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This paper covers the whole of Topic 6. Like a real Cambridge paper, the seven questions range across every sub-topic — photosynthesis and leaf structure — and they are deliberately mixed rather than grouped. All three Topic 6 papers do; they differ in the angle they come at it from, not in what they cover.
Question 1 -- Moss on a Damp Wall
Total: 12 marks
A cushion of moss grows on a damp north-facing wall in a Bangalore garden. In bright conditions it releases small bubbles of gas from its surface. A student scrapes a sample from the wall and finds it is bright green throughout.
(a)[3]
State the word equation for photosynthesis, including the conditions required.
Model Answer -- 1(a)
carbon dioxide + water → glucose + oxygen [1]
the conditions light and chlorophyll written above or below the arrow [1]
reactants and products on the correct sides with a single arrow [1]
⚠ If you missed marks here: The commonest single loss here is writing food or sugar instead of glucose. The second is putting sunlight on the left of the arrow as though it were a reactant — light is energy, not matter, and no atom of glucose comes from it.
(b)[4]
Give the balanced chemical equation for photosynthesis and use it to explain why twelve oxygen atoms leave the plant as gas for every glucose molecule made.
18 oxygen atoms enter, in 6CO₂ (12) and 6H₂O (6) [1]
6 of those oxygen atoms are locked into the glucose molecule C₆H₁₂O₆ [1]
the remaining 12 atoms leave as 6 molecules of O₂ [1]
⚠ If you missed marks here: Counting molecules when the question says atoms is the trap, and it gives the answer 6 instead of 12. Work through the equation atom by atom rather than quoting a remembered number — that is exactly the skill being tested.
(c)[2]
The moss is green throughout, whereas the leaf of a flowering plant has a colourless upper epidermis. Suggest one advantage of the colourless epidermis.
Model Answer -- 1(c)
a layer without chloroplasts is transparent [1]
so light passes through it undiminished to the palisade cells beneath, which contain most of the chloroplasts [1]
⚠ If you missed marks here: Answers that stop at “it has no chloroplasts” describe the structure without explaining the benefit. A two-mark question wants the property and its consequence, in that order.
(d)[3]
Name three substances the moss makes from the glucose produced in photosynthesis, and state a use of each.
Model Answer -- 1(d)
starch — an insoluble energy store [1]
cellulose — to build cell walls [1]
sucrose — for transport in the phloem, or glucose used in respiration to release energy [1]
⚠ If you missed marks here: Naming three substances without a use each halves your score, because the mark scheme awards the pair. “Protein” is not creditworthy here either — proteins need nitrogen from nitrate ions and are not made from glucose alone.
Question 2 -- A Section Through a Dicotyledonous Leaf
Total: 12 marks
A transverse section of a leaf is examined under a microscope. Five regions are labelled: A is a thin transparent layer at the very top of the section, B is a layer of tall closely packed cells beneath A, C is a region of rounded cells with large gaps between them, D is a group of thick-walled dead tubes in the upper half of a vein, and E is a pair of curved cells surrounding a pore in the lowest layer.
(a)[5]
Identify regions A, B, C, D and E.
Model Answer -- 2(a)
A — upper epidermis [1]
B — palisade mesophyll [1]
C — spongy mesophyll, the gaps being air spaces [1]
D — xylem [1]
E — guard cells [1]
⚠ If you missed marks here: D is xylem, not phloem: the clue is thick-walled and dead, and in a leaf vein the xylem lies on the upper side. E is the guard cells, not the stoma — the stoma is the pore between them, which is not a cell.
(b)[2]
Explain two ways in which region B is adapted for photosynthesis.
Model Answer -- 2(b)
contains the greatest number of chloroplasts and lies directly beneath the transparent epidermis, closest to the light, so it absorbs the most light energy [1]
the cells are tall and column-shaped, so light passes through many chloroplasts within a single cell with few cell walls to scatter it [1]
⚠ If you missed marks here: “It is where photosynthesis happens” restates the function instead of explaining the adaptation and scores nothing. Every answer here must connect a structural feature to the capture of light.
(c)[3]
Explain how the arrangement of cells in region C allows carbon dioxide to reach every photosynthesising cell.
Model Answer -- 2(c)
the loose, irregular packing creates interconnected air spaces [1]
carbon dioxide entering through the stomata diffuses through these spaces to every mesophyll cell [1]
the air spaces give a large moist surface area for carbon dioxide to dissolve into before crossing the cell membranes [1]
⚠ If you missed marks here: Saying the air spaces “store” carbon dioxide is wrong — the volume held is negligible. They are a pathway, and the mark is for diffusion, not storage.
(d)[2]
State the tissue that lies immediately below D in the same vein, and name the substances it transports.
Model Answer -- 2(d)
phloem [1]
carries sucrose and amino acids away from the leaf [1]
⚠ If you missed marks here: Xylem above, phloem below — most candidates guess this the wrong way round. And phloem carries sucrose, not glucose and certainly not starch, which is insoluble and cannot be transported at all.
Question 3 -- Testing Whether Light Is Necessary
Total: 12 marks
A student is given a healthy pelargonium plant, aluminium foil, paper clips, iodine solution and the usual laboratory apparatus. She is asked to find out whether light is necessary for photosynthesis. Her first attempt gives a leaf that is blue-black all over, including the region that had been covered.
(a)[3]
Explain why her first attempt gave this result and state what she should have done before starting.
Model Answer -- 3(a)
the leaf already contained starch made before the investigation began [1]
so a blue-black result cannot show that starch was made during the experiment [1]
she should have destarched the plant by keeping it in complete darkness for 24–48 hours [1]
⚠ If you missed marks here: Do not say the iodine was faulty or the foil leaked. The examiner has engineered a specific omission, and the whole of part (a) is about destarching — a step almost no candidate writes down unprompted.
(b)[4]
Describe the four steps used to test a leaf for starch and give a reason for each.
Model Answer -- 3(b)
boil the leaf in water — kills the leaf, stops enzyme reactions and breaks the cell membranes [1]
place in hot ethanol — dissolves out the chlorophyll so the colour change can be seen [1]
rinse in water — softens the brittle leaf so the iodine can spread over it [1]
add iodine solution — turns from orange-brown to blue-black where starch is present [1]
⚠ If you missed marks here: The two liquids get swapped constantly: water kills, ethanol decolourises. Also give the colour change in the right direction — orange-brown to blue-black. Writing only the final colour is worth less than the full change.
(c)[2]
Explain why the ethanol must be heated in a water bath.
Model Answer -- 3(c)
ethanol is highly flammable [1]
so it must not be heated over a naked flame; hot water from a beaker or electric bath heats it safely [1]
⚠ If you missed marks here: A safety mark that very few candidates offer. Note that the reason is not that ethanol boils at a high temperature — it boils at only 78 °C, well below water, which is precisely why a water bath works.
(d)[3]
Her friend argues that the black foil made that part of the leaf cooler, so the experiment does not prove that light is necessary. Evaluate this argument and describe how the design could be improved.
Model Answer -- 3(d)
the criticism is reasonable in principle, because a valid experiment must change only one variable at a time [1]
the foil may alter the temperature or air movement over that region as well as blocking the light [1]
the improvement is to cover an equivalent region with a transparent cover of the same material and thickness, so that everything except light is matched [1]
⚠ If you missed marks here: “Repeat the experiment” is the wrong kind of fix here — repeats address reliability, and this is a problem of validity. Repeating an invalid experiment simply gives you more invalid results.
Question 4 -- Mineral Ions and Barley Seedlings
Total: 12 marks
Three sets of ten barley seedlings were grown for five weeks in aerated culture solutions under identical light and temperature. Solution 1 contained all the mineral ions the plant needs; solutions 2 and 3 each lacked one ion. The mean results are shown below.
set
solution
mean height / cm
mean dry mass / g
leaf appearance
1
complete
23.4
0.52
dark green
2
one ion absent
6.1
0.11
small, pale green
3
a different ion absent
20.8
0.34
normal size, yellow between the veins
(a)[4]
Deduce which mineral ion was absent from solution 2, and explain the biological chain that produces the symptoms recorded.
Model Answer -- 4(a)
nitrate ions [1]
nitrate supplies the nitrogen that carbohydrates do not contain, needed to convert glucose into amino acids [1]
without amino acids the plant cannot make proteins for new cytoplasm, enzymes and cell division [1]
so growth is severely stunted, giving a mean height of only 6.1 cm against 23.4 cm [1]
⚠ If you missed marks here: Saying “nitrate is needed for growth” compresses three marking points into one and scores one. The mark scheme follows the chain: nitrogen → amino acids → proteins → growth. Also quote the figures — a deduction question expects the data to be used.
(b)[3]
Deduce which ion was absent from solution 3 and explain the leaf appearance.
Model Answer -- 4(b)
magnesium ions [1]
magnesium is required to make chlorophyll [1]
without chlorophyll the leaves cannot be green, so they appear yellow (chlorosis) [1]
⚠ If you missed marks here: Nitrate and magnesium are swapped more often than any other pair in this topic. Fix it with the rule small but green → nitrate; normal but yellow → magnesium, and read the data before choosing.
(c)[3]
Set 3 reached almost the same height as set 1 but had only two thirds of its dry mass. Explain this difference.
Model Answer -- 4(c)
with less chlorophyll the rate of photosynthesis is lower, so less glucose is made and less material is built into the plant [1]
height depends largely on cell elongation, which relies on water uptake and turgor and can continue even when little new material is made [1]
dry mass measures the material actually synthesised, so the shortfall appears in mass before it appears in height [1]
⚠ If you missed marks here: This part rewards the Topic 1 definition of growth as a permanent increase in size and dry mass. Whenever a question gives you both height and dry mass, it is inviting you to notice that they can disagree.
(d)[2]
Explain why all three culture solutions were aerated.
Model Answer -- 4(d)
the roots need oxygen for aerobic respiration [1]
respiration releases the energy used for the active transport of mineral ions into the root hair cells against a concentration gradient [1]
⚠ If you missed marks here: “So the roots can breathe” is refused — respiration is a set of chemical reactions in cells, not a movement of air. Notice also that this is a Topic 3 idea placed inside a Topic 6 question, which is exactly what a challenge paper does.
Question 5 -- A Glasshouse in Two Seasons
Total: 10 marks
A grower measures the rate of photosynthesis of her lettuce crop, in arbitrary units, at a range of light intensities. She repeats the measurements under three sets of conditions.
light intensity / arbitrary units
0
2
4
6
8
10
P: 0.04 % CO₂, 15 °C
0
0.8
1.4
1.5
1.5
1.5
Q: 0.04 % CO₂, 25 °C
0
1.1
2.0
2.4
2.4
2.4
R: 0.40 % CO₂, 25 °C
0
1.2
2.3
3.4
4.2
4.6
(a)[2]
Explain what is meant by a limiting factor.
Model Answer -- 5(a)
the environmental factor that is in the shortest supply at that moment [1]
and which therefore limits the rate of the process — if it is increased the rate increases, and increasing any other factor has no effect [1]
⚠ If you missed marks here: “The factor that runs out” is refused. Carbon dioxide is limiting at 0.04 % every day of the year and never runs out. The idea is relative shortage, like the slowest worker on a production line.
(b)[3]
Identify the limiting factor for curve P at a light intensity of 10 arbitrary units, and justify your answer using the data.
Model Answer -- 5(b)
temperature [1]
increasing the light intensity beyond 6 units does not raise the rate above 1.5, so light is not limiting [1]
raising the temperature from 15 °C to 25 °C at the same carbon dioxide concentration raises the plateau from 1.5 to 2.4, so temperature is the factor whose increase moves the line [1]
⚠ If you missed marks here: Naming the factor alone is one mark of three. The justification must be a controlled comparison: P and Q differ only in temperature, so their difference isolates it. Comparing P with R changes two things at once and proves nothing.
(c)[2]
Curve R has not levelled off by 10 arbitrary units. Suggest what this shows and predict what would happen if the light intensity were raised further.
Model Answer -- 5(c)
light is still the limiting factor for R at 10 units, because the rate is still rising [1]
raising the light further would continue to raise the rate until another factor — temperature, or carbon dioxide again — becomes limiting, at which point R would also plateau [1]
⚠ If you missed marks here: Answering “R has no limiting factor” misunderstands the idea — there is always one. A rising line means the factor on the x-axis is the one currently in shortest supply.
(d)[3]
Suggest one practical change the grower could make in winter to raise the yield, and explain why it would work.
Model Answer -- 5(d)
burn a paraffin heater in the glasshouse, or install heating together with carbon dioxide enrichment [1]
this raises the temperature, giving the enzyme and substrate molecules more kinetic energy so more successful collisions occur per second [1]
and it releases carbon dioxide, increasing the supply of a raw material, so two limiting factors are relieved at once [1]
⚠ If you missed marks here: “Install more lamps” is a weak answer in winter, and the data explain why: at 15 °C the plateau is reached at only 6 units of light, so extra light does nothing. Always check the data before recommending a change.
Question 6 -- Four Tubes and an Indicator
Total: 12 marks
Four boiling tubes were each half-filled with red hydrogencarbonate indicator solution and sealed with a rubber bung. Equal masses of the pondweed Elodea were added to tubes 1, 2 and 3. Tube 2 was wrapped completely in aluminium foil. All four tubes were placed the same distance from a bright lamp, standing in a tank of water, for four hours.
tube
pondweed
condition
colour after 4 hours
1
yes
bright light
purple
2
yes
wrapped in foil
yellow
3
yes
bright light through a green filter
red
4
no
bright light
red
(a)[3]
State what the colour of hydrogencarbonate indicator shows, and explain the result in tube 1.
Model Answer -- 6(a)
the indicator responds to the concentration of dissolved carbon dioxide: purple means low, red atmospheric and yellow high [1]
in tube 1 photosynthesis occurred faster than respiration [1]
so there was a net removal of carbon dioxide from the water and the concentration fell below atmospheric [1]
⚠ If you missed marks here: “It turned purple because oxygen was released” scores nothing — the indicator does not detect oxygen at all. And the word net is doing real work: the plant is respiring in tube 1 as well.
(b)[3]
Explain the result in tube 2 and state what it demonstrates about plants.
Model Answer -- 6(b)
wrapped in foil the pondweed receives no light, so no photosynthesis can occur [1]
respiration continues, releasing carbon dioxide into the water, so the concentration rises above atmospheric and the indicator turns yellow [1]
this demonstrates that plants respire continuously, not only at night [1]
⚠ If you missed marks here: Writing “in the dark the plant respires instead” smuggles in a switch that does not exist. Respiration never stops; in the light it is simply outweighed.
(c)[3]
Explain why tube 3 remained red.
Model Answer -- 6(c)
chlorophyll absorbs little green light (most is reflected or passes through), so little energy is transferred and the rate of photosynthesis is low [1]
the low rate of photosynthesis is equal to the rate of respiration, so carbon dioxide is used as fast as it is produced [1]
there is therefore no net change in carbon dioxide concentration — this is the compensation point [1]
⚠ If you missed marks here: “Nothing happened in tube 3” is the answer to avoid. A great deal happened; it balanced. The examiner is testing whether you can distinguish zero rate from zero net change.
(d)[3]
Explain the purpose of tube 4, and explain why all four tubes were stood in a tank of water.
Model Answer -- 6(d)
tube 4 is the control: identical in every way except that it contains no pondweed [1]
it shows that any colour change in the other tubes was caused by the pondweed and not by the lamp, the warmth or the passage of time [1]
the tank of water absorbs heat from the lamp so that the temperature of all four tubes stays the same and does not affect the rate [1]
⚠ If you missed marks here: “Tube 4 is there to compare with” earns nothing — every second tube ever set up is there to compare with. The mark is for naming what it eliminates.
Question 7 -- A Stoma at Dawn
Total: 10 marks
Guard cells are the only cells of the leaf epidermis to contain chloroplasts. Their inner wall, facing the pore, is noticeably thicker and less elastic than their outer wall. A student measures the width of a single stoma through the day and finds it widens shortly after dawn, remains open through the morning, and narrows sharply during a hot dry afternoon.
(a)[3]
Explain how a stoma opens.
Model Answer -- 7(a)
water enters the guard cells by osmosis, moving from a region of higher to lower water potential through the partially permeable membrane [1]
the cells become turgid and turgor pressure rises [1]
because the inner wall is thicker and less elastic, only the outer wall stretches, so each cell curves away from its partner and the pore opens [1]
⚠ If you missed marks here: “The cells swell so the gap between them closes” is the intuitive but wrong picture. The uneven thickening is the whole mechanism, and without it swelling would indeed seal the pore.
(b)[2]
Suggest an advantage of guard cells containing chloroplasts when other epidermal cells do not.
Model Answer -- 7(b)
the chloroplasts allow the guard cells to respond directly to light [1]
so the pore opens at the time when photosynthesis is beginning and carbon dioxide is about to be needed [1]
⚠ If you missed marks here: Do not answer “so the guard cells can photosynthesise for the leaf”. There are far too few guard cells for that to matter; their chloroplasts are a sensor, not a factory.
(c)[3]
Explain why the stoma narrows during a hot dry afternoon, and describe the consequence for the plant.
Model Answer -- 7(c)
water is lost by evaporation faster than it can be replaced from the roots, so the guard cells lose water and become flaccid [1]
the bowing straightens and the two cells fall back together, closing the pore and conserving water [1]
with the pore closed carbon dioxide cannot diffuse in, so it becomes the limiting factor and the rate of photosynthesis falls [1]
⚠ If you missed marks here: The third mark is the one candidates miss, and it is the point of the whole question: closing the stoma is a trade-off, not a free improvement. Notice how it links 6.2 back to the limiting factors of 6.1.
(d)[2]
Suggest why most stomata are found on the lower surface of a leaf rather than the upper surface.
Model Answer -- 7(d)
the lower surface is shaded and therefore cooler, and is sheltered from air movement [1]
so less water is lost by evaporation from the open pores, while carbon dioxide can still diffuse in [1]
⚠ If you missed marks here: “So carbon dioxide can fall into them” is a real misconception: gases do not sink into leaves, they diffuse down a concentration gradient in any direction. The reason is entirely about water conservation.
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