References only, not question text. Open Official paper ↗ where Cambridge publishes the paper,
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school. “Paper 42” means Paper 4, variant 2.
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.
Q1Photosynthesis & limiting factors · Core · Paper 2 style · 1 mark
(a)Which is the balanced chemical equation for photosynthesis?[1]
Show answer
(a) B 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
Six carbon dioxide and six water give one glucose and six oxygen, in the presence of light and chlorophyll. A is the aerobic respiration equation; D is not balanced.
Q2Photosynthesis & limiting factors · Core · Paper 4 style · 5 marks
(a)Write the word equation for photosynthesis.[2]
(b)State the source of energy for photosynthesis and name the pigment that absorbs it.[2]
(c)Describe what happens to the energy absorbed by chlorophyll.[1]
Show answer
(a) carbon dioxide + water → glucose + oxygen (in the presence of light and chlorophyll)
B1 reactants carbon dioxide + water; B1 products glucose + oxygen. Condition (light/chlorophyll) above the arrow is expected but not separately credited.
(b) Light (energy); chlorophyll.
B1 light; B1 chlorophyll.
(c) It is transferred into chemical energy in glucose (and other carbohydrates).
B1 light energy transferred to chemical energy in glucose / carbohydrates.
Q3Photosynthesis & limiting factors · Core · Paper 4 style · 5 marks
A sugarcane plant in a field in Punjab makes glucose by photosynthesis.
(a)State three ways the plant uses the glucose it makes.[3]
(b)Explain why glucose is converted to starch for storage in the leaf.[2]
Show answer
(a) Respiration to release energy; converted to starch for storage; converted to cellulose for cell walls; (also sucrose for transport, amino acids/proteins, fats/oils, nectar).
1 mark each for any three of: respiration; starch (storage); cellulose (cell walls); sucrose (translocation in phloem); amino acids/proteins (with nitrate); fats and oils (storage in seeds); nectar.
(b) Starch is insoluble so it does not affect water potential / osmosis in cells; it is large so does not diffuse out of cells (and is compact).
B1 insoluble; B1 does not affect osmosis / water potential / cannot move out of cells.
Q4Photosynthesis & limiting factors · Core · Paper 4 style · 4 marks
(a)State two factors, other than light intensity, that can limit the rate of photosynthesis.[2]
(b)Explain why the rate of photosynthesis is slow on a cold, sunny winter morning.[2]
Show answer
(a) Carbon dioxide concentration; temperature.
B1 carbon dioxide concentration; B1 temperature. Do not accept light (excluded by the question).
(b) Low temperature means the enzymes controlling photosynthesis work slowly (molecules have less kinetic energy, fewer collisions), so temperature is limiting.
B1 temperature is limiting; B1 enzymes slow / less kinetic energy.
Q5Photosynthesis & limiting factors · Core · Paper 4 style · 5 marks
(a)Name the two raw materials of photosynthesis.[2]
(b)State how each raw material enters a leaf of a neem tree.[2]
(c)Name the gas released as a by-product.[1]
Show answer
(a) Carbon dioxide and water.
B1 each.
(b) Carbon dioxide diffuses in from the air through the stomata; water is absorbed by root hairs and carried to the leaf in the xylem.
B1 CO₂ through stomata by diffusion; B1 water via xylem (from roots).
(c) Oxygen
B1 oxygen.
Q6Photosynthesis & limiting factors · Standard · Paper 2 style · 1 mark
(a)[1]
Show answer
(a) B carbon dioxide decreases, oxygen increases
In bright light the rate of photosynthesis is greater than the rate of respiration, so CO₂ is used up faster than it is made and O₂ is released. Option D would apply only at the compensation point (dim light).
Q7Photosynthesis & limiting factors · Standard · Paper 2 style · 1 mark
(a)Hydrogencarbonate indicator is orange-red at atmospheric carbon dioxide concentration. A leaf is placed in a sealed tube of indicator and kept in bright light for 2 hours. What colour is the indicator at the end?[1]
Show answer
(a) C purple
Photosynthesis is faster than respiration, so CO₂ is removed and the indicator turns purple. Yellow (A) shows a higher CO₂ concentration, e.g. in darkness.
Q8Photosynthesis & limiting factors · Standard · Paper 6 style · 7 marks
A student in Peshawar put a piece of Elodea pondweed in a test-tube of water containing sodium hydrogencarbonate. A lamp was placed at different distances. After 5 minutes at each distance, bubbles were counted for one minute, three times.Distance / cm: 10, 20, 40Bubbles per minute (trial 1, 2, 3): 10 cm – 47, 50, 47; 20 cm – 21, 23, 22; 40 cm – 8, 6, 7
(a)Calculate the mean number of bubbles per minute at each distance.[3]
(b)Describe the relationship between distance from the lamp and the rate of photosynthesis.[2]
(c)Explain why the student allowed the plant to stand for 5 minutes at each new distance before counting.[1]
(d)Suggest one improvement to the method of measuring the gas released.[1]
Q9Photosynthesis & limiting factors · Standard · Paper 4 style · 7 marks
The rate of photosynthesis of tomato plants at 25 °C was measured at two carbon dioxide concentrations.Light intensity / units: 0, 2, 4, 6, 8Rate at 0.04 % CO₂: 0, 9, 16, 18, 18Rate at 0.1 % CO₂: 0, 9, 17, 26, 30
(a)Define the term limiting factor.[1]
(b)Between 0 and 4 units of light, name the factor limiting the rate. Explain your answer.[2]
(c)Above 6 units of light, the rate at 0.04 % CO₂ stays at 18 arbitrary units. State the factor limiting the rate at this point and explain the evidence.[2]
(d)Calculate the percentage increase in rate at 8 units of light when CO₂ is raised from 0.04 % to 0.1 %.[2]
Show answer
(a) Something present in the environment in such short supply that it restricts life processes (here, the rate of photosynthesis).
B1 factor in shortest supply that restricts / limits the rate of a process.
(b) Light intensity; the rate increases as light intensity increases, and the rates at the two CO₂ concentrations are (almost) the same, so CO₂ is not limiting.
B1 light intensity; B1 rate rises as light rises / rates almost identical at both CO₂ concentrations so CO₂ not limiting.
(c) Carbon dioxide concentration; increasing light no longer raises the rate, but the rate is higher (30) at 0.1 % CO₂.
B1 CO₂ concentration; B1 rate higher at 0.1 % CO₂ at the same light intensity / plateau with more light.
(d) 66.7 % (3 s.f.)
M1 (30 − 18)/18 × 100; A1 66.7 %.
Q10Photosynthesis & limiting factors · Standard · Paper 6 style · 8 marks
A student investigates whether chlorophyll is needed for photosynthesis using a potted geranium plant.
(a)Describe how the student should test a leaf for starch.[4]
(b)The plant had been kept in the dark for 48 hours before the experiment. Explain why.[2]
(c)One leaf was variegated (green and white areas). After 6 hours of light, only the green areas turned blue-black. State the conclusion.[1]
(d)State why the ethanol must not be heated directly over the Bunsen flame.[1]
Show answer
(a) Boil the leaf in water (about 1 minute) to kill cells and break membranes; switch off the Bunsen burner and place the leaf in a tube of ethanol in the hot water-bath to remove chlorophyll; dip in water to soften; spread on a white tile and add iodine solution.
B1 boil in water; B1 in ethanol, heated in a water-bath (not over a flame) to remove chlorophyll; B1 rinse in (warm) water to soften; B1 add iodine solution; blue-black = starch.
(b) To destarch the leaves – the starch present is used up (converted to sugar and used in respiration / moved away), so any starch found afterwards was made during the experiment.
B1 removes / uses up starch; B1 so starch found must be produced during the investigation.
(c) Chlorophyll is needed for photosynthesis.
B1 chlorophyll needed for photosynthesis / starch formation.
(d) Ethanol is flammable.
B1 flammable.
Q11Photosynthesis & limiting factors · Standard · Paper 5 style · 8 marks
Leaf discs were cut from a spinach leaf with a cork borer, and the air was removed from them by suction so that they sank in sodium hydrogencarbonate solution. The time for discs to float in light was measured at different temperatures. At 25 °C the mean time for the discs to float was 180 s.
(a)Name the independent variable and the dependent variable in this investigation.[2]
(b)Calculate the rate of photosynthesis at 25 °C as 1000 ÷ mean time (in s). Give your answer to 2 significant figures.[2]
(c)Explain why the leaf discs float.[2]
(d)State two variables that should be kept constant.[2]
Show answer
(a) Independent: temperature of the water. Dependent: time taken for the leaf discs to float (rate of photosynthesis).
B1 temperature; B1 time for discs to float / number of discs floating.
(b) 5.6 (arbitrary units)
M1 1000 ÷ 180; A1 5.6.
(c) Photosynthesis produces oxygen, which collects in the air spaces of the discs, making them less dense than the solution.
B1 oxygen produced by photosynthesis; B1 gas fills air spaces, reducing density.
(d) Light intensity (distance from lamp); concentration of sodium hydrogencarbonate; number / size of discs; type and age of leaf; volume of solution.
1 mark each for any two.
Q12Photosynthesis & limiting factors · Standard · Paper 4 style · 4 marks
The volume of oxygen released by a water plant in a pond in Faisalabad was measured on a summer day.Time: 08:00, 12:00, 16:00Total volume of oxygen collected / cm³: 4.4, 18.4, 24.4
(a)Calculate the rate of oxygen production in cm³ per hour between 08:00 and 12:00.[2]
(b)Suggest why the rate between 12:00 and 16:00 is lower even though the light intensity at midday is high.[2]
Show answer
(a) 3.5 cm³ per hour
M1 (18.4 − 4.4) ÷ 4; A1 3.5.
(b) Another factor became limiting, e.g. the high temperature caused stomata to close so less CO₂ entered, or CO₂ concentration was limiting.
B1 another limiting factor named (CO₂ / temperature too high); B1 linked explanation, e.g. stomata close to reduce water loss so CO₂ uptake falls.
Q13Photosynthesis & limiting factors · Challenge · Paper 2 style · 1 mark
(a)[1]
Show answer
(a) B add sodium hydrogencarbonate to the water
The rate does not respond to light, so light is not limiting; the very low CO₂ is the limiting factor. Sodium hydrogencarbonate releases CO₂ into the water. More light (A) has already been shown to have no effect.
Q14Photosynthesis & limiting factors · Challenge · Paper 6 style · 5 marks
Light intensity from a small lamp is proportional to 1 ÷ (distance)².
(a)Calculate the relative light intensity at 30 cm compared with 10 cm. Give your answer as a fraction.[2]
(b)At 15 cm the plant produced 36 bubbles per minute. Assuming the rate is proportional to light intensity, predict the rate at 30 cm.[2]
(c)The student found that the actual rate at 30 cm was 14 bubbles per minute. Suggest one reason why it was higher than predicted.[1]
Show answer
(a) 1/9
M1 (10/30)² ; A1 1/9.
(b) 9 bubbles per minute
M1 intensity at 30 cm is (15/30)² = 1/4 of that at 15 cm; A1 36 × 1/4 = 9.
(c) Light from the room / window also reached the plant; or the rate is not exactly proportional because another factor (e.g. CO₂) limits at higher intensity.
B1 background light / other light sources; or rate not proportional at 15 cm because CO₂/temperature limiting.
Q15Photosynthesis & limiting factors · Challenge · Paper 4 style · 6 marks
A farmer near Islamabad grows cucumbers in a glasshouse in winter.
(a)Explain why the farmer burns paraffin in heaters inside the glasshouse during winter nights and early mornings.[3]
(b)The farmer also switches on lamps in the glasshouse in the evening. Suggest why this may not increase yield on very cold nights if the heaters are switched off.[2]
(c)Suggest one reason why the farmer should not raise the glasshouse temperature to 45 °C.[1]
Show answer
(a) Burning paraffin releases heat, raising the temperature so enzymes of photosynthesis work faster; it also releases carbon dioxide, which increases CO₂ concentration – a limiting factor; so the rate of photosynthesis increases, giving more glucose and higher yield.
1 mark each for any three of: raises temperature; enzymes work faster / more kinetic energy; releases CO₂; CO₂ (or temperature) is limiting; faster photosynthesis → more growth / yield.
(b) Temperature would then be the limiting factor, so extra light would not increase the rate; enzymes work slowly at low temperature.
B1 temperature limiting; B1 rate cannot increase with more light / low enzyme activity.
(c) Enzymes would be denatured / the rate would fall / it would cost more fuel without extra yield.
B1 enzymes denatured (or cost exceeds benefit).
Q16Photosynthesis & limiting factors · Challenge · Paper 6 style · 8 marks
Four boiling tubes contained equal volumes of orange-red hydrogencarbonate indicator. Tubes A, B and D each contained one spinach leaf; tube C had no leaf. Tube B was wrapped in aluminium foil. All tubes were sealed and placed 20 cm from a lamp.
(a)State the colour of the indicator expected in tubes A, B and C after 4 hours.[3]
(b)Explain the result in tube B.[2]
(c)Explain the purpose of tube C.[1]
(d)Tube D was covered with thin white tissue paper so that it received dim light. After 4 hours the indicator was still orange-red. Explain why.[2]
Show answer
(a) A (light): purple; B (foil-covered, dark): yellow; C (no leaf): orange-red (unchanged).
B1 each.
(b) In the dark there is no photosynthesis; the leaf respires and releases carbon dioxide, so CO₂ concentration rises and the indicator turns yellow.
B1 no photosynthesis (no light); B1 respiration releases CO₂.
(c) It is a control, showing that the colour change is caused by the leaf and not by light or the indicator itself.
B1 control – change is due to leaf.
(d) In dim light the rates of photosynthesis and respiration are equal (compensation point), so the CO₂ taken in equals the CO₂ released and there is no net change.
B1 photosynthesis rate = respiration rate; B1 no net change in CO₂ concentration.
Q17Photosynthesis & limiting factors · Challenge · Paper 4 style · 5 marks
The balanced chemical equation for photosynthesis is 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.
(a)Use the equation to state the ratio of glucose molecules made to oxygen molecules released.[1]
(b)A plant released 0.6 mol of oxygen in one day. Use the equation to calculate how many moles of carbon dioxide it absorbed and how many moles of glucose it made.[2]
(c)Explain why the plant's dry mass at the end of the day increases by less than the mass of glucose it made.[2]
(c) Some of the glucose is broken down in respiration (to release energy), producing CO₂ and water which are lost; so only the remainder is converted into new biomass.
B1 glucose used in respiration; B1 products lost as gases / not stored.
Q18Leaf structure · Core · Paper 2 style · 1 mark
(a)In which tissue of a leaf does most photosynthesis take place?[1]
Show answer
(a) B palisade mesophyll
Palisade cells are tightly packed near the upper surface and contain the most chloroplasts. Spongy mesophyll cells have fewer chloroplasts and many air spaces.
Q19Leaf structure · Core · Paper 4 style · 4 marks
(a)Name the tissue or structure described: (i) a waxy, waterproof layer on the upper surface; (ii) pores, mainly on the lower surface, surrounded by two cells that change shape; (iii) vessels that bring water to the leaf.[3]
(b)State the function of the phloem in a leaf vein.[1]
Show answer
(a) (i) Cuticle; (ii) stomata (guard cells); (iii) xylem.
B1 each.
(b) Transports sucrose and amino acids away from the leaf to other parts of the plant.
Q20Leaf structure · Core · Paper 6 style · 4 marks
(a)A student draws a transverse section of a leaf from a microscope slide. State two rules the student should follow when making a biological drawing.[2]
(b)List the tissues the student would see in order, from the upper surface to the lower surface, in a section that passes between veins.[2]
Show answer
(a) Use a sharp pencil with clear, continuous lines; no shading; draw large, filling the space; correct proportions; label lines ruled and not crossing.
B1 epidermis then palisade then spongy; B1 lower epidermis with guard cells/stomata at the bottom.
Q21Leaf structure · Standard · Paper 2 style · 1 mark
(a)Which leaf adaptation increases the rate of diffusion of carbon dioxide to the palisade cells?[1]
Show answer
(a) B air spaces in the spongy mesophyll
Air spaces allow CO₂ to diffuse quickly from the stomata to the mesophyll cells. A transparent epidermis lets light through, not CO₂.
Q22Leaf structure · Standard · Paper 4 style · 6 marks
Leaves of a banana plant are broad and thin.
(a)Explain how the structure of a leaf is adapted for photosynthesis.[6]
Show answer
(a) Large surface area to absorb light; thin, so short diffusion distance for CO₂; transparent epidermis / no chloroplasts lets light reach mesophyll; palisade cells packed near upper surface with many chloroplasts; air spaces in spongy mesophyll for gas diffusion; stomata let CO₂ in and O₂ out; guard cells control stomata; xylem supplies water; phloem removes sucrose.
1 mark each for any six feature + function pairs from the list. A feature without its function does not score.
Q23Leaf structure · Standard · Paper 5 style · 5 marks
A student painted clear nail varnish on both surfaces of a hibiscus leaf, peeled it off and counted stomata in four fields of view under a microscope.Lower surface: 24, 28, 27, 25Upper surface: 3, 2, 4, 3
(a)Calculate the mean number of stomata per field of view on the lower surface.[1]
(b)The area of the field of view was 0.20 mm². Calculate the mean stomatal density on the lower surface in stomata per mm².[2]
(c)Suggest why the upper surface has far fewer stomata than the lower surface.[2]
Show answer
(a) 26
B1 (24+28+27+25)/4 = 26.
(b) 130 stomata per mm²
M1 26 ÷ 0.20; A1 130.
(c) The upper surface is in direct sunlight and is warmer, so more water would be lost by transpiration; fewer stomata on top reduces water loss.
B1 upper surface hotter / in sunlight; B1 reduces water loss (transpiration).
Q24Leaf structure · Standard · Paper 4 style · 5 marks
(a)State the function of (i) the cuticle and (ii) the upper epidermis of a leaf.[2]
(b)Explain why it is an advantage for every mesophyll cell to be close to a vein.[2]
(c)Suggest why the veins also help the leaf to stay flat and spread out.[1]
Show answer
(a) (i) Reduces water loss (waterproof, waxy); (ii) protects the leaf and lets light pass through to the mesophyll (transparent).
B1 cuticle reduces water loss / evaporation; B1 epidermis transparent – lets light through / protection.
(b) Xylem in the vein supplies water (for photosynthesis and support) over a short distance; phloem removes sucrose quickly, so products do not build up.
B1 water supplied by xylem; B1 sucrose / products removed by phloem.
(c) Xylem walls are strengthened with lignin, giving support so the leaf exposes a large surface to light.
B1 lignin / xylem support keeps large area facing light.
Q25Leaf structure · Challenge · Paper 4 style · 6 marks
(a)Describe the path taken by a carbon dioxide molecule from the air to a chloroplast in a palisade cell.[3]
(b)Explain why palisade mesophyll cells are found near the upper surface of the leaf and contain many chloroplasts.[2]
(c)Suggest why the lower epidermis cells (other than guard cells) have no chloroplasts.[1]
Show answer
(a) Diffuses through an open stoma (between guard cells) into the air spaces of the spongy mesophyll; dissolves in the moist surface / water film of mesophyll cells; diffuses through the cell wall, membrane and cytoplasm into a palisade cell chloroplast.
B1 through stoma; B1 air spaces (spongy mesophyll); B1 dissolves in water on cell surface then diffuses into palisade cell / chloroplast. Process must be diffusion.
(b) The upper surface receives the most light; many chloroplasts absorb as much light as possible for photosynthesis.
B1 most light near upper surface; B1 more chlorophyll absorbs more light / higher rate.
(c) Little light reaches the lower surface, so chloroplasts there would be of little use / the epidermis is mainly protective.
B1 little light reaches / role is protection.
Q26Leaf structure · Challenge · Paper 4 style · 6 marks
(a)Compare the arrangement of cells in the palisade mesophyll and spongy mesophyll.[2]
(b)Explain why a leaf from a plant growing in deep shade in a forest is often thinner and larger than a leaf from a plant growing in full sun.[2]
(c)Name the cells that open and close stomata, and state one advantage to the plant of closing stomata at night.[2]
Show answer
(a) Palisade cells are long, column-shaped and tightly packed in rows; spongy cells are irregular / rounded and loosely packed with large air spaces.
B1 palisade tightly packed / columnar; B1 spongy loosely packed with air spaces.
(b) Light is limiting in shade, so a larger surface area absorbs more of the little light available; a thin leaf has fewer layers so light reaches all cells / shorter diffusion path (fewer palisade layers needed).
B1 larger area to capture more light; B1 thinner – light penetrates to all mesophyll / fewer palisade layers needed.
(c) Guard cells; reduces water loss by transpiration when CO₂ is not needed for photosynthesis (no light).
B1 guard cells; B1 reduce water loss (no photosynthesis at night).
Q27Mineral ions (Mg, nitrate) · Core · Paper 2 style · 1 mark
(a)The leaves of a pot plant turn yellow because the plant cannot make enough chlorophyll. Which mineral ion is the plant most likely to be lacking?[1]
Show answer
(a) C magnesium
Magnesium ions are part of the chlorophyll molecule, so a shortage causes yellowing (chlorosis). Nitrate (not offered) is used for amino acids; the other ions are not used to make chlorophyll.
Q28Mineral ions (Mg, nitrate) · Core · Paper 4 style · 4 marks
(a)State why plants need (i) nitrate ions and (ii) magnesium ions.[2]
(b)Describe the appearance of a plant deficient in nitrate ions.[2]
Show answer
(a) (i) To make amino acids (and so proteins); (ii) to make chlorophyll.
Q29Mineral ions (Mg, nitrate) · Standard · Paper 5 style · 6 marks
Maize seedlings were grown for four weeks in water culture solutions. Mean dry mass per seedling:Complete solution: 2.9 gNo magnesium: 1.6 gNo nitrate: 0.8 g
(a)Calculate the percentage decrease in mean dry mass of the seedlings grown without nitrate compared with complete solution.[2]
(b)Explain why the seedlings grown without nitrate had the lowest dry mass.[3]
(c)Explain why dry mass is a better measure of growth than fresh mass.[1]
Show answer
(a) 72.4 % (3 s.f.)
M1 (2.9 − 0.8)/2.9 × 100; A1 72.4 %.
(b) Nitrate is needed to make amino acids; amino acids are used to make proteins; proteins are needed for new cells / growth (and enzymes), so growth is reduced.
B1 nitrate for amino acids; B1 proteins made from amino acids; B1 proteins needed for growth / new cytoplasm / enzymes.
(c) Water content of the seedlings varies (with time of day / watering), so fresh mass is not reliable; dry mass measures only the biomass.
B1 water content varies.
Q30Mineral ions (Mg, nitrate) · Challenge · Paper 4 style · 5 marks
(a)Explain why the leaves of plants grown without magnesium became yellow and the plants grew less.[3]
(b)A farmer in Sahiwal adds nitrogen fertiliser to a field. Explain how this can increase the yield of a wheat crop.[2]
Show answer
(a) Magnesium is needed to make chlorophyll; without it less chlorophyll forms (leaves yellow); less light is absorbed so less photosynthesis; less glucose is made for growth / respiration.
B1 Mg needed for chlorophyll; B1 less light absorbed / less photosynthesis; B1 less glucose / carbohydrate for growth.
(b) Nitrate is absorbed by root hairs (active transport) and used to make amino acids and proteins, so more new cells and larger plants / more grain protein.
B1 nitrate absorbed by roots; B1 for protein synthesis → growth / higher yield.
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
References only: find each question in its paper (official link where available, otherwise through your school).
