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IGCSE Biology Paper 4 (Theory / Extended)

Topic 8: Transport in Plants -- Challenge Exam 3
1 hour 15 minutes
80
7
75:00
0610

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This paper covers the whole of Topic 8. Like a real Cambridge paper it ranges across every sub-topic — 8.1 xylem and phloem, 8.2 water uptake, 8.3 transpiration and 8.4 translocation — and it mixes them inside single questions. All three Topic 8 papers do; they differ in the angle they come at it from, not in what they cover.
Question 1 — Three Organs, One Rule
Total: 12 marks
Transverse section of a young stem W X Y Z W = the outer part of a vascular bundle, X = the inner part of the same bundle, Y = the tissue filling the centre, Z = the outermost layer of the stem
(a) [3]
Using the diagram, name the tissues at W and at X, and state the rule about their positions that applies to roots, stems and leaves alike.
Model Answer — 1(a)
W is the phloem, on the outer side of the bundle [1]
X is the xylem, on the inner side [1]
the rule: the xylem is always nearer the centre of the plant and the phloem nearer the outside — central star in a root, inner side of each bundle in a stem, upper side of the vein in a leaf [1]
⚠ If you missed marks here: Memorising three separate pictures is three times the work and three times the risk. One rule covers all three organs, and it also lets you answer when a section is drawn at an unfamiliar angle or upside down.
(b) [4]
Compare xylem and phloem under four headings: the substances transported, the direction of transport, whether the cells are living, and whether energy from respiration is required.
Model Answer — 1(b)
xylem carries water and mineral ions; phloem carries sucrose and amino acids [1]
xylem moves material upwards only; phloem moves it from source to sink, so either upwards or downwards [1]
xylem vessels are dead and empty; phloem sieve tubes are living [1]
xylem transport needs no energy from the plant — it is driven by evaporation; phloem transport requires energy from respiration [1]
⚠ If you missed marks here: A comparison must compare. Four separate statements about xylem earn far less than four paired statements, because each mark here is for the contrast rather than for the fact. Use the word “whereas” and the structure looks after itself.
(c) [5]
A student is given three unlabelled transverse sections, one from a root, one from a stem and one from a leaf of the same non-woody plant. Describe how the student could identify which is which, and then locate the xylem in each, explaining the reasoning at every step.
Model Answer — 1(c)
the leaf section is a flat slab with a distinct upper and lower surface, rather than roughly circular [1]
the root section is roughly circular with a single central star of vascular tissue, and often bears root hairs on the outside [1]
the stem section is roughly circular with separate vascular bundles arranged in a ring [1]
having identified the organ, the xylem is the central star in the root, the inner half of each bundle in the stem, and the upper half of the vein in the leaf [1]
in all three the xylem can be confirmed by its thick lignified walls, which stain strongly [1]
⚠ If you missed marks here: Notice the order: identify the organ first, then apply the rule. Trying to spot the xylem directly means relying on a remembered picture, and the moment the section is unfamiliar the method fails. Working from the shape of the whole section is what the syllabus means by “identify in diagrams and images”.
Question 2 — Getting Water and Ions Out of the Soil
Total: 12 marks
A root hair cell One epidermal cell, drawn out into a long thin hair that pushes between the soil particles. soil particles, with a film of soil solution around them large vacuole the hair itself — a huge surface area nucleus many mitochondria release energy from respiration for ACTIVE TRANSPORT of mineral ions cell wall (fully permeable) and, just inside it, the partially permeable cell membrane water in by osmosis No chloroplasts (it is underground) and no waxy cuticle (water must be able to get in).
(a) [3]
Using the diagram, explain how three named features of this cell suit it to its function.
Model Answer — 2(a)
the long thin projection gives a very large surface area, so more water and mineral ions can be absorbed per second [1]
the thin cell wall means a short distance for water to cross, so uptake is faster [1]
the many mitochondria release energy from respiration for the active transport of mineral ions [1]
⚠ If you missed marks here: Each feature needs its consequence attached. Also, do not describe the large vacuole as a store of water for dry weather — it keeps the water potential of the cell sap low, which is what maintains the gradient that drives water in.
(b) [3]
Explain why this cell has no chloroplasts and no waxy cuticle, although both are present in cells of the same plant above ground.
Model Answer — 2(b)
the cell is underground and receives no light, so chloroplasts would be useless — it cannot photosynthesise [1]
a waxy cuticle is waterproof, so it would prevent the entry of water [1]
a cell whose whole function is absorption cannot afford a barrier on its absorbing surface, whereas a leaf needs one to reduce water loss [1]
⚠ If you missed marks here: This part is really about the general principle that structure follows function. The same plant builds a waterproof layer on its leaves and leaves it off its roots, and being able to explain both in one sentence is worth more than remembering either separately.
(c) [6]
A student grows identical barley seedlings in dilute nutrient solution under four conditions for one week and measures the uptake of water and of potassium ions. Air bubbled, 20 °C: water 42 cm³, potassium 100 units. Air bubbled, 5 °C: water 31 cm³, potassium 24 units. Nitrogen bubbled, 20 °C: water 39 cm³, potassium 18 units. Nitrogen bubbled, 5 °C: water 28 cm³, potassium 6 units. Explain these results as fully as you can.
Model Answer — 2(c)
water uptake changed relatively little in all four conditions (42 down to 28 cm³) because water enters by osmosis, which is passive and needs no energy [1]
potassium uptake fell dramatically (100 down to 6 units) because ions are taken up largely by active transport, which does need energy [1]
replacing air with nitrogen removes oxygen, so aerobic respiration and therefore the energy supply falls, and ion uptake drops from 100 to 18 units [1]
lowering the temperature slows the enzyme-controlled reactions of respiration, so again less energy is available and uptake falls to 24 units [1]
the two factors together give the lowest value of all (6 units), because both routes to releasing energy have been restricted [1]
water uptake still falls slightly at 5 °C because molecules have less kinetic energy, so osmosis is slower — and because the cooler seedlings transpire less [1]
⚠ If you missed marks here: Six marks means six separate ideas, so write them as separate sentences rather than one long paragraph. The final mark is the one most often missed: a passive process is not completely unaffected by temperature, and the data show a real if modest fall. Quote figures — a table this detailed is put there to be used.
Question 3 — Wind, Water and Two Species
Total: 12 marks
Rate of water uptake against wind speed, two species 0 1 2 3 4 5 0.0 1.5 3.0 4.5 6.0 wind speed / m per s rate of water uptake / mm per min plant A (shaded woodland) plant B (open hillside)
(a) [3]
Describe the shape of the curve for plant A, quoting figures, and explain why the curve levels off.
Model Answer — 3(a)
the rate rises steeply at first, from about 0.9 to about 3.6 mm per minute between 0 and 2 m per second [1]
it then rises much more slowly and levels off at about 4.8 mm per minute above 4 m per second [1]
once the humid layer of air at the leaf surface has been completely removed, further wind cannot make the water vapour gradient any steeper, so another factor — such as the number and aperture of the stomata — becomes limiting [1]
⚠ If you missed marks here: “A different factor is now limiting” explains almost every plateau you will meet in Biology, and here it must be linked to something specific. Saying only “the wind has no more effect” restates the graph instead of explaining it.
(b) [4]
Plant B was collected from an exposed, dry hillside and plant A from a shaded woodland floor. Use the graph to suggest which structural differences between their leaves would account for the difference in the curves, and explain the advantage to plant B.
Model Answer — 3(b)
plant B loses water more slowly at every wind speed (1.65 against 4.8 mm per minute at the plateau) [1]
this suggests plant B has fewer and/or smaller stomata, or stomata sunk into pits, or a thicker waxy cuticle [1]
these features reduce the area available for the diffusion of water vapour, or keep humid air trapped near the pore [1]
the advantage: on an exposed dry hillside water loss would otherwise exceed uptake, so plant B can avoid wilting where plant A could not [1]
⚠ If you missed marks here: “Suggest” means apply what you know to an unfamiliar case, so the answer must be built from the graph rather than recited. Begin by quoting the comparison the graph gives you; a suggestion with no evidence behind it rarely gains full credit.
(c) [3]
State the cost to plant B of these adaptations, and explain why this cost is unavoidable.
Model Answer — 3(c)
the maximum rate of photosynthesis is lower, so plant B grows more slowly [1]
because carbon dioxide enters the leaf through the same stomata that water vapour leaves by [1]
so anything that restricts the outward diffusion of water vapour also restricts the inward diffusion of carbon dioxide — the two cannot be separated [1]
⚠ If you missed marks here: This trade-off is the single most useful idea in the whole of transpiration. It explains why desert plants grow slowly, why a wilting plant that closes its stomata stops feeding itself, and why transpiration is best described as the price of photosynthesis rather than as a purpose.
(d) [2]
Suggest why the readings for both species were taken at the same temperature and humidity.
Model Answer — 3(d)
both temperature and humidity change the rate of transpiration in their own right — by altering evaporation and the steepness of the vapour gradient [1]
so they must be controlled, otherwise any difference between the two species could not be attributed to wind speed [1]
⚠ If you missed marks here: Naming a controlled variable earns little; the mark is for saying what would go wrong without it. “To make it a fair test” is not a reason and is almost never credited.
Question 4 — From Air Space to Root Tip
Total: 12 marks
(a) [4]
Explain how the internal structure of a leaf leads to a high rate of water loss, even though the leaf is covered by a waterproof cuticle.
Model Answer — 4(a)
the spongy mesophyll is a loose network of cells with large air spaces between them [1]
so the leaf has a very large internal surface area of wet cell walls exposed to air [1]
water evaporates from these walls into the air spaces, which become saturated with water vapour [1]
the vapour then diffuses out through the stomata, which must be open to let carbon dioxide in [1]
⚠ If you missed marks here: The cuticle is doing its job — almost no water crosses it. The loss happens through the pores that have to be open for gas exchange, which is why the answer must reach the stomata rather than stopping at the air spaces.
(b) [5]
Explain how the loss of water vapour from a leaf results in water being drawn out of the soil by a root hair cell several metres below.
Model Answer — 4(b)
evaporation from the mesophyll cells lowers their water potential, so water moves into them from the xylem by osmosis [1]
this puts the water column in the xylem under tension — the transpiration pull [1]
the column does not break because there are forces of attraction between water molecules, so it behaves as one continuous column [1]
the pull is transmitted down the whole column to the root, drawing water out of the root cortex and into the xylem [1]
this lowers the water potential of the cortex cells and then of the root hair cell, so water enters it from the soil solution by osmosis [1]
⚠ If you missed marks here: The chain has to run all the way from the leaf to the soil without a break, and the mark for the root hair is the one usually left off. Notice that every single step is passive: the plant spends no energy at all moving water, which is why a stem killed by steam still conducts.
(c) [3]
Explain fully why a plant wilts, and explain why wilting is partly protective as well as damaging.
Model Answer — 4(c)
wilting occurs when the rate of water loss exceeds the rate of uptake: cells lose water, become flaccid, and turgor pressure no longer supports the soft tissues [1]
it is damaging because the drooping leaves intercept less light and the stomata close, so photosynthesis slows [1]
it is protective because drooping reduces the leaf surface exposed to the sun and closed stomata greatly reduce further water loss, giving uptake a chance to catch up [1]
⚠ If you missed marks here: Wilting is not simply damage. A plant that wilts at midday and recovers by evening has done exactly what it should, and being able to argue both sides is what an “explain fully” command word is asking for.
Question 5 — Where the Sugar Goes
Total: 10 marks
Radioactive label found in each part, 6 hours after supply 0 10 20 30 40 50 label recovered / % labelled leaf other mature leaves stem roots developing fruits One mature leaf was enclosed and supplied with labelled carbon dioxide for 30 minutes.
(a) [3]
Using the chart, identify the two parts of the plant that were acting as sinks, and justify your choice with figures.
Model Answer — 5(a)
the developing fruits, which received 32 % of the label [1]
and the roots, which received 22 % [1]
both received large amounts of labelled sucrose, and a sink is a part that uses or stores what the phloem delivers — both are growing and neither can photosynthesise enough for itself [1]
⚠ If you missed marks here: A justification must use the definition, not just the size of the bar. The stem also holds 6 %, but that is mostly label in transit through the phloem rather than material being used or stored, so it is not counted as a sink.
(b) [3]
Explain why the other mature leaves received only 2 % of the label, and explain what this shows about the direction of translocation.
Model Answer — 5(b)
other mature leaves are sources in their own right: they photosynthesise and export sucrose rather than importing it [1]
so almost nothing is delivered to them, and the 2 % is a trace rather than a transport route [1]
the label reached the roots (below) and the fruits (above), showing translocation can occur in either direction, towards whichever part is a sink [1]
⚠ If you missed marks here: A very small number is a result, not experimental error. It is telling you that a mature leaf is a source, and that is a mark. Treating small values as noise is one of the commonest ways of throwing away data-handling marks.
(c) [4]
Six hours after the experiment, 38 % of the label was still in the leaf it was supplied to. Suggest three reasons for this, and explain how the experiment could be modified to show whether the label eventually leaves.
Model Answer — 5(c)
some of the labelled carbon has been used in respiration in the leaf itself, or built into the leaf’s own structures [1]
some is still stored in the leaf as starch, which must be converted back to sucrose before it can be loaded into the phloem [1]
translocation is continuous but not instant, so six hours is simply not long enough to export it all [1]
modification: repeat the measurements at several times — for example after 6, 12, 24 and 48 hours — and see whether the percentage in the labelled leaf falls [1]
⚠ If you missed marks here: The last mark is for designing a test, not for describing the result you expect. A single time point can never show a trend, which is why “take readings at intervals” is so often the improvement being looked for.
Question 6 — A Field Experiment on Two Crops
Total: 12 marks
(a) [3]
A farmer grows the same variety of maize in two fields. Field 1 is sheltered by a hedge; field 2 is open and exposed to strong wind. The plants in field 2 are shorter and their leaves wilt on hot days even though both fields are irrigated equally. Explain these observations.
Model Answer — 6(a)
wind removes the humid air at the leaf surface, keeping the water vapour gradient steep, so transpiration is faster in field 2 [1]
on hot days the rate of loss exceeds the rate of uptake, so the cells lose turgor and the leaves wilt [1]
wilting leaves droop and their stomata close, so less carbon dioxide enters and photosynthesis is reduced, and the plants grow less [1]
⚠ If you missed marks here: The question links three sub-topics: transpiration, wilting, and the effect on photosynthesis. A complete answer has to get from the wind to the height of the plant, and the middle step — loss exceeding uptake — is what makes the chain work.
(b) [3]
The farmer plants a row of trees as a windbreak along the edge of field 2. Suggest two ways in which this could increase the yield, and one way in which it might reduce it.
Model Answer — 6(b)
it reduces wind speed, so the rate of transpiration falls, less wilting occurs and the stomata stay open longer for photosynthesis [1]
less water is lost, so less irrigation is needed and the plants suffer less water stress [1]
but the trees shade part of the field, and their roots compete with the crop for water and mineral ions, so yield near the trees may fall [1]
⚠ If you missed marks here: A “suggest” question with a disadvantage attached is asking for a balanced judgement. The disadvantage must follow from planting trees specifically, not be a general remark, and shading and root competition are both directly relevant to earlier topics.
(c) [3]
The farmer applies a very heavy dose of soluble fertiliser to one part of field 1. Within two days those plants wilt, although the soil is moist. Explain this result.
Model Answer — 6(c)
the soil solution is now very concentrated, so it has a lower water potential than the cell sap of the root hair cells [1]
water therefore moves out of the root hair cells by osmosis instead of into them [1]
uptake falls while transpiration continues, so the cells lose turgor and the plants wilt — even though there is plenty of water in the soil [1]
⚠ If you missed marks here: This is the counter-intuitive one. Water always moves down the water potential gradient, and if the outside is made more concentrated than the inside, the gradient — and the direction of movement — reverses. The same reasoning explains too much sugar in a vase of cut flowers.
(d) [3]
Later in the season the maize develops cobs. The farmer notices that plants with more leaves produce larger cobs. Explain this using the terms source and sink.
Model Answer — 6(d)
a developing cob is a sink: it cannot photosynthesise enough for itself and imports sucrose and amino acids [1]
the leaves are the sources, releasing sucrose into the phloem after photosynthesis [1]
more leaves means more sucrose translocated to the cob, so more material is available for growth and storage and the cob is larger [1]
⚠ If you missed marks here: Notice that this part uses the vocabulary of 8.4 to explain an observation from the field. The examiner is checking that source and sink are working ideas rather than definitions to be recited, so the answer must connect the number of leaves to the amount translocated.
Question 7 — Following the Water With a Stain
Total: 10 marks
(a) [4]
Describe how you would use a stain to investigate the pathway of water through the above-ground parts of a plant, and state the result you would expect.
Model Answer — 7(a)
cut a leafy shoot (or a stick of celery) under water and stand it in dilute stain such as eosin or methylene blue [1]
leave it for one to two hours in warm, bright, moving air so that transpiration is rapid and the stain is drawn up quickly [1]
then cut thin transverse sections at several heights up the stem and examine them [1]
the stain is found in the xylem only — in a stem, as a ring of coloured patches on the inner side of each vascular bundle, continuing out into the leaf veins [1]
⚠ If you missed marks here: A question about a pathway can only be answered by looking inside, so the sectioning step is essential. Measuring how much stained water disappeared from the beaker is a very common wrong answer: it tells you how much water moved and nothing at all about which tissue carried it.
(b) [3]
Explain why sections are cut at several heights rather than at one, and explain why the shoot is cut under water at the start.
Model Answer — 7(b)
one section could show a local patch of stain; the same tissue stained at every height shows a continuous pathway from the base to the leaves [1]
cutting under water prevents air being drawn into the xylem [1]
because an air bubble would break the continuous water column, so the transpiration pull could not be transmitted and the stain would not rise [1]
⚠ If you missed marks here: Both halves are really the same principle in different clothes: an experiment has to be designed so that it can distinguish between two possible explanations. One section cannot distinguish a pathway from a patch, and a stem cut in air cannot distinguish a working xylem from a blocked one.
(c) [3]
A student repeats the experiment with a shoot from which all the leaves have been removed. Predict the result and explain your prediction.
Model Answer — 7(c)
very little stain would rise, or it would rise far more slowly [1]
because with no leaves there is almost no evaporation from mesophyll cells and so almost no transpiration pull [1]
this also shows that it is the leaves that drive the movement of water, rather than anything happening in the stem or the roots [1]
⚠ If you missed marks here: This is the same logic as the killed-stem experiment approached from the other end: instead of removing the living cells, you remove the evaporating surface. If water still rose freely with no leaves at all, the transpiration pull could not be the explanation — so this is a genuine test of the mechanism, not just a variation.

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IGCSE Biology 0610 — Paper 4 Theory (Extended) — Topic 8: Transport in Plants — Challenge Exam 3
For examination preparation only. Not an official Cambridge Assessment paper.