Last reviewed: November 2026.
Choose one answer for each question, then press Finish to see your score and a full explanation for every answer. The questions follow the order of the paper.
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Question 1 Paper ref 1(a)(i) Knowledge A technician compares mature human red blood cells with onion epidermis cells under a microscope. Which statement about the human red blood cells is correct? A They have no cytoplasm, no nucleus and no cell wall. B They have cytoplasm but no nucleus and no cell wall. C They have a cell wall and cytoplasm but no nucleus. D They have cytoplasm and a nucleus but no cell wall.
Why B is correct: Mature human red blood cells keep cytoplasm (full of haemoglobin) but lose their nucleus, and animal cells never have a cell wall.
Why the other options are wrong A) They have no cytoplasm, no nucleus and no cell wall. – This removes the cytoplasm, but the haemoglobin is carried in the cytoplasm.C) They have a cell wall and cytoplasm but no nucleus. – This gives them a cell wall, which belongs to plant cells such as onion, not animal cells.D) They have cytoplasm and a nucleus but no cell wall. – This gives red blood cells a nucleus, which is true of most other animal cells but not mature red blood cells.Key idea: All cells have cytoplasm and a cell membrane. Plant cells also have a cell wall, and most animal cells have a nucleus. Red blood cells lose their nucleus so there is more room to carry oxygen.
Common mistake: Students think every animal cell has a nucleus, forgetting red blood cells lose theirs.
Exam tip: In ‘which row’ questions, check each structure column one at a time.
Memory hook: Red cells: room for oxygen, so no nucleus. Question 2 Paper ref 1(b)(i) Application A student has a 10% sucrose solution and pure water. She needs 20 cm³ of a 2% sucrose solution. Which method makes it? A Measure 16 cm³ of the 10% solution in a syringe and add 4 cm³ of pure water. B Measure 2 cm³ of the 10% solution in a syringe and add 18 cm³ of pure water. C Measure 10 cm³ of the 10% solution in a syringe and add 10 cm³ of pure water. D Measure 4 cm³ of the 10% solution in a syringe and add 16 cm³ of pure water.
Why D is correct: 2% is one fifth of 10%, so one fifth of the 20 cm³ (4 cm³) must be the stock solution and the rest (16 cm³) water.
Worked steps Volume of stock = (wanted % ÷ stock %) × final volume Volume of stock = (2 ÷ 10) × 20 cm³ 0.2 × 20 = 4 cm³ of 10% solution Water = 20 − 4 = 16 cm³ Check: 4 cm³ of 10% in 20 cm³ gives 4 × 10 ÷ 20 = 2% ✓ Why the other options are wrong A) Measure 16 cm³ of the 10% solution in a syringe and add 4 cm³ of pure water. – This swaps the volumes of solution and water and gives an 8% solution.B) Measure 2 cm³ of the 10% solution in a syringe and add 18 cm³ of pure water. – This copies the 2 from ‘2%’ as a volume, and gives only a 1% solution.C) Measure 10 cm³ of the 10% solution in a syringe and add 10 cm³ of pure water. – This halves the concentration, which gives 5%, not 2%.Key idea: A dilution keeps the amount of solute the same and adds water to increase the volume. The new concentration equals the old concentration × (volume of stock ÷ final volume). Accurate measuring tools such as a syringe or measuring cylinder are needed.
Common mistake: Students add equal volumes of water and stock whatever concentration is wanted, which only halves it.
Exam tip: Always check by working out the concentration of your final mixture.
Memory hook: Fraction wanted of the stock, the rest is water. Question 3 Paper ref 1(b)(ii) Application Red onion epidermis cells are left in a strong sucrose solution for one hour. Under the microscope the coloured cytoplasm has pulled away from the cell wall, which keeps its shape. Which explanation is correct? A Water left the cells by osmosis because the outside solution had a higher water potential than the cytoplasm. B Water left the cells by osmosis, moving from a higher to a lower water potential, so the cytoplasm shrank. C Water left the cells by active transport, using energy to move out against its gradient, so the cytoplasm shrank. D Sucrose entered the cells by diffusion, drawing in water, so the cytoplasm swelled and pressed on the cell wall.
Why B is correct: The strong sucrose solution has a lower water potential than the cell contents, so water moves out by osmosis and the cytoplasm shrinks away from the wall (plasmolysis).
Why the other options are wrong A) Water left the cells by osmosis because the outside solution had a higher water potential than the cytoplasm. – This reverses the gradient: water leaves because the outside solution has a LOWER water potential.C) Water left the cells by active transport, using energy to move out against its gradient, so the cytoplasm shrank. – Water movement in osmosis is passive and needs no energy, so active transport is wrong.D) Sucrose entered the cells by diffusion, drawing in water, so the cytoplasm swelled and pressed on the cell wall. – The cytoplasm shrank, it did not swell, and sucrose is too large to diffuse through the membrane easily.Key idea: Osmosis is the passive net movement of water from higher to lower water potential through a partially permeable membrane. In a concentrated solution a plant cell loses water, becomes flaccid and the membrane pulls away from the cell wall (plasmolysis).
Common mistake: Students say water moves from low to high concentration of water, reversing the direction.
Exam tip: Name ‘osmosis’ and state the direction of water movement, as both gain marks.
Memory hook: Strong solution outside: water walks out, cell shrinks. Question 4 Paper ref 2(a) Knowledge In an insect-pollinated flower one part makes pollen and another part becomes the fruit after fertilisation. Which row is correct? A The filament makes pollen and the ovary becomes the fruit. B The stigma makes pollen and the ovary becomes the fruit. C The anther makes pollen and the ovary becomes the fruit. D The anther makes pollen and the ovule becomes the fruit.
Why C is correct: Pollen is made in the anther, and after fertilisation the ovary wall develops into the fruit.
Why the other options are wrong A) The filament makes pollen and the ovary becomes the fruit. – The filament only holds the anther up; pollen is made in the anther.B) The stigma makes pollen and the ovary becomes the fruit. – The stigma receives pollen; it does not make it.D) The anther makes pollen and the ovule becomes the fruit. – The ovule becomes the seed, not the fruit.Key idea: The male part (stamen) has an anther on a filament. The female part (carpel) has a stigma, style and ovary containing ovules. After fertilisation ovules become seeds and the ovary becomes the fruit.
Common mistake: Students confuse ovule (becomes seed) with ovary (becomes fruit).
Exam tip: Learn the flower parts as a list from top to bottom of the carpel and the stamen.
Memory hook: Ovule to seed, ovary to fruit. Question 5 Paper ref 2(b)(i) Application A pollen grain appears 36 mm wide in a microscope image. Its actual diameter is 240 µm. [1 mm = 1000 µm] What is the magnification, to the nearest whole number? A 0.15 B 8 640 000 C 150 D 1500
Why C is correct: After converting 36 mm to 36 000 µm, dividing by 240 µm gives a magnification of ×150.
Worked steps Magnification = image size ÷ actual size Convert units: 36 mm × 1000 = 36 000 µm 36 000 µm ÷ 240 µm = 150 Magnification = ×150 Check: 150 × 240 µm = 36 000 µm = 36 mm ✓ Why the other options are wrong A) 0.15 – This divides 36 by 240 without converting mm to µm.B) 8 640 000 – This multiplies image size by actual size instead of dividing.D) 1500 – This slips by a factor of ten, for example dividing by 24 instead of 240.Key idea: Magnification = image size ÷ actual size, and both must be in the same unit. 1 mm = 1000 µm. A magnification has no units.
Common mistake: Students forget to convert mm into µm before dividing.
Exam tip: Write the formula triangle (I, A, M) and convert units before you start.
Memory hook: Same units first, then image over actual. Question 6 Paper ref 2(c) Application A gardener has one plant with unusual deep red flowers. She wants many plants with exactly the same flowers, quickly, in winter when no insects are around. Why are cuttings better than growing seeds? A Cuttings are asexual, so the new plants are genetically identical clones and can be made at any time. B Cuttings need pollination first, but give offspring with new allele combinations and so more variety. C Cuttings are made by meiosis, so each new plant has half the parent’s chromosomes and grows faster. D Seeds carry disease from the parent and cuttings do not, so the red colour is passed on.
Why A is correct: Cuttings come from one parent by asexual reproduction, so they have the same genes and can be taken whenever the gardener wants, without pollinators.
Why the other options are wrong B) Cuttings need pollination first, but give offspring with new allele combinations and so more variety. – Cuttings do not need pollination, and they give no new allele combinations; that describes seeds.C) Cuttings are made by meiosis, so each new plant has half the parent’s chromosomes and grows faster. – Cuttings grow by mitosis, giving full chromosome sets identical to the parent.D) Seeds carry disease from the parent and cuttings do not, so the red colour is passed on. – Disease can pass through cuttings too, and disease is not the reason for identical flower colour.Key idea: Asexual reproduction involves one parent and produces genetically identical offspring (clones). It is faster, does not depend on pollinators or season, and keeps useful features. Seeds come from sexual reproduction, which gives variation.
Common mistake: Students say cuttings are ‘better’ without saying they are genetically identical.
Exam tip: For ‘explain why’ questions on cuttings, give ‘asexual/clones’ plus one practical advantage.
Memory hook: Cutting equals clone, copy of mum. Question 7 Paper ref 3(a)(i)-(ii) Application In a bell-jar model of breathing, balloons inside the jar are joined to the outside air by a tube, and a rubber sheet seals the bottom. The sheet is pulled down. Which row says what the sheet represents and what happens inside the jar? A The rib cage; the volume increases and the pressure decreases, so air enters the balloons. B The diaphragm; the volume decreases and the pressure increases, so air is forced into the balloons. C The diaphragm; the volume increases and the pressure increases, so air is pushed out of the balloons. D The diaphragm; the volume increases and the pressure decreases, so air enters the balloons.
Why D is correct: The rubber sheet models the diaphragm. Pulling it down enlarges the jar, lowering the pressure, so higher-pressure outside air flows into the balloons (inhalation).
Why the other options are wrong A) The rib cage; the volume increases and the pressure decreases, so air enters the balloons. – The sheet models the diaphragm, not the ribs; the model has no ribs at all.B) The diaphragm; the volume decreases and the pressure increases, so air is forced into the balloons. – This reverses both changes: pulling the sheet down increases volume and lowers pressure.C) The diaphragm; the volume increases and the pressure increases, so air is pushed out of the balloons. – Volume and pressure change in opposite directions, so pressure cannot rise when volume rises.Key idea: Inhalation: the diaphragm contracts and flattens, the intercostal muscles raise the ribs, the chest volume increases and pressure falls below atmospheric pressure, so air moves in. Exhalation is the reverse.
Common mistake: Students think increased volume raises pressure.
Exam tip: Remember Boyle’s idea in words: bigger space, lower pressure.
Memory hook: Bigger chest, lower pressure, air rushes in. Question 8 Paper ref 3(a)(iii) Analysis Which is a valid reason why the bell-jar model does not accurately show human ventilation? A The model’s pressure rises on inhalation, but real lung pressure falls on inhalation. B The sheet moves downwards, but a real diaphragm moves upwards during inhalation. C The balloons are elastic, but real lungs have no elastic tissue that stretches and recoils. D The jar is rigid and has no ribs or intercostal muscles, so only diaphragm movement is modelled.
Why D is correct: The glass jar cannot change shape and has no rib cage or intercostal muscles, so the real chest wall movement is missing.
Why the other options are wrong A) The model’s pressure rises on inhalation, but real lung pressure falls on inhalation. – In the model the pressure also falls on inhalation, so it matches real lungs on this point.B) The sheet moves downwards, but a real diaphragm moves upwards during inhalation. – The real diaphragm flattens and moves downwards on inhalation, just as the sheet does.C) The balloons are elastic, but real lungs have no elastic tissue that stretches and recoils. – Lungs do contain elastic tissue that stretches and recoils.Key idea: Models simplify reality. The bell jar shows the diaphragm and pressure change but lacks ribs, intercostal muscles, pleural membranes and the air sacs of real alveoli. Evaluating a model means saying what it leaves out.
Common mistake: Students criticise parts the model gets right, such as pressure change.
Exam tip: List what the real system has that the model lacks: ribs, muscles, pleural fluid.
Memory hook: Model has the floor but no walls. Question 9 Paper ref 3(b)(ii) Application A lung has 350 000 000 alveoli. Each alveolus has a surface area of 0.0060 mm². What is the total surface area in mm², in standard form? A 2.1 × 10⁶ B 21 × 10⁵ C 2.1 × 10⁷ D 2.1 × 10⁵
Why A is correct: 350 000 000 × 0.0060 = 2 100 000 mm², which is 2.1 × 10⁶ in standard form.
Worked steps Total area = number of alveoli × area of each alveolus 350 000 000 × 0.0060 mm² 350 000 000 × 0.006 = 2 100 000 mm² Standard form: 2.1 × 10⁶ mm² (decimal moved 6 places) Check: 2.1 × 1 000 000 = 2 100 000 ✓ Why the other options are wrong B) 21 × 10⁵ – This equals the right value but is not standard form because 21 is not between 1 and 10.C) 2.1 × 10⁷ – This moves the decimal one place too many, giving 21 000 000.D) 2.1 × 10⁵ – This has the right digits but moves the decimal one place too few, giving 210 000.Key idea: Standard form is A × 10ⁿ with A between 1 and 10. It is used for very large or small numbers. The huge total surface area of the alveoli makes gas exchange fast.
Common mistake: Students count the wrong number of places when writing the power of ten.
Exam tip: Count how many places the decimal point moves and check by expanding back.
Memory hook: A between one and ten, then count the jumps. Question 10 Paper ref 3(b)(iii) Application Which statement correctly links a feature of an alveolus to how it maximises gas exchange? A Its inner surface is kept dry, so gases can move through the wall more quickly. B The surrounding capillaries carry oxygen-rich blood, which keeps a steep concentration gradient. C Its wall is one cell thick, giving a short diffusion path for oxygen and carbon dioxide. D Its wall is thick and muscular, so it can pump oxygen into the blood by active transport.
Why C is correct: A very thin wall means gases have only a tiny distance to diffuse, so exchange is rapid.
Why the other options are wrong A) Its inner surface is kept dry, so gases can move through the wall more quickly. – The surface must be moist so gases can dissolve; dry surfaces would slow exchange.B) The surrounding capillaries carry oxygen-rich blood, which keeps a steep concentration gradient. – The capillaries bring oxygen-POOR blood, and that is what keeps the gradient steep.D) Its wall is thick and muscular, so it can pump oxygen into the blood by active transport. – Gas exchange is by diffusion, and thick walls would slow it down.Key idea: Alveoli have thin walls (one cell thick), a rich blood supply that carries away oxygen and brings carbon dioxide to keep gradients steep, and a moist lining where gases dissolve. Together with their huge surface area these features make diffusion fast.
Common mistake: Students write ‘large surface area’ alone, but must link each feature to its function.
Exam tip: Give structure plus function in pairs: feature, then ‘so that…’.
Memory hook: Thin, wet, well-fed with blood: perfect swapping shop. Question 11 Paper ref 3(b)(iv) Application A student times how long it takes for limewater to turn cloudy as they breathe out through a straw, first at rest and then after two minutes of running. Which plan gives the fairest comparison of the rate of carbon dioxide release? A Use the same person but time only once, since exercise gives a clear difference. B Use the same volume of limewater and the same person each time, and repeat to find a mean. C Change the room temperature as well as the exercise to see which has more effect. D Use more limewater after exercise because more carbon dioxide is released.
Why B is correct: Keeping the limewater volume and the person the same controls other variables, and repeating improves reliability.
Why the other options are wrong A) Use the same person but time only once, since exercise gives a clear difference. – A single reading cannot show whether the result is reliable.C) Change the room temperature as well as the exercise to see which has more effect. – Changing two things at once means you cannot tell which caused the change.D) Use more limewater after exercise because more carbon dioxide is released. – Changing the volume of limewater adds a second variable, so the times cannot be compared.Key idea: In a fair test, change one independent variable (exercise), measure the dependent variable (time for limewater to go cloudy) and keep control variables such as limewater volume, room temperature and the person constant. Repeating allows a mean to be calculated.
Common mistake: Students change more than one variable at a time.
Exam tip: State what you change, what you measure, and two things you keep the same.
Memory hook: Change one, measure one, keep the rest still. Question 12 Paper ref 4(a)(i)-(ii) Knowledge Amylase digests starch. Which row gives the product of this digestion and the parts of the body that make amylase? A Sucrose; the salivary glands and the pancreas. B Glycerol; the salivary glands and the pancreas. C Maltose; the salivary glands and the pancreas. D Maltose; the colon and the pancreas.
Why C is correct: Amylase breaks starch into maltose, and it is made by the salivary glands and the pancreas.
Why the other options are wrong A) Sucrose; the salivary glands and the pancreas. – Sucrose is not produced when starch is digested.B) Glycerol; the salivary glands and the pancreas. – Glycerol is a product of fat digestion, not starch.D) Maltose; the colon and the pancreas. – The colon does not make amylase.Key idea: Carbohydrases such as amylase break down carbohydrates: starch to maltose. Proteases make amino acids from protein, and lipases make fatty acids and glycerol from fats. Amylase is secreted in saliva and pancreatic juice.
Common mistake: Students list the stomach or colon as sources of amylase.
Exam tip: Match each enzyme to its substrate and product as a trio.
Memory hook: Amylase: starch to maltose, mouth and pancreas. Question 13 Paper ref 4(b)(ii) Analysis Samples of a starch and amylase mixture are tested with iodine solution every five minutes. At 20 minutes the sample turns blue-black. At 25 minutes it stays orange-brown. What does this show? A Maltose was present at 20 minutes and had been used up by 25 minutes. B Starch was absent at 20 minutes and had been made by 25 minutes. C Amylase was denatured at 20 minutes but became active again by 25 minutes. D Starch was still present at 20 minutes and had all been digested by 25 minutes.
Why D is correct: Iodine turns blue-black only when starch is present, so the colour change to orange-brown shows the starch has gone.
Why the other options are wrong A) Maltose was present at 20 minutes and had been used up by 25 minutes. – Iodine does not detect maltose; it only detects starch.B) Starch was absent at 20 minutes and had been made by 25 minutes. – Blue-black means starch is present, so this reverses the meaning of the colours.C) Amylase was denatured at 20 minutes but became active again by 25 minutes. – Denaturation is permanent and iodine does not test enzyme activity.Key idea: The iodine test for starch: add iodine solution and a blue-black colour means starch is present, while orange-brown means absent. In enzyme investigations the time for the blue-black colour to disappear measures the rate of digestion.
Common mistake: Students say iodine turns purple or red; the positive colour is blue-black.
Exam tip: State the reagent and BOTH colours: blue-black for present, orange-brown for absent.
Memory hook: Iodine and starch: blue-black meets. Question 14 Paper ref 4(c)(i) Application In an amylase investigation at 35 °C the time for all starch to be digested was measured in three runs: A 15 B 15.3 C 16 D 46
Why A is correct: The total of 46 minutes divided by 3 runs gives 15.33, which is 15 to two significant figures.
Worked steps Mean = total ÷ number of runs Total = 14 + 16 + 16 = 46 minutes 46 ÷ 3 = 15.33 minutes To 2 significant figures = 15 minutes Check: 15 × 3 = 45, close to 46 ✓ Why the other options are wrong B) 15.3 – This is the right mean but given to three significant figures, not two.C) 16 – This is the most common value (mode), not the mean.D) 46 – This is the total of the three runs without dividing by 3.Key idea: A mean is the sum of repeats divided by the number of repeats. Calculating means makes results more reliable. Significant figures count from the first non-zero digit.
Common mistake: Students round to the wrong number of significant figures.
Exam tip: Write the unrounded value first, then round as the question asks.
Memory hook: Add them up, share them out. Question 15 Paper ref 4(c)(ii) Analysis In an amylase investigation, the time to digest all starch fell from 48 minutes at 10 °C to 12 minutes at 40 °C. Which explanation is correct? A Enzyme and substrate gained kinetic energy, so they collided more often and formed more enzyme-substrate complexes. B Higher temperature increased the amount of amylase in the tube, so more starch was digested. C Higher temperature lowered the activation energy of the reaction, so each collision was more likely to work. D Higher temperature changed the active site so that it fitted the starch more tightly.
Why A is correct: Warmer molecules move faster, so collisions between amylase and starch are more frequent, speeding up digestion.
Why the other options are wrong B) Higher temperature increased the amount of amylase in the tube, so more starch was digested. – The amount of amylase is fixed by the volume added and does not change with temperature.C) Higher temperature lowered the activation energy of the reaction, so each collision was more likely to work. – Enzymes lower activation energy; changing the temperature does not.D) Higher temperature changed the active site so that it fitted the starch more tightly. – A change of active site shape is denaturation, which slows or stops the reaction.Key idea: Up to the optimum, rising temperature increases the kinetic energy of enzyme and substrate molecules, giving more frequent successful collisions and a faster rate. Beyond the optimum the enzyme denatures.
Common mistake: Students say heat ‘activates’ enzymes or gives them energy to work, without mentioning collisions.
Exam tip: Use the chain: more kinetic energy, faster movement, more collisions, more complexes.
Memory hook: Warm molecules race, collide, react. Question 16 Paper ref 4(c)(iii) Application At 50 °C starch was still present in an amylase mixture long after the tests at lower temperatures had finished. What is the best explanation? A The amylase was killed by the heat, so it could not carry out digestion any more. B The starch molecules changed shape at 50 °C, so they no longer fitted the active site. C Too many collisions blocked the active sites, so starch could not reach them. D The amylase was denatured, so its active site changed shape and starch could no longer bind.
Why D is correct: High temperature breaks bonds in the enzyme, changing the shape of the active site so the substrate does not fit.
Why the other options are wrong A) The amylase was killed by the heat, so it could not carry out digestion any more. – Enzymes are proteins, not living things, so they are denatured rather than killed.B) The starch molecules changed shape at 50 °C, so they no longer fitted the active site. – It is the enzyme’s active site that changes shape, not the substrate.C) Too many collisions blocked the active sites, so starch could not reach them. – More collisions would speed the reaction; blocking by collisions is not a cause of denaturation.Key idea: Enzymes are proteins with a specific active site shape. Above the optimum temperature the bonds holding that shape break, so the enzyme is denatured and the substrate no longer fits (it is no longer complementary). Denaturation is permanent.
Common mistake: Students say the enzyme was ‘killed’.
Exam tip: Use the word ‘denatured’ and mention the active site and substrate.
Memory hook: Too hot: active site loses its shape. Question 17 Paper ref 4(c)(iv) Evaluation An amylase investigation gave digestion times of 23 minutes at 30 °C, 12 minutes at 40 °C, and no digestion at 50 °C. How could the student more accurately find the temperature at which amylase works fastest? A Use more amylase solution at each temperature so that digestion is easier to time. B Test more temperatures close to 40 °C, such as 35, 38, 42 and 45 °C, as the fastest may not be 40 °C. C Test 60, 70, 80 and 90 °C as well to see where amylase works fastest. D Repeat 40 °C many more times, since repeats show whether it is the fastest.
Why B is correct: The fastest rate lies somewhere between 30 and 50 °C, so smaller temperature intervals around 40 °C are needed to locate the optimum.
Why the other options are wrong A) Use more amylase solution at each temperature so that digestion is easier to time. – Changing the amount of amylase changes the rate and does not help find the best temperature.C) Test 60, 70, 80 and 90 °C as well to see where amylase works fastest. – Amylase is already not working at 50 °C, so higher temperatures add no useful information.D) Repeat 40 °C many more times, since repeats show whether it is the fastest. – Repeats improve reliability at 40 °C but do not test other temperatures.Key idea: To pinpoint an optimum, use smaller intervals of the independent variable near the peak of the results. Smaller time intervals between tests also help. Wider ranges or more repeats do not locate the optimum more precisely.
Common mistake: Students suggest more repeats when the question asks for greater accuracy of an optimum.
Exam tip: Say ‘smaller temperature intervals between 30 and 50 °C’, not just ‘more temperatures’.
Memory hook: To find the peak, zoom in on the peak. Question 18 Paper ref 5(a)(i) Knowledge Phytoplankton are protoctists that photosynthesise. Which pair of features do they share with plant cells? A A nucleus and a chitin cell wall. B Chloroplasts and a nucleus. C Chloroplasts only, as protoctists have no nucleus. D Chloroplasts and a chitin cell wall.
Why B is correct: Phytoplankton photosynthesise using chloroplasts and, like all protoctists, are eukaryotes with a nucleus.
Why the other options are wrong A) A nucleus and a chitin cell wall. – Chitin is not in plants or phytoplankton, so it cannot be shared.C) Chloroplasts only, as protoctists have no nucleus. – Protoctists are eukaryotes and do have a nucleus.D) Chloroplasts and a chitin cell wall. – Plant cell walls are made of cellulose; chitin is found in fungi.Key idea: Plants have chloroplasts, a nucleus and a cellulose cell wall. Fungi have chitin in their walls. Protoctists are single-celled eukaryotes, and some, like phytoplankton, photosynthesise.
Common mistake: Students think chitin is the plant cell wall material.
Exam tip: Link: cellulose for plants, chitin for fungi.
Memory hook: Plants cellulose, fungi chitin. Question 19 Paper ref 5(a)(ii)-(iii) Application Part of an ocean food web has these feeding links (arrow means ‘is eaten by’): A phytoplankton → zooplankton → plankton-eating fish → seal → polar bear B phytoplankton → plankton-eating fish → predatory fish → seal → polar bear C phytoplankton → zooplankton → plankton-eating fish → grey whale D phytoplankton → zooplankton → plankton-eating fish → predatory fish → seal → polar bear
Why D is correct: This chain passes through six organisms: phytoplankton, zooplankton, plankton-eating fish, predatory fish, seal and polar bear.
Why the other options are wrong A) phytoplankton → zooplankton → plankton-eating fish → seal → polar bear – This skips predatory fish, giving only five trophic levels.B) phytoplankton → plankton-eating fish → predatory fish → seal → polar bear – This skips zooplankton, giving only five trophic levels.C) phytoplankton → zooplankton → plankton-eating fish → grey whale – This stops at the grey whale and has only four trophic levels.Key idea: A food chain starts with a producer and arrows point in the direction of energy flow, from the food to the feeder. Each step is a trophic level. Tertiary consumers eat secondary consumers, such as the predatory fish, seal and polar bear here.
Common mistake: Students draw arrows from predator to prey.
Exam tip: Trace every path from the producer to the top predator and count the organisms.
Memory hook: Arrows follow the energy, from eaten to eater. Question 20 Paper ref 5(b)(i) Application The biomass of phytoplankton in the water increased from 2.5 mg per dm³ to 17.5 mg per dm³ in one month. What is the percentage increase? A 15 B 86 C 600 D 700
Why C is correct: The increase is 15 mg per dm³, which is 6 times the starting value, so 600%.
Worked steps % increase = (new − original) ÷ original × 100 Increase = 17.5 − 2.5 = 15 mg per dm³ 15 ÷ 2.5 = 6 6 × 100 = 600% Check: 2.5 + 600% of 2.5 = 2.5 + 15 = 17.5 ✓ Why the other options are wrong A) 15 – This is the increase in mg per dm³, not a percentage.B) 86 – This divides the increase by the final value instead of the starting value.D) 700 – This divides the final value by the start (7 × 100), giving the new amount as a percentage, not the increase.Key idea: Percentage change = change ÷ original value × 100. Always divide by the starting value. In ecology, big percentage rises in producers such as phytoplankton happen when light and nutrients are plentiful.
Common mistake: Students divide the change by the final value.
Exam tip: Write ‘change ÷ original × 100’ before you substitute.
Memory hook: Change over original, times a hundred. Question 21 Paper ref 5(b)(ii) Analysis In an ocean, phytoplankton biomass peaks sharply in spring. In early summer the biomass drops steeply, while hours of light per day are still high, nitrate concentration is very low and zooplankton biomass is rising. What best explains the fall in phytoplankton? A Zooplankton compete for light, so phytoplankton can no longer photosynthesise. B Day length has fallen sharply, so photosynthesis can no longer make glucose. C Zooplankton have released extra nitrate into the water, which poisons the phytoplankton. D The growing population used up the nitrate, and the increasing zooplankton are eating the phytoplankton.
Why D is correct: Phytoplankton absorb nitrate to make amino acids and chlorophyll; when it runs out they cannot grow, and more zooplankton graze them.
Why the other options are wrong A) Zooplankton compete for light, so phytoplankton can no longer photosynthesise. – Zooplankton are consumers and do not compete with phytoplankton for light.B) Day length has fallen sharply, so photosynthesis can no longer make glucose. – Light hours are still high in early summer, so light is not what falls.C) Zooplankton have released extra nitrate into the water, which poisons the phytoplankton. – Zooplankton do not add poisonous nitrate; the nitrate is low because phytoplankton took it up.Key idea: Phytoplankton biomass increases with light and nitrate for photosynthesis and growth. It falls when nitrate becomes limiting or when zooplankton feed on it. Zooplankton rise after phytoplankton because they have more food.
Common mistake: Students blame light for every fall in phytoplankton, ignoring nitrate and grazing.
Exam tip: Use the data: say which factor is high and which is low at that time.
Memory hook: Plankton boom, nitrate gone, grazers feast. Question 22 Paper ref 6(a) Application A person’s finger touches a hot object and the hand pulls away automatically. Which sequence shows the pathway of the nerve impulse in the reflex arc? A receptor → sensory neurone → relay neurone in the spinal cord → motor neurone → effector muscle B effector muscle → sensory neurone → relay neurone in the spinal cord → motor neurone → receptor C receptor → sensory neurone → brain → motor neurone → effector muscle D receptor → motor neurone → relay neurone in the spinal cord → sensory neurone → effector muscle
Why A is correct: The impulse starts in the receptor, travels along a sensory neurone to the relay neurone in the CNS, then along a motor neurone to the effector.
Why the other options are wrong B) effector muscle → sensory neurone → relay neurone in the spinal cord → motor neurone → receptor – This runs the arc backwards, starting at the effector.C) receptor → sensory neurone → brain → motor neurone → effector muscle – A spinal reflex uses a relay neurone in the spinal cord rather than waiting for the brain.D) receptor → motor neurone → relay neurone in the spinal cord → sensory neurone → effector muscle – This swaps sensory and motor neurones; sensory neurones carry impulses toward the CNS.Key idea: A reflex arc is receptor, sensory neurone, relay neurone (in the spinal cord), motor neurone, effector. Impulses cross synapses between neurones using neurotransmitters. Reflexes are fast and automatic and protect the body.
Common mistake: Students send the impulse through the brain for a simple reflex.
Exam tip: Learn the arc in order, and add synapses and neurotransmitters for extra marks.
Memory hook: Receptor, sensory, relay, motor, effector: R-S-R-M-E. Question 23 Paper ref 6(b)(ii) Analysis A student touches a snail’s eyestalk eight times and records how long the snail stays in its shell after each touch. Which description of the trend in the chart above is correct? Time snail stays in shell
0 5 10 15 1 2 3 4 5 6 7 8 Stimulus number Time (s) Show the chart data as a table Stimulus number Time in shell (s) 1 15 2 15 3 14 4 15 5 12 6 8 7 5 8 3
A The time stays at about 15 seconds for four stimuli, then falls more steeply from stimulus 5. B The time falls by the same amount after every stimulus, from 15 s to 3 s. C The time falls steeply at first and then levels off after the fourth stimulus. D The time rises slightly until the fourth stimulus and then stays at 15 seconds.
Why A is correct: Times are 14–15 s for stimuli 1 to 4, then drop to 12, 8, 5 and 3 s, so the decrease is larger at higher stimulus numbers.
Why the other options are wrong B) The time falls by the same amount after every stimulus, from 15 s to 3 s. – The drops are not equal; the first four changes are tiny while later ones are 3 to 4 s.C) The time falls steeply at first and then levels off after the fourth stimulus. – This reverses the pattern; the fall is small at first and steeper later.D) The time rises slightly until the fourth stimulus and then stays at 15 seconds. – The time does not stay at 15 s; it falls to 3 s by stimulus 8.Key idea: When describing a graph, state the overall trend and then any change in steepness, using values from the axes. Here the snail’s response weakens as the same harmless stimulus is repeated.
Common mistake: Students describe only ‘it goes down’ without noting where the change in rate happens.
Exam tip: Quote data pairs, such as ’15 s at stimulus 4 to 3 s at stimulus 8′.
Memory hook: Flat first, then a slide. Question 24 Paper ref 6(b)(iii) Evaluation A snail withdraws into its shell for a shorter time each time its eyestalk is touched repeatedly with a cotton wool bud. What is a benefit to the snail of this change in response? A It shows damage to the snail’s nerves, so it saves energy because it senses fewer stimuli. B It stops wasting time hiding from a harmless touch, so it can feed and move sooner, while still reacting to new danger. C It stays hidden longer each time, which gives more protection from a predator that keeps attacking. D It lets the snail respond faster, because the reflex pathway skips the effector after many touches.
Why B is correct: A repeated touch with nothing harmful following suggests no danger, so ignoring it (habituation) saves energy and allows feeding and movement.
Why the other options are wrong A) It shows damage to the snail’s nerves, so it saves energy because it senses fewer stimuli. – The change is learned and useful, not nerve damage, and the snail still responds to new stimuli.C) It stays hidden longer each time, which gives more protection from a predator that keeps attacking. – The results show shorter times, not longer ones.D) It lets the snail respond faster, because the reflex pathway skips the effector after many touches. – The reflex always uses an effector; the time in the shell is what shortens.Key idea: Habituation is a learned reduction in response to a repeated harmless stimulus. A first touch might be a predator, so hiding helps survival. Repeated touches without harm show no threat, so the snail saves energy and spends more time feeding.
Common mistake: Students say the snail ‘gets used to it’ without stating the benefit.
Exam tip: Give a survival benefit and a use of time or energy.
Memory hook: Same poke, no harm: stop hiding, get eating. Question 25 Paper ref 7(a)(i) Analysis Which statement correctly compares the wall of an artery with the wall of a vein and gives the reason? A The artery wall has thinner elastic and muscle layers because the blood is at lower pressure. B The artery wall has a wider lumen because blood moves more slowly than in a vein. C The artery wall has thicker elastic and muscle layers because the blood is at higher pressure. D The artery wall has valves in it because blood must be stopped from flowing backwards.
Why C is correct: Arteries carry blood away from the heart at high pressure, so their walls have thick elastic tissue (to stretch and recoil) and muscle.
Why the other options are wrong A) The artery wall has thinner elastic and muscle layers because the blood is at lower pressure. – This is the reverse; thin walls belong to veins where pressure is low.B) The artery wall has a wider lumen because blood moves more slowly than in a vein. – Arteries have a narrow lumen, and blood travels fast, not slowly.D) The artery wall has valves in it because blood must be stopped from flowing backwards. – Valves are a feature of veins, which have low pressure; arteries have no valves.Key idea: Arteries have thick, elastic and muscular walls, a narrow lumen and no valves, because they carry blood at high pressure. Veins have thin walls, a wide lumen and valves to prevent backflow at low pressure.
Common mistake: Students give valves to arteries.
Exam tip: Always explain the structure with ‘because of high or low pressure’.
Memory hook: Arteries: tough and stretchy. Veins: thin with valves. Question 26 Paper ref 7(a)(ii) Analysis During exercise, why does the heart of a person with coronary heart disease respond less well than a healthy heart? A Narrowed coronary arteries mean fewer heart muscle cells are present, so the heart is smaller and weaker. B Narrowed coronary arteries bring less glucose but plenty of oxygen, so the heart muscle makes lactic acid instead. C Narrowed coronary arteries bring less oxygen, so less aerobic respiration and ATP are available and the heart cannot beat faster. D Narrowed coronary arteries stop blood leaving the heart, so the chambers cannot fill and pump harder.
Why C is correct: Less oxygen reaches the heart muscle, limiting aerobic respiration and the energy (ATP) needed for faster or stronger beating.
Why the other options are wrong A) Narrowed coronary arteries mean fewer heart muscle cells are present, so the heart is smaller and weaker. – The disease does not reduce cell number at once; it restricts blood supply.B) Narrowed coronary arteries bring less glucose but plenty of oxygen, so the heart muscle makes lactic acid instead. – Oxygen delivery is reduced as well as glucose, and the key problem is the energy shortage.D) Narrowed coronary arteries stop blood leaving the heart, so the chambers cannot fill and pump harder. – Coronary arteries supply the heart muscle itself; they do not carry blood out of the heart.Key idea: The coronary arteries supply oxygenated blood to heart muscle. In CHD fatty deposits narrow them, so less oxygen and glucose arrive, less ATP is made by respiration, and the heart cannot meet the extra demand during exercise.
Common mistake: Students say the heart gets ‘no blood’ instead of explaining reduced oxygen and energy release.
Exam tip: Chain it: blockage, less oxygen, less respiration, less energy, cannot beat faster.
Memory hook: Narrow pipe, less oxygen, less energy, weak beat. Question 27 Paper ref 7(b)(ii) Analysis The chart above shows the relative risk of coronary heart disease (CHD) for non-smokers and smokers with few, medium or many risk factor alleles. Which conclusion is supported by the chart? Relative risk of CHD
0 1 2 3 4 few medium high Risk alleles Relative risk 1 2 3 3 3.5 4 Non-smokers Smokers Show the chart data as a table Risk alleles Non-smokers Smokers few 1 3 medium 2 3.5 high 3 4
A In smokers, a higher number of risk alleles does not change the relative risk. B Smokers with few risk alleles have the same relative risk as non-smokers with many risk alleles. C Smoking causes the biggest rise in risk in the group with the most risk alleles. D Smoking adds the same amount of extra risk in every risk allele group.
Why B is correct: Smokers with few alleles have a relative risk of 3, equal to the non-smokers with high numbers of alleles (3).
Why the other options are wrong A) In smokers, a higher number of risk alleles does not change the relative risk. – Smokers’ risk rises from 3 to 3.5 to 4 as alleles increase.C) Smoking causes the biggest rise in risk in the group with the most risk alleles. – The biggest increase from smoking (2) is in the few-allele group, not the high-allele group.D) Smoking adds the same amount of extra risk in every risk allele group. – The extra risk from smoking is 2, 1.5 and 1 in the three groups, so it is not constant.Key idea: Both genes (risk alleles) and smoking raise CHD risk, and their effects can be compared with data. Quoting values and calculating differences (for example 3 − 1 = 2) makes a graph conclusion convincing.
Common mistake: Students describe only the general trend and do not quote any numbers from the graph.
Exam tip: Read the exact heights from the chart and subtract to compare effects.
Memory hook: Smoke can match many alleles. Question 28 Paper ref 7(b)(ii) Knowledge Smoking increases the risk of coronary heart disease in all groups. Which statement explains one way smoking does this? A Tar coats the alveoli, so less oxygen reaches the blood and the heart muscle respires more quickly. B Carbon monoxide in smoke damages the artery wall, and fatty plaques build up and block the coronary arteries. C Smoke raises blood glucose, so sugar crystals form and block the coronary arteries. D Nicotine thins the wall of the coronary arteries, so they burst and release blood into the heart muscle.
Why B is correct: Chemicals in cigarette smoke such as carbon monoxide are linked with plaque formation, narrowing the coronary arteries and cutting the oxygen supply to heart muscle.
Why the other options are wrong A) Tar coats the alveoli, so less oxygen reaches the blood and the heart muscle respires more quickly. – Tar harms the lungs, but this option wrongly says the heart respires faster; the problem is reduced supply.C) Smoke raises blood glucose, so sugar crystals form and block the coronary arteries. – Smoke does not make sugar crystals; the blockage is fatty plaque.D) Nicotine thins the wall of the coronary arteries, so they burst and release blood into the heart muscle. – Smoking does not thin artery walls; it leads to build-up of deposits.Key idea: Smoking damages artery walls and promotes build-up of cholesterol plaques (atheroma), narrowing the coronary arteries. This reduces oxygen delivery to the heart muscle and can cause CHD. Other risk factors include a high-fat diet, lack of exercise and high blood pressure.
Common mistake: Students link smoking to lung damage only and miss its effect on arteries.
Exam tip: Link the chemical, the artery effect and the heart muscle outcome.
Memory hook: Smoke scars arteries, fat sticks, flow slows. Question 29 Paper ref 8(a)(i) Application Lactose intolerance is caused by a recessive allele, d. The ability to digest lactose is caused by the dominant allele, D. A woman and a man both digest lactose normally, and each has a parent with lactose intolerance (dd). What is the probability that their child will be lactose intolerant? A 0.25 B 0.33 C 0.5 D 0.75
Why A is correct: Each partner must be Dd (they got d from the dd parent and can digest lactose), and Dd × Dd gives a 1 in 4 chance of dd.
Worked steps Each partner digests lactose (has D) and has a dd parent (gave d), so both are Dd Gametes from each parent: D or d Punnett square gives offspring DD, Dd, Dd, dd Only dd is lactose intolerant: 1 out of 4 Probability = 1 ÷ 4 = 0.25 (25%) Why the other options are wrong B) 0.33 – This treats only the three tolerant genotypes as the chances, ignoring that DD, Dd, Dd and dd are four equally likely outcomes.C) 0.5 – This would be correct for Dd × dd; one parent is wrongly given dd.D) 0.75 – This is the chance of a child who can digest lactose, not the chance of intolerance.Key idea: A recessive condition shows only in dd individuals. Two Dd carriers have a 25% chance of a dd child. Work out parent genotypes from the information given before drawing the cross.
Common mistake: Students give a parent the wrong genotype and use 50% automatically.
Exam tip: Show parent genotypes, gametes, offspring genotypes and then the probability.
Memory hook: Two carriers: one in four. Question 30 Paper ref 8(a)(ii) Analysis Why do people who cannot digest lactose often get watery faeces (diarrhoea) after drinking milk? A Undigested lactose is absorbed into the blood unchanged and makes the kidneys remove extra water in the faeces. B Undigested lactose raises the water potential in the gut, so water moves from the blood into the intestine. C Undigested lactose breaks down the walls of the villi, so the gut cannot absorb any substances. D Undigested lactose stays in the gut and lowers its water potential, so less water is absorbed into the blood by osmosis.
Why D is correct: Lactose that is not digested cannot be absorbed, so it stays in the colon and intestine, keeping water there instead of letting it pass into the blood.
Why the other options are wrong A) Undigested lactose is absorbed into the blood unchanged and makes the kidneys remove extra water in the faeces. – Lactose cannot be absorbed unless it is first digested by lactase.B) Undigested lactose raises the water potential in the gut, so water moves from the blood into the intestine. – Sugar lowers, not raises, the water potential of the gut contents.C) Undigested lactose breaks down the walls of the villi, so the gut cannot absorb any substances. – Lactose does not dissolve the villi walls; the problem is osmotic.Key idea: Only small digestion products such as glucose and galactose can be absorbed. Undigested lactose stays in the gut, lowers water potential, reduces water absorption by osmosis and may feed gut bacteria, so faeces become watery.
Common mistake: Students say lactose ’causes’ water to be added to the gut without explaining osmosis.
Exam tip: Mention not absorbed, water potential, and less water absorbed.
Memory hook: Lactose left in the gut holds the water. Question 31 Paper ref 8(a)(iii) Analysis Fewer adults are lactose intolerant in regions where milk is a major food all through life. Which statement explains this by natural selection? A People who drank milk changed their own genes, so their children inherited the ability to digest lactose. B A mutation created allele D; people with it digested milk, gained nutrients, survived and reproduced more, passing on D. C Everyone with allele d in milk-drinking regions died of diarrhoea before reproducing, so no one carried d. D The allele d changed into allele D in all people because a milk diet made it necessary.
Why B is correct: Natural selection needs variation (from mutation), an advantage for some individuals, more reproduction by them and passing the allele to offspring.
Why the other options are wrong A) People who drank milk changed their own genes, so their children inherited the ability to digest lactose. – Characteristics acquired during life do not alter the genes passed on to children.C) Everyone with allele d in milk-drinking regions died of diarrhoea before reproducing, so no one carried d. – Not everyone with d died; the allele would remain in carriers (Dd), so this exaggerates the effect.D) The allele d changed into allele D in all people because a milk diet made it necessary. – Mutations are random and do not occur because they are needed.Key idea: Natural selection: random mutation creates variation; individuals with an advantage (here digesting milk) survive and reproduce more; they pass the beneficial allele on, so it becomes more common over generations.
Common mistake: Students write that organisms ‘adapt’ or ‘need’ to change.
Exam tip: Use the four steps: mutation, advantage, survive and reproduce, pass on allele.
Memory hook: Variation, advantage, survive, breed, pass it on. Question 32 Paper ref 8(b)(i)-(ii) Application An industrial yoghurt fermenter has a stirrer inside a tank, with a water jacket around the outside. Which row correctly gives the roles of the stirrer and the water jacket? A The stirrer mixes milk and bacteria and spreads heat evenly; the water jacket boils the milk to sterilise it. B The stirrer breaks open bacterial cells to release enzymes; the water jacket removes heat to keep the optimum temperature. C The stirrer mixes milk and bacteria and spreads heat evenly; the water jacket removes heat to keep the optimum temperature. D The stirrer adds oxygen for bacteria to respire aerobically; the water jacket warms the milk to speed up growth.
Why C is correct: Stirring keeps contents evenly mixed with an even temperature, and cool water flowing through the jacket takes away heat made by the bacteria.
Why the other options are wrong A) The stirrer mixes milk and bacteria and spreads heat evenly; the water jacket boils the milk to sterilise it. – Boiling would kill the bacteria needed to make yoghurt.B) The stirrer breaks open bacterial cells to release enzymes; the water jacket removes heat to keep the optimum temperature. – The stirrer mixes; it does not break open bacteria, which would stop fermentation.D) The stirrer adds oxygen for bacteria to respire aerobically; the water jacket warms the milk to speed up growth. – The stirrer is not there to add oxygen, and the jacket cools rather than warms the fermenter.Key idea: In a fermenter the stirrer keeps microorganisms and nutrients in contact and prevents hot spots. A water jacket removes heat made by the microorganisms to keep the optimum temperature, so that enzymes are not denatured and the bacteria do not die.
Common mistake: Students write that the stirrer just ‘stirs’ without saying why it helps.
Exam tip: For each part of a fermenter say what it does and why that helps growth.
Memory hook: Stirrer spreads, jacket cools: bacteria stay happy. Question 33 Paper ref 9(a) Application A gene for efficient photosynthesis is to be placed into tomato plant cells using a plasmid vector. How is the recombinant plasmid made? A Cut the gene and the plasmid with ligase, then join the sticky ends together with a restriction enzyme. B Copy the gene using the polymerase chain reaction and fuse it to the plasmid using restriction enzymes. C Cut the gene and the plasmid with a restriction enzyme, then join the gene into the plasmid with ligase. D Cut the gene out with a restriction enzyme and place it straight into the plasmid with DNA polymerase.
Why C is correct: Restriction enzymes cut DNA at specific sequences and ligase seals the gene into the cut plasmid, forming a recombinant plasmid.
Why the other options are wrong A) Cut the gene and the plasmid with ligase, then join the sticky ends together with a restriction enzyme. – This swaps the enzymes: ligase joins, restriction enzymes cut.B) Copy the gene using the polymerase chain reaction and fuse it to the plasmid using restriction enzymes. – PCR copies DNA but restriction enzymes cut, so ligase is still needed to join.D) Cut the gene out with a restriction enzyme and place it straight into the plasmid with DNA polymerase. – DNA polymerase copies DNA; it does not join a gene into a cut plasmid.Key idea: Genetic engineering uses restriction enzymes to cut out a gene and open a plasmid, and ligase to join the gene into the plasmid. The recombinant plasmid is then put into the host cell, here a tomato cell.
Common mistake: Students swap restriction enzyme and ligase.
Exam tip: Remember: restriction cuts, ligase links.
Memory hook: Restriction cuts, ligase links. Question 34 Paper ref 9(b) Analysis Tomato plants are grown at high light intensity. As temperature rises, the rate of photosynthesis first increases and then levels off at a steady value without falling. What is the most likely reason it levels off? A Temperature has no further effect, because all the enzymes have been used up. B The enzymes have been denatured by the temperature, so the rate has stopped rising. C Light intensity has become limiting, because the light supply was high only at the start. D Another factor, such as carbon dioxide concentration, has become limiting instead of temperature.
Why D is correct: At high light the rate is fastest, but once temperature is no longer limiting, another factor such as carbon dioxide stops it rising further.
Why the other options are wrong A) Temperature has no further effect, because all the enzymes have been used up. – Enzymes are not used up; they are reusable catalysts.B) The enzymes have been denatured by the temperature, so the rate has stopped rising. – Denaturation would make the rate fall, not stay level.C) Light intensity has become limiting, because the light supply was high only at the start. – The light intensity is held high throughout this experiment, so it is not the limiting factor.Key idea: The rate of photosynthesis is controlled by the factor in shortest supply (limiting factor): light, carbon dioxide or temperature. A graph that levels off shows that another factor is now limiting. A fall at high temperature shows denaturation.
Common mistake: Students say enzymes denature wherever a graph levels off.
Exam tip: Level means ‘another factor limiting’; falling means ‘denaturing’.
Memory hook: Flat line: something else is the bottleneck. Question 35 Paper ref 9(c) Evaluation In a UK glasshouse, tomatoes are lit and heated using electricity from a wood-burning power station, and the station’s carbon dioxide is piped into the glasshouse. Which statement gives a correct advantage for farmers and for the environment? A Farmers get high yields all year as no factor limits photosynthesis; the environment gains because wood is renewable and CO₂ is used rather than released. B Farmers save money because plants no longer need light; the environment gains because wood burning produces no carbon dioxide at all. C Farmers avoid buying heaters because the CO₂ keeps the glasshouse warm; the environment gains because no electricity is needed. D Farmers get high yields all year because the plants ignore the seasons; the environment gains because the power station releases more sulfur dioxide.
Why A is correct: Supplying light, CO₂ and warmth removes limiting factors so yields are high, and the CO₂ is captured by plants instead of entering the atmosphere, with wood being renewable.
Why the other options are wrong B) Farmers save money because plants no longer need light; the environment gains because wood burning produces no carbon dioxide at all. – Plants still need light, and burning wood does produce carbon dioxide; it is piped into the glasshouse.C) Farmers avoid buying heaters because the CO₂ keeps the glasshouse warm; the environment gains because no electricity is needed. – The heater and lighting use electricity, so the system still needs energy.D) Farmers get high yields all year because the plants ignore the seasons; the environment gains because the power station releases more sulfur dioxide. – More sulfur dioxide would increase acid rain, which is not an advantage.Key idea: Controlled-environment glasshouses give plants optimum light, CO₂ and temperature, protecting them from pests and bad weather and giving high year-round yield. Using wood (renewable) and recycling CO₂ reduces fossil fuel use and greenhouse gas release.
Common mistake: Students write only about yield and forget the environmental benefit asked for.
Exam tip: When asked for two groups (farmers and environment), give at least one point for each.
Memory hook: Plants eat the power station’s exhaust. Question 36 Paper ref 10 Application A farmer wants to test whether the colour of a scarecrow’s clothes changes how well it keeps birds off a crop. Which plan is the best design? A Put scarecrows of different colours in matching plots of the same crop and measure the height of the crop after a month. B Put scarecrows of identical design but different colours in matching plots of the same crop, count birds over the same time, and repeat. C Put scarecrows of different colours and sizes in plots of different crops, count birds over the same time, and repeat. D Use one scarecrow whose colour is changed each week, then count birds once at the end of the season.
Why B is correct: Only colour changes; scarecrow design, crop, plot and time are controlled, bird numbers are measured and repeats allow a mean.
Why the other options are wrong A) Put scarecrows of different colours in matching plots of the same crop and measure the height of the crop after a month. – Crop height is not a direct measure of bird numbers or crop eaten, so the dependent variable is poorly chosen.C) Put scarecrows of different colours and sizes in plots of different crops, count birds over the same time, and repeat. – Changing size and crop as well as colour means more than one variable is altered.D) Use one scarecrow whose colour is changed each week, then count birds once at the end of the season. – Changing colour week by week mixes up colour with changing weather and season, and one count is not enough.Key idea: A good investigation changes one independent variable (colour), measures a suitable dependent variable (birds counted or crop eaten) for a stated time, controls other variables (scarecrow type, crop, area, weather) and repeats to find averages.
Common mistake: Students forget to say how effectiveness is measured or what is kept the same.
Exam tip: Use a checklist: change, control, repeat, measure, time.
Memory hook: Change colour, count birds, copy the test. Finish
The paper lasts 2 hours and is worth 110 marks in total. The marks shown in brackets beside each question are a guide to how much time to spend on it.
The front page says you must have a ruler and a calculator. Several questions need calculations such as magnification, percentage increase, standard form and a mean, and one asks you to plot a graph.
Yes. The instructions say to answer all questions in the spaces provided. There are 10 questions, and you should try every one and check your answers if you have time at the end.
The paper mixes single-mark multiple-choice questions, short calculations, graph plotting, and longer ‘explain’, ‘discuss’ and ‘design an investigation’ questions worth up to 6 marks.
It covers cells and osmosis, flowers and cuttings, breathing and gas exchange, enzymes, food webs, reflexes, heart disease, genetics and natural selection, fermenters, genetic engineering, photosynthesis and experimental design.
Re-read the explanation for every question you missed, then try the written paper under timed conditions and mark it with the official mark scheme from Pearson Edexcel. This quiz is independent practice material. It is not affiliated with or endorsed by Pearson Edexcel.
