IGCSE Biology Paper 4 (Theory / Extended)

Topic 3: Movement Into and Out of Cells -- Challenge Exam 1

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 – Agar Cubes and Surface Area

Total: 12 marks

A class in Nairobi investigates how the size of a block of tissue affects the movement of substances into it. They make cubes of agar jelly containing a pink indicator that turns colourless in acid, and drop three cubes of different sizes into a large excess of dilute hydrochloric acid at the same time. The time taken for each cube to become completely colourless right through to the centre is recorded.

length of side / mm	10	20	30
time to become completely colourless / s	240	480	720

(a) [3]

For each of the three cubes, calculate the surface area, the volume and the surface area to volume ratio. Show your working.

Model Answer – 1(a)

10 mm cube: surface area = 6 × 10² = 600 mm², volume = 10³ = 1000 mm³, ratio = 0.6 : 1 [1]

20 mm cube: surface area = 6 × 20² = 2400 mm², volume = 8000 mm³, ratio = 0.3 : 1 [1]

30 mm cube: surface area = 6 × 30² = 5400 mm², volume = 27 000 mm³, ratio = 0.2 : 1 [1]

⚠ If you missed marks here: The commonest error is forgetting the six faces and calculating only one face, which divides every surface area by six. The second is assuming that because all three shapes are cubes the ratio must be the same — it is not, because doubling the side multiplies the area by four but the volume by eight. Always write the ratio in the form “something : 1” so the three numbers can be compared directly.

1 mark -- surface area to volume ratio 0.6 : 1 for the 10 mm cube with working 1 mark -- surface area to volume ratio 0.3 : 1 for the 20 mm cube with working 1 mark -- surface area to volume ratio 0.2 : 1 for the 30 mm cube with working

(b) [3]

Use your answers and the idea of diffusion to explain why the smallest cube became completely colourless first.

Model Answer – 1(b)

the 10 mm cube has the largest surface area to volume ratio, so there is more surface for acid to enter through for each unit of volume that has to be decolourised [1]

the distance from the surface to the centre is shortest in the smallest cube (5 mm compared with 15 mm) [1]

diffusion is slower over longer distances, so the acid takes much longer to reach the centre of the large cube [1]

⚠ If you missed marks here: Two separate factors are being tested here — surface area and distance — and an answer that names only one usually scores one mark out of three. Saying “the big cube has more surface area” is true but self-defeating: the large cube really does have the greater total area, and the point is that it has far more volume to supply, so the ratio is what counts.

1 mark -- largest surface area to volume ratio for the smallest cube 1 mark -- shortest distance from the outer surface to the centre 1 mark -- rate of diffusion decreases as distance increases

Predict how long a cube with sides of 25 mm would take to become completely colourless. Justify your prediction using the pattern in the results.

Model Answer – 1(c)

600 s [1]

the time is directly proportional to the length of the side (240 s for 10 mm means 24 s for every millimetre of side), and 25 × 24 = 600 [1]

⚠ If you missed marks here: A prediction without a justification is worth half the marks at most. The justification examiners want is the proportionality shown by the data, not a general remark that “bigger cubes take longer”. Check the proportionality before you use it — here 240, 480 and 720 rise in equal steps for equal steps of side length, which is what makes the linear prediction safe.

1 mark -- prediction of 600 s 1 mark -- justification that time is directly proportional to length of side using the data

(d) [2]

Explain why the acid was used in a large excess and why the same concentration of acid was used for every cube.

Model Answer – 1(d)

a large excess means the acid concentration barely falls, so the concentration gradient stays high and constant throughout the experiment [1]

using the same concentration for every cube makes the concentration gradient a controlled variable, so any difference in time must be caused by the size of the cube [1]

⚠ If you missed marks here: This is a controlled-variable question dressed up as a chemistry question. The mark is not for saying “to make it a fair test” on its own; you must name the variable being controlled — the concentration gradient, which is one of the four factors affecting the rate of diffusion.

1 mark -- excess acid keeps the concentration gradient high and constant 1 mark -- same concentration controls the gradient so size is the only variable

(e) [2]

Use the results of this experiment to explain why a large animal such as an elephant needs a transport system, while a single-celled organism does not.

Model Answer – 1(e)

as an organism gets larger its surface area to volume ratio falls and the distance from the surface to the innermost cells increases [1]

diffusion is then too slow to supply the inner cells with oxygen and nutrients, so a transport system is needed to carry substances close to every cell [1]

⚠ If you missed marks here: The link back to the data is what earns the second mark: the 30 mm cube took three times as long as the 10 mm cube, and a real animal is thousands of times thicker again. Answers that simply state “elephants are big so they need blood” describe the situation without explaining it and are not credited.

1 mark -- surface area to volume ratio falls and diffusion distance increases with size 1 mark -- diffusion alone is too slow to supply inner cells so a transport system is required

Question 2 – Potato Cylinders in Sucrose

Total: 12 marks

A student in Bangalore cuts six cylinders of the same length and diameter from a single potato, blots each one and weighs it, and leaves each in a different concentration of sucrose solution for 30 minutes. Each cylinder is then removed, gently blotted and reweighed.

sucrose concentration / mol dm⁻³	0.0	0.2	0.4	0.6	0.8	1.0
initial mass / g	4.20	4.00	4.10	4.50	4.00	4.40
final mass / g	4.83	4.32	4.18	4.32	3.60	3.74
change in mass / %	+15.0	+8.0	+2.0	to be calculated	−10.0	−15.0

(a) [2]

Calculate the percentage change in mass of the cylinder left in 0.6 mol dm⁻³ sucrose solution. Show your working and give the sign of your answer.

Model Answer – 2(a)

change in mass = 4.32 − 4.50 = −0.18 g; percentage change = (−0.18 ÷ 4.50) × 100 [1]

= −4.0 % (the negative sign is required) [1]

⚠ If you missed marks here: Dividing by the final mass instead of the initial mass is the standard error in this calculation and gives −4.2 %, which loses the accuracy mark. The other frequent loss is the missing minus sign: the sign is not decoration, it is the piece of information that tells the examiner water left the tissue.

1 mark -- working shown dividing the change by the initial mass of 4.50 g 1 mark -- answer of minus 4.0 per cent including the negative sign

(b) [2]

Explain why the results are expressed as a percentage change in mass rather than as a change in grams.

Model Answer – 2(b)

the cylinders had different initial masses, ranging from 4.00 g to 4.50 g [1]

dividing by the initial mass allows a fair comparison between cylinders of different starting sizes [1]

⚠ If you missed marks here: Quote the actual range of starting masses from the table — an answer that says only “percentages are fairer” has not used the data and rarely gets both marks. Note that percentage change does not remove the need for careful blotting or equal timing; it only corrects for differences in starting mass.

1 mark -- cylinders had different initial masses quoted from the table 1 mark -- percentage allows fair comparison between different starting masses

Estimate the concentration of sucrose solution that has the same water potential as the potato cell contents. Explain how you obtained your estimate and what this concentration means.

Model Answer – 2(c)

approximately 0.45–0.50 mol dm⁻³ (between 0.4 and 0.6, nearer 0.5) [1]

this is the concentration at which the percentage change in mass is zero, found where the plotted line crosses the concentration axis [1]

at this concentration the solution and the cell contents have the same water potential, so there is no net movement of water into or out of the cells [1]

⚠ If you missed marks here: Two traps sit in this part. The first is quoting 0.4 because it gives the smallest number in the table — the crossing point lies between 0.4 and 0.6, not at either of them. The second is writing that “osmosis has stopped”; water molecules go on crossing the membrane in both directions, and what is zero is the net movement.

1 mark -- estimate of about 0.45 to 0.50 mol per dm cubed 1 mark -- explained as the point where percentage change in mass is zero 1 mark -- equal water potential inside and outside so no net movement of water

(d) [3]

Explain, in terms of water potential, what happened to the cells of the cylinder placed in 1.0 mol dm⁻³ sucrose solution.

Model Answer – 2(d)

the 1.0 mol dm⁻³ 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 the partially permeable cell membranes [1]

the cells lose water and become flaccid, and some may become plasmolysed, so the cylinder loses mass [1]

⚠ If you missed marks here: This is the part where vocabulary is worth real marks. Write partially permeable, not “semi-permeable”; write net movement of water molecules, not just “water moves”; and write that the solution has a lower water potential, not that it “has more solute so it sucks water out”. Each precise phrase is a mark the vague version does not get.

1 mark -- solution has a lower water potential than the cell sap 1 mark -- net movement of water out by osmosis through partially permeable membranes 1 mark -- cells become flaccid or plasmolysed so mass decreases

(e) [2]

Suggest two changes that would improve the reliability of these results.

Model Answer – 2(e)

repeat each concentration with at least three cylinders and calculate a mean percentage change [1]

control the temperature and the time in solution for every cylinder, and cut all cylinders to identical dimensions from the same potato [1]

⚠ If you missed marks here: Reliability comes from repeats and means; accuracy comes from careful measurement. Suggestions such as “use a better balance” improve accuracy but not reliability, and “do the experiment more carefully” is too vague to be credited. Naming which variable you would control turns a weak answer into a marked one.

1 mark -- repeat each concentration and calculate a mean 1 mark -- control temperature time and cylinder dimensions using one potato

Question 3 – Cells in Different Solutions

Total: 12 marks

A student examines strips of onion epidermis and samples of human cheek cells under a microscope. One sample of each is mounted in distilled water and one in 1.0 mol dm⁻³ sucrose solution. The onion cells in distilled water look full and firm; the onion cells in sucrose show their contents shrunken into the middle of each cell. The cheek cells in distilled water swell and then disappear from view, leaving only debris.

(a) [3]

Name and explain the condition of the onion cells mounted in distilled water.

Model Answer – 3(a)

the cells are turgid [1]

distilled water has a higher water potential than the cell sap, so water enters the cells by osmosis through the partially permeable cell membrane [1]

the vacuole swells and pushes the cytoplasm against the cell wall, producing turgor pressure that presses outwards on the wall [1]

⚠ If you missed marks here: “Turgid” alone is one mark of three; the explanation must include the direction of the water potential gradient and the resulting turgor pressure. Note that it is the vacuole filling that pushes the contents against the wall — answers that say the cell wall swells or stretches describe something that does not happen.

1 mark -- cells identified as turgid 1 mark -- water enters by osmosis down a water potential gradient through a partially permeable membrane 1 mark -- vacuole swells and turgor pressure presses outwards on the cell wall

(b) [3]

Name and explain the condition of the onion cells mounted in the sucrose solution.

Model Answer – 3(b)

the cells are plasmolysed [1]

the sucrose solution has a lower water potential than the cell sap, so there is a net movement of water out of the cells by osmosis [1]

the cell contents shrink and the cell membrane pulls away from the cell wall [1]

⚠ If you missed marks here: The description that earns the third mark is specific: the membrane separates from the wall. Writing that “the cell shrivels” or “the cell wall collapses” does not describe plasmolysis — the wall keeps its shape throughout, which is exactly why a clear gap becomes visible between it and the shrunken protoplast.

1 mark -- cells identified as plasmolysed 1 mark -- lower water potential outside causes net movement of water out by osmosis 1 mark -- cell membrane pulls away from the cell wall

Explain why the cheek cells in distilled water burst, while the onion cells in distilled water did not.

Model Answer – 3(c)

the onion cell has a strong, inelastic cellulose cell wall that resists the swelling and exerts an inward pressure, so net entry of water stops once the cell is turgid [1]

the cheek cell has only a cell surface membrane, which cannot withstand the pressure as water enters, so the cell bursts [1]

⚠ If you missed marks here: This comparison is examined again and again, and the mark depends on naming the cell wall and saying what it does mechanically. It is not enough to say “plant cells are stronger”: the membranes of the two cells are equally fragile, and the wall outside the membrane is the entire difference.

1 mark -- cell wall is strong and inelastic and resists swelling 1 mark -- animal cell has only a membrane which cannot withstand the pressure so it bursts

(d) [2]

Explain the difference between a flaccid plant cell and a plasmolysed plant cell.

Model Answer – 3(d)

a flaccid cell has lost enough water for its turgor pressure to fall to zero, so it is soft, but the cell membrane still lies against the cell wall [1]

a plasmolysed cell has lost so much more water that the cell membrane has pulled away from the cell wall [1]

⚠ If you missed marks here: Treating the two words as synonyms costs both marks. They are consecutive stages of the same process, and the dividing line is precisely whether the membrane is still in contact with the wall. A wilting plant contains flaccid cells; only a cell in a strongly concentrated solution becomes plasmolysed.

1 mark -- flaccid cell has lost turgor but membrane still touches the wall 1 mark -- plasmolysed cell has the membrane pulled away from the wall

(e) [2]

Explain how turgor helps to support a young, non-woody plant, and why such a plant wilts if it is not watered.

Model Answer – 3(e)

water inside turgid cells presses outwards on the cell walls, and this pressure makes the tissues firm and holds the stem and leaves up [1]

if water is lost faster than it is taken up, the cells lose turgor and become flaccid, so the pressure supporting the tissue falls and the plant wilts [1]

⚠ If you missed marks here: The syllabus wording is that plants are supported by “the pressure of water inside the cells pressing outwards on the cell wall”, so use it. Answers that say the plant wilts “because it is dying” or “because the walls collapse” miss the mechanism — the walls are unchanged, and a wilted plant recovers within minutes of being watered.

1 mark -- water pressing outwards on cell walls makes tissues firm and supports the plant 1 mark -- loss of water makes cells flaccid so support is lost and the plant wilts

Question 4 – Ion Uptake by Barley Roots

Total: 12 marks

Barley seedlings are grown with their roots in a dilute solution containing potassium ions. The concentration of potassium ions inside the root cells is already about 100 times greater than the concentration in the solution. The oxygen concentration of the solution is varied and the uptake of potassium ions is measured over one hour.

oxygen concentration / %	0	5	10	21
uptake of potassium ions / arbitrary units	4	22	38	50

(a) [3]

Describe how the uptake of potassium ions changes as the oxygen concentration increases, using figures from the table.

Model Answer – 4(a)

uptake increases as oxygen concentration increases [1]

the increase is large at first and then smaller: from 4 to 22 units between 0 % and 5 % oxygen, but only from 38 to 50 units between 10 % and 21 % [1]

uptake rises more than twelvefold overall, from 4 to 50 arbitrary units [1]

⚠ If you missed marks here: A description mark almost always requires quoted figures with their units, and a second mark is usually available for noticing that the relationship is not a straight line. Writing only “uptake goes up” describes the trend but uses none of the data, and it is the data handling that is being assessed here.

1 mark -- states that uptake increases as oxygen concentration increases 1 mark -- notes the rate of increase falls at higher oxygen concentrations with figures 1 mark -- quotes correct supporting figures with units from the table

(b) [3]

Explain these results in terms of respiration and active transport.

Model Answer – 4(b)

the potassium ions are already more concentrated inside the cells, so they are being moved against the concentration gradient by active transport [1]

active transport needs energy, which is released by respiration; more oxygen allows more aerobic respiration [1]

more energy is therefore available to the protein carriers, so more ions are moved into the cells per hour [1]

⚠ If you missed marks here: The chain has three links — oxygen, respiration, energy for active transport — and each is worth a mark, so a compressed answer such as “more oxygen means more uptake” scores one at most. Establish the direction of movement first: without stating that uptake is against the gradient, there is no reason for the process to need energy at all.

1 mark -- ions move against the concentration gradient by active transport 1 mark -- more oxygen allows more aerobic respiration to release energy 1 mark -- more energy available for the carrier proteins so more ions are moved

Suggest why the uptake of potassium ions is not zero when the oxygen concentration is 0 %.

Model Answer – 4(c)

anaerobic respiration still releases a small amount of energy, so a little active transport can continue [1]

a small amount of movement into the cells may also occur by diffusion, which needs no energy from the cell [1]

⚠ If you missed marks here: A “suggest” command means the answer is not printed in your notes and you are expected to reason. Both credited ideas start from the same principle: active transport needs energy from respiration, and respiration does not stop completely without oxygen. Answering “the experiment is inaccurate” throws away a mark that reasoning would have earned.

1 mark -- anaerobic respiration releases some energy for active transport 1 mark -- some ions may enter by diffusion which needs no energy

(d) [2]

Root hair cells contain many mitochondria. Explain how this is related to their function.

Model Answer – 4(d)

mitochondria are the site of aerobic respiration, which releases energy [1]

root hair cells use a great deal of energy for the active transport of mineral ions from the soil into the cell against a concentration gradient [1]

⚠ If you missed marks here: The commonest wrong answer here is “for osmosis”. Osmosis is passive and needs no energy from the cell at all, so mitochondria have nothing to do with water uptake. Link the organelle to the process that actually costs energy, and name that process precisely.

1 mark -- mitochondria are the site of aerobic respiration releasing energy 1 mark -- energy is needed for active transport of mineral ions against the gradient

(e) [2]

Describe the part played by protein carriers in the uptake of potassium ions.

Model Answer – 4(e)

specific carrier proteins in the cell membrane bind the potassium ions on the outside of the membrane [1]

using energy from respiration the carrier changes shape and releases the ions on the inside of the membrane, so ions can be moved against the gradient [1]

⚠ If you missed marks here: Carrier proteins are often described as “holes” or “channels that let things through”, which describes a passive pore and cannot explain uphill movement. The two features to state are specificity and the use of energy to change shape. Nothing is broken down or rebuilt — the ion that binds is the ion released.

1 mark -- specific carrier proteins in the membrane bind the ions 1 mark -- carrier uses energy to change shape and move ions against the gradient

Question 5 – A Model Gut

Total: 10 marks

A student sets up a model of the small intestine. A length of dialysis (Visking) tubing is filled with a mixture of starch solution and glucose solution, tied at both ends, rinsed on the outside and suspended in a boiling tube of distilled water at 37 °C. Samples of the water in the boiling tube are tested at the start and again after 30 minutes.

(a) [3]

Predict the result of testing the water in the boiling tube after 30 minutes with iodine solution and with Benedict’s solution, and explain both predictions.

Model Answer – 5(a)

iodine solution stays orange-brown, showing no starch is present in the water [1]

Benedict’s solution gives an orange-red precipitate on heating, showing glucose is present [1]

glucose molecules are small enough to pass through the pores of the partially permeable tubing by diffusion, but starch molecules are far too large [1]

⚠ If you missed marks here: The explanation mark hangs entirely on molecule size and the tubing being partially permeable. Answers claiming the tubing “chooses” what to let through, or that starch is held back because it is insoluble, are not credited. Remember that Benedict’s solution must be heated to give its result, and describe the colour as an orange-red precipitate, not just “it goes orange”.

1 mark -- iodine stays orange-brown showing no starch outside 1 mark -- Benedict solution gives an orange-red precipitate showing glucose outside 1 mark -- glucose molecules small enough to diffuse through but starch molecules too large

(b) [2]

Explain why dialysis tubing is a good model of the wall of the small intestine.

Model Answer – 5(b)

the tubing is partially permeable, letting small soluble molecules such as glucose through while holding back large molecules such as starch, just as the gut wall does [1]

this models absorption, in which the products of digestion pass from the gut contents into the blood while undigested material stays in the gut [1]

⚠ If you missed marks here: Models earn marks for the feature they share with the real structure, not for looking similar. A weaker answer says “it is a tube like the intestine”, which is a shape comparison with no biology in it. Note also where the model fails, since that is a common follow-up question: the tubing has no villi, no blood supply and no active transport.

1 mark -- tubing is partially permeable letting small molecules through and holding large ones back 1 mark -- models absorption of the products of digestion into the blood

Describe one control that should be set up in this experiment, and state what it would show.

Model Answer – 5(c)

set up an identical tube of Visking tubing containing distilled water only, suspended in distilled water, and test the surrounding water in the same way [1]

a negative Benedict’s test from the control shows that any glucose detected in the main experiment came from inside the tubing and not from contamination of the apparatus [1]

⚠ If you missed marks here: A control must differ from the experiment in exactly one respect and must be tested in exactly the same way. “Do the experiment again” is a repeat, not a control, and repeats improve reliability rather than eliminating an alternative explanation. Always finish by saying what the control rules out.

1 mark -- control tube of tubing containing distilled water treated identically 1 mark -- negative result shows the glucose detected came from inside the tubing

(d) [3]

After 30 minutes the tubing and its contents have gained mass. Explain this observation.

Model Answer – 5(d)

the contents of the tubing contain dissolved starch and glucose, so they have 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 tubing [1]

the volume of water inside increases, so the tubing and contents gain mass [1]

⚠ If you missed marks here: Two different movements are happening at once in this experiment and it is easy to explain the wrong one. Glucose leaving is diffusion of a solute; the gain in mass is osmosis of water in the opposite direction. An answer that says the tubing gained mass because glucose moved will not score, since glucose was leaving.

1 mark -- contents have a lower water potential than the distilled water outside 1 mark -- water moves in by osmosis through the partially permeable tubing 1 mark -- increased volume of water inside gives the gain in mass

Question 6 – Red Blood Cells and Drip Solutions

Total: 12 marks

In a hospital laboratory, three samples of human blood are each mixed with a different solution and examined under a microscope after ten minutes. Solution J is distilled water, solution K is 0.9 % sodium chloride solution (the concentration used in intravenous drips) and solution L is 5 % sodium chloride solution.

solution	J	K	L
appearance of the red blood cells	very few whole cells remain; the mixture looks clear red	cells look normal and unchanged	cells are smaller than normal with crinkled surfaces

(a) [3]

Explain what has happened to the red blood cells in solution J.

Model Answer – 6(a)

distilled water has a higher water potential than the contents of the red blood cells [1]

there is a net movement of water molecules into the cells by osmosis through the partially permeable cell membrane [1]

the cells swell until the membrane can no longer withstand the pressure and they burst, releasing their contents into the solution [1]

⚠ If you missed marks here: Describing the cells as “plasmolysed” or “turgid” costs marks: both terms belong to walled plant cells only. An animal cell that takes in too much water simply bursts, and the correct explanation must state that there is no cell wall to resist the swelling.

1 mark -- distilled water has a higher water potential than the cell contents 1 mark -- net movement of water into the cells by osmosis through the partially permeable membrane 1 mark -- cells swell and burst because there is no cell wall to resist the pressure

(b) [3]

Explain the appearance of the red blood cells in solution L.

Model Answer – 6(b)

the 5 % sodium chloride solution has a lower water potential than the cell contents [1]

there is a net movement of water out of the cells by osmosis [1]

the cells lose water, shrink and their surfaces become crinkled [1]

⚠ If you missed marks here: The direction of the water potential gradient is the whole answer, so make sure you write the comparison the right way round: adding solute always lowers water potential. The other frequent loss is describing the crinkled cells as plasmolysed — there is no wall for a membrane to pull away from, so plasmolysis is impossible here.

1 mark -- 5 per cent solution has a lower water potential than the cell contents 1 mark -- net movement of water out of the cells by osmosis 1 mark -- cells lose water shrink and become crinkled

Explain why solution K is the one chosen for intravenous drips.

Model Answer – 6(c)

solution K has the same water potential as the contents of the red blood cells, so it is isotonic with them [1]

there is therefore no net movement of water into or out of the cells, so the cells are neither burst nor shrunken and continue to work normally [1]

⚠ If you missed marks here: The mark scheme wants “no net movement”, because water molecules still cross the membrane in both directions in equal numbers. Answers stating that “osmosis stops” repeat the single most common misconception in this topic and are not credited even though the practical conclusion is right.

1 mark -- solution has the same water potential as the cell contents so it is isotonic 1 mark -- no net movement of water so cells are neither burst nor shrunken

(d) [2]

Predict what would happen to plant cells placed in solution J and in solution L, and explain why the outcome in solution J is different from that of the red blood cells.

Model Answer – 6(d)

in solution J the plant cells would become turgid rather than bursting, and in solution L they would become plasmolysed [1]

the plant cell has a strong, inelastic cell wall that resists the swelling and exerts an inward pressure, so net entry of water stops before the cell can burst [1]

⚠ If you missed marks here: This is the comparison examiners return to constantly, and the mark hangs on the cell wall, not on the vacuole or on any supposed strength of the plant cell membrane. Both outcomes must be named with the right technical word: turgid in the dilute solution, plasmolysed in the concentrated one.

1 mark -- plant cells become turgid in J and plasmolysed in L 1 mark -- cell wall resists swelling and prevents bursting

(e) [2]

A student writes: “The red blood cells in solution J were plasmolysed by osmosis.” Identify the two errors in this sentence and give the corrected version.

Model Answer – 6(e)

the term plasmolysed is wrong, because plasmolysis can only happen in a cell with a cell wall; the cells in solution J burst [1]

plasmolysis also describes water loss, whereas water was entering the cells in solution J; the corrected sentence is that the red blood cells in solution J burst because water entered them by osmosis [1]

⚠ If you missed marks here: Both errors are vocabulary errors, and vocabulary is where this topic is won or lost. Keep the four plant words for plant cells only — turgid, turgor pressure, flaccid, plasmolysis — and use burst, shrink or crinkle for animal cells. Naming an error without giving the correction usually earns only half the marks available.

1 mark -- plasmolysis cannot occur without a cell wall so the cells burst 1 mark -- plasmolysis describes water loss but water was entering so the corrected sentence says the cells burst

Question 7 – Telling the Three Processes Apart

Total: 10 marks

The table describes four movements of substances into or out of living cells.

	movement
1	water passing from soil water into a root hair cell whose sap is more concentrated than the soil water
2	oxygen passing from an alveolus into a red blood cell in the lung capillary
3	nitrate ions passing from very dilute soil water into a root hair cell that already contains far more nitrate
4	carbon dioxide passing out of a respiring muscle cell into the blood plasma

(a) [4]

Name the process responsible for each of the four movements in the table.

Model Answer – 7(a)

1 osmosis [1]

2 diffusion [1]

3 active transport [1]

4 diffusion [1]

⚠ If you missed marks here: Decide by asking two questions in order: is the substance water, and which way does the gradient run? Water is always osmosis; anything moving from low to high concentration must be active transport; everything else moving down a gradient is diffusion. The movement most often got wrong is the nitrate one, because candidates assume ions in soil water must be plentiful — the stem says the opposite.

1 mark -- movement 1 identified as osmosis 1 mark -- movement 2 identified as diffusion 1 mark -- movement 3 identified as active transport 1 mark -- movement 4 identified as diffusion

(b) [3]

Give a full definition of diffusion, and state where the energy for it comes from.

Model Answer – 7(b)

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

as a result of the random movement of the particles [1]

the energy comes from the kinetic energy of the randomly moving particles themselves, not from respiration [1]

⚠ If you missed marks here: One word carries the first mark: net. Without it the definition describes a one-way stream of particles, which is not what happens and is regularly refused by examiners. The energy mark is lost by anyone who writes “energy from respiration” out of habit — that phrase belongs exclusively to active transport.

1 mark -- net movement down a concentration gradient from higher to lower concentration 1 mark -- as a result of random movement of particles 1 mark -- energy comes from the kinetic energy of the particles not from respiration

A student writes: “Osmosis is when water moves through a semi-permeable membrane to where there is more solute.” Explain what is wrong with this sentence, and rewrite it so that it would gain full credit on an Extended paper.

Model Answer – 7(c)

the term should be partially permeable, not semi-permeable [1]

the movement should be described as a net movement of water molecules down a water potential gradient, not as water moving towards solute [1]

a full-credit version is: 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: Two habits cost marks here and both are hard to unlearn. “Semi-permeable” is the older term and is not on the mark scheme; and describing water as moving “towards the solute” gives the right direction with the wrong mechanism, because water responds to its own gradient, not to the solute. Learn the full sentence and write it the same way every time.

1 mark -- partially permeable is the correct term rather than semi-permeable 1 mark -- movement should be described as net movement of water down a water potential gradient 1 mark -- full definition written correctly with higher to lower water potential

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