRAMEWORK 1
One question decides everything: which way does the gradient run?
Step 1 — Is the substance water? ▶
If the moving substance is water, the process is osmosis — always, whatever is dissolved in it. Osmosis moves water and nothing else.
Osmosis is the net movement of water molecules from higher to lower water potential, through a partially permeable membrane.
There is no such process as active transport of water. If you have written that phrase, something has gone wrong.
Step 2 — Down the gradient, or up it? ▶
High → low concentration: diffusion. Passive, free, powered by the kinetic energy of the particles themselves.
Low → high concentration: active transport. Needs energy released by respiration and specific protein carriers in the membrane.
The identity of the particle tells you nothing. Glucose diffuses into a muscle cell and is actively transported out of a kidney tubule.
Step 3 — Check it against the evidence ▶
Movement stops without oxygen, or with a respiratory inhibitor, or at 5 °C? It was active transport.
Movement continues under those treatments? It was passive — diffusion or osmosis.
Uptake reaches a plateau as external concentration rises? The carrier proteins are saturated — another signature of active transport.
⭐ CORE FRAMEWORK 2
The four factors, and the adaptations built from them
Surface area and distance — the pathway ▶
Surface area up, rate up: millions of alveoli, villi and microvilli, long thin root hairs, folded gill filaments.
Distance up, rate down — the one factor that works the other way: alveolus walls one cell thick, thin flat leaves, flatworms, insect tracheoles ending beside the muscle.
Together they give the surface area to volume ratio = 6 ÷ l for a cube, which falls as an organism grows — and that is why large organisms need transport systems.
Temperature and gradient — the driving conditions ▶
Temperature up, rate up: the particles gain kinetic energy and move faster. It does not change the gradient.
Concentration gradient up, rate up: maintained by ventilation, by blood flow, and above all by respiration or photosynthesis constantly using a substance up.
The reusable sentence: “respiration continually uses the oxygen up, so its concentration inside stays low and the gradient is maintained”.
⭐ CORE FRAMEWORK 3
Where Topic 3 reappears in the rest of the syllabus
In plants ▶
Support: turgor pressure pressing outwards on cell walls holds up every non-woody plant; lose it and the plant wilts.
Water uptake: osmosis into root hairs down the water potential gradient, then all the way to the leaf.
Mineral uptake: active transport through carriers, which is why waterlogged soil starves a crop of nitrate and magnesium.
Gas exchange and transpiration: carbon dioxide in and water vapour out, both by diffusion through stomata.
In animals ▶
Gas exchange: diffusion across alveoli, maintained by breathing and blood flow.
Absorption: diffusion and active transport of digested food across the villi, which is why gut cells have microvilli and mitochondria.
Excretion and the kidney: urea removed in solution; glucose reabsorbed by active transport until none is left in the urine.
Osmoregulation: isotonic drips, freshwater fish producing dilute urine, marine fish drinking constantly.

❌ "Why Is This Wrong?" Exercises

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Six real student answers. Decide what is wrong and what it would score before you reveal the flaw.

Exercise 1: "Define diffusion and say where the energy comes from. [3]"
Student’s Answer"Diffusion is when particles move from a high concentration to a low concentration until they are spread out evenly and then they stop. The energy comes from respiration."
The FlawThree errors in two sentences, and it scores zero. The word net is missing; “then they stop” is actively wrong; and the energy is attributed to respiration, which belongs to active transport alone.
Correct Answer"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], as a result of their random movement [1]. The energy comes from the kinetic energy of the particles themselves, not from respiration [1]."
Key RuleWhenever you write “energy” in Topic 3, check which process you have attached it to. Only active transport may have energy from respiration.
Exercise 2: "Explain why root hair cells contain many mitochondria. [2]"
Student’s Answer"Because they need a lot of energy to take in water from the soil by osmosis."
The FlawOsmosis is passive and costs the cell nothing, so linking mitochondria to it is a biological error rather than an incomplete answer — which is why it scores zero rather than one.
Correct Answer"Mitochondria are the site of aerobic respiration, which releases energy [1]. Root hair cells need a great deal of energy for the active transport of mineral ions from the dilute soil water into the cell, against a concentration gradient [1]."
Key RuleOsmosis works perfectly well in dead potato tissue. If a process still happens without respiration, it cannot be what the mitochondria are for.
Exercise 3: "Explain what happens to plant cells in concentrated sucrose solution, in terms of water potential. [3]"
Student’s Answer"The sucrose has more solute so it sucks the water out of the cells through the semi-permeable membrane and the cells go floppy and die."
The FlawOne mark at most. “Sucks” is not a mechanism and the question demanded water potential; “semi-permeable” is the wrong term; “floppy” is not flaccid or plasmolysed; and the cells are not dead — plasmolysis is reversible.
Correct Answer"The sucrose solution has a lower water potential than the cell sap [1], so there is a net movement of water molecules out of the cells by osmosis through their partially permeable cell membranes [1]. The cells become flaccid, and in a strongly concentrated solution the membrane pulls away from the cell wall so they become plasmolysed [1]."
Key RuleBoth students understood that water leaves. Five words — lower water potential, net, partially permeable, flaccid, plasmolysed — are the whole difference between one mark and three.
Exercise 4: "A graph of percentage change in mass crosses zero at 0.42 mol dm⁻³. Explain what this shows. [2]"
Student’s Answer"At 0.42 the potato and the solution are the same, so osmosis has stopped."
The Flaw“The same” — the same what? And “osmosis has stopped” is wrong: water molecules keep crossing the membrane in both directions.
Correct Answer"At 0.42 mol dm⁻³ the sucrose solution has the same water potential as the potato cell sap [1], so there is no net movement of water into or out of the cells and the mass is unchanged [1]."
Key Rule“No net movement” is the most valuable three-word phrase in this topic: it answers the isotonic point, the fully turgid cell, the isotonic drip and the equilibrium question.
Exercise 5: "Name the process by which nitrate ions enter a root hair from very dilute soil water, and give a reason. [2]"
Student’s Answer"Diffusion, because nitrate ions are small and dissolved in water so they can pass through the membrane easily."
The FlawThe reason given is about the particle, when the only thing that decides the process is the direction of the gradient. The stem said the soil water is very dilute, which means the ions are already more concentrated inside the cell.
Correct Answer"Active transport [1], because the nitrate ions are already at a higher concentration inside the cell, so they are being moved against the concentration gradient, which requires energy released by respiration [1]."
Key RuleBefore naming any process, work out which way the gradient runs. Small, soluble and dissolved describe almost every particle in biology and prove nothing.
Exercise 6: "State two variables that must be controlled in a potato osmosis experiment and explain why. [2]"
Student’s Answer"Everything must be kept the same so that it is a fair test and the results are reliable."
The FlawNo variable is named and no consequence is given, so there is nothing to credit. “Fair test” and “reliable” are labels, not explanations — and reliability is not even what control variables are for.
Correct Answer"The temperature must be the same for every cylinder, because temperature affects the rate of movement of water molecules [1]. The time in solution must be the same, because a cylinder left longer would move closer to its final mass and the percentage changes would not be comparable [1]."
Key RuleName the variable, then say what it would otherwise change. And keep the distinction sharp: control variables give validity; repeats and means give reliability.

✍️ Ultra-Detailed Practice Questions

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Ten Cambridge-style challenge questions. Write a full answer first, then reveal the model answer with the mark allocation and the examiner’s notes.

Question 1
[6 marks]
Define diffusion, osmosis and active transport, stating in each case the direction of movement and the source of the energy. [6]
Model AnswerDiffusion: the net movement of particles from a region of their higher concentration to a region of their lower concentration, down a concentration gradient, as a result of their random movement [1]; the energy comes from the kinetic energy of the particles themselves, so no energy from the cell is required [1].

Osmosis: 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]; also passive, powered by the random movement of the water molecules [1].

Active transport: the movement of particles through a cell membrane from a region of lower concentration to a region of higher concentration, against a concentration gradient [1], using energy released by respiration, with protein carriers moving the particles across [1].
Examiner’s NotesSix marks, six specific ideas — and each definition is marked as direction plus energy. The most frequently dropped marks are the word net in the first two, and the phrase partially permeable in the second. Writing “energy from respiration” in the diffusion or osmosis definition does not merely fail to score; it contradicts the definition and can cost the energy mark outright.
Question 2
[8 marks]
A student cuts six cylinders from one potato and leaves each in a different sucrose concentration for 30 minutes. In 0.2 mol dm⁻³ a cylinder of initial mass 4.00 g has a final mass of 4.32 g; in 0.8 mol dm⁻³ a cylinder of initial mass 4.00 g has a final mass of 3.60 g. (a) Calculate both percentage changes in mass. [3] (b) Explain, in terms of water potential, why the two results differ. [3] (c) Explain why percentage change is used rather than change in grams. [2]
Model Answer(a) 0.2 mol dm⁻³: change = +0.32 g; (0.32 ÷ 4.00) × 100 = +8.0 % [1]. 0.8 mol dm⁻³: change = −0.40 g; (−0.40 ÷ 4.00) × 100 = −10.0 % [1]. Both divided by the initial mass, with signs shown [1].

(b) The 0.2 mol dm⁻³ solution has a higher water potential than the cell sap, so there is a net movement of water into the cells by osmosis and they gain mass [1]. The 0.8 mol dm⁻³ solution has a lower water potential than the cell sap, so there is a net movement of water out and the cylinder loses mass [1]. In both cases the water crosses the partially permeable cell membranes by osmosis [1].

(c) The cylinders do not all have exactly the same initial mass [1], so dividing by the initial mass makes the results comparable between cylinders of different starting sizes [1].
Examiner’s NotesPart (a) is where the marks disappear: dividing by the final mass gives +7.4 % and −11.1 %, and dropping the minus sign removes the only evidence that water left the tissue. In (b) notice that the same sentence is used twice with one word changed — higher, then lower. Learn it as a template and it writes itself. In (c) an answer saying only “percentages are fairer” usually gets one mark of two; say why they are fairer.
Question 3
[7 marks]
Explain why a red blood cell bursts in distilled water while an onion epidermal cell in the same solution becomes turgid. Then explain what happens to each cell in a strongly concentrated salt solution, using the correct technical terms. [7]
Model AnswerIn distilled water, the water potential outside is higher than inside both cells, so water moves into both by osmosis through their partially permeable membranes [1]. The red blood cell has only a cell surface membrane, which cannot withstand the pressure as it swells, so it bursts [1]. The onion cell has a strong, inelastic cellulose cell wall [1] which resists the swelling and exerts an inward pressure, so net entry of water stops and the cell becomes turgid [1].

In concentrated salt solution, the water potential outside is lower than inside, so water moves out of both cells by osmosis [1]. The red blood cell shrinks and its surface becomes crinkled [1]. The onion cell loses turgor, becomes flaccid, and with further water loss the cell membrane pulls away from the cell wall — it is plasmolysed [1].
Examiner’s NotesThis is the most heavily examined comparison in Topic 3 and the mark scheme is strict about vocabulary. Turgid, turgor pressure, flaccid and plasmolysed require a cell wall and may never be applied to the red blood cell; burst may never be applied to the plant cell. Note also that the osmosis itself is identical in both cells — the entire difference is one structure, and the answer must say what that structure does mechanically.
Question 4
[8 marks]
A student investigates the effect of surface area on the rate at which a substance enters plant tissue. She has one potato, a cork borer, a scalpel, a balance, a ruler, sucrose solutions and paper towels. (a) Describe how she could vary surface area while keeping the mass of tissue constant. [2] (b) State the independent, dependent and three controlled variables. [3] (c) Predict the results and explain them. [3]
Model Answer(a) Cut equal masses of potato and divide each mass into a different number of pieces — one whole cylinder, two halves, four quarters, eight smaller pieces [1]; this increases the total surface area while the total mass of tissue stays the same [1].

(b) Independent variable: the surface area of the tissue [1]. Dependent variable: the percentage change in mass after a fixed time [1]. Controlled: the concentration and volume of sucrose solution, the temperature, and the time in solution, along with consistent blotting [1].

(c) The samples with the largest surface area reach their final percentage change sooner, because a larger surface area allows water to enter or leave faster [1]. However, the final percentage change is about the same for every sample [1], because cutting does not alter the water potential of the tissue or of the solution, so the position of equilibrium is unchanged [1].
Examiner’s NotesPart (c) is the discriminator. Surface area changes the rate at which equilibrium is reached, not where equilibrium lies — and candidates who predict a larger final change have confused the two. In (b), “the mass” is not an acceptable dependent variable because the starting masses differ; it must be the percentage change. Naming the same idea twice in the controlled list, such as “the solution” and “the sucrose”, earns only one mark.
Question 5
[7 marks]
Explain how a root hair cell obtains both water and mineral ions from soil water, and explain why waterlogging the soil reduces the uptake of ions but not of water. [7]
Model AnswerSoil water is very dilute, so it has a higher water potential than the root hair cell sap [1]; water therefore enters by osmosis, down the water potential gradient, through the partially permeable cell membrane, and this requires no energy from the cell [1]. The long, narrow extension of the root hair gives a large surface area for uptake [1].

Mineral ions such as nitrate are already far more concentrated inside the cell than in the soil water, so they must be moved against the concentration gradient by active transport [1], using energy released by respiration in the many mitochondria of the cell, with protein carriers moving the ions across the membrane [1].

Waterlogged soil contains almost no air spaces, so root cells receive little oxygen and aerobic respiration is greatly reduced; less energy is available, so active transport of ions falls sharply [1]. Osmosis is passive and needs no energy from the cell, so water uptake continues largely unaffected [1].
Examiner’s NotesThe whole question turns on the contrast between a passive and an active process happening at the same time across the same membrane. Answers that describe water being “absorbed” without naming osmosis, or that give mitochondria as the reason for water uptake, lose marks even when the rest is correct. The waterlogging section is a chain — oxygen, respiration, energy, active transport — and each missing link costs a mark.
Question 6
[6 marks]
Cubes of agar containing a pink indicator that turns colourless in acid have sides of 5 mm, 10 mm and 20 mm. They become completely colourless after 150 s, 300 s and 600 s. (a) Explain what the pattern in the times shows about the factors affecting diffusion. [3] (b) Use the results to explain why an elephant needs a transport system and Amoeba does not. [3]
Model Answer(a) Doubling the length of the side doubles the time, so the time taken is directly proportional to the side length and therefore to the distance from the surface to the centre [1]. This shows that the rate of diffusion falls as distance increases [1]. The smaller cubes also have a larger surface area to volume ratio (6 : 1 at 5 mm falling to 1.5 : 1 at 20 mm, using ratio = 6 ÷ l in millimetres), so more surface is available per unit of volume to be supplied [1].

(b) As an organism gets larger, its surface area to volume ratio falls and the distance from the surface to the innermost cells increases [1]. In Amoeba no part of the cell is more than a fraction of a millimetre from the surface, so diffusion alone supplies it [1]. In an elephant the inner cells are far too distant for diffusion to supply them quickly enough, so a transport system carries substances close to every cell by mass flow, leaving diffusion to cover only the final short distance [1].
Examiner’s NotesIn (a) the data must be used, not just described: the proportionality is the evidence, and answers that simply state “bigger cubes take longer” score one mark at most. Note that a volume relationship would predict times of 150, 1200 and 9600 s, which the data flatly contradict — testing a suggested relationship against the numbers is a habit worth building. In (b) the phrase mass flow is what lifts the answer from a description to an explanation.
Question 7
[6 marks]
A dialysis tubing bag containing 1.0 mol dm⁻³ sucrose solution is fitted to a capillary tube and stood in distilled water. The liquid rises 26 mm in the first 10 minutes, reaching 64 mm by 30 minutes and 78 mm by 60 minutes, after which it stops rising. (a) Calculate the mean rate of rise over the first 30 minutes, with its unit. [2] (b) Explain why the liquid rises. [2] (c) Explain why it slows and finally stops. [2]
Model Answer(a) 64 ÷ 30 [1] = 2.1 mm/min (or mm min⁻¹) [1].

(b) The sucrose solution has a lower water potential than the distilled water, so water moves into the bag by osmosis through the partially permeable tubing [1]; the extra volume inside cannot escape, so the liquid is pushed up the narrow capillary tube [1].

(c) As water enters, the sucrose solution is diluted, raising its water potential so the water potential gradient becomes less steep [1]; in addition the rising column of liquid exerts an increasing downward pressure opposing further entry, and when the two balance there is no further net movement [1].
Examiner’s NotesRate answers need working, a number and a unit; a rate quoted without units does not gain the second mark. In (c) either the dilution or the pressure argument is creditable, but the answer must say why the gradient changes rather than simply that “osmosis slows down”. Note the phrasing at the end: water molecules never stop crossing the membrane; what reaches zero is the net movement.
Question 8
[6 marks]
Explain fully why salting fish preserves it for months in a warm climate, and why a wilted lettuce leaf becomes crisp again when soaked in cold water. [6]
Model AnswerSalting: a heavy coating of salt gives the liquid film around the bacteria and fungi on the fish a very low water potential [1]. There is therefore a net movement of water out of the microorganisms’ cells by osmosis through their partially permeable membranes [1]. They lose so much water that they cannot grow or reproduce, so the fish does not spoil [1].

Lettuce: the water has a higher water potential than the cell sap of the lettuce cells [1], so there is a net movement of water into the cells by osmosis and they become turgid [1]. The pressure of the water inside pressing outwards on the cell walls makes the tissue firm again, which is what makes the leaf crisp [1].
Examiner’s NotesThe salting half must be about the microorganisms, not the fish — “salt kills bacteria” names no mechanism and scores nothing. The lettuce half must end with the mechanism of support: turgor pressure acting on the cell wall. Saying only that “the cells fill with water” describes the change without explaining why full cells are stiff, and that final link is usually the third mark.
Question 9
[7 marks]
The uptake of a substance by a tissue is measured with and without oxygen as the external concentration is raised. With oxygen the uptake rises steeply and then plateaus; without oxygen it is much lower but continues to rise steadily with no plateau. (a) Identify the process operating in each case, with a reason. [3] (b) Explain the plateau. [2] (c) Describe one further experiment that would confirm your identification. [2]
Model Answer(a) With oxygen the uptake is largely by active transport, because it depends on oxygen and therefore on the energy released by aerobic respiration [1]. Without oxygen the remaining uptake is by diffusion [1], because diffusion is passive and its rate depends only on the steepness of the concentration gradient, which is why it keeps rising [1].

(b) The rate of active transport depends on the number of protein carriers in the cell membrane [1]; once every carrier is working at its maximum rate, adding more of the substance outside cannot increase the rate any further, so the line levels off [1].

(c) Repeat the measurements with a respiratory inhibitor such as cyanide added, or at a low temperature such as 5 °C [1]; if the uptake falls to the level seen without oxygen, that confirms the extra uptake depended on energy from respiration and so was active transport [1].
Examiner’s NotesA plateau always means that something other than the substance supplied has become limiting, and for a carrier-based process that something is the number of carriers. “The cells are full” is not credited, because uptake is a rate and the cells go on using the substance as it arrives. In (c) the experiment must include what result would count as confirmation — a proposed method with no predicted outcome usually scores one mark of two.
Question 10
[6 marks]
A student writes the following three statements. For each, identify the error and give the corrected biology. (a) “Osmosis is the movement of water through a semi-permeable membrane to where there is more solute.” (b) “A plant cell in distilled water bursts.” (c) “Active transport uses energy from the concentration gradient.” [6]
Model Answer(a) Two errors: the term should be partially permeable, not semi-permeable, and the movement should be described as a net movement of water molecules down a water potential gradient rather than as water moving towards solute, since solutes do not attract water [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].

(b) A plant cell has a strong, inelastic cell wall which resists the swelling and exerts an inward pressure, so net entry of water stops and the cell becomes turgid rather than bursting [1]. It is the animal cell, which has no cell wall, that bursts in distilled water [1].

(c) A concentration gradient is not an energy source — active transport works against it [1]. The energy for active transport is released by respiration, mostly aerobic respiration in the mitochondria [1].
Examiner’s NotesAll three errors come from the same habit: borrowing an idea from one part of the topic and using it where it does not belong. Note that (a) predicts the direction of osmosis correctly, which is exactly why it survives so long — it only fails when a question asks why. Marks in questions of this style are usually split evenly between identifying the error and giving the correction, so an answer that only says “this is wrong” scores half.
IGCSE Biology 0610 — Topic 3 Challenge Prep. Work through every section, and reread Traps 1, 2 and 3 the night before the exam — the word net, the phrase partially permeable, and the water potential gradient are worth more marks in this topic than anything else you can revise.