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Step 1: Calculate the amount of decrease. Decrease = 0.24 mm * 0.35 = 0.084 mm. Step 2: Subtract the decrease from the original diameter. New diameter = 0.24 mm - 0.084 mm = 0.156 mm. Alternatively: New diameter = 0.24 mm * (1 - 0.35) = 0.24 * 0.65 = 0.156 mm.

(a) 1. Thermoreceptors in the skin detect the drop in temperature (1 mark); 2. Nerve impulse is generated (1 mark); 3. Impulse travels along sensory neurones to the hypothalamus / brain (1 mark). (b) 1. Arterioles near the skin surface constrict / narrow (1 mark); 2. Shunt vessels dilate (1 mark); 3. Less blood flows through the capillary loops close to the skin surface (1 mark); 4. Less heat is lost by radiation (1 mark). (c) 1 mark for showing the correct calculation method (e.g., 0.24 * 0.65 or 0.24 * 0.35); 1 mark for the correct final answer of 0.156 mm. (d) 1. Rapid involuntary muscle contraction and relaxation (1 mark); 2. Increases the rate of respiration, which releases heat energy (1 mark).

Step 1: Calculate the decrease in rate: 15.6 - 3.9 = 11.7. Step 2: Divide the decrease by the original rate at 10 cm: 11.7 / 15.6 = 0.75. Step 3: Multiply by 100 to get percentage: 0.75 * 100 = 75%.

(a) 1. At 10 cm (bright light), rate of photosynthesis is greater than rate of respiration (1 mark); 2. Net uptake of carbon dioxide causes indicator to turn purple (1 mark); 3. At 80 cm (dim light), rate of respiration is greater than rate of photosynthesis (1 mark); 4. Net release of carbon dioxide causes indicator to turn yellow (1 mark). (b) 1. Broad and flat leaf provides a large surface area for diffusion (1 mark); 2. Thin leaf ensures a short diffusion distance (1 mark); 3. Spongy mesophyll has air spaces to allow gases to circulate (1 mark); 4. Stomata allow gases to enter and leave the leaf (1 mark). (c) 1 mark for calculating the difference (11.7); 1 mark for dividing by 15.6; 1 mark for correct answer (75%).

Step 1: Find the total number of offspring: 312 + 96 = 408. Step 2: Calculate the fraction of green-stemmed plants: 96 / 408. Step 3: Calculate the percentage: (96 / 408) * 100 = 23.53% (or 23.5%).

(a) Allele: An alternative form of a gene (1 mark). Homozygous: Having two identical alleles for a particular gene (1 mark). (b) Parental genotypes: Pp x Pp (1 mark). Gametes: P and p from each parent (1 mark). Offspring genotypes: PP, Pp, Pp, pp (1 mark). Offspring phenotypes and ratio: 3 purple-stemmed : 1 green-stemmed (1 mark). (c)(i) 1 mark for calculating total (408) and setting up division (96/408); 1 mark for correct percentage of 23.5% (accept 24%). (c)(ii) 1 mark for: Fertilization is a random event / independent assortment / chance. (d) 1. In codominance, both alleles are expressed in the phenotype of a heterozygote (1 mark); 2. Neither allele is dominant or recessive over the other (1 mark).

Step 1: Divide the increase in biomass by the total food input: 18 kg / 120 kg = 0.15. Step 2: Convert to percentage: 0.15 * 100 = 15%.

(a) 1. To supply dissolved oxygen for the fish to carry out aerobic respiration (1 mark); 2. To prevent the accumulation of toxic gases / suffocation of the fish (1 mark). (b) Intraspecific: 1. Separate fish of different sizes/ages (1 mark); 2. Feed fish regularly and adequately to prevent cannibalism (1 mark). Interspecific: 3. Cover ponds with nets to protect against birds (1 mark); 4. Use wire mesh/fences to exclude aquatic predators (1 mark). (c)(i) 1 mark for correct fraction (18/120); 1 mark for correct percentage (15%). (c)(ii) 1 mark for: Restrict movement (by keeping fish in cages) / control temperature of water to reduce respiration. (d) 1. Excess nutrients/feces can leak out and cause eutrophication (1 mark); 2. Pests/parasites (e.g. sea lice) can spread easily to wild fish populations (1 mark).

Step 1: Use the formula Q10 = Rate at (T + 10) / Rate at T. Step 2: Substitute values: Q10 = 3.6 / 1.8 = 2.0.

(a) 1. For reducing sugars: Add Benedict's reagent and heat in a hot water bath (1 mark), positive result is green/yellow/orange/brick-red (1 mark). 2. For proteins: Add Biuret reagent (1 mark), positive result is lilac/purple (1 mark). (b) 1. Up to optimum, increasing temperature increases the kinetic energy of enzymes and substrates (1 mark); 2. Leading to more frequent successful collisions and more enzyme-substrate complexes forming (1 mark); 3. Above optimum, high temperatures break bonds holding the tertiary structure of the protein (1 mark); 4. Active site changes shape / denatures, and the substrate can no longer fit (1 mark). (c)(i) 1 mark for 2.0 (or 2). (c)(ii) 1. Test at smaller temperature intervals (1 mark); 2. Specifically between 30 degrees C and 40 degrees C / around the suspected optimum (1 mark).

Step 1: Calculate resting cardiac output: 60 beats/min * 80 cm^3 = 4800 cm^3/min. Step 2: Calculate exercise cardiac output: 150 beats/min * 120 cm^3 = 18000 cm^3/min. Step 3: Calculate the difference: 18000 - 4800 = 13200 cm^3/min. Step 4: Calculate percentage increase: (13200 / 4800) * 100 = 275%.

(a) 1. Thick muscular wall to withstand the high blood pressure (1 mark); 2. Elastic fibers allow the wall to stretch and recoil to maintain blood pressure (1 mark); 3. Narrow lumen helps maintain high pressure (1 mark). (b) 1. Biconcave shape increases the surface area to volume ratio for faster diffusion of oxygen (1 mark); 2. Absence of a nucleus provides more space to pack in hemoglobin (1 mark); 3. Contain hemoglobin which binds reversibly with oxygen (1 mark). (c)(i) 1 mark for correct calculation: 150 * 120 = 18000; 1 mark for correct unit: cm^3/min (or cm^3 per min). (c)(ii) 1 mark for calculating resting cardiac output (4800); 1 mark for dividing increase (13200) by resting (4800); 1 mark for correct answer (275%).

Step 1: Divide the patient's GFR by the healthy GFR: 45 / 120 = 0.375. Step 2: Convert to percentage: 0.375 * 100 = 37.5%.

(a) 1. High blood pressure in the glomerulus forces small molecules out of the blood (1 mark); 2. Liquid is filtered through the capillary walls and basement membrane (1 mark); 3. Small molecules like water, ions, glucose, and urea enter the Bowman's capsule, while large proteins/cells remain in the blood (1 mark). (b) 1. Dehydration is detected by the hypothalamus, causing pituitary gland to release more ADH (1 mark); 2. ADH makes the walls of the collecting duct more permeable to water (1 mark); 3. More water is reabsorbed back into the blood (1 mark); 4. Reabsorption occurs by osmosis down a concentration gradient (1 mark). (c)(i) 1 mark for showing division (45/120); 1 mark for correct answer (37.5%). (c)(ii) 1 mark for glucose; 1 mark for amino acids.

Step 1: Convert the image size from cm to mm: 4.8 cm = 48 mm. Step 2: Convert mm to micrometers: 48 mm * 1000 = 48000 micrometers. Step 3: Divide image size by magnification to find actual size: 48000 / 800 = 60 micrometers.

(a) 1. Plant cell has a nucleus whereas a bacterium has circular DNA/nucleoid (1 mark); 2. Plant cell wall is made of cellulose whereas bacterial cell wall is made of peptidoglycan (1 mark); 3. Plant cell contains mitochondria/chloroplasts whereas a bacterium does not (1 mark). (b)(i) 1 mark for converting 4.8 cm to 48 mm or 48000 micrometers; 1 mark for setting up formula (Actual = Image / Magnification); 1 mark for correct answer of 60 micrometers. (b)(ii) 1. Packed with many chloroplasts to absorb maximum light energy (1 mark); 2. Columnar/elongated shape allows many cells to pack together near the upper epidermis (1 mark); 3. Large vacuole pushes chloroplasts to the edge of the cell to shorten diffusion distance for CO2 (1 mark). (c) Aerobic respiration: Mitochondrion / mitochondria (1 mark); Protein synthesis: Ribosome / ribosomes (1 mark).

(a) During ultrafiltration in the Bowman's capsule, blood is under high pressure. Small molecules pass through the capillary walls and basement membrane, but large molecules such as proteins are too big to pass through, meaning they remain in the blood plasma.

(b) Glucose is a small molecule, so it easily passes through the basement membrane into the nephron lumen during ultrafiltration (appearing in the filtrate). However, as it travels through the proximal convoluted tubule, it is entirely (100%) reabsorbed back into the blood capillaries via active transport against its concentration gradient.

(c) Percentage increase in urea concentration:
\(\text{Percentage increase} = \frac{\text{Concentration in urine} - \text{Concentration in filtrate}}{\text{Concentration in filtrate}} \times 100\)
\(\text{Percentage increase} = \frac{20.0 - 0.5}{0.5} \times 100\)
\(\text{Percentage increase} = \frac{19.5}{0.5} \times 100 = 3900\%\)

(d) Sweating on a hot day causes water loss, lowering the water potential of the blood plasma. This is detected by osmoreceptors in the hypothalamus, which stimulates the pituitary gland to release more ADH (antidiuretic hormone) into the blood. ADH travels to the kidneys and increases the permeability of the collecting duct walls to water. As a result, more water is reabsorbed out of the filtrate back into the blood by osmosis, leaving a lower volume of highly concentrated urine.

Part (a) [2 marks]:
- Protein molecules are too large to pass through (1)
- Cannot pass through the basement membrane / gaps in capillary walls / glomerulus (1)

Part (b) [3 marks]:
- Glucose is a small molecule / filtered during ultrafiltration (1)
- It is selectively reabsorbed back into the blood (1)
- By active transport / in the proximal convoluted tubule (1)

Part (c) [2 marks]:
- Correct substitution: \(\frac{20.0 - 0.5}{0.5} \times 100\) or equivalent (1)
- Correct final answer: \(3900\%\) (1) (No units needed but % must be understood)

Part (d) [4 marks]:
- Low water potential / dehydration detected by hypothalamus / pituitary gland releases more ADH (1)
- ADH travels in blood to kidneys (1)
- ADH increases permeability of the collecting duct (1)
- More water is reabsorbed back into blood by osmosis (1)

(a) The independent variable is the environmental condition (e.g., still air vs. moving air vs. high humidity).

(b) To set up and use the potometer accurately:
1. Cut the leafy shoot underwater to prevent air bubbles entering and blocking the xylem vessels.
2. Assemble the entire potometer apparatus underwater to ensure no air leaks.
3. Apply petroleum jelly (Vaseline) to all joints to make the apparatus completely airtight.
4. Dry the leaves of the shoot before starting so that the stomata are not blocked by water.
5. Introduce a single air bubble into the capillary tube, and record its starting position. Reset the bubble using the reservoir between runs to repeat the test and check for reliability.

(c) In Condition C (high humidity), the air outside the leaf contains a high concentration of water vapour. This decreases the water vapour concentration gradient between the air spaces inside the leaf and the external atmosphere. As a result, the rate of diffusion/evaporation of water vapour out through the stomata decreases, resulting in a lower transpiration rate compared to Condition A.

(d) Calculate the rate of water uptake in Condition B:
- Distance moved in 10 minutes = \(38\text{ mm}\)
- Cross-sectional area of capillary tube = \(0.5\text{ mm}^2\)
- Volume of water absorbed in 10 minutes = \(\text{distance} \times \text{area} = 38\text{ mm} \times 0.5\text{ mm}^2 = 19\text{ mm}^3\)
- Rate of water uptake per minute = \(\frac{19\text{ mm}^3}{10\text{ minutes}} = 1.9\text{ mm}^3/\text{minute}\)

Part (a) [1 mark]:
- Environmental condition / wind and humidity (1)

Part (b) [4 marks - max 4 from standard potometer procedures]:
- Cut shoot underwater (1)
- Assemble apparatus underwater (1)
- Apply Vaseline / sealant to make joints airtight (1)
- Dry leaves before starting (1)
- Introduce an air bubble / reset bubble using reservoir (1)

Part (c) [3 marks]:
- Humid air increases external water vapour concentration / decreases the concentration gradient (1)
- Between the inside of the leaf and the atmosphere (1)
- Less diffusion / evaporation of water occurs out of the stomata (1)

Part (d) [3 marks]:
- Calculation of total volume: \(38 \times 0.5 = 19\text{ mm}^3\) (1)
- Division by time (10 minutes): \(\frac{19}{10}\) (1)
- Correct final answer with units: \(1.9\text{ mm}^3/\text{minute}\) (1)
(Allow full marks for correct answer with units without working)

Auxin is synthesised in the tip of the shoot and diffuses downwards. When unilateral light is applied, auxin is redistributed and moves to the shaded side of the shoot. A higher concentration of auxin on the shaded side stimulates greater cell elongation on that side compared to the illuminated side. This unequal growth causes the shoot to bend towards the light source. Group 2 coleoptiles have their tips removed, so no auxin is produced, resulting in no growth. Group 3 coleoptiles have their tips covered with a lightproof cap, so the tip cannot detect unilateral light. Consequently, auxin diffuses evenly down all sides, leading to equal cell elongation on all sides and vertical growth straight upwards. Roots grow downwards (positive geotropism) because gravity causes auxin to accumulate on the lower side of the root, where it inhibits cell elongation, allowing the upper side to grow faster and bend downwards. Shoots grow upwards (negative geotropism) because the accumulated auxin on the lower side stimulates cell elongation, causing the shoot to bend upwards.

Part (a) [4.0 marks max]:
- Auxin is produced in the tip of the shoot (1)
- Unilateral light causes auxin to diffuse/move to the shaded side of the shoot (1)
- Higher concentration of auxin on the shaded side (1)
- Auxin stimulates cell elongation on the shaded side (1)
- Unequal growth causes the shoot to bend towards the light (1)

Part (b) [4.6 marks max]:
- Group 2: No growth / growth stops (0.6)
- Explanation for Group 2: Auxin is produced in the tip, so removing it prevents cell elongation (1.0)
- Group 3: Growth straight upwards / vertical growth (1.0)
- Explanation for Group 3: Light cannot penetrate the cap, so auxin is not redistributed / remains evenly distributed (1.0)
- Both sides elongate equally (1.0)

Part (c) [3.0 marks max]:
- Roots show positive geotropism / grow downwards, while shoots show negative geotropism / grow upwards (1)
- In roots, auxin accumulates on the lower side and inhibits cell elongation (1)
- In shoots, auxin accumulates on the lower side and stimulates cell elongation (1)

In Tube A (bright light), the rate of photosynthesis is much greater than the rate of respiration, leading to a net uptake of \(CO_2\), reducing the concentration of carbon dioxide in the indicator and turning it purple. In Tube B (darkness), only respiration occurs, producing \(CO_2\) which dissolves in the indicator to form carbonic acid, lowering the pH and turning it yellow. In Tube C (dim light), the compensation point is reached where the rate of photosynthesis equals the rate of respiration, so no net change in \(CO_2\) occurs, and the indicator remains red. Tube D is a control showing that light alone does not affect the indicator. Spongy mesophyll adaptations include air spaces, moist cell walls, thin walls, and a large surface area.

Part (a) [5.6 marks max]:
- Tube A: Purple (0.6)
- Tube B: Yellow (0.6)
- Tube C: Red / unchanged (0.6)
- Explanation for Tube B:
- Respiration occurs, but photosynthesis does not occur / is prevented by the dark (1)
- Carbon dioxide is released/produced by respiration (1)
- Carbon dioxide dissolves to form an acidic solution / lowers pH (1)
- Net increase in \(CO_2\) concentration (1)

Part (b) [2.0 marks max]:
- Acts as a control (1)
- To show that any change in indicator color / pH is due to the presence of the leaf / living tissue (and not due to light, temperature, or leaks) (1)

Part (c) [4.0 marks max]:
- Large intercellular air spaces to allow gases to diffuse easily through the leaf (1)
- Cells have moist surfaces/outer walls so that gases can dissolve before diffusing into/out of cells (1)
- Large surface area of cell walls in contact with air spaces increases rate of diffusion (1)
- Thin cell walls provide a short diffusion pathway (1)

To have a child of blood group O (genotype \(I^O I^O\)), both parents must carry the recessive \(I^O\) allele. Since the mother is group A, her genotype must be \(I^A I^O\). Since the father is group B, his genotype must be \(I^B I^O\). The genetic diagram shows: Parent genotypes: \(I^A I^O\) x \(I^B I^O\). Gametes: \(I^A\), \(I^O\) and \(I^B\), \(I^O\). Offspring genotypes: \(I^A I^B\) (Group AB), \(I^A I^O\) (Group A), \(I^B I^O\) (Group B), \(I^O I^O\) (Group O). The probability of AB is \(1/4\) or 0.25. Codominance means both alleles contribute to the phenotype; in heterozygous individuals, both antigens (A and B) are produced on red blood cells.

Part (a)(i) [3.0 marks max]:
- Mother's genotype is \(I^A I^O\) and Father's genotype is \(I^B I^O\) (1)
- Child must be homozygous recessive / genotype \(I^O I^O\) to have blood group O (1)
- The child inherits one recessive \(I^O\) allele from each parent (1)

Part (a)(ii) [5.0 marks max]:
- Parental genotypes shown: \(I^A I^O\) and \(I^B I^O\) (1)
- Gametes correctly identified: \(I^A\), \(I^O\) from one parent and \(I^B\), \(I^O\) from the other (1)
- Four correct offspring genotypes shown: \(I^A I^B\), \(I^A I^O\), \(I^B I^O\), \(I^O I^O\) (1)
- Corresponding phenotypes correctly linked to genotypes: Group AB, Group A, Group B, Group O (1)
- Correct probability of blood group AB stated as 0.25 / 25% / \(1/4\) (1)

Part (b) [2.0 marks max]:
- Both alleles are expressed / active (1)
- In the phenotype of a heterozygous individual (1)

Part (c) [1.6 marks max]:
- The physical characteristic / appearance / observable feature of an organism (1)
- Determined by the genotype / genes / alleles (and environment) (0.6)

(a)(i) Separating fish by size reduces intraspecific competition and intraspecific predation where larger fish eat smaller ones, which increases the survival rate of young fish. (a)(ii) Frequent small feeds ensure that most food is consumed, reducing waste. Uneaten food would sink to the bottom, where bacteria decompose it, leading to eutrophication/deoxygenation of the water. It also prevents dominant individuals from hogging all the food. (b) Aeration/movement maintains high dissolved oxygen concentrations, which fish require for aerobic respiration. It also helps disperse metabolic wastes such as ammonia, preventing them from reaching toxic levels. (c) Farmers can use chemical treatments (antibiotics, pesticides for lice), disinfect equipment, isolate new fish, or reduce stocking densities to limit disease transmission.

Part (a)(i) [3.0 marks max]:
- Prevents intraspecific predation / larger fish eating smaller fish (1)
- Reduces competition for food/resources between different sizes of fish (1)
- Increases overall survival rate of younger/smaller fish (1)

Part (a)(ii) [3.0 marks max]:
- Ensures all food is eaten / reduces uneaten food sinking to the bottom (1)
- Prevents decay of uneaten food by aerobic bacteria (1)
- Which would deplete dissolved oxygen levels in the water (1)
- Reduces competition at feeding times, allowing smaller fish to feed (1)

Part (b) [3.6 marks max]:
- Provides oxygen for aerobic respiration of the fish (1)
- Prevents oxygen depletion / suffocation of fish (1)
- Moves / disperses toxic waste products / ammonia / urea excreted by fish (1)
- Prevents buildup of decomposing organic matter at the bottom (0.6)

Part (c) [2.0 marks max]:
- Any two from:
- Add antibiotics / fungicides / pesticides to the water/food (1)
- Keep stocking density low / do not overcrowd fish (1)
- Remove dead or visibly diseased fish immediately (1)
- Use biological control (e.g., wrasse to eat sea lice) (1)
- Ensure clean/filtered water supply (1)

Ultrafiltration: The afferent arteriole is wider than the efferent arteriole, creating high hydrostatic pressure in the glomerulus. This pressure forces small molecules (water, ions, glucose, amino acids, urea) through the capillary walls, basement membrane, and podocytes into the Bowman's capsule space, forming the glomerular filtrate. Large proteins and blood cells are too big to pass through the basement membrane. Osmoregulation: Low water potential in blood is detected by osmoreceptors in the hypothalamus. The pituitary gland is stimulated to secrete more ADH (antidiuretic hormone). ADH travels in the blood to the kidneys, where it increases the permeability of the collecting duct walls to water. More water is reabsorbed by osmosis back into the blood capillaries. Consequently, a small, concentrated volume of urine is produced. Glucose in urine: In healthy people, all glucose is selectively reabsorbed in the proximal convoluted tubule by active transport. In diabetes, blood glucose concentration is very high, so the amount of glucose in the filtrate exceeds the transport capacity of the carrier proteins, leaving some glucose to be excreted in the urine.

Part (a) [4.6 marks max]:
- High pressure in the glomerulus / capillary network (1)
- Caused by the wider afferent arteriole than the efferent arteriole (1)
- Forces small molecules / water, glucose, salts, urea out of blood (1)
- Through basement membrane / capillary wall (which acts as a filter) (1)
- Large molecules / proteins / red blood cells are too large to pass through (0.6)

Part (b) [5.0 marks max]:
- Hypothalamus detects low water potential of blood (1)
- Pituitary gland releases more ADH (antidiuretic hormone) (1)
- ADH travels in blood to kidney / collecting ducts (1)
- Increases permeability of collecting duct walls to water (1)
- More water is reabsorbed back into blood by osmosis (1)
- Producing a small volume of concentrated urine (1)

Part (c) [2.0 marks max]:
- Normally, all glucose is selectively reabsorbed in the proximal convoluted tubule (1)
- By active transport / carrier proteins (1)
- In diabetes, blood glucose is too high, so not all can be reabsorbed / carriers are saturated (1)

To set up a potometer accurately: cut the shoot at an angle underwater (prevents air locks in xylem vessels, increases surface area for water uptake), seal all joints with vaseline/petroleum jelly (ensures airtight system so no external air disrupts water flow), dry the leaves (water on leaves blocks stomata and reduces transpiration). The potometer measures water uptake, but not all water taken up is transpired: some water is used in photosynthesis, some is used for cell turgor/support, and some water is produced during aerobic respiration. An increase in temperature increases transpiration: 1. Water molecules gain more kinetic energy, causing faster evaporation from mesophyll cell walls into the air spaces. 2. Diffusion of water vapour out of stomata occurs faster. 3. Warm air holds more moisture, reducing the relative humidity of the surrounding air, which maintains a steeper concentration/water potential gradient between the inside and outside of the leaf.

Part (a) [3.0 marks max]:
- Cut the shoot under water (1)
- Cut the shoot at an angle (1)
- Seal all joints/connections with petroleum jelly / grease (to ensure it is airtight) (1)
- Dry the leaves before starting measurements (1)
- Allow time for the plant to equilibrate before taking readings (1)

Part (b) [2.6 marks max]:
- Potometer measures water uptake, which is not exactly equal to transpiration (0.6)
- Some water is used in photosynthesis (1)
- Some water is used to maintain cell turgidity / support (1)
- Some water is produced during respiration (1)

Part (c) [4.0 marks max]:
- Increasing temperature increases the rate of transpiration (1)
- Water molecules gain more kinetic energy (1)
- Leads to faster evaporation from the surfaces of mesophyll cells (1)
- Increases diffusion rate of water vapour out through stomata (1)
- Warm air can hold more water vapour / reduces external relative humidity, maintaining a steep concentration gradient (1)