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(a) Amylase is produced in the salivary glands, pancreas, and small intestine. (b) To test for glucose, Benedict's reagent is added to the food sample, and the mixture is heated in a water bath at around 80 degrees Celsius for 5 minutes. If glucose is present, the color changes from blue to green, yellow, orange, or brick-red depending on the concentration. (c)(i) The optimum pH is pH 7 because the time taken for starch to be completely broken down is the lowest (30 seconds), meaning the rate of reaction is highest. (c)(ii) At pH 9, the pH is too extreme (alkaline) for the enzyme, causing the shape of its active site to change (denature). The starch substrate is no longer complementary in shape and cannot bind. (c)(iii) At pH 5, it takes 120 seconds. At pH 7, it takes 30 seconds. To find how many times faster: 120 / 30 = 4 times faster.

(a) 1 mark for salivary glands, 1 mark for pancreas (accept small intestine). Max 2 marks. (b) 1 mark for adding Benedict's reagent/solution. 1 mark for heating/warming in a water bath. 1 mark for stating the color changes from blue to green/yellow/orange/brick-red. (c)(i) 1 mark for identifying pH 7. 1 mark for stating it took the shortest time/30 seconds for starch to be broken down. (c)(ii) 1 mark for stating the active site changed shape / enzyme denatured. 1 mark for stating that the substrate/starch can no longer fit/bind. 1 mark for stating this is because the pH is too high/alkaline. (c)(iii) 1 mark for identifying times of 120s and 30s. 1 mark for dividing 120 by 30. 0.5 marks for correct final answer of 4 (times faster).

(a) The vena cava carries deoxygenated blood back from the body to the right atrium. The pulmonary artery carries deoxygenated blood from the right ventricle to the lungs. Capillaries allow diffusion of gases and nutrients due to thin walls. (b)(i) CHD leads to fatty deposits narrowing the lumen of the coronary arteries. This reduces blood flow and oxygen delivery to the heart muscle. Lacking oxygen, heart muscle cells cannot respire aerobically, leading to anaerobic respiration, lactic acid buildup, or cell death (heart attack). (b)(ii) Any two from: smoking, poor diet (high saturated fat), high blood pressure, lack of exercise, high blood cholesterol, diabetes, or genetic factors. (c)(i) A stent is a metal mesh tube inserted into the blocked artery to physically hold it open, restoring normal blood flow. (c)(ii) Statins are drugs that lower blood cholesterol levels. (c)(iii) Stents have a long-term effect and offer an immediate solution, unlike statins which must be taken daily and take time to be effective.

(a) 1 mark for Vena cava - carries deoxygenated blood from body to heart. 1 mark for Pulmonary artery - carries deoxygenated blood from heart to lungs. 1 mark for Capillary - allows substances to diffuse. (b)(i) 1 mark for fatty material narrowing coronary arteries / reducing blood flow. 1 mark for less oxygen/glucose reaching heart muscle. 1 mark for heart muscle cells cannot respire / die. (b)(ii) 2 marks for any two correct risk factors (e.g., smoking, high-fat diet, obesity, lack of exercise, high stress, high blood pressure). (c)(i) 1 mark for stating it holds/opens the artery. 1 mark for stating it allows blood to flow (to heart muscle). (c)(ii) 1 mark for statins. (c)(iii) 1 mark for any valid advantage: immediate effect, long-lasting, patient doesn't need to remember daily medication; 0.5 marks for clear phrasing.

(a)(i) The primary consumer is the herbivore that eats the producer, which is the caterpillar. (a)(ii) The oak tree is a producer and uses sunlight / light energy for photosynthesis. (b)(i) Percentage biomass transferred = (Biomass of blue tit / Biomass of caterpillar) * 100 = (40 / 400) * 100 = 10%. (b)(ii) Biomass is lost because: not all parts of the caterpillar are eaten (e.g., exoskeleton), some biomass is lost as waste (egestion of feces), and some is broken down in respiration to release energy, producing carbon dioxide and water which are lost to the surroundings. (c)(i) Photosynthesis removes carbon dioxide. (c)(ii) Respiration (by plants, animals, or decomposers) and combustion of organic materials or fossil fuels release carbon dioxide. (c)(iii) Decomposers digest the dead leaves by secreting extracellular digestive enzymes onto them. They absorb the small, soluble food molecules (like glucose). These decomposers then use these molecules for aerobic respiration, which produces carbon dioxide as a waste product, releasing it back into the atmosphere.

(a)(i) 1 mark for caterpillar. (a)(ii) 1 mark for light / sunlight (do not accept sun alone without reference to light/energy). (b)(i) 1 mark for correct working: 40 / 400 * 100. 1 mark for correct answer of 10(%). (b)(ii) 2 marks for any two reasons from: not all of the organism is eaten, egestion/feces/waste, excretion/urine, lost as carbon dioxide/water during respiration. (c)(i) 1 mark for photosynthesis. (c)(ii) 2 marks for any two from: respiration, combustion/burning, decay. (c)(iii) 1 mark for decomposers breaking down/feeding on dead leaves. 1 mark for secreting enzymes (extracellular digestion). 1 mark for decomposers absorbing digested nutrients. 0.5 marks for decomposers respiring, which releases carbon dioxide.

(a)(i) The growth response of plant shoots to light is phototropism (specifically positive phototropism). (a)(ii) Auxin is produced at the shoot tip. When light shines from one side, auxin diffuses to the shaded side of the shoot. Auxin stimulates cell elongation in shoots. Since there is more auxin on the shaded side, the cells on the shaded side elongate more than the cells on the lit side. This unequal growth rate causes the shoot to bend towards the light. (b)(i) Roots grow towards gravity (positive gravitropism). (b)(ii) Growing downwards ensures the roots anchor the plant securely in the ground. It also places the roots deeper in the soil where they are more likely to find water and essential mineral ions. (c)(i) Gibberellins are used to initiate seed germination. (c)(ii) Ethene gas is used to control fruit ripening. (c)(iii) Auxins are used as selective weedkillers. (c)(iv) Selective weedkillers affect only broad-leaved weeds (not narrow-leaved crops), reducing competition for resources like space, light, water, and minerals, which helps crop yield.

(a)(i) 1 mark for phototropism (accept positive phototropism). (a)(ii) 1 mark for auxin produced in the tip and diffuses down. 1 mark for auxin moving to/accumulating on the shaded side. 1 mark for auxin causing cell elongation (in shoots). 0.5 marks for shaded side growing faster, causing the bend. (b)(i) 1 mark for towards. (b)(ii) 1 mark for anchoring the plant. 1 mark for absorbing water / mineral ions. (c)(i) 1 mark for gibberellins. (c)(ii) 1 mark for ethene. (c)(iii) 1 mark for auxin(s). (c)(iv) 2 marks for reducing competition (for light, water, space) allowing crop plants to grow better.

(a)(i) Rose black spot is a fungal disease of plants. (a)(ii) Measles is a viral disease in humans. (a)(iii) Salmonella food poisoning is caused by bacteria. (b)(i) The skin acts as a physical barrier to prevent entry and produces antimicrobial secretions to kill microbes. (b)(ii) The stomach produces concentrated hydrochloric acid which kills pathogens present in swallowed food and mucus. (b)(iii) The trachea and bronchi secrete sticky mucus which traps inhaled pathogens. They are lined with ciliated cells whose cilia waft the mucus up to the esophagus where it is swallowed. (c)(i) In phagocytosis, the phagocyte (white blood cell) detects the foreign pathogen, moves towards it, engulfs it into a vesicle, and then fuses with lysosomes containing digestive enzymes to break down and destroy the pathogen. (c)(ii) Vaccination involves introducing small quantities of dead or inactive forms of a pathogen into the body. This stimulates white blood cells (lymphocytes) to produce specific antibodies against the antigens. Memory cells remain in the blood. If the active pathogen infects the person in the future, the white blood cells respond quickly, producing antibodies in much larger quantities and much faster, destroying the pathogen before the person feels ill.

(a) 1 mark for Fungus - Rose black spot. 1 mark for Virus - Measles. 1 mark for Bacterium - Salmonella. (b)(i) 1 mark for physical barrier / antimicrobial secretions. (b)(ii) 1 mark for producing acid that kills pathogens. (b)(iii) 1 mark for mucus trapping pathogens. 1 mark for cilia wafting mucus (upwards/to throat). (c)(i) 1 mark for engulfing/surrounding the pathogen. 1 mark for digesting/breaking down the pathogen (using enzymes). 0.5 marks for reference to lysosomes / cell membrane movement. (c)(ii) 1 mark for introducing dead/inactive pathogen. 1 mark for white blood cells producing specific antibodies. 1 mark for rapid/large-scale production of antibodies upon reinfection.

(a)(i) Reflex actions are rapid, automatic, and do not involve conscious parts of the brain, protecting the body from immediate damage. (a)(ii) Pupil constricting in bright light, withdrawing hand from a hot surface, or the knee-jerk reflex. (b)(i) Temperature or pain receptors in the skin. (b)(ii) The pathway is: Sensory neurone -> Relay neurone (in spinal cord/CNS) -> Motor neurone. (b)(iii) When an electrical impulse arrives at the pre-synaptic neurone terminal, it causes chemical messenger molecules (neurotransmitters) to be released into the synaptic cleft. These chemicals diffuse across the tiny gap down a concentration gradient. When they reach the post-synaptic neurone, they bind to specific receptor molecules, which initiates a new electrical impulse in that neurone. (c)(i) Sum of trials = 22 + 18 + 15 + 12 + 13 = 80 cm. Mean distance = 80 cm / 5 trials = 16 cm. (c)(ii) The distance the ruler fell generally decreased with each trial (from 22 cm down to 12-13 cm). Since a shorter distance indicates a faster reaction, this shows that practice improves / speeds up reaction time.

(a)(i) 1 mark for protection / preventing injury. (a)(ii) 1 mark for any correct example (e.g. eye blink, hand withdrawal). (b)(i) 1 mark for temperature/pain receptor (accept skin receptor). (b)(ii) 1 mark for correct order of neurones. 1 mark for correct spelling of sensory, relay, and motor neurones. (b)(iii) 1 mark for chemical/neurotransmitter released. 1 mark for chemical diffusing across the gap/synapse. 1 mark for chemical binding to receptors on the next neurone, triggering an impulse. (c)(i) 1 mark for sum = 80 or correct working (80 / 5). 1 mark for correct mean of 16 (cm). (c)(ii) 1 mark for stating that distance fell decreases with practice. 1 mark for explaining that decreasing distance means reaction time decreases / becomes faster. 0.5 marks for structured conclusion.

(a)(i) Chromosomes are located inside the nucleus of eukaryotic cells. (a)(ii) Human body cells contain 46 chromosomes (or 23 pairs). (a)(iii) A chromosome is a single, tightly coiled molecule of DNA. Along this DNA molecule, there are many smaller sections called genes, which contain the genetic code for producing specific proteins. (b)(i) Mitosis is critical for growth (increasing cell count) and repair (replacing damaged or dead tissues). It also enables asexual reproduction in some organisms. (b)(ii) Stage 1 (Interphase): The cell grows in size, increases sub-cellular structures (like ribosomes and mitochondria), and replicates its DNA to make two copies of each chromosome. Stage 2 (Mitosis): The chromosomes line up at the center of the cell, and one set of chromosomes is pulled to each end of the cell. The nucleus then divides. Stage 3 (Cytokinesis): The cytoplasm and cell membrane split, forming two genetically identical diploid daughter cells. (c)(i) The root tip contains meristem tissue, which is actively growing and undergoing mitosis. (c)(ii) Proportion of cells in mitosis = 30 / 200 = 0.15. Total time of cell cycle in minutes = 16 hours * 60 minutes/hour = 960 minutes. Time spent in mitosis = 0.15 * 960 minutes = 144 minutes.

(a)(i) 1 mark for nucleus. (a)(ii) 1 mark for 46 (or 23 pairs). (a)(iii) 1 mark for made of DNA. 1 mark for containing genes / genes are sections of DNA. (b)(i) 2 marks for any two from: growth, repair/replacement of damaged cells, asexual reproduction. (b)(ii) 1 mark for Stage 1: cell growth / DNA replication. 1.5 marks for Stage 2: chromosomes separate/move to opposite ends and nucleus divides. 1 mark for Stage 3: cell membrane/cytoplasm divides. (c)(i) 1 mark for rapid growth / meristem tissue / active cell division. (c)(ii) 1 mark for correct proportion (30/200 = 0.15) OR total minutes (16 * 60 = 960). 1 mark for correct final answer of 144 (minutes).

(a)(i) Osmosis is the movement of water from dilute to concentrated solution across a partially permeable membrane. (a)(ii) Diffusion is the passive movement of particles down a concentration gradient. (a)(iii) Active transport is the movement of substances against a concentration gradient, requiring energy from respiration. (b)(i) The 0.4 mol/dm3 sugar solution is hypertonic to the potato cells (it has a lower water concentration). Therefore, water moves out of the cells by osmosis through the partially permeable cell membrane, reducing the mass. (b)(ii) The starting mass of each potato chip varies. Using percentage change normalizes the results, allowing a direct and fair comparison between different potato chips. (b)(iii) When there is 0% change in mass, the water potential (concentration of solutes) inside the potato cells is equal to the concentration of the external sugar solution. There is no net movement of water into or out of the cells because the system is in equilibrium. (c) Gills have several adaptations: 1. They are made of many gill filaments and lamellae, creating a very large surface area for diffusion. 2. They have very thin walls (only one cell thick) which shortens the diffusion pathway. 3. They have a rich network of blood capillaries and a countercurrent flow system that maintains a steep concentration gradient for oxygen and carbon dioxide.

(a)(i) 1 mark for Osmosis. (a)(ii) 1 mark for Diffusion. (a)(iii) 1 mark for Active transport. (b)(i) 1 mark for stating water moved out of the cells (by osmosis). 1 mark for explaining the sugar solution is more concentrated / has a lower water concentration than the cytoplasm. (b)(ii) 1 mark for stating starting masses were different. 1 mark for stating percentage change allows direct / fair comparison. (b)(iii) 1 mark for stating that the concentration outside equals the concentration inside. 1 mark for stating there is no net movement of water. (c) 1 mark for large surface area (filaments/lamellae). 1 mark for thin walls / one-cell thick (short diffusion distance). 1 mark for good blood supply / capillaries (maintains concentration gradient). 0.5 marks for linking adaptations to efficient rate of diffusion.

(a) Independent variable is the factor changed: drinking coffee. (b) The anomalous result is 12 cm because it is significantly lower than the other trials (17-19 cm). (c) Sum of valid trials = 19 + 17 + 18 + 18 = 72. Divide by 4 = 18 cm. (d) Computer programs provide more precise measurements with smaller intervals (milliseconds) and eliminate the manual delay of the person dropping the ruler. (e) Control variables include: same person catching, same hand, starting distance of fingers from the ruler.

Total: 9.1 marks. (a) 1 mark for identifying coffee intake / presence of caffeine. (b) 1 mark for identifying 12. (c) 1 mark for identifying and excluding 12, 1 mark for correct sum of remaining trials (72) divided by 4, 1 mark for correct mean of 18 (allow error carried forward if 12 is included: 84 / 5 = 16.8 for max 2 marks). (d) 1 mark for higher precision/resolution (milliseconds) and 1 mark for removing human error/reaction delay of the dropper. (e) 2.1 marks: 1 mark for each valid control variable (e.g., using same hand, same person dropping, same starting finger-gap, same volume of drink) up to two, plus 0.1 bonus for complete clarity.

(a) Seedlings in unidirectional light (Group A) show positive phototropism and bend towards the light, whereas those lit from above (Group B) grow straight upwards. (b) Auxin is the hormone that regulates phototropism. (c) Light destroys or causes redistribution of auxin to the shaded side. This higher concentration of auxin causes cells on the shaded side to elongate more than cells on the illuminated side, resulting in a bend. (d) Matching uses: Auxins are used as selective weedkillers; Ethene controls fruit ripening; Gibberellins are used to trigger germination/end seed dormancy.

Total: 9.1 marks. (a) 1 mark for Group A bending towards light, 1 mark for Group B growing straight up. (b) 1 mark for Auxin. (c) 1 mark for auxin moving to the shaded side, 1 mark for causing cell elongation on the shaded side, 1 mark for causing the shoot to bend towards light. (d) 3.1 marks: 1 mark for each correct match: Auxins to weedkiller, Ethene to ripening, Gibberellins to ending seed dormancy, plus 0.1 for all three correct.

(a) The pancreas regulates blood glucose levels. (b) When glucose rises, the pancreas secretes the hormone insulin. Insulin stimulates liver and muscle cells to absorb glucose from the blood and convert it into stored glycogen, reducing blood glucose. (c) Peak for X = 14.0, peak for Y = 7.0. Increase = 14.0 - 7.0 = 7.0. Percentage increase of X compared to Y = (7.0 / 7.0) x 100 = 100%. (d) Managing diet (eating fewer simple carbohydrates) and regular exercise help manage Type 1 diabetes. (e) Glucagon is released when blood glucose is low.

Total: 9.1 marks. (a) 1 mark for pancreas. (b) 1 mark for pancreas releasing insulin, 1 mark for glucose moving into cells/liver/muscles, 1 mark for glucose being converted into glycogen. (c) 1 mark for finding difference (7.0), 1 mark for division by base value (e.g. 7.0 or 14.0), 1 mark for final percentage (100% or 50% depending on direction specified). (d) 1 mark for diet control / regular exercise. (e) 1.1 marks for glucagon.

(a) Differences: Sexual reproduction requires meiosis, fusion of male and female gametes, and leads to variation; asexual reproduction involves mitosis, only one parent, and produces genetically identical offspring. (b) Sexual reproduction introduces genetic variation, which provides a survival advantage if environmental conditions change (natural selection). (c) Meiosis is the cell division type. (d) Steps of meiosis: 1. DNA replicates. 2. The cell divides into two cells. 3. These cells divide again to form four gametes. 4. Each gamete is haploid (half the DNA) and genetically unique.

Total: 9.1 marks. (a) 2 marks: 1 mark for mentioning parents/gametes difference, 1 mark for variation/clones difference. (b) 1 mark for genetic variation, 1 mark for enabling survival/adaptation to environmental changes. (c) 1 mark for meiosis. (d) 4.1 marks: 1 mark for DNA replication, 1 mark for first division, 1 mark for second division (resulting in four cells), 1.1 marks for gametes being genetically different and haploid.

(a) Selective breeding steps: 1. Identify individuals with desired traits (high milk yield). 2. Breed them. 3. Select offspring showing the trait and breed them. 4. Repeat over generations. (b) Inbreeding reduces the gene pool. This increases the likelihood of expressing harmful recessive genetic diseases and lowers disease resistance. (c) Genetic engineering inserts genes from one species into another at an embryonic stage, whereas selective breeding works with existing alleles within the species via breeding. (d) Concerns about GM crops include: horizontal gene transfer to weeds, reduction in biodiversity (affecting insects), and unknown long-term health effects on humans.

Total: 9.1 marks. (a) 1 mark for choosing high-yielding parents, 1 mark for breeding them, 1 mark for repeating the process with offspring over multiple generations. (b) 1 mark for inbreeding/reduced gene pool, 1 mark for increased risk of genetic defects or disease vulnerability. (c) 1 mark for transferring/inserting a gene, 1 mark for it being between different organisms/faster compared to natural breeding. (d) 2.1 marks: 1 mark for each valid concern (e.g., impact on wild ecosystems, health concerns, herbicide resistance spreading) up to two, plus 0.1 bonus.

(a) To avoid bias, the students should map the field as a grid and use a random number generator to select coordinates. (b) Sum = 4 + 3 + 5 + 1 + 2 + 4 + 3 + 5 + 2 + 1 = 30. Mean = 30 / 10 = 3 dandelions per quadrat. (c) Area of quadrat = 0.5m x 0.5m = 0.25 m². Field area = 30m x 20m = 600 m². (d) Estimated population = (Field Area / Quadrat Area) x Mean = (600 / 0.25) x 3 = 2400 x 3 = 7200 dandelions. (e) Accuracy can be improved by using more quadrats (e.g., covering at least 10% of the total area) to get a more representative mean.

Total: 9.1 marks. (a) 1 mark for grid system, 1 mark for random number generator / coordinates. (b) 1 mark for 3. (c) 1 mark for quadrat area (0.25 m²), 1 mark for field area (600 m²). (d) 1 mark for dividing field area by quadrat area (2400), 1 mark for multiplying by mean, 1 mark for correct final population of 7200. (e) 1.1 marks for sampling more quadrats / increasing sample size.

(a) Peat bogs store vast amounts of carbon. When destroyed, the peat is either burnt as fuel or decays when used as compost. Microorganisms decompose the peat, releasing stored carbon as carbon dioxide, which is a greenhouse gas. (b) Deforestation in tropical regions is mainly done to clear land for cattle ranching, growing crops (like soy or palm oil), or growing crops for biofuels. (c) Clearing forests removes habitats, causing species extinction and biodiversity loss. It also leads to soil erosion and increased risk of flooding due to lack of roots holding the soil and absorbing water. (d) Conservation strategies include: breeding programs for endangered species, protection/regeneration of habitats, and reintroduction of field margins/hedgerows.

Total: 9.1 marks. (a) 1 mark for stating peat contains locked-up carbon, 1 mark for stating that burning/decay releases carbon dioxide, 1 mark for identifying carbon dioxide as a greenhouse gas that traps heat. (b) 2 marks: 1 mark for each valid reason (agriculture/crops, cattle ranching, timber, biofuel crops) up to two. (c) 2 marks: 1 mark for loss of habitats/biodiversity, 1 mark for soil erosion/landslides/flooding. (d) 2.1 marks: 1 mark for each valid conservation measure (nature reserves, breeding programs, recycling, replanting/reforestation) up to two, plus 0.1 bonus.

(a) The correct sequence is: Kingdom, Phylum, Class, Order, Family, Genus, Species. (b) The binomial system uses Genus and Species names. It is universal and avoids translation issues or confusion caused by regional common names. (c) The three domains are: Archaea (primitive bacteria), Bacteria (true bacteria), and Eukaryota (organisms with cells containing a nucleus). (d) Technological advances allowed chemical analysis, specifically ribosome structure, RNA sequencing, and DNA analysis, showing deep genetic differences.

Total: 9.1 marks. (a) 3 marks: 1 mark for Class, 1 mark for Order, 1 mark for Genus. (b) 1 mark for being a universal language (Latin/globally understood), 1 mark for preventing confusion from local common names. (c) 3.1 marks: 1 mark for Archaea, 1 mark for Bacteria, 1 mark for Eukaryota, plus 0.1 bonus for complete correctness. (d) 1 mark for chemical analysis / DNA sequencing / RNA analysis / molecular biology.

For (a), the independent variable is the one changed by the investigator (Trial number or amount of practice) and the dependent variable is what is measured (Reaction time in milliseconds). For (b), the mean of the last three trials is calculated as: \((270 + 250 + 260) / 3 = 780 / 3 = 260\) ms. For (c), the data shows that as trial number increases from 1 to 4, the reaction time decreases from 320 ms to 250 ms, and then levels off at 260 ms, indicating that practice improves reaction speed. For (d), the effector is the muscle in the hand or finger which contracts to cause the movement. For (e), a computer program is more accurate because it can measure time precisely in milliseconds (higher resolution), whereas a ruler drop test has lower resolution and involves human reading errors.

(a) [2 marks] 1 mark for identifying the independent variable as Trial number (or practice). 1 mark for identifying the dependent variable as Reaction time. (b) [2 marks] 1 mark for showing correct working: \((270 + 250 + 260) / 3\). 1 mark for correct final answer: 260 (accept with or without ms). (c) [2 marks] 1 mark for stating that reaction time decreases as trial number/practice increases. 1 mark for mentioning that it levels off/reaches a plateau. (d) [1 mark] 1 mark for stating muscle (accept hand/finger muscle; reject motor neurone). (e) [2.1 marks] 1 mark for stating that the computer measures with higher resolution (e.g. milliseconds instead of centimetres). 1.1 marks for stating that it reduces human error (e.g. no manual reading of a scale or anticipation of the drop).

For (a), systematic sampling along a line is called a line transect. For (b), the area of the quadrat is calculated as 0.5 m x 0.5 m = 0.25 m^2. For (c), at 8 m there are 12 dandelions in 0.25 m^2. To find the count per 1 m^2, multiply by 4: 12 x 4 = 48 dandelions per m^2. For (d), there is a positive correlation: as distance increases, the number of dandelions increases. This is because abiotic factors like light intensity are lower near the tree due to shading, which reduces photosynthesis. For (e), random sampling is achieved by setting up a grid over the area, generating random coordinates using a calculator or computer, and placing quadrats at these coordinates to ensure an unbiased sample.

(a) [1 mark] 1 mark for line transect / transect line. (b) [1 mark] 1 mark for 0.25 (accept m^2). (c) [2 marks] 1 mark for correct working: 12 / 0.25 or 12 x 4. 1 mark for correct answer: 48. (d) [3.1 marks] 1 mark for identifying the relationship (as distance increases, abundance increases). 1 mark for naming a relevant abiotic factor (light intensity / water / mineral ions). 1.1 marks for explaining the effect (e.g. more light allows more photosynthesis, or less competition with tree roots for water/minerals). (e) [2 marks] 1 mark for using random numbers / random coordinate generator. 1 mark for placing quadrats at these coordinates to prevent bias.

For (a), the increase in Person B's blood glucose is 7.2 - 4.5 = 2.7 mmol/L. The percentage increase is (2.7 / 4.5) * 100 = 60%. For (b), two key differences are: 1) Person A's glucose levels reach a much higher maximum level (14.2 mmol/L) than Person B's (7.2 mmol/L), and 2) Person A's glucose levels remain elevated at 120 minutes, whereas Person B's levels return to the baseline of 4.5 mmol/L. For (c), the pancreas of a person with Type 1 diabetes cannot produce enough insulin. For (d), insulin enters the blood stream and binds to target cells (liver/muscle), causing them to absorb glucose from the blood and convert it into insoluble glycogen, lowering blood glucose levels. For (e), Type 2 diabetes is often managed through diet, exercise, or medications that increase insulin sensitivity, whereas Type 1 diabetes requires direct insulin therapy (injections).

(a) [2 marks] 1 mark for calculating the increase: 7.2 - 4.5 = 2.7. 1 mark for correct percentage: (2.7 / 4.5) x 100 = 60%. (b) [2 marks] 1 mark for stating that Person A's blood glucose levels peak higher than Person B's (or start higher). 1 mark for stating that Person A's blood glucose levels remain high / do not return to normal while Person B's return to the baseline. (c) [1 mark] 1 mark for insulin. (d) [2.1 marks] 1 mark for stating insulin causes glucose to move from blood into cells (accept liver or muscle cells). 1.1 marks for stating glucose is converted to glycogen (for storage). (e) [2 marks] 1 mark for identifying lifestyle management (e.g., carbohydrate-controlled diet, weight loss, or regular exercise). 1 mark for stating that Type 2 does not usually require daily insulin injections (or is treated with drugs that improve insulin sensitivity rather than insulin itself).