IGCSE Biology Paper 4 (Theory / Extended)

Topic 8: Transport in Plants -- 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 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 — Two Tissues, Two Cargoes, and an Experiment That Separates Them

Total: 12 marks

(a) [4]

State the substances transported by xylem and by phloem, and state one further function of xylem.

Model Answer — 1(a)

xylem transports water [1]

and mineral ions [1]

phloem transports sucrose and amino acids (both needed) [1]

xylem also provides support for the plant [1]

⚠ If you missed marks here: “Minerals” and “nutrients” are not accepted — the syllabus wording is mineral ions. “Sugar” and “glucose” are not accepted either: the leaf converts glucose to sucrose before loading it into the phloem, so glucose is wrong rather than merely vague.

1 mark — xylem transports water 1 mark — xylem transports mineral ions 1 mark — phloem transports sucrose and amino acids 1 mark — support named as a function of xylem

(b) [3]

Describe how you would identify the xylem in a transverse section of a young non-woody stem, and give the reasoning you used.

Model Answer — 1(b)

the bundles are arranged in a ring near the outside, which identifies the section as a stem [1]

within each bundle the xylem is the tissue on the inner side, nearer the centre of the stem [1]

the xylem also has thick walls containing lignin, which usually stain strongly [1]

⚠ If you missed marks here: A description is not a justification. The mark for “reasoning” is for the rule you applied — xylem is nearer the centre — not for a memory of one particular textbook drawing. Identify the organ first and the tissue follows automatically, even in an unfamiliar section.

1 mark — ring of bundles identifies the section as a stem 1 mark — xylem is on the inner side of each bundle 1 mark — thick lignified walls as supporting evidence

A leafy stem is held in steam for one minute so that all its cells are killed. It is then stood in dilute red dye in a warm, bright, breezy place. Two hours later the dye has risen and is visible in the leaf veins. A second, living stem is treated with a respiratory poison and stood in the same dye. In this stem the dye also rises, but no sucrose moves out of the leaves. Explain what these two results show about the two transport tissues.

Model Answer — 1(c)

water still moves in the killed stem, so movement in the xylem does not require living cells [1]

this is because the water is pulled by evaporation from the leaves (transpiration pull) rather than pushed by the plant [1]

the energy comes from outside the plant — from the sun evaporating water — so no respiration is needed [1]

in the poisoned stem sucrose does not move, so translocation does require energy from respiration [1]

this shows the phloem must be made of living cells doing work, unlike the dead, empty xylem vessels [1]

⚠ If you missed marks here: A common answer says the killed stem proves “water moves by diffusion”, which is not the point — what it proves is that no living process is involved. Notice also that the two experiments are a matched pair: each removes a different requirement, and only by comparing them can you attribute one result to the xylem and the other to the phloem.

1 mark — water movement does not require living cells 1 mark — water is pulled by transpiration pull, not pushed 1 mark — the energy comes from evaporation, so no respiration is needed 1 mark — translocation stops without respiration, so it needs energy 1 mark — therefore phloem is living and xylem is dead

Question 2 — A Root Hair Cell, and What Happens When the Soil Floods

Total: 12 marks

(a) [3]

Describe three ways in which a root hair cell is adapted for the uptake of water and mineral ions.

Model Answer — 2(a)

a long, thin projection giving a very large surface area for absorption [1]

a thin cell wall, giving a short distance for water and ions to cross [1]

many mitochondria, releasing energy from respiration for the active transport of mineral ions [1]

⚠ If you missed marks here: Each adaptation must be finished with what it achieves. “Large surface area” on its own is half an answer — large surface area for absorption of water and mineral ions is the whole one. Do not offer chloroplasts: a root hair is underground and has none.

1 mark — large surface area from the long thin projection, for absorption 1 mark — thin wall giving a short distance to cross 1 mark — many mitochondria for the energy needed for active transport

(b)(i) [3]

Explain how water enters a root hair cell.

Model Answer — 2(b)(i)

the soil solution is dilute and so has a higher water potential than the cell sap [1]

water therefore moves down the water potential gradient, through the partially permeable cell membrane [1]

this is osmosis, and it requires no energy from the plant [1]

⚠ If you missed marks here: “Semi-permeable” is not the syllabus word — write partially permeable. And the direction must be stated in terms of water potential, not in terms of “there is more water outside”, which is loose enough to be marked wrong when the outside solution is concentrated.

1 mark — soil solution has the higher water potential 1 mark — water moves down the gradient through a partially permeable membrane 1 mark — named as osmosis, requiring no energy

(b)(ii) [2]

Explain why mineral ions usually enter by a different process.

Model Answer — 2(b)(ii)

the concentration of ions in the soil solution is often lower than the concentration inside the cell [1]

so ions must be moved against the concentration gradient by protein carriers, using energy from respiration — active transport [1]

⚠ If you missed marks here: The reason is the direction of the gradient, not the size of the particles. Ions are small; they are simply being moved uphill, and uphill costs energy.

1 mark — ion concentration is lower outside than inside 1 mark — moved against the gradient by active transport, using energy from respiration

A field of wheat is flooded for four days. When the water drains away, the plants are yellow and show signs of mineral ion deficiency, although the soil is known to be rich in mineral ions and the plants have not been short of water. Explain these observations.

Model Answer — 2(c)

flooding fills the air spaces in the soil, so the roots cannot obtain enough oxygen [1]

the rate of aerobic respiration in the root cells falls, so less energy is released [1]

less energy means less active transport, so fewer mineral ions are taken up even though plenty are present [1]

water uptake is not affected in the same way because osmosis is passive and needs no energy, which is why the plants are not short of water [1]

⚠ If you missed marks here: The last mark is the one most often missed. The question deliberately tells you the plants were not short of water, and a full answer has to explain why one kind of uptake failed and the other did not. An answer that only says “the roots drowned” explains neither.

1 mark — flooded soil has no air spaces, so little oxygen reaches the roots 1 mark — aerobic respiration falls, so less energy is released 1 mark — less energy means less active transport of mineral ions 1 mark — water uptake is unaffected because osmosis is passive

Question 3 — Transpiration, Measured and Explained

Total: 12 marks

(a) [3]

Define transpiration, and describe the two stages by which water leaves a leaf.

Model Answer — 3(a)

transpiration is the loss of water vapour from the leaves [1]

water evaporates from the wet cell walls of the mesophyll cells into the air spaces [1]

the water vapour then diffuses out through the stomata, down a concentration gradient [1]

⚠ If you missed marks here: Two words decide all three marks: vapour in the definition, and the pair evaporates then diffuses for the stages. Using one verb for both stages, or writing “osmosis” for the exit, throws away marks on the highest-frequency question in the topic. A stoma is a pore, and nothing crossing a pore is doing osmosis.

1 mark — loss of water vapour from the leaves 1 mark — evaporation from the mesophyll cells into the air spaces 1 mark — diffusion of water vapour out through the stomata

(b) [2]

A potometer bubble moved 84 mm in 12 minutes. Calculate the rate of water uptake and give the unit.

Model Answer — 3(b)

84 ÷ 12 [1]

= 7.0 mm per minute (unit required) [1]

⚠ If you missed marks here: The working is worth a mark of its own, so write the division out even when you can do it in your head. A number with no unit is not a rate, and the unit mark is lost more often than the arithmetic one.

1 mark — correct working: 84 divided by 12 1 mark — 7.0 mm per minute, with the unit

The same shoot was then tested under four conditions, each for 12 minutes. Still air at 20 °C: 84 mm. Moving air at 20 °C: 158 mm. Still air at 30 °C: 121 mm. Still air at 20 °C with the shoot enclosed in a clear plastic bag: 19 mm. Explain the results for the moving air and for the plastic bag.

Model Answer — 3(c)

moving air removes the humid layer of air just outside the stomata [1]

so the water vapour concentration gradient between the air spaces and the outside stays steep and diffusion out is faster [1]

the bag traps the water vapour lost from the leaves, so the air around the shoot becomes very humid [1]

this makes the gradient much shallower, so less water vapour diffuses out and uptake falls (from 84 to 19 mm) [1]

⚠ If you missed marks here: Both explanations must go through the gradient. Answers that say wind “blows the water out of the leaf” or that the bag “stops the water escaping” describe the result rather than the mechanism. Quoting a pair of figures is usually worth a mark on its own in a question like this.

1 mark — moving air removes the humid layer at the leaf surface 1 mark — the gradient stays steep, so diffusion is faster 1 mark — the bag traps water vapour, making the air humid 1 mark — the shallower gradient reduces diffusion out of the stomata

(d) [3]

A student concludes that the potometer has measured the rate of transpiration exactly. Evaluate this conclusion, and suggest one improvement to the method.

Model Answer — 3(d)

a potometer measures the rate of water uptake, not the rate of water loss [1]

a small proportion of the water taken up is used in photosynthesis and retained in the cells to keep them turgid, so uptake slightly exceeds transpiration [1]

improvement, any one: allow the apparatus to equilibrate for several minutes before the first reading / repeat each condition and take a mean / check the joints are airtight with petroleum jelly / keep the shoot the same size and leaf area throughout [1]

⚠ If you missed marks here: An evaluation is not a complaint. The mark is for saying what is being measured and by how much it differs, not for saying the experiment is unreliable. Over a short period uptake is a good estimate of loss — that honest qualification is what is being paid for.

1 mark — a potometer measures uptake, not loss 1 mark — some water is used in photosynthesis and retained, so uptake is slightly greater 1 mark — a valid, specific improvement

Question 4 — Sources, Sinks and the Evidence for Them

Total: 12 marks

(a) [3]

Define translocation, and define a source and a sink.

Model Answer — 4(a)

translocation is the movement of sucrose and amino acids in the phloem [1]

a source is a part of the plant that releases sucrose or amino acids [1]

a sink is a part that uses or stores them [1]

⚠ If you missed marks here: Define source and sink by what the tissue is doing, not by which organ it is. “A source is a leaf” fails the moment the question is about a potato tuber in spring, which is exactly the question examiners set.

1 mark — translocation of sucrose and amino acids in the phloem 1 mark — source releases sucrose or amino acids 1 mark — sink uses or stores them

(b) [4]

A potato plant is dug up in July and again in the following March. In July the leaves are large and the tubers are swelling. In March there are no leaves, but a shoot is growing rapidly from one tuber. For each date, state whether the tuber is a source or a sink, and explain the direction of movement in the phloem.

Model Answer — 4(b)

in July the tuber is a sink [1]

the leaves are photosynthesising and export sucrose, which moves down the phloem to be stored in the tuber [1]

in March the tuber is a source [1]

the stored material is converted to sucrose and moves up the phloem to the growing shoot, which is the sink [1]

⚠ If you missed marks here: “Translocation is the movement of sugar downwards” is one of the most common wrong sentences in this topic, and this question exists to catch it. Direction is decided by where the sink is, and in spring the sink is above the source.

1 mark — July: tuber is a sink 1 mark — July: sucrose moves down from the photosynthesising leaves 1 mark — March: tuber is a source 1 mark — March: sucrose moves up to the growing shoot

One mature leaf of a bean plant was enclosed and supplied with radioactively labelled carbon dioxide for 30 minutes. Six hours later the label recovered was distributed as follows: labelled leaf 38 %, other mature leaves 2 %, stem 6 %, roots 22 %, developing pods 32 %. Explain what these results show about translocation.

Model Answer — 4(c)

the labelled carbon entered as carbon dioxide and was fixed in photosynthesis, then converted to sucrose and loaded into the phloem [1]

the roots (22 %) and the developing pods (32 %) received large amounts, so both were acting as sinks [1]

the label moved both downwards (to the roots) and upwards (to the pods), showing that translocation is not restricted to one direction [1]

the other mature leaves received almost none (2 %) because they are sources themselves and make their own sucrose [1]

38 % remained in the labelled leaf because export takes time and some of the labelled carbon has been used or stored there [1]

⚠ If you missed marks here: The 2 % is a result, not experimental error. Every number in a table like this is telling you something: a large value in a growing organ means sink, a near-zero value in a mature leaf means source, and a large value at the origin means the process is continuous rather than instant.

1 mark — label fixed in photosynthesis and moved as sucrose in the phloem 1 mark — roots and developing pods identified as sinks, with figures 1 mark — movement occurred in both directions 1 mark — other mature leaves are sources, so receive almost none 1 mark — label remaining in the labelled leaf explained

Question 5 — A Leaf in Section, and the Force That Lifts the Water

Total: 10 marks

(a) [3]

The diagram shows a transverse section of a leaf. Identify which lettered structure, J or K, is the xylem, give its two transport functions, and explain why this arrangement is what you would expect from the arrangement of tissues in the stem.

Model Answer — 5(a)

J — the xylem is on the upper side of the vein [1]

it transports water and mineral ions [1]

in a stem the xylem lies on the inner side of each bundle, nearer the centre of the plant; a leaf vein is that bundle turned outwards, so the inner side now faces upwards [1]

⚠ If you missed marks here: The third mark is for the reasoning, not the fact. Knowing why the xylem is on top is what lets you answer the question when the diagram is printed upside down, which examiners occasionally do.

1 mark — J identified as the xylem, on the upper side 1 mark — transports water and mineral ions 1 mark — explained by the position of xylem in the stem bundle

(b) [5]

Explain how water moves upwards from the roots to the leaves of a tree 30 m tall.

Model Answer — 5(b)

water evaporates from the mesophyll cells and the vapour diffuses out of the stomata [1]

the mesophyll cells then have a lower water potential, so water moves into them from the xylem [1]

this puts the water in the xylem under tension, creating the transpiration pull [1]

because there are forces of attraction between water molecules, the water forms a continuous column that does not break [1]

so the pull is transmitted all the way down and water is drawn up from the roots [1]

⚠ If you missed marks here: Two things are being marked that are easy to leave out: the exact phrase transpiration pull, and the reason the column holds together. “The plant sucks the water up” names no mechanism and earns nothing. Root pressure is not the answer — it is far too weak to reach 30 m, and a stem killed by steam still conducts water.

1 mark — evaporation from mesophyll and diffusion out of the stomata 1 mark — mesophyll cells have a lower water potential, so water leaves the xylem 1 mark — tension in the xylem, named as transpiration pull 1 mark — forces of attraction between water molecules hold the column together 1 mark — the pull is transmitted down a continuous column to the roots

Explain why the walls of xylem vessels must contain lignin if this mechanism is to work.

Model Answer — 5(c)

the water is being pulled, so it is under tension and the vessel would tend to collapse inwards [1]

lignin makes the wall rigid and strong, so the tube stays open (and the same rigidity supports the plant) [1]

⚠ If you missed marks here: This part is testing whether you can connect two syllabus statements that are usually learned separately. Lignin is not there simply because xylem happens to have thick walls — it is there because a tube under tension needs it.

1 mark — water under tension would collapse an unsupported tube 1 mark — lignin makes the wall rigid so the vessel stays open

Question 6 — Designing and Evaluating an Investigation Into Water Loss

Total: 12 marks

(a) [4]

Four leaves of equal area were taken from the same plant and their cut stalks sealed with wax. Petroleum jelly was applied as follows and the leaves were hung in the same room for 24 hours. Leaf W: upper surface only, mass lost 0.62 g. Leaf X: lower surface only, 0.14 g. Leaf Y: both surfaces, 0.03 g. Leaf Z: neither surface, 0.71 g. Using the data, explain what these results show about the distribution of the stomata.

Model Answer — 6(a)

blocking the lower surface (X) reduced the loss from 0.71 to 0.14 g, a fall of about 80 % [1]

blocking the upper surface (W) reduced it only from 0.71 to 0.62 g [1]

so most of the water is lost through the lower surface [1]

therefore most of the stomata are on the lower surface [1]

⚠ If you missed marks here: Quote the figures. “Leaf X lost less” is an observation the examiner can already see; the mark is for using the numbers. And finish the chain — the data show where the water was lost, and only the last step turns that into a conclusion about where the stomata are.

1 mark — figures quoted for the lower surface being blocked 1 mark — figures quoted for the upper surface being blocked 1 mark — conclusion that most water is lost from the lower surface 1 mark — therefore most stomata are on the lower surface

(b) [3]

Explain the purpose of leaf Y, and state what its result tells you.

Model Answer — 6(b)

leaf Y is the control, in which both possible stomatal routes are blocked [1]

it measures the water lost by other routes — through the cuticle, and through the sealed stalk if the seal was imperfect [1]

the value of 0.03 g shows this background loss is small, so the differences between the other leaves can safely be attributed to the stomata [1]

⚠ If you missed marks here: A control is not just “something to compare with”. Say what it removes and what its result lets you conclude. Here it also disproves the claim that a leaf loses no water at all through its cuticle — 0.03 g is small, but it is not zero.

1 mark — identified as the control with both surfaces blocked 1 mark — measures loss through the cuticle and the sealed stalk 1 mark — small value means the other differences can be attributed to stomata

Suggest three variables that must be kept the same for this to be a fair test, and in each case explain why.

Model Answer — 6(c)

leaf area (and the same species and age of leaf) — a larger leaf has more stomata and more surface, so it would lose more whatever the treatment [1]

temperature — it changes the rate of evaporation from the mesophyll and so the rate of loss [1]

humidity and air movement — both change the steepness of the water vapour gradient at the stomata; hanging all four leaves in the same room controls them [1]

⚠ If you missed marks here: Every variable must come with a reason, and the reason has to be about this experiment. “To make it fair” is not a reason. Note also that all four leaves came from the same plant, which is itself a control of species, age and growing conditions.

1 mark — leaf area, with a reason 1 mark — temperature, with a reason 1 mark — humidity or air movement, with a reason

(d) [2]

The student repeated the whole investigation on a hot, windy day and obtained much larger masses lost, but the same pattern. Explain why the pattern was unchanged.

Model Answer — 6(d)

heat and wind increase the rate of transpiration for every leaf, by speeding evaporation and keeping the vapour gradient steep, so all four masses rise [1]

but the distribution of the stomata is a feature of the leaf and is unchanged, so the relative differences between the four treatments stay the same [1]

⚠ If you missed marks here: This part tests the difference between a variable that scales every result and one that changes the shape of the result. Recognising that a condition can change the size of an effect without changing the conclusion is a genuine data-handling skill, and it appears in every subject.

1 mark — heat and wind raise the rate for all four leaves 1 mark — stomatal distribution is unchanged, so the pattern is the same

Question 7 — One Tomato Plant in a Greenhouse, From Root Hair to Fruit

Total: 10 marks

(a) [3]

A grower waters a tomato plant well but on a hot afternoon the leaves droop. By the following morning they have recovered. Explain the drooping and the recovery.

Model Answer — 7(a)

in the afternoon the rate of water loss exceeded the rate of uptake, so the cells lost water [1]

the cells became flaccid and there was no longer enough turgor pressure pushing outwards on the cell walls to support the soft tissue [1]

overnight the rate of loss falls, uptake continues, the cells regain water by osmosis and become turgid again [1]

⚠ If you missed marks here: Wilting is a balance problem, not a shortage in the soil — the question says the plant was watered well. The marks are in the vocabulary you already have from osmosis: flaccid, turgor pressure, turgid.

1 mark — rate of loss exceeded rate of uptake 1 mark — cells became flaccid, so turgor pressure no longer supported the tissue 1 mark — recovery by osmosis overnight as cells become turgid again

(b) [3]

To reduce the wilting the grower closes the greenhouse vents and sprays a fine mist of water into the air. Explain how this reduces the rate of transpiration, and suggest one disadvantage of doing it for several days.

Model Answer — 7(b)

misting and closing the vents raise the humidity of the air around the leaves [1]

this makes the water vapour concentration gradient between the air spaces and the outside air shallower, so less water vapour diffuses out through the stomata [1]

a disadvantage, any one: still humid air favours fungal disease / with the vents closed the carbon dioxide concentration falls, reducing photosynthesis / the temperature may rise further [1]

⚠ If you missed marks here: The explanation must go through the gradient. “The air is wet so the leaf cannot lose water” states the result. The disadvantage mark is a “suggest” mark — any sensible, clearly explained consequence is credited, but it must follow from closing the greenhouse rather than being a general remark about plants.

1 mark — humidity of the surrounding air is increased 1 mark — shallower water vapour gradient, so less diffusion out of the stomata 1 mark — a sensible disadvantage, explained

The grower removes many of the leaves near the developing tomatoes so that more light reaches the fruit. The fruits ripen more slowly and are smaller than usual. Using your knowledge of translocation, explain this result.

Model Answer — 7(c)

a developing fruit is a sink: it cannot make enough sugar for itself and imports sucrose [1]

the leaves nearest to it are its main sources, releasing sucrose into the phloem [1]

removing those leaves reduces the amount of sucrose translocated to the fruit [1]

so less material is available for growth and storage in the fruit, and it stays small and ripens slowly — more light does not help, because a fruit contributes very little photosynthesis of its own [1]

⚠ If you missed marks here: The trap is the phrase “so that more light reaches the fruit”, which invites an answer about photosynthesis in the tomato. The limiting factor here is supply, not light. The same reasoning explains why growers thin fruit: fewer sinks sharing one source means more for each.

1 mark — developing fruit identified as a sink 1 mark — nearby leaves identified as its sources 1 mark — removing them reduces the sucrose translocated to the fruit 1 mark — less material for growth, so the fruit is small and slow to ripen

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