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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 -- Designing the Carbon Dioxide Investigation
Total: 12 marks
A student is asked to find out whether carbon dioxide is necessary for photosynthesis. She is given a healthy potted geranium, two identical transparent polythene bags with ties, soda lime, sodium hydrogencarbonate solution in a small open dish, and the usual laboratory apparatus.
(a)[3]
Describe how she should prepare the plant before setting up the bags, and explain why this preparation is essential.
Model Answer -- 1(a)
keep the whole plant in complete darkness for 24–48 hours to destarch it [1]
in the dark no photosynthesis occurs but respiration continues, so the starch already stored in the leaves is used up [1]
therefore any starch present at the end of the investigation must have been made during the investigation [1]
⚠ If you missed marks here: This is the mark almost nobody writes down, and without it the entire experiment is uninterpretable. If you learn one sentence from Topic 6, make it this one.
(b)[3]
Describe how she should set up the test leaf and the control leaf.
Model Answer -- 1(b)
enclose leaf A, still attached to the plant, in a bag containing the soda lime, which absorbs carbon dioxide, and tie the neck of the bag [1]
enclose leaf B, on the same plant, in an identical bag containing the dish of sodium hydrogencarbonate solution, which releases carbon dioxide, and tie it in the same way [1]
leave the plant in bright light for six to eight hours [1]
⚠ If you missed marks here: Using leaves from two different plants would introduce dozens of uncontrolled differences. Keeping both leaves on the same plant controls age, genetics, water supply and light exposure without any extra effort.
(c)[3]
State three variables she must keep the same for the two leaves, and explain why the control leaf must also be enclosed in a bag.
Model Answer -- 1(c)
any three of: light intensity and duration, temperature, size and age of leaf, transparency and size of bag, position on the plant [1]
the control leaf must be bagged because a bag alters humidity, temperature and air movement around the leaf [1]
if only one leaf were bagged, she would be testing the effect of being in a bag as well as the effect of carbon dioxide, so the result would not be valid [1]
⚠ If you missed marks here: Most candidates name variables but never explain the bag. The examiner has put the control leaf in a bag deliberately; explaining why is where the reasoning marks are.
(d)[3]
State the results she should expect, and explain how they support the conclusion that carbon dioxide is necessary for photosynthesis.
Model Answer -- 1(d)
leaf A, with soda lime, remains orange-brown with iodine: no starch, so no photosynthesis occurred [1]
leaf B, with sodium hydrogencarbonate, turns blue-black: starch present, so photosynthesis occurred [1]
because the two leaves were identical in every respect except the availability of carbon dioxide, that difference must be the cause, so carbon dioxide is necessary [1]
⚠ If you missed marks here: Stopping at the colours gives you two marks out of three. The final sentence — naming the single difference and drawing the causal conclusion — is what turns an observation into evidence.
Question 2 -- Counting Stomata
Total: 12 marks
A student wants to compare the number of stomata on the upper and lower surfaces of three species of leaf. She paints a thin layer of clear nail varnish onto each surface, allows it to dry, peels off the film with tape and examines it under a microscope, counting the stomata in the field of view. Her results, converted to a density, are shown below.
species
habitat
upper surface / mm²
lower surface / mm²
oak
woodland
0
340
water lily
floating on a pond
460
0
oleander
dry hillside
0
95
(a)[3]
Explain why the nail varnish method allows stomata to be counted, and state two things she must keep the same so that the three species can be compared fairly.
Model Answer -- 2(a)
the varnish sets against the epidermis and takes an impression of its surface, including the outline of each stoma and its guard cells [1]
the same area must be examined for every sample, since a density is a number per unit area [1]
the same magnification and the same region of the leaf, avoiding veins and the leaf margin, and the same number of fields of view counted [1]
⚠ If you missed marks here: “She should repeat it” is a reliability point and is not what part (a) asks for. The question is about making the three species comparable, which is standardisation.
(b)[3]
Explain the pattern of stomatal distribution shown by the oak leaf.
Model Answer -- 2(b)
all 340 stomata per mm² are on the lower surface, which is shaded and cooler and sheltered from air movement [1]
this reduces the rate of evaporation of water from the open pores while still allowing carbon dioxide to diffuse in to the air spaces [1]
the upper surface, which is exposed to direct sun, carries the thicker waxy cuticle instead [1]
⚠ If you missed marks here: Avoid the idea that carbon dioxide “sinks” into the lower surface. Gases diffuse down a concentration gradient in any direction, and the reason for the arrangement is entirely about conserving water.
(c)[3]
Explain the pattern shown by the water lily leaf.
Model Answer -- 2(c)
the lower surface of a floating leaf is in contact with the water, where gas exchange with the atmosphere is impossible [1]
so the stomata are on the upper surface, which is exposed to the air [1]
there is no danger of drying out on the upper surface either, because the leaf is continuously supplied with water, so a high density of 460 per mm² is not a liability [1]
⚠ If you missed marks here: Candidates often assume the rule “stomata are on the lower surface” is universal. It is a consequence of a trade-off, and when the environment changes, the trade-off — and the answer — changes with it.
(d)[3]
Suggest why the oleander, from a dry hillside, has so few stomata, and state one further structural feature you would expect its leaves to have.
Model Answer -- 2(d)
fewer stomata mean fewer openings through which water vapour can be lost, so water is conserved in a habitat where it is scarce [1]
the cost is a slower supply of carbon dioxide, so the rate of photosynthesis is lower — the plant trades growth for survival [1]
a further expected feature, for example a much thicker waxy cuticle, sunken stomata, or a smaller leaf area [1]
⚠ If you missed marks here: The best answers always name the cost as well as the benefit. Adaptations are compromises, and a question that says “suggest why” in an unfamiliar habitat is usually asking you to identify the compromise.
Question 3 -- A Flawed Method
Total: 12 marks
A student writes the following method. “Put some pondweed in a beaker of tap water. Put a lamp 10 cm away and count the bubbles for one minute. Move the lamp to 20 cm and count again. Repeat for 30, 40 and 50 cm. Plot bubbles per minute against distance. Conclude that light intensity affects the rate of photosynthesis.”
(a)[3]
Identify two variables that have not been controlled in this method and explain, for each, how it would affect the results.
Model Answer -- 3(a)
temperature: the lamp emits heat as well as light, so bringing it closer warms the water, raising the rate for a second reason and exaggerating the effect of light [1]
carbon dioxide concentration: tap water contains only a small amount of dissolved carbon dioxide, which may become limiting at high light intensity and cause an early plateau [1]
background light from the room adds an unmeasured and roughly constant amount of light at every distance [1]
⚠ If you missed marks here: Naming variables without saying what each would do to the results earns half the marks at best. The command word is explain, so every variable needs a consequence attached.
(b)[3]
Describe three changes to the apparatus or procedure that would improve the method, giving a reason for each.
Model Answer -- 3(b)
place a glass tank of water between the lamp and the beaker to absorb heat, so that temperature stays constant as the lamp is moved [1]
add sodium hydrogencarbonate solution to the water so that carbon dioxide does not become the limiting factor [1]
collect the gas and measure its volume per minute rather than counting bubbles, because bubbles vary in size and are not proportional to the volume of oxygen [1]
⚠ If you missed marks here: “Use a better lamp” and “be more careful” are not creditworthy. Every improvement must name a specific piece of apparatus or procedure and the specific problem it solves.
(c)[3]
Explain why the student should allow a period of time after each move of the lamp before starting to count, and state one further change that would improve the reliability of the results.
Model Answer -- 3(c)
the plant needs time to equilibrate: the rate of photosynthesis takes several minutes to settle to a steady value at a new light intensity, and the water needs time to reach a steady temperature [1]
counting immediately would record a rate that is still changing, so the reading would not represent that light intensity [1]
for reliability, take at least three readings at each distance and calculate a mean, identifying and excluding any anomalies [1]
⚠ If you missed marks here: Equilibration is one of the two details candidates most often omit from this practical, the other being the heat shield. Both are straightforward marks once you know to look for them.
(d)[3]
The student concludes that light intensity affects the rate of photosynthesis. Explain why plotting bubbles against distance does not fully justify this conclusion.
Model Answer -- 3(d)
distance is only a proxy for light intensity, and intensity is proportional to 1 ÷ distance², so equal steps in distance are not equal steps in intensity [1]
because temperature was not controlled, any change in rate could have been caused by heat from the lamp rather than by light, so the conclusion is not valid [1]
she should measure light intensity directly with a light meter and plot rate against measured intensity [1]
⚠ If you missed marks here: This part is asking about validity, not accuracy or reliability. The question is whether the experiment measured what it claims to have measured, and with two variables changing together the honest answer is that it did not.
Question 4 -- Following One Carbon Atom
Total: 12 marks
A researcher supplies a wheat plant with carbon dioxide in which the carbon is a heavy isotope that can be detected in any molecule containing it. The plant is in bright light. Samples of leaf, phloem sap, root and grain are taken at intervals and tested for the isotope. A grazing insect that feeds on the leaves is also sampled.
(a)[3]
State the molecule in which the labelled carbon would first appear, where in the cell it would be found, and explain why.
Model Answer -- 4(a)
it would first appear in glucose [1]
in the chloroplasts of the mesophyll cells [1]
because glucose is the immediate product of photosynthesis, and everything else the plant makes from carbon is made from it afterwards [1]
⚠ If you missed marks here: “Starch” is the most tempting wrong answer, and it is wrong only because of timing: starch is made from glucose, so the label reaches it later. Read tracer questions as questions about the order of events.
(b)[3]
The label appears in the phloem sap and later in the grain. Name the molecule carrying it in the phloem and explain why the plant converts glucose into this molecule for transport.
Model Answer -- 4(b)
sucrose [1]
sucrose is soluble, so it can be transported in solution, whereas starch is insoluble and cannot move [1]
sucrose is not used directly in respiration in the way glucose is, so it can be moved across the plant without being consumed on the way [1]
⚠ If you missed marks here: This part rewards understanding why a plant bothers to make three different carbohydrates from one. Solubility is an advantage for transport and a disadvantage for storage, which is why starch and sucrose both exist.
(c)[3]
The label is later found in the tissues of the grazing insect, and eventually in carbon dioxide in the air. Explain how the labelled carbon returns to the atmosphere.
Model Answer -- 4(c)
the insect eats the leaf, digesting and absorbing the carbohydrate, so the labelled carbon is built into its own tissues [1]
the insect and the plant both respire, breaking down glucose and releasing carbon dioxide as a product [1]
the labelled carbon dioxide diffuses out of the organism into the air, completing the cycle [1]
⚠ If you missed marks here: Remember that the plant respires too, so some of the labelled carbon returns to the air without ever reaching an animal. Answers that route all of it through the food chain miss a marking point.
(d)[3]
Explain why photosynthesis is described as the process on which almost all food chains depend.
Model Answer -- 4(d)
photosynthesis is the process that transfers light energy into chemical energy in glucose, which is the only large-scale entry point for energy into living systems [1]
the carbohydrates, and the proteins and fats made from them, are then passed along the food chain from producers to consumers when organisms feed [1]
consumers cannot photosynthesise, so every organism in the chain ultimately depends on the producer [1]
⚠ If you missed marks here: Strike the phrase “plants make energy” from your writing. Energy is transferred, never created, and examiners treat the difference as a genuine error rather than loose wording.
Question 5 -- One Leaf, Two Conclusions
Total: 10 marks
A destarched variegated pelargonium leaf, still attached to the plant, has a strip of aluminium foil pinned across it so that the foil covers part of the green region and part of the white region. The plant is left in bright light for eight hours and the leaf is then tested for starch.
(a)[4]
State the four regions of the leaf created by this arrangement, and the result expected in each.
Model Answer -- 5(a)
green and uncovered → blue-black, since both light and chlorophyll are present [1]
green and covered → orange-brown, since light is absent [1]
white and uncovered → orange-brown, since chlorophyll is absent [1]
white and covered → orange-brown, since both are absent [1]
⚠ If you missed marks here: Only one of the four regions turns blue-black, because photosynthesis requires both light and chlorophyll and the absence of either is enough to stop it. Candidates who expect two positive regions have not thought the logic through.
(b)[3]
State which two regions must be compared to show that light is necessary, and explain why that particular comparison is valid.
Model Answer -- 5(b)
compare green and uncovered with green and covered [1]
chlorophyll, water supply, temperature and age are identical in the two regions because they lie on the same leaf, so light is the only variable that differs [1]
therefore any difference in the result must be caused by light [1]
⚠ If you missed marks here: Naming the regions is one mark of three. The reasoning — that one variable and only one variable differs — is what the question is really testing, and it is the same reasoning behind every control in biology.
(c)[2]
Explain why the white covered region provides no useful evidence.
Model Answer -- 5(c)
two factors are absent at once in that region, light and chlorophyll [1]
so a negative result cannot show which of the two was responsible, and no conclusion can be drawn from it [1]
⚠ If you missed marks here: This is the principle of changing one variable at a time, met from an unusual direction. A region with two things missing is not a stronger test; it is an uninterpretable one.
(d)[1]
Before the leaf is tested, the ethanol removes all its colour. Suggest what the student must therefore do before starting the investigation.
Model Answer -- 5(d)
make a labelled drawing, or a photograph, recording exactly which regions of the leaf were green and which were white, and where the foil was pinned [1]
⚠ If you missed marks here: A single-mark part that catches almost everyone. Once the chlorophyll has been dissolved out you can no longer tell which regions were green, so the record has to be made in advance.
Question 6 -- Predicting Structure From Habitat
Total: 12 marks
Three plants grow in very different places: a marram grass on an exposed coastal dune, an Elodea shoot fully submerged in a pond, and a shade-tolerant fern on the floor of a dense forest. Each has leaves adapted to its own habitat.
(a)[3]
Predict two ways in which the leaf of the marram grass would differ from a typical broad leaf, and explain each prediction.
Model Answer -- 6(a)
a much thicker waxy cuticle, because the exposed dune is dry and windy and evaporation must be reduced [1]
fewer stomata, or stomata sunken in pits or on a rolled inner surface, so that water vapour is trapped and the diffusion gradient for water loss is reduced [1]
a smaller exposed surface area, for example by rolling the leaf, reducing the area from which water evaporates [1]
⚠ If you missed marks here: Predictions must come with reasons in the same sentence. Note also that each of these adaptations costs the plant carbon dioxide, so a good answer acknowledges the compromise rather than presenting them as pure gains.
(b)[3]
The submerged Elodea leaf has almost no cuticle, no stomata and is only two cells thick. Explain why none of these is a disadvantage.
Model Answer -- 6(b)
under water the plant cannot dry out, so a waterproof cuticle would serve no purpose and would only obstruct the entry of gases [1]
carbon dioxide is dissolved in the surrounding water and can diffuse in across the whole leaf surface, so pores are unnecessary [1]
being only two cells thick gives an extremely short diffusion distance to every cell, and the leaf does not need to be self-supporting because the water supports it [1]
⚠ If you missed marks here: The recurring idea in this question is that a structure is only an adaptation relative to an environment. A cuticle is essential on a dune and a hindrance under water, and the same is true of stomata.
(c)[3]
Predict how the fern’s leaves would differ from those of a plant of the same species grown in full sun, and explain the advantage.
Model Answer -- 6(c)
the shade leaves would have a larger surface area, intercepting more of the little light that reaches the forest floor [1]
they would be thinner with fewer palisade layers, because too little light penetrates for a second layer of chloroplasts to repay the cost of building it [1]
light is the limiting factor in deep shade, so the plant invests in capturing light rather than in the machinery to use large amounts of it [1]
⚠ If you missed marks here: The unifying principle across all three habitats is worth stating explicitly: a plant invests in whatever is in shortest supply. Answers that list features without that logic read like memorised facts.
(d)[3]
All three plants carry out photosynthesis using the same equation. Explain why their leaves are nonetheless so different.
Model Answer -- 6(d)
the chemistry of photosynthesis is identical in all plants, but the supply of the raw materials, of light and of water differs enormously between habitats [1]
each leaf is a structural solution to the particular shortage its habitat imposes, so the same process is served by very different engineering [1]
every adaptation involves a compromise, most often between admitting carbon dioxide and conserving water [1]
⚠ If you missed marks here: This is a synthesis question, so it wants the general principle rather than more examples. Reaching for the words limiting factor and compromise is what lifts an answer into the top band.
Question 7 -- Culture Solutions and Root Hairs
Total: 10 marks
A student plans to investigate the effect of nitrate ion concentration on the growth of maize seedlings. She has identical seedlings, a range of culture solutions containing every ion except that the nitrate concentration differs, aquarium pumps, and access to a drying oven and a balance.
(a)[3]
State the independent variable, the dependent variable and two variables she must control.
Model Answer -- 7(a)
independent variable: the concentration of nitrate ions in the culture solution [1]
dependent variable: the dry mass of each seedling after a fixed period [1]
two controlled variables from: light intensity and duration, temperature, volume of solution, concentration of all other ions, aeration rate, initial size and age of the seedlings [1]
⚠ If you missed marks here: Height is a weaker dependent variable than dry mass, because a plant can elongate by taking up water without building any new material. Whenever growth is being measured, dry mass is the honest quantity.
(b)[2]
Explain why dry mass is measured rather than fresh mass, and describe how she should obtain it.
Model Answer -- 7(b)
fresh mass includes water, which varies with the time of day and the water supply and is not material the plant has built [1]
dry the seedlings in an oven at a low temperature and weigh repeatedly until the mass is constant, showing all the water has gone [1]
⚠ If you missed marks here: “Dry it for an hour” is not enough: the mark is for drying to constant mass, since a fixed time leaves an unknown quantity of water behind. Note also that this measurement destroys the plant, so it can only be made once.
(c)[3]
Explain why the culture solutions must be aerated, referring to how nitrate ions enter a root hair cell.
Model Answer -- 7(c)
nitrate is usually far more concentrated inside the root hair cell than in the solution, so it must be taken in by active transport, against the concentration gradient, using protein carriers in the cell surface membrane [1]
active transport requires energy released by aerobic respiration, so the roots need a supply of oxygen [1]
without aeration the roots would respire anaerobically, less energy would be released, ion uptake would fall and every seedling would show deficiency symptoms regardless of the nitrate concentration [1]
⚠ If you missed marks here: Never decide the transport process from the particle: it is the direction of the gradient that decides. And the third mark rewards seeing that a poorly aerated experiment would produce the same symptoms in every treatment, hiding the effect being studied.
(d)[2]
Predict the shape of a graph of dry mass against nitrate concentration, and explain the shape.
Model Answer -- 7(d)
dry mass would increase as nitrate concentration increases and then level off, because at higher concentrations nitrate is no longer limiting and another factor such as light intensity or carbon dioxide concentration sets the rate [1]
on the rising part nitrate is the factor in shortest supply, so more nitrate allows more amino acids to be made and therefore more protein and more growth [1]
⚠ If you missed marks here: Limiting-factor reasoning is not confined to light, carbon dioxide and temperature — it applies to any resource. Recognising the familiar plateau in an unfamiliar setting is exactly what a challenge paper is testing.
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