2022 May/June · Paper 61 · Q1 (a(ii), c, d(i), e(i), e(ii), f)18 marksLimiting factors & rate investigations · Starch test on leaves
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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.
Q1Limiting factors & rate investigations · Core · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) B carbon dioxide concentration
Once more light no longer increases the rate, something other than light is in short supply; with temperature fixed, carbon dioxide concentration is the likely limiting factor. A is wrong because extra light has no effect on the plateau.
Q2Limiting factors & rate investigations · Core · Paper 2 style · 5 marks
(a)State three environmental factors that can limit the rate of photosynthesis.[3]
(b)Explain what is meant by a limiting factor.[2]
Show answer
(a) Light intensity; carbon dioxide concentration; temperature
1 mark each: light intensity; carbon dioxide concentration; temperature
(b) The factor that is in shortest supply; increasing it increases the rate of the process.
B1 factor in shortest supply / that restricts the rate; B1 increasing it increases the rate
Q3Limiting factors & rate investigations · Core · Paper 1 style · 1 mark
(a)Pondweed is kept in water baths at different temperatures. The rate of bubbling increases from 10 °C to 35 °C but falls sharply above 45 °C. What is the best explanation for the fall?[1]
Show answer
(a) A enzymes controlling photosynthesis are denatured
High temperatures change the shape of the active sites of photosynthesis enzymes, so the rate falls. C is wrong: gases are less soluble, not more, in warmer water, and this would not explain a sharp fall.
Q4Limiting factors & rate investigations · Standard · Paper 4 style · 9 marks
A student placed a piece of pondweed in a test tube of water containing sodium hydrogencarbonate. A lamp was placed at different distances from the tube, and a beaker of water was placed between the lamp and the tube. After 5 minutes at each distance, the student counted the bubbles released in one minute, three times.Distance / cm: 10, 20, 30, 40, 50Trial 1: 52, 38, 22, 14, 8Trial 2: 48, 35, 25, 12, 7Trial 3: 50, 35, 22, 13, 9
(a)Calculate the mean number of bubbles per minute at 20 cm and at 40 cm.[2]
(b)Describe the relationship between the distance of the lamp and the rate of bubbling.[2]
(c)Explain why the rate of bubbling changes as the lamp is moved further away.[2]
(d)State two variables that the student should keep the same.[2]
(e)Suggest why the student placed a beaker of water between the lamp and the test tube.[1]
Show answer
(a) 20 cm: 36; 40 cm: 13 bubbles per minute
M1 add the three readings and divide by 3; A1 both means correct (36 and 13)
(b) As the distance increases the number of bubbles per minute decreases; from 50 per minute at 10 cm to 8 per minute at 50 cm.
B1 further lamp → fewer bubbles (inverse relationship); B1 quotes paired data with units
(c) Light intensity falls with distance; light is the limiting factor, so less photosynthesis and less oxygen is released.
B1 lower light intensity further away; B1 less photosynthesis so less oxygen (gas) produced / light is limiting
(d) Any two of: temperature of water; carbon dioxide (hydrogencarbonate) concentration; same piece/length of pondweed; same lamp/bulb power; time allowed to acclimatise.
1 mark each for any two valid controlled variables
(e) To absorb heat from the lamp so the temperature of the pondweed stays constant.
B1 acts as a heat shield / keeps temperature constant
Q5Limiting factors & rate investigations · Standard · Paper 2 style · 6 marks
A grower near Murree produces tomatoes in a glasshouse during winter. Paraffin heaters are burned inside the glasshouse.
(a)Explain how burning paraffin in the heaters can increase the yield of tomatoes.[3]
(b)Suggest why the grower switches on extra lamps on cloudy winter days.[1]
(c)Suggest why the grower does not keep increasing the carbon dioxide concentration in the glasshouse.[2]
Show answer
(a) Burning releases heat, raising the temperature towards the optimum for photosynthesis enzymes; it also releases carbon dioxide, a raw material; so photosynthesis is faster and more sugar is made for growth.
B1 raises temperature (enzymes work faster); B1 adds carbon dioxide (raw material/limiting factor); B1 faster photosynthesis → more glucose/growth/yield
(b) Light intensity is low, so light is the limiting factor; lamps increase the rate of photosynthesis.
B1 light limiting on cloudy days
(c) Above a certain concentration another factor (light or temperature) becomes limiting so yield does not increase further; the extra cost would then be wasted.
B1 another factor becomes limiting / rate levels off; B1 extra cost gives no extra profit
Q6Limiting factors & rate investigations · Standard · Paper 4 style · 8 marks
Hydrogencarbonate indicator is red when in equilibrium with air, yellow when carbon dioxide concentration rises and purple when it falls. Four tubes were set up and left for 3 hours:Tube A: leaf, bright lightTube B: leaf, tube covered in black paperTube C: no leaf, bright lightTube D: leaf, tube covered with thin gauze (dim light)All tubes started red. After 3 hours the indicator in tube D was still red.
(a)State the colour of the indicator in tubes A, B and C after 3 hours.[2]
(b)Explain the colour in tube A and in tube B.[3]
(c)Explain why tube C was set up.[1]
(d)Suggest why the indicator in tube D stayed the same colour.[2]
Show answer
(a) A purple; B yellow; C red (unchanged)
2 marks all three correct; 1 mark for two correct
(b) A: the leaf photosynthesised faster than it respired, taking in carbon dioxide, so CO₂ fell and the indicator turned purple. B: in the dark there is no photosynthesis; respiration released carbon dioxide so the indicator turned yellow.
B1 A photosynthesis > respiration; B1 CO₂ removed → purple; B1 B only respiration, CO₂ added → yellow
(c) As a control, to show that the leaf caused the colour change (light alone does not change the indicator).
B1 control to show the change is due to the leaf
(d) In dim light the rate of photosynthesis equalled the rate of respiration, so there was no net change in carbon dioxide concentration.
B1 photosynthesis rate = respiration rate (compensation point); B1 no net uptake or release of CO₂
Q7Limiting factors & rate investigations · Challenge · Paper 4 style · 5 marks
In a pondweed investigation, relative light intensity is calculated using the formula: relative light intensity = 10 000 ÷ d², where d is the distance of the lamp from the plant in cm.
(a)Calculate the relative light intensity at 25 cm and at 50 cm.[2]
(b)State how many times greater the light intensity is at 25 cm than at 50 cm.[1]
(c)The volume of oxygen collected at 25 cm was only 2.5 times the volume collected at 50 cm. Suggest why it was not 4 times greater.[2]
Show answer
(a) 25 cm: 16.0 units; 50 cm: 4.0 units
M1 10 000 ÷ 25² = 10 000 ÷ 625; A1 16.0 and 4.0
(b) 4 times
B1 16 ÷ 4 = 4 (halving the distance gives four times the light)
(c) At the higher light intensity another factor, such as carbon dioxide concentration or temperature, was limiting, so the rate could not rise in proportion to light.
B1 another factor limiting at high light (CO₂ / temperature named); B1 so rate is not proportional to light intensity / light no longer the only limiting factor
Q8Limiting factors & rate investigations · Challenge · Paper 2 style · 7 marks
The rate of photosynthesis of a crop plant (arbitrary units) was measured at 25 °C.Light intensity / units: 0, 2, 4, 6, 8, 10Rate at 0.04% CO₂: 0, 12, 20, 24, 24, 24Rate at 0.10% CO₂: 0, 12, 24, 34, 40, 40
(a)Name the factor limiting the rate at a light intensity of 2 units for both concentrations of carbon dioxide. Give a reason.[2]
(b)Name the factor limiting the rate at 0.04% carbon dioxide and a light intensity of 8 units. Give a reason.[2]
(c)Calculate the percentage increase in rate at a light intensity of 8 units when the carbon dioxide concentration is raised from 0.04% to 0.10%. Give your answer to 3 significant figures.[2]
(d)Suggest one factor that could be limiting the rate at 0.10% carbon dioxide and a light intensity of 10 units.[1]
Show answer
(a) Light intensity; the rate is the same at both carbon dioxide concentrations (and rises when light increases).
B1 light intensity; B1 same rate at both CO₂ levels
(b) Carbon dioxide concentration; increasing CO₂ to 0.10% raises the rate, while more light does not.
B1 carbon dioxide concentration; B1 higher CO₂ gives higher rate / plateau with more light
(c) 66.7%
M1 (40 − 24) ÷ 24 × 100; A1 66.7%
(d) Temperature (or amount of chlorophyll / number of chloroplasts / enzymes).
B1 temperature or another valid internal factor; not light or CO₂
Q9Limiting factors & rate investigations · Challenge · Paper 4 style · 6 marks
(a)Plan an investigation to find out how temperature affects the rate of photosynthesis of pondweed. Your plan should include the apparatus, the method, the variables to control and how the results would be recorded.[6]
Show answer
(a) Pondweed in a boiling tube of hydrogencarbonate solution placed in water baths at a range of temperatures (e.g. 10, 20, 30, 40, 50 °C); acclimatise 5 minutes; count bubbles or collect gas volume in a set time (e.g. 1 minute); keep light intensity (lamp at fixed distance), CO₂ concentration and the same piece/length of pondweed constant; repeat 3 times and calculate a mean; record in a table of temperature against bubbles per minute; plot rate against temperature.
1 mark each for any six of: water baths/thermostatic control; at least 5 temperatures with a stated range; acclimatisation time; measure bubbles/gas volume in fixed time; light at fixed distance/same lamp; same CO₂/hydrogencarbonate concentration; same plant/length of pondweed; repeats and mean; results table with units / graph
Q10Mineral ions · Core · Paper 1 style · 1 mark
(a)Magnesium ions taken up by a plant are needed to make[1]
Show answer
(a) A chlorophyll
Magnesium is part of the chlorophyll molecule. Protein (B) needs nitrogen from nitrate ions, not magnesium.
Q11Mineral ions · Core · Paper 2 style · 3 marks
(a)State why plants need nitrate ions.[1]
(b)Describe two visible signs of nitrate ion deficiency in a plant.[2]
Show answer
(a) To make amino acids (and so proteins).
B1 amino acids / proteins
(b) Stunted (poor) growth; yellowing of the (older, lower) leaves.
1 mark each: stunted/weak growth; yellow leaves
Q12Mineral ions · Standard · Paper 2 style · 4 marks
(a)Explain why a plant growing in soil that lacks magnesium ions grows slowly.[4]
Show answer
(a) Magnesium is needed to make chlorophyll; less chlorophyll is made, so leaves turn yellow; less light is absorbed so the rate of photosynthesis falls; less glucose is available for respiration and for making new materials (e.g. cellulose, proteins) for growth.
B1 Mg for chlorophyll; B1 less chlorophyll/yellow leaves; B1 less light absorbed/less photosynthesis; B1 less glucose for growth/energy/new cells
Q13Mineral ions · Standard · Paper 4 style · 8 marks
Maize seedlings were grown for 4 weeks in jars of different solutions.Solution 1: complete nutrient solution – mean height 32 cm, dark green leavesSolution 2: no nitrate – mean height 11 cm, pale yellow older leavesSolution 3: no magnesium – mean height 18 cm, yellow between the leaf veinsSolution 4: distilled water – mean height 6 cm, very pale leaves
(a)State which solution acts as the control.[1]
(b)Calculate the mean height of seedlings in solution 2 as a percentage of the mean height in solution 1. Give your answer to 3 significant figures.[2]
(c)Explain the appearance of the seedlings in solution 3.[2]
(d)Air was bubbled through each jar. Suggest why.[2]
(e)Each jar was covered with black paper. Suggest why.[1]
Show answer
(a) Solution 1 (complete nutrient solution)
B1
(b) 34.4%
M1 11 ÷ 32 × 100; A1 34.4%
(c) No magnesium so less chlorophyll is made; leaves turn yellow (between the veins) and less photosynthesis means slower growth.
B1 no Mg → little chlorophyll / yellow leaves; B1 less photosynthesis → less growth
(d) To supply oxygen to the roots for aerobic respiration; this releases energy for active transport (uptake) of ions.
B1 oxygen for root respiration; B1 energy for active uptake of mineral ions
(e) To stop algae growing in the solution (algae would compete for ions).
B1 prevent growth of algae
Q14Mineral ions · Challenge · Paper 2 style · 8 marks
(a)Nitrate ions are often at a higher concentration inside root hair cells than in the soil water. Explain how root hair cells absorb nitrate ions.[3]
(b)Explain why a crop growing in waterlogged soil absorbs fewer nitrate ions.[3]
(c)Explain how adding nitrate fertiliser increases the yield of a wheat crop.[2]
Show answer
(a) By active transport; ions move against the concentration gradient (from lower to higher concentration) using carrier proteins in the membrane; energy is released by respiration in the root cells.
B1 active transport; B1 against the concentration gradient; B1 energy from respiration (carrier proteins)
(b) Waterlogged soil contains little oxygen (air spaces filled with water); root cells carry out less aerobic respiration; so less energy is released for active transport of ions.
B1 less oxygen in soil; B1 less aerobic respiration in roots; B1 less energy for active transport
(c) Nitrate is used to make amino acids and proteins; these are needed for growth of new cells (e.g. enzymes, more leaves), so the plant grows larger and produces more grain.
B1 amino acids/proteins made; B1 more growth / more leaves / more photosynthesis → greater yield
Q15Mineral ions · Challenge · Paper 2 style · 6 marks
A farmer in Punjab tested different amounts of nitrate fertiliser on wheat.Fertiliser / kg per ha: 0, 50, 100, 150, 200Yield / tonnes per ha: 2.8, 4.1, 5.0, 5.4, 5.5
(a)Calculate the increase in yield for each of these steps: 0 to 50 kg/ha, and 150 to 200 kg/ha.[2]
(b)Calculate the percentage increase in yield when the fertiliser is raised from 0 to 150 kg/ha. Give your answer to 3 significant figures.[2]
(c)The farmer decides to use 150 kg/ha rather than 200 kg/ha. Suggest two reasons for this decision.[2]
Show answer
(a) 0 to 50: 1.3 tonnes/ha; 150 to 200: 0.1 tonnes/ha
M1 subtraction from table; A1 both values
(b) 92.9%
M1 (5.4 − 2.8) ÷ 2.8 × 100; A1 92.9%
(c) The extra 50 kg gives very little extra yield, so the cost of the fertiliser is greater than the value of the extra crop; excess fertiliser may be washed (leached) into rivers and cause eutrophication.
1 mark each for any two: small extra yield; cost not repaid/profit; leaching/run-off; eutrophication/water pollution; another factor now limiting
Q16Leaf structure · Core · Paper 1 style · 1 mark
(a)In which tissue of a leaf do the cells contain the most chloroplasts?[1]
Show answer
(a) B palisade mesophyll
Palisade mesophyll cells are packed with chloroplasts near the upper surface where light is brightest. Spongy mesophyll (C) has fewer chloroplasts and more air spaces.
Q17Leaf structure · Core · Paper 2 style · 6 marks
A student describes a section through a leaf, from top to bottom:W – a thin, waxy layer with no cellsX – a single layer of transparent cells with no chloroplastsY – tall, column-shaped cells packed with chloroplastsZ – rounded, loosely packed cells with large air spaces between themThe lowest layer of cells contains many small pores, each surrounded by a pair of bean-shaped cells.
(a)Name layers W, X, Y and Z.[4]
(b)Name the pores in the lower layer and the cells on either side of each pore.[2]
Show answer
(a) W cuticle; X upper epidermis; Y palisade mesophyll; Z spongy mesophyll
1 mark each
(b) Stomata; guard cells
1 mark each
Q18Leaf structure · Standard · Paper 2 style · 6 marks
(a)Explain how palisade mesophyll cells are adapted for photosynthesis.[3]
(b)State the function of the air spaces in the spongy mesophyll.[2]
(c)Explain why the upper epidermis has no chloroplasts and is transparent.[1]
Show answer
(a) They contain many chloroplasts to absorb light; they are near the upper surface where light intensity is greatest; they are tall/closely packed, so many fit in the layer and absorb most light.
1 mark each for any three: many chloroplasts; near upper surface; closely packed/column shape; thin walls/large vacuole pushes chloroplasts to edge
(b) Allow gases (carbon dioxide and oxygen) to diffuse quickly through the leaf to and from the cells; provide a large surface area for gas exchange.
B1 diffusion of CO₂ to cells / O₂ away; B1 large surface area for gas exchange
(c) So that light passes through to the palisade cells below.
B1
Q19Leaf structure · Standard · Paper 2 style · 5 marks
(a)Describe the pathway taken by a carbon dioxide molecule from the air outside a leaf to a chloroplast in a palisade cell.[3]
(b)Explain why carbon dioxide diffuses into the leaf during the day.[2]
Show answer
(a) Through an open stoma; diffuses through the air spaces of the spongy mesophyll; dissolves in the film of water on the cell surface and diffuses through the cell wall and membrane into the chloroplast.
B1 enters via stoma; B1 through air spaces; B1 dissolves in moisture / through cell wall and membrane into chloroplast
(b) Carbon dioxide is used up in photosynthesis, so its concentration inside the leaf is lower than in the air; it diffuses down the concentration gradient.
B1 used in photosynthesis; B1 lower concentration inside / down concentration gradient
Q20Leaf structure · Challenge · Paper 4 style · 7 marks
A student painted clear nail varnish on both surfaces of a hibiscus leaf, peeled it off and counted stomata in five fields of view under a microscope. The diameter of the field of view was 0.4 mm.Lower surface: 22, 25, 19, 24, 20Upper surface: 3, 5, 4, 2, 6
(a)Calculate the mean number of stomata per field of view for each surface.[2]
(b)Calculate the area of the field of view in mm². Use area = πr². Give your answer to 3 significant figures.[1]
(c)Calculate the number of stomata per mm² on the lower surface. Give your answer to the nearest whole number.[2]
(d)Suggest why there are more stomata on the lower surface of this leaf.[2]
Show answer
(a) Lower: 22; upper: 4
M1 total ÷ 5; A1 both
(b) 0.126 mm²
B1 π × 0.2² = 0.126
(c) 175 stomata per mm²
M1 22 ÷ 0.1257; A1 175
(d) The lower surface is shaded and cooler; so less water is lost by transpiration/evaporation than if stomata were on the upper surface.
B1 lower surface cooler/out of direct sunlight; B1 reduces water loss
Q21Leaf structure · Challenge · Paper 2 style · 6 marks
A student measured leaves from the top (sunny) and the lower (shaded) branches of a mango tree.Sun leaves: palisade layer 120 µm thick, mean leaf area 42 cm²Shade leaves: palisade layer 60 µm thick, mean leaf area 68 cm²
(a)Calculate the ratio of the palisade thickness of sun leaves to shade leaves.[1]
(b)Suggest an advantage to the tree of having a thicker palisade layer in sun leaves.[2]
(c)Suggest why shade leaves have a larger surface area.[2]
(d)Suggest why shade leaves are thinner.[1]
Show answer
(a) 2 : 1 (twice as thick)
B1 120 ÷ 60 = 2
(b) More layers of palisade cells / more chloroplasts; so more of the bright light is absorbed and used, giving more photosynthesis.
B1 more palisade cells/chloroplasts; B1 absorb more light → more photosynthesis
(c) Light is dim in the shade; a larger area traps more of the limited light for photosynthesis.
B1 low light intensity in shade; B1 larger area absorbs more light
(d) Little light reaches deep layers, so extra layers of cells would not photosynthesise much (and would use materials/respire).
B1 little light penetrates / thick leaf not needed
Q22Photosynthesis equation & products · Core · Paper 1 style · 1 mark
(a)Which is the balanced symbol equation for photosynthesis?[1]
Show answer
(a) A 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
A has 6 C, 18 O and 12 H on each side. C is the equation for aerobic respiration, the reverse of photosynthesis.
Q23Photosynthesis equation & products · Core · Paper 2 style · 5 marks
(a)Write the word equation for photosynthesis.[2]
(b)State the source of energy for photosynthesis and name the substance that absorbs it.[2]
(c)State where water used in photosynthesis enters the plant.[1]
Show answer
(a) carbon dioxide + water → glucose + oxygen (in the presence of light and chlorophyll)
B1 reactants; B1 products
(b) Light (from the Sun); chlorophyll
1 mark each
(c) Through the root hair cells (roots)
B1
Q24Photosynthesis equation & products · Standard · Paper 2 style · 6 marks
(a)State four ways in which a plant uses the glucose made in photosynthesis.[4]
(b)Explain why glucose is converted to starch for storage in leaf cells.[2]
Show answer
(a) Any four of: respiration to release energy; converted to starch for storage; cellulose for cell walls; sucrose for transport in phloem; amino acids/proteins (with nitrate ions); fats and oils for storage (e.g. in seeds); nectar to attract insects.
1 mark each for any four
(b) Starch is insoluble; so it does not affect the water potential of the cell / cause osmosis (and does not diffuse out of the cell).
B1 starch insoluble; B1 no effect on osmosis / water potential, or compact and stays in cell
Q25Photosynthesis equation & products · Standard · Paper 2 style · 6 marks
(a)Define photosynthesis.[2]
(b)Describe the role of chlorophyll in photosynthesis.[2]
(c)State two things that can happen to the oxygen made in a palisade cell.[2]
Show answer
(a) The process by which plants make carbohydrates (glucose) from raw materials (carbon dioxide and water) using energy from light.
B1 makes carbohydrates/glucose from CO₂ and water; B1 using light energy
(b) Chlorophyll absorbs light energy; and transfers it into chemical energy (for making carbohydrates).
B1 absorbs light; B1 converts/transfers it to chemical energy
(c) Used in respiration in the plant's own cells; diffuses out of the leaf through the stomata into the air.
1 mark each
Q26Photosynthesis equation & products · Challenge · Paper 2 style · 6 marks
A maize crop was kept in a sealed chamber with 12 hours of light and 12 hours of darkness. In the light, the crop took in carbon dioxide at a rate of 30 units per hour. In the dark, it released carbon dioxide at a rate of 6 units per hour.
(a)Calculate the net uptake of carbon dioxide by the crop over the whole 24 hours.[2]
(b)Explain why the crop releases carbon dioxide at night.[2]
(c)The measured uptake of 30 units per hour in the light is less than the true rate of photosynthesis. Explain why.[2]
Show answer
(a) 288 units
M1 (30 × 12) − (6 × 12) = 360 − 72; A1 288
(b) There is no light so no photosynthesis; respiration continues all the time and releases carbon dioxide.
B1 no photosynthesis in dark; B1 respiration releases CO₂
(c) During the day the plant also respires; some of the carbon dioxide used in photosynthesis comes from respiration inside the plant, so the measured uptake = photosynthesis − respiration.
B1 respiration also happens in the light; B1 CO₂ from respiration used, so uptake underestimates gross photosynthesis
Q27Starch test on leaves · Standard · Paper 4 style · 7 marks
A student tests a leaf for starch.Step 1: the leaf is placed in boiling water for 1 minute.Step 2: the leaf is placed in a tube of ethanol in a hot water bath until it turns white.Step 3: the leaf is dipped into warm water.Step 4: the leaf is spread on a white tile and covered with iodine solution.
(a)Give the reason for each of steps 1 to 4.[4]
(b)Explain why the ethanol is heated in a water bath and not directly over a Bunsen burner.[1]
(c)State the colour seen if the leaf contains starch and if it contains no starch.[2]
Show answer
(a) 1: kills the cells / stops reactions and breaks down membranes so iodine can enter; 2: removes the chlorophyll (decolourises the leaf) so the colour change can be seen; 3: softens the brittle leaf (and removes ethanol); 4: tests for starch.
1 mark per step
(b) Ethanol is flammable.
B1
(c) Starch present: blue-black; no starch: orange-brown (colour of iodine).
1 mark each
Q28Starch test on leaves · Challenge · Paper 4 style · 8 marks
A potted plant with variegated leaves (green areas with chlorophyll and white areas without) was kept in the dark for 48 hours. A strip of aluminium foil was then fixed across part of one leaf and the plant was placed in sunlight for 6 hours. The leaf was then tested for starch.Region P: green, not coveredRegion Q: white, not coveredRegion R: green, covered by foil
(a)Explain why the plant was kept in the dark for 48 hours before the experiment.[2]
(b)Predict the colour after the iodine test in each of regions P, Q and R. Explain each prediction.[5]
(c)State which two regions show that light is needed for photosynthesis.[1]
Show answer
(a) To destarch the leaves; the plant uses up its stored starch (converted to sugar and respired/translocated), so any starch found later must have been made during the experiment.
B1 removes starch / destarching; B1 so starch present must come from photosynthesis during the experiment
(b) P (green, uncovered): blue-black – it had chlorophyll and light so photosynthesised and made starch. Q (white, uncovered): orange-brown – no chlorophyll so no photosynthesis. R (green, covered): orange-brown – no light reached it so no photosynthesis.
B1 P blue-black; B1 Q orange-brown; B1 R orange-brown; B1 Q explained (no chlorophyll); B1 R explained (no light)
(c) P and R
B1
Q29Leaf structure (option) · Standard · Paper 2 style · 7 marks
(a)Describe how the arrangement of tissues in a leaf allows as much light as possible to be absorbed.[4]
(b)Describe how guard cells control the opening of a stoma during the day and explain why this is important for photosynthesis.[3]
Show answer
(a) Leaf is broad/flat with large surface area; thin so light reaches all cells; upper epidermis and cuticle are transparent; palisade layer with most chloroplasts is just below the upper surface; spongy cells also contain chloroplasts to absorb light passing through.
1 mark each for any four: large surface area; thin; transparent cuticle/epidermis; palisade near top; many chloroplasts in palisade; chloroplasts can move/spongy layer absorbs remaining light
(b) In light, guard cells take in water and become turgid; this makes them curve apart and the stoma opens; carbon dioxide can then diffuse in for photosynthesis.
Q30Leaf structure (option) · Challenge · Paper 2 style · 7 marks
(a)Explain how the vascular bundles in a leaf support photosynthesis and transport its products.[4]
(b)Explain why most leaves are thin.[2]
(c)Suggest why a leaf has a waxy cuticle even though this reduces gas exchange through the surface.[1]
Show answer
(a) Xylem brings water (a raw material) and mineral ions such as magnesium to the mesophyll; phloem carries sucrose (and amino acids) made in photosynthesis away from the leaf to other parts of the plant; the network of veins means every cell is close to a vein; xylem also helps to support the leaf so it is held out flat to the light.
1 mark each for any four: xylem supplies water; xylem supplies mineral ions; phloem removes sucrose/amino acids; translocation to growing/storage parts; branching veins close to all cells; support/lignin keeps blade spread
(b) Short diffusion distance for gases (CO₂ in, O₂ out) to and from all cells; light can reach all the cells.
B1 short diffusion distance for gases; B1 light penetrates to all cells
(c) It reduces water loss by evaporation; gases can still enter through the stomata.
B1 reduces water loss (gases use stomata)
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