IGCSE Biology Paper 4 (Theory / Extended)

Topic 8: Transport in Plants -- Challenge Exam 2

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0610

Instructions

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

Question Navigation

This paper covers the whole of Topic 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 — Reading a Root, and Getting Water Into It

Total: 12 marks

(a) [3]

Using the diagram, name the tissue at P and the tissue at Q, and state what each transports.

Model Answer — 1(a)

P is the xylem, forming the central star [1]

which transports water and mineral ions [1]

Q is the phloem, which transports sucrose and amino acids [1]

⚠ If you missed marks here: In a root the xylem is central and star-shaped, with the phloem in the gaps between the arms. Anyone who has memorised the ring-of-bundles picture from a stem will place them the wrong way round here, which is exactly why examiners give roots as often as stems.

1 mark — P identified as xylem 1 mark — xylem transports water and mineral ions 1 mark — Q identified as phloem, transporting sucrose and amino acids

(b) [4]

Structure S bears many fine projections. Explain how these projections increase the uptake of water and of mineral ions, naming the process involved in each case.

Model Answer — 1(b)

the projections are root hairs, and they give a very large surface area for absorption [1]

water enters by osmosis, because the dilute soil solution has a higher water potential than the cell sap [1]

mineral ions are taken up largely by active transport, against their concentration gradient [1]

using energy released by respiration in the many mitochondria, and protein carriers in the cell membrane [1]

⚠ If you missed marks here: Two processes, two mechanisms, and they must not be swapped. Water is never taken up by active transport, and mineral ions are never taken up by osmosis. If the two are confused, every later question about flooding, cold soil or respiratory poisons becomes unanswerable.

1 mark — root hairs give a large surface area for absorption 1 mark — water enters by osmosis, down a water potential gradient 1 mark — mineral ions by active transport against the gradient 1 mark — energy from respiration, via protein carriers

Describe the pathway taken by a water molecule from the soil solution until it evaporates inside a leaf, naming every tissue it passes through and the process at each stage.

Model Answer — 1(c)

enters the root hair cell by osmosis [1]

passes across the root cortex cells, still by osmosis, down a gradient of decreasing water potential [1]

enters the xylem in the centre of the root [1]

is drawn up the xylem in a continuous column by the transpiration pull [1]

leaves the xylem in the leaf, reaches the mesophyll cells and evaporates from their walls into the air spaces [1]

⚠ If you missed marks here: The order in the syllabus is root hair, cortex, xylem, mesophyll — and the cortex comes before the xylem because in a root the xylem is right in the centre. Any route that involves the phloem is wrong: phloem never carries water anywhere.

1 mark — root hair cell, by osmosis 1 mark — root cortex cells, by osmosis down a water potential gradient 1 mark — into the xylem 1 mark — up the xylem by transpiration pull 1 mark — to the mesophyll cells, where it evaporates into the air spaces

Question 2 — Why a Xylem Vessel Looks the Way It Does

Total: 12 marks

(a) [3]

State the three structural features of a xylem vessel required by the syllabus.

Model Answer — 2(a)

thick walls containing lignin [1]

no cell contents — no cytoplasm and no nucleus [1]

cells joined end to end with no cross walls, forming one continuous tube [1]

⚠ If you missed marks here: “Thick, strong walls” is usually one mark short of “thick walls containing lignin”. The syllabus says the details of lignification are not required, but the word itself is.

1 mark — thick walls containing lignin 1 mark — no cell contents 1 mark — no cross walls, forming a continuous tube

(b) [3]

For each feature named in (a), explain what it achieves.

Model Answer — 2(b)

lignin makes the wall rigid and waterproof, so the vessel does not collapse under the tension of the transpiration pull, and it supports the plant [1]

with no contents the lumen is an open pipe, so there is very little resistance to the flow of water [1]

with no cross walls the vessels form one continuous column from root to leaf, with nothing to cross [1]

⚠ If you missed marks here: A structure-and-function question is only half answered by naming the structures. Each consequence has to be a consequence of that particular feature — writing “so water can move quickly” three times scores once.

1 mark — lignin: rigid, prevents collapse, gives support 1 mark — no contents: open pipe, little resistance 1 mark — no cross walls: continuous column

A scientist finds a mutant plant whose xylem vessels contain only about half the normal amount of lignin. Suggest, with reasons, two problems this plant is likely to have on a hot, windy day.

Model Answer — 2(c)

on a hot windy day transpiration is fast, so the tension in the xylem is high [1]

with less lignin the vessels are more likely to collapse or be damaged, so water delivery to the leaves falls and the plant wilts [1]

less lignin also means less support, so the stem is weaker and more likely to bend or break in the wind [1]

⚠ If you missed marks here: This is a “suggest” part: the mutant is not in any textbook, and the marks are for reasoning from the two known functions of lignin. Answers that simply repeat what lignin does, without applying it to a hot windy day, do not gain the application marks.

1 mark — fast transpiration means high tension in the xylem 1 mark — vessels may collapse, so water delivery falls and the plant wilts 1 mark — less support, so the stem bends or breaks

(d) [3]

Explain why a xylem vessel being dead is an advantage, and state one process in a plant that could not be carried out by a dead tissue.

Model Answer — 2(d)

the cell contents have broken down, so the tube is empty and offers no obstruction to the flow of water [1]

no energy has to be spent maintaining living cytoplasm along the whole length of the plant, and the flow is driven from outside by evaporation [1]

translocation in the phloem could not be done by a dead tissue, because loading sucrose requires energy from respiration in living cells [1]

⚠ If you missed marks here: Being dead sounds like a defect, and calling xylem “dead but still working” misses the point that the death is what makes it work. The contrast with phloem is the cleanest way to show you understand the difference.

1 mark — empty tube, no obstruction to flow 1 mark — no energy needed to maintain it; the flow is driven by evaporation 1 mark — translocation named as a process needing living cells

Question 3 — A Day in the Life of One Plant

Total: 12 marks

(a) [3]

Describe how the rate of water loss changes between 00:00 and 24:00, quoting figures from the graph.

Model Answer — 3(a)

the rate stays very low (about 2 units) from midnight until about 04:00 [1]

it then rises steeply to a peak of about 92 units at around 10:00 [1]

it falls through the afternoon and is back to about 2 units by 20:00 [1]

⚠ If you missed marks here: “Describe” means say what the line does, with figures — not explain why. Marks are routinely lost by launching straight into an explanation, and by describing the shape without quoting a single number from the axes.

1 mark — low and steady overnight, with a figure 1 mark — rises steeply to a peak, with the value and the time 1 mark — falls through the afternoon and evening, with a figure

(b) [4]

Explain the shape of the water loss curve between 04:00 and 10:00.

Model Answer — 3(b)

light intensity rises after dawn and the stomata open, providing an exit route for water vapour [1]

the temperature rises, so water molecules have more kinetic energy and evaporation from the mesophyll cell walls is faster [1]

the air also becomes drier as it warms, so the water vapour concentration gradient between the air spaces and the outside becomes steeper [1]

so more water vapour diffuses out through the stomata and the rate of loss rises [1]

⚠ If you missed marks here: The stomata open because the plant needs carbon dioxide for photosynthesis; the water loss is the price. An answer that says the plant opens its stomata “in order to transpire” will not be credited, because it treats a consequence as a purpose.

1 mark — stomata open as light intensity rises 1 mark — higher temperature gives more kinetic energy, so faster evaporation 1 mark — the water vapour gradient becomes steeper 1 mark — so more water vapour diffuses out through the stomata

Between about 06:00 and 14:00 the rate of loss is above the rate of uptake. Explain what this means for the plant, and explain why the rate of uptake stays above the rate of loss in the late afternoon.

Model Answer — 3(c)

loss greater than uptake means the plant is developing a water deficit: cells lose water, become flaccid and the plant may wilt [1]

in the late afternoon the rate of loss falls as the temperature drops and the stomata begin to close [1]

uptake continues, so the deficit is repaid and the cells become turgid again — which is why a plant that wilts at midday recovers by evening [1]

⚠ If you missed marks here: The mark here is for reading the gap between two curves rather than either curve alone. Choosing the time of maximum loss, instead of the time of maximum difference, is the standard error on this graph.

1 mark — a water deficit develops, so the plant may wilt 1 mark — loss falls in the afternoon as temperature drops and stomata close 1 mark — uptake continues, so the deficit is repaid and cells regain turgor

(d) [2]

Suggest why the rate of water loss does not fall to exactly zero at night.

Model Answer — 3(d)

a small amount of water vapour is still lost through the cuticle, which is not completely waterproof [1]

and not every stoma closes completely, so a little diffusion continues while there is any temperature difference [1]

⚠ If you missed marks here: “The stomata are closed so no water is lost” contradicts the graph, which is the point of the question. When data and a remembered rule disagree, the data win — and here the non-zero value is telling you the cuticle is a real, if minor, route.

1 mark — some loss through the cuticle 1 mark — not all stomata close completely

Question 4 — Two Investigations Into Transpiration

Total: 12 marks

(a) [4]

Describe how you would use a potometer to compare the rate of water uptake of a leafy shoot in still air and in moving air. Include the precautions you would take when setting the apparatus up.

Model Answer — 4(a)

cut the shoot under water and assemble the apparatus under water, so no air enters the xylem and breaks the water column [1]

seal the joint at the bung with petroleum jelly so the apparatus is airtight, and introduce a single air bubble into the capillary tube [1]

allow the apparatus to equilibrate, then measure the distance the bubble moves in a fixed time and calculate the rate [1]

repeat with a fan at a set distance and speed, keeping temperature, humidity, light and the shoot itself the same; repeat each reading and take a mean [1]

⚠ If you missed marks here: Cutting under water is not a detail — it is the difference between a working potometer and a meaningless one, because a broken water column cannot transmit the pull. The final mark is for the fair test, and it must name what is kept constant, not merely say “keep everything else the same”.

1 mark — cut and assemble under water to keep air out of the xylem 1 mark — airtight seal and a single bubble introduced 1 mark — allow to equilibrate, then time the movement over a set distance 1 mark — controlled variables named and repeats taken

(b) [3]

In still air the bubble moved 45 mm in 15 minutes; with a fan it moved 126 mm in 15 minutes. Calculate both rates and the percentage increase caused by the fan.

Model Answer — 4(b)

still air: 45 ÷ 15 = 3.0 mm per minute [1]

with fan: 126 ÷ 15 = 8.4 mm per minute [1]

percentage increase = (8.4 − 3.0) ÷ 3.0 × 100 = 180 % [1]

⚠ If you missed marks here: The percentage increase uses the difference on the top, not the new value. 8.4 ÷ 3.0 × 100 gives 280 %, which is the new rate expressed as a percentage of the old one — a different quantity, and the commonest mistake in the whole of data handling.

1 mark — still air rate of 3.0 mm per minute 1 mark — fan rate of 8.4 mm per minute 1 mark — percentage increase of 180 %

A second student measures transpiration instead by standing a potted plant on a balance and recording the loss in mass. Explain why the pot and soil must be sealed inside a plastic bag, and state one advantage this method has over a potometer.

Model Answer — 4(c)

without the bag, water would also evaporate from the soil surface and be counted as if it had been lost by the plant [1]

sealing the pot means all the mass lost is water lost from the leaves [1]

advantage: it measures water loss directly rather than uptake, and the plant is intact with its roots, so it is closer to a plant growing normally [1]

⚠ If you missed marks here: Notice that this method fixes exactly the weakness identified in the potometer — it measures loss rather than uptake. Being able to say what one method does better than another, rather than just listing faults, is what an evaluation question is for.

1 mark — without the bag, evaporation from the soil would be included 1 mark — with it sealed, the mass lost is water lost from the leaves 1 mark — a valid advantage over the potometer

(d) [2]

Suggest why the mass method is unsuitable for measuring changes in the rate of transpiration over periods of a few minutes.

Model Answer — 4(d)

the mass lost in a few minutes is very small compared with the total mass of the pot, soil and plant [1]

so the change may be smaller than the resolution of the balance, and any error is a large proportion of the reading [1]

⚠ If you missed marks here: The issue is resolution, not accuracy. A balance that reads to 0.1 g is fine for a whole day and useless for five minutes, and being able to say why is a genuine practical skill rather than a piece of recall.

1 mark — the mass change is very small compared with the total mass 1 mark — smaller than the resolution of the balance, so the error is proportionally large

Question 5 — Stomata, Guard Cells and the Cost of Saving Water

Total: 10 marks

(a) [3]

Explain how the size and number of the stomata affect the rate of transpiration, and explain why most stomata are on the lower surface of a leaf.

Model Answer — 5(a)

all the water vapour must leave through the stomata, so more stomata and larger stomata give a faster rate of diffusion out of the leaf [1]

fewer or smaller stomata reduce the rate, which is an adaptation in dry habitats [1]

the lower surface is shaded and cooler, so evaporation at an open pore is slower and less water is lost for the same carbon dioxide gained [1]

⚠ If you missed marks here: “Stomata are on the bottom so rain does not get in” is a common invented answer. The real reason is temperature: the shaded surface is cooler, so the same open pore loses less water.

1 mark — more and larger stomata increase the rate of diffusion out 1 mark — fewer or smaller stomata reduce the rate 1 mark — the lower surface is shaded and cooler, so less is lost

(b) [3]

Describe what happens to the guard cells when a stoma opens, and explain how the structure of a guard cell makes this possible.

Model Answer — 5(b)

the guard cells take in water by osmosis and become turgid [1]

the inner wall (next to the pore) is thicker and does not stretch, while the outer wall does [1]

so each cell curves away from its partner and a pore opens between them [1]

⚠ If you missed marks here: “The guard cells get bigger so the hole appears” does not explain anything — two swelling sausages could just as easily close a gap. The uneven wall thickening is what turns a change in volume into a change in shape.

1 mark — guard cells take in water by osmosis and become turgid 1 mark — the inner wall is thicker and does not stretch 1 mark — the cells curve away from each other, opening the pore

A species growing on an exposed hillside has fewer and smaller stomata, and each stoma is sunk into a small pit, compared with a related species from a shaded woodland. Explain how each of these features reduces water loss, and state the cost to the plant.

Model Answer — 5(c)

fewer and smaller stomata reduce the total area through which water vapour can diffuse out [1]

a pit traps a pocket of humid air immediately outside the pore [1]

which the wind cannot easily remove, so the water vapour gradient stays shallow and diffusion out is slower [1]

the cost: the same features restrict the entry of carbon dioxide, so the maximum rate of photosynthesis is lower and the plant grows more slowly [1]

⚠ If you missed marks here: The last mark is the one that separates a good answer from a complete one. Every water-saving adaptation in a plant is paid for in sugar, because the stomata are the route for carbon dioxide as well as for water vapour.

1 mark — fewer and smaller stomata reduce the area for diffusion 1 mark — the pit traps humid air outside the pore 1 mark — the gradient stays shallow, so diffusion is slower even in wind 1 mark — the cost is reduced carbon dioxide entry and slower photosynthesis

Question 6 — Ringing a Tree, and What It Proves

Total: 12 marks

(a) [4]

A complete ring of bark, including the phloem, is removed from the trunk of a young tree in spring. Predict and explain what will be seen at the ring, and what will happen to the leaves, over the following weeks.

Model Answer — 6(a)

a swelling develops above the ring [1]

because sucrose being translocated down from the leaves cannot pass the gap and accumulates there [1]

the leaves stay green and healthy for several weeks [1]

because the xylem lies deeper in the trunk and was not removed, so water and mineral ions still reach them [1]

⚠ If you missed marks here: The prediction is worth little without the reason, and both halves — the swelling and the healthy leaves — are needed because together they identify two tissues. Getting the swelling on the wrong side of the ring means the direction of translocation in summer has not been understood.

1 mark — swelling forms above the ring 1 mark — because sucrose moving down accumulates there 1 mark — the leaves remain healthy 1 mark — because the xylem is deeper and still supplies water

(b) [3]

Explain why the tree eventually dies, and suggest why a tree with only a narrow vertical strip of bark removed usually survives.

Model Answer — 6(b)

the roots below the ring receive no sucrose, so they cannot respire enough to grow or to carry out active transport, and they eventually die [1]

once the roots die, water and mineral ion uptake stops and the whole tree dies [1]

a narrow vertical strip leaves the rest of the ring of phloem intact, so sucrose can still travel past the wound [1]

⚠ If you missed marks here: The tree dies from the bottom up, which surprises people who expect the leaves to suffer first. The second part tests whether you picture the phloem as a complete cylinder rather than as a single tube — a good check on whether the transverse sections in 8.1 have actually been understood.

1 mark — roots are starved of sucrose and die 1 mark — without roots, water and mineral ion uptake stops and the tree dies 1 mark — a vertical strip leaves the rest of the phloem ring intact

Define a source and a sink, and explain how the same storage organ can be each of them at different times of year.

Model Answer — 6(c)

a source releases sucrose or amino acids; a sink uses or stores them [1]

in summer a tuber or bulb is a sink, receiving sucrose from photosynthesising leaves and storing it [1]

in spring, before the leaves open, the store is converted to sucrose and exported to the growing shoot, so the same organ is now a source [1]

⚠ If you missed marks here: Define them by what the tissue is doing, not by which organ it is. This is also the reason translocation cannot be described as “downwards”: in spring the sink is at the top of the plant and the source is underground.

1 mark — correct definitions of source and sink 1 mark — storage organ as a sink in summer 1 mark — the same organ as a source in spring

(d) [2]

Aphids feed by pushing a fine mouthpart into a single phloem tube. Sap flows out under pressure and is found to contain a high concentration of sucrose. Explain what this shows about phloem, and why the same technique would not work on xylem.

Model Answer — 6(d)

it shows directly that phloem sap contains dissolved sucrose and is under pressure [1]

xylem sap is under tension, not pressure, because the water is being pulled from above — so a tube pushed into a xylem vessel would draw air in rather than let sap out [1]

⚠ If you missed marks here: The aphid is a nice reminder that the strongest evidence is often a direct sample rather than an inference. The contrast with xylem also checks that transpiration pull has been understood as a pull: a liquid being pulled is under tension, and tension does not squirt.

1 mark — phloem sap contains sucrose and is under pressure 1 mark — xylem sap is under tension, so nothing would flow out

Question 7 — A Cutting, a Vase and a Wilting Bunch of Flowers

Total: 10 marks

(a) [3]

A cut flower stem is left lying on a bench for ten minutes before it is put into water. It wilts within an hour, while a stem cut under water and transferred immediately does not. Explain the difference.

Model Answer — 7(a)

air was drawn into the xylem at the cut end while the stem lay in the air [1]

the air bubble breaks the continuous column of water, so the transpiration pull can no longer be transmitted past it [1]

water therefore does not reach the leaves, loss exceeds uptake, the cells lose turgor and the stem wilts [1]

⚠ If you missed marks here: This is the same physics as the potometer instruction to cut under water, met in an everyday setting. If the water were being pushed from below a bubble would simply be pushed along; it matters only because the water is being pulled and the column must stay unbroken.

1 mark — air drawn into the xylem at the cut end 1 mark — the continuous water column is broken, so the pull is not transmitted 1 mark — water does not reach the leaves, so the cells lose turgor and it wilts

(b) [4]

A florist gives two pieces of advice for making cut flowers last: keep them out of draughts and direct sunlight, and add a small amount of sugar to the water. Explain the biology behind each piece of advice.

Model Answer — 7(b)

draughts remove the humid air at the leaf surface, keeping the water vapour gradient steep, and sunlight warms the leaves and opens the stomata [1]

both therefore increase transpiration, so loss is more likely to exceed uptake and the flowers wilt sooner [1]

the cut stem has been separated from its roots and from the leaves that were its source, so it can no longer be supplied with sucrose by translocation [1]

sugar in the water provides a substitute supply that the flower can use in respiration, so it stays alive longer [1]

⚠ If you missed marks here: The second half is the interesting one, and it is a translocation question in disguise: a cut flower is a sink that has lost its source. Answers that say the sugar “feeds the flower by osmosis” confuse a transport process with a supply of energy.

1 mark — draughts and sunlight raise the rate of transpiration, with a reason 1 mark — loss exceeds uptake, so the flowers wilt sooner 1 mark — the cut stem has lost its source of sucrose 1 mark — added sugar provides a substrate for respiration

Adding too much sugar to the vase water makes the flowers wilt faster. Explain why.

Model Answer — 7(c)

a concentrated sugar solution has a lower water potential than the contents of the cells at the cut end [1]

so water moves out of the stem by osmosis instead of into it [1]

uptake falls while loss continues, so the cells lose turgor and the flowers wilt sooner [1]

⚠ If you missed marks here: The same reasoning explains why over-fertilised soil makes a plant wilt even though it is wet. Whenever a solution outside a cell is made more concentrated than the inside, the direction of osmosis reverses — and that is worth recognising instantly, because examiners set it in a new disguise every year.

1 mark — the concentrated solution has a lower water potential than the cell contents 1 mark — water leaves the stem by osmosis instead of entering 1 mark — uptake falls while loss continues, so the flowers wilt

Self-Assessment

Tick marks earned, then click Calculate Grade.

Your Score

Out Of

Percentage