⚡ Challenge Paper Preparation

Challenge Prep: Plant Nutrition

IGCSE Biology 0610 — Topic 6

Every exam paper in this topic is a challenge paper, so this is where the ramping happens. Topic 6 is not conceptually difficult — and that is exactly why it is dangerous. The marks are lost in a short list of predictable places: the forgotten destarching step, the missing or badly chosen control, “the plant makes food” where the word glucose was wanted, a limiting factor described as “the thing that runs out” instead of the factor in shortest supply at that moment, enzymes said to be killed rather than denatured, and the belief that plants photosynthesise by day and respire by night. Add the leaf-structure questions that stop at a label when two marks were on offer, and you have almost the whole mark distribution of this topic. Twelve traps, six walkthroughs, six pairs to separate, a concept map, six wrong answers to dissect and ten full challenge questions — all below.

⚠️ Common Traps & Misconceptions

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Twelve traps that cost marks on Topic 6 questions. The first three are the ones that cost the most, so they are dealt with in the most depth.

⚠️ TRAP 1

Trap 1: Forgetting to destarch the plant

The Trap“Cover half the leaf with foil, leave the plant in sunlight for six hours, then test for starch.” It sounds like a complete method. It is missing the step that makes the whole experiment mean anything, and Cambridge prints a separate mark for it.

The TruthBefore any photosynthesis experiment, the plant is kept in complete darkness for 24–48 hours. This is destarching. In the dark the plant cannot photosynthesise but goes on respiring, so the starch already stored in the leaves is used up. Only then can you claim that any starch present at the end was made during the investigation. Learn the sentence whole: “the plant is destarched by leaving it in the dark for 48 hours, so that any starch found at the end must have been made during the experiment.”

Why It MattersIt is a mark on three different experiments (light, chlorophyll, carbon dioxide) and it also appears as an “evaluate this method” question, where the whole answer is that the conclusion is unsafe without it. Very few candidates write it. It is close to a free mark for you.

Example Question“Describe how you would show that light is necessary for photosynthesis. [5]”

⚠️ TRAP 2

Trap 2: “A limiting factor is the one that runs out”

The Trap“Carbon dioxide is the limiting factor because it has all been used up.” The word limiting sounds like finished, so the definition drifts towards exhaustion — and the mark is refused.

The TruthA limiting factor is the factor in shortest supply at that moment, which therefore limits the rate. It need never reach zero. Carbon dioxide sits at 0.04 % in ordinary air all day, never runs out, and is limiting for most land plants for most of the day. Think of a production line: the slowest worker sets the output even though nobody has stopped working. The operational test is the sentence to memorise: a factor is limiting if and only if increasing it increases the rate.

Why It MattersLimiting factors is a Supplement objective and therefore certain to appear on Extended papers, usually attached to a graph with two or three curves. The definition mark, the identification mark and the “what should the grower do” mark all collapse if the definition is wrong.

Example Question“Explain what is meant by a limiting factor and identify the limiting factor at point X on the graph. [3]”

⚠️ TRAP 3

Trap 3: Writing “the plant makes food”

The Trap“In photosynthesis the plant uses carbon dioxide, water and sunlight to make food and oxygen.” Every word of that sentence feels right, and it can score as little as zero.

The Truth“Food” is not a chemical. The product is glucose, a carbohydrate. “Sugar” is usually refused too, because sucrose and starch are sugars in the loose sense and they are made later, from the glucose. And the same sentence contains a second error: light is not a raw material. The raw materials are carbon dioxide and water; light supplies the energy. Listing light among the raw materials can cost the mark on its own.

Why It MattersThe word equation, the balanced equation and the definition all hang on naming glucose, and “state the raw materials” is a routine two-mark opener. Both halves of this trap show up in the first question of almost every Topic 6 paper.

Example Question“State the word equation for photosynthesis, including the conditions required. [3]”

⚠️ TRAP 4

Trap 4: Believing plants respire only at night

The Trap“During the day the plant photosynthesises and at night it respires.” It is symmetrical, it is memorable, and it is false.

The TruthRespiration happens in every living cell, continuously, day and night — the plant needs energy for active transport, protein synthesis and growth around the clock. In daylight photosynthesis happens as well, and usually faster, so there is a net uptake of carbon dioxide and a net release of oxygen. At the compensation point the two rates are equal and there is no net gas exchange. In the dark only respiration occurs.

Why It MattersEvery hydrogencarbonate-indicator question is built on this, and so is every “explain why the tube in the dark turned yellow”. The word that earns the marks is net, exactly as it was in Topic 3.

Example Question“Explain why the indicator in the foil-wrapped tube containing pondweed turned yellow. [3]”

⚠️ TRAP 5

Trap 5: “The control is there to compare with”

The TrapAsked why a second flask was set up, the answer given is “so she has something to compare her results to”. That is the purpose of every second flask ever set up, and it tells the examiner nothing.

The TruthA control is a set-up that is identical in every respect except the one factor being tested, so that any difference in the results must be caused by that factor. The mark is for saying what it rules out. In the carbon dioxide experiment, the control leaf must be sealed in an identical bag containing sodium hydrogencarbonate — otherwise you have tested “being in a bag” rather than “having carbon dioxide”. In the pondweed indicator experiment, the control tube has no pondweed, ruling out colour change caused by the lamp, the warmth or time.

Why It MattersControls carry marks on every experimental question in this topic, and challenge papers love asking you to design or criticise one rather than just name it.

Example Question“Describe the control that should be used and explain what it shows. [3]”

⚠️ TRAP 6

Trap 6: Confusing soda lime with sodium hydrogencarbonate

The TrapTwo long chemical names, both encountered on the same page, and they do opposite things. Under time pressure the wrong one gets written into the wrong flask and the answer becomes nonsense.

The TruthSoda lime absorbs carbon dioxide — it is used to remove CO₂ from the air around the test leaf. Sodium hydrogencarbonate releases carbon dioxide — it is used to supply CO₂ to the control leaf, or to the water around pondweed in a rate experiment. A third name to keep separate: hydrogencarbonate indicator, which neither adds nor removes CO₂, it merely reports the concentration by changing colour.

Why It MattersGet them the wrong way round and every subsequent prediction in the question is inverted, so a single slip can cost three or four marks in one part.

Example Question“Name a chemical that could be placed in flask B to remove carbon dioxide, and state what should be placed in the control flask. [2]”

⚠️ TRAP 7

Trap 7: Saying “the enzymes are killed” at high temperature

The Trap“Above 45 °C the rate falls because the enzymes are killed / die / are destroyed.” The biology being described is right; the vocabulary makes it unmarkable.

The TruthEnzymes are protein molecules, not organisms. The correct word is denatured, and the full answer says what denaturing is: high temperature breaks the bonds holding the protein in shape, so the active site changes shape and the substrate no longer fits, so fewer enzyme–substrate complexes form and the rate falls.

Why It MattersThe temperature graph is the odd one out among the three factors precisely because of denaturing — it peaks and falls where the other two plateau. That shape is examined every year, and the explanation mark depends on this one word.

Example Question“Explain the shape of the graph above 35 °C. [3]”

⚠️ TRAP 8

Trap 8: Swapping nitrate and magnesium

The Trap“Nitrate ions are needed to make chlorophyll.” Both ions are absorbed from soil, both cause deficiency symptoms, both appear in the same sentence of the syllabus — so they get swapped constantly.

The TruthNitrate → nitrogen → amino acids → proteins → growth. Magnesium → chlorophyll → green colour. The symptoms follow directly: nitrate deficiency gives a small plant, magnesium deficiency gives a yellow one. Read the data before you choose: badly stunted, often with older leaves yellowing → nitrate; normal size but yellow between the veins → magnesium.

Why It MattersThis is usually set as a deduction question with two seedlings and no ions named, so the mark depends entirely on matching symptom to ion. It also links to Topic 3: these are mineral ions taken up by active transport in root hair cells, not “food” and not “plant vitamins”.

Example Question“Deduce which mineral ion was missing from each culture solution and justify your answer. [4]”

⚠️ TRAP 9

Trap 9: Confusing the ethanol step with the boiling-water step

The Trap“The leaf is boiled in ethanol to kill it.” Or “the leaf is boiled in water to remove the chlorophyll.” Four steps, four reasons, and the middle two get exchanged.

The TruthBoiling water kills the leaf, stops the enzyme reactions and breaks the cell membranes so that reagents can get in. Hot ethanol dissolves out the chlorophyll, so the leaf turns pale and the blue-black of a positive iodine test can actually be seen. Then rinse in water, because ethanol leaves the leaf brittle and it must be softened before iodine will spread over it. Then iodine solution: orange-brown to blue-black means starch. And the safety mark: the ethanol is heated in a water bath, never over a naked flame, because ethanol is highly flammable.

Why It Matters“Explain why the leaf is placed in hot ethanol” is a one-mark gift that half the candidature gets wrong, and the safety point is a second easy mark that almost nobody offers.

Example Question“Describe how you would test a leaf for starch, giving a reason for each step. [5]”

⚠️ TRAP 10

Trap 10: Labelling a leaf diagram and stopping there

The TrapA two-mark question says “identify structure X and explain how it is adapted for photosynthesis”. The answer given is “palisade mesophyll”. One noun, one mark, and the second mark was sitting right there.

The TruthUse a fixed sentence pattern: “[Structure] is [description], which means [property], so [consequence for photosynthesis].” For example: “The upper epidermis is a single layer of cells containing no chloroplasts, which makes it transparent, so light passes straight through to the palisade cells below.” That will score two marks for any structure in the leaf, every time.

Why It MattersLeaf structure is half of Topic 6 and it is almost always examined as identify-plus-explain. Counting the marks before you write is the single cheapest habit you can build.

Example Question“Identify layer B and explain two ways in which it is adapted for its function. [3]”

⚠️ TRAP 11

Trap 11: Confusing repeats with controls

The Trap“To improve the validity she should repeat the experiment.” Or “to make it more reliable she should add a control.” Two words, swapped, and the mark goes.

The TruthRepeats address reliability — doing the same thing several times and taking a mean reduces the effect of random error and lets you spot anomalies. Controls and controlled variables address validity — making sure the thing you measured really was caused by the factor you changed. Repeating an invalid experiment ten times gives you ten invalid results.

Why It MattersChallenge papers ask for improvements constantly, and they signal which one they want by using the words “reliable” or “valid” in the question. Read that word and answer the matching one.

Example Question“Suggest two changes that would improve the validity of this investigation. [2]”

⚠️ TRAP 12

Trap 12: Treating the lamp as if only its light changed

The Trap“Move the lamp from 50 cm to 10 cm and count the bubbles.” Two things have changed, not one — and the graph you draw is then a graph of nothing in particular.

The TruthA lamp emits heat as well as light, so bringing it closer raises the temperature of the water too. Place a beaker or glass tank of water between the lamp and the plant to absorb the heat, and check the temperature with a thermometer at each distance. There is a second, subtler point worth a mark on the best answers: light intensity is proportional to 1/distance², so equal steps in distance are not equal steps in intensity — ideally use a light meter or a lamp with adjustable output rather than distance at all.

Why It MattersEvery rate-of-photosynthesis practical uses this apparatus, and “identify one variable that has not been controlled” is a standard question with this exact answer.

Example Question“Suggest why the student placed a glass tank of water between the lamp and the boiling tube. [2]”

🧩 Multi-Step Reasoning Walkthroughs

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Six challenge questions broken into steps. Try each step yourself before revealing the next — the reasoning is the point, not the answer.

Walkthrough 1 — Three Curves, One Limiting FactorA student measures the rate of photosynthesis of pondweed as the volume of oxygen collected in cm³ min⁻¹. Curve P: 0.04 % CO₂, 15 °C, plateau at 1.2. Curve Q: 0.04 % CO₂, 25 °C, plateau at 1.9. Curve R: 0.40 % CO₂, 25 °C, plateau at 3.6. All three rise from the origin and all three level off. (a) Name the limiting factor on the steep part of every curve. [1] (b) Name the limiting factor at the plateau of curve Q and justify it with the data. [3] (c) Predict the plateau value for 0.40 % CO₂ at 15 °C and explain your reasoning. [2]

On the slope, the x-axis factor is limiting

Every curve rises steeply from the origin, which means that in that region increasing light intensity increases the rate. Apply the test — a factor is limiting if and only if increasing it increases the rate — and the answer to (a) is light intensity. Note that this is true for all three curves at once, which is why they are indistinguishable near the origin.

Which change still moves the line?

At the plateau of Q, more light does nothing, so light is not limiting. Compare Q with P: the only difference is temperature, and raising it from 15 to 25 °C lifted the plateau from 1.2 to 1.9 — but Q has now plateaued too, so temperature is no longer limiting either. Compare Q with R: the only difference is carbon dioxide, and raising it from 0.04 % to 0.40 % lifted the plateau from 1.9 to 3.6.

The justification must quote the comparison

“The limiting factor at the plateau of Q is carbon dioxide concentration [1]. Increasing the light intensity does not increase the rate above 1.9 cm³ min⁻¹, so light is not limiting [1]. Increasing the carbon dioxide concentration from 0.04 % to 0.40 % at the same temperature raises the plateau from 1.9 to 3.6 cm³ min⁻¹, so carbon dioxide is the factor in shortest supply [1].” Naming the factor alone is one mark out of three.

Reason from the pattern, do not guess a number blindly

Going from 15 °C to 25 °C at 0.04 % CO₂ raised the plateau from 1.2 to 1.9. So dropping from 25 °C to 15 °C at 0.40 % CO₂ should lower the plateau of R below 3.6 — something in the region of 2.0–2.5 cm³ min⁻¹ is creditworthy [1]. The explanation is the mark that matters: at the lower temperature the enzymes controlling photosynthesis have less kinetic energy, so there are fewer successful collisions per second and temperature becomes the limiting factor even though carbon dioxide is now plentiful [1].

Never read a limiting factor off one point

You identify a limiting factor by asking which change moves the line. That is why multi-curve graphs are the standard vehicle for this question: they hand you three controlled comparisons and expect you to use them. If you only ever look at one curve you can say what is not limiting but never what is.

Walkthrough 2 — Designing the Carbon Dioxide Experiment From ScratchDesign an investigation to show that carbon dioxide is necessary for photosynthesis, using a destarched potted plant, two clear polythene bags, soda lime, sodium hydrogencarbonate solution and the usual laboratory apparatus. Include the control, state the expected results and explain how the results support the conclusion. [8]

Destarch, and say why

“Keep the whole plant in complete darkness for 48 hours so that the starch already in the leaves is used up in respiration, and any starch found later must have been made during the investigation.” One mark, and it is the first sentence of the answer.

Two bags, one difference

Enclose leaf A in a clear polythene bag containing soda lime, which absorbs carbon dioxide, and seal the neck of the bag. Enclose leaf B, on the same plant, in an identical clear bag containing sodium hydrogencarbonate solution, which supplies carbon dioxide, and seal it the same way. Using the same plant controls age, water supply, genetics and light exposure at a stroke.

Controlled variables earn their own mark

Both bags must be transparent and the same size, both leaves must receive the same light intensity for the same time (for example six hours in bright light), and both are at the same temperature. If one bag were opaque you would be testing light as well, and the experiment would prove nothing.

The four testing steps, briefly

Boil each leaf in water to kill it and break the membranes; place it in hot ethanol in a water bath to remove the chlorophyll; rinse in water to soften it; add iodine solution.

State the prediction, then state what it rules out

Leaf A (soda lime) stays orange-brown — no starch, so no photosynthesis. Leaf B (sodium hydrogencarbonate) turns blue-black — starch present, so photosynthesis occurred. Because the two leaves were identical in every way except the availability of carbon dioxide, the difference must have been caused by carbon dioxide, so carbon dioxide is necessary for photosynthesis. That final sentence is where the reasoning marks live — do not stop at the colours.

Walkthrough 3 — Reading a Hydrogencarbonate Indicator TableFour sealed tubes of red hydrogencarbonate indicator are left for four hours. 1: pondweed, bright light → purple. 2: pondweed, wrapped in foil → yellow. 3: pondweed, dim light → red. 4: no pondweed, bright light → red. Explain each result and state what tube 4 shows. [7]

The indicator reports CO₂ and nothing else

Purple = low CO₂. Red = atmospheric CO₂, unchanged. Yellow = high CO₂. It does not detect oxygen, starch or photosynthesis directly — you infer those. Writing “purple because oxygen was released” is a guaranteed zero.

Purple — photosynthesis exceeds respiration

In bright light the pondweed photosynthesises rapidly, taking carbon dioxide from the water faster than respiration returns it. There is a net removal of CO₂, the concentration falls below atmospheric and the indicator turns purple. The word net is the mark: the plant is respiring the whole time.

Yellow — and it proves plants respire in the dark

Wrapped in foil the pondweed receives no light, so no photosynthesis occurs. Respiration continues, releasing carbon dioxide into the water, so the concentration rises above atmospheric and the indicator turns yellow. This tube is the direct evidence that plants respire continuously — a favourite follow-up question.

Red — the compensation point

In dim light the rate of photosynthesis exactly equals the rate of respiration, so carbon dioxide is used as fast as it is produced. There is no net change in CO₂ concentration and the indicator stays red. Do not write “nothing happened” — a great deal happened; it balanced.

It rules out everything except the plant

Tube 4 has no pondweed but everything else is identical. It stays red, showing that the colour changes in tubes 1–3 were caused by the pondweed and not by the light itself, by warmth from the lamp, by the passage of time or by carbon dioxide leaking in. Saying “it is there to compare with” scores nothing; saying what it eliminates scores the mark.

Walkthrough 4 — The Variegated Leaf With a Foil StripA destarched variegated pelargonium leaf 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. After eight hours in bright light the leaf is tested for starch. Sketch and explain the pattern of colours you would expect, and state what the leaf shows about the requirements for photosynthesis. [6]

Two variables crossed gives four regions, not two

Green + uncovered. Green + covered. White + uncovered. White + covered. The question is testing whether you can see that this single leaf is a two-factor experiment, and the answer is a table with four rows, not a sentence.

Photosynthesis needs BOTH; one missing is enough to stop it

Green + uncovered → blue-black, both requirements met. Green + covered → orange-brown, no light. White + uncovered → orange-brown, no chlorophyll. White + covered → orange-brown, neither. Only one of the four regions goes blue-black.

Each conclusion needs its own controlled pair

Comparing green-uncovered with green-covered holds chlorophyll constant and varies light, so it shows light is necessary. Comparing green-uncovered with white-uncovered holds light constant and varies chlorophyll, so it shows chlorophyll is necessary. Notice that each region acts as the control for another — which is why one leaf can prove two things.

White + covered proves nothing on its own

The fourth region has two factors missing at once, so it cannot tell you which one mattered. It is a good illustration of why an experiment must change one variable at a time — and a challenge paper may well ask you why that region is of no use, which is exactly this point.

Walkthrough 5 — From Leaf Section to RateThe table gives measurements from two leaves of the same species, one grown in full sun and one in deep shade. Sun leaf: thickness 320 µm, palisade layers 2, stomata 340 per mm², area 18 cm². Shade leaf: thickness 140 µm, palisade layers 1, stomata 190 per mm², area 41 cm². (a) Describe two differences and suggest an explanation for each. [4] (b) Explain which leaf would photosynthesise faster in bright light and why. [3]

Numbers are marks

“The shade leaf has a much larger area (41 cm² against 18 cm²) but is less than half as thick (140 µm against 320 µm).” A “describe” instruction with a table in front of you always means use the figures.

In shade, light is the limiting factor

A larger surface area intercepts more of the little light available. There is no point investing in a second palisade layer, because too little light penetrates for it to pay for itself — so the leaf is thin with a single palisade layer. Fewer stomata are needed because the rate of photosynthesis, and therefore the demand for carbon dioxide, is lower.

In sun, light is plentiful and CO₂ becomes limiting

Light is abundant, so a second palisade layer is worth building: it gives more chloroplasts under the same area of surface, and there is enough light to reach them. That raises the demand for carbon dioxide, which is why the sun leaf has nearly twice as many stomata per mm². A smaller area also reduces water loss and overheating in full sun.

Per unit area, and say so

In bright light the sun leaf photosynthesises faster per unit area, because it has two palisade layers and therefore more chloroplasts beneath each square millimetre of surface [1], and more stomata per mm² so carbon dioxide can diffuse in fast enough to keep up [1]. The shade leaf would be light-saturated at a low intensity and its single palisade layer would limit the rate [1]. Watch the wording: the shade leaf is larger overall, so “which leaf photosynthesises more in total” is a different question with a possibly different answer.

Walkthrough 6 — The Mineral Deficiency DeductionThree identical barley seedlings are grown in aerated culture solutions for five weeks. Solution 1 is complete. Solution 2 lacks one ion; the seedling reaches 6 cm with pale small leaves and a dry mass of 0.11 g. Solution 3 lacks a different ion; the seedling reaches 21 cm with yellow leaves and a dry mass of 0.34 g. The seedling in solution 1 reaches 23 cm with a dry mass of 0.52 g. (a) Deduce which ion is missing from each. [4] (b) Explain why the solutions are aerated. [2] (c) Suggest why solution 3 gave a lower dry mass than solution 1 even though the seedlings are nearly the same height. [3]

Small but pale, or normal but yellow?

Solution 2: severe stunting (6 cm against 23 cm) → nitrate is missing. Nitrate supplies nitrogen for amino acids and hence proteins, without which the plant cannot make new cytoplasm or enzymes, so growth almost stops. Solution 3: near-normal height but yellow leaves → magnesium is missing, because magnesium is needed to make chlorophyll.

This is a Topic 3 question hiding in Topic 6

Roots need oxygen for aerobic respiration [1], which releases the energy used for the active transport of mineral ions into the root hair cells against a concentration gradient [1]. Without aeration the roots respire anaerobically, less energy is released and ion uptake falls — which would give deficiency symptoms in every solution and wreck the experiment.

Follow the chain: no magnesium → no chlorophyll → less glucose

Without magnesium the seedling makes little chlorophyll [1], so less light energy is transferred and the rate of photosynthesis falls, meaning 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 for a while, but dry mass measures the material actually synthesised — so the shortfall shows up in mass long before it shows up in height [1].

Dry mass is the honest measure of growth

You met this in Topic 1: growth is 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 — and the dry mass is the one that tells you what the plant actually built.

🔍 Spot the Difference

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Six pairs of questions that look almost identical and have different answers. Find the distinction before you read the key difference.

Question A

A leaf is left in bright light for six hours and tested for starch. Result?

Blue-black — provided the plant was destarched first. Photosynthesis made glucose, which was converted to starch and stored in the leaf.

Question B

A leaf is left in bright light for six hours and tested for starch, but the plant was not destarched. Result?

Also blue-black — and it proves nothing whatever. The starch could have been present before the experiment began, so the result is uninterpretable.

Key DifferenceIdentical observation, completely different value as evidence. The colour is not the answer; the logic that connects the colour to the conclusion is, and destarching is what supplies that logic.

Question A

Soda lime is placed in a sealed flask with a leaf. What does it do?

It absorbs carbon dioxide, so the leaf has no CO₂ and cannot photosynthesise. This is the test flask.

Question B

Sodium hydrogencarbonate is placed in a sealed flask with a leaf. What does it do?

It releases carbon dioxide, so the leaf has plenty of CO₂ and photosynthesises normally. This is the control flask.

Key DifferenceTwo long names, opposite jobs. Soda lime takes CO₂ away; sodium hydrogencarbonate gives CO₂. Put them in the wrong flasks and every prediction that follows is inverted.

Question A

A graph of rate against light intensity at constant temperature. Shape?

Rises steeply, then plateaus. The rate stays high — another factor has become limiting.

Question B

A graph of rate against temperature at constant light. Shape?

Rises to an optimum and then falls steeply, because above the optimum the enzymes are denatured and the active site changes shape.

Key DifferenceA plateau means a different factor is now limiting. A fall means something has been damaged. Only temperature can produce a fall, and only denaturing explains it.

Question A

Why does a plant need nitrate ions?

To supply nitrogen for making amino acids, and therefore proteins for growth and enzymes. Deficiency gives stunted growth.

Question B

Why does a plant need magnesium ions?

To make chlorophyll. Deficiency gives yellow leaves (chlorosis) on a plant of near-normal size.

Key DifferenceBadly stunted → nitrate. Normal size but yellow → magnesium. Read the symptom in the data before you choose; the question is a deduction, not a recall.

Question A

A sealed tube of red hydrogencarbonate indicator with pondweed goes purple in the light. Why?

Photosynthesis is faster than respiration, so there is a net removal of carbon dioxide and the concentration falls below atmospheric.

Question B

A sealed tube of red hydrogencarbonate indicator with pondweed goes yellow in the dark. Why?

There is no photosynthesis, but respiration continues, so carbon dioxide is added to the water and the concentration rises above atmospheric.

Key DifferenceThe plant respires in both tubes. The colour reports the net effect, not which process is happening. Writing “the plant photosynthesises in the light and respires in the dark” earns nothing because the second half implies it stops respiring in the light.

Question A

Why is the carbohydrate stored as starch?

Starch is insoluble, so it cannot diffuse out of the cell and does not lower the water potential and draw water in by osmosis.

Question B

Why is the carbohydrate transported as sucrose?

Sucrose is soluble, so it can move in the phloem, but it is not used directly in respiration the way glucose is.

Key DifferenceThe same property — solubility — is an advantage for one job and a disadvantage for the other. That is why the plant makes two different molecules from the same glucose.

🔗 Topic 6 Concept Map

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Click each node. Three frameworks: how the topic hangs together, how to decide a limiting factor, and how a leaf solves four problems at once.

⭐ CORE FRAMEWORK 1

Follow the carbon atom: from air, to glucose, to everything else

Step 1 — It arrives as carbon dioxide ▶

CO₂ diffuses in through the stomata, travels through the air spaces of the spongy mesophyll, dissolves in the moisture on the cell surfaces and diffuses into the mesophyll cells.

Nothing carries it. There is no vessel for gases — this is pure diffusion down a concentration gradient, which is why the leaf is thin and full of holes.

The gradient is maintained because photosynthesis keeps using the CO₂ up, so its concentration inside stays low. Same logic as oxygen entering a respiring muscle in Topic 3.

Step 2 — It is built into glucose using light energy ▶

In the chloroplast, chlorophyll transfers energy from light into energy in chemicals. Water arrives from the xylem. 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.

The oxygen is a waste product. Some is used in the plant’s own respiration; the rest diffuses out through the stomata — which is also, incidentally, an example of excretion from Topic 1.

Step 3 — The glucose immediately becomes something else ▶

Starch for storage (insoluble, no osmotic effect). Cellulose for cell walls. Sucrose for transport in the phloem. Nectar to attract pollinators. Or broken down again in respiration to release energy.

With nitrate ions from the soil it becomes amino acids, then proteins. That is the only route by which nitrogen enters the food chain, and it is why the whole topic is called plant nutrition.

With magnesium ions the plant makes more chlorophyll — so the ions feed back into the very process that captured the carbon.

Step 4 — Where it goes next in the syllabus ▶

The glucose you made here is the glucose that is digested and absorbed in Topic 7, transported in Topic 9 and respired in Topic 12. Every food chain in Topic 19 starts with a producer doing exactly what this topic describes.

If you can trace one carbon atom from the air, into a leaf, into a plant, into an animal and back into the air, you understand more than half the syllabus already.

⭐ CORE FRAMEWORK 2

One question decides every limiting-factor answer: which change moves the line?

Step 1 — Are you on a slope or a plateau? ▶

On the rising part, increasing the factor on the x-axis increases the rate, so that factor is limiting.

On the flat part, increasing it does nothing, so that factor is no longer limiting and something else is. You must then name the something else.

If instead the curve falls, you are not looking at a limiting factor at all — you are looking at denaturing, and the x-axis must be temperature.

Step 2 — Use the other curves as controlled comparisons ▶

Two curves that differ in one condition only are a ready-made experiment. If raising that condition lifts the plateau, it was limiting; if it does not, it was not.

This is why multi-curve graphs are the standard vehicle for this topic. Reading one curve tells you what is not limiting; comparing curves tells you what is.

Step 3 — Translate it into the real world ▶

Just after dawn: light is limiting. Noon on a bright summer day: carbon dioxide is usually limiting, because 0.04 % is not much. Unheated glasshouse in winter: temperature.

The commercial answer is a paraffin heater, which raises temperature and carbon dioxide at once. That is a real horticultural practice, not an exam invention.

And the definition to keep saying: the factor in shortest supply at that moment, not the factor that has run out.

⭐ CORE FRAMEWORK 3

A leaf solves four problems, and every structure belongs to one of them

Problem 1 — Catch as much light as possible ▶

Large surface area to intercept light. Transparent cuticle and upper epidermis with no chloroplasts so the light is not blocked on its way in.

Palisade mesophyll directly under the epidermis: tall cells, tightly packed, the most chloroplasts, positioned nearest the light. This is the main site of photosynthesis.

Problem 2 — Get carbon dioxide to every cell ▶

Stomata let CO₂ in; air spaces in the spongy mesophyll carry it to every cell by diffusion and provide a large moist surface for it to dissolve into.

Thin leaf = short diffusion distance. Straight out of Topic 3: rate of diffusion falls as distance increases.

Problem 3 — Deliver water, remove sugar ▶

Xylem (upper half of the vascular bundle, dead, hollow) brings water and mineral ions up from the roots — including the nitrate and magnesium this topic needs.

Phloem (lower half, living) carries sucrose and amino acids away. Removing the product keeps the process going, exactly as removing oxygen keeps diffusion going in a respiring cell.

The bundle also supports the blade, holding it flat and out in the light.

Problem 4 — Do all of that without drying out ▶

Waxy cuticle, thickest on the exposed upper surface, is waterproof but transparent. Stomata mostly on the lower, shaded surface, where evaporation is slower.

Guard cells close the pore when water is short — and that is the trade-off at the heart of plant life: close the stoma and you save water but starve the chloroplasts of carbon dioxide.

The mechanism is osmosis (Topic 3): turgid guard cells bow apart because the inner wall is thicker than the outer; flaccid ones sag together and close.

❌ "Why Is This Wrong?" Exercises

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Six real student answers. Decide what is wrong and what it would score before you reveal the flaw.

Exercise 1: "State the word equation for photosynthesis, including the conditions. [3]"

Student’s Answer"Carbon dioxide + water + sunlight → food + oxygen."

The FlawThree errors in one line, and it can score zero. Sunlight is not a raw material and must not appear on the left of the arrow; “food” is not a chemical; and the conditions the question explicitly asked for are missing.

Correct Answer"carbon dioxide + water → glucose + oxygen [1], in the presence of light [1] and chlorophyll [1]" — with the conditions written above or below the arrow, not as reactants.

Key RuleLight supplies energy, not matter. Anything on the left of the arrow must end up inside a product molecule, and no atom of glucose came from a photon.

Exercise 2: "Explain what is meant by a limiting factor. [2]"

Student’s Answer"It is the factor that runs out first and stops photosynthesis happening."

The FlawTwo mistakes. A limiting factor does not have to run out — carbon dioxide is limiting at 0.04 % all day and never reaches zero. And it does not stop photosynthesis; it limits the rate, which continues perfectly well at a lower value.

Correct Answer"A limiting factor is the environmental factor that is in the shortest supply at that moment [1] and therefore limits the rate of photosynthesis — if it is increased the rate increases, and increasing any other factor has no effect [1]."

Key RuleWhenever you write the word “stops” in this topic, check whether you mean “slows”. Rates are almost never zero in biology; they are just lower than they could be.

Exercise 3: "Describe how you would show that light is needed for photosynthesis. [5]"

Student’s Answer"Cover half a leaf with foil, leave the plant in the sun all day, then test the leaf with iodine. The covered part will stay brown and the uncovered part will go blue-black, which shows light is needed."

The FlawThe observations and the conclusion are right, so this scores something — but it drops the two marks that separate a good answer from an average one. Destarching is missing, so the starch found could have been there beforehand. And the testing procedure is compressed into “test with iodine”, skipping the boiling water and the ethanol.

Correct Answer"Destarch the plant by leaving it in complete darkness for 48 hours [1]. Cover part of one leaf with aluminium foil and leave the plant in bright light for six hours [1]. Remove the leaf, boil in water to kill it and break the membranes, then place it in hot ethanol in a water bath to remove the chlorophyll, then rinse in water to soften it [1]. Add iodine solution [1]. The covered region remains orange-brown and the uncovered region turns blue-black; since the two regions were on the same leaf and differed only in light, light is necessary for photosynthesis [1]."

Key RuleWhen a question is worth five marks, count five separate ideas before you stop writing. Method questions in this topic almost always want destarch, set-up, control, test, result-plus-conclusion.

Exercise 4: "Explain why the rate of photosynthesis falls above 40 °C. [3]"

Student’s Answer"Because it is too hot and the enzymes get killed so the plant dies."

The Flaw“Killed” is the fatal word — enzymes are protein molecules, not organisms. “The plant dies” is also unwarranted: a plant survives 45 °C perfectly well for a short time; it is the rate that falls, not the organism that expires.

Correct Answer"The enzymes controlling photosynthesis are denatured [1]. The high temperature breaks the bonds holding the protein in its precise shape, so the active site changes shape [1] and the substrate no longer fits, so fewer enzyme–substrate complexes form and the rate falls [1]."

Key RuleThe examiner is testing one word and one mechanism. Denatured, plus active site changes shape so the substrate no longer fits. Say both every time and this three-mark question becomes automatic.

Exercise 5: "Explain why the tube containing pondweed in the dark turned the indicator yellow. [3]"

Student’s Answer"Because in the dark the plant cannot photosynthesise so it respires instead, and respiration produces carbon dioxide which turns the indicator yellow."

The FlawThis would probably gain two of the three marks, and it contains a real misconception worth rooting out. The phrase “respires instead” implies the plant switches between the two processes. It does not: respiration runs continuously, day and night. In the dark, photosynthesis simply stops, leaving respiration unopposed.

Correct Answer"In the dark there is no light energy, so photosynthesis cannot occur [1]. Respiration continues, as it does at all times, releasing carbon dioxide into the water [1]. There is therefore a net increase in carbon dioxide concentration, and the indicator turns yellow [1]."

Key RuleWatch for the word “instead” in your own writing. In biology it usually smuggles in a switch that does not exist. The honest word here is net.

Exercise 6: "The diagram shows a section through a leaf. Identify layer B and explain how it is adapted for photosynthesis. [3]"

Student’s Answer"B is the palisade layer. It is where photosynthesis happens."

The FlawOne mark out of three. The identification is correct; the “explanation” restates the function instead of explaining the adaptation. Saying that the site of photosynthesis is where photosynthesis happens tells the examiner nothing about why the structure suits the job.

Correct Answer"B is the palisade mesophyll [1]. Its cells contain many chloroplasts and are positioned directly beneath the transparent upper epidermis, closest to the light, so they absorb the maximum amount of light energy [1]. The cells are tall and tightly packed, so light passes through a long column of chloroplasts inside a single cell with few gaps for it to escape through [1]."

Key RuleUse the sentence pattern: [structure] is [description], which means [property], so [consequence for photosynthesis]. If your explanation could be written without ever mentioning light, carbon dioxide or water, it is not an explanation.

✍️ Ultra-Detailed Practice Questions

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Ten Cambridge-style challenge questions. Write a full answer first, then reveal the model answer with the mark allocation and the examiner’s notes.

Question 1

[6 marks]

State the word equation for photosynthesis including the conditions required, give the balanced chemical equation, and explain why light is not listed as a raw material. [6]

Model AnswerWord equation: carbon dioxide + water → glucose + oxygen [1], in the presence of light and chlorophyll written above or below the arrow [1].

Balanced equation: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ [1], correctly balanced with the state of each formula recognisable [1].

Why light is not a raw material: a raw material is a substance that is built into the product [1]; light is a form of energy, not matter, so no atom of glucose comes from it — chlorophyll transfers the energy from light into energy in chemicals to drive the reaction [1].

Examiner’s NotesThe third part is the challenge element and it separates candidates who have memorised the equation from those who understand it. Two very common losses: writing “food” or “sugar” instead of glucose, and putting sunlight on the left-hand side of the arrow as if it were a reactant.

Question 2

[8 marks]

A student investigates the effect of light intensity on the rate of photosynthesis of pondweed by moving a lamp to different distances and counting bubbles per minute. Identify three weaknesses in this method and describe an improvement for each. [8]

Model AnswerWeakness 1 — heat from the lamp: moving the lamp closer raises the temperature of the water as well as the light intensity, so two variables change at once [1]. Improvement: place a glass tank or beaker of water between the lamp and the boiling tube to absorb the heat, and check the temperature with a thermometer at each distance [1].

Weakness 2 — bubbles are a crude measure: bubbles vary in size, so the number of bubbles is not proportional to the volume of oxygen released [1]. Improvement: collect the gas in a capillary tube or gas syringe and measure its volume per minute [1].

Weakness 3 — distance is not intensity: light intensity is proportional to 1/distance², so equal steps in distance are not equal steps in intensity, and background light from the room is not accounted for [1]. Improvement: measure the intensity directly with a light meter, and darken the room so the lamp is the only source [1].

Any two further creditworthy points, one mark each: allow the plant several minutes to equilibrate at each new setting before counting; repeat at each distance and take a mean; add sodium hydrogencarbonate to the water so carbon dioxide does not become limiting; use the same piece of pondweed throughout [2].

Examiner’s NotesEight marks means eight ideas, and the structure of the question tells you they come in pairs. Weaknesses on their own are worth half; each must be followed by a matching improvement. Notice that “repeat the experiment” addresses reliability, not validity — useful, but it is not the answer to any of the first three weaknesses.

Question 3

[9 marks]

Describe how you would use a destarched variegated plant to demonstrate that chlorophyll is necessary for photosynthesis, and explain how the result supports your conclusion. Include a labelled sketch of the expected result in words. [9]

Model AnswerMethod: destarch the plant by keeping it in complete darkness for 48 hours so that any starch found later must have been made during the investigation [1]. Record or sketch which parts of a chosen leaf are green and which are white before the experiment [1]. Leave the plant in bright light for 6–8 hours [1]. Remove the leaf and boil it in water to kill it and break the cell membranes [1]. Transfer it to hot ethanol in a water bath (not over a naked flame, as ethanol is flammable) to dissolve out the chlorophyll [1]. Rinse in water to soften the brittle leaf [1]. Cover with iodine solution [1].

Expected result: the regions that were green turn blue-black; the regions that were white remain orange-brown [1].

Conclusion and reasoning: the white regions contain no chlorophyll but were exposed to the same light, temperature, water supply and carbon dioxide as the green regions, being part of the same leaf on the same plant — therefore the only difference between them is chlorophyll, so chlorophyll must be necessary for photosynthesis [1].

Examiner’s NotesThe mark for recording the pattern of variegation beforehand is the one that almost nobody offers, and it matters: the ethanol removes the colour, so after testing you can no longer tell which regions were green. The final reasoning mark requires the phrase “the only difference” or an equivalent — simply stating the colours is a description, not a conclusion.

Question 4

[8 marks]

The table shows the carbon dioxide concentration inside a sealed glasshouse of tomato plants over one summer day.

time	04:00	08:00	12:00	16:00	20:00	00:00
CO₂ / %	0.051	0.038	0.019	0.022	0.041	0.049

(a) Describe the pattern shown. [3] (b) Explain the value at 04:00. [2] (c) Explain what the value at 12:00 tells the grower and suggest what she should do. [3]

Model Answer(a) The concentration falls from 0.051 % at 04:00 to a minimum of 0.019 % at 12:00 [1]; it then rises through the afternoon and evening, reaching 0.049 % at midnight [1]; the fall is steeper than the rise, and the value at midnight has almost returned to that at 04:00 [1].

(b) At 04:00 it is dark, so there is no photosynthesis [1], but the plants have been respiring all night, releasing carbon dioxide into the sealed air, so the concentration has risen well above the 0.04 % of ordinary air [1].

(c) At midday the concentration has fallen to less than half its night-time value, so carbon dioxide has become the limiting factor — the plants are removing it faster than respiration replaces it, and extra light or warmth would no longer raise the rate [1]. She should enrich the air with carbon dioxide, for example from a cylinder or a paraffin burner, or ventilate the glasshouse so that outside air at 0.04 % replaces the depleted air [1]. This would raise the rate of photosynthesis and therefore the yield [1].

Examiner’s NotesPart (a) is pure description and the marks are for quoted figures plus the shape — candidates who write “it goes down then up” score one mark at most. Part (c) is the challenge: notice that the grower has identified her limiting factor without performing any experiment on the plants, purely by monitoring the air. A paraffin burner is the elegant answer because it supplies heat and carbon dioxide together.

Question 5

[8 marks]

Explain how the following features of a leaf adapt it for photosynthesis: the waxy cuticle, the upper epidermis, the palisade mesophyll, the air spaces of the spongy mesophyll, the stomata and the xylem. [8]

Model AnswerWaxy cuticle: waterproof, so it reduces water loss by evaporation [1]; it is also transparent, so it does not block light reaching the mesophyll [1].
Upper epidermis: a single layer of cells containing no chloroplasts, so it is transparent and light passes straight through to the palisade layer beneath [1].
Palisade mesophyll: contains the most chloroplasts and lies closest to the light, so it absorbs the maximum light energy [1]; the cells are tall and tightly packed so light passes through a long column of chloroplasts within one cell [1].
Air spaces: allow rapid diffusion of carbon dioxide from the stomata to every mesophyll cell, and provide a large moist surface area for gases to dissolve into before crossing the cell membranes [1].
Stomata: pores that allow carbon dioxide to diffuse in and oxygen to diffuse out; their position on the shaded lower surface reduces water loss [1].
Xylem: carries water (a raw material) and mineral ions such as nitrate and magnesium to the mesophyll cells [1].

Examiner’s NotesSix structures, eight marks — so two structures carry two marks each, and you have to decide which. The cuticle and the palisade layer are the two with an obvious second point. Every line here follows the same pattern: description → property → consequence for photosynthesis. Answers that stop at the description score nothing at all in this style of question.

Question 6

[7 marks]

A student claims that because a plant releases oxygen during the day, it must be respiring only at night. Evaluate this claim, and describe an experiment using hydrogencarbonate indicator that would settle the question. [7]

Model AnswerEvaluation: the claim is incorrect [1]. Respiration occurs in every living cell continuously, day and night, because the plant needs energy for active transport, protein synthesis and growth at all times [1]. During the day photosynthesis also occurs, and it is faster than respiration, so there is a net release of oxygen and a net uptake of carbon dioxide — the release of oxygen shows the balance, not the absence of respiration [1].

Experiment: set up sealed tubes of red hydrogencarbonate indicator, each containing an equal mass of pondweed: tube A in bright light, tube B wrapped in aluminium foil so that it receives no light, and tube C with no pondweed as a control [1]. Keep temperature and time the same for all three, using a water bath [1].

Expected results and interpretation: tube A turns purple (net removal of carbon dioxide, photosynthesis faster than respiration); tube B turns yellow, showing carbon dioxide is being released, which can only be from respiration [1]; tube C stays red, showing the colour changes were caused by the pondweed and not by light, warmth or time [1].

Examiner’s NotesThe experiment must be capable of disproving the claim, which is why the dark tube is the essential one — it isolates respiration. Weak answers describe only the light tube, which cannot settle the question at all. Note also that this experiment can never show respiration is absent in the light; it shows the net effect. Saying so explicitly is what a top-band answer looks like.

Question 7

[6 marks]

Two barley seedlings are grown in aerated culture solutions. Seedling X reaches 7 cm with small pale leaves; seedling Y reaches 20 cm with yellowing between the veins of its older leaves. Deduce which mineral ion is deficient in each and explain the biological chain that produces each symptom. [6]

Model AnswerSeedling X — nitrate ions [1]. Nitrate supplies the nitrogen that carbohydrates lack [1]; without it the plant cannot convert glucose into amino acids and therefore cannot make proteins, which are needed for new cytoplasm, enzymes and cell division — so growth is severely stunted [1].

Seedling Y — magnesium ions [1]. Magnesium is required to make chlorophyll [1]; without chlorophyll the leaves cannot be green and appear yellow (chlorosis), and less light energy can be transferred, so the rate of photosynthesis falls and growth will eventually slow too [1].

Examiner’s NotesTwo marks each are for the chain, not for the ion. “Nitrate is for growth” is one step; the mark scheme wants nitrogen → amino acids → proteins. Remember also that these are mineral ions absorbed by active transport, not food — describing them as the plant’s food is a genuine error, because they supply no energy.

Question 8

[7 marks]

Guard cells are the only cells of the epidermis to contain chloroplasts. Explain how a stoma opens, why this is advantageous during the day, and what problem it creates for the plant in dry conditions. [7]

Model AnswerOpening: water enters the guard cells by osmosis, moving down a water potential gradient through the partially permeable membrane [1], so the cells become turgid and turgor pressure rises [1]. The inner wall facing the pore is thicker and less elastic than the outer wall, so only the outer wall stretches and each cell curves away from its partner, opening the pore [1].

Advantage by day: the open stoma allows carbon dioxide to diffuse into the air spaces and reach the mesophyll cells, and allows oxygen to diffuse out [1]; the chloroplasts in the guard cells let them respond to light directly, so the pore opens when photosynthesis is about to require carbon dioxide [1].

Problem in dry conditions: an open stoma also allows water vapour to be lost from the moist surfaces of the air spaces, so the plant may wilt [1]. Closing the stoma conserves water but cuts off the supply of carbon dioxide, which then becomes the limiting factor and the rate of photosynthesis falls [1].

Examiner’s NotesThis is the most synoptic question in the topic: osmosis and turgor from Topic 3, structure from 6.2 and limiting factors from 6.1, all in seven marks. The final idea — that closing the stoma makes carbon dioxide limiting — is the one that lifts an answer into the top band, because it connects the two halves of Topic 6 to each other.

Question 9

[6 marks]

A gardener grows lettuce in an unheated glasshouse in winter and in the same glasshouse in summer. In winter, adding extra lighting increases the yield very little. In summer, adding the same extra lighting increases the yield substantially. Explain these observations. [6]

Model AnswerWinter: the glasshouse is unheated and the outside temperature is low, so the temperature inside is low [1]. At low temperature the enzymes controlling photosynthesis have less kinetic energy, so there are fewer successful collisions per second and the rate is low [1]. Temperature is therefore the limiting factor, so increasing the light intensity has almost no effect — increasing a factor that is not limiting does not increase the rate [1].

Summer: the temperature inside the glasshouse is much higher, closer to the optimum for the enzymes, so temperature is no longer limiting [1]. Light then becomes the factor in shortest supply, particularly early and late in the day and on dull days, so increasing the light intensity increases the rate of photosynthesis and therefore the yield [1]. Any valid extension for the final mark: if lighting is increased far enough in summer the rate will plateau again as carbon dioxide becomes limiting, so the gardener should also consider CO₂ enrichment [1].

Examiner’s NotesThe whole question turns on a single principle: increasing a factor that is not limiting changes nothing. Candidates who know that sentence can answer any version of this question. The sixth mark is deliberately open-ended, and predicting the next limiting factor is the sort of forward reasoning challenge papers reward.

Question 10

[9 marks]

Design an investigation to find the optimum temperature for photosynthesis in a species of pondweed. Your answer should identify the independent, dependent and controlled variables, describe the method, explain how you would obtain valid and reliable results, and state the shape of graph you would expect and why. [9]

Model AnswerVariables: independent = temperature of the water, at a minimum of five values, for example 10, 15, 20, 25, 30 and 35 °C [1]. Dependent = volume of oxygen collected per minute [1]. Controlled = light intensity and distance of lamp, carbon dioxide concentration (add sodium hydrogencarbonate), the same piece of pondweed, its mass or length, the volume of water, and the time allowed at each temperature [2].

Method: place the pondweed in a boiling tube of water containing sodium hydrogencarbonate solution, inverted funnel and gas collection tube, and stand the tube in a thermostatically controlled water bath [1]. Place a glass tank of water between the lamp and the tube so the lamp does not alter the temperature [1]. Allow the plant five minutes to equilibrate at each temperature before collecting gas, then measure the volume of oxygen collected in a fixed time [1].

Validity and reliability: validity comes from changing only temperature and controlling everything else; reliability comes from repeating three times at each temperature and calculating a mean, discarding anomalous results [1].

Expected graph: the rate rises with temperature to an optimum and then falls steeply. It rises because the molecules have more kinetic energy so there are more successful collisions between enzyme and substrate; it falls above the optimum because the enzymes are denatured — the active site changes shape and the substrate no longer fits [1].

Examiner’s NotesNine marks, and the structure of the question is a checklist — answer it in the order asked. Two marks are set aside for controlled variables, so list at least four. The most frequently forgotten details are the equilibration time and the heat shield between lamp and plant; the most frequently confused pair is validity (controls) and reliability (repeats).