AS & A Level Biology (9700)
Photosynthesis Banker
39 questions · about 6% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Chloroplast structure & pigments · Core · Paper 4 style · 7 marksAn electron micrograph of a chloroplast from a mango leaf shows:A: two membranes surrounding the whole organelleB: stacks of flattened membrane discs, joined to each other by single membranes (intergranal lamellae)C: a fluid-filled matrix around the membranesD: large, pale, oval granules in the matrix, made of the plant's main storage polysaccharide
- (a) Name structures A, B, C and D. [4]
- (b) State where in the chloroplast (i) the light-dependent stage and (ii) the Calvin cycle take place. [2]
- (c) The stroma also contains circular DNA and 70S ribosomes. Suggest the function of these. [1]
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- (a) A chloroplast envelope (double membrane); B granum (stack of thylakoids); C stroma; D starch grain.
1 mark each. Accept 'thylakoids' for B.
- (b) (i) thylakoid membranes (grana); (ii) stroma.
1 mark each.
- (c) To make some of the chloroplast's own proteins (e.g. part of rubisco / photosystem proteins).
1 mark: protein synthesis within the chloroplast / codes for and makes chloroplast proteins.
A student extracted pigments from spinach leaves and separated them by paper chromatography. The solvent front moved 8.0 cm from the origin. Four spots were seen, which had moved 3.6 cm, 5.2 cm, 5.7 cm and 7.6 cm.Reference Rf values in this solvent: carotene 0.95; xanthophyll 0.71; chlorophyll a 0.65; chlorophyll b 0.45
- (a) Calculate the Rf value of each spot. Give each answer to 2 decimal places. [2]
- (b) Use the reference table to identify the pigment in each spot. [2]
- (c) Explain why the pigments separate on the chromatography paper. [2]
Show answer
- (a) Spot 1: 0.45; spot 2: 0.65; spot 3: 0.71; spot 4: 0.95
Rf = distance moved by spot ÷ distance moved by solvent front. 2 marks all four correct; 1 mark for two or three.
- (b) Spot 1 chlorophyll b; spot 2 chlorophyll a; spot 3 xanthophyll; spot 4 carotene.
2 marks all four; 1 mark for two or three.
- (c) The pigments have different solubilities in the solvent and different attraction (adsorption) to the paper; more soluble/less attracted pigments are carried further by the solvent.
1 mark: differ in solubility in solvent. 1 mark: differ in affinity for/adsorption to paper, so travel different distances.
- (a) Distinguish between an absorption spectrum and an action spectrum. [2]
- (b) Chlorophyll a absorbs very little light between 450 nm and 550 nm, yet the rate of photosynthesis at 480 nm is still quite high. Explain why. [3]
- (c) Explain why most leaves appear green. [1]
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- (a) An absorption spectrum shows how much light of each wavelength is absorbed by a pigment (or pigments); an action spectrum shows the rate of photosynthesis at each wavelength of light.
1 mark: absorption = amount of light absorbed at each wavelength. 1 mark: action = rate of photosynthesis at each wavelength.
- (b) Accessory pigments such as carotene and xanthophylls (and chlorophyll b) absorb light at these wavelengths (blue-green); they pass the energy to chlorophyll a in the reaction centre of the photosystem; so light at these wavelengths is still used for photosynthesis.
1 mark: accessory pigments named (carotenoids/chlorophyll b). 1 mark: absorb blue/blue-green light around 450–500 nm. 1 mark: transfer energy to primary pigment / reaction centre chlorophyll a.
- (c) Green light (about 500–560 nm) is absorbed least by the pigments, so it is reflected or transmitted to the eye.
1 mark: green wavelengths reflected/not absorbed.
- (a) Describe the structure of a photosystem and explain how it captures light energy. [4]
- (b) Some red seaweeds live in coastal seas at depths where most red and blue light has been absorbed by the water and mainly green light remains. These seaweeds contain a red accessory pigment called phycoerythrin. Suggest how phycoerythrin helps them to grow at these depths. [2]
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- (a) A photosystem is a cluster of pigment molecules held by proteins in the thylakoid membrane; a light-harvesting system (antenna complex) of accessory pigments and chlorophyll molecules surrounds a reaction centre; pigments absorb photons and pass energy from molecule to molecule towards the reaction centre; the reaction centre contains a pair of chlorophyll a molecules (P680 in PSII, P700 in PSI) which, when excited, emit an electron to an electron acceptor.
1 mark each for any four of: in thylakoid membrane, pigments + proteins; light-harvesting/antenna complex of many pigments; energy transferred to reaction centre; reaction centre = chlorophyll a (P680/P700); excited electron emitted / passed to electron acceptor (photoactivation).
- (b) Phycoerythrin absorbs green light (it looks red because it reflects red); the absorbed energy is passed to chlorophyll a in the reaction centres, so the seaweed can still photosynthesise using the light available at depth.
1 mark: absorbs green light / wavelengths that reach that depth. 1 mark: energy passed to reaction centre chlorophyll so photosynthesis continues.
Chloroplasts were broken open and separated into a thylakoid fraction and a stroma fraction. Each was incubated under different conditions.Tube 1: thylakoids, light, water — oxygen releasedTube 2: stroma, light, water — no oxygen releasedTube 3: stroma, dark, CO₂ + ATP + reduced NADP — CO₂ fixed into sugarsTube 4: stroma, dark, CO₂ only — almost no CO₂ fixed
- (a) Explain the results for tubes 1 and 2. [3]
- (b) Explain the results for tubes 3 and 4. [3]
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- (a) Thylakoids contain the photosystems and electron transport chains; in light, photolysis of water releases oxygen (tube 1); the stroma has no photosystems, so no oxygen is produced (tube 2); without light, no photoactivation and no oxygen.
1 mark: thylakoids are site of light-dependent stage / photosystems. 1 mark: photolysis of water in light releases O₂. 1 mark: stroma lacks photosystems/chlorophyll so no O₂.
- (b) The stroma contains the Calvin cycle enzymes (rubisco); ATP and reduced NADP, normally supplied by the light-dependent stage, were added, so CO₂ could be fixed even in the dark (tube 3); without ATP and reduced NADP, GP cannot be reduced to triose phosphate and RuBP cannot be regenerated, so fixation stops quickly (tube 4).
1 mark: stroma contains Calvin cycle enzymes/rubisco. 1 mark: ATP and reduced NADP are the products of the light-dependent stage needed, so light not directly required. 1 mark: without them GP not reduced / RuBP not regenerated so fixation stops.
The rate of photosynthesis of a cucumber plant was measured at two CO₂ concentrations. Temperature was kept at 25 °C.Light intensity / arbitrary units: 0, 2, 4, 6, 8, 10Rate at 0.04 % CO₂ / arbitrary units: 0, 3.2, 5.4, 5.9, 6.0, 6.0Rate at 0.10 % CO₂ / arbitrary units: 0, 3.2, 6.1, 8.8, 10.4, 10.8
- (a) Explain what is meant by a limiting factor. [1]
- (b) Identify the factor limiting the rate of photosynthesis at 2 arbitrary units of light intensity at both CO₂ concentrations. Explain your answer. [2]
- (c) Identify the factor limiting the rate at 10 arbitrary units of light intensity in 0.04 % CO₂. Explain your answer. [2]
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- (a) The factor that is in shortest supply (furthest from its optimum) and so restricts the rate of a process; increasing it increases the rate.
1 mark.
- (b) Light intensity; at low light intensity the rate is the same at both CO₂ concentrations and rises as light intensity increases.
1 mark: light intensity. 1 mark: rate same for both CO₂ levels / rate increases with light intensity.
- (c) Carbon dioxide concentration; the rate has levelled off so increasing light has no effect, but the rate is higher at 0.10 % CO₂.
1 mark: CO₂ concentration. 1 mark: plateau with light, but higher rate at higher CO₂ (data quote: 6.0 vs 10.8).
A tomato grower compared yields in two greenhouses. One had normal air (0.04 % CO₂) and the other was enriched to 0.10 % CO₂. Light and temperature were the same.Normal air: 9.6 kg of tomatoes per m²Enriched: 14.2 kg of tomatoes per m²
- (a) Calculate the percentage increase in yield in the enriched greenhouse. Give your answer to 3 significant figures. [2]
- (b) Explain why increasing the CO₂ concentration increases the yield. [3]
- (c) Suggest why the grower also keeps the greenhouse warm in winter, and why there is no benefit in raising the CO₂ concentration further on dull, cloudy days. [2]
Show answer
- (a) 47.9 %
M1: (14.2 − 9.6) ÷ 9.6 × 100. A1: 47.9 %.
- (b) CO₂ is often the limiting factor in bright light; more CO₂ means rubisco fixes more CO₂ with RuBP; more GP and triose phosphate are made, so more sugars/biomass are produced (faster growth, more fruit).
1 mark: CO₂ was limiting. 1 mark: more CO₂ combines with RuBP (via rubisco) / faster Calvin cycle. 1 mark: more triose phosphate → more sugars, starch, growth.
- (c) Warmth increases the rate of enzyme-controlled reactions (Calvin cycle), so temperature does not limit the rate; on dull days light intensity is the limiting factor, so extra CO₂ cannot increase the rate.
1 mark: warmth so temperature is not limiting / enzymes faster. 1 mark: on dull days light is limiting.
Rates of photosynthesis of a cotton leaf were measured at different temperatures in high light and high CO₂.Temperature / °C: 10, 15, 20, 25, 30, 35, 40Rate / arbitrary units: 2.1, 3.4, 5.0, 6.6, 7.8, 5.2, 1.5
- (a) Describe the effect of temperature on the rate of photosynthesis shown by these results. [2]
- (b) Explain the effect of temperature on the rate of photosynthesis between 10 °C and 30 °C and above 30 °C. [4]
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- (a) The rate increases from 10 °C to a maximum (optimum) at 30 °C (from 2.1 to 7.8 units), then decreases sharply at 35 °C and 40 °C (to 1.5 units at 40 °C).
1 mark: increases to optimum at 30 °C. 1 mark: then decreases above 30 °C; data quote with units.
- (b) Below the optimum, rising temperature gives enzymes (e.g. rubisco) and substrates more kinetic energy; more frequent collisions / more enzyme–substrate complexes form, so the Calvin cycle is faster. Above the optimum, enzymes begin to denature: hydrogen bonds break and the active site changes shape, so fewer enzyme–substrate complexes form; stomata may close at high temperature, reducing CO₂ uptake.
1 mark: Calvin cycle/light-independent stage is enzyme-controlled. 1 mark: more kinetic energy, more collisions / E–S complexes. 1 mark: above optimum enzymes denature, active site shape changes. 1 mark: stomatal closure reduces CO₂ entry (or increased photorespiration not required).
A piece of pondweed was placed in hydrogencarbonate solution and a lamp was placed at different distances. The number of bubbles released per minute was counted.Distance / cm: 10, 15, 20, 30, 40Bubbles per minute: 17, 16, 15, 8, 4
- (a) Light intensity is proportional to 1/d², where d is the distance from the lamp. Calculate the relative light intensity (10 000/d²) for each distance. [2]
- (b) Describe the relationship between relative light intensity and the rate of oxygen production. [2]
- (c) Suggest two factors that may be limiting the rate when the lamp is 10 cm away and explain why the lamp's heat might be a source of error. [3]
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- (a) 10 cm: 100; 15 cm: 44.4; 20 cm: 25.0; 30 cm: 11.1; 40 cm: 6.25 (6.3)
2 marks all correct; 1 mark for three or four.
- (b) Rate increases as light intensity increases from 6.25 to 25 (from 4 to 15 bubbles per minute) but levels off at higher intensities (16–17 bubbles per minute at 44.4 and 100).
1 mark: increases at low intensity. 1 mark: levels off/plateaus at high intensity with data.
- (c) CO₂ (hydrogencarbonate) concentration; temperature; (number of chloroplasts/amount of chlorophyll or enzyme). A lamp close to the plant warms the water, so temperature changes as well as light intensity, which also affects the rate (the independent variable is no longer the only factor changing).
1 mark each for two limiting factors (CO₂ concentration, temperature, amount of chlorophyll/rubisco). 1 mark: heat raises water temperature, which changes rate – use a heat shield/LED/water bath.
- (a) State what is meant by photolysis of water and give the equation for it. [2]
- (b) State what happens to each product of photolysis. [3]
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- (a) The splitting of water using light energy (at photosystem II): 2H₂O → 4H⁺ + 4e⁻ + O₂
1 mark: splitting of water by light (energy). 1 mark: correct balanced equation (accept H₂O → 2H⁺ + 2e⁻ + ½O₂).
- (b) Electrons replace those lost from the chlorophyll in photosystem II; protons (H⁺) build up in the thylakoid space and are then used (with electrons) to reduce NADP; oxygen diffuses out of the leaf (or is used in respiration).
1 mark each: e⁻ replace electrons lost by PSII; H⁺ reduce NADP (after contributing to gradient for ATP synthesis); O₂ by-product released / used in respiration.
- (a) Compare cyclic and non-cyclic photophosphorylation by giving four differences. [4]
- (b) State one similarity between cyclic and non-cyclic photophosphorylation. [1]
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- (a) Cyclic involves PSI only; non-cyclic involves PSI and PSII. Cyclic produces ATP only; non-cyclic produces ATP and reduced NADP. Cyclic involves no photolysis of water / no O₂ released; non-cyclic involves photolysis and releases O₂. In cyclic, electrons return to PSI; in non-cyclic, electrons end up on NADP (electrons from water replace those lost from PSII).
1 mark for each difference, max 4: photosystems involved; products (ATP only vs ATP + reduced NADP); photolysis/oxygen; fate of electrons.
- (b) Both make ATP by chemiosmosis using ATP synthase / both use an electron transport chain in the thylakoid membrane / both involve PSI and photoactivation of chlorophyll.
1 mark.
- (a) Describe the events of non-cyclic photophosphorylation in the thylakoid membrane. [6]
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- (a) Light absorbed by photosystem II; chlorophyll a in the reaction centre is photoactivated and emits excited electrons to an electron acceptor; electrons are passed along an electron transport chain of carriers to photosystem I, losing energy; this energy is used to pump H⁺ into the thylakoid space; light excites PSI, whose electrons pass to a second carrier and then reduce NADP (with H⁺) to form reduced NADP; electrons from PSII replace those lost by PSI; photolysis of water supplies electrons to PSII, releasing H⁺ and O₂; H⁺ diffuse back through ATP synthase (chemiosmosis), producing ATP from ADP and Pi.
1 mark each for any six of: light absorbed by PSII; photoactivation – electrons excited and emitted; electron acceptor; ETC, electrons lose energy; H⁺ pumped into thylakoid space / proton gradient; PSI excited; electrons reduce NADP (NADP reductase) to reduced NADP; PSII electrons replace those lost from PSI; photolysis replaces electrons in PSII, O₂ released; chemiosmosis through ATP synthase; ATP formed from ADP + Pi.
Isolated thylakoids were kept in the dark in an acidic buffer at pH 4.5 until the thylakoid space reached pH 4.5 (step 1). They were then transferred, still in the dark, to a buffer at pH 8.0 containing ADP and phosphate (step 2). ATP was produced for a short time.
- (a) Calculate how many times greater the concentration of H⁺ is in a solution at pH 4.5 than at pH 8.0. (pH = −log₁₀[H⁺]). Give your answer to 2 significant figures. [1]
- (b) Explain why ATP was made in step 2 even though the thylakoids were in the dark. [3]
- (c) A chemical that makes the thylakoid membrane permeable to H⁺ was added to illuminated chloroplasts. Predict and explain the effect on ATP production and on the reduction of NADP. [2]
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- (a) About 3200 times (3.2 × 10³)
10^(8.0 − 4.5) = 10^3.5 = 3162 → 3200.
- (b) The thylakoid space had a high H⁺ concentration (pH 4.5) compared with the outside (pH 8.0), so there was a proton gradient across the membrane; H⁺ diffused out through ATP synthase; the flow of protons drove the phosphorylation of ADP to ATP (chemiosmosis); light is normally needed only to generate the gradient.
1 mark: proton/H⁺ gradient across thylakoid membrane. 1 mark: H⁺ diffuse through ATP synthase (channel). 1 mark: energy from proton flow used to make ATP from ADP + Pi – chemiosmosis, light not directly needed.
- (c) ATP production falls (stops) because H⁺ leak back across the membrane, so no proton gradient forms to drive ATP synthase; reduction of NADP continues because electron flow along the chain still occurs.
1 mark: ATP production decreases, no proton gradient. 1 mark: NADP still reduced / electron transport continues.
- (a) Name the enzyme that catalyses the fixation of carbon dioxide in the Calvin cycle. [1]
- (b) Copy and complete the table to show the number of carbon atoms in each molecule.RuBP | GP | triose phosphate [2]
- (c) Outline how GP is converted to triose phosphate, naming the products of the light-dependent stage that are needed. [2]
- (d) State two uses of triose phosphate in the plant. [2]
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- (a) Rubisco (ribulose bisphosphate carboxylase/oxygenase).
1 mark.
- (b) RuBP 5; GP 3; triose phosphate 3
2 marks all correct; 1 mark for two.
- (c) GP is reduced to triose phosphate using hydrogen from reduced NADP and energy from ATP.
1 mark: reduction (by reduced NADP). 1 mark: ATP provides energy / phosphate.
- (d) Any two: regenerating RuBP; making glucose/sucrose/starch/cellulose; making lipids (glycerol and fatty acids); making amino acids.
1 mark each.
Cells of a single-celled green alga were kept in bright light with plenty of CO₂, and the concentrations of glycerate 3-phosphate (GP) and ribulose bisphosphate (RuBP) were measured. The light was then switched off. During the next minute the GP concentration increased and the RuBP concentration decreased.
- (a) Explain the changes in the concentrations of GP and RuBP after the light was switched off. [4]
- (b) In a second experiment the light was kept on, but the CO₂ concentration was suddenly reduced. Predict the changes in GP and RuBP concentrations and explain your prediction. [2]
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- (a) No light means no ATP or reduced NADP from the light-dependent stage; GP cannot be reduced to triose phosphate, so GP accumulates; RuBP is still used up as it combines with CO₂ (to form GP); RuBP cannot be regenerated from triose phosphate because there is no ATP (and no TP being made), so RuBP falls.
1 mark: no ATP/reduced NADP. 1 mark: GP not converted to TP so GP rises. 1 mark: RuBP still used in fixation of CO₂. 1 mark: RuBP not regenerated (needs ATP/TP) so falls.
- (b) RuBP rises and GP falls; less CO₂ is available to combine with RuBP, so RuBP is not used but is still regenerated from TP; GP is still converted to TP but less is formed.
1 mark: RuBP increases and GP decreases (both needed). 1 mark: less CO₂ fixed with RuBP while regeneration / GP reduction continue.
- (a) In the Calvin cycle, fixing one molecule of CO₂ and regenerating RuBP uses 3 molecules of ATP and 2 molecules of reduced NADP. Calculate the number of molecules of ATP and of reduced NADP needed to make one molecule of sucrose (C₁₂H₂₂O₁₁). [2]
- (b) Describe the Calvin cycle, including the regeneration of RuBP. [5]
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- (a) 36 ATP and 24 reduced NADP
M1: sucrose has 12 carbon atoms, so 12 CO₂ must be fixed. A1: 12 × 3 = 36 ATP and 12 × 2 = 24 reduced NADP.
- (b) In the stroma, CO₂ combines with RuBP (5C); catalysed by rubisco; forming an unstable 6C compound that splits into two molecules of GP (3C); GP is reduced to triose phosphate using reduced NADP; and ATP; five out of every six triose phosphates are used to regenerate RuBP; using ATP; one in six is used to make glucose and other organic compounds.
1 mark each for any five of: CO₂ + RuBP; rubisco; two GP formed (via unstable 6C intermediate); GP reduced to TP; reduced NADP and ATP used; most TP (5/6) regenerates RuBP; ATP used in regeneration; some TP leaves to make sugars, amino acids, lipids.
- (a) Plan an investigation to find the effect of light intensity on the rate of photosynthesis of an aquatic plant such as Elodea. Your plan should include how you would measure the rate and how you would control other variables. [6]
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- (a) Cut a piece of Elodea of fixed length (e.g. 8 cm) and place it upside down in a boiling tube of 1 % sodium hydrogencarbonate solution (excess CO₂); stand the tube in a beaker of water at 25 °C as a heat shield/water bath; place a lamp (LED) at distances 5, 10, 15, 20, 25 and 30 cm (or measure light intensity with a light meter); leave 5 minutes at each distance to acclimatise; collect the gas in a photosynthometer/capillary tube for a fixed time (e.g. 5 min) and measure the volume/length of bubble; repeat three times at each distance and calculate means; carry out in a darkened room so the lamp is the only light source.
1 mark each for any six of: IV light intensity varied by lamp distance (≥5 values) or light meter; same plant piece/length, same species; NaHCO₃ in excess for CO₂; temperature controlled with water bath/heat shield/LED; acclimatisation time at each new intensity; DV measured as volume of gas (photosynthometer/syringe) in fixed time (not just bubble count); darkened room / no other light; at least 3 repeats and mean; rate = volume ÷ time.
- (a) Outline a method using leaf discs to investigate the effect of the colour (wavelength) of light on the rate of photosynthesis. [6]
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- (a) Cut discs from the same leaf with a cork borer, avoiding large veins; infiltrate the discs with 0.2 % sodium hydrogencarbonate solution by drawing them into a syringe and pulling the plunger to remove air, so they sink; place ten discs in the same volume of hydrogencarbonate solution in each of several beakers; illuminate each with a different colour of LED (red, blue, green, white), each at the same light intensity measured with a light meter and the same distance; keep temperature constant with a water bath; record the time for five of the ten discs to float; repeat three times and use 1/time as a measure of rate.
1 mark each for any six of: discs same size and from same plant/leaf; infiltration with NaHCO₃ using syringe so discs sink; same number of discs and volume of solution; different colours via filters/LEDs; same light intensity (light meter) or same distance and lamp; constant temperature; DV time for (e.g.) half of the discs to float; rate = 1/t; repeats and mean; control in dark.
To study the Hill reaction, a chloroplast suspension is prepared by blending spinach leaves in an ice-cold buffer solution that has the same water potential as the leaf cells. DCPIP is a blue dye that becomes colourless when it is reduced.
- (a) Explain why the leaves are blended in a buffer solution that is ice-cold and has the same water potential as the leaf cells. [2]
- (b) Plan an investigation into the effect of temperature on the rate of the Hill reaction, using a chloroplast suspension and DCPIP. [6]
Show answer
- (a) Ice-cold: slows enzyme activity (e.g. enzymes that damage chloroplasts) / reduces damage. Same water potential: prevents water entering or leaving chloroplasts by osmosis, so they do not burst or shrink. Buffer keeps pH constant so proteins are not denatured.
1 mark for ice-cold reason. 1 mark for same water potential or buffer reason.
- (b) Prepare chloroplast suspension by blending spinach leaves in ice-cold buffer, filtering and centrifuging, then resuspending the pellet; set up tubes with the same volume of chloroplast suspension and the same volume and concentration of DCPIP; equilibrate tubes in water baths at five temperatures, e.g. 10, 20, 30, 40, 50 °C; place each tube at the same distance from the same lamp; measure the time for the blue DCPIP to become colourless, or measure absorbance with a colorimeter (red filter) every 30 s; rate = 1/time; controls: tube kept in the dark and tube with boiled chloroplasts; three repeats at each temperature and calculate means.
1 mark each for any six of: method for extracting chloroplasts (blend, filter, centrifuge); same volumes/concentrations of DCPIP and suspension; ≥5 temperatures with thermostatic water bath; equilibrate before mixing; constant light intensity (same lamp, same distance); DV time to decolourise or colorimeter reading against colour standard; rate = 1/t; control in dark or with boiled chloroplasts; repeats and means.
An aquatic plant with a fresh mass of 2.5 g was placed in a photosynthometer. In 5 minutes, the gas collected formed a bubble 18 mm long in a capillary tube with an internal radius of 0.4 mm.
- (a) Calculate the volume of gas collected, using volume = π r² l. Give your answer in mm³ to 3 significant figures. [2]
- (b) Calculate the rate of gas production per gram of plant in mm³ min⁻¹ g⁻¹. Give your answer to 2 significant figures. [2]
- (c) Suggest why the rate of gas production underestimates the true rate of photosynthesis. [1]
Show answer
- (a) 9.05 mm³
M1: π × 0.4² × 18. A1: 9.05 mm³.
- (b) 0.72 mm³ min⁻¹ g⁻¹
M1: 9.05 ÷ 5 = 1.81 mm³ min⁻¹. A1: 1.81 ÷ 2.5 = 0.72 mm³ min⁻¹ g⁻¹.
- (c) Some oxygen produced is used in respiration by the plant (or dissolves in the water), so not all is collected.
1 mark: oxygen used in respiration / dissolves in water.
A Hill-reaction experiment measured the time for DCPIP to be decolourised by a chloroplast suspension at two temperatures.20 °C: 240 s, 252 s, 246 s30 °C: 150 s, 156 s, 300 s
- (a) Identify the anomalous result and explain why it should not be included in the mean. [1]
- (b) Calculate the mean time at each temperature, leaving out the anomalous result, and then the rate of reaction as 1000/mean time. Give rates to 2 decimal places. [3]
- (c) Calculate the percentage increase in rate from 20 °C to 30 °C. Give your answer to 3 significant figures. [1]
Show answer
- (a) 300 s at 30 °C; it is far from the other two readings (150 and 156 s), so it is probably due to an error (e.g. wrong volume or tube shaded) and would distort the mean.
1 mark: 300 s at 30 °C identified with reason.
- (b) 20 °C: mean 246 s, rate 4.07; 30 °C: mean 153 s, rate 6.54 (rates as 1000/s)
1 mark: mean 246 s. 1 mark: mean 153 s (anomaly excluded). 1 mark: rates 4.07 and 6.54.
- (c) 60.8 %
(6.536 − 4.065) ÷ 4.065 × 100 = 60.8 %.
A student uses a photosynthometer to measure the volume of gas given off by a piece of pondweed in sodium hydrogencarbonate solution when a lamp is placed at different distances.
- (a) State the independent variable and the dependent variable in this investigation. [2]
- (b) State three variables that should be standardised and explain how each could be kept constant. [3]
- (c) Explain why the sodium hydrogencarbonate solution should be in excess. [1]
Show answer
- (a) Independent: light intensity (distance of lamp). Dependent: volume of gas (oxygen) produced per minute.
1 mark each.
- (b) Temperature – water bath / heat shield / LED lamp; CO₂ concentration – same concentration and volume of sodium hydrogencarbonate solution, in excess; plant – same piece/length of the same species; time of collection – stopwatch; background light – darkened room.
1 mark for each variable with a valid method (max 3).
- (c) So that CO₂ concentration is not a limiting factor and does not fall during the experiment; only light intensity limits the rate.
1 mark.
A student investigates the effect of sodium hydrogencarbonate concentration (0.0 %, 0.1 %, 0.2 %, 0.4 %, 0.8 %) on the time taken for infiltrated leaf discs to float in bright light.
- (a) Suggest why all the leaf discs should be cut from the same plant and should avoid large veins. [2]
- (b) Describe a suitable control and explain what it shows. [2]
- (c) Explain why a change in temperature during the experiment would make the results less valid. [2]
Show answer
- (a) Same plant: discs have similar chlorophyll content / genotype / age, so differences are due to CO₂ concentration. Avoid veins: veins have fewer chloroplasts and different density/air spaces, which would affect floating time.
1 mark: same plant – reduces variation in chlorophyll/genetics. 1 mark: veins differ in chloroplasts/buoyancy.
- (b) Discs in distilled water with no sodium hydrogencarbonate (no added CO₂) under the same light; shows that floating depends on CO₂ for photosynthesis (discs should not float or float very slowly). Or discs in hydrogencarbonate kept in the dark – shows light is needed.
1 mark: valid control (no NaHCO₃ or dark). 1 mark: what it shows.
- (c) Temperature affects the rate of the enzyme-controlled reactions of photosynthesis (Calvin cycle), so the floating time would no longer depend only on CO₂ concentration.
1 mark: enzymes/rubisco affected by temperature. 1 mark: rate would change for a reason other than the independent variable.
A student will add DCPIP to chloroplast suspensions kept at 10, 15, 20, 25, 30 and 35 °C in light and record how quickly the blue colour disappears.
- (a) State a hypothesis for this investigation. [1]
- (b) State the null hypothesis. [1]
- (c) Predict and explain the results you would expect if the range were extended to 60 °C. [3]
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- (a) As temperature increases from 10 °C to 35 °C, the rate of DCPIP reduction by chloroplasts increases.
1 mark: testable, directional statement linking temperature and rate.
- (b) There is no relationship between temperature and the rate of DCPIP reduction.
1 mark.
- (c) Rate increases up to an optimum (around 35–40 °C) because molecules have more kinetic energy and electron carrier/enzyme proteins work faster; above the optimum the rate decreases sharply because proteins in the photosystems/electron carriers are denatured and thylakoid membranes are damaged.
1 mark: increase then decrease (optimum). 1 mark: more kinetic energy / faster reactions below optimum. 1 mark: denaturation of proteins (photosystems, carriers) / membrane damage above optimum.
The rate of photosynthesis (arbitrary units) of a tomato leaf at 25 °C was measured at different CO₂ concentrations in low and in high light.CO₂ / %: 0.02, 0.04, 0.08, 0.12Low light: 2.9, 3.6, 3.6, 3.6High light: 4.1, 7.9, 12.6, 12.8
- (a) A student concluded: 'Increasing CO₂ concentration always increases the rate of photosynthesis.' Use the data to evaluate this conclusion. [3]
- (b) Suggest which factor is limiting the rate at 0.12 % CO₂ in high light. [1]
Show answer
- (a) Supported at high light: the rate rises from 4.1 to 12.6 as CO₂ increases from 0.02 to 0.08 %. Not supported at low light: the rate rises only from 2.9 to 3.6 and stays at 3.6 above 0.04 %, because light is limiting. Also at high light the rise slows (12.6 to 12.8 from 0.08 to 0.12 %), so another factor becomes limiting – 'always' is not justified; no replicates or variability shown.
1 mark: support at high light with data. 1 mark: at low light rate levels off at 3.6 – light limiting. 1 mark: plateau at high CO₂ in bright light / no repeats, so 'always' not valid.
- (b) Temperature (or the amount of rubisco/chlorophyll; possibly light intensity).
1 mark: temperature (accept enzyme/chlorophyll amount).
Four tubes each contained 1 cm³ of blue DCPIP solution.A: chloroplast suspension, in light – colourless after 6 minutesB: chloroplast suspension, in the dark – still blue after 30 minutesC: boiled chloroplast suspension, in light – still blue after 30 minutesD: buffer only (no chloroplasts), in light – still blue after 30 minutes
- (a) Explain the result for tube A. [3]
- (b) Explain the results for tubes B and C. [2]
- (c) Explain the purpose of tube D. [1]
Show answer
- (a) In light, the chloroplasts carry out the light-dependent stage; electrons from photolysis/photoactivated chlorophyll pass along the electron transport chain; DCPIP acts as an electron acceptor (instead of NADP) and is reduced, so it turns from blue to colourless.
1 mark: light-dependent stage / photoactivation. 1 mark: electrons released. 1 mark: DCPIP accepts electrons and is reduced (blue → colourless).
- (b) B: no light, so no photoactivation of chlorophyll and no electrons released – DCPIP stays blue. C: boiling denatured the proteins of the photosystems/electron carriers (and damaged membranes), so no electron flow.
1 mark for B. 1 mark for C.
- (c) To show that the colour change is due to the chloroplasts and not to light acting on DCPIP alone.
1 mark.
A student counted the bubbles released from a cut stem of pondweed in a beaker of tap water for one minute at each of five lamp distances. A filament bulb was used and the lab blinds were open. He moved the lamp and started counting immediately.
- (a) Identify three sources of error in this method and suggest an improvement for each. [6]
Show answer
- (a) Bubbles differ in size, so counting is inaccurate – collect gas in a photosynthometer/gas syringe and measure volume. Lamp heats the water – use an LED or a water bath/heat shield and check temperature. No time for the plant to adjust to each new light intensity – leave 5 minutes before counting. Other light in the room – use a darkened room. Only one count – repeat three times and calculate the mean. CO₂ in the tap water may run out – use sodium hydrogencarbonate solution in excess.
1 mark per error and 1 mark per matching improvement, for three pairs (max 6).
The rate of oxygen production (µmol m⁻² s⁻¹) was measured for five leaves from plants of the same species grown in full sunlight and five grown in shade, all measured under identical conditions.Full sunlight: 12.4, 13.1, 11.8, 12.9, 12.6Shade: 8.2, 9.0, 7.6, 8.8, 8.4
- (a) Calculate the mean and the standard deviation for the plants grown in full sunlight. Give the standard deviation to 2 decimal places. (s = √[Σ(x − x̄)² ÷ (n − 1)]) [2]
- (b) The standard deviation for the shaded plants is 0.55 and the mean is 8.40. Calculate the standard error (SE = s ÷ √n) and the 95 % confidence interval (mean ± 2 × SE) for both groups. Give values to 2 decimal places. [3]
- (c) Explain what the confidence intervals show. [1]
Show answer
- (a) Mean 12.56; standard deviation 0.50
1 mark: mean 12.56. 1 mark: SD 0.50.
- (b) Sun: SE 0.22, 95 % CI 12.11–13.01. Shade: SE 0.25, 95 % CI 7.91–8.89.
1 mark: SE sun = 0.503 ÷ √5 = 0.22. 1 mark: SE shade = 0.55 ÷ √5 = 0.246 → 0.25. 1 mark: both intervals correct: 12.56 ± 0.45 and 8.40 ± 0.49 (allow 7.90–8.90 from rounded SE).
- (c) The intervals do not overlap, so the difference in mean rate between sun and shade plants is significant (not due to chance).
1 mark.
Mean rates of photosynthesis of a pondweed (arbitrary units) at different temperatures, with standard deviations:Temperature / °C: 15, 20, 25, 30, 35, 40Mean rate: 2.0, 3.3, 4.8, 6.4, 6.9, 4.1Standard deviation: 0.2, 0.3, 0.4, 0.4, 0.6, 0.7
- (a) Describe how the student should present these data as a graph, including how to show the variability of the results. [3]
- (b) Estimate the rate at 27 °C by linear interpolation. [1]
- (c) Suggest why the student cannot state the exact optimum temperature from these results. [1]
Show answer
- (a) Line graph with temperature / °C on the x-axis and mean rate / arbitrary units on the y-axis, both with linear scales using most of the grid; plot the mean values as crosses and join with ruled straight lines point to point; draw error bars of ± standard deviation (or 95 % CI) through each point.
1 mark: temperature on x-axis, rate on y-axis, labelled with units. 1 mark: points plotted and joined by ruled lines (not extrapolated). 1 mark: error bars showing SD/range for each mean.
- (b) 5.4 arbitrary units (5.44)
4.8 + (6.4 − 4.8) × 2/5 = 5.44.
- (c) Readings are 5 °C apart; the true optimum could lie anywhere between 30 °C and 40 °C – more readings at intervals of 1–2 °C are needed around 35 °C.
1 mark.
Spinach leaves will be blended in ice-cold buffer with a kitchen blender, filtered and centrifuged. The chloroplast suspension will be mixed with DCPIP solution in test-tubes placed in a water bath next to a bench lamp.
- (a) Identify three hazards in this procedure and state a precaution for each. [3]
- (b) State the overall level of risk and justify your answer. [1]
Show answer
- (a) Blender blades are sharp – switch off and unplug before handling, do not put fingers inside; centrifuge spinning at high speed – balance tubes and keep the lid closed until it stops; lamp gets hot and is electrical near water – keep water away from the lamp and plug, do not touch the bulb; DCPIP may irritate eyes/skin – wear eye protection and gloves; glassware may break – handle carefully.
1 mark for each hazard with a matching precaution (max 3).
- (b) Low (to medium) – the chemicals are of low hazard in small volumes and all hazards can be controlled with simple precautions.
1 mark: level with justification.
