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a) Hydrogencarbonate indicator is sensitive to pH changes caused by CO2 concentration. It is orange-red at atmospheric CO2 levels. High CO2 concentrations make it acidic, turning it yellow. Low CO2 concentrations make it alkaline, turning it purple/magenta. By monitoring these colour changes, we can detect net release (yellow) or net uptake (purple) of CO2. b) Tube D acts as a control to prove that any colour change in the indicator is caused by the biological activity of the leaf and not by the physical environment (light/temperature). c) In Tube A, the rate of photosynthesis is higher than the rate of respiration in bright light, leading to a net uptake of CO2, turning the indicator purple/magenta. In Tube C, photosynthesis cannot occur in the dark, but respiration continues, releasing CO2 and turning the indicator yellow.

a) Max 4 marks: - Indicator is pH-sensitive / changes colour based on carbon dioxide concentration (1) - Red/orange indicates normal atmospheric CO2 level (1) - Yellow indicates high CO2 / acidic pH due to net respiration (1) - Purple/magenta indicates low CO2 / alkaline pH due to net photosynthesis (1) b) Max 2 marks: - Acts as a control (1) - To show that light/temperature alone do not change the colour / change is due to the presence of the leaf (1) c) Max 5 marks: - Tube A turns purple / magenta / blue (1) - Because rate of photosynthesis is greater than rate of respiration (1) - Leading to a net uptake of carbon dioxide (1) - Tube C turns yellow (1) - Because only respiration occurs / no photosynthesis (1) - Leading to a net release of carbon dioxide (1)

a) Nitrates leach into water and cause rapid growth of algae (algal bloom). This blocks light from reaching submerged aquatic plants, preventing photosynthesis and causing them to die. Decomposers/bacteria multiply rapidly as they feed on dead plants, respiring aerobically and using up dissolved oxygen. This lack of oxygen causes fish and other aquatic animals to suffocate and die. b) Fish require dissolved oxygen for aerobic respiration, which releases energy (ATP) needed for vital metabolic processes like muscle contraction and active transport. c) Farmers can leave an uncultivated buffer zone of grass/vegetation near waterways to absorb excess nutrients, and apply fertilizers only during dry weather to prevent runoff.

a) Max 5 marks: - Nitrate runoff causes rapid growth of algae / algal bloom (1) - Algae block light from reaching plants deeper in the water (1) - Plants die due to lack of photosynthesis (1) - Bacteria/decomposers multiply rapidly as they feed on dead plants (1) - Decomposers respire aerobically / use up dissolved oxygen (1) - Low oxygen levels cause fish to suffocate and die (1) b) Max 2 marks: - Dissolved oxygen is needed for aerobic respiration (1) - To release energy / ATP for vital processes (1) c) Max 4 marks (2 marks per described measure): - Apply fertilizer in dry weather / avoid wet days (1) to prevent wash-off (1) - Leave a buffer zone / strip of grass next to rivers (1) to absorb excess nutrients before they reach the water (1) - Apply fertilizer only when crops are growing (1) so plants can absorb it quickly (1) - Use organic fertilizer/manure (1) which releases nutrients more slowly (1)

a) The sharp stimulus is detected by receptors in the skin. An electrical impulse is generated and travels along a sensory neurone to the spinal cord (CNS). In the spinal cord, the impulse passes across a synapse to a relay neurone, and then across another synapse to a motor neurone. The motor neurone carries the impulse to the effector (muscle), which contracts to pull the hand away. b) Reflexes do not involve conscious parts of the brain, bypass voluntary pathways, and have fewer synapses, making them extremely fast to minimize tissue damage. c) When an electrical impulse reaches the axon terminal of the pre-synaptic neurone, it causes vesicles to release neurotransmitter molecules. These diffuse across the synaptic gap and bind to specific receptors on the post-synaptic membrane, triggering a new electrical impulse.

a) Max 5 marks: - Stimulus detected by receptor (1) - Impulse travels along sensory neurone to spinal cord / CNS (1) - Passes to relay neurone (1) - Passes to motor neurone (1) - Reaches effector / muscle which contracts to cause response (1) b) Max 2 marks: - Involuntary / does not involve conscious thoughts from the brain (1) - Fast response protects body from harm / minimizes tissue damage (1) c) Max 4 marks: - Impulse triggers release of neurotransmitters from pre-synaptic neurone (1) - Neurotransmitter molecules diffuse across synaptic gap / cleft (1) - Bind to specific receptors on post-synaptic membrane (1) - Initiates new electrical impulse in the post-synaptic neurone (1)

a) Cut the leafy shoot underwater at an angle to prevent air bubbles from entering the xylem and block conduction. Assemble the potometer underwater to ensure no air bubbles are trapped. Seal all joints with petroleum jelly to make the system airtight. Introduce a single air bubble into the capillary tube, and record the distance it travels over a measured time period. b) High wind speed increases transpiration because it blows away water vapour around the stomata, maintaining a steep water vapour concentration gradient between the inside of the leaf and the atmosphere, which speeds up diffusion. c) (i) The potometer measures the water absorbed by the shoot to replace water lost. (ii) Not all absorbed water is transpired; some is used in photosynthesis, and some is kept inside cells to maintain turgidity.

a) Max 5 marks: - Cut shoot underwater (1) - Cut shoot at an angle (1) - Assemble apparatus underwater / ensure no air bubbles in system (1) - Seal joints with Vaseline / petroleum jelly to make airtight (1) - Introduce a single air bubble (1) - Record distance moved by bubble per unit time (1) b) Max 3 marks: - Increases transpiration rate (1) - Moves water vapour away from leaf surface / stomata (1) - Increases concentration gradient of water vapour (between leaf and air) (1) c) (i) Max 1 mark: - It measures the water absorbed/taken up by the stem to replace lost water (1) c) (ii) Max 2 marks (any two): - Water is used in photosynthesis (1) - Water is used to maintain turgidity of cells / vacuole storage (1) - Water is produced during respiration (1)

a) Starch digestion begins in the mouth by salivary amylase, which breaks starch into maltose. In the duodenum, pancreatic amylase continues this breakdown. In the ileum, maltase on the cell membranes digests maltose into glucose, which is absorbed. b) (i) Temperature, amylase concentration, and starch concentration must be controlled. (ii) At pH 3, the acidic environment denatures amylase. The ionic/hydrogen bonds holding the enzyme's structure are disrupted, changing the shape of the active site so that starch can no longer bind. c) The ileum is adapted with villi and microvilli to increase surface area for diffusion, and has walls only one-cell thick to provide a short diffusion pathway.

a) Max 4 marks: - Digestion starts in mouth by salivary amylase (1) - Continued in duodenum by pancreatic amylase (1) - Starch is broken down into maltose (1) - Maltose is broken down into glucose by maltase (1) b) (i) Max 2 marks (any two): - Temperature (1) - Volume/concentration of amylase (1) - Volume/concentration of starch solution (1) b) (ii) Max 3 marks: - At pH 3, amylase is denatured (1) - Active site changes shape (1) - Substrate / starch can no longer bind / fit / form enzyme-substrate complexes (1) c) Max 2 marks (any two): - Villi / microvilli increase surface area (1) - Capillaries / good blood supply maintains concentration gradient (1) - One cell thick wall / short diffusion distance (1)

a) Palisade mesophyll cells have a nucleus, cellulose cell wall, and membrane-bound organelles (chloroplasts and mitochondria). Bacteria lack a nucleus (have circular DNA/nucleoid), have a peptidoglycan cell wall, and lack membrane-bound organelles. b) Red blood cells lack a nucleus to allow more space for haemoglobin, have a biconcave disc shape to increase the surface-area-to-volume ratio for faster diffusion of oxygen, and contain haemoglobin to bind and transport oxygen. c) (i) Formula: Actual size = Image size / Magnification. Convert 45 mm to micrometres: 45 mm = 45,000 \mu\text{m}. Actual size = 45,000 / 500 = 90 \mu\text{m}. (ii) Ribosomes are the site of protein synthesis, where amino acids are linked during translation.

a) Max 3 marks (any three): - Palisade cell has a nucleus, bacterium does not / has circular DNA (1) - Palisade cell has chloroplasts/mitochondria, bacterium does not (1) - Palisade cell wall is made of cellulose, bacterium cell wall is peptidoglycan (1) - Palisade cell has a large permanent vacuole, bacterium does not (1) b) Max 3 marks: - No nucleus (1) to pack more haemoglobin (1) - Biconcave disc shape (1) to increase surface-area-to-volume ratio (1) - Contains haemoglobin (1) to bind reversibly to oxygen (1) c) (i) Max 3 marks: - Convert mm to \mu\text{m}: 45 mm = 45,000 \mu\text{m} (1) - Use of formula: Actual = Image / Magnification or 45,000 / 500 (1) - Final answer: 90 \mu\text{m} (1) c) (ii) Max 2 marks: - Site of protein synthesis (1) - Joining of amino acids / translation (1)

a) (i) Phenotype is the physical characteristic or expressed trait of an organism. (ii) Heterozygous means having two different alleles for a particular gene (e.g., Ff). b) Parents must be carriers (Ff) to be unaffected but pass on the allele. Parental genotypes: Ff x Ff. Gametes: F and f from each parent. Offspring genotypes: FF, Ff, Ff, ff. Offspring phenotypes: FF and Ff are unaffected; ff has cystic fibrosis. Probability of having an affected child is 1 in 4 (25% or 0.25). c) In a pedigree diagram, if two unaffected parents have an affected child, the condition must be recessive because the parents are unaffected carriers. If the condition were dominant, at least one parent of the affected child would have to be affected.

a) (i) Max 1 mark: - Physical characteristics / expressed feature of an organism (1) a) (ii) Max 1 mark: - Having two different alleles of a gene (1) b) Max 5 marks: - Parent genotypes: Both Ff (1) - Gametes clearly identified: F and f from each parent (1) - Correct offspring genotypes in Punnett square/diagram: FF, Ff, Ff, ff (1) - Offspring phenotypes correctly linked to genotypes: FF/Ff as unaffected, ff as affected (1) - Probability stated as 1 in 4 / 25% / 0.25 / 1/4 (1) c) Max 4 marks: - If two unaffected parents have an affected child, the disease is recessive (1) - Because the parents are carriers / hold the allele but do not show the phenotype (1) - If the disease is dominant, every affected individual must have at least one affected parent (1) - Recessive conditions can skip generations, dominant conditions cannot (1)

a) \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}. b) (i) Yeast cells produce ethanol and carbon dioxide, while human muscle cells produce lactic acid. Neither process requires oxygen. (ii) Aerobic respiration involves complete oxidation of the glucose molecule, releasing all stored chemical energy. In anaerobic respiration, glucose is only partially broken down, leaving much of the energy locked up in lactic acid or ethanol. c) Mix yeast and glucose solution in a tube, and place a layer of oil on top to prevent oxygen from dissolving. Place the tube in a water bath at a controlled temperature. Connect a delivery tube to another tube to collect or count gas bubbles (carbon dioxide) produced. Repeat the experiment at different temperatures, keeping yeast/glucose volume and concentration constant.

a) Max 2 marks: - Correct reactants and products: \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} (1) - Correct balancing: 6O2, 6CO2, 6H2O (1) b) (i) Max 3 marks: - Yeast produces ethanol and carbon dioxide (1) - Muscle cells produce lactic acid (1) - Both do not require oxygen (1) b) (ii) Max 2 marks: - Aerobic respiration involves complete oxidation of glucose (1) - Anaerobic respiration involves incomplete oxidation / energy remains locked in products (1) c) Max 4 marks: - Yeast in glucose with a layer of oil/liquid paraffin on top to exclude oxygen (1) - Use of a water bath to control and vary the temperature (1) - Measure the rate by counting gas bubbles per unit time / measuring volume of gas (1) - Control variables: yeast concentration / glucose volume (1)

(a) To measure breathing rate at rest, count the number of breaths (one inhalation and exhalation) in a timed interval using a stopwatch. For example, count breaths for 1 minute, or count for 30 seconds and multiply by 2. (b)(i) The independent variable is the running speed (or speed of the treadmill). (ii) Controlled variables include: the duration of running (5 minutes), room temperature, humidity, fitness level of the volunteer, or gender. (c) Minute ventilation is calculated as: Breathing Rate \(\times\) Tidal Volume. At 12 \(\text{km/h}\), this is \(36 \text{ breaths/min} \times 1.8 \text{ dm}^3 = 64.8 \text{ dm}^3\text{/min}\). (d) During exercise, muscles contract more rapidly and require more energy. This increases the rate of aerobic respiration, which requires more oxygen and produces more carbon dioxide as waste. The increased level of carbon dioxide in the blood is detected by receptors, signaling the brain to increase breathing rate and depth (tidal volume). This brings more oxygen into the alveoli to diffuse into the blood, and faster removes the excess carbon dioxide, avoiding muscle fatigue and minimizing anaerobic respiration.

(a) 1 mark: Count number of breaths in a set time period (e.g., 1 minute). 1 mark: Use a stopwatch/timer to measure the time period OR repeat and calculate an average. (b)(i) 1 mark: Running speed / speed of treadmill. (b)(ii) 1 mark: Any valid control variable, e.g., duration of run (5 minutes), room temperature, fitness level of subject, type of clothing. Reject: age (stated in question). (c) 1 mark: Correct working shown: \(36 \times 1.8\). 1 mark: Correct calculation of \(64.8\) with correct units (\(\text{dm}^3\text{/min}\) or \(\text{dm}^3\text{ min}^{-1}\)). Accept 64.8 without units if correct working is shown. (d) Any 5 marks from: 1. Muscle cells contract more during exercise; 2. Require more energy; 3. Increased rate of aerobic respiration; 4. More oxygen needed / supplied to muscle cells; 5. More carbon dioxide produced; 6. High blood carbon dioxide levels detected by receptors/brain; 7. Increased depth and rate of breathing increases rate of gas exchange (oxygen in / carbon dioxide out); 8. Prevents/reduces anaerobic respiration / buildup of lactic acid.

(a) Phototropism is a plant growth response determined by the direction of a light stimulus. (b) Group D acts as a control to demonstrate that the physical presence of a cap on the tip does not mechanically prevent the coleoptile from bending. It shows that light must be able to pass through the cap for bending to occur. (c)(i) Group B had their tips removed. Because auxin is produced in the coleoptile tip, removing the tip means no auxin is present to diffuse downwards to stimulate cell elongation, resulting in no growth or bending. (ii) Group C had opaque caps. Light is detected by the tip; the opaque cap blocks light from reaching the tip. Therefore, auxin remains evenly distributed across the stem, causing equal cell elongation on all sides, resulting in straight growth. (d) In Group A, unilateral light causes auxin produced in the tip to migrate/redistribute to the shaded side of the coleoptile. Auxin diffuses down the shaded side, where it stimulates cell elongation. The cells on the shaded side grow longer than the cells on the light side, causing the coleoptile to bend towards the light.

(a) 1 mark: Growth response of a plant. 1 mark: In response to a directional light stimulus. (b) 1 mark: To act as a control / to show that the physical presence/weight of a cap does not prevent bending / to show that light transmission through the cap is necessary for the response. (c)(i) 1 mark: Auxin is produced in the tip of the coleoptile. 1 mark: Without the tip, there is no auxin to stimulate cell elongation/growth. (c)(ii) 1 mark: The tip detects the light stimulus / light is blocked from the tip. 1 mark: Auxin remains evenly distributed, so cells on both sides elongate equally (causing straight growth). (d) Any 4 marks from: 1. Auxin is produced in the tip; 2. Auxin moves to / accumulates on the shaded side of the coleoptile; 3. Auxin diffuses down the stem; 4. Auxin stimulates cell elongation; 5. Cells on the shaded side elongate more than cells on the light side; 6. This unequal growth causes the coleoptile to bend towards the light.

(a) Mass loss in Condition B = \(250.0\text{ g} - 224.8\text{ g} = 25.2\text{ g}\).
Rate of water loss = \(25.2\text{ g} / 4\text{ hours} = 6.3\text{ g/h}\).

(b) The rate of water loss is higher in Condition B (windy) because the fan blows away the boundary layer of moist air/water vapour surrounding the stomata. This maintains a steeper water potential/concentration gradient between the humid air inside the spongy mesophyll and the drier air outside. Consequently, diffusion of water vapour out of the stomata occurs at a faster rate.

(c) The oil creates an impermeable barrier on top of the water surface. This prevents any evaporation of water directly from the flask, ensuring that the measured reduction in mass is solely due to uptake and transpiration by the plant shoot.

(d) Controlled variables: Light intensity (keep the light source constant), temperature (if using a different room, keep constant), and relative humidity of the room.

(e) A potometer measures the rate of water uptake by the plant. Not all water taken up is transpired; a small percentage is used in chemical reactions (such as photosynthesis) and some remains within the plant cells to maintain turgidity and cell growth.

(a) [2 marks]
- 1 mark for correct calculation of mass loss (\(25.2\text{ g}\)) or correct formula.
- 1 mark for correct final rate (\(6.3\text{ g/h}\)) with correct units.

(b) [4 marks]
- 1 mark for noting wind removes water vapour / moist air surrounding the leaves.
- 1 mark for mentioning this maintains a steep concentration/water potential gradient.
- 1 mark for linking this to faster diffusion.
- 1 mark for stating this increases transpiration rate overall.

(c) [2 marks]
- 1 mark for stating that it prevents evaporation of water from the flask.
- 1 mark for explaining this ensures only water loss via the plant is measured.

(d) [2 marks]
- 1 mark for each correct controlled variable (e.g., light intensity, temperature, or relative humidity) up to 2 marks. Reject: 'amount of water'.

(e) [1.66 marks]
- 1 mark for stating that water is used for photosynthesis / metabolic processes.
- 0.66 marks for mentioning water is stored to keep cells turgid.

(a) The three neurones in order are:
1. Sensory neurone
2. Relay neurone
3. Motor neurone

(b) When an electrical impulse reaches the end of the pre-synaptic neurone, it stimulates vesicles containing neurotransmitters to fuse with the pre-synaptic membrane. This releases neurotransmitters into the synaptic cleft by exocytosis. The neurotransmitters diffuse across the gap and bind to complementary, specific receptor molecules on the post-synaptic membrane. This triggers sodium channels to open, generating a new electrical impulse in the post-synaptic neurone.

(c)(i) If sodium ion channels are blocked, sodium ions cannot diffuse into the neurone. This means depolarization cannot occur, preventing the generation of an action potential/nerve impulse. Thus, no impulse can propagate along the axon.

(c)(ii) Because nerve impulses cannot be initiated or conducted along the sensory pathway, pain signals from receptors cannot reach the sensory areas of the brain (cerebrum). If the brain does not receive these impulses, the individual cannot process or feel the sensation of pain.

(a) [3 marks]
- 1 mark for sensory neurone first.
- 1 mark for relay neurone second.
- 1 mark for motor neurone third.

(b) [4 marks]
- 1 mark for vesicle fusion/release of neurotransmitter into the cleft.
- 1 mark for diffusion of neurotransmitters across the synaptic cleft.
- 1 mark for binding to specific receptors on the post-synaptic membrane.
- 1 mark for triggering a new electrical impulse/action potential in the post-synaptic neurone.

(c)(i) [2 marks]
- 1 mark for stating that sodium ions cannot enter the axon/neurone.
- 1 mark for stating that depolarization/action potential cannot occur.

(c)(ii) [2.66 marks]
- 1 mark for stating that impulses do not travel to/reach the brain.
- 1.66 marks for explaining that without brain processing, pain is not felt/perceived.

(a) The sequence of events is as follows:
1. Nitrate and phosphate fertilizers leach into the river.
2. This causes rapid growth of algae on the surface (algal bloom).
3. The dense algae layer blocks sunlight from reaching aquatic plants deeper in the river.
4. These submerged plants cannot photosynthesise and therefore die.
5. Decomposers/bacteria feed on and break down the dead plant matter.
6. The population of these bacteria increases rapidly, and they respire aerobically, consuming vast amounts of dissolved oxygen.
7. Dissolved oxygen levels drop sharply, causing fish and other aquatic animals to suffocate and die.

(b) This process is called eutrophication.

(c) Alternatives include:
- Using organic fertilizers (manure or compost) which release nutrients more slowly.
- Planting buffer zones of vegetation along watercourses to absorb runoff.
- Applying fertilizers only during dry periods when heavy rain is not forecasted.

(d) Dissolved oxygen drops dramatically as the bacterial population peaks. Scientists can monitor this using indicator species: the presence of species like stonefly nymphs or mayfly nymphs indicates clean, highly oxygenated water, whereas species like bloodworms or sludgeworms thrive in low-oxygen, polluted environments.

(a) [6 marks]
- 1 mark for algal bloom/rapid growth of surface algae.
- 1 mark for blocking sunlight from reaching plants below.
- 1 mark for plants being unable to photosynthesise and dying.
- 1 mark for bacteria/decomposers multiplying and feeding on dead plants.
- 1 mark for bacteria respiring aerobically / consuming oxygen.
- 1 mark for fish dying due to lack of oxygen.

(b) [1 mark]
- 1 mark for eutrophication (spelling must be reasonably accurate).

(c) [2 marks]
- 1 mark for each sensible suggestion (e.g., planting buffer zones, using organic manure, timing application to dry weather, or crop rotation) up to 2 marks.

(d) [2.66 marks]
- 1 mark for stating that oxygen levels fall/decrease.
- 1.66 marks for explaining that clean-water species (e.g. stonefly) are absent in polluted water, or low-oxygen tolerant species (e.g. bloodworms) are present in high numbers.

(a) Codominance occurs when both alleles in a heterozygous organism are expressed fully and affect the phenotype. In the ABO system, the \(I^A\) and \(I^B\) alleles are codominant. An individual with the genotype \(I^AI^B\) has blood group AB because both A and B antigens are produced on the red blood cells.

(b)(i)
- Mother: \(I^AI^O\) (she must carry the recessive \(I^O\) allele because her child has group O)
- Father: \(I^BI^O\) (he must also carry the \(I^O\) allele)
- Child: \(I^OI^O\)

(b)(ii)
Parents' Genotypes: \(I^AI^O\) x \(I^BI^O\)
Gametes: Mother: \(I^A\) or \(I^O\); Father: \(I^B\) or \(I^O\)

Offspring Genotypes & Phenotypes:
- \(I^AI^B\) : Blood group AB
- \(I^AI^O\) : Blood group A
- \(I^BI^O\) : Blood group B
- \(I^OI^O\) : Blood group O

Probability of having a child with blood group AB = \(0.25\) (or \(25\%\) or \(1\text{ in }4\)).

(c) The \(I^O\) allele is recessive because its effect is masked/hidden when a dominant allele (\(I^A\) or \(I^B\)) is present in the genotype. It is only expressed in the phenotype when homozygous (\(I^OI^O\)).

(a) [3 marks]
- 1 mark for stating both alleles are expressed in the phenotype / neither is dominant over the other.
- 1 mark for mentioning the heterozygous state.
- 1 mark for linking to blood group AB, where both A and B antigens are expressed.

(b)(i) [3 marks]
- 1 mark for Mother: \(I^AI^O\).
- 1 mark for Father: \(I^BI^O\).
- 1 mark for Child: \(I^OI^O\).

(b)(ii) [3.66 marks]
- 1 mark for showing correct gametes in a Punnett square or cross format.
- 1 mark for listing the four correct offspring genotypes: \(I^AI^B\), \(I^AI^O\), \(I^BI^O\), \(I^OI^O\).
- 1 mark for matching each genotype to its correct phenotype.
- 0.66 marks for the correct probability of 0.25 (or 25% or 1/4).

(c) [2 marks]
- 1 mark for stating it is only expressed if homozygous / two copies are present.
- 1 mark for stating its effect is masked by dominant alleles (\(I^A\) / \(I^B\)).

(a) The completed table is:
- **Cell wall:** Onion epidermal cell (\(\checkmark\)), Human cheek cell (\(\times\)), Bacterium (\(\checkmark\))
- **Nucleus:** Onion epidermal cell (\(\checkmark\)), Human cheek cell (\(\checkmark\)), Bacterium (\(\times\))
- **Chloroplast:** Onion epidermal cell (\(\times\)), Human cheek cell (\(\times\)), Bacterium (\(\times\)) *(Note: Onion epidermal cells do not have chloroplasts as they are underground, and bacteria are prokaryotic and lack membrane-bound organelles like chloroplasts)*
- **Plasmid:** Onion epidermal cell (\(\times\)), Human cheek cell (\(\times\)), Bacterium (\(\checkmark\))

(b) Onion bulbs are storage organs that grow underground where there is no light. Since photosynthesis cannot take place without light, chloroplasts are not formed or needed in these cells.

(c)
Formula: \(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\)
- Convert image size from millimetres to micrometres:
\(80\text{ mm} \times 1000 = 80,000\text{ }\mu\text{m}\)
- Calculate actual size:
\(\text{Actual size} = \frac{80,000\text{ }\mu\text{m}}{400} = 200\text{ }\mu\text{m}\)

(d) Adaptations:
- Biconcave shape increases surface area to volume ratio for faster diffusion of oxygen.
- Packed with haemoglobin to transport oxygen.
- No nucleus to provide more space for haemoglobin.

Why it cannot undergo mitosis:
- Red blood cells do not have a nucleus (and thus lack chromosomes/DNA), which is required to replicate and divide during mitosis.

(a) [3 marks]
- 3 marks if all 12 cells in the table are correct.
- 2 marks if 9 to 11 cells are correct.
- 1 mark if 6 to 8 cells are correct.
- 0 marks if fewer than 6 are correct.

(b) [2 marks]
- 1 mark for stating onion bulbs grow underground / in the dark.
- 1 mark for explaining that photosynthesis cannot occur, so chloroplasts are not needed.

(c) [3.66 marks]
- 1 mark for showing correct formula rearrangement: \(\text{Actual} = \text{Image} / \text{Magnification}\).
- 1 mark for correct unit conversion (\(80\text{ mm} = 80,000\text{ }\mu\text{m}\)).
- 1.66 marks for the correct final calculation of \(200\text{ }\mu\text{m}\).

(d) [3 marks]
- 1 mark for mentioning biconcave shape / lack of nucleus to carry more oxygen.
- 1 mark for linking lack of nucleus to lack of DNA/chromosomes.
- 1 mark for explaining that without DNA, mitosis/division cannot occur.

(a) Alveoli adaptations include:
- **Large surface area:** Millions of microscopic alveoli provide an enormous total surface area for diffusion.
- **Very thin walls:** The alveolar walls and capillary walls are only one cell thick, providing a very short diffusion pathway for gases.
- **Rich blood supply:** A dense network of capillaries surrounds each alveolus, carrying away oxygen and bringing carbon dioxide, maintaining steep concentration gradients.
- **Moist lining:** A thin layer of moisture allows oxygen to dissolve before diffusing across the membrane.

(b)(i) Emphysema causes the delicate elastic walls of the alveoli to break down and fuse together. This creates larger, irregular air spaces with a greatly reduced surface area. During exercise, the decreased surface area limits the rate at which oxygen can diffuse into the blood. Consequently, body cells receive less oxygen, reducing aerobic respiration and causing rapid fatigue and shortness of breath.

(b)(ii) Tar in tobacco smoke paralyzes and destroys the specialized ciliated epithelial cells lining the trachea and bronchi. Normally, these cilia beat to sweep mucus (containing trapped dust and pathogens) up towards the throat to be swallowed. When cilia are damaged, excess mucus accumulates in the airways. This mucus blocks air passages and provides an ideal breeding ground for bacteria, leading to chronic infections (bronchitis) and a persistent 'smoker's cough' to clear the airways.

(a) [4 marks]
- 1 mark for each correct adaptation linked to its explanation (up to 4 marks):
- Large surface area to volume ratio for faster diffusion.
- Thin walls (one cell thick) for short diffusion distance.
- Rich capillary network/blood flow to maintain concentration gradient.
- Moisture for gases to dissolve.

(b)(i) [3.66 marks]
- 1 mark for describing that alveolar walls break down / fuse together.
- 1 mark for stating this reduces the total surface area for gas exchange.
- 1.66 marks for linking reduced diffusion of oxygen to less aerobic respiration in muscles/tissues during exercise.

(b)(ii) [4 marks]
- 1 mark for identifying that tar paralyzes/destroys cilia.
- 1 mark for stating that mucus cannot be swept upwards/out of lungs.
- 1 mark for explaining that mucus builds up and blocks airways (causing cough).
- 1 mark for noting that trapped bacteria in mucus lead to lung infections / chronic bronchitis.