2025 Edexcel IGCSE Biology Practice Paper with Answers

(a) Surface area of 2 cm cube = \( 2 \text{ cm} \times 2 \text{ cm} \times 6 = 24 \text{ cm}^2 \). Volume = \( 2 \text{ cm} \times 2 \text{ cm} \times 2 \text{ cm} = 8 \text{ cm}^3 \). SA:Vol ratio = \( 24 : 8 = 3 : 1 \) (or 3). (2 marks) (b) A larger cube has a smaller surface area to volume ratio, meaning there is less surface area relative to its volume for the acid to enter. Additionally, the diffusion distance from the outer surface to the center of the cube is greater, so it takes longer for the acid to reach the center. (4 marks) (c) Variables to control include: the concentration of the acid, the volume of the acid used, the temperature of the solution, the concentration of phenolphthalein/sodium hydroxide in the agar, and the shape of the agar blocks. (Any three: 3 marks) (d) Unicellular organisms have a very large surface area to volume ratio and a very short diffusion distance, so simple diffusion is fast enough to supply all their nutrient and gas exchange needs. (1 mark)

(a) 1 mark for correct calculation of SA (24) and Volume (8), 1 mark for correct simplified ratio of 3:1 or 3. (b) 1 mark for larger cube has a smaller SA:Vol ratio; 1 mark for less surface area per unit volume; 1 mark for longer diffusion distance to the center; 1 mark for slower overall rate of penetration to the center. (c) 1 mark for each correct control variable up to 3 (e.g. temperature, concentration of acid, volume of acid). Reject 'amount of acid'. (d) 1 mark for recognizing that a large SA:Vol ratio allows diffusion to be fast enough to meet metabolic demands.

(a) When moving into a bright room, the photoreceptors detect bright light. Impulses are sent to the brain, causing the circular muscles of the iris to contract and the radial muscles to relax. This causes the pupil to constrict (become smaller), reducing the amount of light entering the eye. (4 marks) (b) The retina contains light-sensitive photoreceptor cells that detect the light stimulus and convert it into electrical impulses. The optic nerve transmits these electrical impulses from the retina to the brain. (2 marks) (c) It is involuntary because it occurs automatically without conscious thought or decision-making. This benefits the organism by rapidly protecting the sensitive retina from being damaged by excessively bright light. (2 marks) (d) The hormone is adrenaline, which is secreted by the adrenal glands. (2 marks)

(a) 1 mark for circular muscles contract; 1 mark for radial muscles relax; 1 mark for pupil constricts/becomes smaller; 1 mark for this reduces the amount of light entering. (b) 1 mark for retina contains photoreceptors / detects light; 1 mark for optic nerve carries impulses to the brain. (c) 1 mark for involuntary meaning automatic / no conscious control / does not involve the brain's conscious hemispheres; 1 mark for benefit of protecting the retina from damage / blinding. (d) 1 mark for adrenaline; 1 mark for adrenal gland(s).

(a) Agricultural fertilisers contain high concentrations of soluble mineral ions like nitrates and phosphates. When it rains, these ions leach (dissolve and wash) into nearby lakes. The increased availability of these nutrients removes a limiting factor, stimulating the rapid growth and reproduction of algae at the surface. (2 marks) (b) The rapid growth of algae creates an algal bloom on the water surface, blocking sunlight from reaching submerged aquatic plants. These submerged plants cannot photosynthesise and die. Decomposing bacteria multiply rapidly as they feed on the dead plant matter. These bacteria respire aerobically, using up the dissolved oxygen in the water. As a result, the water becomes severely depleted of oxygen (anoxic), causing fish and other aquatic organisms to die from lack of oxygen for respiration. (5 marks) (c) Nitrates: needed to make amino acids, which are polymerised into proteins for growth. Phosphates: needed to make DNA, RNA, and cell membranes (phospholipids). (3 marks)

(a) 1 mark for leaching of soluble fertilisers into water; 1 mark for nutrients stimulating rapid growth of algae / algal bloom. (b) 1 mark for algal bloom blocking sunlight; 1 mark for submerged plants being unable to photosynthesise and dying; 1 mark for bacteria decomposing dead plant material; 1 mark for bacteria respiring aerobically / using up dissolved oxygen; 1 mark for fish dying due to lack of oxygen for respiration. (c) 1 mark for identifying nitrates AND phosphates; 1 mark for function of nitrates (making amino acids/proteins); 1 mark for function of phosphates (making DNA/membranes/ATP).

(a) Restriction enzymes are used to cut out the human insulin gene from human DNA and also to cut open the bacterial plasmid. They cut the DNA at specific base sequences, leaving matching complementary single-stranded ends called sticky ends. DNA ligase enzymes are then used to join the insulin gene and the plasmid together by forming phosphodiester bonds, creating recombinant DNA. (4 marks) (b) Plasmids are vectors because they are vehicles used to transfer foreign genetic material (the human insulin gene) into another cell (the bacterium) where it can be expressed. (2 marks) (c) To maximise yield in a fermenter, several factors are controlled: 1. Nutrients (like glucose, amino acids, and mineral ions) are constantly supplied for growth and reproduction. 2. A water cooling jacket maintains the optimum temperature for bacterial enzymes, preventing overheating from respiration. 3. Oxygen is pumped in to allow aerobic respiration. 4. Stirring paddles keep the mixture agitated, ensuring even distribution of nutrients, oxygen, and temperature, and preventing settling. 5. pH is monitored and adjusted with acids or alkalis to maintain the optimum pH for enzymes. (4 marks for any four points)

(a) 1 mark for restriction enzymes cut the human gene AND plasmid; 1 mark for restriction enzymes leaving matching sticky ends; 1 mark for ligase joining the human DNA and plasmid DNA together; 1 mark for forming recombinant DNA. (b) 1 mark for vector meaning a carrier/vehicle; 1 mark for transferring the gene into the host cell. (c) 1 mark per correct factor with explanation (up to 4 marks): - Cooling jacket controls temperature so enzymes do not denature. - Stirring paddles distribute oxygen/nutrients/heat evenly. - Sterile air/oxygen supplied for aerobic respiration. - pH probe and buffer used to maintain optimum pH. - Nutrient inlet to supply glucose/amino acids.

(a) LH is secreted by the pituitary gland. Its role is to trigger ovulation (the release of a mature egg from the ovary) around day 14 of the cycle. (2 marks) (b) Progesterone keeps the uterus lining thick, vascularised, and ready for implantation of an embryo. If fertilisation does not occur, the corpus luteum degenerates, causing progesterone levels to fall. This drop in progesterone causes the uterus lining to break down, resulting in menstruation. (3 marks) (c) Wind-pollinated flowers have small, dull petals (often green) whereas insect-pollinated flowers have large, brightly coloured petals. Wind-pollinated flowers have long filaments with exposed anthers hanging outside the flower, while insect-pollinated flowers have anthers inside the flower. Wind-pollinated flowers have feathery stigmas to catch pollen from the air, while insect-pollinated flowers have sticky, compact stigmas. (3 marks for three distinct structural comparisons) (d) The sperm cell has a flagellum (tail) which allows it to swim towards the egg, many mitochondria in its middle piece to release energy via respiration for swimming, and an acrosome in its head containing digestive enzymes to penetrate the outer jelly coat of the egg. (2 marks for any two points)

(a) 1 mark for pituitary gland; 1 mark for triggers ovulation / release of egg. (b) 1 mark for maintaining the thickness/blood supply of the uterus lining; 1 mark for progesterone levels fall if fertilisation does not occur; 1 mark for causing menstruation / shedding of lining. (c) 1 mark per correct structural comparison (up to 3 marks), e.g., large vs small petals, scent/nectar present vs absent, feathery vs sticky stigma, stamens hanging outside vs enclosed. (d) 1 mark per adaptation and explanation (up to 2 marks): flagellum for swimming, mitochondria for energy/ATP from respiration, acrosome containing enzymes to digest egg membrane.

(a) The palisade mesophyll cells are located near the upper surface of the leaf where they receive the maximum amount of light. The cells are column-shaped and tightly packed together vertically to maximise light absorption. They also contain a very high density of chloroplasts, which contain chlorophyll to absorb light energy. (3 marks) (b) First, place the leaf in boiling water for about 1 minute to kill the cells and stop all chemical reactions. Next, place the leaf in a tube of ethanol heated in a water bath (do not use a direct flame as ethanol is flammable) to dissolve and remove the chlorophyll, which decolourises the leaf so that colour changes can be clearly seen. Rinse the leaf in warm water to soften it. Spread the leaf on a white tile and add a few drops of iodine solution. If starch is present, the iodine will turn from orange-brown to blue-black. (5 marks) (c) Carbon dioxide + water -> glucose + oxygen (2 marks)

(a) 1 mark for location near the top of the leaf to receive maximum light; 1 mark for cells packed closely together to absorb more light; 1 mark for high concentration of chloroplasts. (b) 1 mark for boiling in water to kill the leaf / break down cell membranes; 1 mark for boiling in ethanol to remove chlorophyll; 1 mark for safety precaution (using water bath, not naked flame); 1 mark for softening leaf in water and adding iodine solution; 1 mark for correct colour change (orange-brown to blue-black shows starch). (c) 1 mark for left side of equation (Carbon dioxide + Water); 1 mark for right side of equation (Glucose + Oxygen). Reject chemical formulas if incorrect or unbalanced, but allow correct balanced symbol equation: 6CO2 + 6H2O -> C6H12O6 + 6O2.

(a) Carbon dioxide is a raw material/reactant in photosynthesis. Increasing its concentration increases the rate of photosynthesis by removing it as a limiting factor. Heaters raise the temperature, providing more kinetic energy to molecules, which increases the frequency of successful collisions between enzymes and substrates in photosynthetic reactions, raising the rate of photosynthesis. This leads to more glucose production, which is used for growth, increasing overall crop yield. (4 marks) (b) In bread-making, yeast is mixed with flour, water, and sugar. The yeast uses glucose (the substrate) from the dough and respires it anaerobically. The products of this anaerobic respiration (fermentation) are carbon dioxide and ethanol. (4 marks) (c) The carbon dioxide gas produced by yeast is trapped as bubbles in the dough, causing the dough to expand and rise. During the baking process, the high temperature causes the yeast to die, and any ethanol (alcohol) produced evaporates. (2 marks)

(a) 1 mark for CO2 is a reactant in photosynthesis; 1 mark for increasing CO2 removes it as a limiting factor; 1 mark for temperature increases kinetic energy of enzymes/substrates; 1 mark for more glucose produced which is used for growth / biomass. (b) 1 mark for yeast respires anaerobically; 1 mark for substrate is glucose / starch / sugars; 1 mark for product is carbon dioxide; 1 mark for product is ethanol. (c) 1 mark for carbon dioxide gas bubbles expand, causing dough to rise; 1 mark for ethanol evaporates at high baking temperatures.

(a) Arteries have thick walls containing muscle and elastic fibres to withstand high blood pressure from the heart. They have a narrow lumen to maintain high blood pressure. They do not have valves (except semi-lunar valves at the heart exit). Veins have thin walls with less muscle and elastic tissue because blood is under low pressure. They have a wide lumen to reduce resistance to blood flow and contain valves to prevent the backflow of blood, ensuring it flows towards the heart. (4 marks) (b) On a warm day, water molecules have more kinetic energy, so they evaporate faster from the cell walls of the spongy mesophyll cells into the air spaces, and diffuse faster out of the stomata. On a windy day, the wind blows away the water vapour that has accumulated outside the stomata, maintaining a steep water vapour concentration gradient between the inside of the leaf and the outside air. Both factors increase the rate of transpiration. (4 marks) (c) The tissue is the phloem. It transports substances from where they are made (the source, e.g., leaves) to where they are used or stored (the sink, e.g., roots/fruits) in both upwards and downwards directions. (2 marks)

(a) 1 mark for artery has thick wall with muscle/elastic fibers AND vein has thin wall; 1 mark for artery has narrow lumen AND vein has wide lumen; 1 mark for veins have valves to prevent backflow of blood (arteries do not); 1 mark for relating structure to pressure (arteries withstand high pressure, veins transport under low pressure). (b) 1 mark for warm temperature increases kinetic energy of water molecules; 1 mark for increased evaporation of water from mesophyll cell walls; 1 mark for wind removes saturated air / water vapour from around the leaf surface; 1 mark for maintaining a steep concentration gradient of water vapour. (c) 1 mark for phloem; 1 mark for transport is both upwards and downwards / from source to sink.

(a) Untreated sewage contains organic waste such as faeces, urine/urea, and undigested food molecules.

(b) When sewage is discharged into a river, it provides a large food source for decomposers, mainly bacteria. These bacteria feed on the organic material and multiply rapidly. As they decompose the organic waste, they respire aerobically, extracting dissolved oxygen from the water, which rapidly depletes the oxygen levels.

(c) Mayfly nymphs are pollution-sensitive organisms that require high oxygen concentrations to survive, which explains why they are present at Site A but completely absent at Site B. In contrast, bloodworms are highly tolerant to low oxygen and organic pollution, explaining their high abundance at Site B. This confirms Site B is heavily polluted.

(d) A rapid increase in algae population in response to elevated mineral nutrients is known as an algal bloom.

(e) Inorganic ions like nitrates or phosphates present in sewage act as artificial fertilizers, stimulating rapid algal growth.

(a) [2 marks] Any two from: Urea / urine, faeces, organic waste, proteins, pathogens / bacteria, domestic detergents.
(b) [4 marks]
- Sewage contains organic matter / dead organic material [1]
- This organic matter is broken down / decomposed by bacteria / decomposers [1]
- The population of bacteria increases rapidly [1]
- These bacteria respire aerobically [1]
- This uses up / depletes the dissolved oxygen in the river water [1]
(c) [2 marks]
- Site B is highly polluted / has very low dissolved oxygen levels [1]
- Because mayfly nymphs (sensitive to pollution / require high oxygen) cannot survive there, whereas bloodworms (tolerant to pollution / low oxygen) thrive [1]
(d) [1 mark] Algal bloom (accept eutrophication)
(e) [1 mark] Nitrate / phosphate / ammonium [1]

(a) A potometer measures the rate of water absorption by the cut shoot. Because more than 99% of the water absorbed by a plant is lost through transpiration from the leaves, water uptake is a very reliable proxy/estimate for the rate of transpiration.

(b) Distance moved in still air = 12 mm.
Time taken = 10 minutes.
\text{Rate} = \frac{\text{Distance}}{\text{Time}} = \frac{12 \text{ mm}}{10 \text{ minutes}} = 1.2 \text{ mm/min}.

(c) As wind speed increases, moving air physically blows away the accumulated water vapour from directly outside the stomata on the leaf surface. This prevents the boundary layer from becoming saturated. As a result, a steep water vapour concentration gradient is maintained between the moist air spaces inside the leaf and the drier air outside, which significantly accelerates the rate of diffusion of water vapour out through the stomata.

(d) Key precautions include: cutting the stem underwater at an angle (to prevent air bubbles entering the xylem vessels and blocking water transport) and sealing all connections with petroleum jelly / Vaseline (to ensure the entire system is airtight and water is drawn up effectively).

(a) [2 marks]
- Potometer measures water absorption / water uptake [1]
- Water uptake is approximately equal to water loss by transpiration from leaves [1]
(b) [2 marks]
- Correct working: 12 / 10 [1]
- Correct final answer with unit: 1.2 mm/min [1] (Accept 1.2, but reject alternative units if incorrect)
(c) [4 marks]
- Increasing wind speed increases the rate of transpiration [1]
- Wind blows away water vapour / humid air from around the leaf surface [1]
- This maintains / increases the water vapour concentration gradient [1]
- Between the inside of the leaf (air spaces) and the outside atmosphere [1]
- Leading to faster diffusion of water vapour out of stomata [1]
(d) [2 marks] Any two from:
- Cut the shoot/stem underwater [1]
- Cut the stem at an angle [1]
- Seal all joints/connections with petroleum jelly / Vaseline / make airtight [1]
- Dry the leaves before starting the measurements [1]
- Allow the plant time to equilibrate before starting [1]

(a) First, the human insulin gene is isolated and cut from human DNA using a specific restriction enzyme. The bacterial plasmid vector is cut open using the same restriction enzyme, ensuring both DNA fragments have matching, complementary sticky ends. The gene and the plasmid are joined together using the enzyme DNA ligase to form a recombinant plasmid. Finally, this recombinant plasmid is inserted back into a bacterial host cell (transformation).

(b) An industrial fermenter is used to provide perfectly controlled, optimal conditions that allow the genetically modified bacteria to reproduce rapidly by asexual binary fission. This ensures that large populations of bacteria are maintained to produce highly commercial yields of human insulin.

(c) Temperature must be controlled (e.g., using a cooling water jacket) because respiration releases heat, which could otherwise raise the temperature, denaturing the bacterial enzymes. pH must be monitored and adjusted using acids/alkalis to ensure bacterial enzymes operate at their optimal pH, preventing denaturation and maintaining high metabolic activity.

(a) [5 marks]
- Use restriction enzyme to cut out/isolate the human insulin gene [1]
- Cut open the plasmid vector using the same restriction enzyme [1]
- This produces complementary sticky ends [1]
- Use DNA ligase enzyme to join the gene and plasmid together [1]
- Creating a recombinant plasmid / recombinant DNA [1]
- Transfer / insert the recombinant plasmid into the host bacterium [1]
(b) [2 marks]
- Provides optimal / controlled conditions for growth [1]
- Allows rapid reproduction / cloning of bacteria to produce a large scale / yield of insulin [1]
(c) [3 marks]
- Identify any two conditions [1 mark for both]: Temperature, pH, Oxygen concentration, Nutrient levels.
- Explanation [2 marks; 1 mark for each corresponding explanation]:
- Temperature: maintained to prevent denaturation of bacterial enzymes / keep at optimum temperature [1]
- pH: kept constant to maintain optimum enzyme activity / prevent denaturation [1]
- Oxygen / aeration: supplied for aerobic respiration to provide energy for growth/synthesis [1]
- Nutrients / glucose: supplied to provide respiratory substrate / amino acids for protein (insulin) synthesis [1]

(a) A transgenic organism contains genetic material that has been artificially introduced from an entirely different species.
(b) A promoter is a region of DNA that acts as an 'on/off switch' for transcription by allowing RNA polymerase to bind. The ocean pout promoter is constitutively active, meaning it is transcribed year-round without requiring the environmental or hormonal cues that normally restrict wild salmon growth hormone production to spring and summer.
(c) Sterile triploid females cannot produce viable gametes. If they escape, they cannot interbreed with wild species, meaning the modified transgene cannot enter the wild gene pool. They also cannot form a self-sustaining population that would compete with native wildlife indefinitely.
(d) If \(2n = 58\), then the haploid number \(n = 58 / 2 = 29\). Triploid cells contain three complete sets of chromosomes, so \(3n = 29 \times 3 = 87\).
(e) Since sterile fish do not undergo sexual maturation, they do not expend metabolic energy or resources on egg production and reproductive behaviors. This energy is conserved and redirected into muscle mass development, yielding faster growth and higher feed conversion efficiency.

(a) [2 marks] 1 mark for mentioning the transfer of DNA/gene; 1 mark for specifying that the donor DNA is from a different species.
(b) [3 marks] 1 mark for explaining the role of a promoter as the binding site for RNA polymerase; 1 mark for explaining that the ocean pout promoter does not depend on seasonal cues; 1 mark for stating this allows continuous transcription/translation of the growth hormone.
(c) [3 marks] 1 mark for stating that they cannot reproduce or interbreed with wild populations; 1 mark for explaining that this prevents gene flow/transfer of the transgene; 1 mark for stating that it prevents the establishment of a self-sustaining invasive population.
(d) [2 marks] 1 mark for calculating the haploid number of chromosomes (29); 1 mark for the final answer of 87.
(e) [1.66 marks] 1.66 marks for explaining that resources/energy are redirected from reproduction to muscle/flesh growth, improving economic yield/growth rate.

(a) Cutting the shoot underwater is vital to prevent air locks from entering the exposed xylem vessels. Sealing the joints with petroleum jelly ensures that the system is completely airtight, preventing any external pressure leaks from disrupting bubble movement.
(b) A potometer measures the rate of water uptake rather than water lost directly through transpiration. Some of the absorbed water is utilized by the plant for photosynthesis, and some is kept in vacuoles to maintain cell turgor and structural support.
(c) Increased wind speed sweeps away the humid layer of air surrounding the stomata (the boundary layer). This decreases the humidity outside the leaf, making the water vapor concentration gradient steeper between the air spaces inside the leaf and the atmosphere. Thus, transpiration rates increase dramatically due to faster diffusion.
(d) The diameter is 0.8 mm, which means the radius \(r = 0.4\text{ mm}\). The distance moved is \(h = 35\text{ mm}\). Applying the volume formula: \(V = \pi \times (0.4)^2 \times 35 \approx 3.14159265 \times 0.16 \times 35 \approx 17.59\text{ mm}^3\). Rounding to 2 significant figures yields 18 \(\text{mm}^3\).

(a) [2 marks] 1 mark for cutting shoot underwater (prevents air locks in xylem); 1 mark for sealing joints with vaseline/paraffin (ensures system is airtight).
(b) [3 marks] 1 mark for pointing out water is used in photosynthesis; 1 mark for pointing out water is used for cell turgidity/support; 1 mark for noting water is produced by aerobic respiration.
(c) [4 marks] 1 mark for wind removing the humid air boundary layer; 1 mark for increasing the water potential gradient/concentration gradient; 1 mark for faster diffusion out of stomata; 1 mark for overall increase in transpiration rate.
(d) [2.66 marks] 1 mark for identifying radius \(r = 0.4\text{ mm}\); 1 mark for substituting values correctly into the cylinder volume formula; 0.66 marks for correct calculation and rounding to 2 significant figures (18 \(\text{mm}^3\)).

(a) Ultrafiltration occurs due to high blood pressure in the glomerulus. The afferent arteriole feeding into the glomerulus is wider than the efferent arteriole exiting it, causing blood pressure to build up. This high pressure forces small molecules (water, glucose, salts, amino acids, and urea) through the capillary walls and the semi-permeable basement membrane into the Bowman's capsule. Large molecules (proteins and blood cells) are too large to pass and stay within the capillaries.
(b) The proximal convoluted tubule is adapted for selective reabsorption by having a lining of epithelial cells covered in microvilli, which greatly increase surface area. These cells contain many mitochondria to generate ATP for active transport of substances such as glucose and amino acids against their concentration gradients. They also possess transport proteins specific to these substances.
(c) Healthy individuals filter glucose but reabsorb 100% of it in the proximal convoluted tubule through active transport. In diabetic patients, insulin deficiency or resistance leads to exceptionally high blood glucose levels, which translates to massive amounts of glucose in the glomerular filtrate. The carrier proteins responsible for glucose active transport in the nephron become saturated (exceeding the renal threshold), meaning some glucose escapes reabsorption and remains in the tubule to be excreted in the urine.

(a) [4 marks] 1 mark for high blood pressure in the glomerulus; 1 mark for pressure created by afferent arteriole being wider than efferent arteriole; 1 mark for small molecules (water, urea, glucose, salts) being forced through glomerular capillary wall/basement membrane; 1 mark for large molecules (proteins, blood cells) remaining behind as they cannot pass through the basement membrane.
(b) [4 marks] Max 4 marks: 1 mark for microvilli which increase surface area; 1 mark for many mitochondria to provide ATP/energy; 1 mark for active transport of glucose/amino acids; 1 mark for carrier proteins in membrane; 1 mark for thin membrane/short diffusion distance.
(c) [3.66 marks] 1.66 marks for stating that in healthy people, all glucose is actively reabsorbed in the PCT; 1 mark for explaining that high blood glucose in diabetics leads to high glucose in the filtrate; 1 mark for explaining that carrier proteins in the PCT become saturated/exceeded, leaving remaining glucose to pass to urine.

(a) Groups A and D will show phototropic bending because they both possess tips capable of receiving the light stimulus, and they permit the chemical messenger (auxin) to move downward from the tip to the zone of elongation.
(b) Gelatin agar is an aqueous gel through which water-soluble chemicals can easily diffuse. Thus, auxins produced in the tip of Group D pass freely into the lower shoot to stimulate unequal growth. Mica is an impermeable mineral barrier that physically blocks chemical diffusion; therefore, auxins cannot reach the shoot in Group E, preventing any growth response.
(c) In Group A, auxin synthesized in the tip is distributed laterally in response to light, migrating to the shaded (darker) side of the coleoptile. As auxin diffuses down the shaded side, it stimulates cell wall loosening and water uptake, causing cell elongation. Because the cells on the shaded side elongate faster than those on the illuminated side, the coleoptile curves toward the light.
(d) Gravity causes auxins to accumulate on the lower side of a horizontally placed root. Unlike in shoots where auxin promotes growth, high auxin concentrations in roots inhibit cell elongation. Consequently, cells on the upper side (where auxin concentration is lower) elongate faster, causing the root to bend downward, toward gravity.

(a) [2 marks] 1 mark for Group A; 1 mark for Group D. (Deduct 1 mark for each incorrect group included, minimum 0).
(b) [4 marks] 1 mark for stating gelatin is permeable/porous; 1 mark for explaining auxin can diffuse through gelatin; 1 mark for stating mica is impermeable; 1 mark for explaining auxin cannot diffuse through mica, preventing growth/response.
(c) [4 marks] 1 mark for auxin production in the tip and diffusion downwards; 1 mark for light causing redistribution of auxin to the shaded side; 1 mark for auxin stimulating cell elongation in shoots; 1 mark for uneven growth (shaded side grows faster than light side) leading to bending.
(d) [1.66 marks] 0.66 marks for auxin accumulating on the lower side of the root due to gravity; 1 mark for auxin inhibiting cell elongation in roots, causing the upper side to elongate faster and bend downwards.

(a) Intensive fish farming utilizes several strategies to manage disease. Farmers add therapeutic agents such as antibiotics, fungicides, and pesticides directly to the water or feed. Biological controls, such as introducing wrasse (cleaner fish) to consume parasitic sea lice, are also common. Minimizing stocking densities prevents overcrowding and reduces physical contact, limiting the transmission of pathogens. Water is continually filtered, oxygenated, and circulated to flush out biological waste.
(b) Wild ecosystems feature long and complex food chains. Human pollutants (such as heavy metals like mercury or microplastics) undergo biomagnification, increasing in concentration at successive trophic levels. Wild apex predator fish accumulate high concentrations of these toxins. Farmed fish, however, are fed a highly standardized, artificial diet of pellets formulated from screened, uncontaminated sources, and live in cleaner water systems with a much shorter, controlled food chain.
(c) First, calculate percentage increase for the control group:
\(((320\text{ g} - 150\text{ g}) / 150\text{ g}) \times 100 = (170 / 150) \times 100 = 113.33\%\).
Second, calculate percentage increase for the probiotic-fed group:
\(((410\text{ g} - 150\text{ g}) / 150\text{ g}) \times 100 = (260 / 150) \times 100 = 173.33\%\).
Third, find the percentage difference between their final masses:
Using the control mass as the baseline: \(((410 - 320) / 320) \times 100 = (90 / 320) \times 100 = 28.125\%\) (or relative to the probiotic mass: \(21.95\%\)).

(a) [4 marks] Max 4 marks: 1 mark for use of antibiotics/pesticides/fungicides; 1 mark for biological control (e.g. wrasse/cleaner fish); 1 mark for lower stocking density to prevent transmission; 1 mark for removing diseased individuals immediately; 1 mark for keeping species separate or cleaning/filtering water.
(b) [3 marks] 1 mark for wild food chains being longer/involving biomagnification across high trophic levels; 1 mark for wild environments containing industrial runoff/pollutants that accumulate in wild prey; 1 mark for farmed fish having controlled pellet diets that are free of bioaccumulated toxins.
(c) [4.66 marks] 1.66 marks for calculating Control group increase correctly (113% or 113.3%); 1.5 marks for calculating Probiotic group increase correctly (173% or 173.3%); 1.5 marks for calculating the percentage difference between final masses (e.g., 28.1% greater final mass for probiotic, or 21.95% smaller for control, or 24.7% using average). Allow full marks for correct calculations with working.

(a) When nitrates enter a lake, they act as an abundant plant nutrient, causing a massive population explosion of algae called an algal bloom. This dense layer of algae blocks sunlight from entering the deeper layers of the lake, preventing submerged aquatic plants from photosynthesizing. Consequently, these submerged plants die. Decomposers, such as bacteria, feed on the abundant dead plant tissues and reproduce rapidly. As these aerobic bacteria metabolize and multiply, they use up the dissolved oxygen in the water through aerobic respiration. The lake becomes anoxic (deprived of oxygen), leading to the suffocation and death of fish and other aquatic animals.
(b) Raw sewage is rich in organic compounds and already contains large populations of microbes. Once discharged, bacteria immediately begin breaking down the organic matter, actively respiring and consuming oxygen right away. Fertilizer, however, is inorganic. It does not provide direct food for decomposers, so oxygen levels only drop after an indirect ecological pathway (algal growth, blocking light, plant death, and finally decomposition).
(c) Farmers can use crop rotation involving leguminous crops (such as clover or peas) which have mutualistic nitrogen-fixing bacteria (Rhizobium) in their root nodules to enrich soil nitrogen. Alternatively, they can apply organic manures (compost, slurry) which release nutrients more slowly, reducing leaching and maintaining soil structure, or plant catch crops to absorb excess nutrients.

(a) [6 marks] 1 mark for algal bloom / rapid growth of algae; 1 mark for blocking of sunlight; 1 mark for death of submerged plants due to lack of photosynthesis; 1 mark for increase in decomposers/bacteria feeding on dead matter; 1 mark for bacteria respiring aerobically; 1 mark for depletion of dissolved oxygen leading to death of fish.
(b) [3 marks] 1 mark for identifying that sewage contains organic matter/waste; 1 mark for stating bacteria directly feed on and decompose this organic waste immediately (no delay for plant death); 1 mark for contrasting with fertilizer which is inorganic and requires an indirect chain of events (algal bloom, death, then decomposition) to cause oxygen drop.
(c) [2.66 marks] 1.66 marks for first correct method (e.g. crop rotation with nitrogen-fixing plants/legumes, or use of manure/compost); 1 mark for second correct method.