2023 Edexcel IGCSE Biology 模擬試題連答案詳解

At \(0.3\text{ mol/dm}^3\) sucrose solution, the net movement of water is zero because there is no water potential gradient. This means the water potential of the solution is equal to the water potential of the potato cell sap. Water molecules still move in and out of the cells, but at equal rates (no net movement).

1 mark for the correct answer (B). Reject other options because water potential is equal (no gradient), and dynamic equilibrium involves equal rates of movement, not stopped movement.

During inspiration, the external intercostal muscles contract (pulling the ribs up and out) and the diaphragm contracts and flattens. This increases the volume of the thorax, which decreases the air pressure inside the lungs relative to atmospheric pressure, causing air to enter.

1 mark for identifying the correct sequence of muscle action, volume change, and pressure change during inspiration (A).

The pulmonary artery is an artery, meaning it has a thick muscular and elastic wall and a narrow lumen to withstand high pressure. It carries deoxygenated blood away from the right ventricle to the lungs. The pulmonary vein carries blood to the heart, vena cava has thin walls, and aorta has a very thick muscular wall.

1 mark for selecting the correct row that matches structural features and function (A).

Crossing two heterozygous pink plants (RW x RW) produces offspring with genotypes RR, RW, and WW in a 1:2:1 ratio. Since R and W are codominant, RR is red, RW is pink, and WW is white. Thus, the phenotypic ratio is 1 red : 2 pink : 1 white.

1 mark for the correct phenotypic ratio of codominant offspring (C).

To test for non-reducing sugars, you must first hydrolyse them into monosaccharides by heating with dilute acid, neutralise the mixture (e.g., with sodium hydrogencarbonate), and then perform the standard Benedict's test (heating). To test for lipids, the emulsion test is used: dissolve the sample in ethanol and pour it into water, which yields a cloudy white emulsion if lipids are present.

1 mark for identifying the correct protocol for both the non-reducing sugar test and the lipid emulsion test (C).

Total magnification = eyepiece magnification \(\times\) objective magnification = \(10 \times 40 = 400\). Actual size = Image size / Magnification. Image size = \(4.0\text{ mm} = 4000\text{ }\mu\text{m}\). Actual size = \(4000\text{ }\mu\text{m} / 400 = 10\text{ }\mu\text{m}\).

1 mark for calculating the correct actual length of the vacuole in micrometres (B).

Nitrifying bacteria carry out nitrification, which is the oxidation of ammonium ions to nitrites, and then nitrites to nitrates, making nitrogen available for plant absorption.

1 mark for identifying nitrifying bacteria as the organism responsible for this conversion (D).

Lipase is secreted by the pancreas (and acts in the small intestine) to break down lipids into fatty acids and glycerol. Amylase breaks starch down to maltose (not glucose). Pepsin breaks down proteins to peptides (not free amino acids). Maltase breaks down maltose to glucose (not sucrose).

1 mark for selecting the correct enzyme, source, substrate, and product combination (C).

When a plant cell is placed in a highly concentrated sucrose solution, the water potential outside the cell is lower than inside the cell. Therefore, water leaves the cell by osmosis (moving down its water potential gradient). As water leaves the vacuole and cytoplasm, the volume decreases, causing the cell membrane to pull away from the cell wall. This state is known as plasmolysis.

[1 mark] C is correct. Accept C only. Reject A, B and D because water moves out of the cell (not in) down a water potential gradient by osmosis (not active transport or diffusion), causing plasmolysis.

Vessel X represents an artery, which has a thick layer of muscle and elastic fibres to withstand and maintain high blood pressure. Vessel Y represents a capillary, whose walls are one cell thick to provide a short diffusion distance for the exchange of substances. Vessel Z represents a vein, which has valves to ensure blood flows in one direction back to the heart under low pressure.

[1 mark] A is correct. Accept A only. Reject B, C, and D as they incorrect match the structural features to the blood vessels.

1. Runoff/leaching of nitrates into the river triggers rapid algal growth (algal bloom).
2. This blocks sunlight from reaching aquatic plants below, causing them to die, and the algae themselves eventually die.
3. Aerobic bacteria decompose the dead organic matter, multiplying rapidly and consuming dissolved oxygen from the water during respiration. This lack of oxygen causes fish to die.

- Mark 1: Mention of leaching/fertiliser runoff causing rapid growth of algae / algal bloom.
- Mark 2: Mention of dead plants/algae being decomposed by bacteria (aerobically).
- Mark 3: Mention of bacteria using up dissolved oxygen for respiration, leading to the suffocation/death of fish.

1. The placenta is covered with villi and microvilli, which significantly increase the surface area available for diffusion.
2. The barrier between the maternal and fetal blood is very thin (only one cell thick), ensuring a very short diffusion pathway.
3. A rich blood supply on both sides (maternal and fetal circulation) constantly moves blood, maintaining steep concentration gradients for substances like oxygen, glucose, and urea.

- Mark 1: Large surface area due to presence of villi/microvilli.
- Mark 2: Short diffusion distance/pathway because the separating membrane is thin (one cell thick).
- Mark 3: Rich/constant blood supply maintains a steep concentration gradient.

1. At high altitudes, there is a lower partial pressure of oxygen in the atmosphere.
2. A higher concentration of red blood cells means more haemoglobin molecules are available to load/bind oxygen in the lungs.
3. This ensures that sufficient oxygen is still transported to tissues and organs to support a normal rate of aerobic respiration, preventing fatigue and hypoxia.

- Mark 1: More red blood cells provide more haemoglobin.
- Mark 2: This allows more oxygen to be absorbed/bound/transported per unit volume of blood (to compensate for lower air pressure).
- Mark 3: This maintains a sufficient rate of aerobic respiration in tissues/cells.

1. Auxin is synthesized at the growing tip of the shoot and diffuses downwards.
2. Unidirectional light causes auxin to move and accumulate on the shaded side of the shoot.
3. The higher concentration of auxin on the shaded side stimulates faster cell elongation on that side compared to the illuminated side, causing the shoot to bend towards the light.

- Mark 1: Auxin is made in the tip and diffuses down the shoot.
- Mark 2: Auxin redistributes/accumulates on the shaded side of the shoot.
- Mark 3: Auxin stimulates cell elongation on the shaded side, causing the shoot to bend towards the light.

1. The concentrated sodium chloride solution has a lower water potential than the cytoplasm/vacuoles of the potato cells.
2. Water moves out of the potato cells by osmosis down a water potential gradient (from a higher water potential to a lower water potential).
3. Water passes across the selectively permeable cell membranes, leading to a loss of turgor pressure and mass, causing the cells to become flaccid.

- Mark 1: Reference to the external salt solution having a lower water potential than the potato cells.
- Mark 2: Water moves out of the potato cells by osmosis.
- Mark 3: Water crosses the selectively permeable membrane down a water potential gradient, causing loss of turgor/mass.

1. During the day, guard cells carry out photosynthesis and actively accumulate ions (like potassium), which lowers their internal water potential.
2. Water enters the guard cells from surrounding epidermal cells by osmosis, causing them to swell and become turgid.
3. The cell walls of guard cells are unevenly thickened (the inner wall near the pore is thicker and less elastic than the outer wall). When turgid, the outer walls stretch more, causing the cells to curve outward and open the stomatal pore.

- Mark 1: Mention of water entering guard cells by osmosis during the day (due to solute accumulation/photosynthesis).
- Mark 2: Guard cells become turgid/swell.
- Mark 3: Description of uneven cell wall thickness (thick inner wall, thin outer wall) causing the cells to curve outward to open the pore.

1. The genotype of a pink-flowered plant must be heterozygous: \(C^R C^W\).
2. Crossing two pink-flowered plants (\(C^R C^W \times C^R C^W\)) produces gametes of both types from each parent: \(C^R\) and \(C^W\).
3. The offspring genotypes will be:
- \(1 \times C^R C^R\) (red)
- \(2 \times C^R C^W\) (pink)
- \(1 \times C^W C^W\) (white)
Therefore, the probability of obtaining white-flowered offspring (\(C^W C^W\)) is 1 out of 4, or 0.25 (25%).

- Mark 1: Correct genotype of the pink flower stated as \(C^R C^W\).
- Mark 2: Correct genetic diagram / Punnett square showing the possible offspring genotypes: \(C^R C^R\), \(C^R C^W\), and \(C^W C^W\).
- Mark 3: Correct probability identified as 0.25, 25%, or 1/4.

1. Measure a volume of the liquid food sample into a test tube.
2. Add an equal volume of Biuret reagent (or add sodium hydroxide solution followed by a few drops of copper sulfate solution) and gently shake the test tube to mix.
3. Observe the colour change: if protein is present, the solution changes from its initial blue colour to a lilac, purple, or violet colour.

- Mark 1: Add Biuret reagent (or sodium hydroxide AND copper sulfate) to the food sample.
- Mark 2: Mix/shake the solution (no heating required).
- Mark 3: Correct colour change identified from blue to lilac/purple/violet.

1. Thin alveolus wall (one cell thick) which reduces the diffusion distance. 2. Large total surface area provided by millions of alveoli, allowing more gas molecules to diffuse simultaneously. 3. Rich capillary network which continuously carries oxygen away and brings carbon dioxide, maintaining a steep concentration gradient.

1 mark for each correct adaptation linked to its explanation, up to a maximum of 3 marks:
- Thin wall / one cell thick AND reduces diffusion distance (1)
- Large surface area AND increases rate / area of diffusion (1)
- Good blood supply / capillary network AND maintains concentration gradient (1)
- Moist lining AND allows gases to dissolve (1)

1. The cross of Nn x Nn produces offspring genotypes NN, Nn, Nn, and nn. The vestigial wing phenotype (nn) occurs in 1 out of 4 offspring, which is a 25% probability.
2. The actual ratio of phenotypes can differ from the theoretical 3:1 ratio because:
- Fertilisation is a random process (which gametes fuse is down to chance).
- The sample size of offspring might be small, making random variations more significant.

- Correct percentage: 25% (1)
- Explanation that fertilisation / fusion of gametes is random / due to chance (1)
- Explanation that the sample size of offspring is small (1)

1. Biconcave disc shape increases the surface area to volume ratio, facilitating faster diffusion of oxygen. 2. No nucleus (or other organelles) maximizes the internal volume available to hold haemoglobin. 3. Contains haemoglobin, a protein that binds reversibly with oxygen to form oxyhaemoglobin, allowing large amounts of oxygen to be transported.

Any three of the following for 1 mark each (must pair feature with its physiological benefit):
- Biconcave shape AND increases surface area / surface area to volume ratio for faster diffusion (1)
- No nucleus / lack of organelles AND allows more space for haemoglobin / oxygen (1)
- Contains haemoglobin AND binds to oxygen / forms oxyhaemoglobin (1)
- Flexible membrane AND can squeeze through narrow capillaries without bursting (1)

1. Receptors in the skin detect the stimulus (heat) and generate an electrical impulse. 2. The impulse travels along a sensory neurone to the spinal cord (CNS). 3. It passes across synapses to a relay neurone, and then to a motor neurone. 4. The motor neurone carries the impulse to the effector (the muscle in the arm), causing it to contract and pull the hand away.

- Sensory neurone transmits impulse from receptor to spinal cord / CNS / relay neurone (1)
- Impulse passes across synapse(s) to relay neurone and then to motor neurone (1)
- Motor neurone transmits impulse to effector (muscle) causing it to contract (1)
Note: Accept correct description of the sequential pathway: Receptor -> Sensory neurone -> Relay neurone (in spinal cord/CNS) -> Motor neurone -> Effector (muscle).

1. The concentrated salt solution has a lower water potential (is hypertonic) compared to the cytoplasm of the red blood cells. 2. Water moves out of the cells by osmosis, which is the movement of water from an area of higher water potential (inside the cell) to an area of lower water potential (outside the cell). 3. This movement occurs across the partially permeable cell membrane. As water is lost, the cell volume decreases, causing the red blood cells to shrink (crenation).

- Solution has a lower water potential than the inside of the red blood cell / cytoplasm (1)
- Water moves out of the cell by osmosis (1)
- Across the partially permeable cell membrane (1)
(Reject: active transport, water potential gradient described the wrong way)

1. When the blood is too concentrated (dehydrated), the hypothalamus detects this change, causing the pituitary gland to release more ADH into the bloodstream. 2. ADH targets the collecting ducts in the nephrons of the kidneys, making their walls more permeable to water. 3. This allows more water to be reabsorbed back into the blood by osmosis. Consequently, the blood water levels return to normal, and a smaller, more concentrated volume of urine is produced.

- (More) ADH is released by the pituitary gland (1)
- ADH increases the permeability of the collecting duct (walls) to water (1)
- More water is reabsorbed (by osmosis) back into the blood (leading to concentrated/less urine) (1)

In the nitrogen cycle, bacteria play vital roles in changing the chemical state of nitrogen: 1. Nitrifying bacteria convert ammonia or ammonium ions (from decomposed organic matter) into nitrites (\(\text{NO}_2^-\)), and then oxidize nitrites into nitrates (\(\text{NO}_3^-\)), which plants can easily absorb. 2. Denitrifying bacteria convert nitrates back into nitrogen gas (\(\text{N}_2\)), which is released into the atmosphere, typically in anaerobic conditions such as waterlogged soils.

- Nitrifying bacteria convert ammonium / ammonia to nitrites and/or nitrates (1)
- Denitrifying bacteria convert nitrates into nitrogen gas (1)
- Nitrification is an aerobic process / denitrification occurs in anaerobic conditions (e.g., waterlogged soil) OR denitrifying bacteria reduce soil fertility while nitrifying bacteria increase it (1)

1. Add an equal volume of Benedict's reagent/solution to the liquid food sample in a test tube. 2. Heat the mixture in a water bath set to at least \(70^\circ\text{C}\) (or boiling water bath) for about 5 minutes. 3. Observe the colour change: if reducing sugars are present, the solution will change from its original blue color to green, yellow, orange, or brick-red (depending on the concentration of reducing sugars).

- Add Benedict's reagent / solution (1)
- Heat / warm in a water bath (above \(60^\circ\text{C}\) / boiling) (1)
- (Positive result is a) colour change from blue to green / yellow / orange / brick-red (1)
(Reject: heating directly with a Bunsen burner flame without water bath, or just 'heating' without safety/method context unless water bath is mentioned)

Waterlogged soil lacks oxygen because the spaces between soil particles are filled with water. Draining the field allows air containing oxygen to enter these spaces. Nitrifying bacteria require aerobic conditions to convert ammonia and ammonium ions into nitrates. Conversely, denitrifying bacteria thrive in anaerobic conditions, breaking down nitrates to release nitrogen gas. Thus, oxygenation increases nitrification and decreases denitrification, raising soil nitrate levels.

1. Draining allows air/oxygen to enter soil pores, creating aerobic conditions (1 mark). 2. Nitrifying bacteria can actively convert ammonium/nitrite into nitrates (1 mark). 3. Activity of anaerobic denitrifying bacteria is reduced, meaning less nitrate is converted back to nitrogen gas (1 mark).

Amylase is adapted to work at an optimum pH of around 7 in the mouth. When food enters the stomach, it encounters hydrochloric acid, which lowers the pH to approximately 1.5 to 2.0. This extreme acidity disrupts the ionic and hydrogen bonds holding the amylase protein structure together, changing the shape of its active site (denaturation). Consequently, the substrate (starch) can no longer fit, and digestion stops.

1. Stomach contains hydrochloric acid which has a low pH / is highly acidic (1 mark). 2. The low pH denatures the amylase enzyme (1 mark). 3. The shape of the active site is altered so the substrate/starch can no longer bind (1 mark).

Transpiration involves the diffusion of water vapour out of the stomata of a leaf. This process depends on a concentration gradient of water vapour. When the surrounding air is highly humid, it contains a high concentration of water vapour. This reduces the difference in water vapour concentration between the inside of the leaf air spaces and the atmosphere outside, resulting in a slower rate of diffusion.

1. High humidity means there is a high concentration of water vapour in the air surrounding the leaf (1 mark). 2. This reduces the concentration gradient of water vapour between the inside and the outside of the leaf (1 mark). 3. Therefore, the rate of diffusion of water vapour out of the stomata decreases (1 mark).

Sweating decreases the water potential of the blood, which is detected by osmoreceptors in the hypothalamus. This stimulates the pituitary gland to secrete more antidiuretic hormone (ADH) into the bloodstream. When ADH reaches the kidney nephrons, it increases the permeability of the collecting ducts to water. More water is reabsorbed by osmosis back into the capillaries surrounding the tubule, resulting in the production of a low volume of concentrated (darker) urine, conserving water in the body.

1. Pituitary gland releases more ADH (1 mark). 2. Collecting ducts of the nephrons become more permeable to water (1 mark). 3. More water is reabsorbed back into the blood, resulting in a smaller volume of highly concentrated urine (1 mark).

In genetics, alleles are alternative forms of a gene. When alleles are codominant, they are both expressed equally in the phenotype of a heterozygote (an organism with two different alleles). Neither allele masks the expression of the other. For example, a flower with one allele for red pigment and one allele for white pigment produces pink flowers because both pigments (or lack thereof) are represented in the phenotype.

1. Both alleles are expressed in the phenotype (1 mark). 2. Applies to a heterozygous individual / individual with two different alleles (1 mark). 3. Neither allele is dominant or recessive / both alleles contribute to the observed characteristic (1 mark).

Sewage contains organic compounds and nutrients. When released into a river, it provides a large food source for decomposers, particularly aerobic bacteria. These bacteria multiply exponentially. As they break down the organic waste, they carry out aerobic respiration, which uses up the dissolved oxygen in the river water. This rapid fall in oxygen levels (eutrophication effect) makes it impossible for fish to obtain enough oxygen for respiration, leading to their death.

1. Sewage provides organic matter/nutrients for decomposers/bacteria (1 mark). 2. Bacteria multiply rapidly and respire aerobically, consuming dissolved oxygen (1 mark). 3. Oxygen concentration in the water drops, so fish cannot respire and suffocate (1 mark).

The alveolus is adapted to maximise diffusion in three main ways: 1) High Surface Area: Spherical structure and sheer numbers create a huge surface area for gas exchange. 2) Short Diffusion Path: Both the alveolar wall and the capillary wall are one cell thick, meaning gases only have to diffuse a very short distance. 3) Steep Concentration Gradient: The rich capillary blood supply constantly brings deoxygenated blood and carries away oxygenated blood, while ventilation keeps fresh air in the alveoli.

1. Thin wall / only one cell thick, which provides a short diffusion distance (1 mark). 2. Surrounded by a rich/dense network of capillaries / constant blood flow, maintaining a steep concentration gradient (1 mark). 3. Large surface area for diffusion of gases (due to spherical shape/large numbers) (1 mark).

Mineral ions like nitrates are present in low concentrations in the soil compared to inside the root hair cells. Consequently, they must be absorbed via active transport, a process that moves substances against their concentration gradient. Active transport is an active process that requires energy in the form of ATP. This ATP is produced during aerobic respiration in the mitochondria of root cells. Waterlogged soils lack oxygen, so root cells cannot respire aerobically and cannot generate enough ATP for active transport.

1. Waterlogged soil lacks oxygen/air spaces (1 mark). 2. Root cells must absorb mineral ions/nitrates by active transport against the concentration gradient (1 mark). 3. Active transport requires ATP/energy from aerobic respiration, which is restricted without oxygen (1 mark).

To find the volume of water taken up, multiply the distance the bubble moved by the cross-sectional area of the capillary tube: \(45\text{ mm} \times 0.8\text{ mm}^2 = 36\text{ mm}^3\). To find the rate per minute, divide this volume by the time in minutes: \(36\text{ mm}^3 / 10\text{ minutes} = 3.6\text{ mm}^3\text{ min}^{-1}\).

Award 1 mark for calculating the total volume of water absorbed: \(36\text{ mm}^3\) (allow working showing \(45 \times 0.8\)). Award 1 mark for dividing the volume by the time of 10 minutes. Award 1 mark for the correct final answer of 3.6 with correct units (\(\text{mm}^3\text{ min}^{-1}\) or \(\text{mm}^3/\text{min}\)). Accept 3.6 for full marks even if no working is shown.

Ultrafiltration requires high pressure and a selective barrier. The high blood pressure is generated because the afferent arteriole entering the glomerulus is wider than the efferent arteriole leaving it. The capillary walls of the glomerulus have tiny pores (fenestrations) and are surrounded by a basement membrane that acts as a fine mesh/filter. This allows small molecules (water, glucose, urea, salts) to be forced out of the blood into the Bowman's capsule, while larger structures (proteins and red blood cells) remain in the blood.

Award 1 mark for mentioning high pressure in the glomerulus due to the afferent arteriole being wider than the efferent arteriole. Award 1 mark for stating that the capillary walls have pores and a basement membrane that acts as a filter. Award 1 mark for stating that small molecules (such as glucose, water, urea, or ions) are forced through into the Bowman's capsule, while large molecules (proteins or blood cells) are held back.

To test for lipids using the emulsion test: 1. Add ethanol to the crushed sunflower seeds and shake/stir to dissolve any lipids present. 2. Pour or decant the liquid into a second test tube containing cold water. 3. Observe the appearance of a cloudy, milky-white emulsion, which indicates a positive result for lipids.

Award 1 mark for adding ethanol and mixing/shaking. Award 1 mark for adding the mixture/liquid to water. Award 1 mark for stating the correct positive result: a cloudy/milky emulsion forms. Reject 'white precipitate' or 'cloudy precipitate' as it is an emulsion.

Fertilisers contain high levels of nitrates and phosphates that leach into the river, causing rapid growth of algae (algal bloom) at the water surface. This layer of algae blocks sunlight from reaching submerged aquatic plants, preventing photosynthesis and causing them to die. Decomposing bacteria feed on the dead plant material and multiply rapidly, using up dissolved oxygen in the water for aerobic respiration. This leaves insufficient oxygen for fish, causing them to suffocate and die.

Award 1 mark for algal bloom / rapid growth of algae blocking light, leading to death of submerged plants. Award 1 mark for decomposers/bacteria breaking down dead plant material and multiplying. Award 1 mark for bacteria consuming dissolved oxygen during aerobic respiration, causing fish to die of suffocation/lack of oxygen.

To design this investigation, use the CORMS framework:
- **C (Change):** Use at least five different concentrations of sodium chloride solution (e.g., 0.0, 0.2, 0.4, 0.6, and 0.8 mol/dm³).
- **O (Organism):** Cut cylinders of tissue from the same carrot to ensure the cells have the same initial osmotic potential.
- **R (Repeat):** Repeat the test at least three times at each concentration to calculate a mean and identify anomalous results.
- **M1 (Measure 1):** Use an electronic balance to measure the mass of each carrot cylinder before and after immersion, calculating the percentage change in mass.
- **M2 (Measure 2):** Leave the cylinders in the solutions for a set time, such as 2 hours.
- **S (Same):** Control the temperature of the solutions using a temperature-controlled water bath, and control the volume of solution in each test tube (e.g., 20 cm³). Blot the cylinders dry gently with a paper towel before weighing.

Award 1 mark for each point up to a maximum of 6 marks:
- **C (Change):** Range of at least 5 different sodium chloride concentrations (1)
- **O (Organism):** Carrot cylinders from the same carrot / same variety / same age (1)
- **R (Repeat):** Repeat at least 3 times for each concentration / calculate a mean / average (1)
- **M1 (Measure):** Measure the starting and final mass of the carrot cylinders / calculate percentage change in mass (1)
- **M2 (Measure):** Leave the cylinders in solution for a specified minimum time period (between 1 hour and 24 hours) (1)
- **S1 (Same):** Control the temperature of the solution / use a water bath (1)
- **S2 (Same):** Control the volume of sodium chloride solution / the starting dimensions or surface area of the carrot cylinders (1)

*Note: Do not accept 'measure mass' without reference to initial and final mass, or change in mass.*

When a person is dehydrated, the high solute concentration of the blood is detected by the hypothalamus. This stimulates the pituitary gland to release more ADH (antidiuretic hormone). ADH increases the permeability of the collecting duct walls in the nephrons to water. Consequently, more water is reabsorbed by osmosis from the filtrate back into the surrounding capillaries, resulting in a smaller volume of highly concentrated urine.

Award 1 mark for identifying option B as the correct answer.
Reject other options as they incorrectly state ADH levels, permeability, or urine volume outcomes.

During digestion, carbohydrates are broken down into glucose and absorbed in the ileum (small intestine). The blood capillaries in the villi absorb this glucose and drain into the hepatic portal vein, which carries this nutrient-rich, deoxygenated blood directly to the liver. The liver then regulates blood glucose levels by storing excess as glycogen, meaning the hepatic vein leaving the liver will have a lower concentration of glucose.

Award 1 mark for identifying option B as the correct answer.
Reject Option A (hepatic vein has regulated glucose levels).
Reject Option C (hepatic artery carries oxygenated blood from the aorta).
Reject Option D (renal vein carries deoxygenated blood to the vena cava with normal systemic glucose concentrations).

In waterlogged soils, the lack of oxygen creates anaerobic conditions. Under these conditions, anaerobic denitrifying bacteria thrive and convert nitrates in the soil into nitrogen gas \(\text{N}_2\), which escapes into the atmosphere. This depletes the soil of nitrate ions needed by plants for healthy growth and protein synthesis.

Award 1 mark for identifying option C as the correct answer.
Reject Option A (nitrogen fixation increases soil nitrogen content).
Reject Option B (nitrification is an aerobic process that produces nitrates).
Reject Option D (decomposition breaks down dead matter but does not convert nitrates directly to gas).

First, calculate the rate of digestion at each pH (or use the mass directly since time is constant). Rate at pH 2 = \(10.0 \text{ g} / 120 \text{ s} = 0.0833 \text{ g/s}\). Rate at pH 4 = \(3.5 \text{ g} / 120 \text{ s} = 0.0292 \text{ g/s}\). Calculate the decrease in rate: \(0.0833 - 0.0292 = 0.0541 \text{ g/s}\). Calculate the percentage decrease relative to the starting rate at pH 2: \((0.0541 / 0.0833) \times 100\% = 65\%\). Alternatively, using the masses: \(((10.0 - 3.5) / 10.0) \times 100\% = 65\%\).

1.35 marks: Correct calculation of rates or setting up the correct formula using masses: \((10.0 - 3.5) / 10.0\). 1 mark: Calculation of the change in rate (6.5 g or 0.054 g/s). 1 mark: Correct final percentage decrease of 65%.

Mutualism is a relationship where both organisms benefit. The legume plant cannot fix atmospheric nitrogen directly, so it relies on the nitrogen-fixing Rhizobium bacteria in its root nodules to convert nitrogen gas into ammonium/ammonia, which the plant uses to make amino acids and proteins. In return, the plant provides the heterotrophic bacteria with sucrose and other carbohydrates from photosynthesis, which the bacteria use as an energy source, as well as providing a highly specialized, protective home in the root nodules.

1 mark: Explaining the benefit to the plant (obtaining fixed nitrogen/ammonium/nitrates to synthesise amino acids/proteins). 1 mark: Explaining the benefit to the bacteria (obtaining carbohydrates/sugars/sucrose from the plant's photosynthesis). 1.35 marks: Identifying the source of energy/respiration for bacteria or describing the protective/low-oxygen environment provided by the root nodules.

The 0.1% solution has a higher water potential than the red blood cell cytoplasm (hypotonic). Water moves into the cells by osmosis, causing them to swell and burst (lysis) because animal cells lack a rigid cell wall to resist the turgor pressure. The 2.0% solution has a lower water potential than the cytoplasm (hypertonic). Water moves out of the cells by osmosis into the surrounding solution, causing the cells to lose volume, shrink, and become shrivelled (crenated).

1.35 marks: Detail on 0.1% solution: water enters cells by osmosis, causing swelling and lysis/bursting (accept lack of cell wall). 1 mark: Detail on 2.0% solution: water leaves cells by osmosis, causing shrinkage/crenation. 1 mark: Explaining the direction of movement in terms of water potential gradients (high to low water potential).

Ultrafiltration is driven by high blood pressure in the glomerulus. This high pressure is created because the afferent arteriole carrying blood into the glomerulus is wider than the efferent arteriole carrying blood out. The capillary walls of the glomerulus are fenestrated (have pores), and the basement membrane acts as a fine mesh/filter. Additionally, the inner wall of the Bowman's capsule is made of specialized cells called podocytes, which have gaps called filtration slits. Large components like plasma proteins (e.g., fibrinogen, albumin) and blood cells are too large to pass through this barrier and remain in the capillary blood.

1 mark: Explaining the generation of high blood pressure (afferent arteriole wider than efferent arteriole). 1 mark: Describing the role of the basement membrane as a molecular sieve/filter. 1 mark: Explaining the role of podocytes or glomerular capillary pores. 0.35 marks: Correctly naming a substance not filtered (large proteins, red blood cells, white blood cells, platelets).

The placenta acts as an exchange organ. It is adapted for diffusion in three main ways: 1) Large surface area: The presence of numerous finger-like projections called chorionic villi increases the surface area for diffusion. 2) Short diffusion distance: The barrier separating maternal and fetal blood is very thin (only a couple of layers of cells). 3) Steep concentration gradient: Continuous flow of maternal blood in the intervillous spaces and fetal blood in the umbilical capillaries ensures that oxygen and nutrients are carried away, and waste products are delivered, maintaining a concentration gradient.

1 mark: Explaining the role of chorionic villi in providing a large surface area. 1 mark: Explaining the thin membrane/barrier providing a short diffusion pathway. 1.35 marks: Explaining how continuous blood flow / extensive capillary network maintains a steep concentration gradient for diffusion.

The parental genotypes are \(C^R C^W\) (pink) and \(C^R C^W\) (pink). The gametes produced by each parent are \(C^R\) and \(C^W\). Crossing these yields: \(C^R C^R\) (red, 25%), \(C^R C^W\) (pink, 50%), and \(C^W C^W\) (white, 25%). The probability of having pink offspring is \(2/4 = 0.5\) (or 50%). The expected phenotypic ratio of the offspring is 1 Red : 2 Pink : 1 White.

1.35 marks: Showing the correct genetic cross (e.g. Punnett square) with offspring genotypes: \(C^R C^R\), \(C^R C^W\), \(C^W C^W\). 1 mark: Correct probability of pink flowers: 0.5, 50%, or 1/2. 1 mark: Correct phenotypic ratio specified with matching phenotypes (1 red : 2 pink : 1 white).

In winter, light intensity, carbon dioxide, and temperature can all be limiting factors for photosynthesis. By burning paraffin or using gas heaters, the grower can increase the carbon dioxide concentration. Since carbon dioxide is a crucial reactant for the light-independent stage of photosynthesis, increasing its levels prevents it from limiting the rate of reaction. Additionally, heaters maintain an optimum temperature. Photosynthesis is controlled by enzymes; raising the temperature increases the kinetic energy of enzymes and substrate molecules, leading to more frequent successful collisions. However, temperature must not be raised too high to avoid denaturing the enzymes.

1.5 marks: Explaining that carbon dioxide is a reactant/limiting factor, and increasing it increases the rate of carbohydrate production. 1.5 marks: Explaining the effect of temperature on the kinetic energy of enzymes and substrate molecules, leading to more successful collisions. 0.35 marks: Reference to avoiding enzyme denaturation by keeping temperature at an optimum.

Auxin is a plant hormone synthesized in the apical meristem of the shoot tip. It normally diffuses down the shoot evenly. When the shoot is exposed to light from one side (unilateral light), auxin is actively transported to the shaded side of the shoot. The higher concentration of auxin on the shaded side stimulates cells in that region to elongate more than the cells on the illuminated side. This asymmetrical cell elongation causes the shaded side to grow longer, forcing the stem to bend towards the source of light, showing positive phototropism.

1 mark: Auxin is synthesized in the tip and diffuses downwards. 1 mark: Auxin redistributes / moves to the shaded side of the shoot. 1 mark: Auxin stimulates cell elongation on the shaded side. 0.35 marks: Unequal elongation/growth causes the shoot to bend towards the light source.

When nitrate fertilisers leach into a river, they cause a rapid increase in the growth of algae, forming an algal bloom. This layer of algae blocks light from reaching plants deeper in the water, preventing photosynthesis and causing these plants to die. Decomposers, such as bacteria, break down the dead organic matter. These bacteria multiply rapidly and respire aerobically, consuming a large amount of dissolved oxygen from the water. Consequently, the oxygen concentration falls so low that fish cannot respire and die of suffocation.

1. Award 1 mark for mentioning algal bloom blocks light, causing underwater plants to die/stop photosynthesising. 2. Award 1 mark for stating that bacteria decompose the dead plants. 3. Award 1 mark for explanation that bacteria respire aerobically, using up dissolved oxygen, leading to fish death.

In wind-pollinated flowers, petals are small and dull (green/brown) as there is no need to attract insects, and anthers hang loosely outside the flower where they can be easily shaken by the wind to release pollen grains. In contrast, insect-pollinated flowers have large, brightly coloured petals to attract pollinators, with anthers firmly fixed inside the flower so that insects must brush against them, transferring pollen.

1. Award 1 mark for contrasting petals (small/dull in wind-pollinated vs large/brightly coloured in insect-pollinated). 2. Award 1 mark for contrasting anther position (hanging outside vs enclosed inside). 3. Award 1 mark for explaining the functional advantage of either arrangement (e.g., hanging anthers facilitate wind release of pollen, or enclosed anthers ensure contact with insect vectors).

During ventricular systole (contraction), pressure within the ventricles rises above that in the aorta and pulmonary artery, forcing the semi-lunar valves open so blood can exit. During ventricular diastole (relaxation), pressure in the ventricles drops below the arterial pressure. This pressure difference causes blood to start flowing backward, which catches the pockets of the semi-lunar valves, forcing them shut and preventing backflow.

1. Award 1 mark for stating that semi-lunar valves prevent the backflow of blood from arteries (aorta/pulmonary artery) into the ventricles. 2. Award 1 mark for explaining that high pressure during ventricular contraction opens the valves. 3. Award 1 mark for explaining that lower pressure in ventricles during relaxation causes the valves to shut.

When a mammal is exposed to a cold environment, nerve impulses cause the arterioles supplying the capillaries near the skin surface to narrow (vasoconstrict). This reduces the volume of blood flowing close to the skin surface, redirecting it through shunt vessels deeper in the body instead. As a result, less heat energy is transferred from the warm blood to the cooler surroundings via radiation, conduction, and convection.

1. Award 1 mark for stating that arterioles near the skin surface constrict/narrow. 2. Award 1 mark for explaining that this reduces blood flow through capillaries close to the skin surface. 3. Award 1 mark for stating that this reduces heat loss by radiation/convection/conduction.

A concentrated sucrose solution has a lower water potential than the cytoplasm and cell vacuole. Consequently, water moves out of the plant cell down a water potential gradient by osmosis, through the selectively permeable cell membrane. As water is lost, the vacuole shrinks and the cell membrane and cytoplasm pull away from the cellulose cell wall, causing the cell to become plasmolysed (flaccid).

1. Award 1 mark for describing the cell as plasmolysed or flaccid (or stating cell membrane pulls away from cell wall). 2. Award 1 mark for stating water moves out of the cell by osmosis. 3. Award 1 mark for explaining this is due to a water potential gradient (from a region of higher water potential inside the cell to lower water potential outside).

Alveoli possess several key adaptations to maximise gas exchange: 1) There are millions of alveoli in the lungs, creating an enormous total surface area for diffusion. 2) The walls of the alveoli (and the surrounding capillaries) are only one cell thick, which minimizes the distance gases must diffuse. 3) A rich network of capillaries surrounds each alveolus, carrying oxygenated blood away and bringing deoxygenated blood, which maintains a steep concentration gradient for oxygen and carbon dioxide.

1. Award 1 mark for stating that a large total surface area increases the rate of diffusion. 2. Award 1 mark for stating that the walls are one cell thick to provide a short diffusion path. 3. Award 1 mark for stating that a rich capillary network/blood supply maintains a steep concentration gradient.

The gametes produced by the mother are \(I^A\) and \(I^O\). The gametes produced by the father are \(I^B\) and \(I^O\). Crossing these gametes yields the following offspring genotypes: \(I^A I^B\) (blood group AB), \(I^A I^O\) (blood group A), \(I^B I^O\) (blood group B), and \(I^O I^O\) (blood group O). The probability of an offspring inheriting the homozygous recessive genotype \(I^O I^O\) is 1 out of 4, which is 0.25 or 25%.

1. Award 1 mark for showing correct parental gametes (mother: \(I^A\) and \(I^O\); father: \(I^B\) and \(I^O\)). 2. Award 1 mark for correct offspring genotypes in a genetic diagram or Punnett square (\(I^A I^B\), \(I^A I^O\), \(I^B I^O\), \(I^O I^O\)). 3. Award 1 mark for correct final probability: 0.25, 25%, or 1/4.

Nitrifying bacteria convert ammonium ions or ammonia present in the soil into nitrites, and then into nitrates. This process occurs under aerobic conditions (requires oxygen). In contrast, denitrifying bacteria convert nitrates in the soil back into nitrogen gas, releasing it into the atmosphere. This process occurs under anaerobic conditions, typically in waterlogged, poorly-aerated soils.

1. Award 1 mark for stating that nitrifying bacteria convert ammonium/ammonia into nitrites/nitrates. 2. Award 1 mark for stating that denitrifying bacteria convert nitrates into nitrogen gas. 3. Award 1 mark for distinguishing their oxygen requirements (nitrifiers require aerobic conditions, denitrifiers operate in anaerobic/waterlogged conditions).

1. Biconcave shape: This shape increases the surface area to volume (SA:Vol) ratio of the cell, allowing oxygen to diffuse into and out of the cytoplasm more rapidly. 2. Absence of a nucleus: By not having a nucleus, the mature red blood cell has significantly more internal space. This space is filled with haemoglobin, the oxygen-binding protein, meaning each cell can transport a much larger quantity of oxygen.

Award 1 mark for explaining that the biconcave shape increases the surface area to volume ratio. Award 1 mark for linking the increased surface area to volume ratio to a faster rate of diffusion of oxygen. Award 1 mark for explaining that the absence of a nucleus provides more space for haemoglobin to transport more oxygen.

1. Rhizobium bacteria carry out nitrogen fixation, converting inert nitrogen gas (N2) from the soil air into ammonia or ammonium ions. 2. The plant absorbs these nitrogen compounds and uses them to manufacture organic molecules such as amino acids, proteins, and nucleotides (DNA). 3. In return, the plant provides the bacteria with carbohydrates (sugars) produced via photosynthesis, representing a mutualistic relationship.

Award 1 mark for stating that Rhizobium bacteria convert nitrogen gas into ammonia, ammonium, or nitrates (nitrogen fixation). Award 1 mark for explaining that the plant uses these nitrogen compounds to produce amino acids, proteins, or nucleic acids. Award 1 mark for identifying the relationship as mutualistic, where the plant provides carbohydrates or sugars to the bacteria.

1. Entry into Bowman's capsule: High pressure in the glomerulus (ultrafiltration) forces small molecules, including glucose, water, urea, and ions, through the basement membrane into the Bowman's capsule, while large molecules like proteins remain in the blood. 2. Removal from filtrate: As the filtrate passes through the proximal convoluted tubule, all the glucose is selectively reabsorbed back into the surrounding blood capillaries. 3. Mechanism: This selective reabsorption occurs against a concentration gradient via active transport, requiring energy from respiration.

Award 1 mark for explaining that high pressure in the glomerulus forces small glucose molecules through the filter into the Bowman's capsule (ultrafiltration). Award 1 mark for stating that glucose is selectively reabsorbed in the proximal convoluted tubule. Award 1 mark for stating that this reabsorption occurs via active transport (using energy).

1. Role in rising: Yeast respires anaerobically (fermentation), converting glucose into ethanol and carbon dioxide. The carbon dioxide gas is trapped as pockets of air within the gluten network of the dough, causing the dough to expand and rise. 2. Effect of oven: At 200 degrees Celsius, the extreme heat denatures the respiratory enzymes within the yeast cells. This halts metabolic processes, killing the yeast and ending carbon dioxide production.

Award 1 mark for stating that yeast respires anaerobically to produce carbon dioxide. Award 1 mark for explaining that the carbon dioxide gas bubbles get trapped in the dough, causing it to expand and rise. Award 1 mark for explaining that the high oven temperature denatures enzymes in the yeast, halting respiration and killing the yeast cells.